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ELEMENTS OF
PROPERTIES OF MATTER
WITH TYPICAL NUMERICALS SOLVED
I FOR DEGREE CLASSES ]
by
D. S, MATHUR
S . C H AWD & C
DELHI NEW DELHI JULLUNDUR
LTJCKNOW  BOMBAY
SPECIAL FEATURES
1. Detailed and simple treatment, with each step fully explain
ed.
2. 336 illustrative diagrams given.
3 A large number of typical numerical problems solved, (in
cluding those set in the various University Examinations),
covering 150 pages or more, of the book,
4 Illustrative solutions, with the use of logarithms, shown on
the margin to the left in the first two chapters.
5. Useful appendices, on Differential and Integral Calculus,
together with those on important Trigonometrical Relations
and the use of Logarithms, included, as also Logarithmic
Tables and Tables of Important Constants.
Published by
S. CHAND & CO.
for Shyam Lai Charitable Trust,
16B/4, Asaf AH Road, New Delhi
( All profits from this book are spent on charities)
S. CHAND & CO.
Ram Nagar NEW DBLKI
Fountain DBLHI
M*i Hiran Gate JULLUNDUB
Hazrat Ganj LUOKNQW
Lamington Road BOMBAY
First published October, 1949
Seventh Edition July, J962
Price : Rs, 900
PublMed by 0. S. Sharma, for a. unana <s uo. t Kam Wagar, New Delhi and
Printed a* Rajendra Ravindra Printer*, (P) Ltd., Ram Nagar, New Delhi 1.
CONTENTS
PA<
Chapter I Units and Dimensions. 1
Units Fundamental and Derived Units Principal Systems of Units
Dimensions Dimensional Formulae and Equations Uses of Dimen
sional Equations Limitations of Dimensional Analysis Solved Ex
amples Exercise I.
Chapter II Motion along a Curve The Projectile. 20
Rotation Angular Velocity Angular Acceleration Couple Work
done by a Couple Relation between Couple and Angular Acceleration
The Hodograph Velocity in the Hodograph Uniform Circular
Motion Centripetal Force Centrifugal Force Practical Applications
of Centripetal and Centrifugal Forces Other Effects and Applications
of Centrifugal Force The Projectile Motion of a Projectile in a non
resisting medium Horizontal Range of a Projectile Maximum Height
attained by a Projectile Angle of Projection for Maximum Range
Range on an Inclined Plane Resultant Velocity of a Projectile at a
given instant Solved Examples Exercise II.
Chapter III Moment of Inertia Energy of Rotation. 48
Moment of Inertia and its Physical Significance Radius of Gyration
Etpression for Moment of Inertia Torque General Theorems on
Moment of Inertia Calculation of the Moment of Inertia of a Body
Its Units etc. Particular Cases of Moments of Inertia Table of Mo
ments of Inertia Routh's Rule Practical Methods for the Determina
tion of Moments of Inertia Angular Moment and Angular Impulse
Law of Conservation of Angular Momentum Laws of Rotation Kine
tic Energy of RoUtiQri Acceleration of a body rolling down an in
clined" riline uraphical Representation of Plane Vectors Precession
The Gyrostat Gyroscope The Gyrostatic Pendulum Case of a
Rolling Disc or Hoop Gyrostatic and Gyroscopic Applications
Solved Examples Exercise III.
Chapter IV Simple Harmonic Motion. Ill
Definition Characteristics oj#a Linear S.H.M. Equation of Simple
Harmohic Motion ComposKrfcm of Two Simple Harmonic Motions
(Graphicat^Qd Analytical JtXEftoas) Composition of two equal circular
motions in oppis^ite directing Energy of a Particle in simple Harmo
nic Motion A vehkge Kinetv and Potential Energies of a Particle in
S. H.M. Solved ExaSlpl^s Exercise IV.
Chapter V Measurement of Mass The Balance. 146
Mass and Weight The Common Balance Essentials or Requisites of a
Good Balance Faults in a Balance Determination of True Weight
Correction for Buoyancy Solved Examples Exercise V.
Chapter VI Acceleration due to Gravity. 160
Acceleration due to Gravity The Simple Pendulum Borda's Pendu
lum^ Compound Pendulum fnterchangeability of the Centres of Sus
pension and Oscillation Centre of Percussion Other points, collinear
with the e.g., about which the timeperiod is the same Conditions for
Maximum and Minimum Time periods Bar Pendulum Owen's modi
fication of the bar pendulum Kater *s Reversible Pendulum Kater' s
Method of Coincidences Computed Time BesseVs Contribution Errors
in the Compound Pendulunfand their Remedies Other Improvements
due to Bessel Conical Pendulum Steam Eogine Governor Other
methods for the determination of # Variation of the value of g~~ 
Determination of the value of g at Sea Local and Temporal Changes in
the value of g Gravity SurveyGeophysical Prospecting Solved
Examples Exercise VI.
(v/)
1'AOES
Chapter VII Gravitation. 224273
Historical Kepler's Laws Note on Newton's deductions from Kepler's
laws Newton's Law of Gravitation Determination of the Gravita
tional Corstant Density of the Earth Qualities of Gravitation Law
of Gravitation and the Theory of Relativity Gravitational Field In
tensity of the Field Gravitational Potential Potential Energy Gra
vitational potential at a point distant r from a body of mass m Velo
city of Escape Equipotential Surface Potential at a point Outside
and Inside a Spherical Shell Gravitational Field Inside a Spherical
Shell or a Hollow Sphere Potential and Field Intensity due to a Solid
Sphere at a point (p Inside the Sphere and (//) Outside the Sphere
Intensity and Potential of the Gravitational Field at a Point due to a
Circular Disc Intensity and Potential of the Gravitational Field at a
point due to an Infinite Plane Inertial and Gravitational Mass
Earthquakes Seismic Waves and Seismographs Seismology Seismo
graphs GG litzin's Seismograph Determination of the Epicentre and
the Focus Modern Applica ions of Seismology Solved Examples
Exercise VII.
Chapter VIII Elasticity. 274341
Introductory Stress and Strain Hook 's Law Three Types of Elasti
city Equivalence of a shear to a Compression and an Extension at
right angles to each other Shearing stress equivalent to an equal
linear tensile stress and an equal compression stress at right
angles to each other Work done per Unit Volume in a Strain
Deformation of a Cube Bulk Modulus Modulus of Rigidity Young's
Modulus Relation connecting the Elastic Constants Poisson's Ratio
Determination of Young's Modulus Determination of Poisson's
Ratio for Rubber Resilience Effect of a suddenly applied load
Twisting Couple on a Cylinder (or wire) Variation of stress in a
twisted cylinder (or wire) strain energy in a twisted cylinder (or wire)
Alternative expression for strain energy in terms of stress Torsional
PendulumDetermination of the Coefficient of Rigidity (r\) for a Wire
Determination of Moment of Inertia with the help of a Torsional
Pendulum Bending of Beams Bending Moment The Cantilever (/)
Loaded at the free end (/*) Loaded uniformly Limitations of the Simple
Theory of Bending Strongly bent beams Transverse vibrations of a
loaded cantilever Depression of a beam supported at the ends (/) when
the beam is loaded at the centre 07) when the beam is loaded uniformly
Searle's Method for the comparison of Young's Modulus and coefficient
of Rigidity for a given material. Strain energy in a bent beam
Resilience of bent beams Columns, Pillars and Struts Critical load
for long columns (/') When the two ends of the column are rounded or
hinged (//) When the two ends of the column are fixed (///) When one
end of the column is fixed and the other loaded. Elastic waves (/) Com
pressional waves Impact coefficient of Restitution loss of kinetic
Energy on Impact Relative masses of colliding bo dies Solved
Examples. Exercise VIII.
Chapter IX Hydrostatics. 342366
Fluids Liquids and Gases Hydrostatic Pressure Hyprostatic Press
ure due to a liquid Column The Hydrostatic Paradox A liquid
transmits Pressure equally in all directions Pascal's Law Thrust on
an Immersed Plane Centre of Pressure Particular Cases of Centre of
Pressure Change of Depth of Centre of Pressure Principle of Archi
medesEquilibrium of Floating Bodies Stability of Equilibrium
Roll ing and Pitching of a Ship Determination of Metacentric Height
Pressure due to a Compressible Fluid or a Gas Measurement of
Atmospheric Pressure Correction of Barometric Reading Change of
Pressure with Altitude Solved Examples Exercise IX.
Chapter X Flying machines Jet planes, Rockets and Satellites 367393
Flying machinesThe kite The Airplane Different parts of an Air
plane and thiif functions Jet propulsion Thrust supplied by the jet
(wV)
PAGES
Efficiency of the jetEffect of smaller crosssection of the jet Rocket
planes Rocket fuel Specific impluse Shape of the Rocket The
Multistage Rocket Take off of ttie rocket Salvaging of the various
stage rockets Satellites Conditions for a satellite to be placed in
orbit Launching of the satelliteStability of the rocket during flight
Form of the satellite Weight and size of the satellite Material of the
frame of the satellite Duration of satellite's existence Other essentials
Return of Artificial satellite uses of an artificial satellite Exercise X.
Chapter XI Friction and Lubrication Principle of Virtual Work and its
Simple Applications. 394417
Static Friction Laws of Friction Sliding Friction Angle of Fric
tion Cone of Friction Acceleration down an Inclined PUne Rolling
Friction Friction and Stability Friction, a necessity Simple Prac
tical Applications of Friction Rope Machines (/) The Prony Brake
(//) The Rope Brake 'Hi) The Band Brakes Mechanism of Friction*
Lubricants Principle of Virtual Work (f> Case of a body in equi
librium on a smooth Inclined Plane undet the action of a force (ii) Cast
of equilibrium of a body on a rough Inclined Plane (Hi) Case of equili
irium of a system of two or mare connected bodies (/v) Relation between
Equilibrium and potential energy (v) Tension in a Fhwheel Solved
Examples Exercise XL
Chapter XII Flow of Liquids YiS&i& &^ < *&S 453
Rate of Flow of a liquid Lines and Tubes of Flow Energy of tnlP
Liquid Bernoulli's Theorem and its important Anjpiications^Impor
tant Applications of Bernoulli's Equation Viscosity Coefficient of
Viscosity Fugitive Elasticity  Critical ^VclochyPoiseuille's Equation
for flow of liquid through a tube Experimental Determination of rj for
a liquid Poheuillfs method Motion in a Viscous Medium Determi
nation of Coefficient of Viscosity of a Liquid Stoics' Method
Rotation Viscomster Variation of Viscosityxrf a Liquid with Tempera
tureComparison of Viscosities Ostwald Viscometer Determination
of Viscosity of Gases Rankine's Method for the determination of the
Viscosity of a Gas Solved Examples Exercise XII.
Chapter XIII Diffusion and Osmosis. 454 474
Diffusion Pick's law Relation between Time of Diffusion and Length
of Column Experimental Measurement of Diffusivity Graham's Law
for Diffusion of Gaie s Effusion Transpiration and Transfusion
Osmosis and Osmotic Pressure La^s of Osmotic Pressure Kinetic
Theory of Solutions Osmosis and Vapour Pressure of a Solution
Osmosis and Boiling Point of a Solution Osmosis and Freezing Point
of a Solution Determination of Percentage of Dissociation of an
Electrolyte Determination of Molecular Weight of a Substance from
Elevation of Boiling point or Depression of Freezingpoint of a Solution
of the substance Solved Examples Exercise XIII.
^tapter XIV Surf ace Tension  Capillarity. 475
Molecular Force Molecular Range Sphere of Influence
Tension Explanation of Surface Tension SurfaceFilm and Surface
Energy Free Energy of a Surface and Surface Tension Pressure""
umerence across a Liquid Surface Drops and Bubbles Excess Pres
sure inside a Liquid Drop Excess Pressure inside a Soap Bubble
Determination of the Surface" tension pf "g BubbleWork done in
blowing a Bubble Curvature, Pressure and Surface Tension Layer of
Liquid between two plates Shape of Liquid Meniscus in a Capillary
Tube Angle of Contact Measurement of the Angle of Gontact Rise of
Liquid in a Capillary Tube Rise of liquid in \ i uoe of insufficien t
Length Rise of liquid in a Conical Capillary Tube Energy required
to raised liquid in a Capillary Tube Rise of a liquid between two
Parallel Plates Force between Bodies Partly" Immersed in a Liquid
Shape of Liquid Drop on a Horizontal Plate Experimental Deter
mination of Surface Tension, (Different Methods)  Surface Tension of
(vm)
PAGSB
Liquid Interfaces Factors affecting Surface TensionExperimental
Study of the Variation of Surface Tension with Temperature Surface
Tension and Vapour Pressure over a Liquid Surface Effect on Evapora
tion and Condensation Solved Examples Exercise XIV.
Chapter XV Gases Kinetic Theory. 532575
The Kinetic Theory Introduction Kinetic Theory of Gases Pressure
Exerted by a Perfect Gas Value of c Relation between c and I 1
Deduction of Gas Laws on the basis of the Kinetic Theory Kinetic
Energy of a Molecule Value of the Gas Constant Van der Waal's
Equation Mean Free Path of a Molecule Viscosity of Gases Produc
tion of Low Pressure Exhaust Pumps Exhaust Pumps and their charac
teristics Different Types of Pumps The Common Air Pump Rotary
Oil Pumps (Gaede and Hyvac types) Molecular Pumps Diffusion Con
densation Pumps (Gaede and Waran types)  Other methods of Producing
Vacua Measurement of Low Pressures Manometers and Gauges
Common Mercury Manometers The Bourdon Gauge Mcleod Vacuum
Gauge Improved modifications of Mcleod Gauge The Pirani Resis
tance Gauge Thermocouple Gauge lonisation Gauges aray lonisa
tion Gauge The Knudsen Gauge Solved Examples Exercise XV.
APPENDICES
Appendix I Important Trigonometrical Relations 576577
Appendix 1 1 Logarithms 578 580
Appendix III Differential Calculus 581588
Appendix IV Integral Calculus 589596
Constant Tables 597 600
/. Densities of Common Substances 597
//. Elastic Constants 598
///. Coefficients of Restitution 598
/ V. Coefficients of Viscosity 598
V. Molecular Elevation of Boiling Points of Solvents 599
r/, Molecular Depression of Freezing Point of Solvents 599
F/7. Surface_Tensions of Important liquids 599
VIII." Molecular Constants (.00
Logarithmic and Antilogarithmic Tables 602605
Index 606
CHAPTER I
UNITS AND DIMENSIONS
1. limits. The Physicist always seeks to reduce his physical
concepts and conclusions to measurable quantities, in the spirit of
Lord Kelvin's very shrewd and apt remark that 'when you can mea
sure what you are speaking about and express it in numbers, you know
something about it, but when you cannot measure it in numbers, your
knowledge is meagre and unsatisfactory 9 , a remark which is at once a
challenge and an inspiration to men of science to sift and clarify
their ideas and notions until they become quite precise and clearcut.
Now, measurement inevitably involves comparison with a
chosen standard or unit of a similar kind ; so that, the first essential
step to be taken is the selection of a suitable standard or unit in
.accordance with the nature of the physical quantity to be measured,
und the second, to determine its value in terms of the chosen unit.
In other words, to form an exact idea of the magnitude of a physical
quantity, it is neq^sary to express (/) the standard or . unit in which
the quantity is metipured and (ii) the number of ti^s the quantity
contains that unit. *
Thus, for example, when we speak of a distance as being equal
to 5 miles, we mean that the standard or unit in which it is measured
is the mile, and that the distance in question is five times this unit.
If we choose the yard (which is I/ 1760 of 1 mile) or thfe foot (which
is 1/1760x3 of 1 mile) as our unit, the same distance will be equal
to 8800 yards or 26,400 feet respectively, i.e., its numerical value
will be 1760 times or 1760x3 times 5. Thus, the larger the unit, the
smaller the numerical value of the quantity ; and the smaller the unit, the
larger its value. Or, the numerical value of a quantity is* inversely pro
portional to the magnitude of the unit selected as the standard. It
follows, therefore, that the product of the numerical value of the
quantity and the magnitude of the unit in which it is expressed is a
constant. Thus,
5 X I^
Or, in general, if n t and /I 8 , be the numerical values of a given
physical quantity, corresponding to the units x t and x, respectively,
we have
2. Fundamental aijtf Derived Units. For measuring different
kinds of qu^$Jrefes, ^^itmst obyiously have different kinds of units. Ij
these be selected in any arbitrary manners they will be quite unrelated
to each other, and their use will create difficulties and complication)
in actual practice. They are, therefore, all based on some funda
mental units, so as to be interdependent and properly related t<
each other, the guiding principle in their choice being to D
(a} they are welldefined and of a suitable size,
2 PBOPBBTIES Off MATTER
(b) they are easily reproducible at all places,
(c) they are not subject to any secular changes (ie^ to changes
with time),
(d) they do not readily or appreciably vary with varying physica*
conditions, like temperature, pressure etc., and, if they do r
their manner of variation is perfectly correctly known.
The fundamental units chosen, and internationally employed f
are those of mass, length and time which C. F. Gauss, in 1832, termed
as absolute units*. The reason why these alone are chosen as the
'fundamental' units, and not any others, would seem to be that they
represent our elementary scientific notions and cannot be derived
from one another ; nor can they be resolved into anything more basic
or fundamental. All other units in Mechanics can be derived from
them and are, therefore, called 'derived units 9 . Thus, the units of
area and volume are derived units, for they can both be derived from
the unit of length, the former being the area of a square, and the
latter, the volume of a cube, each of unit length. Similarly, the
unit of velocity is a derived unit and is the velocity of a body
which covers unit distance, or length, in unit time, and so en.
3. Principal Systems of Units. There are three principal
systems of units in vogue, viz.,
(/) the CentimetreGrammeSecond system or the C G.S. system,
(ii) the FootPoundSecond system or the F. P. S. system and (/w) the
MetreKilogrammeSecond system or the M. K. S. system.
(i) The C. G. S. System. In this system, the unit of length is
the centimetre, that of mass, the gramme and that of time, the
second.
The Centimetre is onehundredth part of a metre, 'which is the
distance, at a temperature ofOC, between two lines on a platinumiridium
bar, preserved at the International Bureau of Metric Weights and
Measures at Sevres, near Paris. Originally intended to be one
thousand millionth part of the longitude of the earth from the north
pole to the equator, passing through Paris, it is found, however, to
be slightly smaller.
The International Bureau of Weights and Measures has con
structed a line standard metre, known as the Prototype Metre, copies
or replicas of which have been supplied to various Governments.
The Gramme is onethousandth part of a lump of platinumindium,
called a Kilogramme, made by Borda, in accordance wjth a decree of
the French Republic, and also preserved at Sevres. It is equal to the
mass of water, whose volume is one cubic centimetre, at 4C, when it has
its maximum density, (viz., I gm./c.c.)
The Second, or the mean solar second, as it is called, ig
1/24 x 00 X 60/A, or 1/86400//? part of the mean solar day, which is tht
average value, for one year, of the solar day, or the time which elapse*
between two consecutive transits of the Sun across the meridian, at any
place on the Earth's surface.
*In connection with the measurements of the earth's maenetk field
carried out by him at Gottingen.
TJUTTS ATffD DIMENSIONS 3
Another unit of time, used in Astronomy, is the mean siderial
second, which is 1/86400*// part of the siderial day, or the true period
of revolution of the Earth on its axis, i.e., the interval which elapses
between two consecutive passages of a fixed star across the meridian.
(ft*) The F. P. S. System. Here, the unit of length is the foot,
the unit of mass, the pound and the unit of time, the second.
The Foot w one third of the distance between two transverse lines,
at a temperature of62F, on two goldplygs in a bronze bar*, kept at the
Standards Office of the Board of Trade, London.
The Pound (avoirdupois) is the mass of a platinumindium cylinder,
marked "P S., 1844, I Ib." also kept at the Standards Office of the
Board of Trade, London. %
And, the Second, or the mean solar second, is the same as defined:
above.
Other units, derived from , those given above, are called the
Board of Trade units or the B. O. T. units.
It may as well be mentioned here that we generally choose our
units to smt the quantity to be measured. Thus, for example, for the
measurement of very small lengths or distances, we have successively
smaller units of length, v/z., the micron (//) = 10~ 3 mm., the milli
micron (m fl ) = 10 6 mm. and the Angstrom unit (A. U. or, simply, A}
= U) 7 mm. ; and, for the measurement of very large distances, like
those of interbteller space, we have correspondingly larger units, like
the light year, or the distance covered by light in vacuo, (with a
velocity of 2 9.) x 10 10 cm. /sec.) in one full year. Similar being the case
with the units of mass and time.
(iii) The M.K.S. System. This is a comparatively new system,
very much akin to the C.G.S. system, in which the units of length,
mass and time are the Metre, the Kilogramme and the Second
respectively.
The fir.^t system is the one invariably used in scientific work all
over, the second is more or less confined in its use to ojily Great
Britain and the third is now being increasingly adopted m electrical
engineering etc., where it is found to be more convenient and useful
4. Dimensions. Dimensional Formulae and Equations.
(a) Dimensions. The units of mass, length and tiine are
denoted by the capioal letters, [M], [t] and [T}\, which merely iri&h
cate their nature and not their magnitude. And, since the unit of are*a
is. the product of two unit length*, we have the unit of area repre
sented by [L] x fJL] or [L 2 ] ; and, similarly, the unit of volume, being
the product of three unit lengths, is represented by [L] x [L] x [L] or
[L 8 ]. We express this by saying that the unit of area is of two dimen
sionsjf. in length, and the unit of volume, of three dimensions in length.
This bar has now also been replaced by a platinumindium one.
tThe square brackets merely indicate 'dimension of\ Once this is under
stood, they may as well be lt orrutted, as we shall quite often do.
J Which is the abbreviated form of * exponent of dimension', but is now
commonly used and well understood.
4 PROPERTIES OF MATTER
Since neither the unit of area nor that of volume depends upon mass
and time, their dimensions are said to be zero in both mass and time
and we may, therefore, represent these units as M L 1 T* and M L*
Z* f respectively.
"The dimensions of a derived unit may thus be defined as the
powers to which the fundamental units of mass, length and time must be
raised to represent it" Thus, if a derived unit depends upon the wth
power of a fundamental unit, it is said to be of n dimensions in that
fundamental unit. For example,
, . distance or length f L 1
velocity =  ^  \ T J  LT .
and hence the dimensions of the unit of velocity are 1 in length and 1
in time. Since it is independent of mass, its dimension in mass is
zero, and we may, therefore, represent it by MLT~ l .
Again, since deceleration = ~ the dimensions of the unit
f M L 7 1 , , r rr . ,
of acceleration are  ^ == M LI*, and so on.
It will thus be seen that the dimensions of a physical quantity
are obtained by simply defining it in terms of those physical quanti
ties whose dimensions in mass, length and time are known, the value
of a derived unit depending upon the values of the fundamental units
from which it is derived. Thus, if we take a yard as our unit of length,*
instead of a/oof, the units of area and volume will respectively be 3 2
and 3* times as big as their uptits itt the ordinary system. So that,
the dimensions of a physical quantity show how its nature and the value
of its unit depend upon the fundamental units chosen.
(b) Dimensional Formulae and Equations. A dimensional formula
is an expression, showing how and which of the fundamental units enter
into the unit of a physical quantity. Thus, all the expressions in the
Table opposite, indicating the relation between the derived and
fundamental units, are dimensional formulae. For example, the
dimensional formula for work is ML*T~*. But when we put it in the
form, W = ML 2 r~ f , it is called a dimensional equation for work.
This idea of dimensional formulae for physical quantities, as
we know it today, was first clearly given by Fourier, in the year
1822, although it originated initially with Newton, who refers to the
principle of similitude in his famous and well celebrated Principia,
(II, Proposition 32).
The student is no doubt aware that in Physics we come acrooo
two types of quantities, viz., variables and constants, which may
both be dimensional or nondimensional (i.e., dimensionless). Thus,
we have
(/) Dimensional Variables. These are quantities like accelera
tion, velocity, force and most of the others which the Physicist has to
deal with, at every step. These are, so to speak, his 'current coin'.
(0^ Dimensional ConstantsQuantities which have a constant
value jmd yet have dimensions are called dimensional constants. As
TOITS AKD DIMENSIONS
5
examples of these may be cited G, the Gravitational Constant, and c,
the Velocity of Light in vacuo, whose dimensions are M~ 1 L*T~** and
MLT~ l respectively.
(Hi) Nondimensional Variables. These are quantities which are
variables and yet have no dimensions ; as, for example, specific
gravity, strain or an angle, (see Table below).
Here, we also meet with groups of dimensional variables (with
or without dimensisnal constants) such that their dimension is zero in
each of the fundamental quantities, i.e., in length, mass and time. Thus,
for example, the quantity t\/ gjl has no dimensions ; and so also the
quantity up//?, called Reynold's number, can be shown to have zero
dimensions in mass, length and time. Such quantities were given
the name 'numerics' by James Thomson.
(to) NonDimensional Constants. These are mere numbers like
3, 2, TT etc. 
Thus, numerics, pure numbers and quantities like heat, electri
city, temperature and dielectric constant have no dimensions in MLT.
The following Table shows at a glance the dimensional
formulae for some important physical quantities.
Physical quantity
1.
2.
3.
4.
5.
Area
Volume
Velocity
Acceleration
= (length) 5
 (length) 8
= length/time
= velocity/ time j
Dimensional formula
Momentum
= (mass x velocity)
6. Force = (mass x acceleration)
= rate of change of
momentum
7. Work* (force x distance
or length)
8. Couple* = (force x length)
9. Kinetic Energy*
(i mass x velocity 2 )
10. Potential Energy*
(mass x acceleration
due to gravity x
distance)
11. Power, (or rate of doing work)
= work/time
12. Density = mass /volume
13. Specific gravity=a mere ratio.
1 4. Pressure = force/area
15. Stress = force /area
M*L*T, or simply [L 8 ]
ML*T, or simply [*]
r, or MLT\ or [LT~ l \
^ * or M*LT * or [LT~*]
MxL/T, or [MLT 1 ]
MX (LIT 2 ) XL [ML*T~*] '
MX IL*IT*] = [AfL*/T*] or [ML*T~*]
MIL*, or [ML~*T] or [ML~]
No dimensions
MLT~*IL*, or
MLT*IL*. or
*See Solved Example 1 (6), page 13.
*It will be noted that the demensions of couple, kinetic energy and
potential energy are the same as those for work, because they arc mutually
convertible and energy is just work. Same is the case witn pressure and s/re$5.
6
PROPERTIES OF MATTER
Physical quantity
16. Strain
change of length or volume
** original lengttTor volume
^length volume
length volume
=a mere number.
1*1. Coefficient of Elasticity
= stress/strain
18. Coefficient of Viscosity
= fo rce velocity
~~ area "distance
19. Surface tension =force/length,
or, = energy /area
Dimensional formula
No dimensions.
20. Frequency
21. Angle
I/time
' length /length
 a number.
MLT* LT*
L* ' L
MLT~*IL, or
= [M LT~*] or [MT*]
1/T T 1 , or [AfLT 1 ]
No dimensions.
5. Uses of Dimensional Equations. A careful examination
of the dimensional equations of the various physical quantities
involved in a relation, i.e., an analysis of their dimensions, is of great
help to us in more ways than one, the process beim* known as
distnensionai analysis. Its three chief uses are the following :
(a) conversion of one system of units into another,
(b) checking the results arrived at,
and (c) deriving a correct relationship between different physical
quantities.
Let us consider these in some detail.
(a) Conversion of one system of units into another. It is seen
that a physical quantity is expressed in terms of an appropriate unit
of the same nature, its value being equal to the product of a number
and that particular unit. Further, as shown in 1, its value remains
the same on all systems of units. This affords us an easy method
of changing over from one system of units to another.
Thus, suppose there is a physical quantity of dimensions a, b
and c in mass length and time respectively, /.e., whose dimensional
formula is M a L b T c . Then, if its numerical* value be HJ in one lystem
in vvhich the fundamental units are M Lt L x and T 19 it is clearly equal
tonAM'LfTf].
Also, if its numerical value be w a * n Another system of funda
mental units M 2 , L 2 and T 2 , it is equal to n^MJLfTJ] in this
ystem.
So that, n&
whence,
l ~M
AKB DIMENSIONS
=
So thad;, knowing the fundamental units in the two systems
ind the numerical value of the quantity in one of them, its numerical
value in the other system can be easily determined. Care must,
however, be taken to apply relation (i) above, after expressing
the given quantity in absolute units. Let us consider an example
or two.
(1) To convert a poundal, (the unit of force in the F.P.S. system),
into dynes, (the unit of force in the C.G.S. system). /
We know that force has dimensions MLT~* and that
i Ib. =4536 gms., and 1 ft. = 12 x 254 <w.=3048 cms. So that,
M units in F.P.S. system = 453 6 M units in C.G.S. system,
L units in F.P.S. system = 30'48 L units in C.G.S. system,
and T units in F.P.S. system= T units in C.G.S. system,
the fundamental unit of time being the same, viz., the second, in the
two systems,
.. MLT* poundals = (4536M)(3048 L)T~*.
~ . , 7 4536MX 3048
.r*^ <, tcti Ur, 1 poundal =* ,> r 
log 453*6 = 26567 ' ^ MxLx
log 3048  1 4840
Antilog 4 1407
= l'382xl0 4 s=s 1382 x 10* units in the
C.G.S. system.
Thus, 1 poundal = l382x 10* dynes. </
(2) To convert one Horse Power, (F. P. S. system), into Watt*
(C.G.S. system). We know that
1 H. P. == 550 ft. Ibs.jsec. = 550 X 322 ft. poundals I sec.
and g = 322 /*. /sec. 2 y
Again, as shown in Ex. (1),
M units in F.P.S. system = 4536 M units in C.G.S. system,
L units = 3048 L
and T units . = T 
Since the dimensional formula for power is AfL 2 r~ 8 , we have
H.P. = 550x322(4536M)x(3048L) 2 xr.
F P  550 x 322 x 453 ' 6M X ( 30 ' 48L ) 8 x r  .
H.P. 550 x 32 2 x    
log 550 = 27404
= 550 x 322 x 4536 x (3048) 2 ergs/see.
log 32 2 = T5079
log 453'6 SB 2*6567
550 X 322 x 4536 x (30'48) 2 J^fySL
2 log 30 48  2 9680
10 7
98730"
= 7464 watts.
7 i g 10 = 70000
Antilog 28730
7464
[Because, as we know, 1 joule/sec. =* 1 watt.]
Thus, 1 Horse Power = 7464 watts.]
*This ratio MJM^ if ^ be the unit of mass in the C.Q.S. system
at in the F.P.S. system, i.e., the ratio 'gram to the pound" is called
ion factor.
M t , that " iv
conYcrsion factor,
8 PROPERTIES OF MATTER
(b) Checking the results arrived at. This depends upon what
10 called the principle of homogeneity of dimensions, again due to
Fourier, according to which the dimensions of all the terms on the two
sides of an equation must be the same. This follows at once from the
fact that it is not possible to compare twa physical quantities of
different natures, and that only quantities of the same nature can be
added up together, their resultant being also of the same nature.
If, therefore, in a given relation the terms on either side have the same
dimensions, the relation is a correct one, but if they have not, there is
a flaw somewhere, which must be diligently sought out.
Let us again take a couple of examples :
(1) To check the accuracy of the relation, t = 2ir^i]if 9 for a
simple pendulum.
Here, the term / on the lef fc hand side has only one dimension in
time, or the dimension of t is [ T], its dimensions in both mass and
length being zero.
And, on the right hand side, 2ir has na dimensions, being just a.
number ; / has one dimension in length, or its dimension is [L], those
in mass and time being zero ; and the dimensions of g, the accelera
tion due to gravity, are LT~ 2 , that in mass being zero. Hence the
dimensions of the term, 2n y7// = ^HUlF* or \/~f* = [T], i.e.,
it has only one dimension in time, the same as the term on the left
hand side. The relation / = 2n\/l/g is, therefore, a correct one.
(2) To check the relation S = ut+\ at*, for the distance covered
in t seconds by a body, having an initial velocity u and an acceleration a.
Here, the dimension of the term S on the left hand side is one
in length, or [L], and taking the terms on the right hand side, we have
(/) dimensions of u (velocity) = LT* 1
(U) dimensions oft (time) = T
(Hi) dimensions of J (a number) = Nil
(iv) dimensions of a (acceleration) = LT~*
and (v) dimensions of f 2 (time 2 ) = I* 2 .
., dimensions of the term, ut f \at* = LT~* x T+LT* x T* r
= L+L,
i.e., the dimension of each term on the right hand side is the same as
that oj the term on the left hand side ; hence the given relation i
correct.
A similar dimensional homogeneity will be observed in the case
of any other relation, representing a physical phenomenon. The
method of dimensions has thus a very definite mnemonical value*
and enables the beginner to resolve his confusion between two alter
native possibilities occurring to him regarding a particular half for
gotten formula, as, for example^whether the timeperiod of a simple
pendulum is given by t 2?r\/^ or by f = 2n^/l/g, or whether the
formula iirr 1 gives the surface area or the volume of a sphere etc., etc.
* i.e., value at an aid to memory.
UNITS AND DIMXHIONS 9
(c) Deriving a correct relationship between different pftysical
quantities. The principle of homogeneity of dimensions also enables
us to deduce a relationship between different physical quantities, or,
at any rate, a preliminary form of such a relationship For, knowing
the factors on which a physical quantity may possibly depend*, and
this requires a little physical insight and a certain amount of 'horse
sense' an expression for it can be obtained in terms of these factors,
such that the dimensions of the terms on the two sides of the
expression are the same, the only acceptable form of the relationship
being the one which remains true irrespective of the system of units
employed. A few examples will illustrate the point.
(1) To deduce an expression for the timeperiod of a simple pendu
lum.
The factors on which the timeperiod (/) may possibly depend^
are the following :
(i) the mass of the bob (m),
(ii) the length of the pendulum (/),
(Hi) acceleration due to gravity (g) and
(iv) the angle of swing of the pendulum (6).
Let / be proportional to m a , /*, g c and 6 d . So that,
t=K.m a l*y 6 d , where Kis a constant of proportionality.
Taking dimensions of the terms on either side of the sign of
equality, we have
[T] = [M a ][L*][LT~*Y = M*L*L e T*<. r* and having no
Or, T = M*L d + c T 2e . 1 dimensions.
Since the dimensions of the terms on the two sides must be the
same, we have, equating the indices of M , L and T,
a = 0, b+c = and 2c = 1,
whence, c =  and . b \ =0, or b = .
Therefore, t = K.I*. g~~*.
Or/ t^KVlfg. +J
The value of K can be found out experimentally f, and comes to
2ir ; so that, the required relation is t = 2?r \fTfg.
It will easily be noted, from the above, that
(/) the timeperiod of the pendulum is independent of its mass,
a fact we know to be true by actual experience ;
and (ii) the expression t^/gjl 9 has no dimensions, as it is equal to the
dimensionless constant K 9 and is thus a numeric.
An important deduction emerges from this latter point, viz.,
that if two pendulums having different lengths, (^ and / t ), oscillating
* It is absolutely necessary to take into account all possible major
factors on which our result may reasonably be expected to depend, though one
or more of these factors may get eliminated later. The method, however, ceases
to give any worthwhile result if the number of variables included is more than
six.
tThe value of K can be determined easily by substituting in the relation
obtained, the observed value of /, for known values of / and #.
10 PROPERTIES OF MATTER
at two different places, where the values of the acceleration due to
gravity are o l and g+ respegtively, take time TA and T a to describe
equal arcs, they may have
i.e., the value of the nondimensional expression or the numeric
may be the same for both.
And, if this be so, it means that the two pendulums pass
through exactly the same phase for the same value of r\/g]T. This is a
case of what is called dynamical similarity, and all moving systems of
this type are said to be dynamically similar.
A very interesting and a classic example of this principle is the compari
son of the speeds of fully grown animals with those of their young.
Very reasonably, taking the density of the two animals to be the same and
their muscular strengths directly proportional to the crosssection of their limbs,
we have the ratio between their densities equal to one and similarly that between
their strengths per unit area of crosssection of their limbs, also equal to one ; so
that, if subscripts 1 and 2 refer to the adult animal and to its young respective
ly, we have
ratio of their densities, i.e., ^ /^ f =1
L>i I L, z
md also ratio between their muscular strengths per unit area, i.e.,
From these two relations then, we easily get
L\ L"i
X = 17'
where L^IT^ is the speed of the full grown animal and L 2 /T t9 that of its young.
The speeds of the two animals are thus the same, a result which, at first
sight, appears to.be simply ridiculous. And yet it is an actual fact, the shorter
strides of the young being taken faster than the longer ones of the adult.
(2) To deduce a relationship for the velocity of sound in a material
medium, the temperature of the medium remaining constant.
The velocity K may depend upon (/) the elasticity of the medium
E and (ii) the density of the medium, p ; so that,
V = K.E a ^ b > where K is a constant.
Again, taking dimensions of the terms on both sides, we have
> (" v elasticity stress/ strain
MOLT* I r*r* . I 1 /". I force /area
m j^j. = ro ii rn \ ^rilrJzl
J ' a ratio
l^and density mass/ volume.
Since the dimensions on the two sides must be the same , it
clear that a+b = ; a3b = 1, and 2a = 1,
whence, a = \ and b = a =  J.
Hence, V =
Or, 'V^
UNITS AND DIMENSIONS
The value of K is again determined by experiment, and i*
found to be 1 in this case ; so that, V 5= \/~Ejp
So far only simple cases have been considered. In other cases
the method used above may not always be applicable. Let us con
sider one such typical example by way of illustration of the method
adopted in such cases
(3) To obtain a relation between the distance travelled by a bod)
in time t, if its initial velocity be u and acceleration a.
Let the distance covered by the body in time t be represented
by S = K.u a .a b .t c Then, taking dimensions, we have
J ^
Or, [L] = L a T~* x L b T~* b x T c =
Since the dimensions on the two sides must be the same, we
have
a+b == l...(i) ; a2b+c = ; or, a+2b~~~c = ...(&']
These two equations alone are not enough to give us the values
of a, b and c. Hence we proceed as follows :
Suppose the body has no acceleration. Then,
S = K'u a t c , where K' is another constant.
Taking dimensions, we have
L = L a T~*T c = L a T c ~ a ,
whence, * a = 1 ; and ca = 0, or c = a = 1,
S = K' . ut. ...(A)
Now, suppose the body has no initial velocity. Then,
S = K" a b t c , where K" is yet another constant,
Again, taking dimensions, we have
L == L b T 26 T c = L b T c * b .
b = 1 ; and c 2b = 0, or c = 26 => 2.
Hence S = K".at*. ...(B)
If, therefore, a body has both, initial velocity as well as
acceleration, its equation of motion contains both the expressions,
(A) as well as (B) ; so that, we have
svhere the constants K' and K" can be determined experimentally,
and are found to be equal to 1 and \ respectively. Thus, the re
quired relation comes to be S = t//fa/ 2 .
In addition to the three chief uses of dimensional analysis, dis
cussed above, mention may also be made here of a couple of others.
Thus,
(iv) it is helpful in selecting experiments likely to give some useful
information and avoiding others. In this connection, Lord Rayleigtfs
remark is worth quoting. Says he, 'I have often been impressed by
the scanty attention paid even by original workers in Physics to the
great principle of similitude. It happens not infrequently that
results in the form of 'laws' are put forward as novelties on th*
12 PROPERTIES OF MATTER
basis of elaborate experiments which might have been predicted
a priori after a few minutes' consideration.' How true, indeed !
(v) Then, again, it is a powerful aid to mathematical analysis,
when the problem .happens to be a, complex one and when no experi
ments to solve it are possible. Even if the number of variables
involved in the problem be a large one, dimensional analysis does
help obtain at least a partial solution of it.
^Limitations of Dimensional Analysis. It will be readily seen
from the examples, given above, that the method of dimensional
analysis is after all not quite so simple or straight in its application,
except in obviously easy cases. Very helpful, as far as it goes, it has
also its own limitations. Thus, for example :
(i) Its one obvious drawback is that it gives little or no informa
tion about pure numerics (like t^/yjlin Ex. 1) and nondimensional con
stants (like K in Ex. 2), involved in various physical relations, and
which, therefore, have to be determined by separate calculation or
experiment.
(ii) Then, again, since at best only three equations can be ob
tained by equating the dimensions of [Af], [L] and [T}\ the method is
of no avail in deducing the exact form of a physical relation which
happens to depend upon more than three quantities. For, clearly, of a
given number of quantities involved, the indices of only three can be
expressed in terms of the rest, thus leaving us with a relation between?
the remaining number* of nondimensional groups of terras ; so that,
what we may ultimately succeed in obtaining is just an equation in
terms of an undetermined function.
It will thus be clear that, while the method of dimensional
analysis remains unrivalled and almost unique, in so far as conversion
from one system of units into another and checking the correctness
of physical relations are concerned, its use is not quite so safe or
certain when it comes to establishing a definite or exact relationship
between a given set of physical quantities and, particularly so, in the
hands of beginners.
More often than not, the success of the method depends upon
the proper choice of dimensional constants (like G or c), which have to
be introduced as additional variables. And, it needs a trained,
subtle and intuitive mind, with the solid background of a mature
and a comprehensive knowledge of the subject, to decide, on the basis
of analysis or experience or perhaps just on that of some sort of
inspiration of the moment, what particular variables to select, and
how, when and where to introduce them. A very apt illustration in
support of these remarks is perhaps Raleigh's explanation, by the
method of dimensions, as to why the sky is blue.
That the colour of the sky is due to the scattering of light by
suspended drops of moisture and dust particles etc. (of molecular
size) in the atmosphere is fairly well known. From this basic fact,
Raleigh proceeds as follows :
*v/z., the given number of quantities minus three.
UNITS AND DIMENSIONS 13
Let A t be the amplitude of the scattered wave. Then, the possi
ble factors on which it may depend are
(/) Af the amplitude of the incident wave of light,
(it) I, the linear dimension of the scattering particle,
(Hi) r, the distance from the particle,
And (iv) \, the wavelength of light.
So that, expressing A s in terms of all these variables, we have
A g **K.Afl*r*K,
where K is a constant of proportionality.
Or, taking dimensions, we have
L = L* . U . If . U,
for, obviously, the dimensions of all these quantities are the ame.
viz., L, and those of K = 0,
We, therefore, have
; = a+b+c+f.
Now, we know that the araflfjtude of the scattered light is
</) directly proportional to that of t$e incident light and (//) inversely
proportional to its distance from the scattering particle. This at
once gives us a = 1 and c = 1. And, therefore,
1 ._ i+bl+d, whence, d = 16.
So that, A, = K . A, V r* A 1 "* = K.
Now, as Rayleigh remarks, 'from what we know of the dyna
mics of the situation** / varies directly as the volume of the scatter
ing particle. And, therefore, 6=3.
Hence A s = K . ' 2 . Or, A s oc I/ A 2 .
And since intensity oc (amplitude)*, we have
intensity of scattered light, I s oc I/ A 4 .
It thus follows, as a natural consequence, that the wavelength of blue
light being roughly half that of red light, the in tensity of scattered blue light
is sixteen times that of scattered red light and that the sky, therefore, appears to
us to be blue.
The student will appreciate how, in capable hands, the method of dimen
sional analysis can be made to yield results beyond the pale of elementary
analysis.
SOLVED EXAMPLES
1. Deduce the dimensions of (a) the Coefficient of Viscosity, and ( the
Constant of Gravitation (G).
Obtain a formula for the time of swing af a simple pendulum from a know
Ledge of the dimensions of the physical quantfp Involved. (Punjab)
(a) We know that the coefficient of viscosity (17) of a liquid is given by
the relation, *? = w jpr/8v/,
vhere P is the pressure difference between the two ends of the capillary tube ;
% its radius ; /, its length and v, the rate of flow of the liquid through it, or the
>olume of liquid flowing out per second
*v/z., the ratio of the respective amplitudes of the incident and reflected
ight.
n PEOPEETIES OF MATTER
Therefore, taking dimensions of all these quantities, we have
dimensions of P = ML" 1 !"* [see Table on page Si
M r* ^
v = IT" 1 [v rate of flow = volume/time.
/  L
and and, Bare z^ro. [Both being numbers.
Hence, dimensions of coefficient of viscosity TJ, are
(b) We know that the value of G, the Garvitational Constant, is given by
the relation, G^CQd*IM.m.l. ,
where C is the restoring couple per unit twist of the wire; B, the angle of twist oj
the wire ; d, the distance between the centres of the near large and small balls ; M
andm, the masses of the large and small balls respectively and /, the length of the
torsion rod, (Cavendish's experiment).
Therefore, taking the dimensions of the quantities involved, we have
dimensions of C (couple) = ML 2 T~* [See Table on page 5
9 (angle) =
d 1 = L 2
M = M
m = M
Hence dimensions of G are or
For answer to the second part of the question, see page 9, (Ex. 1).
2. Find the unit of length if one minute be the unit of time ; one stone,
the unit of mass, and one poundweight, the unit of force. (g 32'2 ft per sec 2 ).
We know that 1 Ib. wt. = g poun dais 32'2 poundals, and that the di
mensions of force are MLT~~ 2 .
Now, M units in the new system (14M) units in the ordinary system,
[v 1 stone *= 14 Ibs.
T ,, ,, ,, = (607) units in the ordinary system,
and let L ,, ,, ,, ,, = (xL) units in the ordinary system.
log 32*2  1 5079
2 log 60 = 35564
50643
Jog 14 = 1J461
Antilog 3*9182"
 8283
Then, 32'2 MLT~ Z =14MxxLx[6QT]~* units in this system
= 14M.*L.60~ 2 r 2 units
Or, 322 JJf. And/. x= 14 ^8283.
Or, the unit of length in the new system would be xL
=8283xL, i.e., equal to 8283/h [since [I] = l ft.
3. If the acceleration due to gravity be represented by unity and one
second be the unit of time, what must be the unit of length ?
In the ordinary system, in which the unit of length [L] = 1 cm., and that
of time [71 = 1 sec , we have
unit of acceleration, [dimensions LT~*] = 1 cm. /.sec. 2 , and acceleration!
due to gravity equal to 981 cm./sec*.=9B\ LT~*.
If the unit of length, in the new system, be LI, we have
L acceleration due to gravity, on this system, == 1 xL x sec 2 . = L{T~* 9 '
the unit of time being the same, i.e., 1 second, in this system also.
I^T 2 981 IT 2 . [1.
Or, LI 981 L ; that is L l 981 cms., since L 1 cm.
Thus, the unit of length in the new system is equal to 981 cms.
4. Given that the unit of power is one million ergs per minute, the unit of
force is 1000 dynes and the unit of time, 1/10 sec., what are the units of ma**
and length?
Here (a), unit of power, [dimensions ML*T~*] 1000,000 Srgslmt.
= 1000,000 16Q ergs per $rc~
(b) unit of force, [dimensions MLT~*] IQQQ dynes,
and (c) unit of time [dimension T] 1/10 sec.
UNITS AND DIMENSIONS
/. multiplying (a) by (c), we have
unit of power x unit of time = unit of work.
 [ML 2 T*][T] = ML*T~*.
_ 1000,000 J_ _ 10*
60 x 10"""" 6 '
Dividing this unit of work by the unit of force, we have
unit of distance or length ~ , x JQQQ , = ~~cms.
Now, from (b) we have
unit of mass MIT' 2 f ^ = MLT~* x ^ F v maw
r v iii7 of force =1000 dyne&
.  1000x(l/10) 2 1000x3 r r ' *,,
Or.iniro/mai* 5/3 lOQxS* 6 ^! and im/^/arc.
Therefore, the units of mass and length, in the given system, are 6 ,gms. and"
5/3 cms., respectively,
5. If the fundamental units are the velocity of light in air, the acceleration
of gravity at Greenwich, and the density of mercury at 0C, find the units or
mass, length and time. (Velocity of light  3 x i0 10 cm \see ; acceleration of gravity
at Greenwich = 9 81 x 10 2 cm.isec 2 ; density of mercury = 13*6 gm. per c.c.).
Here, (a) unit of velocity, (dimensions LT~ l )=3 x 10 10 cm.fsec.
(b) acceleration, ( Lr 2 )=9*81 x 10 2 cm.jsec.*
(c) ,. density, ( ML~ 3 ) = 13*6 gm.jcm*
log 3 *=0'4771 j /. dividing (a) by (b), we have
8 log 10 =8000p t 3xl0 10 3xl0 8
8 477 1 j mlt f time > ( 7) = 9 X 1 x 1 2 ^ 9" 8"P '
log 9*81 09917 1 !!L . _!.
Antilog 7*4854 j Substituting this value in (a), we have
3'058 x 10 7  mit O f length, (L) ^LT~ l .T.
j =3 x 10 10 x 3058 x 10 7 =9 174x 10" cms.
log 136= 11 335 j And, from (c), we have
log(9'174x 10 17 ) ; unit of mass, (M) = ML 3 xL 8 =13'6x(9174x 10 17 ) 3 .
=53 8878 
Antilog 550213
rosixio 65
Thus, the required units of mass, length and time are
1 051 x 10 65 gms., 9'174x 10 17 cms. and 3'058x 10 7 seconds,
respectively.
6. If the units of length and force he each increased four times, show that
the unit of energy is increased sixteen times.
We have unit of energy = unit of force x unit of distance.
If now, the units of force and distance be made four times each, they
would be 4[MLT~*] and 4L respectively, and, therefore, the new unit of energy
would be 4Afr a x4L=16AfL J T 2 , which is sixteen times ML 2 r~ 2 , the ordinary 1
unit.
Thus, we see that by increasing the unit of force ^md length four times each,
the unit of energy is increased sixteen times.
1. Show by the method of dimensions that the relation, C=nnr 4 j2l for
the couple per unit twist of a wire of length /, radius r and cefficient of rigidity
, is a correct one.
Let us take the dimensions of the terms on the two sides of the sign of
equality and see if they are the same. Thus,
dimensions of C (couple) =* ML 2 r~*
n (rigidity} = ML^ 1 T" Z [same as for elasticity ~
r* (radius? L*
/ (length) L
and. " and 2, being numbers, have no dimensions.
16 PROPERTIES OF MATTER
Therefore, the dimensions of the term wirr*/2J are
\AT IT* v f *
ML 1 XL
_
JL
*he same as for C, on the left hancTside.
Hence, the relation C wrr*/2/ is a correct one.
8. Test by the method of dimensions the accuracy of the relation
tf = 2 \Afc* +?*)/# f r tne timeperiod of a compound pendulum.
If the relation be correct, the dimensions of the terms on either side of
the sign of equality must be tne same.
Let us put the relation as t = 2A / _ 4.
V lg g
Now, the dimensions of / = [T]
K* = [L 2 ], K being the radius of gyration.
I  [L]
Therefore, the dimensions of the term on the right hand side are
V2^" vri+5 *"
Thus, we see that the dimensions of the terms on either side are the same,
viz., [T]. The relation is, therefore, a correct one.
9. Find the dimensions of velocity and acceleration. Assuming that
when a body falls from rest under gravity the velocity v is given by Kg*W 9 where
h is the distance fallen through, g, the acceleration of gravity and K, p and q are
Constants. Show, by a consideration of the dimensions involved, that v^K\/gh.
(London Higher School Certificate)
'For answer to part one of the question, see Table on page 5.
We are given that v = Kg*hP. Taking dimensions, therefore, we have
dimensions of v = ^ =* LT~ l .
dimensions of h 9
,, K = 0, for it is a constant or a mere number.
Therefore, the dimensions of the term,
Since dimensions on both sides of the sign cf equality must be the tame,
AVC have
Or, p+q 1 and 2p = 1, whence, p i and 0= i.
KgW Kg* . A* JSTV^
Or, v = K^/giT
10. The frequency of vibration (n) of a stretched string is a function
of the tension (T), the length (/) and the mass per unit length (p). Prove that
"~ /
JLet n DC T a /V. Then, taking dimensions, we have
dimensions of !// ** > M*L*T~* 9 or T~\
TTKITS AND DIMENSIONS 17
Dimensions of J a , (force)* = [MLT~*]* M*L*T** 9
, t /*, (length)*  /A
r rami* rM"\ c ., fr f
" pe > U^/J LrJ =JV/a " c 
dimensions of the term T a L*P c = M a l*T Za .L b .M Lr c t
Or, yi . jv/f +
Since dimensions of n must be the same on both sides, we have a 4 c ;
c = Oand 2a = 1, or 2a = 1 i.e , a = i, and hence if c=0, or c= .
Also 4A(~i) = 0, or fh&~h4 = 0, i.e., lt6 = 0, or 6 1.
i _i 1 A / F
Therefore, T a l b c T 1 . / 1 . p f = / V "p
_
And hence n oc
11. The time of oscillation (n of a small drop of liquid under sur
face tension depends only on the density (/>), the radius (a), and the surface
1 _l
tension (T). Show that the period of oscillation is KP^ . a 2 .T *, where K
Is a numeric.
Let t = K? a a b T*. So that, taking dimensions, we have
dimensions of t T,
a* = \L\ b = ZA
7^ = [/V/r 2 ]^ = M C T~ ZC , [See page 5.
and K has wo dimensions, being a numeric.
dimensions of the term K?a b T c ~= ML* a U>McT* c .
Or. = M a + c L* a +*T 2 c.
Since the dimensions of the terms on both sides must be the same, we
have
/jhc^O; 3a+Z> = 0, and 2c 1.
Or, c = J and /. a = i and ^ f .
Hence //iff f/m^ c/ oscillation of the drop, t = X"p2 . a* . T *
12. Explain the Principle of Homogeneity of dimensions in a physical
equation.
Assuming that the mass M of the largest stone that can be moved by a
flowing river depends on K, the velocity, p, the densitv of water and on g, show
that M varies with the sixth power of the velocity, of flow in the river.
(Punjab)
Let M depend upon K a , P* and g c .
So that, M = KV a &&g c . [K being a constant*
Taking dimensions, we have
dimensions of M = [MJ
[ L T =
1 T I
and *T has no dimensions, being a mere number.
18 PROPERTIES OF MATTER
Af L a :r a Af*Zr' & r~ w . Or, M = ,
5/fK* ffo dimensions on both sides must be the same, we have
a36+c 0, 1) = 1, and a2c = 0, or a\ 2c = D,
So that, 0+c = 3 and 0+2c = 0. And .*. c = 3 and & ~ 6 r
Hence M = JCKV* 8 .
Or, M oc K f .
Le. 9 the mass M varies with the sixth power of the velocity of flow.
EXERCISE 1
1. If 10000 gms. be the unit of mass, 60 sees., the unit of time, and the
acceleration due to gravity (981 cms.lsec*.), the unit of acceleration, what
would be the unit of energy in ergs 1 Ans. 3'465 x 1C 12 ergs.
\j) Convert by the method of dimensions, 4*2 x 10 7 ergs into foot poun*
dals (1 //. = 30'48 cms., and 1 Ib. =453'6 gms.). Ans. * 96 6ft. poundals*
3. Deduce the dimensions of (/) specific gravity, (a) surface density and
(Hi) angular velocity.
Show that the kinetic energy of a body of mass m, moving with a velo
city v, is given by kmv 2 , where k is a constant.
4. Test, by the method of dimensions, the accuracy of the following
relations :
(i) v 2 u 2 2aS, connecting initial velocity u, final velocity v, accelera
tion a and distance S covered by a body.
(//) S = ut + Jflf a connecting distance S with initial velocity u, time t
and acceptation a of the body,
/\ _ V where v is the w^aw density of the earth t r, its radius
(in) p= 4^0 * g y the acceleration due to gravity and (7, the gravita
tional constant.
5. Assuming that the excels pressure (p) inside a soap bubble depends
on (() the surface tension (T) of the soap film and (//) its radius (n, show, by
your knowledge of dimensions, that it is directly proportional to the former and
inversely proportional to the latter.
[Hint Simply show that p^k.Tfr, whence it follows that p varies directly
as T and inversely as r.j
6. A drop of liquid is suspended in another liquid of the same density
but with which it is immiscible. If the drop is distorted fiom the spherical
shape and released, deduce, by dimensional methods, a formula for its period
of oscillation (/), given that the latter depends on surface tension T, density?
and dropradius r. _
Ans. tk \I P JL~, where fc is a constant.
/Ch Convert, by the method of dimensions, a pressure of Impounds wt.
per square inch into dynes per sq. cm. Ans. 7*912 x 10 4 dynes Jem 21 .
8. Show that when bodies of geometrically similar form and of the same
material, differing only in dimensions, vibrate in the same manner, the vibra
tions being due to,the elasticity of the material, their periods are proportional
to their dimensions.
_
. Proceeding in the usual manner, show that t klj tfE, where I /*
the linear dimension of the body, p, 1/5 density and E, the elasticity of the material.
Since p and E are the same for all bodies, t varies directly as /.]
^ Calculate, by the method of dimensions, the number of footpounds
in me calorie. (Given that 1 ca/0rie=4'2x 10 7 ergs ; #=32/r./,sec 2 . ; 1 /^.=453'6'
gms., and 1 iwcA=2'54 cms.). Ans. 3'1 15.
10. If in a system of units, the unit of length be 1 mile and that of time,.
1 hour, what will be the value of y ? Ans. 14*88 miles Isec.*
11. The time of oscillation / of a small drop of a liquid under surface
tension depends upon the density d, radius r and surface tension S. Prove
dimensionally that t oc \J *. (Punjab, 1947),
V S
UNITS AND DIMENSIONS 19
12. Explain what you mean by the dimensions of a physical quantity;
calculate the dimensions of Young's modulus.
Assuming that the period of vibration of a tuning fork depends upon the
length of the prongs, and on the density and Young's modulus of the material,
find, by the method of dimensions, a formula for the period of vibration.
(Calcutta, 1950)
Ans. [ML" 1 ! 1  2 ] ; t oc iVdIY,
(where / is the period of vibration ; /, the length ; d t the density and Y, the value
of Young's Modulus for the material of the fork.)
*13. Using the method of dimensions, obtain an expression for
(/) the acceleration of a particle moving with a uniform speed v, in a
circle of radius r ;
(//') the tension Tin a uniform circular wire of radius r and mass m per
unit length, rotating in its own plane with an angular velocity o>, about am axis
passing through its centre and perpendicular to its plane ;
(*ii) the mass M of a planet round which a satellite completes its orbit of
radius r, in a timeinterval T.
Ans. (i) K.v*lr ; (ii) K.mrW, where Ki$ a constant ; (///) M oc r*!GT 2 .
*14. Obtain an expression for the height h to which a liquid, of density p
and surface tension Twill rise in a capillary tube, of radius r, given that /zocl/r.
T
Ans. h~k.  , (k being a constant).
r  P g
*15. Assuming that the viscosity 73 of a gas is proportional to the mean
free path X of us molecules, show that, if the temperature be kept constant, it is
independent of the density p of the gas.
[Hint. First obtain an expression for ?), in terms of p, X, c, (the root
mean square velocity of the molecules) and />, the diameter of a molecule.
Then, since r, oc x, we shall have 73 fc.p.c.X, (where A: is a constant). Again,
since p is inversely proportional to X, / e., p=A;'/X, (where &' is another cons
tant), we shall have f\=k.k'c^ showing that ?j is independent of p.]
*16. Show that if the linear dimensions of the whole of Cavendish's or
Boys' method f jr the determination of G be changed, the sensitiveness of the
apparatus remains the same.
*17 Show that the volume of a liquid, of coefficient of viscosity *j, flowing
per second through a tube of circular crosssection is given by K=wpr 4 /8r</. where
p is the excess pressure between the ends of the tube, r, its radius and /, its
length.
*18. If the resistance of a liquid to the motion of a body through it with a
velocity v, be proportional to v 2 , show that it is quite independent of the visco
sity of the liquid.
*19. A Nicholson's hydrometer of mass w, floating in a liquid of density p,
is given a slight downward displacement and then released. Obtain an expres
sion for the timeperiod Tof its oscillation. (Assume the area of crosssection
of its neck to be a.) Ans 7=2" Vm/tW
*20. A (/tube of uniform crosssection contains mercury up to a height
h in either limb. The mercury in one limb is depressed a little and then released.
Obtain an expression for its timeperiod of oscillation. Ans r
[Hint Just put T=*k.d*tfg and show that
where T is the timeperiod of oscillation of mercury and d, its initial displace
ment, K being the usual const airtof proportionality. For small values of rf, a=0
and experiment gives K=* w\/2. Substitute their values and obtain the
result.]
Note. The questions marked^ith an asterisk are of rather an advanced
character and may be attempted whin some confidence has been gained with
others.
CHAPTER II
iviOTION ALONG A CURVE THE PROJECTILE
7. Rotation Suppose we have a rigid body, with a fixed
axis, within or without it. Then, if a force be applied to it, it cannot
move bodily, as a whole, relatively to the axis, i.e., no motion of
translation is possible ; but it simply moves round or rotates about
the axis, such that every particle of it undergoes the same angular
displacement. A body, so rotating about a fixed axis, is said to per
form rotatory or circular motion.
The force, producing rotatory motion about the fixed axis,
called the axis of rotation, is said to have a moment about that axis,
which is measured by the product of the force and the perpendicular
distance between its line of actioi and the axis of rotation. Obviously,
therefore, if either of these be zero, the moment, or the turning tendency
of the force, will be zero, for the prod'ict of the force and perpendi
cular distance between the axis and the line of action of the force is,
then, zero.
It* fie rotation produced bs anticlockwise, the moment of the
force is said to ba positive, *ad if it b3 in the clockwise d ration, the
moment is said to be negative. And, since th^ m >m3nt of a force is
a vector quantity, it follows that if a number of forces act simultane
ously on a body, the algebraic sum of their individual msmints about the
given axis of rotation will be equal to the moment of their resultant
about it.
8. Angular Velocity. Let a body rotate about a fixed axis
through 0, (Fig. 1). Then, the particles composing it, at any distance
from 0, such as at A, B 9 C, etc., complete
one rotation in the same time i.e., they
describe the same angle in the same time,
and, therefore, the angle described by
them per unit time is the same. This
angle described by a rotating body per
unit time is called its angular velocity and
is usually denoted by the Greek letter a>.
Thus, if the rate of rotation of a body be
uniform, i.e., if its angular velocity be
constant, and it describes an angle 9
(radians) in tima / (seconds), we have
angular velocity of the body, a> = 6/t.
If the body makes n rotations in time /, the angle described by
it is equal to 2irn. Or, = 27T/I.
And, therefore, its angular velocity 01 = 2irn/t.
If, however, the velocity be not constant, it may, at a given
instant, be expressed in the fornotai = d0/dt, where d0 is the small
angle described by it in the small iAryal of time dt<
*
MOTION ALONG A CUBVE THE PBOJECTILE 21
Now, although the angle described by all the particles of the
body in a given time t is the same, the linear distances ^travelled by
them are different. Thus, the particles at A, B, and C, (Fig. 1), cover
the linear distances AA', BB' and CC' respectively, (which are arc$
of radii OA, OB, OC), depending upon their respective distances from
the axis of rotation through O.
If OA =r l5 OB = r ? and OC = r t ,
clearly, arc A A' = r l 9 ~v arc = radius* angle
' _ r Q subtended by it.
,, uu '2 U u
and CC = r 3 8.
.. linear velocity of A = rrf't, that of B = r t 0/f, and that of
C = v //.
Or, in general, linear velocity v of a particle at a distance r from
the axis of rotation is r6/t.
Or, v = roj, [v 0\t = o>.
i.e., linear velocity = distance from the axis of rotation x angular
velocity.
9. ^Angular Acceleration. If the angular velocity of a rotat
ing body be not constant, it is said to have an angular acceleration,
which is defined as the rate of change of angular velocity. It is
usually denoted by the symbol dwjdt. Thus, if the angular velocity
of a particle about a given axis changes from > to a/ in time /, its
rate of change of angular velocity, or its angular acceleration is,
clearly, (a/ co)/f, or dw/dt, ifdfo>be the change in angular velocity
in time dt.
Now, if the distance of the particle from the axis of rotation be
r, its linear velocity changes from ru> to ro/ in time t, and, therefore,
rate of change of its linear velocity, or its linear acceleration, is given
by
_/o/ rw roj( r w\ da>
a ~ ~ t = " t ~~ T ' ~dt '
Thus, linear acceleration = distance from axis of rotation
X angular acceleration.
10. Couple. When two equal, opposite, parallel and non
collinear forces act on a body, (Fig. 2), bringing about rotation, (with
no motion of translation), they are said to
constitute a couple, the turning moment of
the couple be.ng measured by the product of
one of the forces and the perpendicular distance
between them, or the arm of the couple, as it is
called.
Thus, moment of a couple, C
= one of the forces x arm of the couple. Fig. 2.
The moment of the couple (also sometimes referred to as the
torque), acting upon a body , is quite independent of the position of the
axis of rotation. For, if the t w ^P*ces F and F, (Fig. 2), constitut
ing a couple, act at points P andK, and if the axis of rotation passes
through P, there is no moment IS the force acting at P about it and
the moment oi the force acting mt Q is FxPQ, and therefore, the
22 PROPERTIES OF MATTER
moment of the couple is FxPQ. And, if the axis of rotation passes
through any ojher point 0, the moment of the couple about it is
equal to the algebraic sum of the moments of the forces P and Q
about it, i.e., equal to (FxOQ)(FxOP)=FxPQ, as before.
The same will be true for any other position of the axis.
11, Work done by a Couple. Work is done by a couple in
rotating the body on which it acts, the amount of work done being
equal to the product of the couple and the angle of rotation "of the
body, as will be clear from the following :
Let the axis of rotation of a body, acted upon by a couple,
pass through P, the point of application of one of the forces, consti
tuting the couple, (Fig. 3).
Now, if the body rotates through an
angle d&, the point Q moves through a
distance PQ.dQ, where PQ is the perpendi
cular from P on to the line of action of the
force T 7 , acting at Q. Therefore, the work
done by this force is equal to FxPQ.dB.
And, since the point P does not move, no
Fig. 3. work is done by the force at P. Thus,
the work done by the two forces, i.e., by
the couple, in rotating the body through an angle dO, is equal to
Hence, work done by the couple in rotating the body through
the whole angle is obtained by integrating this expression, for the
limits 6 = and = 0.
Or, work done by the couple in rotating the body through
the whole angle Q is given by
W  P F.PQ . dd =F.P0 1 d0.
Or, W = F.PQ f e T = F.PQ.8.
Now, F.PQ is the moment of the couple C, acting on the body.
" ^ix work done by the couple in rotating the body through angle $, i.e.,
W as C.0 = couple x angle of rotation.
Now, in one complete rotation, the body describes an angle 2tr;
co that,
work done by the couple in one full rotation of the body =2?rC.
And .. work done by the couple in nfull rotations of the body*=*ZvnC.
12. Relation between Couple and Angular Acceleration. When
the resultant couple acting on a body is not zero, it produces an
angular acceleration in the body. Let us deduce the relation bet
ween the two.
In Fig. 3, the couple C, acting on the body, causes it to rotate
about the axis of rotation through P.
Breaking up the couple and tHjk body into small elements, let
ah element SC of the couple cause tilrotation of an element of mass
8m of the body situated at Q. Thei^ince couple = force x distance,
MOTION ALONG A CURVE THE PROJECTILE ^
the farce acting on the mass w at Q is == 8C/r, where r is the arm 01
the couple. And, since a couple consists of two equal, opposite and
parallel forces, it follows that an equal, opposite and parallel force if
also acting at P.
Again, since force mass X acceleration, the linear acceleration of
the particle 8m at Q~8C/r.Sm. But, if angular acceleration of th
particle be dw/dt, its linear acceleration is also equal to
[see 9].
SC da) n ~ dot ,
.c. sr =r ,j~. Ur, SC ~ ~j.r z .dw.
Or, C = (da>ldt)Z.r*.8m.
Now, J?r a .Sm == /, the moment of inertia of the body about the
axis of rotation, (see 27).
Or, Couple = moment of 'inertia X angular acceleration.
13. The Hodograph. When a body describes a curvilinear
path, so that its motion is accelerated and also changes in direction,
its acceleration and its path may easily be determined by means of
what is called the hodograph of its motion.
The hodograph may be defined as an auxiliary curve, obtained by
joining the free ends of a moving vector representing the velocity of a
moving particle along any path.
For instance, if a point P moves along a curve ABC, [Fig. 4 (a)]
such that its velocities are v,, v 2 , and v s . ..respectively at A, B and
C etc., then, if we take any point O and draw straight lines, i.e.,
vectors, Oa. Ob and Oc, [Fig. 4 (&)], representing the velocities of Pat
A, B and C, in magnitude as well as in direction, the curve passing
through a, b and c is the hodograph of the motion of P,
(a)
Fig. 4.
Now, different cases arise :
(/) If the point P be moving with a uniform velocity along tfa
same direction, the points a, b, c, etc. will all lie in the same plac<
and the hodograph will, therefore, be a single point.
(ii) If the point P be moving with a variable velocity, but in the
sapie direction, the hodograph will be a straight line, passing through
0, For example, in the case of a body falling freely under the action
of gravity, the hodograph will b a vertical line, passing through O
24
PROPERTIES OF MATTER
(tit) IfP be projected with a horizontal velocity, the patfo
described will be a parabola, (see 20), and both the direction and
magnitude of the velocity will change. The horizontal velocity will
throughout remain constant and equal to the initial horizontal velo
city, because the acceleration due to gravity acts vertically down
wards. The points, a, &, , etc. will, thfrefore, always be at the
same horizontal distance from O, and the hodograph, in this case,
will thus be a vertical line, not passing through O.
(iv) If the path of P be a closed curve, the hodograph will also
be a closed curve. For example, if P moves in a circle with a uniform
speed v, the hodograph will also be a circle of radius v, because all the
lines, Oa, Ob, Oc< etc. will be of the same length v. If on, the other
hand, it moves in a circle with a variable speed, the hodograph might
be an oval curve about the point O.
14. Velocity in the Hodograph. An important property of the
hodograph is that the acceleration of P at any point on the curve
ABC is represented, in magnitude as well as in direction, by the velo
city of the corresponding point on the hodograph, as can be seen fronu
the following :
Let A and B be two points, close together, [Fig. 4 (a)], and let
P move from A to B in time t lt such that its velocity v v at A is changed
to V 2 at B. l h
Further, let another point/? describe the hodograph abc, [Fig. 4
(b)}, while P describes the curve ABC.
Then, clearly, the point p moves from a to b in time t, and its
velocity is, therefore, equal to ab/t.
But, since oa represents the velocity of P at A and ob, that at
B, ab represents, in accordance with the law of triangle of velocities,
the change in velocity of P in time t, and, therefore, the we of change
of velocity, or the acceleration of P, is represented by abjt i.e., by the
velocity ofp in the hodograph.
We thus see that, at any instant, the acceleration of P is given by
the velocitv,ofp in the hodograph of its motion.
IS/ Uniform Circular Motion. The above affords us a very
simple method of determining the acceleration of a body, moving in a
circle.
Let P move in a
circle, with cei.treOand?
radius r, with a uniform
speed v, [Fig. 5 (a)].
Then, the hodograph is
also a circle, of radius v,
[Fig. 5 (b)}.
Now, the velocity
of P at any instant is
at right angles to the
radium of its circular
Fig. 5. path, passing through P,
Therefore, oa is perpendicular to OA aaJ ob is perpendicular to O
*nH / AOB as / aob as B fin circular measure).
MOTION ALONG A CTTBVE THE PROJECTILE
If P takes time t to describe the arc AB, its velocity v
= r0/t, whence, 6 = vt/r.
And, the velocity of the corresponding point p, in the hodo
graph, is abjt = v&jt.
Since the velocity of/; in the hodograph gives the acceleration 
of P in its actual path, we have acceleration ofP
v0 __ v vt __ v 2
= t t x r = j
And, since a6 is small, it is, in the limit, perpendicular to oa, or
parallel to AO.
 Thus, the acceleration of Pis v 2 /r and is directed along the radius
or towards the centre of the circular path in which it is moving.
Further, since v = r.aj, (where o> is the angular velocity of P),
we have
acceleration of P, also = r 2 .o> 2 /r = roA
Alternative Method. The acceleration of a body, executing
a uniform circular motion may also be found out directly as follows :
Let a particle move with a uniform linear velocity v, in a circle
of radius r, (Fig. 6), and let it cover the small distance from A to B
in a small interval of time bt, describing an angle
80. Then, clearly, its angular velocity, o = 86 //. $"
The direction of the linear velocity is at
every point, tangential to the circle at that point
and is, therefore, represented by the tangent AC
at the point^, and by the tangent BD at the
point B, whtre AC BD.
Now at A, the entire linear velocity is
along AC, there being no component of it along
AO, which is at right angles to AC. And,
revolving the velocity at B into two rectangular
components, one along AO and the other, at
right angles to it, we have the component along
or parallel to AO, represented by BE = v sin 8n,
and the component at right angles to AO, pj g 5.
represented by BF= v cosSti.
If 80 be very small, sin 80 = 86 (in radians), and cos 8$ = 1.
So that, component BE, parallel to AO = v.S0,
and component BF, perpendicular to AO = v.
Thus, if B be very close to A, there is no change in the velocity
of the particle along the perpendicular to AO, for it remains the same
v, but an additional velocity v 80 is acquired by it along AO. And,
since this velocity is acquired in time 8t, the acceleration imparted
to the particle is v.80/8t } = Vo>, where Ml** =,.. the angular velocity
of the particle.
:26 PROPERTIES OF MATTER
V V^
Or, acceleration of the particle s= tto = v, = .
Now, because the magnitude of the velocity remains the same
at every point on the circular path of the particle, it follows that the
acceleration must be acting in a direction perpendicular to the direc
tion of the velocity at that point, i.e , along the radius of the circle, or
else it will also have a component along the tangent at the poiit, or
along the direction of the velocity at that point, which will, there
fore, no longer remain constant.
Since this acceleration acts along the radius of the .circle, or
towards the centre of the circle, it is called radial or centripetal
meaning centreseeking acceleration, (from 'peto* I seek).
Thus, centripetal acceleration = , or, =  = rco 1 .
And, if n be the number of revolutions made by the particle
per unit time, we have w = 2irn.
centripetal acceleration, also = r.(27r/i) a = 47rVr.
Even if the path be not exactly a circle, but any other curve,
the value of the acceleration is v 2 /r, where v is the linear velocity, and
r, the radius of curvature of the path at the point considered.
16. CentrigetgLEttcce. According to Newton's first law of
motion, a body must continue to move with a uniform velocity in a
straight line, unless acted upon by a force. It follows, therefore, that
when a body moves along a circle, some force is acting upon it, which
continually deflects it from its straight or linear path ; and, since the
body has an acceleration towards the centre, it is obvrous that the
force must also be acting in the direction of this acceleration, i.e.,
along the radius, or towards the centre of its circular path. It is called
the centripetal force, and its value is given by the product of the mass
of the body and its centripetal acceleration. Thus, if m be the mass
of the body, we have
centripetal force = mv<o = wv 2 /r, or, = mrof *=* AnWrnr.
Numerous examples of centripetal force are met with in daily
life. Thus, (/) in the case of a stone, whirled round at the end of a
string whose other end is held in the hand, the centripetal force is
supplied by the tension of the string ; (') in the case of a motor car or
a railway train, negotiating a curve, it is supplied by the push due to
the rails on the wheels of the train and (Hi) in the case of (a) the
planets revolving round the sun, or (b) the moon revolving round the
earth, by the gravitational attraction between them.
If this force somehow vanishes at any point in its circular path,
the body will fly off tangentially to it at that point, for it will no
longer be compelled to move in the circular path.
17. Centrifugal Force. The equal and opposite reaction to
the centripetal force is called ihe centrifugal force, because it tends to
bake the body away from the centre, (from fugo* I flee). Centripetal
force and centrifugal force being just action and reaction in the sense
MOTION ALONG A OTTBVH THJB PBOJBOTILB 2T
of Newton's third law of motion, tfio>4mmerical values of the two are
the same, viz., mv*/r = mrof = 47rWftr.
Thus, in the case of a stone, whirled round at the end of a string,
not only is the stone acted upon by a force, (the centripetal force),
along the string towards the centre, but the stone also exerts an equal
and opposite forc3, (the ce^tjfcfugal force), on the hand, away from the
centre, also along the string.
 18. Practical Applications of Centripetal and CentrjfugalJorcS>
1. Road Curves. The centripetal force being directly propor
tional to the square of the linear velocity of the body and inversely
proportional to the radius of its circular path, the radii of curvature
of road curves must be large and the speed of the vehicles negotiating
them slowed down, in order to keep the value of the centripetal force
required within reasonable limits.
2. Rotating Machinery. The centrifugal force being proportional
to w 2 , where n is the number of rotations made by the body per
second, the spokes of a wheel, joining its outer revolving parts to
the axis of rotation, experience an outward force, away from the
centre, and are, therefore, in a state of tension, and may give way if
the value of n is very large. So is the case with the parts of other
rotating machinery, connecting its outer revolving parts to its axis
of rotation. In other words, there is a limit set to the value of n by
the tension these connecting parts can withstand. This fact is
always kept in view while designing highly rotating machinery, like
armatures of motors and dynamos etc.
Let us, as a specific example, discuss the case of a belt or a
string rotating at a high speed over a pulley etc.
Let the string rotate in a circle of radius 'r , (Fig. 7), and Jet its angular
velocity be o>. Consider a small portion AB of the string, of length / and
subtending an angle 20 at the centre O
of the circle. This portion is obviously
subjected to a tension T, at either end, by the
rest of the string as shown. Resolving these
tensions T and T at A and B into two rec
tangular components along and at right
angles to PO, (where PO passes through the
midpoint of AB), we find that the compo
nents T cos at right angles to PO are equal
and opposite and thus neutralise e?ch other,
but the components T sin $ along PO act in
the same direction. So that, we have
resultant tension on portion AB of the string Fig. 7.
= 2Tsin in the direction PO.
And, the centrifugal force acting on portion AB of the string
mass of AB x r 8 , in the direction OP.
If m be the mass per unit length of the string, clearly,
mass of AB = mx /.
And .*. centrifugal force acting on portion AB of the string
mx/xr<w*, in the direction OP 9
For equilibrium, therefore, 2T sin 9 m/ro = rn.2r0.ri* 1 . p.* clearly,
If B be small, we have sin B 0. So that,
2T & m.2rQ.r<** 9
whence, T mrV.
28 PBOPBBTIES OF MATTER
It will thus be seen that due to the centrifugal force, the tension in the
string is very Hgh. Indeed, if the rapidJy rotating chain or belt be pushed off
the pulley, it will run along like a rigid hoop. *
The same is true about other rotating bodies which are always under
a state of elastic stress. It is this stress which sets a limit to the speed up to
which the flywheels can be rotated safely. Again, it is as a consequence of
this stress that the tyres of racing cars get stretched and there is a danger of
their being cast off the rims and flung out, at very high speeds.
3. Revolution of Planets and the Length of the Year. In the case
of a planet revolving round the sun, it is the gravitational force of
attraction between the two which supplies the centripetal force,
necessary to keep it moving in its neatly circular orbit. Now, the
gravitational force between two bodies is directly proportional to the
product of their masses and inversely proportional to the square of
the distance] bet ween them ; so that, if m and M be the masses of the
planet and the sun respactively and r, thj distance between them (or
the radius'of the planet's orbit round the sun), we have
M, '
grarfationa! pull = '"f .G = *. pttin '
r^ r 2 La constant.
k k
Or, 4n 2 n 2 rm = , whence, n 2 == .
' 2 23
^
Or,
1 A / T
n =  A/
2?r V m
A /~mr^
= 2ir \ , =
V K:
o
mr*
"mr*
,^
MG
where t is the time taken hy one revolution of the planet round the sun,
or the length of the year for that planet.
Thus, / varies as \/ r 3 , i.e., the smaller the value of r, or the
smaller the distance of the planet from the sun, the smaller th3 valuo
of /, or the length of the year, for it. A planet will, therefore, have a
shorter year if nearer to the sun than when at a distance from it.
4. Banking of Railway Lines and Roads. When a railway train
goes round a level curve on a railway track, the necefesary ceiitiipetal
force is provided only by the force between the flanges or the 'rims of
the wheels and the raits, the normal reaction Of the ground or the
track acting vertically upwards and supporting its weight. This
results in a grinding action between the wheels and the rails, result
ing in their wear and tear. Not only that, it may also prove dangerous
in the sense that it may bring about a displacement of the rails and
hence a derailment of the train.
To avoid these eventualities, the level of the outside rail is raised
a little above that of the inside one. This is known as the banking of
railway lines, and the angle that the track makes with the horizontal
is called the angle of banking.
With the track thus banked, i.e., with the outer rail thus raised
above the level of the inner one, the reaction R acts perpendicularly
to the track, as before, but is now inclined to tlie vertical at an angle
MOTION ALONG A CURVETHE PBOJECTILE
29
equal to the angle of banking and its horizontal component (and not
the lateral thrust of the wheel flanges on the outside rail) now supp
lies the necessary contripetal force to keep the train moving along the
curve, thereby eliminating all unnecessary wear and tear.
Thus, if 6 b3 tha angle of banking (Fig. 8), and R, the normal
reaction acting psrpendiculaily to it, we have
vertical componet of R = R cos $,
and horizontal component ofR = R sin 6.
The former component balances the weight mg of the train and
the latt ?r supplies th'j required ceritripstal force mv 2 /r t where v is
the speed of the train atid r the radius of tha curve it negotiates., So
that, Rsin 8 = mv 2 jr
and R cos $ = mg.
R sin Q wv 2 /r
v 8
Or, tan 9 = ~
rg
Or, =
rg
The angle of banking thus
dopends upon the speed (v) of the
train and the radius (r) of the curve
of the track. Obviously, therefore, a
track can be banked correctly only
for a particular speed of the train, in practice, naturally for its
average sp3od. At higher or lower spe3ds than this, thore is again a
lateral thrust due to the wheel flanges on the outer or the inner rail
of the track respectively.
Cleanly, the angle that the track makes with the horizontal is
equal to 0, i.e., equal to th3 a*igb of inclination of the train with the
vertical, (Fig. 8).
Further, it will be readily seen that if the distance between the
Tails be d and the height of the outer rail above the inner one be A,
we also have
sin 9 = y .
a
Or, sine of the angle of banking
__ height of the outer rail over the inner one .
~~ distance between the rails
Similarly, in the case of a car moving round a level corner, the
centrifugal force is largely provided by the friction between the road
and the tyres of the wheel. That is why, when the road is slippery
and the frictional force not enough the car begins to slide or skid.
Here, too, therefore, ,the roads are *banked\ the slope being generally
steeper outwards, more or less like a saucer the outer parts being
meant to be used at higher speeds and the inner ones, at lower speeds.
30
PROPERTIES OF MATTER
Again, an aeroplane, in order to turn, must also bank, the
centripetal force here being supplied by the horizontal component of
the lift L, (Fi. 9).
The same applies to a cyclist,
when negotiating a curve or a
corner, and he has to lean inwards,
(/.e., towards the centre of the
curve), by an angle = tan 1 v 2 /rg ;
so that, the faster his speed and
the sharper the curve, the more
must he lean over. This will be
clear from the following :
Let Fig. 10 represent a cyclist
turning to the left in a circle of
radius r, at a speed v. Then, the
normal reaction R of the ground
acts vertically upwards, with the force of friction F between the
ground and the tyres and the centrifugal force wv 2 /r in the direc
tions shown,
where
R = mg and F = /??v 2 /r.
Then, for equilibrium, clearly, we have
moment of mg about P equal and opposite
to moment of /nv 2 yr about P.
Or,
mgxPQ =
mv
Or, mg x PG.sin = . PG.cos 6,
whence,
sin 9 A v 1
 ~ = tan =
cos 6 rg
In other words, in order to keep himself in equilibrium,,
the cyclist must lean inwards from the vertical at an angle
= tan 1 (v 2 /rg).
If he were to remain vertical, his weight would act through P,
having no moment about it, so that the moment of mv 2 /r about P
would remain unbalanced. In fact it will be readily seen that the
system of forces acting on the cyclist form two pairs of couples, one
due to F and mv 2 /r and the other due to R and mg. So that, in the
event of the latter couple vanishing (i.e , if the cyclist were vertical),
the former alone will remain operative, resulting in the cyclist toppling
over.
Further, since the maximum value of F = itfng, (where M is the
coefficient of friction between the ground and the tyres), the cyclist
will skid when mv 2 /r > nmg, or when v*>urg.
Thus, skidding will occur (i) ifv is large, i.e., if the speed of the
cyclist is large. (ii) if n is small f i.e., if the road is slippery and (Hi]
tfris small, i.e., if the curve is sharp.
MOTION ALONG A CT7BVE THH PROJECTILE
31
Similar conditions apply in the case of a motor car or any other*
vehicle. For, here too, if we imagine it to be turning to the left r
(Fig. 11), the various forces acting
on it are the normal reactions R l
and R t , the frictional forces F l and
F 2t its weight mg and the centrifugal
force mv*/r, as shown, the whole
system being in equilibrium.
Obviously, in the event of
the car being about to be upset, it
will be moving on the wheels on
one side only, so that the normal
reaction on the wheels on the other
side will be zero. ; say R l = 0. So
that, it will overturn as soon as the
moment of mv 2 /r about P is greater
than the opposing moment of mg
about P.
i.e.,
Or,
mv z
as soon as .GQ > mg . PQ,
when
mv 2
Jt > mg.d,
where h is the height of the e.g., <7, of the car above the ground and?
2d, the distance between the two wheels.
For the car to be upset, therefore, we have v 2 > '.
The car is, therefore, not likely to bo upset if 2d, the distance
between the two wheels is large and if /i, tho height of the e.g.
from the ground is small.
Again, the maximum value of the total frictional, force
So that, as before, skidding will occur when wv 2 /r > iimg.
Or, when v 2 > urg.
To avoid skidding, therefore, while taking a turn at a fast
speed, the corner must be cut so as to move along a comparatively
flatter curve than that of the actual turning.
J&. Other Effects and Applications of Centrifugal Force.
1. Rotation of the Earth Its Effect. As we know already,
the earth rotates or spins about its axis once during a day. It is
this rotation of it which is responsible for its getting flattened at the
poles and its bulging out at the equator, a direct consequence of the
centrifugal force m<u*R acting on each particle of mass m of it, where
to is its angular velocity about the axis of rotation and R, the distance
of the particle from this axis. The value of a> is obviously the same
for each: particle, but the distance R increases from zero for particles
at the poles to a maximum for those at the equator. The centrifugal
force pulling the earth outwards, as it were, is thus zero at the
32 PBOPBBTIES OF MATTER
and the maximum at the equator and it is this force which has made
the earth (behaving like a plastic body) to bulge out at the equator
and to flatten at the poles, thus bringing about an incr3ase of about
13 miles in its equatorial, as compared with its polar radius.
This effect of the rotation of the earth had been first predicted
by Newton and was duly verified by a French expedition to Lapland
under the leadership of Maupertius. whose undue pomposity provok
ed Voltaire into making the caustic remark that he * behaved as
though he had flattened the poles himself/
2. The Centrifuges. These are simple devices used to separate
ou f / substances of different densities suspended in a liquid, by rapidly
rotating the liquid, when particles, whose density is greater than that
of the liquid, are driven away from the axis of rotation, whereas
those, with a density lower than that of the liquid, are drawn inwards
towards it. Thus, for example, in the familiar cream separator, when
the vessel containing milk is rotated fast, the cream, being lighter,
collects in a cylindrical layer round about the axis, whence it can be
easily drawn off.
Since the centrifugal force (wo>V) increases with r, the pressure
on the rotating liquid progressively increases as we move away from
the axis, with the result that, on a heavier particle, the centrifugal
force outwards is greater than the inward thrust of the liquid,
whereas, on a lighter one, the reverse is the case. The problem is
more or less akin to the sinking or floating of a body in a liquid at rest,
depending upon the difference in the magnitude of the forces acting
on the two sid^s of the body, the inward thrust on it corresponding to
.the upthrust in the case of a stationary liquid.
Since centrifuges have as h><*h speeds of rotation as 40,000
revolutions (or more) per minute, the difference between the outward
and inward forces acting on the heavier and lighter particles exceeds
more than a thousand times the difference between their weights, so
that quick and effective separation results. Sediments, precipitates
and bacteria etc., may all be thus separated speedily.
Another familiar example is the centrifugal drying machine,
which is just a cylindrical vessel, with perforations in its walls. When,
with damp clothes placed inside it. it is rotated fast, the centrifugal
force acting on them forces the water out through the perforations
.and the clothes thus get dried up quickly.
3. The Centrifugal Pump. Also known as the Turbine Pump,
it consists of three essential p^rts, viz., (i) an outer drumshaped
'casing\ having an inlet near its axis and an outlet near its periphery,
(//) a paddle wh*el (i.e., a hollow wheel, fitted with vanos) called the
'impeller', which can be rotated inside the casing, and (Hi) the
'spindle'* through which energy is transmitted from the driving motor
to the impeller.
If the casing be filled with water and the impeller rapidly rotat
ed, it sets the water into similar rapid rotation, which, due to its
centrifugal force, exerts high pressure a<?a ; nst the outer wall of the
casing, forcing the water out through the outlet at the periphery into
MOTION ALONO A CtJBVl THU PROJBOTILB 33
the rising dischargetube connected to it. At the same time, there
is decreased pressure near the axis, so that the atmospheric pressure
forces fresh water in, from the reservoir, through the inlet. This is
then again flung out through the "outlet, in the manner explained, and
the process goes on repeating itself over and over again.
The pump starts working only when the casing is full of water,
but, once it has started working, it gi ves a continuous supply of water,
unlike the ordinary piston pump, where we get only an intermittent
supply.
Further, as there are no valves to operate, the pump can be used
safely even if the water contains sledge or any other suspended mat
ter, including sand or smallsized stones etc,,
20. The Projectile Motion of a Projectile in a nonresisting
medium. Before Galileo's time, it was supposed that a body thrown
horizontcally, travelled in a straight line until it had exhausted its
force and then fell vertically down. It was he who first showed that
it must take a parabolic path*, realizing, as he did, die physical in
dependence of its horizontal and vertical motions, so that each could
be considered separately.
Such a body, subjected simultaneously to a uniform horizontal
motion and a vertical uniform acceleration, is called a projectile, &nd
the path it describes is called its trajectory. Let us study its motion
in some detail.
Let a body be projected upwards with a velocity w, at an angle
6 with the horizontal. Then, resolving u into two rectangular com
ponents, along the verticalf and along the horizontal, we have (/) t he
vertical component (along (7*7) = us in 9 and (//) the horizontal com
ponent (along OX) = u cos 6. The latter component, being perpendi
cular to the direction of gravity, is not accelerated, and hence
dx
= u cos 0.
at
Since at / = 0, x = 0, we have
x =s. ut cos 9 ... (/)
And, because the vertical component is subjected to a downward
acceleration due to gravity, we have
dt* ** ~~~ gt
integrating which, we have  ~ gf+Cj,
where C l is a constant of integration.
dy
Now, at t = 0, , s= u sin 6 ; so that C 1 = u sin 9.
dv
J = u sin 6gt.
*He dropped objects from masts of moving ships, which fell vertically iq
relation to the ship but along parabolic paths in relation to the sea.
f /.<?,, along the dirftfjpg in which th<? force due to gravity acts,
34 PROPERTIES OF MATTBB
Integrating this again, we have y ut sin 8 ]
where C 8 is another constant of integration.
Since y =* at t = 0, we have C t = 0.
Or, y^utsine Igt*. ... (i7)
Now, from relation (/), we have t
v y w cos 6
Substituting this value of t in relation (), we have
j = w, *  .sin Q\g. ( ~ A }
U CO^ J \ tl C05 1 0/
V 2
This is clearly an equation of the second degree in x and the
first degree in y and thus represents a parabola, with its axis vertical,
The trajectory of the particle is thus & parabola.
21. Horizontal Range of a Projectile. Clearly, the time taken
by the body to reach the maximum height = u sin 0/g*, its vertical
velocity being u sin 0.
And, since time of ascent is equal to time of descent , the total
time taken by the body for the whole flight = 2u sin dig.
During this time, the horizontal distance covered by the body,
with its uniform horizontal velocity u cos Q is given by
A 2u sin 6 2u 2 .sin Q.cos 6 u*.sin 20 r . n . ..
U COS 6. = = ~ ['' 2 sin cos Q=*sin 29.
g g g
This horizontal distance covered by a projectile is called its hori
zontal range, or, more usually, simply, its range. Denoting it by R,
therefore, we have R = w 2 sin 20/g.
22. Maximum Height attained by a Projectile. We have the
kinematic relation, v a w 8 = 2aS t where the symbols have their usual
meanings.
Here, a g (the body being projected upwards), and, at the
highest point, obviously, v = 0. So that, if the maximum height
attained by the projectile be h, (i.e., S~h). we have
0( sin 6)* = 2.(~g)./j, f v the initial, upward
7v2 c/2 /) velocity here is u sin $
whence, h ** ^ * . L and not u.
2g
23. Angle of Projection for Maximum Range. It is obvious
that for a given initial velocity (u) of the body, its horizontal range
(R) will depend upon its angle of projection (d).
Now, the horizontal range ^R, as we know, is given by
D u*.sin 20
~^ g .
Putting x for R, we have x
g
*We have the kinematic relation, v wf at, where u is the initial velocity
(here equal to u sin 9), a== g and v, the final velocity, equal to 0, (at the highest
pointj. So that,  u sin Q~gt ; or, gt *u sin fl. Or, t * u sin Qlg.
MOTION ALONG A CTJBVE THB PBOJJROTILB
35
It is thus clear that the value of for maximum (horizontal) range
of the projectile would be that for which sin 20 = 1, i.e., when
28 =90, and, therefore, =45.
Thus, for maximum range, the angle of projection should be 45
N.B. The following interesting result follows, howevw. from the relation
R = u* sin 2$/g,
We know that the sine of
an angle is the same as that of
its supplement. And, therefore,
sin 2$ ^ sin (18025),
from which it is clear that the
projectile will have the same
range (not the maximum), for
the angles of projection $ and
(90~0), the two paths taken
being, however, different, and
called the high (H) and the low
(L) trajectories respectively, as
shown in Fig 12, in view of the
different maximum heights at*
tained by the body.
Fig. 12.
24. Range on an Inclined Plane. We have seen above, (20),
that the equation to the trajectory of a projectile is
y=x tan 0$g. a ~.~ f _ .
Let us consider a plane inclined to the horizontal at an angle a,
(Fig. 13) ; so that, y =. x tan a.
Now, to obtain the range on the inclined plane, we must deter
mine the point where the trajectory of the projectile will meet the
plane ; and to do this, we must
solve the above two equations. So
that, substituting y = x tan a in
the equation for the trajectory, we
have
v
x tan a = x tan J. ^^
M COS 8
Or,
g*
Fig. 13.
M^J = tane  tan 
2(tqn 6 tan a).u* cos* 6
_ ^
And, therefore, substituting Ms value of x irry = x tan a,
, 2(tan 0tan a),w 2 cos 2 B
wre have y   i 
o
,
.tan a.
pp tbat f the range R is clearly given by the relation,
36 PROPERTIES OF MATTER
ac P!^J?ITJ^ a l <w2 c *_6~]* r2(tan Qtan y.)tan a.u*co$* 6 1*
r 2(tan Q tan a).u 2 cos 2 "I 2 n , . ^
=  ~ [1 tan* a]
Now, (14 fan 2 a) = sec 2 a = n .
cos z a
_ T2(/^ 0tan a) w 2 ros* I 2 /
jv '     I , CL/o (A*
L Jy
a p _ 2 (' flW ^ tan a ) " 2 coja *
<Ji XV ~  ' ~  * ~  ~
g cos a.
25. Resultant Velocity of a Projectile at a given instant. The
vertical and horizontal components of the velocity of a projectile, t
sees, after its projection, are clearly dyjdt and dxjdt. So that, its
resultant velocity v, at the instant, is given by the relation,
dt \dt
Or, putting the values of dy/dt and dxjdt from above, ( 20), we have
v = i/usmtigty+fa casoy.
Or, v = ^u*2ugt.sinT+g*t 2 . [/ jw 1 tf+co^ 2 = 1.
And, the angle, p, say, that this resultant velocity makes with
the horizontal, is obviously given by
dy dx u sin 6~gt , . gt
tan p * , *  /  tan e  4?
Or, P=^ 3
N.B. For small values of t, tan (3, and, therefore, p is positive ; but, for
large values of /, it acquires a negative value. >4/H/, obviously, when (3=0, //re
/><?*/>> /5 moving horizontally, i.e., at the very peak of the parabola. In this case,
since tan p = 0, v/e have
Let this value of f be denoted by /'. Then, we have
gt' t sin
6  aw tan 0= 
cos
Or, ^' j/ 0, whence, /' ~ .
Clearly, the vertical distance covered by the body at an instant t is given
by y** ut + la/*.
I/ C/H /Q
.y = A, the maximum height attained by the body, when f * f ' ~ ~ K
Substituting these valuss in the above relation, we have
h = w j/ $.r'~t^' 3 , F ' initia  P ward velodt y
I SB M5i fl.and a = j?,
MOTION ALO&Q A CtTEVE T?H1S P&OJEdTlLU 37
u sin 9
.
= u sin
w 2 stn 2 Q u 2 sin* ^ i sin 2 $
a,. _   . QJ fi ssr
2# 2#
the same result, as obtained above in 22 (page 34).
Again, the horizontal range may be easily obtained by equating the \alue
of>toO.
Let f  /*, when y 0. Then, we have, from above,
= u sin Q.t"lgt"*.
Or, u sin 0.f " = \gt"\ Or, w j/ = \gt".
Or, */ *  2u sin 0, whence, / * =, 2w J//l * .
#
Now, as we know, the initial horizontal velocity u cos 0.
And, therefore, the horizontal range R is given by the relation,
D A
R = u cos O./* u cos
u 2 2 sin cos 6 u' 2 . sin 20 .
=      , [v 2 cos = Mfl 20.
o o
the same result as obtained before in 21, (page 34).
And, finally, if we substitute this value of /, (i.e , t" = 2u sin 0/0), in the
expression for tan p above, we have
tan f) = (on 6  ^r to  ff 2 " '"" 9 
14 COS #
Or, tan p = ran 02 /an ^tan o,
showing that the projectile comes back to the horizontal surface at the same
angle at which it was projected upwards. And, it is a further simple deduction
that its tangential velocity at this moment is the same as at the instant of projection.
It ijmst be emphasized again, however, that the above treat
ment applies to the motion of a projectile, only in a nonresisting
medium, i.e., in vacuo. The presence of a medium, like air, offers a
frictional resistance to its motion, which depends, to a great extent,
upon the velocity of the body and is, for moderate velocitiesf , direct
ly proportional to the square of the velocity, in accordance with the
law of resistance given by Newton, in the year 1687. This alters the
very character of the trajectory of a projectile, which no longer
remains a parabola but becomes what is called a ballistic curve**, with
its descending part much steeper than its ascending part and the height
and range of the projectile considerably reduced, particularly at high
initial velocities.
At higher altitudes, well above the average level of the earth's
surface, however, the air pressure, and hence also the air resistance,
becomes much smaller and, therefore, if a projectile be shot up to
such great maximum heights, it is quite possible to obtain a high
range for it. This is perhaps the most probable explanation of the
long range German canon, usad with such conspicuous success in the
historic Great War, and which could fire shots to a maximum height of
about 54 kilometres and hence had a range of about 130 kilometres.
* Another possible value of t" is 0. We reject it, however, as it refers to
the time when the body is just starting on its trajectory.
fFor example, from a velocity of a few metres per second to about the
velocity of sound, in the case of air,
** Ballistics is the special name given to the science of the motion of
38 PROPERTIES OF MATTfiR
SOLVED EXAMPLES
1. A particle, moving in a circle of radius 105 cms., has its velpcit
increased in one minute by 120 rotations per minute. Calculate (i) its linea
acceleration (//) its angular acceleration.
Change in velocity of the particle in 1 min. = 120 rotations per min.
= 120/60 = 2 rotations/5^*
Now, in 1 rotation, distance covered = 2nr.
,, 2 rotations ,, = 2x2*r = 2x2x*x IQScms.jsec.
change in velocity of the particle per minute=420 x n cms /sec.
Or, change in velocity per second 420x^/60 = 7rr cms. /sec.
Or, rate of change of velocity = 7 x 22/7 cms. /sec*. 22 cms. jsec 1 .
i.e., linear acceleration, a 22 cms. I sec*.
Now, linear acceleration, i.e., a r.du/dt,
where r is the radius, and dajdt, the angular acceleration.
So that, angular acceleration, d&jdt = a/r.
log 22 = 13424  . ,. 22 _ A __ ,. ,
log 105 = 2 0212 ^ r ' d^jdt = = *2095 radian/ sec.
Antilog 1*32 1 2 Hence, the linear and angular accelerations of the particl<
= 0*2095 are 22'0 cms. /sec*, and '2095 radianjsec 2 ., respectively.
2. If the Earth be onehalf its present distance from the Sun, how man:
days will there be in one year ?
Let the present distance of the Earth from the Sun (i.e., the radius of its
orbit round the Sun) be = R.
So that, half its present distance from the sun would be = R/2.
If a t and a* be the accelerations (linear) of the earth in the two cases res
pectively, we have
If v t and v 2 be the linear velocities of the Earth, corresponding to" these
two values of the acceleration, we have
= * and a =  1   ?&!
80 that ' 2^7*  T ' Or ' %" x 5? = I"
Vl a 1 _ a
XT 2"R , f. circumference of Earth's orbit
JNOW, i = oZc~ P* aay. \ i.e., = *~*   ~  ~.   .
365 L time taken
And, therefore, Vt = ^2v t = V2. ^ per day.
Now, circumference of the Earth's orbit, in the second case = 2n.R/2
And, since one year is the time taken by the Earth to go once round the
Sun, we have
log 365  25623 ..(/)
log 2  03010
 log 2  OJ_505
6'45f5 ..(//)
Subtracting (//) from (/)
we have 2*1108
Antilog 2*1108
 129*0
one year, in the second cave
Hence, the number of days in one year, when
the earth is half its present distance from the Sun,
to 129,
MOTION AL6tfG A OURVli TfiE PEOJECTlLli
3. Assuming that the Moon revolves uniformly in a circle round the
Earth s centre, calculate the acceleration due to gravity at the Earth's surface
from the following data :
Radius of the Earth = 6'4 x 10 8 cms.
Radius of the Moon's orbit = 3*84 x 10 10 cms
Period of rotation of the Moon = 27*3 days.
(The force of gravity on a particle is inversely proportional to the square
of its distance from the Earth's centre).
(Oxford & Cambridge Higher School Certificate)
Here, velocity of the Moon, v = ^ ircui ? feence of he orbit
time taken
cms. Isec.
384 xlO 10
time taken
& 2*x384xl0 10
& 273x24x60x60
.. acceleration of the Moon towards the centre of the Earth
where r is the radius of the Moon's orbit
r2*rx3;84xl0 10 ] 2 1
L27 ; 3x24x366oJ X
4*2 x 384 xlO 10
& (273 x 24 x 3600)' ~
If the acceleration due to gravity on the earth's surface be g cms.Jsei
(radius of" Earth)* and *m ** (radius of Moon's orbit) 8
" 6021 i ^ (radius of Moon's orbit) 2
(radius of Earth) 2
v 2 /r
iay.
we have
log 4
2 log rr
3 log 384
3 log 10 10
09944
=17529
=300000
333494 (')
log 64 08062
8 log 10 =80000
log 273 = 14362 \
log 24 = 13802
log 3600 = ^5563 ;
15 : 1789 I
2x151789 =30'3578..(/0
Subtracting (//) from (/),
we have 29916
Antilog 29916
= 9809
*
Or, g
F
L
(64 xlO 8 ) 2 J'
K
64 x 10* J A L(27'3x24x3600) 2 J
__ 4ir^x_(384xl0 1( V
** (6*4 x "iOx 27 : 3x 24 x 3600) 2 '
9809 cms./sec 2 .
Hence, acceleration due to gravity at the Earth's
surface = 980 9 cms.lsec.
4. The radius of curvature of a railway line at a place when the train is
moving with a speed of 20 miles per hour is 800yds., distance between the rails
being 5 ft. Find the elevation of the outer rail above the inner rail so that there
may be no sidepressure on the rails. (Bombay)
20x1760x3 88 f ,
Here, we have v 20 m.p.h. =
and,
log 121
log 10800
Antilog
And,
Or,
= 20828
40344
r  800 yds.
So that, tan
Or,  tt
Now, d**5
tin 9  h\
.5X'0112'C
2*0484
 0112
A
60x60 ~ T '
800x3 = 2400 ft.
j^ ^ _ 88 x 88 _
rg 3x3x2400x3T
i 0112 3833'.
5ft. and sin 38*33' 0112.
And .*. h == dsinQ.
_
10800
0112.
[0 being imall.
Tbtrefore, tbt wtw rail ihpul4 b ralitd '6120 fecftt* above tbt innw
*U PROPERTIES OF MAtff fift
5. A stone of mass 10 Ibs. is revolving in a vertical circle at the end of 2
string, 8 ft. bag, the other end of which is fixed. When the stone is at the top o]
the circle, the velocity is 16ft. per sec. Assuming g to be 32 ft./sec 2 ., find the
stretching force in the string when the stone is (/) at the top, (//) at the bottom
(iiV) at a level with the centre.
Here, mass of the stone, m = 10 Ibs. ; radius of the circle, r = 8 ft..
and velocity at the top of the circle 16 ft. (sec.
(/) Therefore, force acting outwards, i.e., upwards along the string, or alone
OA, (Fig. 14),
A mv 2 10x16x16 _ A , ;
* = = . ,_ =320 poundah.
r o
And, downward force due to weight of the stone
= mg 10x32 = 320 poundah.
.*. resultant stretching force along the string
. 320320 = 0.
(/"/) By the time the stone reaches the bottom
of the ciicle, it has acquired additional kinetic
energy due to its having fallen a vertical distance
16/f , (the diameter of the circle).
Therefore, kinetic eneigy ot the stone at the
bottom of the circle mi 2 4 tngh.
= [xlOxl6xl6i+[10x32xl6] [v h = 16/r,
=128015120 = WWft.poundals.
This should be equal to \mvf, where V L is the velocity at the bottom of
the circle. So that,
 mVl 2 = 6400. Or, mv t 2 = 6400 x 2 = 12800.
Or, r z 2 = 12800/10 = 1280. [v m = 10 Ibs
Now, centrifugal force acting downwards along the string, />., along OB
 mv f, and, therefore,  10x g 1280 1600 poundah.
And the downward force due to the weight of the stone is equal to
10x32 = 320 poundah (as before).
.. total stretching force in the string = 1600^320 = \92Qpoundals.
= 1920/32 = 6Qpounds weight.
(///) Here, the additional K.E. acquired by the stone is due to the fall
through a distance 8/f., the radius of the circle, and is
= 10x32x8 = 2560 ft. poundah.
.. total K.E. of the stone = i/wv 2 h2560 = Jx lOx I6x 1642560.
= 1230+2560 = WWft.poundals*
Obviously, this must be equal to imr a 2 , where v, is the velocity of the
stone, in this position.
Jmv a 2 = 3840, whence, v t 2 ** A 5 ^?/ 768.
centrifugal force acting outwards along the string, i.e., along OC
= ?>". "xl . 10x96 = 960po U ndah.
r o
960/32 = 30 Ibs. wt.
And, since the weight of the stone is acting vertically downwards, it is
acting at right angles to this centrifugal force and has no component along it.
Therefore, the stretching force in the string, in this case, is equal to 960 poundals
or 30 Ibs. wt.
6. Four masses, each of 10 Ibs., are fastened together with four strings,
each 3 ft. long, so as to form a square. This square rotates in a horizontal plane
on a smooth table at a speed of one revolution per second ; find in poundsweight
tbt tension in tte string. (Cambridge Higher School Certificate)
MOTION ALONG A CTTRVE THE PROJECTILE
41
Let /fh, mi, wa and m, be the four masses of 10 Ibs. each, forming a
square, as shown in Fig, 15.
The radius of the circle in which they rotate will obviously be equal to
Om l = 0w 2 = Om$ Ow 4 r, where is the
point of intersection of the diagonals of the square. /^3/ 2 , p %
If P be the midpoint of m l w 4 , we have '
OP = Pnii = 3/2 //.
So that,
 V( v
+
Let rbe the tension in each string, when the
square is rotating on the smooth table.
Then, representing T, in magnitude and
direction, by the straight line Pm^ we have
force acting along Om^ =\/T*~+'f 2 ~ \/2T 2
Fig. 15.
= V2T.
This, clearly, represents the centrifugal force acting along Otn^
NO.V, the centrifugal force acting along m^O is also equal to mra\ i.e.,
So that,
ITT
^21 =
lo ? 15 = M76i
2 log n = 0^9944
2 : 1 705
log 8 = 09031
Antilog 1 2674
= 18*51
Or,
whence,
** ^ 120 n 2
2T == 120X* 2 ,
T 60?r 2 poundals.
=  Ha. wt.
where m is the mass of
wij  10 Ibs., r  3/V2 //.,
and to = 2n, because the
square makes one revolu
tion per sec., i.e., des
cribes an angle of 2n
radians per sec.
7.
= 8  = 18'51 Ws. wt.
The tension in each string is, therefore, equal to 18 51 Ibs. wt.
A certain string will break under a load of 50 k.gms. A mass of 1
k.gm. is attached to the end of a piece of the string, 10 metres
long, and is rotated in a horizontal circle. Find the greatest
number of revolutions per minute which the weight can make
without breaking the string.
Here two cases arise, viz.,
(/) when the fixed end of the string, is itself the
centre of the circle in which the load is rotating, i.e., the
radius of the circle is the length of the string, [Fig. 16 (/)], and
(/O when the string hangs vertically and a circular
motion is given to the load at its end, the circle described
being in the horizontal plane, [Fig. 16 (//)].
First am?.The centrifugal force, acting on the
string is given by
(W
Fig. 16. r
Now, the maximum value of F is 50 k. gm. wt. (given).
= 50 x 1000 x 980 dynes.
So that, 50 x 1000X980  y^
Therefore,
lOOOxv 2
____ V*
7000 cms. I sec.
7000x60cws./
420000 cm$./wto/e.
where m is the
mass of the load
= 1 k.gm.
and v
10 metres*
1000 cm*.,
PROPERTIES OP MATTER
Mow, the distance covered by the load in 1 rotation
log 42000 = 5*6232
log 6283 =17982
Antilog 18250"
6683
2nxlOOO = 2x31416x1000.
62832 cms. 6283 cms., say.
number of revolutions made by the load per minute
without breaking the string = v/2*r = 420000/6283
= 66 83.
Or, the number of complete rotations made by the load is 66 per minute.
Second case . Here, two forces are acting on the load, viz.,
(/) wv 2 /r, (centrifugal force) horizontally=mr^\ since v = rco, where v
and co are the linear and angular velocities
of the load respectively, and r, the
radius of the horizontal circle in which it
rotates.
(//) mg, (weight of the load) verti
cally downwards.
Obviously, for the load to be in
steady motion, the resultant of these two
must act along the string, as shown in
Fig. 17. Fig. 17.
Let $ be the angle, that the resul
tant force F (the tension in the string) makes with the vertical. Then,
tan Q=~
Now,
rco'

8
..(1)
tng
r=AB sin = 1000 sin e,
m 1 k. gm. = 1000 gms.
The maximum value of the tension of the string, i.e.,
T=50x 1000x980 dynes.
If n be the number of rotations made by the load per minute, the angular
velocity = 2nnper minute.
2nn nn
Or,
'60
 per sec.
Substituting the values of r and w in relation (1) above, we have
,^ fl * 2 2 10005/*0
Or,
Or,
sinO
900 g
n z n*.1000 sin 9
cos
co*t
900 g
,._*.JL_
1 10 *V '
10
:V sin
9g '
But, from the figure, we ha
 cos
10 '
log 9=09542
log 5=06990
log 980=29912
Or,
whence,
Or,
46444
X 46444
log w 04972
Antilog T8250
=6683
F
g_
__
1000 g
50x1000x980
g ___
50x980 '
1
"50x980 '
9x50x980
9
= 50x980*
V9x5x980
Hence, Dumber of rotations 66*83 per minute ; or, number of complete
rotttioni made by the loed per minute is 66.
MOTION ALONG A CtfRtE THI PROJECTILB
S. A particle of mass m is attached to a fixed point by means of a string
bf length / and hangs freely. Show that if it is projected horizontally with a velo
city greater than V*#/ it will completely describe a vertical circle.
Let OB, (Fig. 18), be the string, fixed at Oand suspended freely, with
a mass m at B. Let it be given a horizontal velocity u, when at rest at B. It
will naturally move along an arc of radius /,
the length of the string. Let B' be its position
on the arc at a given instant, when its velocity
is v.
Then, clearly, v 2 = u*2gh,
where h is the vertical distance through which
the mass has been raised up.
The weight mg of the mass is acting verti
cally downwards at B', and the centripetal force
mv 2 //, along the string, in the direction B'O.
The component of mg, acting along the
string in the direction OB', i.e , opposite to that
of th? centripetal force, is thus clearly = mg.cos 0,
.'. If T be the tension of the string
we have
Or,
Now,
  = T mg cos 0.
B
Fig. 18.
mv*
I
cos 8
oc
OB
f mg cos Q.
OBCB
OB'
(I)
lh
I
substituting the values of v* and cos in relation (I) above, we have
m(u*~2gh)
1 I
,
f **
Or,
(II)
At the highest point on the circle, i.e., at A, when the mass completes
half the vertical circle, we have h = 21.
This will clearly become equal to zero, if u* = 5#/.
Now, at A, v 2 u*4gl. ['. h 2/, here.
Obviously, therefore, if the mass is to continue in motion along the circle,
the tension T should not vanish, i.e., should not become zero, which means that
u 2 >5gl. For this value of u 2 , its velocity at A will also not vanish, and hence the
mass will describe a complete circle of radius /.
Thus, the condition necessary for the mass to complete a vertical circle is
that if>5gl, or that u
9. Assuming the law of Gravitation, and taking the orbit of the Earth
round the Sun, and of the Moon round the Earth as circular, compare the masses of
the Sun and the Earth, given that the Moon makes 13 revolutions per year and that
the Sun is 390 times as distant as the Moon.
Let mass of the Sun be M t > mass of the Earth Af*.
and mass of the Moon = M m ,
And, let distance of the Moon from the Earth be ~ JR,
*ft that _. __ Sun _ . wm 300 IL
** PROPERTIES OF MATTEtt
Then, force of attraction between the Sun and the Earth
where G is the gravitational constant.
centripetal acceleration of the Earth = /? c ?  t
mais (6
__ Ms_ G
~ (390 R)*' '
Similarly, force of attraction between the Earth and the Moon
And .'. centripetal acceleration of the Moon ~ 2 u ** G ^ . G.
J\ . jrJ /TJ </v
Let co^ and co w be the angular velocities of the Earth and the Moon res
pectively. Then, clearly,
centripetal acceleration of the Earth is also = 390 R.& e *.
and ,, ,, 3t Moon ,, = Rw OT 2 .
[Mj/(390 R)*]G ^ 390.tf.ov 2
Or, (390)2 ~M e " c) *z '
Now, the Earth goes round the Sun only once in one year ; and, therefore,
angular velocity of the Earth 2x per year.
And the Moon goes round the Earth 13 times in one year ; so that,
angular velocity of the Moon = 13 x 2rr /?er year.
Mr
if* Or > 2ri3v
r.e., Mass of the sun : Mass of the Earth : : (390) 3 : (13) .
10. show that in the case of a liquid, rotating with a uniform angular
velocity, (/) the pressure varies directly as the distance from the axis of rotation
and (//') the free surface of the liquid is a paraboloid of revolution.
(0 Imagine a closed, vertical^ cylindrical vessel, just full of a liquid of
density p, to be rotating about its axis with a uniform angular velocity co.
Now, consider a ring of the liquid,
of radius x, width $x and vertical height
8/r, with its centre at O on the axis of
f>+6f> rotation, (Fig. 19). Then, if the pressures
in the liquid at distances x and x+8x
from the axis of rotation be p and (p+8p)
outwards and inwards respectively, we
have
resultant inward thrust on the ring
and centrifugal force outwards on the
where m is the mass of the liquid in the ring.
Clearly, m= volume of liquid in the ring x p=2
So that, centrifugal force outward on the ring = 2KX$x.$h.p.(**.x [v r=;c, here,
And, therefore, />.2rrx.8/z = 2nx$x S/j.p.w 2 .*.
whence, &p = pcAx.&e.
Integrating this expression, we have
MOTION ALONG A CURVE THE PBOJBOriLl
Or, p = ip
where C is a constant of integration.
This is then the expression for the pressure at a point distant x from the
axis of rotation in a rotating liquid.
The value of C is obviously equal to that of p at x = 0, i.e., equal to the
pressure at 0, which, as we know, is zero at the surface of the liquid, but
increases with depth as in the case of a liquid at rest.
(//) Consider a particle P in the liquid surface, of mass m, whose co
ordinates are (x, y) with respect to the axes OX and OY t (Fig. 20;, the liquid
being supposed to be rotating with a
uniform angular velocity co in the direction
shown, about OY.
The forces acting on this particle
P are, clearly, (/) its weight mg, vertically
downwards, and (//) 'the centrifugal force
wto 2 *, outwards. The resultant R of
these two forces must act at right angles
to the liquid surface, since there is no
flow of the liquid taking place, and,
obviously, it is counterbalanced by the
thrust due to the rest of the liquid on P.
If, therefore, be the angle that the
centrifugal force makes with the tangent at P, we have
Fig. 20.
Now, obviously,
tan = =
mg
tan = slope at P
dyldx.
.
dx
. Or,
Integrating this, we have
/*=/
,^.dx.
~~ \x.dx.
Or, *~2T +Cf
where C is again a constant of integration.
y = 0, when x = 0, we have C = 0.
Since
And, therefore,
y***
whicli is the equation to a parabola.
The free surface of a uniformly rotating liquid is thus clearly a para
boloid of revolution.
EXERCISE II
!. A particle of mass 1 Ib. is whirled uniformly at the end of a string,
2ft. long, and makes 3 revolutions in 1 2 sees. Find the tension in the string.
Ahs. 1543 Ib. wt.
2. A halfpound weight is being whirled in a horizontal circle at the end
of a string, 2 feet long, the o'her end of the string being fixed. If the breaking
tension of the string is 112 Ib. w/., find the greatest speed which can be given to
the weight. Ans. l\91ft./sec.
3. At what angle should a cyclist lean over, when negotiating a curve of
132 //. radius at 15 miles p*r hour. Ana. 6* 32'.
4. A person skating on ice at the rate of 20 ft. per se< ond describes a cir
cle of 20 ft. radius. What is, his inclination to the vertical ? Ans. 32*0,
46 PROPERTIES OF MATTER
5. A train is going round a curve of 1000 ft, radius and the distance
between the rails is 5 feet. By how much should the outer rail be raised above
the inner one, so that a train running at 45 miles per hour may exert no lateral
thrust on the outer rail ? Ans. 8'094*.
6. A stone is suspended from the roof of a railway carriage by means of
a string 5 //. long. The angle through which the mass moves from the vertical
is 10, when fhe train moves along a curve of radius 600 //. Calculate the speed
of the train. Ans. SSl/f./iec.
7. Assuming that the Moon describes a circular orbit of radius 3*84x 10
metres in 27'3 days and the outer satellite of Mars describes a circular orbit of
radius = 2*35 x 10 7 metres in 1*26 days, find the ratio of the mass of Mars to the
mass of the Earth. (Cambridge University)
Ans. '1076: 1.
8. A curve on a railway line has a radius 1 600 //., and tne distance
between the inner and outer rails is 5 //. If the outer rail be 6" above the inner
one, calculate the maximum speed of a train going along the curve, so that no
side thrust is exerted on the outer rail. Ans. 48 89 mileslhr.
9. Calculate the increase in leagth of an elastic string of original length
10 /f., at the end of which a stone of mass 5 Ib. is whirled at the rate of 4 re
volutions per second, if a load of 25 Ibs. increases the length of the string by 2%.
Ans. '8576 //.
10. A merrygoround is revolving in a horizontal circle of radius 3ft. at
the rate of 7 revolutions in 11 seconds. A child of weight 20 Ibs. rides a wooden
horse suspended by a vertical string. Find the tension in the string and its
inclination to the vertical. Ans. (i) 36 Ib. vr/., nearly, (//) tan' 1 3/2.
[Hint : See solved example 1 (second case, page 42.)]
11. A seaplane of total mass 1000 Ib. (including the pilot) rounds a
pylon in a circular arc of radius half a mile at a speed of 300 mp.h. Draw a
diagram showing the forces acting on the seaplane, and calculate the resultant
force at right angles to its direction of motion exerted upon it by the air.
Assuming that the pilot weighs 12 stones, calculate the force with which he is
pressed against his seat during the "turn/ 1 (Cambridge Higher School Certificate)
Ans. 8x 10* poundals : 30 stonewt.
12. Calculate the angle at which a curve of radius 352 //. should be banked
so as to avoid sideslip when a motor car is travelling round it at a speed of
W m.p.h. Ans. 9 45'.
13. A road over a bridge has the form of a vertical arc of radius 60 //.
What is the greatest speed in m.p.h. at which a car can cross the bridge without
leaving the ground at the crest of the road ? Ans. 30 m.p.h.
14. A skater is moving on one foot in a circle of radius 20//. at 10 m.p.h.
At what angle with the vertical will the line passing through his centre of gravity
and the edge of his skate be inclined ? Ans. 18 35'.
15. In a 'looptheloop* railway, the cars, after descending a steep
incline, run round the inside of a vertical circular track, 20 ft. in diameter,
making a complete turn over. Assuming there is no friction, find the minimum
height above the top of the circular track from which the cars must start.
Ans. 5ft.
16. A symmetrically loaded lorry weighs 5 tons, and the height of its
centre of gravity is 5ft. above the ground in a vertical plane midway between
the wheels. The breadth of the wheel base may be taken to be 6ft. 3 in. If
there is no sideslip, what is the maximum speed at which the lorry can take a
curve of msan radius 6 yards without beginning to overturn ? Ans. 1 3 m.p.h.
[Hint : It will overturn only when the moment of the centrifugal force about
the wheels on one side is greater than the moment of the weight about them, (see
13, case 4, page 31).]
17. , An India rubber band has a mass of 4 gm* per metre when* stretched
on the circumference of a wheel of 10cm. radius, the stretching force being
MOTION ALONG A OTTEVJB THB PROJHCTILB 47
20,000 dynes. Find how many revolutions per second the wheel must make to
that the band may not press upon the wheel. Ans. 1 1*3.
[Hint : See 18, case 2 t page 27.]
18. Discuss the possibility of a motor cyclist riding round the inside
surface of a vertical cylinder. (Cambridge Scholarship)
19. Explain why a motorcycle combination (side car on left) is liable to
overturn when taking a left hand corner at speed. Assuming that the centre of
mass of the combination is 2 ft. from the ground and 1 ft. to the left of the
motor cycle, calculate the maximum speed 6f the combination in a circle of
radius 50 ft . Assume that the road surface is horizontal and that there is no
skidding. (Oxford Scholarship)
Ans. 19" 3 m.p.h.
20. A closed cylindrical can of radius a and height h is first filled with a
fluid of density p. It is then rotated with angular velocity < about its own
axis, held vertical. Prove that the total thrust on the top of the. can will he
CHAPTER III
MOMENT OF INERTIA ENERGY OF ROTATION
26. Moment of Inertia and its Physical Significance Radius of
Gyration. We know that, according to Newton's first law of motion,
a body must continue in its state of rest or of uniform motion along a
straight line, unless acted upon by an external force. This inertness
or inability of a body to change by itself, its position of rest, or of
uniform motion, is called inertia*, and is a fundamental property of
matter. Thus, it is by virtue of its inertia that a body, at rest,
resists or opposes being put into motion, and a body, in linear or
translatory motion, opposes not only being brought to rest but also
any change in the magnitude and direction of its motion And, we
know, by experience, that thegreater the Tpass_of a bqdy^ the greater
its _inertiaj)r opposition to the desired _chaiige ; for, the greater is the
force requireJToTTa "appFed for the purposeT^ The mass of a body is
thus taken to b2 a measure of its 'inertia for translatory motion', as it
is this that opposes the acceleration, (positive or negative), desired to
be produced in it by the applied force.
Exactly in the same manner, in the case of rotational motion
also, we find that a body, free to rotate about an axis, opposes any
change desired to be produced in its state of rest or rotation, showing
that it possesses 'inertia* for this type of motion And, obviously, the
greater the couple or torque, (see 28), required to be applied to a
body to change its state of rotation, i.e., to produce in it a desired
angular acceleration, the greater its opposition to the desired change,
or the greater its 'inertia for rotational motion' . It is this 'rotational
inertia' of the body which is called its moment of inertia** about the
axis of rotation, Him name being given to it on the analogy of the
moment of the couple, which it opposes.
It will thus be seen that the moment of inertia of a body, in the
case of rotational motion, plays the same part as, or is the analogue
of the mass of a body in he case of translatory motion ; and we may,
therefore, for purposes of analogy, describe the moment of inertia of
a body, in rotational motion as the 'effectiveness of its mass.' Or,
pushing the analogy a little further, we may define mass as the
'coefficient of inertia^ for translatory motion', and the moment of
inertia, as the 'coefficient of rotational inertia'.**
Yet, with all this seaming similarity, there is all the difference
between the two cases. For, in the case of translatory motion, the
*That is why the comparative slackness or sluggishness of the people SF
Eastern countries, a consequence of climatic conditions is dubbed by the
Westerners as the 'Inertia of the East. 9
**It is also sometimes referred to as the 'Spin inertia' of the body
its axis of rotation.
fThe m^ss of a body being usually referred to as its inertia coefficient,
48
MOMENT Off INBHTIA BNEEGY Of ROTATION 49
inertia of the body depends wholly upon its mass and is, therefore,
measured in terms of it alone. In the case of rotational motion, on
the other hand, the rotational inertia, or the moment of inertia, of
the body, depends not only upon the mass (M) of the body but also
upon the ^Jfr^jv? ditifw' (K) of its particles from the axis of rota
tion, and is measured by the expression MK*, (see next Article).
This 'effective distance' (K) of the particles of a body from its
axis of rotation is called its radius of gyration about that axis, and
is equal to the root mean square distance of the particles from the axis,
i.e., equal to the square root of their mean square distance (not the
square of their mean distance) from it. Or, to give it a clear cut .
definition, the radius of gyration of a body, about a given axis of nota
tion, may be defined as the distance from the axis, at which, if the whole
mass of the body were to be concentrated, the moment of inertia of the
body about the given axis of rotation would be the same as with its
actual distribution of mass.
Now, it is obvious that a change in the position or inclination of
the axis of rotation of a body will bring about a corresponding
change in tho relative disfcancas of its particles, and hence in their
'effective distance 9 , from the axis, i.e., in the value of the radius of
gyration of the body about the axis And, so will the transference of a
portion of the matter (or mass) of the body from one part of it to
another, or a change in the distribution of the mass about the axis,
the total mass of the body remaining the same, in either case.
Thus, whereas the mass of a body remains the same, irrespective
of the location or inclination of the axis of rotation, the value of its
radius of gyration about the axis depends upon
(/) the position and direction of the axis of rotation, and
(ii the distribution of the mass of the body about this axis ; so that,
its value for the same body is diifererit for different axes of rotation.
Further, it follows, as a converse of the above, that the radius
of gyration of a body about a given axis of rotation gives an indica
tion of the distribution of the mass of the body about it and hence,
also, the effect of this distribution of mass on the moment of inertia
of the body about that axis.
"""27. Expression for the Moment of Inertia. Suppose we have a
body of mass M, (Fig. 21), and any axis YY'.
Imagine the body to be composed of a large
number of particles of masses m v m 2 , m 3
etc., at distances r,, r 2 , r s ...etc from the axis
YY'. Then, the moment of inertia of the
particle m l about YY is mj^, that of the
particle m a is JW 2 r a a , and so on ; and, there
fore, the moment of inertia, /, of the whole
body, about the axis YY', is equal to the
sum of w^ 2 , /w 2 r a 8 , W 3 r 8 2 etc,
Thus, / sss m^f + #y a * +W 3 r 3 2 f . . .
ss= JEVwr*. Twhere M is the mass and Ml
r\ r MY* the summation 2'Mr 1 for tL
Ur > * ** M & Lwhoie body,
K beinjj tlie radius of gyration of the body about the axis YY',
5C PBOPBKTIBS Ot MATTER
28. Torque. If we wish to accelerate the rotation of a body,
free to rotate about an axis, we have to apply to it a couple. The
moment of the couple, so applied, is called torque, and we say that a
torque is applied to the body.
Obviously, the angular acceleration of all the particles, irrespec
tive of their distances from the axis of rotation, is the same, but because
their distances are different from the axis, their linear accelerations
are different, (the linear acceleration of a particle being the product
of the angular acceleration and the distance of the particle from the
axis of rotation).
If, therefore, dwfdt be the angular acceleration of the body, or
its particles, we have
linear acceleration of the particle distant r t from the axis r^dwldt,
,, ,, ,, r a ,, r^dcoldt.
and so on.
Hence, if m be the mass of each particle of the body, the forces
on the different particles are mr^d&jdt, mr 2 .d^jdt t etc., and the
moments o these forces about the axis of rotation will, therefore, be
x r l9 (mr 2 .da>ldt) x r z and so on.
Therefore, total moment for the whole body
= (mr v da)ldt) X ^ + (mr 2 .da}jdt) x r f + ......
= (d&ldt).mr*. [d&fdt being constant.
But Zmr 2 = /, the moment of inertia of the body about the
axis of rotation. And, therefore,
moment for the whole body = I.d^ldt.
This must be equal to the torque applied to the body.
So that, torque = Ldaj/dt.
It will at once be clear that this relation corresponds to the
familiar relation, force = m x a, in the case of linear motion, where
m is the mass and a, the acceleration of the body.
Thus, in the case of rotatory motion, torque, moment of inertia
and angular acceleration are the analogues of force, mass and linear
acceleration respectively in the case of linear or translatory motion.
Now, if dwjdt =s 1, clearly, torque = /.
Or, the moment of inertia of a body about an axis is equal to the
torque, producing unit angular acceleration in it about that axis.
Incidentally, the expression for torque, obtained above, fur
nishes us with a method of deducing an expression for the moment of
inertia of a particle of mass m, about an axis, distant r from it.
For, if F be the force applied, we have
torque = F x r.
torque is also
MOMENT OF INEBTIA ENERGY 0V BOTATlOH
torque F x r \
81
... (0
And, therefore,
Now, F a m x a,
where a is the linear acceleration of the particle.
And, since a = dvjdt, (where v is the linear velocity of the
particle), we have F = m.dvjdt.
Again, since v == ro>, where co is the angular velocity of the
particle, we have
Now, the component, (drldt).a), plays no part in the rotation of
the body and may, therefore, ba ignored ; so that, F = mr.dco/dt.
Substituting this value of F in relation (/) above, we have
T mr.(dcoldt).r t
/== \ ,' L = mr*.
dwjdt
Thus, the moment of inertia of a particle of mass m, about an
axis distant r from it, is equal to wr 2 .
29. General Theorems on Moment of Inertia. There are two
general theorems of great importance on moment of inertia, which,
in some cases, enable us to determine the moment of inertia of a
body about an axis, if its moment of inertia about some other axis
be known. We shall now proceed to discuss these.
(a) The Principle or Theorem of Perpendicular Axes.
() For a Plane Laminar Body. According to this theorem, the
moment of inertia of a plane lamina about an axis, perpendicular to the
plane of the lamina, is equal to the sum of the moments of inertia of
the lamina about two axes at right angles to each other, in its own
plane, and intersecting each other at the point where the perpendicular
axis passes through it.
Thus, if /,, and / be the moments of inertia of a plane lamina
about the perpendicular axes, OX and OF, which lie in the plane of
the lamina and intersect each other at '0,
(Fig. 22), the moment of inertia about an
axis passing through O and perpendicular to
the plane of the lamina, is given by
/  4+V
For, considering a particle of mass m
at P, at distances x and y from OY and OX
respectively, and at distance r from 0, we
have
/ *= Zmr 8 , l m = Zmy* and 7 y
So that, /.+/, *= my*+Zmx*.
Rg. 22.
Zmr*. [v y*+x*
Or, /.+/,:=/.
52
PROPERTIES OF MATTER
(ii) For a ThreeDimensional Body*. Suppose we have a cubical
or a threedimensional body, shown dotted in Fig. 23, with OX, OY
and OZ as its three mutually
perpendicular axes, represent
ing its length, breadth and
height respectively.
Consider a mass m of the
body, at a point P, somewhere
inside it. Drop a perpendi
cular PM from P on the xy
plane to meet it in M . Join
OM and OP, and from M
draw MQ parallel to the xaxis
Fig. 23. and MN, parallel to the jaxis ;
also, from P draw PjR, parallel to OM. Then, clearly, the coordinates
of the point P are
x ;= ON = QM ; y = OQ = NM and z = M P = OR.
Since the plane xy is perpendicular to the zaxis, any straight
line drawn in this pLane is also perpendicular to it, and, therefore,
OM and PR are bath perpendicular to the zaxis, (v PR is drawn
parallel to OM).
Obviously, therefore, /. OMP is a right angle, because OM is
parallel to PR and PM is parallel to OR. Hence, we have
OM 2 +MP* = OP*.
Or,
But
Or,
= r
where, OP = r.
OAP = QM>+OQ*.
OM 1 = * 2 +>> 2 .
(0
is a right
angle, being the
angle between
the axes x and y.
Therefore, substituting the value of OJf 2 in relation (/) above,
we have x*+y*+z 2 = A ... (ii)
Join PN and PQ. Then, PN and PQ are the respective normals
to the axes of x and y. For, /_PMN is a right angle, being the
angle between the axes y and z, and, therefore,
PN 2 = MN*+PM* = j> 2 +z 2 .
So that, x*+PN* = x 2 +j> 2 +z 2 = r 2 . [From (//) above
Or, ON 2 +PN* = r 2 , l\ x  OM
from which it is clear that /.PM? is a right angle, and, therefore,
PN is perpendicular to the xaxis.
Similarly, in the rightangled &PMQ, we have
But
x*+y* +z a SB r 2 .
.*.
PQ*+y 2 *= r 2 .
Or,
P$ 2 +0<2 2 *= OP 2 . f v y  OQ.
L and r * OP.
Or,
.PQO is a right angle, i\e., PQ is perpendicular to the^axis.
*Not strictly included in the Q.c. (Pass gr Geacral) course.
MOMEtf 1? 0# INERTIAlfilfBlItGUr Off ROfATIOl*
Now, moment of inertia of mass m at P, about the zaxis
=*mxPR 2 =* m.OM\
because PR = OM is the perpendicular distance of the mass from
the axis.
.*. moment of inertia of the whole body about the zaxis, i.e.,
I = Zm.OM 2 .
Or, /, = Zm(x*+y*).
Similarly, the moment of inertia of the body about the yaxis, i.e.,
L = Zm.PQ 2 . p' PQ is the J_ dis
Or / Sm (x 2 4z 2 \ tance between ^e
ur ' '*  Zm '\ X + Z > L mass and the axis.
And, the moment of inertia of the body about the xaxis, i.e. 9
4 = Zm.PN\ p. PV is the 1 dis
Or T _ ymlv*\7%\ tance between the
Ur ' '* ~~ * m(y ^ Z > L mass and the axis.
/. adding up the moments of inertia of the body about the
three axes, we have
Or, I x +lv+Ia 
Hence the sum of the moments of inertia of a threedimensional
body about its three mutually perpendicular axes, is equal to twice the
summation Z"mr 2 about the origin.
^f) The Principle or Theorem of Parallel Axes. This theorem
(due to Steiner) is true both for a plane laminar body as well as a
threedimensional body and states that the moment of inlertia of J
body about any axis is equal to its moment of inertia about a parallel
axis, through its centre of mass, plus the product of the mass of the body
and the square of the distance between the two axes.
(/) Case of a Plane Laminar Body. Let C be the center of mass
of a body of mass M, (Fig. 24). and I c , its moment of inertia
about an axis through (7, perpendicular to
the plane of the paper.
Now, let it be required to deter
mine the moment of inertia / of the body
about a parallel axis through 0, distant
r from C.
Consider any particle P of the body,
of mass m 9 at a distance x from 0.
Then, the moment of inertia of the Fi S 2 *
body about is given by
I = Zwjc 2 . [Since OP 1 = .
From P drop a perpendicular PQ on to OC produced, and join
PC. Theri,
OP 1 ss= CP*+OC*+20C.CQ. [By simple geometry.
And .% m.OP 2 = w.CP*+w.0C 2 +2m.0C.C#,
Emx* = 2m.CP*+2m**+%rZm.CQ. [v OP * a? & OC  r.
Hence / I c +Mr*+2r2m.CQ. [v
64
PROPERTIES OF MATTBft
Now, since a body always balances about an dxis passiiig
through its centre of mass, it is obvious that the algebraic sum of the
moments of the weights of its individual particles about the centre of
mass must be zero. Hence, here, Emg.CQ, (the algebraic sum of
such moments about C) and, therefore, the expression Sm.CQ is
equal to 0, g being constant at a given place. Consequently,
2r.Zm.CQ = 0.
So that, / = I c +Mr*.
(ii) Case of a ThreeDimensional Body. Let AB be the axis
about which the moment of inertia of a body (shown dotted)
is to be determined, (Fig. 25).
Draw a parallel axis CD through
the centre of mass G of the body,
at a distance r from it.
Imagine a particle of mass
m at any point P, outside the
plane of the axes A B and CD and
let PK and PL be perpendiculars
drawn from P on to AB and CD
respectively and PT, the per
pendicular dropped from P on to
KL produced.
Put PL = d, LK = r, LT = x and Z.PLK = 6.
Then, if / be the moment of inertia of the body about the axi
AB and I c its moment of inertia about the axis CD (through G), wa
clearly have
/ = Zm.PK* and I f = 27w.PI 2 = Zm.d*.
Now, from the geometry of the Figure, we have
PK* = PL*+LK*2PL.LK cos PLK.
= d*+r*2d.r cos 0.
And, in the rightangled &PTL, we have
cos PLT = LT/PL,
where /_PLT == (180 PLK) = (180
cos (180 0) =*= x/d.
Or, ^ , cos & = x)d,
whence, d cos Q = x.
Substituting this value of d cos in the expression ftor PXT
above, we "
Fig. 25.
). So that,
And, therefore, / = Em.PK* = Zm(d*} r 2 +2rx).
= J c +Mr*+2rZmx,
because mr* = Mr 2 , where M is the mass of tHe whole body and r,,
the distance between the two parallel axea and hence a constant..
Clearly, Zmx = 0, being the total moment about an axis through
the centre of mass of the body.
We 4 ^therefore, have / a /^f Mr*,
the same result as obtained above in case (/) for a plane laminar
body*
MOMENT OF INERTIA ENERGY OF ROTATION 00
r 30. Calculation of the Moment of Inertia of a Body. Its Units
etc. In the case of a continuous, homogeneous body of a definite
geometrical shape, its moment of inertia is calculated by first obtain
ing an expression for the moment of inertia of an infinitesimal mass
of it about the given axis by multiplying this mass (m) by the square
of its distance (r)from the axis, (see page 51) , and then integrating
this expression over the appropriate range, depending upon the shape
of the body concerned making full use of the theorems of perpendicular
and parallel axes, wherever necessary.
In case, however, the body is not homogeneous or of a definite
geometrical shape, the safest thing to do is to determine its moment
of inertia by actual experiment, as explained later, in 34 and in
Chapter VIII.
Now, it will be seen that since the moment of inertia of a body
about a given axis is equal to MK* 9 where M is its mass and K, its
radius of gyration about that axis, its demensions are 1 in mass and 2
in length, its dimensional formula being [ML 2 ]. If the mass of the
body and its radius of gyration be measured in the C.G.S. units, i.e.,
its mass in grams and radius of gyration in centimetres, the moment
of inertia of the body is expressed in gramcentimetre 2 , (i.e., in gm.
cm 2 .). And, if the two quantities be measured in the F.P.S. units,
i.e., the mass of the body in pounds and its radius of gyration in
feet, the moment of inertia is expressed in Poundfeet*, (i.e., in Ib.fP)
And, finally, it must be carefully noted that since the moment
of inertia of a body, about a given axis, remains unaffected by reversing
its direction of rotation about that axis, it is just a scalar quantity.*
Thus, the total moment of inertia of a number of bodies, about a
given axis, will be equal to the sum of their individual moments oi
inertia about that axis, in exactly the same manner as the tota'
mass of a number of bodies will be equal to the sum of their indi
vidual masses.
Note. The argument is sometimes advanced that since the moment o
inertia of a body changes with the direction of the axis of rotation, it is not i
scalar quantity; and, since it is independent of the sense or direction of rota
tion about that axis, it is not a vector quantity either, and that it is what i
called a 'tensor'.
The author begs to differ. For, the term, 'moment of inertia of a bod)
has hardly any meaning unless clear mention is also made of the axis of rotati
of that body. And, once the axis of rotation is fixed, the moment of inertia
the body, about that particular axis* becomes a scalar quantity, being independe
of the sense of rotation about that axis. Indeed, it would be misleading to cz
it a tensor ; for, the fact is that the moment of inertia and the products of inert
(see below), at a point, together constitute the components of a symmetric tens
of the second order, which simply means that, knowing the system of momer
and products of inertia at a point about any three mutually perpendicular axe
we can, by means of certain simple, transformations, obtain their values for ai
other set of three mutually perpendicular axes at tbat very point.
A general tensor, of the second order, in threedimensional space, has,
general, nine components, say, C n , C,,, C, C tl , C M , C M , C S1 , C t? , C tl . But, f
a symmetric tensor, C, a = C 21 , C 18 = C M and C S i = C )8 , so that it has only s
distinct components, viz., three moments of inertia and three products of inert
about the three perpendicular axes. \\ \\ ( O ^ J '*
* Scalar quantities are those which possess only magnitude \ but no direc
tlon,e.g. 9 mass, time etc. On the other hand, vector quantities are those whicl
possess both magnitude as well as direction, f ., acceleration, velocity, force, etc.
66 pjtopUBTiEis otf
us, if x, y t 2 be the coordinates of a particle of mass w, at /*, in Fig. 23.
We have
(/) moments of inertia about these three perpendicular axes respectively
given by
I x =" 2m(y 2 + z) 2 , I v = 2m(z*+x 2 ), /0 = 2Vtt[# 2 h}> 2 ), and
(//) the products of inertia about these axes defined by
Pys 2tnyz, P*x 2mzx, P xy = 2mxy,
Then, /, /, 70, P v g, Pey and P xv are the six components of the
symmetric tensor at point P.
It will thus be seen that it is, at best, only a halftruth to say that the
moment of inertia of a body about a given axis is a tensor.
11. Particular Cases of Moments of Inertia.
Cv^/1. Moment of Inertia of a Thin Uniform Rod :
(i) about an axis through its centre and perpendicular to its
length. Let AB, (Fig. 26), be a thin uniform rod of length /and mass
M, free to rotate about an axis CD through its centre O and per
pendicular to its length. Then, its mass per unit length is MIL
Consider a small element of length dx of it, at a distance x from O.
Its mass is clearly equal to (M/l).dx, and its moment of inertia about
the axis through O = (M/l).dx.x z .
The moment of inertia / of the whole rod about the axis is,
therefore, obtained by integrating the above expression between the
limits x = 7/2 and x = +//2 ; or between .v=0 and jc=//2 and
multiplying the result by 2, to include both halves of the rod.
C f//2 M
, Thus, 7=2 *. T  t x*.dx.
! JO /
_ Mr *n//;
"~ / L 3 Jo
i
i ~TL
 F f 26. 01= 2M /3 = M8
lg F) ~" / 24 12 '
(//) about an ixis passing through one end of the rod and per
pendicular to its length. The treatment is the same as above, except
that, since the axis CD here
passes through one end B of the * y
rod, (Fig. 27), the expression for '
the moment of inertia of the ele ('"  T
ment dx of the rod is now to be %
integrated between the limits,
\x = 0, at B and x = /, at A.
Thus, if 7 be the moment of F te 27 
inertia of the rod about CD, we have
F M * A Af pc 7
==  r .* f .lfit = ^ v~
Jo / / LS Jo
Af / 3 Ml 2
r * y " 3^"
MOMENT Otf UN1BT1A JEHB&Glf OF ROTATION ft<9
Or, we could have arrived at the same result by an application of the
principle of parallel axes, according to which the moment of inertia of the rod
about the axis through B is equal to the sum of its moment of inertia about a
parallel axis through its centre of mass and the product of its mass and the
square of the distance between the two axes.
TK T M /2 L ,./ / Y MI 2 ,M/ 2
Thus, I  + M~  +
2. Moment of Inertia of a Rectangle.
(/) about an axis through its centre and parallel to one of its
sides. Lot A BCD be a rectangle, and let / and b be its length and
breadth respectively, (Fig. 28). Let the axis of rotation YY' pass
through its centre and be parallel to the side AD or BC.
If M be the mass of the rectangle, (supposed uniform), its mass
per unit length will be MIL
Consider a small strip, of width dx of the rectangle, parallel to
the axis. The mass of the strip will obviously be (mjl).dx t an,
therefore, the moment of inertia of }
the strip, about the axis YY' will be /\. , ,.. j ,;. c ;r ... *jo.. a#
The whole rectangle may be
supposed to be composed of such like
strips, parallel to the axis, and there
fore, the moment of inertia / of the
whole rectangle about the axis YY' is
obtained by integrating the expression Fig. 28.
(Af//).dx.JC 2 , for the limits x=0 and x=//2 and multiplying the
result by 2.
r A f ; / 2 M , , 2Af f//2 2 , 2M r x 3 ~//2
i.e., 7=2 . .x*.dx = H x *.dx ~i  ^r~
Jo / /Jo / L 3 Jo
_2M / 3 _ Ml 2
Ur, 1  ;  2   12 
It will be seen at once that if b be small, the rectangle becomes
a rod, of length /, whose moment of inertia, about the axis YY'
passing through its centre and perpsnclicular to its length, would be
M/ 2 /12, fas obtained above, 31, case 1, (/')].
(//") about one side. We majr proceed as above (in case we want
an independent proof) except that the expression (Mjl).dx.x z may
here be integrated for x = and x = /. Thus, the moment of
inertia of the rectangle about the side AD or BC is given by
(V M 2 , M [ I 2 .
SB  ,~,x*.dx = , 1 x 2 .dx.
Jo / , /Jo
M r x 3 iC i/ / 3 Mi 2
Or, i^
Alternatively, proceeding on the basis of the previous article, we may
apply the principle of parallel axes, according to which the moment of inertia of
the rectangle abouc side AD or BC is given by
/ / V
/ =* MJ. about. a I! axis through its cent re +M (^ ^ )
, M 8 r Ml* Ml 2
or, '+"' '
Off MATTES
(Hi) about an axis passing through its t$nit& and perpendicular to
its plane. This may be obtained by an application of the principle of
perpendicular axes to case (i) above, whence the moment of inertia
of the rectangle about an axis through its centre 0, perpendicular to
rts plane, is equal to its moment of inertia about an axis through O r
parallel to its breadth b, plus its moment of inertia about a perpendi
cular axis through O, parallel to its length /,
_ Ml* _
i.e., 1 +~ 12
The above relation is equally valid in the case of thin (/..,
laminar) or thick rectangular plates or bars, no stipulation with
regard to its thickness having been made in deducing it. And, after
all, a thick rectangular plate or bar may be regarded as just a pile of
thin (or laminar) plates or bars, placed one above the other.
* The same argument will hold good in all other cases of a similar
type, [see cases (iv) and (v) below],
(/v) about an axis passing through the midpoint of one side and
perpendicular to its plane.
(a) Suppose the axis of rotation passes through the midpoint
of Ap or EC, (Pig. 28). Then clearly, in accordance with the prin
ciple of parallel axes, we have
moment of inertia about this axis, i.e.,
I = moment of inertia about a parallel axis
through its centre + M x (//2) 2 ,
where //2 is the distance between the two parallel axes.
M (
* """12" " 4 ~ 12
_
 12 "= 12
or. '= M (T+H>
(b) Similarly, if the axis of rotation passes through the mid
point of AB or DC > we have
b ^ a ' f v the distance between
"2 } L ^e two fl axes is now 6/2.
Or,
3. Moment of inertia of a solid uniform bar of rectangular crosssection,
about an axis, perpendicular to its length and passing through its middle point,*
Let ABCDEFGH (Fig. 29) be the rectangular uniform bat of
length /, breadth b and thickness d, whose moment of inertia about the
am XX', passing through its centre and perpendicular to its length, is desired.
"This is really covered by case 2 above, but is given here for a clearer
understanding of the student.
OF INERTIA EttEftOY 0*
Imagine the whole bar to be made up of a large number of thin rectan
feular sheets, parallel to the face CDEHand perpendicular to the axis XX', pass
ing through the centre of mass
of each sheet Consider one JT
such sheet, (shown dotted),
of mass m, of length and ^ ^_
MfffUJlJ'K / and d respectively, 4*
and centre of mass O, through
which the axis XX' is passing
perpendicular to its plane.
Then, the M.I. of this
sheet about XX' = its mass
x (/ 2 f</ 2 )/12, as can be seen
from the following :
Let PQ be an axis through
O\ in the plane of this sheet, and
parallel to its breadth CH or DE.
H
, Fig. 29.
Take a thin strip of width dx of this strip, parallel to, and at distance x
Jrom, the axis PQ Then, mass of the strip = (w//).</Jcand, therefore, its moment
of inertia about the axis PQ is
(mil) dx.x\
moment of inertia / of the whole sheet about PQ is given by
2f 7 / 2 m __ 2m [U2
if//2
Jo
2m
I
.
7/2
2m ftf
' Jo
x\dx
# V
T Jo "
2m
r
8x3'
Or,
Ml 2
12
Similarly, the moment of inertia of the sheet about an axis through O,
in its own plane and perpendicular to PQ, i.e., parallel to its length DC or EH wiL
be Md 2 /12.
Therefore, by the principle of perpendicular axes, the moment of inertia
of the sheet about the axis XX' through O and perpendicular to its plane
ml 2 md 2 _ /Pd
."" 12 + 12" "~ m V 12
Hence, moment of inertia of the whole bar about the axis XX'
mass of the bar x
f
V
12
Or.
M being the mas.
of the bar.
1. Moment of Inertia of a Thin Triangular Plate or Lamina about
one side. iittABC, (Pig. 30), be a th?n, triangular plate or lamina,
of surface density or mass pet
unit area, p, whose moment
of inertia is to be determined
about the side BC.
Then, if the altitude
the plate be AP = H,
area =* J base x altitude
[v BC
it*
Fig. 30.
and, therefore,
its mass M
*.. (i
Off MAfTUft
Now, let us imagine the triangular plate to be made up of a
number of thin strips, parallel to BC, and placed side by side ; and,
let us consider one such strip DE t of width dx, at a distance x from
the base BC. Then, clearly, the area of this strip, (which may be
considered to be almost rectangular, its width being infinitesimally
small) =5 DE,dx. And, therefore,
mass of the strip = DE.dx.p ....... (#)
New, in the similar triangles AQD and APB, we have
whence, DQ = BP. f r .
ri
Similarly, from the similar triangles AQE and APC t we have
M=^ = 4' whence Q* *>..
And, therefore, DQ+QE = BP.^+PC. 4
ri ri
Or, = (BP+PC). . = .
A
mass of the strip = a~ H ~.dx.p. [From (//) above]
Now, clearly, moment of inertia of strip DE about the side 5C
h
= mass of the strip xx*=a.  .dx.f.x*.
Hx
g
And, therefore, moment of inertia of the whole triangular
plate about BO is equal to the integral of this expression, between the
limits x = and x = //. So that,
M.L of the plate about BC, i.e., I = Pa. (*~\*.&.dx.
_ a.p rff.x 3 x 4 ~ H _ a.p / /?* H \
~ #l~3 ~4"Jo ~ "jtfA T~T"y
~ff( i2~~) ^ 7T"w = ""12""
_H
But ^ a./f p. = Af, ffte maw of the plate. [Sec (/) above.
.. M. I. of the triangular plate about side BC, i.e , I = 'r
MOMENT OF INBBTIA BK1BGY OF BOTATIOH
61
$. Moment of Inertia of an Elliptical Disc or Lamina. (?) about
oneof its axes. Let XYX f Y' be a thin elliptical plate or lamina, of
mass M, and surface density (ic. 9
mass per unit area), P, and let its
major axis XX f and minor axis YY'
be equal to 2a and 2b respectively,
(Fig. 31).
Consider a strip PQSR of the
plate of width dx, parallel to the
minor axis YY 1 and at a distance x
from it. Then, if 2y be the length of ~y
the strip, its area is clearly equal to  Fig. 31.
2y.dx and, therefore, its mass equal to 2y>dx.?.
Obviously, then, M, I. of the strip about the minor axis
YY' = 2y.dx.p.x* ; and, therefore, M. L of the whole elliptical plate
about the axis YY' is equal to twice the integral of the above
expression, between the limits x = 0, and x = a. Or, denoting it by
I v , we have
2y.*x*.dx = 4 P [ a y.x*.dx (I)
Now, with the centre of the ellipse as the origin, and with the
coordinate axes coinciding with its major and minor axes respectively,
we have 2+ /,> = 1> as the equation to the ellipse ; whence,
fa = ! ii. or y*  l
b z a*
So that, y = b ^/i^x^ja^T
Substituting this value of y in relation (I) above, we have
= 4P Jo
...(II)
Now, putting x = a sin 8, we have T = a cos 6.
aQ
Or, dx = a cos 0. d9.
Substituting these values of x and dx in expression (II) above,
we have
/2 .a cos 6.dd.
Or,
= 4 P
L f/ 2
'Mo **
gr~ 4
, fw/2
P ' Jo a '
/, = 4/=
L o fw/2
( 6.a 3 .
Jo
cos
or MATTIE
pfi.t;
I,
*2
1
2
PTC/2
lo
f 2 1 cos 1 40
2
sm
.dg.
12
'Jo
TT 7T
2" ^ ~
Now, Tr.a.b.p = Af, the mass of the elliptical plate.
I, = M.a 2 /4.
Similarly, the moment of inertia (I x ) of the elliptical plate about
the major axis XX' is given by the expression,
I, = Mb*/4.
(//) about an axis passing through the centre of the plate or lamina
and perpendicular to its own plane The axis in this case will pass
through O, (Fig. 31), and will be perpendicular to the plane of the
paper, (or the plane of its two axes, XX' and 77') . Hence, if /be
the moment of inertia of the elliptical plate about this axis, we have,
by the principle of perpendicular axes,
0, !
^fff Moment of Inertia of a Hoop or a Circular Ring.
(i) about an axis through its centre and perpendicular to its
plane. Let the radius of the hoop or circular ring be 7?, and its mass,
M, (Fig. 32).
Consider a particle of it, of mass m. Then, the moment of
inertia of this particle about an axis through the centre O of the hoop,
and perpendicular to its plane, will obviously be mR 2 .
And, therefore, the moment of inertia /of the entire hoop about
the axis will be ZmR*.
^ r m* P' 2'^=Mand R is the same
Or, I = MR 2 . I for all particles.
(ii) about its diameter. Let it be required to determine the
moment of inertia of the hoop about the diameter AB, (Fig.
Obviously, the moment of inertia of the hoop
will be the same about all the diameters.
Thus, if / be the moment of inertia of the
hoop about the diameter AB, it will also be
its moment of inertia about the diameter
CD, perpendicular to AB.
Then, by the principle of perpendicular
axes, its moment of inertia about the axis
through the centre O, and perpendicular to
its plane, is equal to the sum of its moments
of inertia about the perpendicular axes AB
and CD, in its own plane, and intersecting
MOMENT of INBBTU BNBEQY of BOTATKW
63
And, therefore, 7+7 MR 2 . Or, 2 7 MR*. [Seecaie (0.
Or I = MR 2 /2.
' 1. Moment of inertia of a Circular Lamina or Disc.
(i) about an axis through its centre and perpendicular to its
plane. Let M be the mass of the disc and R, its radius. Then, since
the area of the disc is 7T.R 2 , its mass per unit
area will be
Consider a ring of the disc, distant x
from O, i.e., of radius x and of width dx,
(Fig. 33). Its area is clearly equal to its
circumference, multiplied by its width, or
equal to 2wx x dx, and its mass is thus
Hence, moment of inertia of this ring
about an axis through O and perpendicular
to its plane
Fig. 33.
M.2<xx.dx
2Mx*dx
Since the whole disc may be supposed to be made up of such
like concentric rings of radii ranging from to jR, we can get the
moment of inertia / of the disc by integrating the above expression
for the moment of inertia of the ring, for the limits x=0, and x=/Z.
*,T r*u A
... MJ. of the dzsc .
Or,
(//)
diameters
I = MR 2 /2.
Or,
about its diameter. Let AB and CD be two perpendicular
of a circular disc of radius R and mass Af, (Fig. 34).
Since the moment of inertia of the disc
about one diameter is the same as about
any other diameter, the moment of inertia
about the diameter AB is equal to the
moment of inertia about the diameter CD,
perpendicular to AB. Let it be /.
Now, we have, by the principle of
perpendicular axes, M.L of the disc about
AB+its M.L about CD
= its M L about an axis through
O and perpendicular to its plane.
n r . . MR* T MR 2
Or, ,/+/  %OT, 27 = g.
MR*
(ii) about a tangent to the disc in itsi&vn plane. Let AB be
tangent to the circular disc of radius R and mass M, about which it*
64
PROPERTIES Off MATTER
moment of inertia is to be determined, (Fig. 35). Let CD be a
diameter of the disc, parallel to the tangent AB. The moment of
A inertia of the disc about this diameter is, clearly,
equal to MRJ4.
So that, by the principle of parallel axes, we
have
MJ. of the disc about AB = MJ. of the
disc about CD+MR*.
Or,
I =
MR 2
+MR* =
MR 2
(iv) about a tangent to the disc and perpendi
D
Fig. 35. cular to its plane This tangent will obviously be
parallel to the axis through the centre of the disc and perpendicular
to its plane, the distance between the two being equal to the radius
of the disc. Hence, by the principle of parallel axes, we have
M.L about the tangent = M.L about the perpendicular axis+MR*.
Or,
= ;: MR 2 .
Moment of Inertia of an Annular Ring or Disc.
(/) about an axis passing through its centre and perpendicular to
its plane. An annular disc or ring is just an ordinary disc from
which a smaller co axial disc is removed, so
that there is a concentric circular hole in
it. Let 7? and r be the outer and inner radii
of the disc, (Fig. 36), and M, its mass. Then,
clearly,
facearea of the annular disc = facearea
of disc of radius R face area of disc of radius
r,
And .. mass per unit area of the disc
= M/7r(# 2 r 2 ). p. 2$
Imagine the disc to be made up of a
number of thin circular rings, and consider one such ring of radius x
and of width dx.
Then, facearea of this ring =27ix.dx,
M
__
and
its mass =
And, therefore, its moment of inertia about an axis through O
, ,. , x . x . 2Mx , 9 2Mx 3 
and perpendicular to its plane =   ax.x* = rnz^TW
The moment of inertia of the whole annular disc may, therefore,
be obtained by integrating the above expression for the limits x = r
and x = R, Or, moment of inertia of the disc about the axis through
O and perpendicular to its plane
* 2^*> 2M f*
r *r*) dX ~ (W=
f
jr
MOMENT OF OTBBTU ENERGY 0! ROTATION

r  (IPr)
2M mp+ r *).(/l_rn
^IrSjL r ~J
Or, I  M
It follows at once from the above that if r = 0, i.e., if there is
no hole in the disc, or that it is just a plane, (and not an annular)
disc, its moment of inertia is MR 2 /2. [Case 7 (/), above.
.Again, if r = R, i.e, t we have a hoop or a circular ring, of radius
R, and its M.L = ^ ^ MR * t [Case 6 (/), above.
A *
(it) about its diameter. Obviously, due to its symmetrical shape,
the moment of inertia of the annular disc about one diameter will be
the same as that about any other diameter. Let it be 7. Then, the
sum of its moments of inertia about two perpendicular diameters will,
by the principle of perpendicular axes, be equal to its moment of inertia
about an axis through O and perpendicular to its plane, i.e., equal to
MGR 2 +r 2 )/2.
Or 747 M( * 2 + r *> /* 2/ Aff/P+l * )
ur, /j7 =  , i.e., LL   ,
whence, =   
Now, if r = 0, i.e., if the disc be a plane one, we have
M.L of the disc about a diameter = MR 2 /4:. [Case 7 (//), above.
Or, if r = R, we have a hoop or circular ring of radius R, and
its moment of inertia about its diameter
MR*
. ..... u
=    . = r. [Case 6 (//) above.
442
(Hi) about a 'tangent, in its own plane. The tangent being
parallel to the diameter of the ring or disc, and at a distance R from
it, we have, applying the principle of parallel axes,
M.L about the tangent = M.L about the diameter {MR*.
Or, I = M +M R* = M
(iv) about a tangent, perpendicular to its own plane. The tan
gent, in this case, is parallel to the axis through the centre of the ring
or disc and perpendicular to its plane, the distance between the two
being equal to jR. Hence, by the principle of parallel axes, we have
M.L about the tangent = M.L about the perpendicular axis+M/? 2 .
Or I **
9. Moment of inertia of a Solid Cylinder.
(i) about its own axis, or its axis of cylindrical symmetry. A
cylinder is just a thick circular disc, or a number of thin circular
86
PROPERTIES OF MATTER
discs, piled one upon the other, and, therefore, its moment of inertia
about its axis is the same as that of a circular disc or lamina &bout an
axis through its centre and perpendicular to its plane, i.e., equal to
MR 2 /2, where M is its mass and R, its radius. [Case 7 (i) above.
(ii) about an axis passing through its centre and perpendicular to
its own axis of cylindrical symmetry. Let M be the mass of the cylin
der, R its radius and /, its length, (Fig. 37). Then, obviously, if it be
homogeneous, its mass per unit length will be M/l. Let YY' be the
axis, passing through its centre and perpendicular to its own axis
XX' , about which the moment of inertia is to be determined.
Imagine the cylinder to be made up of a number of thin idiscs
and consider one such disc at a distance x from O, and of thickness dx.
Obviously, the mass of the disc is (Mjl).dx and its radius, equal
to R ; so that, its
moment of inertia
about its diameter
is equal to mass of
/.Jj U~. ,.l~l. SI 4J the disc x(radius)^.
?*' f '
And, its mo
iftent of inertia
about the axis YY' 9
Y
/_ . . _ t/g _a>
?\
4V4
/
""\i )
. . _    /  ,
* {
/
\r
Fig. 37.
M
by the principle of parallel axes, = ,
Therefore, the moment of inertia of the whole cylinder about
the axis YY' may be obtained by integrating this expression for the
limits x = and x = //2 and multiplying the result by 2. Thus,
MJ. of the cylinder about the axis YY f
2M
i
+*J =
J!_l
Jx3J
Or,
(HI) afeowf a diameter of one of its faces. It is an easy deduction
from the above ; for, by the principle of parallel axes, we have
M.I. of the cylinder about the diameter of one face
MR* .Ml* .Ml*
\*
MR* Ml*
4,
I.
M
4T M
MOMENT Off INERTIA ENERGY Of ROTATION
67
Moment of Inertia of a Solid Cone.
(i) about its vertical axis. Let mass of the cone be M , its vertical
height, h and radius of its base, R, (Fig. 38).
Then, its volume = ^nR z h.
And, if p be the density of its material, its
mass M = knR*hp, whence, p = ..
TT/v II
Imagine the cone to be made up of a
number of discs, parallel to the base, and
placed one above the other. Consider one
such disc at a distance x from the vertex,
and of thickness dx.
If r be the radius of this disc, we have
r = x tan a,
[where a is the semivertical angle of the cone.
And, volume of the disc = 7ir z .dx. Fig. 38.
its mass = 7rr 2 .c/x.p.
Now, moment of inertia of the disc about the axis AO y pass
ing through its centre and perpendicular to its plane, i.e., about the
vertical axis of the cone, is clearly equal to its mass ^ ts radius ? .
And, therefore, the moment of inertia of the whole cone about its
vertical axis AO will be the integral of this expression, for the limits
x = and x = /?.
/. ., J/.7. of bhe cone about its vertical axis is given by
 [ h
Jo
2
7TP./? 4
'*]*
.5 Jo ~
A 5
5
Or, substituting the value of p from above, we have
1 = 7t~R*.h.2h* ' T == " 10
(ii) about an axis through its vertex and parallel to its base.
Again, considering the disc at a distance x from the vertex of the cone,
we have
r 2
M.I. of the disc about its diameter = 7rr 2 .Jx.p. . .
4
And, therefore, by the principle of parallel axes, its moment of
inertia about a parallel axis XX 1 ', passing through the vertex of th
cone
r*
08 *ftoEfeTiB$ oir
Therefore, the moment of inertia of the whole cone about the
axis passing through the vertex and parallel to the base, i.e., about
XX', is obtained by integrating this expression for the limitSj x =
and x = h. Thus,
M.I. of the cone about XX*
> tan* a fh 4 ^ , 9 fh ^ ^
f* **.</* +irPtoi of*
'0 J
4 A*
^ TTP f /? 4 A 5 , R* h b
Or, substituting the value of P, we have
M.L of the cone about XX' = ^ , , , .  
/i* 5
. _ 3MR* 3Mh
Or, 1= +
r Moment of Inertia of a Hollow Cylinder.
((/) aftowf its own axis. A hollow cylinder may be considered to
consist of a large number of annular discs or rings of the given inter
nal and external radii, placed one above the other, the axis of the
cylinder passing through their centre and being perpendicular to
their planes, (Fig. 36).
The moment of inertia of the hollow cylinder about its own
axis is, therefore, the same as that of an annular disc of the given
external and internal radii
about an axis through its cen
tre and perpendicular to its
plane, i.e., equal to M(R 2 fr 2 )/2,
where M is the mass of the
cylinder, R and r, its external
and internal radii respectively.
Alternative Proof. Let M Fig. 39.
be the mass of the cylinder ; R and r, its external and internal radii
and /, its length, (Fig. 39).
Then, facearea of the cylinder = Tr^ 3 r f ).
And, volume of the cylinder = 7r(7? 2 r 2 )/ ;
M
BO that, its mass per unit volume = ^ ^
Now, imagine the cylinder to be made up of a large number of
thin co axial cylinders, and consider one such cyluvfer of radius x and
thickness dx.
MOMENT 0* INERTIA ENERGY OF ROTATION
69
Then, its face area = 27tx.dx 9 its volume =
M , ,
and its mass = /j ? 2_^/x27rx.a.x./ /m~ jr
Since all its particles are equidistant from the axis, its moment
. .. , . ., . 2Mx.dx f 2 Jf **.</*
of inertia about the axis = , ~ 2  .x 3 == D2  2
(A r) (K r )
And, therefore, the moment of inertia of the whole cylinder may
be obtained by integrating the above expression for the limits, x r
and x s= jR.
Or,
J/.7. of the cylinder about its axis, i.e.,
CR 2M.X 8 , _ *M_[ R a
 r (j?2Z7 a ) ~ (^^H7^J r x
5Jf r^iJ 2jj/ n^ 14 )"
VOL 4 Jr ^ (^ 2 r2)
t
Jr
(R*
']
Or,
an axis passing through its centre and perpendicular to
x
V
Fig 40.
its own axis. As be
fore, let M be the
mass of the cylin
der, /, its length,
and/? andr, its ex
ternal and internal
radii respectively ;
and let YOY'bz the
axis through its cen
tre 0, and perpendi
cular to its own axis
XX', (Fig. 40).
Then, facearea of the cylinder = 7r(jR a r f ),
and its volume = 7t(R z r 2 )/.
its mass per unit volume == Jf/7r(J? 8 / a )/.
Imagine the cylinder to be made up of a large number of
annular discs of external and internal radii R and r, placed one by
the side of the other, and consider one such disc at a distance x from
the axis YY f , and of thickness dx. Then, clearly,
surface area of the disc = 7r(/? 2 ~~r a ),
its volume = 7r(jR 2 r*).t/x, and ..its mass = Jf.rfx//.
Now, moment of inertia of an annular disc of external and
internal radii, jR and r, about its diameter, is equal to its massx
+R* + r 2 )/4. [Case 8 (),
.. M.I. of the disc about its (}i$iter
70
PROPERTIES OT MATTJBH
And, therefore, its moment of inertia about the parallel axis
YT' is, by the principle of parallel axes, given by
M . (R*+r l ) M ,
,w^c x * ~ ~l ,dx.x.
And, clearly, therefore, moment of inertia /, of the whole cylin
der, about the axis IT', is twice the integral of this expression, for
the limits, x = and x = 7/2.
tip.
i.e.,
2Jff//2P
i \
/ JO L
2P 7?2 + ,a

dx+x*.dx \~ .
J

L
"3 Jo
Or,
I =
2Afr(/? 2 fr 2 )/
/ L 4x2 "
It follows, therefore, that if r = 0, i.e.> if the cylinder bo a
solid one, we have
M.L of the cylinder, (solid), about an axis through its centre
and perpendicular to its own axisM(R 2 /4 + F/12). [Case 9, (//) above.
^2r Moment of Inertia of a Spherical Shell.
(/) about its diameter.
First Method. Let ABCD be the section of a spherical shell
through its centre O and let the mass of the shell be M , and its
radius R, (Fig. 41).
Then, area of the shell is equal to 47T/? 2 ,
and
.*. mass per unit area of the shell = MI&nR 2 *
Let it be required to determine its
moment of inertia about the diameter AB.
Consider a thin slice of the shell, lying
between two planes EF and OH, perpendi
cular to the diameter AB 9 and at distances x
and (x\~dx) respectively from its centre O.
This slice is obviously a ring of radius
PE, and width EG, (not PQ, which is equal
to dx).
area of this ring => its circumference X width.
=*= 27T.PEXEG,
and, hence its mass == its area XM tin R*.
=c 2n.PExEGxM/4>7rR*. ... (/)
Join OE and OG, and let COE . and l_EOG * d0.
Then, PE = OE.cos OEP R cos
[v LOEP  iCOE  e.
Similarly, OP = R sin Q
Or. x * J? $m e, ['' OE  J? ni/ O/  *.
Fig. 41.
MOMENT OF INERTIA ENERGY OF ROTATION 71
Now, differentiating x with respect to 0, we have
dxjde = R cos g t
Or, dx = R cos Q.dB *= PE.df). [v X cos o JP&
And, G == 0".rf0 = #.^0. I '' flrc = radius* angl*
L subtended by the arc.
mass of the ring = 27r.P.^.rf0.M/47r# 2 . [from (/).
Hence, moment of inertia of the ring about AB, (an axis passing
through its centre and perpendicular to its plane), is equal to its
mass x (its radius) 2 , i.e., = M *.PE\ where PE 2 = (R*x*).
ZK
/. moment of inertia of the ring about AB = > V .(R 2 x\
AJ\
And, therefore, the moment of inertia /, of the whole spherical
shell about AB = twice the integral of " D .(R 2 x 2 ), between the
AjK
limits, x = and x = R.
'
ie I  M 2 R 3 ~~ 2
i.e., l ~ ^ . ^ y<  y
Second Methad.~Let M be the mass of the shell and R, its radius.
Consider a particle, of mass m, anywhere on the shell. Then, since the
thickness of the shell is negligible, the distance of the particle from the centre
of the shell is the same as the radius of the shell, i e., R.
Obviously, therefore, the summation / , for all the particles of the shell,
about its centre 0, is given by the relation,
TAll particles being at distance R
/o  2mR*. Or, /.  MK a [j rom Q an d Jm = M.
Now, if /be the moment of inertia of the shell about one diameter, it
will be the same about any other diameter also, from the sheer symmetry of tha
shell. Hence, in accordance with the principle of perpendicular axes for a three
dimensional body, [29 (a), (//), page 52] the sum of the moments of inertia of
the shell about its three mutually perpendicular diameters must be equal to
twice the summation / , for all its particles, about their point of intersection,
i.e., the centre of the shell O ; so that,
/+/+ /  2/o, " Or, 3/  2MR*, [V /  MR 9 .
whence, I  MR*.
(ii) about a tangent. Obviously, a tangent, drawn to the shell
at any point, must be parallel to one of its diameters, and at a
distance from it equal to J?, the radius of the shell. Hence, applying
the principle of parallel axes, we have
A/./, of the shell about a tangent
= its M.L about a diameter {MR 2 .
Or, /  MR*+MR* =* JMR*.
71
OF MATTBH
13. Moment of Inertia of a Solid Sphere.
(i) about its diameter. Let Fig. 42 represent a section of the
sphere through its centre O.
Let mass of the sphere be M t and its radius, R.
Then, clearly, its volume = 47T# 3 /3.
And .*. its mass per unit volume
Consider a thin circular slice of the
sphere at a distance x from the centre O t
and of thickness dx.
This slice is obviously a disc of
radius \/ R* x 2 , and of thickness dx.
Now, the moment of inertia of /Af5 disc about ^4.
passing through its centre and perpendicular to its plane)
= its mass x (radius) 2 1 2.
moment of inertia of the disc AB
.. surface area of the slice = TT^/^X^ = Tr(R* x 2 ), and its
volume area x thickness == 7t(R* x 2 ).dx.
And, .. its mass = its volume x mass per unit volume of the
sphere
(an axis
;#"' * ."(I)
/. moment of inertia /, of the sphere about the diameter AB is
equal to twice the integral of expression (1) between the limits x =
and x = R.
3 " "*" 5 Jo
Or,
Now, the moment of inertia of the sphere about one diameter
is the same as about another diameter, so that we have the moment
of inertia of a solid sphere about any diameter given by I =c MR*.
Alternative Meth6d. Let M be the mass of the sphere and p, the density
of its material.
Imagining the whole sphere to be made up of a number of thin, concen
tric spherical shells, one inside the other, and considering one such shell of
radius x and thickness dx, we have
surface arm of the shell
3M
MOMBNt OP INERTIA INjBRQY OF ROTATION 73
and /. volume of the shell = 4*rx*.djc and its mass = 4*x*.dx.p.
.'. moment of inertia of the shell about a diameter
x(its tfttm)x(its radius)* $.4nx*.dx.pxx* npx* dx, [case 11 (/),
And, therefore, the moment of inertia /, of the whole sphere, about its
diameter, is obtained by integrating the above expression between the limits.
x and x = R.
( R 8 L
J r w
8 f x ~]R 8 R* 8 _.
T" P U Jo 3 "'TTrP*
t 437^3/3 = the v0///w? of the sphere ; and, therefore, 4rtR*?l3 = M, its
M.L of the sphere about its diameter, /.*., I = 2,MR/5.
(n) fl&0w/ a tangent. A tangent, drawn to the sphere at any
point, will obviously be parallel to one of its diameters and at a
distance from it equal to R, the radius of the sphere.
Therefore, in accordance with the principle of parallel axes, we
have M.L of the sphere about a tangent
= its M.L about a diameter + MR*.
Or, 12 MR*/!> f MR* = 7MR 2 /5.
v x x ^14. Moment of Inertia of a Hollow Sphere or a Thick Shell.
(/) about its diameter A hollow sphere is just a solid sphere
from the inside of which a smaller concentric solid sphere has been
removed. And so, the moment of inertia of the hollow sphere is
equal to the moment of inertia of the bigger solid sphere minus the
moment of inertia of the smaller solid sphere removed from it, (both
about the same diameter). If R bs the radius of the bigger sphere
and r, that of the smaller sphere, i.e., if R and r be the external and
internal radii of the hollow sphere, and p, the density of its material,
we have
volume of the bigger sphere = ^TtR 3 , and .. its mass =
and, smaller =a JTrr 3 and =
.. volume of hollow sphere = ^(J? 3  r 8 ) and its mass
And /. M.L of the bigger sphere about its diameter
and M.L of the smaller sphere about the same diameter
 H7rr*.p).r*.
.. M.L of the hollow sphere about that diameter
rr.p).^
... (1)
Now, mass of the hollow sphere, M = i.7r(jR 8 r s ).p.
Or, 3Af 47r(JR !l r s ),p, And .. p
rftOFEBTlBS Of MATTER
Substituting this value of p in relation (1) above, wo have
moment of inertia / of the hollow sphere about its diameter
_ 1 *
6 ' 3 w '
Or i_A M J^'.
Ur ' l 5 iV1 * (Rr)
Alternative Method. As in the case of the solid sphere, so also here, we
can imagine the sphere to be made up of a number of thin, concentric spherical
shells, and considering one such spherical shell of radius x and thickness dx, we
have, as before, mass of the shell = 4nx*.dx.p. fp being the
.. ML of the shell about a diameter = &Anx*.dxj.&. ( density of the
6 material of
.7r. P .**.</*. [ the sphere.
Hence, the moment of inertia of the whole sphere about its diameter is
the integral of the above expression, between the limits, x = r and x = R.
fR o
Or, M.L of the sphere about a diameter i.e., I = I  TC p.x*.dx.
8 I"* 8 r x *R
But ^(R* r 8 ).p Af, the mass of the sphere. [See case (/) above.]
I = .
(ii) about a tangent. Again, as in the case of a solid sphere, the
tangent to the sphere, at any point, will be parallel to one of its dia
meters, and at a distance equal to its external radius R from it.
Hence, by the principle of parallel axes, we have
M.I. of the sphere about a tangent
= its M.L about a diameter {MR*.
Or, I = ["M(R 5 ~r 5 )/(R 8 r 3 )"l+MR 2 .
15. Moment of Inertia of a Flywheel and Axle. A flywheel
is just a targe heavy wheel, with a long, cylindrical axle, passing
through its centre. Its centre of gravity lies on its axis of rotation^
so that, when properly mounted over ballbearings (to minimise
friction), it may continue to be at rest in any desired position.
Let M be the mass of the flywheel, and m, that of the axle ;
and let R and r be their respective radii.
Then, for our present purpose, we may ' regard the flywheel to
be a disc, or a small cylinder, from which a smaller, concentric disc or
cylinder, equal in radius to that of the axle, has been cut off. In
sther words, we may take it to be an annular ring, (or hollow cylin
ier) with an outer radius equal to R> and an inner radius equal to r,
iose moment of inertia is to be determined about an axis passing
ough its centre and perpendicular to its plan*.
MOMENT OF INERTIA ENERGY OF EOTATION
75
The face area of this wheel or annular disc is clearly equal to
the area of the whole disc of radius R minus the area of the disc of
radius r.
i.e., face area of the wheel ==7r# a 7rr a =7r(jR 2 r 2 ).
And, if its mass be M, clearly,
mass per unit area of the w/zee/=Jf/7r(JR 2 r 2 ).
Now, consider a thin circular ring at a distance x from tho
centre, and of width dx.
Then, face area of the ring=its circumference x its width=27rx.dx.
And, therefore, its mass == 27rx.dx.M/7r(R*r 2 ).
Now, s.ince the moment of inertia of a ring about an axis
through its centre and perpendicular to its plane is equal to its
mass x (radius) 2 , we have
f R M *
M. I. of the wheel about its axis = j _.  , ^ .2nx.dx.x*.
TT(R*1
2M
2 r') 4
M
Or, ALL of the wheel about its axis v ., , . , M .
z> JL
The axle, again, is just a disc, (or solid cylinder), and its
moment of inertia about its axis is, therefore, just the same as that
of a disc or a cylinder about its axis, i.e., = its massx(radius) 2 /2.
So that, M.L of the axle = w.r 2 / 2 
Hence, M.L of the. wheel and axle = M.L of the wheel \M.L of
the axle.
Or, I  [M(R+r)/2]+iM 2 /2.
32. Table of Moments of Inertia. The values of moments of
inertia for the cases discussed above, together with some other impor
tant ones are given in the Table below for ready reference of the
student, the mass of the body being taken to be M, in all cases.
BODY
AXIS
"(Position and Direction)
MOMENT
OF
INERTIA
1. Tbin uniform rod, of
(/) Through its centre
length /.
and perpendicular to its M/ 3 /12
length.
(ii) Through one end
and perpendicular to its
A// 2 /3
length.
2. Thin and rectangular
(/) Through Its centre
sheet or lamina, of sides
and parallel to one side
M> 2 /12 or M/ 2 /12
/ and >.
(//) About one side.
M6 2 /3 or M/ 2 /3
(Hi) Through its centre
and perpendicular to its
M(/ 2 + 2 )/42
plane.
(i v) Through the midpoint
of one side (/ or b) and per
peedicula to it* plan* *
M(6*/3/ 2 /U^
or M(/ 2 /3J
FBOPKETIBS OF MATTE*
4.
5.
6.
1.
8.
BODY
3. Thick uniform rectangu
lar bar, of length / and
thickness d.
Thin triangular plate or
lamina, of altitude H.
Elliptical disc or lamina,
of major and minor axes
2a and 2b.
Hoop or circular ring,
of radius R.
Circular lamina
of radius R.
or disc,
Annular ring or disc of
outer and inner radii R
and r.
9.
Solid cylinder of length
/ and radius R.
10.
Solid cone, of altitude h
and base radius R.
11. Hollow cylinder, of
length / and external and
internal radii R and r.
AXIS
(Position and Direction)
Through it* midpoint
and perpendicular to its
length.
About one side.
(/) About one of the
axes, (major or minor).
(ii) Through its centre
and perpendicular to its
plane.
(/) Through its centre
and perpendicular to its
plane.
(ii) About a diameter.
(i//j About a tangent in
its own plane.
(iv) About a tangent, per
pendicular to its plane.
(/) Through its centre
and perpendicular to its
plane.
(//) About a diameter.
(///) About a tangent, in
its own plane.
(iv) About a tangent per
pendicular to its plane.
(/) Through its centre
and perpendicular to its
plane.
07) About a diameter.
(/i7) About a tangent, in its
own plane.
(iv) About a tangent per
pendicular to its plane.
(/) About its axis of cy
lindrical symmetry.
(ii) Through its centre
and perpendicular to its axis
of cylindrical symmetry.
(///) About a diameter of
one face.
(/) About its v*rtical
axis.
(//) Through its vertex
and parallel to its base.
(/) About its own axis,
(i.e., about its axis of cylin
drical symmetry).
(i7) Through its centre
and perpendicular to its
own
MOMENT
OF
INERTIA
or
2MR*
5MR*I4
MR*/2
3MJK* 3MfP
Of INE&TiAENBRGY 6f
77
AXIS
MOMENT
BODY
(Position and Direction)
OF
INERTIA
12, Spherical shell, of radius
(0 About a diameter.
2MJ? 2 /3
R.
(11) About a tangent.
3MJP/3
13. Solid sphere, of radius
(i) About a dimeter.
2MR Z I5
R.
(ii) About a tangent.
1MR*I5
14. Thick shell or hollow
(i) About a diameter.
2 /R 6 ~~r*\
sphere, of external and
5 V jR 8  r*/
internal radii 7? and r.
(11) About a tangent
2 M fR* r 5 \
5 \^R*r*/
+MR 2
15. Flywheel, with radii of
*
wheel and axle, R and
About its own axis.
M(R f r ) mr
/(mass of axle, m).
~2 ' 2 *
16. Spheroid of revolution,
About polar axil.
2MR 1 5
of equatorial radius R.
17. Ellipsoid, of axes 20,26
About one axis, (2a)
M(6 2 f^)
and 2r.
5
18. Rectangular parallelepip
ed of edges /, 6 and d.
Through its centre and
perpendicular to one face,
M(/ 2 f6 3 )
12"
(say, face / 6)
19. Rectangular prism, of
About axis 21.
(^ 2 ^ 2 )
3
dimensions 2/, 26 and 2d.
20. Very thin hollow cylin
Through its centre and
Mf /3 4 ^
der, of length / and
perpendicular to its own
\ 12 2 /
mean radius R
axis
33. Routh s Rule. This rule states that the moment of inertia
of a body about any one of the three perpendicular axes of symmetry
passing through its centre of mass is given by
(i) the product of its mass and onethird of the sum of the squares
of the other two semiaxes, in the case of a rectangular lamina or para
llelopiped ;
(//) the products of its mass and onefourth of the sum of the
squares of the other two semiaxes, in the case of a circular or an ellipti
cal lamina ;
(til) the product of its mass and onefifth of the sum of the squares
of the other two semiaxes, in the case of a sphere or a spheroid.
Quite a few of the cases, dealt with in the proceeding pages, may be
easily deduced by an application of this rule. Thus, for example,
(/) moment of inertia of a uniform rec
(angular lamina (of mass M, length /and breadth
) about an axis passing through its centre O
and perpendicular to its plane
12
for, here, the two semiaxes of the lamina are
clearly, //2 and 6/2 respectively, (Fig. 43).
78
PKOPERTIJES OF MAtTBR
(it) Moment of inertia of a uniform circular lamina or disc, (of mass M and
"*"' radius R), about an axi, passing through its
enire and perpendicular to its plane is equal to
#;'"
Fig. 44.
because (a) and (6) are the two semiaxes of the
lamina, (Fig. 45).
(///) moment of inertia of a solid sphere, (of
mass A/and radius R) about its diameter is equal
to
because here the two semiapes of the lamina or
disc are obviously R and R> (Fig. 44).
And,~again, moment of inertia of a uniform
elliptical lamina, (of mass M, and with 2a and 2b t
as its major and minor axes respectively), about
a perpendicular axis passing through its centre,
is equal to
because here the two semiaxes of the sphere are
R and R.
Fig. 45.
34. Practical methods for the Determination of Moments of
Inertia. The principle underlying the experimental determination of
the moment of inertia / of a body, about a given axis, is to apply a
known couple C to it and to measure the angular acceleration doj/dt
produced in it. Then, from the relation,
(2
C = Ldwldt, we have / = , , . ,
** u to I at
whence, /may be easily calculated.
(/) Moment of Inertia of a Flywheel.
First Method. The flywheel, whose moment of inertia is to be
determined, is mounted on ballbearings (to minimise friction), and
its axle is arranged to be in the hori
zontal position at a convenient height
from the ground, (Fig. 46).
A small loop at the end of a
small piece of fine cord is then slipped
on to a tiny pag on the axle and the
entire length of the cord wound evenly
round the latter, with a suitable mass
; m suspended from its free end, and
properly held in position,
released and allowed to fall under the
j~:...J:r _. r T_
trn
mg
Fig. 46.
As the mass is
action of its own weight, the cord starts unwinding itself round the
axle, thereby setting the wheel in rotation. The length of the cord
is so adjusted that the moment the mass reaches the ground, the
of it gets just unwound from the axle arid slips off the
Hbviously, the rotation of the wheel, (with the descent of
MOMENT OF INERTIA ENERGY OF BOTATION 79
the mass), is due to a couple T.r. where T is the tension in the cord
and r, the radius of the axle*.
If, therefore, / be the moment of inertia of the flywheel about
its axis of rotation and dwjdt, the angular acceleration produced in
it, we have Ldw/dt = T.r.
The downward force due to the weight of the mass, when it has
no acceleration, is mg ; but when it has a vertical acceleration a, the
force due to it is equal to m.a., and this must clearly be equal to
mgT.
Or, m.a = mgT 9 whence, T = m.(g a),
/.dot/at m.(gd)r.
But dw/dt = tf/r, And .. I.a\r = m(gd)r. [v a = r.da>jdt]
Or, / =, mr\ (g ~~ = wr  1 ...(1)
The timeinterval between the release of the mass and the slipping of
the cord from the axle is r rpfully noted. Let it be , and let the
distance through which the m i falls down during this interval be S.
Then, since the mass starts from rest, we have
S = I at 2 , whence, a = 2S/t 2 .
So that, substituting this value of a in relation (1) above, we
have
whence /, the moment of inertia of the flywheel about its axis of
rotation, can be easily calculated.
Second Method. Proceeding as above, the loss of potential
energy af the falling mass is equated against the gain in kinetic energy
of the wheel, the K. E. of the mass itself and the work done against
friction. Thus, \vheri the mass falls through distance S, the potential
energy lost by it is equal to tng.S. And, if a> bo the angular velocity
of the wheel at the time, the K.E. gained by it is  7oj 2 , the K.E.
acquired by the mass being \ wv 2 , where v is its velocity on descend
ing through distance S.
.. mg.S = J 7o> 2 f lmv*+the work done against friction. . . (2)
To determine the work done against friction, we note the num
ber of tunn made by the whoel before coming to rest, after the mass
has been detached from the axle. Then, obviously, the kinetic energy
 7o> 2 , of the wheel, is used up ia overcoming; the fricuional forces at
the bearings. If the couple due to friction t>3 G and the number of
turns made by the wheel before coming to rest be n, work done by
Ms couple is equal to STTH xC, (v work done = couple xangle t and
the angle, described by the wheel in one rotation is equal to 2v). So
that, i 7o> a = 2nnC. Or, C = 7co 2 /47rfl.
The couple due to friction being thus determined, we can easily
calculate the work done against friction during the descent of the
*If the cord be appreciably thick, half of its thickness, added to t!
radius of the axle, gives the effective value of r.
90
mass through distance 5. For, clearly, the number of turns madt
by the wheel during the fall of the mass through this distance is
S/27rr ; and, therefore, the total angle turned through by it is equal
to 27r.5/27rr = S/r.
Hence, work done against friction is equal to C.S/r
.*. our energy equation (2) now becomes
Or
Now, if/ be the time taken by the mass to fall through the dis
tance S, its average velocity = Sjt ; and since average velocity =
(initial velocity {final velocity) l'2 t we have
final velocity, v = 2#/f. Or, v* = 45 2 // a .
[ the initial velocity is zero, the mass starting from rest.)
Substituting this value of v 2 in expression (3) above, we have
(
Or
1
_
"~ 2S( 1
Alternative Calculation. Let the number of rotations made by the"wheel,
before the cord and the mass slip off from the axle, (i.e., after the mass has
fallen through a distance S), be N.* Then, taking the fractional force to be uni
form, and the work done against it p? r rotation of the wheel to be w, we have
werk done against friction during AT rotations of the wheel = N.w.
Thus, our energy equation (2) becomes
mg.S = J It**+ Jwv 2 +JV.w. . .(5)
Now, after the detachment of the mass from the axle, the wheel cornea
to rest after n rotations, and, therefore, work done against friction during these
n rotations of the wheel n.w and this must obviously be equal to i /a> , the
K.E of the wheel at the instant that the mass gets detached from it. Thus,
n w = i /eo 2 , whence, w J 7w 2 /.
Substituting this value of w in equation (5) above, we have
mg.S  J
Or.
__ _
*This is obviously equal to the number of turns of the cord on tb* axle
it the very start.
MOMENT OF INERTIA ENERG* OF ROTATION
Of,
whence,
..
[Smce v "
Or, by dividing both the numerator and the denominator of this expression by
w 8 , we have
(2mg 5/6>*)~ _
Now, the angular velocity of the wheel at the instant that the mass gets
detached from it is , and becomes zero , when the wheel conies to rest, after
time t'> say. Hence, if the fractional force uniformly retards the rotation of the
wheel, its average angular velocity, during this interval of time f, may be taken
to be
tions
, , ,
equal to (to40)/2, i.e., equal to co/2. And, since the wheel makes n rota
before coming to rest, it describes an angle equal to 2w in time t',
co/2 2rc/i//', whence, co = 4nnjt'.
So that, substituting this value of co in relation (6) above, we have
(n+N)ln
wX'S?" 1 )
Or,
..(7)
whence /, the moment of inertia of the flywheel, about its axis of rotation, can
be easily calculated.
Accurate value ofu>. In the above treatment, the angular velocity w of
the wheel has been obtained on the supposition that the factional force remains
constant during the time t' that the value of o> falls to zero, after the detach
ment of the mass from the axle. Obviously, this is by no means a valid assump
tion, because, as we know, the frictional force decreases with increase of velo
city ; so that, the value of/, the moment of inertia of the wheel, deduced on
the basis of the above calculations, cannot possibly be quite accurate.
If we aim at accuracy, therefore, we must adopt a sensitive method for
determining the value of w, and the one method, which at once suggests itself,
is to make use of a tuning fork, as explained below :
A tuning fork, of a k no wit frequency , is arranged horizontally, (Fig.
47), with a slightly bent metallic style, attached to one of its prongs, such that,
when desired, it can be made to
lightly press against, or taken
off, a strip of smoked paper,
wrapped round 'the rim of the
wheel.
Now, with the style
kept off the paperstrip, the
mass m is allowed to fall down,
thus setting the wheel in rota
tion, and just a second or so
before the mass is due to get
detached from the axle, the
tuning fork is set into vibration
Fig. 47.
{by smartly drawing a bow across it), and the style pressed lightly on to the
strip, taking care to take it off soon after the detachment of the mass. A long
wavy curve is thus traced out by the style on the smoked strip. The mean wave
length A of this wave is then determined by dividing the tota distance occupied
by the wavy curve by the total number of waves constituting it.
Since one wave is traced out by the style_j!uring one vibration of the
prong or the fork, we have linear distance covered by the wheel during on*
vibration of the fork x. So that, distance covered by the wheel during *
vibrations of the fork  if*.
Again, since n vibrations are made by the fork in one second, it foliowi
that distance covered by the wheel in 1 second, i.e., the linear velocity v = n\.
But v ~ Rco, where R is the radius of the wheel and , its angular velo
city ; so that, we have ,/fo ~ n\; whence, to = n\jR.
Thus, knowing /t, X and R, we can easily calculate the value of w for the
wheel.
This value of co, substituted in relation (6) above, then gives a much
more accurate value of /, the moment of inertia of the flywheel about its axis of
rotation,
Note. The student may, as an interesting exercise, show that expression
(4) above can also be reduced to the same form as expression (7). This may be
easily done by remembering (/) that when the wheel makes one full turn, the
mass descends through a distance 2^r, tne circumference of the axle, and,
therefore, when the mass descends through a distance 5, the number of rotations
made by the wheel is equal to S/2rcr ; so that, S/2nr = N ; and further (11) that
t = 25/v 25/ro, where o> = 4nn/t', (see page 81).
(//) Moment of inertia of a disc about an axis passing through
its centre and perpendicular to its plane ~
(a) Disc suspended by two parallel threads. The disc, with a
metal axle, is supported on two cords, wound uniformly on the axle
on either side, (Fig. 48), On releasing
the disc, it begins to fall down until the
whole cord is unwound from the axle, say
through a distance S.
Then clearly, P E. lost by the disc
mg.S, where m is the mass of the disc and
the axle. This energy will obviously be
gained by the disc in the form of kinetic
energy of rotation and translation. Fig 48.
If a} be the angular velocity acquired by it after falling through
this distance *S, its K.E. of rotation will clearly be /o> 2 , where 7 is
its moment of inertia about an axis passing through its centre and
parallel to the axle, (i.e., perpendicular to its plane) ; and its kinetic
energy of translation will be \mv 2 .
mg.S = i/o> 2 + Jwi> 2 , (v being its final linear velocity),
[v o> 2 = v 2 /r 2 , where r = radius of the disc.
Or, /v 2 /r 2 = mg.Slmv*,
whence, / = (mgS Jmv 2 ).2r 2 /v 2 .
Now, average velocity = S/t, where t is the time taken by the
disc in falling through distance S ; and, therefore, velocity v of the
disc =z 2S]t, and .. v 2 = 45 2 // 2 . So that,
__
~~ VS
Or, / SB m
. _
mr
MOMENT Off ItfERTlA EtfEBOY Off BOrATtON 83
(b) Di c mounted on axle t rolling on inclined rails. Here, the
disc, of mass M and moment of inertia I, is allowed to roll down along
inclined rails, as shown in Fig. 49. Let it
acquire a linear velocity v and an angular
velocity o>, when it descends a vertical
distance h, as it rolls down a distance S along
the rails. / '<&*' h
Then, clearly, loss ofP.E. of the disc
= K.E. of translation gained by disc
+ K E. of rotation gained by disc.
Or, Mgh = \M v 2 f /oA Fig. 49.
So that, Mgh = ~ Mv *+ r L ^
2 r*
r where r= radius of the axle
LAnd .. o,=v 2 /r 2 .
Or, /. 2 = M(gh\v*) 9 whence, / = ~ . (*AJv)
Or, /  ^ (*ghv*).
Or, substituting the value of v =~ 2S/t, (see page 82), where / is
the time taken by the disc to cover the distance S, we have
whence the value of /, th3 moment of inertia of the disc can be easily
calculated.
Note : For other methods for the determination of moment of inertia, see
underjbrsional Pendulum, (Chapter VIII)
35. Angular Moment and Angular Impulse. In the case of
linear motion, the momentum of a body, as we know, is the product
of its mass and velocity. On the same analogy, we have, in the
case of rotational motion, the product of the moment of inertia and the
angular velocity as the angular momentum bfa rotating body.
Thus, angular momentum = /.<o,
where I is the moment of inertia and o>, the angular velocity of the body
about the axis of Dotation.
For, suppose we have a body, rotating about an axis with a
velocity w. Then, all its particles will have the same angular velocity
o>, but their linear velocities will depend upon their respective dis
tances from the axis of rotation, being equal to the product of the
angular velocity and the distance from the axis. Thus, the linear
velocity of a particle, distant r x from the axis, will be r^ ; of that
distant r g from the axis will be r 2 cu and so on.
And, therefore, if m be the mass of each particle, we have,
linear momentum of the particle, distant r t from the axis, equal to
m.^w and, therefore, the moment of its mttmentim about the axi
would be m.r l .cuxr=m.r l 2 .oi. Similarly, the moment of momentum
of the particle, distant r g from the axis, would b3 w,r 4 a .cu and so on
84 PBQPERTltS.Oir MATMft
Therefore, the moment of momentum of the whole rotating body
about the axis = wr 1 2 a>+wr 2 2 ai+' .........
i:mr*a>=/.a>, [v ZVwr 2 = /.
where / is the moment of inertia of the body about the axis of
rotation.
Thus, the angular momentum of a rotating body about its axis
of rotation is the sum of the moments of the linear momenta of its
particles about that axis. For this reason, it is also referred to as it*
moment of momentum about the axis.
Now, we have Ldt*>\dt = C. Or, I.dw = C.dt,
where C is the torque or the couple acting on the body.
Integrating this with respect to t, between the limits and t,
we have
angular momentum, I co = 1 C.dt,
JO
an expression which is true, however C may vary with time.
If C be constant, we have 7.o> = I C.dt =~ C.t t
which gives the angular momentum acquired by the body in time t,
If t be very small and C quite large, the expression I C.dt
stands for the angular impulse given to the body, which again be*
comes equal to C.t, if C be constant.
36. Law of Conservation of Angular Momentum. Just as we
have the law of conservation of momentum for linear motion, we
have, for rotational motion also, the law of conservation of angular
momentum, which states that the angular momentum of a rotating
body about an axis remains constant, if no external torque be applied
to it.
For, suppose the angular velocity of a body is changed by d<# 9
by a torque C, applied to it for a very small interval of time dt.
Then, we have C.dt I.d<*>,
where / ia the moment of inertia of the body about the given axis.
Hence, C = l.dw/dt, assuming /to remain constant.
If, however, /also change*, we have C = d(Ia>)ldt,
i.e., the torque is equal to the rate of change of angular momentum.
Obviously, therefore, if C 0, i.e., if there be no external
torque applied to the body, dw/dt or d(Ia>)jdt is also equal to zero, or
the rate of change of angular momentum remains constant.
It is obrious from tha above that in the case when / is not con*
stant, and no external torque ia applied to the body, the angular
velocity must change in the inverse ratio to /, in order to keep its
angular momentum constant.
This may be clearly seen by whirling round a stone tied to one
end of a string, whose other end is held in the hand. On stopping the
Application of any force to it, /.*., on removing the external torque,
MOMENT Of IHBBTIA ENERGY OF EOTATION
85
the string besrins to wind its If on the hand, with continuously in
creasing velocity, because as the distance of the stone from the hand
decreases, its moment of inertia about its axis of rotation also
decreases, resulting in a proportionate increase in its angular velocity.
Another good illustration is provided by an acrobat executing a
somersault. For, as we know, he instinctively curh himself up in
air, thereby decreasing his moment of inertia and consequently
incr .Basing his speed of rotation. But, before his feet touch the ground,
he slows it down by straightening himself up and increasing his mo
ment of ineitia.
37. Laws of Rotation. Corresponding to Newton's three laws
in the case of linear motion, we have also three laws of rotational
motion, viz.,
1. Unless an external torque be applied to it, the rate of rotation
of a rigid body, about a fixed axis in it, remains unaltered.
An obvious example of this is the constant rotation of the Earth
about its axis. The force of attraction due to the Sun is certainly
there, but it acts at the centre of the earth and hence produces no
effect on its rotation.
2. The rate of change of rotation of a body, about a fixed axis
in it, is directly proportional to the external torque applied and takes
place in the direction of the torque.
3. If a torque be applied by one body upon another, an equal
and opposite torque is applied by the latter upon the former, about the
same axis of rotation. In other words, a change in the angular
momentum of one body brings about an equal and opposite change in
the angular momentum of the other body.
It is useful to remember that the moment of inertia (I), in
rotational motion, corresponds to mass, (m), and the angular velocity (w)
to linear velocity (v), in the case of linear motion.
The following Table gives the linear and rotational analogues at
a glance :
Linear Motion
Rotational Motion
1
Mass covered w.
Moment of Inertia /.
2
Distance S.
Angle described 0.
3
Velocity v or dS\dt.
Angular velocity ... co or dftldt
4
Acceleration a or d^S/dl**
Angular acceleration . d<&ldt
or d*$ldt*.
5.
Force, F m.a.
Torque C t or (moment of the couple).
6.
Linear K.E. = iwv 1 .
Angular KE. =* i/w 8 .
7.
Linear momentum = mv.
Angular momentum or moment of
momentum /w.
8.
Work done by force, I.e.,
Work done by couple, i.e.,
W ** F.S
W  C.0.
9.
v = u+at.
(,J 8 SB (Oj+yW/t//)./ 1 .
10.
5* =* ut\~$at*
Q == tojt+Kfltoldt)?*.
11.
f~u\ * 2^S.
toi 1 *^)!* 2(d(j[dnQ,
where 6> a and w, are the initial and
final angular volocities and d<*\dt> the
angular acceleration.
86
PROPERTIES OF MATTEB
38. Kinetic Energy of Rotation __
(a) Kinetic energy of a body about an axis through its centre of
mass. Suppose u r e h ive a body of mass M rotating about an axis AB,
parsing through its centre of mass O, (Fig. 50).
It, obviously, possesses kinetic energy due to
its motion ; this energy of the body is called its
energy of rotation, because it is due to its motion
of rotation.
Imagine the body to be divided up into a
large number of small particles, of masses m lt
w 2 , w 3 , etc., at distances r l9 r 2 , r s .... etc.,
respectively from the axis AB. Then, we have
linear velocity ofm = r lW = v 1 ;
= r 2 co
= v
of w 3 =
of w a
so on,
.. kinetic energy of mass m 1 =
MI = J W 2 v 2 2 ; of mass w 3 = Jw 3 v 3 2 *nd so on.
. of the body =
= v and
of
mass
Or,
= W[w/ 1 +"V i a4"V'3 ......
iw*mr* = \<JMK\ [v 27mr 2 MK*.
Or, K.E. of the body = MK 2 co 2 = pa> 2 , [ . MK*  /.
where /is the moment of inertia of the body about axis AB.
Now, if aj = 1, then, obviously, K.E. of the body = \ /.
Or, /  2 #..
rAw, /A^ moment of inertia of a body, rotating with unit angular
velocity, is equal to twice its kinetic energy of rotation.
(b) K.E. of body which is not only rotating but whose centre of
mass has also a linear velocity v. A body which is rotating as well
as moving forwards with a velocity v, has both types of kinetic
energy, viz., {/) energy of rotation, because of its motion of rotation
about a perpendicular axis through its centre of mass, and (//) energy
of translation ^bez&use of its linear motion. And, clearly, therefore,
we have K.E. of rotation of the body == $ /w 2 ,
and its K.E. of translation = } Mv*.
.. total K.E. of the body  K.E. ofrotation+K.E. of translation,
because w 2 = v^/r 1 where r is the radius of the body.
Or, total kinetic energy of the body = Jf v s [(X ? /r*)+l].
39. Acceleration of a body rolling down an inclined plane.
Let a body of mass M roll freely down an inclined plane, of incli
nation a to the horizontal, (Fig. 51), The plane is supposed to
be rough enough, so that thero may be no slipping, and hence
no vork done by frict ; on.
MOMENT OF INERTIA ENERGY OF ROTATION 87
Then, if v be the velocity acquired by the body after traversing
a distance S along the plane, we have
vertical distance through which
it has descended = S.sin a.
And, therefore,
P.E. lost by the body = Mg.S. sin a.
This must, obviously, be equal to
the K.E. gained by the body.
Now, K.E. of rotation of the body
= i/ w2 Fig. 51.
where w is its angular velocity about a perpendicular axis through its
centre of mass.
And, its K E. of translation = {Mv*^
because its centre of mass has a linear velocity v.
total K.E. gained by the body = f 7o> 2 + Mv*.
= I Mv\(K 2 jr*) + l]. ISee 38.
Since gain in K.E o r the body is equal to the loss in its P.E., we
have
iMV<[(K 2 lr 2 )+l] = Mg.S sin a.
Or, Mv*[(K 2 lr 2 ) + }] =r 2Mg.sin a S.
Or, v*(K 2 +r")lr* = 2g.sina.S,
whence, v' 2 = 2(r 2 /K 2 +r 2 ).g sin a.S.
Comparing this with the kinematic relation, v 2 = 2aS, for a
body starting from rest, we have
acceleration of the body down the plane, i e. t
a = (r 2 /K 2 +r 2 ).g sina.
Or, the acceleration is proportional to r l j(K 2 +r 2 ) for a given angle
of inclination a.
This show that
(/) the greater the value of K, as compared with r, the smaller the
acceleration of the body coming down the plane and, therefore, the
greater the time it takes in rolling down along it and vice versa.
(//} the acceleration and, therefore, the time of descent is indepen
dent of the mass of the body.
Thus, a solid sphere, for which K 2 = 2r 2 /5, will roll down
faster than a disc, for which K 2 = r 2 /2, and, similarly, a disc will
roll down faster than a hoop, for which K 2 is equal to r 2 .
Since K 2 for a hollow sphere about the diameter is greater than
that for a solid sphere of tho same mass and radius, they can be dis
tinguished from each other by allowing them to roll down the plane,
Obviously, the solid sphere will roll down faster than the hollow one.
The same test may be applied in the case of a hollow and a solid
cylinder etc.
Some particular cases :
(0 Case of a Spherical Shell. Let a be the angle of inclination of the
plane, down which* the spherical shell is rolling and let the velocity be v when U
has moved a distance S along the plane, (Fig. 51) ,
88 PROPERTIES OF MATTER
Clearly, the vertical distance covered by the shell 5 s i n .
loss in P.E. of the shell == Mg.S sin a. %
This loss in \?E must be equal to the gain in K E. of the shell, for no
work is done by friction, as there is no slipping.
Now, K.E of the shell  j h^ + \Mv* = JMff'w'f J Mv.
 W.f r'^+iMv 2 . [v A:* = r 8 for a shell.
}MrV + iMv 2 = JAfv 2 + JAfv a .
2Mv*3Afv t 5Mv*
__ ^ .
Since #fl/Vi m K.E. of shell = loss in its P.E.,
we have 5Mv~/6 Mg.S.sln a, Or, 5r* = 6# 5.J//I a.
Or, v 2 = ... fltf a =* 2( j/w a)^
Comparing it with the relation, v 2 = 2a S, (when // = 0), we find that the valua
of acceleration a of the shell, down the pla*ie = ^g sin a.
(ii) Case of a SaMd.) Shere. We know that the acceleration (a) of a body
down an inclined plane =(/ ~JK 2 + r 2 ) g sin a, where A' is the radius of gyration of
the body, and a, the angle oi inclination of the plane.
,*. a = (r/4'' 2 4r 8 ) g sin a. [v K*  r 9 , in this case*
= (r z ! 7 r r z )g sin a = *..g sin a.
Thus, the acceleration of a solid sphere down the inclined plane is equal
to  sin .
40. Graphical Representation of Plane Vectors. We are
already familiar with the two types ot % physical quantities, viz.,
(i) scalar and (//) vector, the former poswssmij only magnitude, but
no direction ; and the latter, possessing both magnitude and direction,
(see foot note on pai^e 55). Theso latter can, as we know, be
represented by a straight line, drawn to a chosen scale, whose length
and direction respectively represent the magnitude and direction of
the quantity.
Any other quantity, either derived from a vector, or obtained
by combining a vector with a scalar quantity, is also veetorial in
nature. Thus, for example, the acceleration of a body, depending
upon the velocity of the body, (a vector quantity), is also a vector
quantity.
The vector quantities referred to above are, strictly speaking,
linear vectors, and must be clearly distinguished from what are called
plane vectors, a term applied, in rotational dynamics, to such
quantities as angular velocity, angular momentum and torque etc ,
which are all directional in the sanso that they are confined to one
plane.
Such a plane or two dimensional vector is also represented by a
straight line, drawn normal to its plane of rotation, or parallel to its
axis of rotation, its clockwise or anticlockwise rotation being indi
cated, according to an agreed and established convention, by the
straight line being directed towards, or away from, the observer
respectively.
Further, corresponding to the parallelogram law for the compo
sition of linear vectors, we have, here, a modified form of it to deter
mine the resultant of two plane vectors, viz., that
"if there be two plane vectors acting simultaneonsly on a body in
two different planes, such that they can be represented in magnitude and
MOMENT OF INERTIA ENERGY OF ROTATION
Fig. 52.
direction* by the two adjacent sides of a parallelogram, drawn perpendi
cular to those planes, their resultant is represented completely by the
diagonal of the parallelogram, passing through their point of intersec
tion, this diagonal representing a plane vector in a third plane, per
pendicular to itself."
Thus, if OA and OB, (Fig. 52), represent two couples, in two
different planes, acting simultaneously on a body, whero OA and OB
are drawn perpendicular to those
planes, their resultant is given
completely by the diagonal OC of
the parallelogram, which repre
sents a couple in a third plane, per
pendicular to itself.
And, obviously, what is true
about the composition of couples
is equally true for the composition
of any other plane or twodimen
sional vector quantities.
41. Precession. Just as in the case of linear motion, we may
have a constant acceleration acting on a body, without changing its
constant speed, (e g. t
the centripetal accele
ration acting on a
body, moving with a
uniform speed in its
circular orbit), so also,
in rotational motion,
we may have a constant
~~Jt an S^ ar acceleration
acting on a body, hav
ing a constant angular
speed. This is rendered
possible bv the plane of
rotation changing direc
tion at a given rcte ,
without, in any way,
Fig. 53. affecting the rate of
rotation of the body about its axis of rotation, or axis of spin, as it is
also sometimes referred to. This change in the plane of rotation is
called 'precession', and is caused by a couple or torque, called the
precessional torque, acting: in a plane, perpendicular to the immediate
or instantaneous plane of rotation (or spin) of the body. In other
words, the axis of the torque is, at any given instant, perpendicular to
the rotationaxis of the body, as will be clear from the following :
Let DD, (Fig. 53), be the edge of a disc, with its plane revolv
ing about its geometric axis, with an angular velocity w. Then, if its
moment of inertia about this axis be /, its angular momentum will
clearly be lw. Let this be represented by the straight line OA,
drawn perpendicular to the plane of rotation of the disc.
rr  ;
*{n accordance with the convention, stated abpv$.
fO
PROPERTIES OP MATTER
Now, let the axle of the disc also rotate, i.e., let there be a pre
cessional motion, about an axis, perpendicular to the plane of the
paper at a (prece:?sional) rate ; so that, after a small interval of
time dt, the disc takes up the position D'D', making an angle <f>.dt
with its original position Its angular momentum, again equal to
/co, is now represented by the straight line OA\
The change in the angular momentum of the disc is thus repre
sented vectorially by A A' = /to <j>.dt. [ arc = radius x angle.
This change has, clearly, been brought about in time dt, and
therefore,
rate of change of momentum of the disc = I w.<f>.dtjdt = 7o>.0.
And, since the r ite of change of momentum of a rotating body
is equal to the torque applied to it, we have
TI  I CO (f),
where 7^ is the torque applied to the disc.
So that, the rate of precession, <f>  TJfw.
Now, since the change in the angular momentum of the disc is
along AA', it is clearly parallel to its plane of rotation, or perpendi
cular t ( > its axis of rotation, and A A' is thus the axis of the torque
applied. In other words, the axis of the torque lies along OX.
Thus, we see that if the axis of rotation of a body be along OY
and the axis of the applied torque along OX, the body 'precesses'
about the third mutually
perpendicular axis OZ. This
will be readily understood
from Fig. 5*, which shows
the disc 7/2 perspective.
Here, OY is the axis
of rotation arid, therefore,
XOZ is the plane of rota
tion ; OX is the axis of the
torque or couple applied,
and, therefore, YOZ is the
plane of the torque and, since
the axle of the disc turns to
wards OX, i.e., about the
axis OZ, the plane of pre
cession is XOY. In other
words, the axis of rotation
(OY) turns in tins plane,
which is, clearly, perpendi
cular to the first two planes, its direction of rotation (towards
OX, here), depending upon the direction of rotation of the disc and
that of the torque or the couple applied.
42, The Gyrostat. A gyrostat is just a disc or a flywheel, having
a large moment of inertia, rotating at a high speed about an axle,
passing through its centre of mass, and mounted, as shown in Fig. 54,
so that the wheel and the axle are both free to turn, as a whole, obout
any axis, perpendicular to the axle itself,
MOMENT OF INERTIA ENERGY OF ROTATION
91
As explained above in 41, if a torque or couple be applied to
the wheel, with its axis perpendicular to the axis of rotation of the
wheel, the wheel 'precesses' about the third mutually perpendicular
axis, at a processional rata, given by </> T } /Ia), where T l is the
torque or couple applied, /, the moment of inertia of the wheel about
it rotationaxis and o>, its angular velocity about this axis.
Clearly, therefore, for a given torque (Tj) applied to the wheel,
the precessional rate is inversely proportional (?) to the moment of iner
tia of the wheel about its axle, and (//) to the angular velocity of the
wheel ; so that, the larger the moment of inertia of the wheel about
its axle, and the higher its angular velocity, the smaller the rate of
precession of the axle, and vise versa.
The following simple experiment will beautifully illustrate the
above results :
Take a fairly larse awl heavy disc, (Fig. 55), free to rotate
about its axis YY' passing through its centre, and fitted inside two
sockets at the ends of the horizontal diameter of a bigger ring, sus
pended by nvans of a string vertically above its centre of gravity.
(/) Now, if with the disc quite stationary, a weight Mg be
suspend fd at Y', the torque due to it will tilt the ring, the end Y'
moving down and the
end Y moving up, / e ,
the ring will turn about
OX.
But , if instead of
suspending the weight,
the ring bo simply
pushed horizontally at
y, from in front or
behind, it will turn
about OZ.
(//') Let the disc
be now set into rota
tion about its axle, in
the direction shown,
with the weight Mg
kept properly supported,
so as to exert no down
ward pull at y. It will
be found that the ring
remains quite steady and a twist, given to the string either \\ ay,
hardly produces any tendency in it to rotate about OZ, as it certainly
would, if the disc were stationary.
(///) \With the disc in motion let the weight Mg be released, so
as to exert a downward pull at Y r 9 thus producing a torque about
OX. It will be found that the ring at once rotates about OZ, with
the end Y' slightly tilted downwards. On pushing the ring horizon
tally at y, as before, the axle, instead of turning more rapidly about
OZ, as might be expected, simply gets tilted a little, raising the
weight Mg slightly upwards, clearly showing thereby that the horizontal
y
Fig 55.
92
PROPERTIES OF MATTER
rotation of the axle YY' opposes the torque due to Mg, which, therefore,
descends comparatively slowly now. So long, however, as the down
ward descent of the weight continues, just so long does the rate of
rotation of the axle about OZ also continue to increase, thereby in
creasingly opposing the torque duo to the weight, until a stage is
reached where the two exactly balance each other. After this, the
downward descent of the weight naturally ceases, and the ring con
tinues to turn about OZ at a constant rate, with the axle YY f slight
ly tilted.
(iv) It will be found that the greater the moment of inertia of
the disc about its rotationaxis or the axle, and the higher its angular
velocity about it, the smaller the rate of rotation of the axle about
OZ, i.e.. the smaller the processional motion about it.
(v) Since the torque or a couple is needed to produce this proces
sional motion of the rotationaxis of (he disc, it is clear that a rotat
ing body offers resistance to a processional motion of its axis. This
resistance to a processional motion is called gyrostat ic resistance, and
is equal and opposite to the prccessional torque.
43. Gyroscope. In a majority of cases, a body, subject to
preccssional motion, is supported at a point, away from the vertical
line through its centre of gravity.
A gravitational torque or couple thus acts upon the body, which,
in its stationary condition, simply tends to rotate it into a position of
a lower potential energy, ie., simply tends to lower its centre of
gravity. But, if the body be rotating obout some axis, this gravita
tional torque supplies the necessary processional torque equal in value
to its own, provided there is no other couple acting on the body. The
rate of precession <, maintained by this gravitational torque jP 2 , is
given by the relation,
where /and o> stand, as usual, for the moment of inertia of the body
and its angular velocity about its axis of rotation.
Such a body is called a gyroscope, its motion being appropriately
ter m ed 'gyroscop ic ' .
Thus, consider a heavy disc D, revolving with a high angular
velocity o> about its physical axis POQ, itself resting on a vertical
pivot at P, (Fig. 56).
Then clearly, its weight Mg, acting vertically downwards at its
ft c g'i O, exerts a gravitational torque T z on it,
JL y given by T = Mg.OP = Mg.l. [Putting OP  /].
//Q\\  9 So that, if (/> be the rate of precession of the
; disc maintained by it, we have
P
I .
1 ***
A
Fjg. 56.
~ "LaT ""
putting 7 MK 2 , where K is the radius of
gyration of the disc about the axis POQ.
Hence, if t be the timeperiod of its pre
cessional motion, i e. 9 if it takes time t to
complete its one full cycle of processional mo
tion, we have
MOMENT OF INERTIA ENERGY OF ROTATION
93
' " "f  gltK'.w * 2ir gl' '
This precession, once started, can be maintained, at this very
rate, by the gravitational torque alone. A higher rate of precession
than this will make axis POQ rise and a lower rate will make it fall.
This rise and fall of the axis of rotation, or its oscillation up and
down about its position of dynamic equilibrium, accompanied by a
correspondingly changing preccssional rate, is termed nutation.
Further, there is a centrifugal force acting on the disc along
POQ and an equal centripetal force in the opposite direction QOP t
their net effect, if they act along the same line, being to increase the
fricticnal resistance at the pivot P. If, however, their lines of action
be different, we have yet another couple T 3 , formed by them, aptly
known as the centrifugal torque.
In order to prevent the disc, or a precessing body, in general,
from moving outwards from the centre of precession, it is necessary
that the centrifugal torque on it must be balanced by an equal and
opposite centripetal torque, this balancing effect being supplied by
part of the gravitational torque, the remaining pan of it producing
precession. Thus, if T 3 be the centripetal torque and r l\ and T a ,
the gyrostatic and gravitational torques, we have
T T T (11
* 2 3 1 " " \ /
where the different torques are given their proper sings, (i.e., anti
clockwise, positive and clockwise, negative), all acting in the same
direction in the case shown.
A general rule to determine the sense of the torque, producing
precession in a given direction, is given by Lanchester's rule, which
may be stated as follows :
If the gyrostat be viewed from a point in its own plane, with the
line of sight perpendicular to the axis of the given precession* it is seen
to describe an ellipse, the sense of whose path gives the direction of the
precessional torque, with the line of sight as its axis.
44. The Gyrostatic Pendulum. A gyrostatic pendulum is a
small and heavy disc or gyrostat (Z)), revolving with uniform angular
velocity (co) about a light rigid rod,
(SD) as axis and precessing about the
vertical (SO) at a uniform rate (<^) as
shown in Pig. 57.
Obviously, there are the three
following torques acting on the pen
dulum.
(/) A gyrostatic torque, Tj, duo
to the gyrostat or disc D possessing two
simultaneous rotatory motions.
Since the plane of rotation of D /'
is always perpendicular to the rod (
SD t its rate of precession < is the same *
as the angular velocity of SD 9 i.e^
equal to vji For, in time dt, SD
traoas out an are v.dt, where v is its
velocity in the horizontal circle pt, 57.
OF
of radius r, which is described by it in its processional motion about
SO ; or, the angle described by it in time dt is equal to v.rfr//, and
hence the angle described by it in unit time is clearly equal to
(v.dtfidt == v//. Thus, < v//.
But, we know that < = ^ So that > T" 1 = ~7 '
T la) Iw I
..
whence, 1/1 =  , =  * 
where # is the ra^to of gyration of the gyrostat about the axis SD.
To determine the direction of this torque, let us apply
Lanchester's rule, i.e., let us look at a point B on the edge of the disc
along LB, where LB is perpendicular to both OD and #Z), when B
clearly appears to move in the anticlockwise direction, indicating that
the direction of the torque T l is anticlockwise.
If t be the periodic time of precession of the disc or the gyrostat,
we have v.t = 27rr, or v = 2irr//. So that, substituting this value
of v in the relation for T 1 above, we have
T l = MK*a>.2vrjt.L
And, since r/l = MI 0, we have
T! : +Jf 2 .o>. sin 0.(27r//),
the fve sign indicating that its direction is anticlockwise.
(ii) A gravitational torque T ? , clue to the weight Mg of the
gyrostat acting vertically downwards at D, (where its whole mass is
supposed to be concentrated).
Clearly, the moment of this gravitational torque = MgxBO =
MgJSD sin = Mg.l sin 0, where BO is the perpendicular distance
between Mg and an equal and opposite reaction at S ; /, the length
of the rod SD and 0. the angle that it makes with the vertical.
So that, T 2 = Mg. I sin 0,
the ve sign indicating that its direction is clockwise.
(Hi) A centripetal torque T 3 , due to the centrifugal force Jfv 2 /r,
acting on the gyrostat, outwards along OD.
And, the moment of this torque T 3 is obviously equal to
where SO is the perpendicular distance between Af v a /r and an equal
and opposite reaction at S. Or, since SO=l cos Q, we have
 Mv * Icosfi
3 =   ~ . i cos v ,
the ve sign again indicating that the direction of the torque is
clockwise.
Or, substituting the value of v=27rr// [see (/) above], we have
T t ~*g)\ I cos , * f (?1 )'. I co, 9 =
^=sin i or r=/ sin 0.
* t
MOMENT OF INERTIA ENERGY OF ROTATION
Hence 3 = M 2 .jm 0. cos
Now, from relation (1), (page 93), we have
Or, T 2 +T 3 =T 1 .
So that, substituting thoir values, we have
 Mgl sin e +M1* . sin e . cos 6 ( ^ ) =MK 2 w.sin (y
Or, /+/ 2 cos e ( ^Wo, (^L).
[Dividing by Af s//i throughout.*
Now, putting (2irjt)p, we have
glip 2 ! 2 cos0^pK 2 w.
Or, /? 2 / 2 coy ~pK 2 w  /= 0,
which is a quadratic equation in p.
T , , n
Therefore, P
which, obviously, gives two values of J9.
To decide between the two values, we put w=0, so that there
is no rotation of the disc about SD t and the whole arrangement
reduces to a conical pendulum, with
_^ 4 l_^ c ^l~a.A /* gi ^ COs IA.\ I 7 g 1
P ~~ ~ 2/ 2 . coils V ^/ 4 cos* ^" ^/ / cw fl*
2?r
"^
But, since 2?r// naust necessarily be positive for a conical
pendulum, the negative value^becomes inadmissible and, we, therefore,
have
It follows, therefore, that, in the expression for p above, only
the positive value must be taken. So that,
p ~~~~
whence '
45. Case of a Rolling Disc or Hoop. A simple and a familiar
example of gyroscopic motion is that of a thin circular disc or a hoop,
set rolling over a plane horizontal surface. If its velocity be large
enough, it continues to roll along a straight path in a stable vertical
position. But, as its velocity decreases, due to friction between it
and the horizontal surface, its plane inclines progressively to one side
and its path becomes curved towards the 'side of few', the curva
fcure of the path constantly increasing with the decrease in its velocity
PROPERTIES 0? MATTER
so that it follows a spiral path, until, finally, it f&llajtat on the surface.
Let us study this motion of the disc in some detail.
Let D be the
circular disc, (Fig.
58), of mass M and
radius r, rolling
along a horizontal
surface with v, as
the linear velocity
of its centre 0, and
with its plane AB
inclined at an
Fig. 58. angle 6 to the
vertical. Then, the three torques acting on it are :
(i) Gyrostatic torque T ls due to its simultaneous rotation about
its point of contact and about E t such that 7^=7.60. $, where 7 is its
moment of inertia about the axis OE through its centre and per
pendicular to its plane ; co, its angular velocity about the same axis
and <f>, its rate of precession.
Now, I=MK*, where K is its radius of gyration about OE,
o>=v/r and <=v//. [See 44 (/).
So that T=if*: 2 V V =zMK* v> T ~MK* V ? tan
where r//=tan 0, and the ~ve sign of T x shows that the torque is
clockwise.
(ii) Gravitational torque T 2 , due to its weight Mg, acting verti
cally downwards at O, such that
"v in the rightangled
&OCB
_,_ CB CB
also acting in the clockwise direction.
(Hi) Centrifugal torque T a , due to its rotation about , such
that
v in the rt.angled
AOCE
cos Qa/l.
and in the rt, angled
AOCB
cos 0=OC/r.
Mv*
.. OC=, . r cos $,
I COS B
where EC=a.
Or,
Mv* . ~ = Jl/v 2 .
the ve sign again indicating the clockwise direction of the torque.
Substituting these values of T lf T, and T s in relation (1), (page
93) we have, for equilibrium,
v 8
Jtf.r sin $(Mv*.tan g)*=MK*.  r . tan 0.
Or,
Or,
Mg.r sin o+Mv*.tan JfA^.r. tan 9
v*
*F'
Jlfjf.r
MOMENT OF INERTIA ENERGY OF ROTATION 97
Or, v* tan g +K*. . tan 0= gr. sin g
Or, v*tang(l+=gr.sing. Or, v
Or, vl+*r.o0. Or, c
whence the angle of 'lean 9 of the disc for a given velocity v of it is
clearly given by
p/" c
and its velocity by v2 =i rv/ a for equilibrium in the leaning
position.
Now, for the critical velocity v c , i.e., the minimum velocity at
which the disc can move along a straight path, with its plane vertical,
clearly, 0=0, so that cos = 1.
And, .. in this case, v^=_ Or, v, = Y~ ja/TT
For a value of v less than v c , the upright position would ob
viously be unstable ; for, on the slightest displacement, it will be
tilted over by the force of gravity until attains the value given
above, (by expression A), corresponding to the leaning position.
Now, for a (uniform) disc, K z =r 2 /2. Hence,
for a disc,
?/2r V =
And, for a hoop, v c
Let us now calculate the radius of curvature R of the path of
the disc on the horizontal surface. It is clearly equal to EBR.
And, in the rightangled triangle EOB, we have
sin g = OBjEB = r/R ; so that, r = R sin g.
R = == ^7 ^~~. TV sin g = \/icos 2 g.
sin g y lcos 2 o L v
Or, substituting the value of cos deduced above, we have
V
n ^_
which, with the substitution of the appropriate value of AT, gives the
radius of curvature of the path of the disc or the hoop along the
horizontal surface.
46. Gyrostatic and Gyroscopic Applications. The tendency of a rapidly ro
tating disc or wheel (and, in fact, any rigid body), to preserve its axis of rotation
endows a gyrostat with a stability of direction, which is made use of in a
PROPERTIES OF MATTER
number of ways for the steadying of motions. Among the more important
and familiar applications of this may be mentioned the following :
(0 The Gyrostatic or the GyroCompass. It is a special type of compass
used in aeroplanes and ships, and, more particularly, in submarines. In
essentials, it consists of a disc or a flywheel, of a large moment of inertia, (/ <?., a
gyrostat), suspended in fnctionlcss gimbals inside a supporting frame, which is
kept rotating at a high speed by means of an electric motor about a horizontal
ax's, lying in t^e geographic meridian (i.e., in the vertical plane passing through
the geographic north and south of the earth) Its directional stability and the
conservation of its angular momentum make its axis always lie in the direction
of the metndian, i.e., along the geographic north and south. And. since the
arrangement is such that the disc or the flywheel has three degrees of freedom,
irrespective of any porsition of the supporting frame, a movement of the latter
produces no deflecting toque or couple on it, and this particular direction of its
axis continues to be maiatain^d in space all the time, despite any changes in the
direction of the ship or the submarine, or any tossings or pitchings of it. It is,
therefore, preferred to the ordinary magnetic compass and is more dependable
than the latter, in view of the additional advantage of its remaining altogether
unaffected by any type of magnetic diturbanccs.
The Pendulum GyroCompass. The above arrangement, with a small
rnass, suitably suspended below the rota
ting disc or flywheel, constitutes what
is called the pendulum Gyrocompass,
the small mass supplying the necessary
restoring torque to bring its axis back:
to its original direction, should it get
displaced due to some disturbance. In
the absence of this simple but ingenious
device, the instrument would lack its
restorative action, due to the inherent
G stability of a gyrostat in any position.
The essential features of the
construction of the Pendulum Gyro
compass will be clear from Fig 59,
where the rotating disc or gyrostat D
has its axle PQ mounted in a horizontal
ring R, free to rotate about the axis EF
inside a vertical ring C which, in its
** turn, rotates freely about the axis AB
Fig. 59. within a frame work M, carried on
horizontal gimbals, (of which GG forms one pair), to ensure the fullest freedom
of movement.
The horizontal ring R has a stirrup S, fixed rigidly to it, which is loaded
with a weight W, immediately below O, the centre of the disc or the gyrostat.
It can be shown that this arrangement would be stable, at any given
place, only along true north and south, i.e., when the end P points truly north,
any accidental displacement of it calling into play a directive force, restoring it
back to its original direction.
(//') Rifling of barrels of Guns and Rifles. This is another wellknown
application of the directional stability of a rapidly revolving body. For, it is
found that if a shot or a bullet be given a rapid *spin\ about an axis along its
direction of motion, its uniformity of flight is greatly improved by making it
less responsive to small deflective forces during its passage through air. This
is achieved by 'rifling* the barrel, i.e., by cutting spiral grooves inside it so that
the shot or bullet is first forced to move along these, before it emerges out into
the air, thus acquiring the necessary 'spin* to ensure an almost uniform linear
motion.
(///) Riding of Bicycle and Rolling of Hoops or Discs. These are both
cases of what is called 'statical instability ; for, neither of the two, at rest, can
possibly remain in equilibrium in the position in which it does, when it is in
motion. Here, again, it is the gyroscopic action that does the trick, by appro*
priately deflecting their axes of rotation and thereby changing their planes of
rotation, to counterbalance the disturbing effect due to gravity.
MOMENT OF INERTIA ENERGY OF ROTATION 99
Thus, when a person rides a bicycle, without holding its handle, he has
simply to tilt to one side in order to turn to that side ; for, by so doing, ho
produces a couple about the horizontal direction of motion of the front wheel
of his bicycle, which, here, acts as a rotating gyrostat. This couple, then, turns
the axle of the wheel about the vertical, and hence its plaie of rotation, into the
desired direction.
The same is true about a hoop or a disc, projected, with its plane vertical,
to roll over a horizontal surface, which we have discussed fully in $45, above. As
explained there, so long as its linear or translational velocity remains above a
certain critical value, it continues to advance along a straight path, but as soon
as its velocity falls below this critical value, its plane gets inclined to the vertical,
or it begins to 'lean' from the veitical and its path gets curved towards
its 'side of lean'. And, then, as its velocity goes on progressively decreas
ing, due to friction, the curvature of its path goes on increasing corresponding
ly, so that it follows a more or less spiral path until, finally, it falls flat on the
surface.
(iv) Precession of the Equinoxes. The earth, as we know, is not an
exact sphere, but bulges out slightly at the equator, (or has the shape of a
'flattened ellipsoid of revolution") Further, the Sun and the Moon do not usually
He in its equatorial plane but rather in the plane of the ecliptic, which is inclined
at an angle of 23 5 to the former, with the result that the gravitational attrac
tion due to the Sun and the Moon, on this equatorial bulge gives rise to a
torque, bringing about the precession of the axis of the earth, which, acting
as a gigantic top*, describes a comrr, relative to the fixed stars, e.g., the pole star,
similar in manner to the cone described by the axis of a precessing top, due to its
M>e/>/tf, the phenomenon being spoken of as the 'precession of the equinoxes'.
Tins couple on the earth due to the attractive force of the Sun and the Moon
is, however, very small, so that it takes 25,800 years for the earth's axis to des
cribe the complete cone, at which rate of rotation, the star Vega will be the pole
star in about 12,000 years hence.
It is interesting to observe that atoms too have the mechanical proper
ties of tops, and, at least in one special case, their gyrostatij moment has been
demonstrated experimentally by Einstein and De Haas.
(v) Other Recent Applications. The modern aircraft appliances, like
the automatic pilot, the artificial horizon and the turn and bank indicators etc., all
depend for their ction on gytoslatic principles.
The function of all these instruments is to record the effects of a change
of orientation between a relatively fixed plane, provided by a fast rotating gyro
stat, serving as the reference or the datum plane, and some other movable plane
in the machine, and this they do with a degree of precision which makes their
indication far more safe to rely upon than mere human judgement, howsoever
trained or mature.
SOLVED EXAMPLES
1. A flywheel of mass 500 k. gins, and 2 metres diameter, makes 500 re
volutions per minute. Assuming the mass to be concentrated at the rim, calculate
the angular velocity, the energy and the moment of inertia of the flywheel.
(/) No. of revolutions made by the flywheel = 500 per minute.
Angle described in one rotation = 2 n radians.
.'. angle described by the wheel per minute = 2 Tr.500 ,,
And ,, ,, ,, , second 2 ir.500/60.
= 50ir/3 radians.
Or, the angular velocity (o>) of the flywheel = 50:r/3 radiansfsec*
(ii) Moment of inertia / =* MK 2 .
Here, mass M = 500 x 1000 gms.
And K, the radius of gyration = 1 metre or 100 cms.
*A *top\ in Physics, is the name given to a rotating body, either com
pletely free to move, or fixed at the most at just one point with absolute freedom
of rotation, and it must not, therefore, be confused with the toy that goes by
that name.
fin the clockwise direction, as seen from the north*
100 PROPERTIES OF MATTER
because the mass is concentrated at a distance, equal to its radius from the axil
of rotation, which passes through its centre.
moment of inertia of the flywheel == 500 x 1000 x(100) 2 .
= 500 x 1000 x 10000 5 x 10 gmcm*.
(iil) The energy possessed by the flywheel is due to its rotation, i.e., it
possesses only rotational energy, which is equal to i/w 2 .
Or > ener *y of the flywheel =
log 10 9 = 90000
log 1250 = 30969
2 log n = 9944 . '
137903 y
log 9 =0 9542 _ 5 X 1 X 1 250^
Antilog 12 T 8361 ' 9
6857x10* I 6857xlO u ^r^.
 68 57 xlO 11 '
2. A flywheel weighs 10 tons, and the whole of the weight may be con
sidered as concentrated at a distance 3 ft. from the axis. What is the amount of
energy stored in the flywheel when rotating at a speed of 100 revolutions per
minute ? (Punjab, 1934)
log 5 ==0 6990 ' Here, M = 10 tons = 10 x 2240 Ibs.
log 2240 = 3*3502 i to = 100 x 2r> radian f/min.
log 100 =2*0000 "' ' ""
2 log * =09944
7 : 0436
100x2rr/60 = 107T/3 radians/sec.
and K = 3ft. (Given)
Since K.E. of rotation of a body
log 32 = 15051 = * x ^ 2 = 4 M/C 2 co 2 , we have
Af;ir> ^"^8^" JRT.E. of the flywheel
Antilog 5 5385 , = ^ x ]0x 2240x 3 2 x (lOTr/3) 2 .
~ 34 55 x 10 I " 5 x 224 x 9 x 100 x 7i 2 /9 fi.poijgals.
5x2240x9xlOOxnr 2
  " 9x32 
Or, kinetic energy stored up in the flywheel 34*55 x 10* ft. Ibs.
3. A flywheel of mass 100 k. gins, and radius of gyration 20 cms. is
mounted on a light horizontal axle of radius 2 cms., and is free to rotate on bearings
whose friction may be neglected. A light string wound, on the axle carries at its
free end a mass of 5 k. gins. The system is released from rest with the 5 k. gms.
mass hanging freely. Prove that the acceleration of this mass is g/2001 cm. /sec 2 .
If the string slips off the axle after the weisht has descended 2 metres, prove
that a couple of moment 31*8 k. gms. wt.cm. (approximately) must be applied in
order to bring the flywheel to rest in 5 revolutions (Cambridge H. S. Certificate)
(0 The mass of the flywheel (M) = 100 k. gms. =100 x 1000 gms.
and its radius of gyration (K) = 20 cms.
So that, its moment of inertia (I) = MK 2 = 100xlOOOx20 2 .
= 100x1000x400 4 X 10 7 gm.cm*.
Let angular acceleration of the flywheel be d&l dt.
Then, linear acceleration of the mass of 5 k. gm.^r.d^Idt t uhere r is the
radius of the axle.
Or, acceleration of the mass, i.e., a r.da>ldt,
whence, d^jdt a\r = 0/2.
rotational couple acting on the flywheel =* Ld^fdt.
= 4xl0 7 xa/2w.H>/.cw. 2xl0 7 xa^m. wt.em.
This must, obviously, be equal to the couple applied to the wheel by the
tension in the string.
If Tbe the tension in the string, the couple due to it =7>.
If the mass of 5 1 /r. gms. had no acceleration, the tension in the string
would be equal to its weight == 5 x 1000 x^ dynes.
But, since it has an acceleration a, we have
ma mg T. ' Or T m(g ).
MOMENT ot IHBBTU ENEBGY off ROTATION 101
couple applied to the wheel T.r m(ga)r.
= 5 x 1000(# a) x2 dynescm.
Now, /.rfco/cfr = Tr.
2xl0 7 a 5xlOOOtea)x2  10000^100000.
Or f 2xlO T aHOOOOa = 10000^. Or, a(2x!0 7 + 10000) = #x!0 4 ,
i?x 10* j?x 10*
whence, a   (Tx f Q7 = T' Or " */ 2001
Thus, the acceleration of the mass =* #/2001 cw./sec*.
(//) When the weight has descended 2 metres, it has lost some P.E. Thii
must be equal to the gain in K.E. of the wheel and the weight ; so that,
K.E. of the wheel +K.E. of the weight = P.E. lost by the weight.
Now P.E. lost by the weight = work done by it in falling through 2 metres
distance = mgh, (where h = 2 metres = 200 cms ) = 5 x 1000 x 981 x 200 ergs.
.*. K.E. of the wheel and the weight = 1000000x981=981 x 10 6 ergs.
This must, therefore, be the work that the couple applied to the wheel
must do in order to stop it. If C be the couple required for the purpose, we
have
work done by the couple = CO, (where o s the angle of rotation).
Since the wheel comes to rest after 5 revolutions, it describes an angle
= 2^x5 radians.
work done by the couple = 2ir x 5 x C. And, .'. 2rc X 5 x C =981 x 10*.
,.  981x10 981 ,_, . 981 xlO 5 10 5
Or C * '" * "
2 log 10 = 20000 , 10 * io a
log ir  04972,  1000x7r =  k.gm.wt.cm.
ADtilog^ ^15028 j =, 31 . 83 ^ m .^ cm .
Hence, a couple of moment 31 '83 k.gm.wtcm. will bring the flywheel to
rest in 5 revolutions.
4. A flywheel of weight 200 Ibs. which may be regarded as a uniform disc
of radius 1 ft. is set rotating about its axis with an angular velocity of 5 revolutions
per second. At the end of 40 sees., this velocity, owing to the action of a
constant frictional couple, has dropped to 4 revolutions per second. What constant
couple must now be applied so that in further 20 sees., the angular velocity will be
8 revolutions per second.
Find the total angle turned through during the minute.
(Cambridge Higher Secondary School Certificate)
Here mass of the flywheel, M =2000 Ibs. and its radius r lft.
Since it is a uniform circular disc, its moment of inertia about its axis,
/.*.,
/ = j Mr a  i.200./ 2 100 Ib.ft*.
Us angular velocity, to start with = 5 revolutionslsec. = 2rrx5 radians/sec*
and ,, i, after 40 sec. 4 revolutions [sec. = 2::x4 radiansjsec.
.. change in angular velocity in 40 ,, = 2nx52n x4 = 2n radians/ sec*.
rate of change of angular velocity = 2rc/40 = 7t/20 radiansjsec 2 .
Or, angular retardation, i.e., dujdt = re/20 radiansjsec*.
Now, * couple = I.d^ldt.
.*. frictional couple acting on the wheel = 100x7r/20 = STT poundalft.
Again, the velocity of the wheel is now desired to be raised from 4 to I
revolutions per sec., in 20 seconds.
f % initial angular velocity * 2rcx 4 radians/ sec*
102 fBOKBRTlES Oft
And, final angular velocity in 20 sees. = 2rc x 8 radians! sec.
change in velocity in 20 sees. = 2n x 8 2* x 4 = STT radians fsec.
And, .. rate of change of velocity = 8?r/20 2r;/5 radiansl sec 1 .
Or, angular acceleration, </co /<# = 2^/5 radians J sec.
And, the couple required to produce this acceleration
= 100x2^/5 = 40npoundal'ft. [ v couple = Ldujdt.
total couple required to be applied
log 45 16532 \ = this couple of 40rc 4 a couple of SK (to overcome fric
log TT = 04972 tional couple)*
Antilog^2T504 i Or, total couple required = 40:r}5rr = 45?r ^14\'4 poundalft.
= 1414 '
Now, let the angle described by the wheel in the first 40 sec. be Oj
Then, from the relation d  r f 1 ^./ 2 , we have [~ S " J g^ C n
G! == (27ix5)x40~}. 2 * . (40) 2  400nr407r = 360rr radians.
And, if 0, be the angle described by the flywheel in the next 20 sees., we
have, as above,
0,  (27tx4)x20f * 2 ^ (20) 2  160^f 80rr = 2407T radians.
.*. the total angle turned through by the flyweel in one full minute
=, X 4 3 = 3607T+2407T = 600 radians.
Now, since a rotation through 2* radians means one revolution, a rota
tion through 600:r radians means 600^/2^ = 300 revolutions.
Thus, the flyweel makes 300 revolutions during the minute.
5. A pulley of radius 2 ft. has hanging from it, a rope with masses of 60
Ibs. and 52 Ibs. attached to its two ends, the masses being kept at rest initially by
holding one of them. If the moment of inertia of the pulley
be 320 Ib ft 2 ., what will be the velocity of the masses, when
they have moved a distance of 6 ft. from their position of
rest ? It may be assumed that there is no slip between the
rope and the pulley and that friction at the axle of the
pulley is negligible.
Here, obviously, the motive force, i.e., the force
which makes the masses and the pulley move, is the
weight of the excess mass of (60 52) or 8 Ibs. wt. at one
end of the rope = 8x32 = 256 poundals. (Fig. 60).
When the masses have moved through a distance
of 6//., the loss oj potential energy suffered by this excess
mass is clearly == 256x6 = 1536 ft. poundals.
This loss of P.. of the excess mass is equal to the
gain in the K.E. of the system consisting of the two
masses and the pulley.
Let v be the velocity of the masses at this instant.
Fig. 60. Then, K.E. of the two masses
= i (60{52)v 2 = 4.112.V 2 = 56v 2 ft. poundals.
And, K.E. of the pulley = i/w 2 = Jx320xv 2 /r 2  ix320xv 2 /4
= 40v 2 ft. poundals. [v co  v/r and r 1ft.
.. total'gain in K.E. of the system = 56v 2  40v 2 = 96v z ft. poundals.
Since gain in K.E. of the system loss in P E. of the excess mass.
we have 96v 2 = 1536. Or, v 2 = 1536/96 =16.
whence, v = ^16*" 4 /'/ 5ec 
Thus, the velocity of the masses when they have moved through a dis
tance of 6ft. will be 4ft.jsec.
MOMENT OF IN&RTII ENERGY off fcofATlON 103
6. A flywheel of mass 65*4 K. gms. is made in the form of a circular disc
of radius of 18 cms. ; it is driven by a belt whose tensions at the points where it runs
on and off the rim of the wheel are 2 K. gms. and 5 K. gms. weignt respectively. If
the wheel is rotating at a certain instant at 60 revolutions per minute, tincl how long
will it be before the speed has reached 210 revolutions per minute. While the fly
wheel is rotating at this latter speed, the belt is slipped off and a brake applied.
Find the constant braking couple required to stop the wheel in 7 revolutions.
(Cambndge Higher School Certificate)
(/) Here, obviously,
tension 7\, where the belt runs on the rim = 2K. gms. w/.
and r,, ,, off ,, ,, = 5 ,,
/. the resultant tension in the belt = T t 7\ = (52) K. gms. wt.
=3 K. gms. wt. = 3 x 1000x981 dynes.
.*. moment of the couple due to this tension ~ 3 x 1000 x 981 x 18 dynescm.
And, if dujdt be the angular acceleration of the wheel, the couple acting on the
wheel . = l.d<*ldt.
Now, / = M.r 2 /2 = (65'4 X 1000 x!8 2 ;/2 = 65400x18x9 gm cnr.
.'. moment of the couple = 65400 x 1 8 x 9 x dujdt dynescm.
This must be equal to the moment of the couple due to tension in the
belt. Hence,
65400xl8x9x</o>/</f 3x1000x981x18.
~ , ,. 3x1000x981x18 < , a
Or, dldt~ 65400xJ8x9  S radians Isec*.
Now, we ave the relation, <.> a = c^f (r/co/J/)/ . .(/') [See page 85.
where w 2 is the final angular velocity ; t^, the initial angular velocity ; c/u>/Jf ,
the angular acceleration and t t the time.
Here, a> a = 210 rev. I mm. = 210x27t/60 = Jr. radians/ sec.,
^ = 60 = 60x2rc/60 = 2rr
and d&ldt 5 radians/sec 2 .
.'. from relation (/) above, we have In = 2x\~5t. Or, 5* = 5/.
whence, t == K ~ 3*142 sees.
So that, the flywheel will obtain a speed of 210 revolutions per minute
after 3' 142 seconds.
(//) Let the angular retardation produced in the wheel by the braking
couple be dte/Ut, the angle turned through by it before coming to rest being
equal to 7x2rc ~ 14^ radians.
Then, applymgthe relation oj^wj 2 = 2(d>ldt)$, ISee page 85.
we have O a (77r) 2 = 2(d<*fdt) x 14nr. Or, 2^.d^dt = ~49^ 2 . ['/ = 14*.
Or, d<*Idt  49r; 2 /287T == 7^/4 radians I sec\
Or, the angular retardation required = 77T/4 radiansjsec*.
Now, since couple = Ld<^ldt, we have
braking couple required, C = J x 65400 x 18 x 18 x 7rr/4 dynescm.
log 32725145 ! ^ n 65400x18x18x77:
log 6317993 ! Or ' C== 2x4x981
I 65400x18x18x7
log ,o 0= " 4 i 2X4X981X1000 ~ "'" "'"^
327 x 63rt
^; oi 17736
= 5937
1090
_ 59 37 j^ g tnSt
Hence, the required braking couple 59 37 K. gms. wt.cm.
7. A flywheel, which can revolve on a horizontal axis weighs 900 ibs. and
its radius is r ft, A rope is coiled round its rim and a weight of 90 Ibs. hung from
its free end, turns the wheel by its descent. Find the speed at which the weight is
moving after descending 20 ft. from rest.
104
PROPERTIES OF MATTER
Let the acceleration of the weight be a ft. per sec*., and let M .be the
mass of the weight suspended, (Fig. 61).
Then, if T be the tension in the string, we have
M.a=(MgT). Or, T^M(ga) = 90(32 a) pom dais
Now, moment of the couple acting on the wheel due to
tension Tin the rope = T.r.
' T.r = 90(32ra). r poundalft.
Also, rotational couple on the flywheel =* I.d^/dt.
Here. 7 = JAfr 8 = J.900.r 2 . f Considering the flywheel to
= 450r 2 . ^ be a uniform circular disc of
And dujdt =a\r. \jnass Mand radius r.
.'. rotational couple on the wheel=450r 2 (a/r)=45Q a.r poundalft
We have couple due to tension T rotational couple.
90(320) r = 450 a.r. Or, 90(32) = 450 a.
Or, 288090 a = 450 a. Or, 540 a = 2880.
whence, a = 2880/540  16/3 // /sec*.
Fig. 61.
log
log
If v be the velocity of the weight after it has descended 20//., we have
6402K062
3=0 4771
1 x 23291
Antilog 11645
1461
[v S = 20 ft.
v 2  2 = 2.aS.
Or, v'O =2x(16/3)x20.
Or, v 2  640/3,_
whence, v = ^640/3 = 1461 ft. /sec.
Therefore, the speed of the weight would be 14'61 ft. /sec.
8. A sphere of mass 50 gms., diameter 2 cms., rolls without slipping
with a velocity of 5 cms. per sec. Calculate its total kinetic energy in ergs.
Here, mass of the sphere, M = 50 gms. 9 and radius of the sphere, / = 1 cm.
Now, moment of inertia of the sphere (solid) is given by
/ = J.Mr 2 = x50xl = 2Qgm.cm*.
As the sphere rolls, it rotates about its own diameter as axis as well as
its centre of mass moves with a velocity of 5 cms /sec. It has, therefore, both
kinetic energy of rotation as well as kinetic energy of translation ; and, there
fore, its total energy is the sum of both.
Now, K.E. of rotation = }/ <o 2 = J/.v 2 //' 2  x 20x 5 2 /l 2 = 250 ergs, [v o>=v/r.
And KE of translation = iMv 2 = x50x5 2 25 x25 = 625 ergs.
.*. total kinetic energy of the sphere =250 + 625 = 875 ergs.
9. A flywheel of mass 10 K. gms. and radius 20 cms. is mounted on an
axle of mass 8 K. gms. and radius 5 cms. A rope is wound round the axle and
carries a weight of 10 K. gms. The flywheel and the axle are set into rotation by
releasing the weight. Calculate f ) the angular velocity and the kinetic energy of
the wheel and axle and (//) the velocity and kinetic energy of the weight, when the
weight has descended 20 cms. from its original position.
The flywheel, here, (Fig. 62), is just a hollow circular disc or cylinder, (as
it has been cut in the centre for the axle to pass
through) ; its moment of inertia about its axis, there
fore, is equal to MtR 2 + /' 2 )/2, where M is its mass and R
and r its outer and inner radii, (r being the radius of
the axle). (See page 78).
Now, M for the flywheel, is equal to 80 K.gms.
or 80 x 10 8 gms., and R and r, equal to 20 cms. and 5 cms.
respectively ; so that, the moment of inertia of the wheel
=80xl03x(20M5 2 )/2 = 80xl0 3 x425/2.
= 1 7000x10"= llxlW gm.cm*.
And, the axle is just a disc or cylinder whose
moment of inertia about its axis is equal to Mr*,
where M is its mass and r, its radius. [See pp. 63 & 66.
So that, moment of inertia of the axle
= ix8x!0 3 x5 2 = 10'xl0 2 = 10 5 w.cm 8 .
/. total moment of inertia of the wheel and axle, / e ,
I  17xlO+ 10 s . Or, * 171 x lO'^m.cw 1 .
MOMENT 01 INERTIA ENERGY OF ROTATION
105
When the weight descends through a distance h, it loses potential energy
* mgh, and this loss in P E. of the weight is, obviously, equal to the gain in
K.E. of the wheel and axle and the weight itself.
Now, since m = 10 x 10 3 gm. t g 981 cm.jsec 2 . and h = 20 cms.,
we have loss in P.E. of the weight = lOx 10 8 x981 xlO = 1962xl0 5 ergs.
If to be the angular velocity of the wheel and axle, when the weight has
descended through 20 cms., the velocity v (linear) of the weight will be rco,
where r is the radius of the axle ; i.e., v = 5w. [ . r = 5 cms.
.'. K.E. of the wheel and axle = I L o> 2 = i x 1 71 x 10 5 x w 2 =* 855 x 10* x w 2 ergs.
and, K.E. of the weight = Jmv 2 = Jx 10 x 10 3 x(5o>) 2 =125x 10 2 Xo> 2 ergs.
.*. total gain in K . of the wheel and the axle and the weight.
=855xlO*Xco 2 J125xl0 3 xco 2 = 8550x I0 3 xo> 2 fl25x 10 2 x<A
= (8550+ 12:>) x 10 2 x w 2 = 8675 x 10 8 .o> 2 ergs.
Since total gain in K.E. of the wheel and axle and the weight is equal to
the loss in P.E. of the weight, we have
Or,
whence,
8675 xlO 3 . co 2 = 1962x10'.
2 _ 1962 xlO 6 __ 1962*10*
~~ 8675 XlO 3 """"8675 ""
 10 x v/1 962/8675".
= 10x4755.
4 755 radianslscc.
Or, angular velocity of the wheel and axle, i.e., co = 4'755 radians/sec.
and linear velocity of the weight, i.e., v = rw = 5o> = 5 X 4 755 = 23'
log 1962 = 3'2927
log 8675 = 3_9383
JxT'3544
Antilog T : 6772
= 4755
Now,
log 855 =>
29320
4 log 10 =
40000
2 log 4755 =
"13544
Antilog
82864
B
1934X10 7
log 5 
6990
3 log 10 =
30000
log 25 =
13979
2 log 4*755 =
T354I
Antilog
64513
=
2827 xlO 5
of the wheel and axle is given by
i/.w a = }xl71xl0 5 xo)*.
= 855xl0 4 x(4755) 2  19'34x 10 7 ergs.
And, K E. of the weight is given by
i /iiv 2 = i m(5w) 2  Jx 10*.(5x4 755)*
= 5xl0 3 x25x(4755) 2 .
= 28 '27 xlO 5 <?/'#*.
Thus, (i) the angular velocity and kinetic
energy of the wheel and axle are 4 755 radians/sec.
and 19 34 xlO 7 ergs and (//) the velocity and
kinetic energy of the weight are 23775 cms. I sec.
and 28'27x 10 5 ergs respectively.
10. If the pulley in an Atwood's machine be of moment of inertia 1500
e.g.s. units and radius 5 cms., what should be the acceleration of the system in
which the weights at the two ends of the string passing over the pulley be 200 and
250 gms. respectively ? (Given that g = 981 cm. /sec 2 .)
Let a be the acceleration of the system and v, the velocity of the
weights, when they have moved a distance S cms. from the starting position,
(Fig. 63).
Then, clearly, v 8 w a = 2aS. Or, v a = 2aS. [v w=0.
And /. a = v 2 /25.
Now, loss in P.E of the heavier weight M, falling
through distance S = Mg.S = 250 x 981 x S ergs,
and, gain in P.E. of the lighter weight m
= mg.S = 200 x 981 x S ergs. ^^
o?n
200 gm.
.*. net loss in P.E. of the system
= (250x981x5) (200x981x5) 50x981x5er^.
This must be equal to the gain in K.E. of the pulley
as well as the weights themselves.
Clearly, gain in K.E. of the pulley = } 7w* =  /.v 2 //* 2 .
v v = rw, where w is the angular velocity and r, the
radius of the pulley.
M
ZSOgm.
Fig. 63.
106
PROPERTIES Otf
Or, gain in K.E. of the pulley = } 7.v 8 /25 * Jx 1500x v 2 /25 ~ 30v 2 ergs.
And, gain in K.E. of the two weights = iMv 2 fimv 2 = i(Aff m) v 2 ,
= ix450xv 2 = 225
.*. total gain in #.. of the pulley and the weights,
Since this gain in K.E.
we have
log 327 = 2 5145
log 5 = OJ5990
32135
log 17= 12304
Antilog l : 983f
= 9618
255v 2 = 50x981x5. Or, v
loss in P.E. of the system and the weights,
50x981x5
255
Now, the acceleration of the system, i.e., a
50x981x5^327x5
* "25"5x25 17
9618 cms. I sec*.
v 2 /25.
Or, the acceleration of the system of weights is 9618 cms I sec*.
11. A narrow uniform metal bar, 1 metre long, weighing 3 K. gms., rotates
once per second. What is the K.E., if the axis of rotation passes through (i) its
centre of gravity, (//') one extreme end ?
Here, M =3 K.gms. = 3 x 1000 = 3000 #mj., and / = 1 metre = 100 cms.
/. M.L of the bar about an axis through its e.g.
 M/ 2 /12 3000x(100) 8 /12^m. cm 1 .
And, its M.L about an axis through one end f
 M/ 2 /3  3000x(100) f /3#. cm 2 .
Angular velocity of the bar o> = 2n radians/ sec. ['.' it is 1 rotation/sec.
Now,
Aad,
. of a rotating body = Jf o>*.
. of the bar in case (/) = ix[3000x (100)*/12]x (2n)\
Jx[30DOx 10000/12] X4n 2 500X 10000 Xn z = 5xl g X
its ^T.E. in case (//) = J(3000x 1 0000/3) x4* 8 .
2000 x 10000 xn 2 = 20 x 10 6 xn ergs.
12. Find the moment of inertia of a homogeneous circular cylinder of
length 2/, radius of crosssection r, about (/) the axis of the cylindrical symmetry ;
(//) a generating line ; (*//) a diameter of crosssection at a distance x I, and 21 (or
0) from one base.
(/) The moment of inertia of the cylinder about the axis of cylindrical
symmetry is the same as that of a disc about an axis
passing through its centre and perpendicular to its plane,
(for a cylinder is nothing but a thick disc), and is equal
to MR*I2, where M is the mass of the disc or cylinder and
R, its radius.
.'. ifMbc the mass of the cylinder, and r, its
radius of crosssection, (Fig. 64), we have moment of inertia
of the cylinder about its axis of cylindrical symmetry equal to
A/r 2 /2.
^ (11) The generating line is parallel to the axis of
symmetry, passing through the e.g. of the cylinder, and
is at a distance r from it. Therefore, by the principle of
parallel axes, moment of inertia of the cylinder about the
generating line is equal to its moment of inertia about the
axis of symmetry plus mass of the cylinder x (distance from
the axis) 1 , i.e.,  iMr a + Mr a  3Mr 2 /2.
MOMENT OF INERTIA ENERGY OF ROTATION 107
(Hi) (a) The moment of inertia of the cylinder about an axis, passirg through
its centre and perpendicular to its length g\
= M[(4/ 2 /12) f(r a /4)] [v length = 2/.
(6) Since an axis at a distance x from
one base is at a distance (lx) from the axis
through the centre, we have, by the principle
of parallel axes,
M.I. about this axis =* M I. about a
parallel axis through the centref M(l x)*.
 Mf(/ 2 /3)i
A
k
(/.r
A
M[(4/ 2 /3)f (r 2 /4H (x*2lx}\. Fig, 65.
(c) Similaily, by the principle of parallel axes,
M.I. about the diameter of crosssection
= M[(/ 2 /3) + (r 2 /4)]f M/ 2  A4[(4/ 2 /3) + (r f /4)].
13. Find the moment of inertia of a sphere about a diameter.
You are given two spheres of the same mass and size and appearance, but
one of them is hollow at the centre and the other is solid throughout. How will
you find which is hollow and which is solid ? (Delhi)
For answer to first part, see 31, (case 13), pages 72.
The moment of inertia of a solid sphere about its diameter is, as we
know, 2M/T/5, where M is its mass and R, its radius, and that of a hollow sphere
2/5[A/(7? 5 / 5 )/(/ 3 r 8 )], where R and r are Us outer and inner radii respectively,
R and M being the same in the two cases.
.". the radius of gyration for the solid sphere ~ \ / 2R t /5 t and the radius of
gyration for the hollow sphere is
e= \/2/5[/2 6 r ) >/\rt J /)], their masses being the same.
Since acceleration of a body rolling down an inclined plane is given by
/ n > na . v\ f a being the inclination
a = (R*IR*+K~) g sm a, [ of lhc plane> (pagCf . .)
it is clear that the greater the value of A' 2 as compared with R*, the less the
acceleration of the body.
Now, K\ for the solid sphere 27? a /5,
2 rT^j 5 *1
and j* t for the hollow sphere = . i> 3 ~__~"Y
2 * 5 [lr 5 /K 5 ]_ 2 rir*/*'!
~ 5 >[lr 3 /#<J " 5 U 1 '' 3 /* 8 J
Obviously, r a /K 3 < r 3 //^ 3  And /. (1 r 5 //? 6 ) > (lr
j _ r 5 //^ 5
So that, the fraction, ._ 3 ' , > 1.
And .'. K* ^ . 7? 2 /fl quantity greater than 1.
Or, K* for a hollow sphere is greater than 2/? 2 /5.
Thus, the fraction (RjR'+K 2 ) is less for a hollow sphere than for a solid sphere
and, therefore, the acceleration of the hollow ^sphere is less than that of a solid
sphere. In other words, the solid sphere will come down the inclined plane
faster than the hollow sphere, and the two can thus be easily distinguished from
each other.
EXERCISE III
1. Define Moment of Inertia and Radius of Gyration. Explain their
physical significance. State the laws of (i) parallel and (//') perpendicular axes
and prove any one of them. (Bombay, 1945)
108 JteottBfcTite oir
2. Calculate the moment of inertia of a thin circular disc of mass M add
radius r (/) about its diameter (ii) about a parallel axis to the diameter and
tangential to the disc. Ans. Mr 2 /4 ; 5Afr"/4 f
3. Define Moment of Inertia of a body about an axis. Show that the
moment of inertia of a body about an axis through the centre of gravity is less
than that about any other parallel axis.
A uniform circular disc of radius r is free to oscillate in a vertical plane
about an axis perpendicular to it and distant x from its centre. Calculate the
periodic time. (Madras, 1950)
Ans. /  2^(r* + x*)l2gx.
<4) A flywheel of mass 2 I tons and diameter 8//. makes 250 revolutions
per minute. Find (/) its angular velocity (ii) its energy (Hi) its moment of inertia.
Assume its mass to be concentrated at the rim.
Ans. (/) 25n/3 radians/sec, (ii) 3 07x 10 7 ft. poundals. (Hi) 89600 Ib.ft*.
5. Show that the acceleration of a disc rolling down an inclined plane
of angle is 2g sin 0/3, while that of a ball is 5g sin Q/7.
( oP A uniform rod 4 //. long and weighing 9 Ibs., revolves 60 times a
minute^about one end. Calculate its kinetic energy. Ans. 29 61 //. Ibs.
I. A hoop of mass 5 k. gins, and radius 50 cms. rolls along the ground
at the rate of 10 metres per second. Calculate its kinetic energy in ergs.
Ans. 5x10* ergs.
8. Explain clearly what you understand by 'Moment of Inertia' and
'Angular momentum*. State the principle of conservation of angular momen
tum, illustrating your answer by an example.
Find the moment of inertia of a circular lamina about a tangent in its
plane. (Patna, 1949)
Ans. 5Afr/4.
9. A solid spherical ball rolls on a table. What fraction of its total
kinetic energy is rotational ? Ans. 2/7th.
10. Show that the K.E. of a uniform cylinder or disc of mass Af, rolling
so that its centre has a velocity v ii f Mv 2 . In the case of a sphere, show that
the K.E. would be 7Mv 2 /10.
II. A thin hollow cylinder, open at both ends and of mass M (a) slide t
with a velocity v without rotati tg, (b) rolls without slipping, with the same speed
Compare the kinetic energies it possesses in the two cases. Ans. 1 : 2.
12. Define (/) radius of gyration, (//) moment of inertia. Find the
moment of inertia of a circular dire about the axis perpendicular to its plane.
A circular disc of mass m and radius r is set rolling on a table. If <o is
its angular velocity, show that its total energy E is given byj
E = i mr 2 .co a . (Punjab, 1950)
Derive an expression for the kinetic energy of a body rotating about
an axis.
A flywheel is in the form of a uniform circular disc ; its radius is 2 ft.,
and mass 2 Ibs. Find the work which must be done on the flywheel to increase
its speed of rotation from 10 to 20 revolutions per second. (Madras B.A., 1947).
Ans.
14. Five masses, each of 2 k. gms., are placed on a horizontal circular
disc, (of negligible mass) which can be rotated about a vertical axis passing
through its centre. If all the masses be equidistant from the axis and at a dis
tance 10 cms. from it, what is the moment of inertia of the whole system ?
( Hint.: M.L of each mass about the axis = Mr 9 , }
i and .. total M.I. of the system r
C = sum of the MJ. of the masses. ) Ans. 10 8 gm. cm*.
15. Define 'Moment of Inertia' and 'Radius of Gyration.' State the law
of parallel axes* and prove it.
IIOMIHT 01 IHERTIJL ENERGY OF ROTATION 109
A wheel of radius 6 cms. is mounted so as to rotate about art horizontal
axis through its centre. A string of negligible mass, wrapped round its circum
ference carries a mass o! 2W gms. attached to its free end. When let fall, the
mass descends through 100 cms. in the firsts seconds. Calculate the angilar
acceleration of the wheel and its moment of inertia. (Bombay, 1947)
Ans. Angular acceleration *. radians I sec 2 . ; ML 8*748 x 10 4 gm.cm*.
16. What is meant by moment of inertia of a body ? Show with neces
sary theory how the moment of inertia of a flywheel may be determined.
(Allahabad, 1948)
17. The free end of a string wrapped round the axle of a flywheel, of
moment of inertia 27*61 xl0 5 #w.cw 2 ., carries a weight of 5 k.gms., which is
allowed to fall. What is the number of revolutions per second made by the
wheel, when the weight has fallen through 1 metre ? The kinetic energy of the
weight may be neglected. Ans, 3.
18. If in question 17, the wheel be mounted on an axle of half its
moment of inertia (i.e., 1380 xWgm. cm 2 .) and radius 5 cms., and the K.E. of
the weight be taken into account, what will be the number of revolutions per
second made by the wheel ? Ans. 2413.
19. Masses of 95 gm. and 105 gm., hanging freely are connected by a
light string which passes over a pulley of mass 20 gm. when icleased, the system
moves with an acceleration of 46 7 cm. per sec 2 . Calculate a value of g if the
mass of the pulley is (a) neglected, (b) taken into account. Regard the pulley as
a simple disc of moment of inertia i Mr 2 , and assume that no kinetic energy is
lost in friction. (Northern Universities Higher School Certificate)
Ans. (a) 934 cm.se<r 2 ., (b) 9807 cm.sec~*.
20. (a) Four spheres, each of diameter 2a and mass m, are placed with
their centres on the four corners of a square of side b. Calculate the moment of
inertia of the system about one side of the square. (Punjab, 1951)
(b) A flat thin uniform disc of radius a has a hole of radius b in it at a
distance c from the centre of the disc, [c <(/>)]. If the disc were free to ro
tate about a smooth circular rod of radius b passing through the hole, calculate
its moment oi inertia about the axis of rotation. (Punjab,
Ans. (a) m(4a' + 5i> 2 ) ; (b) M
where M is the mass of the disc.
21. Describe the experiment to determine the moment of inertia of a
flywheel. Derive the formula used in the experiment, without neglecting the
friction at the bearings of the flywheel. (Allahabd, 1948 ; Gujrat, 1951)
22. A flywheel, which can turn about a horizontal axis, is set in motion
by a 500 gm. weight hanging from a thin string that passes round the angle.
After the wheel has made 5 revolutions, the string is detached from the axle
and the weight drops off. The wheel then makes 7 revolutions before being
brought to rest by friction. The radius of the axle is 20 cm., and at the instant
when the weight drops off the angular velocity of the wheel is 10 radians per
sec. Assuming that the work done against friction in each revolution is always
the same, calculate the moment of inertia of the flywheel about its axis of
rotation. (Oxford and Cambridge Higher School Certificate)
Ans. 3 59xlO*#m.cm 8 .
23. A pair of rails is supported in a horizontal position and the axle of
a wheel rests on the rails. A thread is wrapped round the axle and a weight
hung on the end of the thread. As the weight falls the wheel moves along the
rails. How would you determine the moment of inertia of the wheel with thii
arrangement ?
24. A circular disc, starting from rest, rolls (without slipping) down an
inclined plane of 1 in 8, and covers a distance of 5*32 //. in 2 sees. Calculate the
value of V. Ans. 31*92 ft [sec*.
25. Two gear Wheels, of equal thickness, of the same material and having
radii in the ratio 2 : 1, are mounted on parallel frictionless spindles, but are
separated so as not to metii with ono another. The larger vbeel is sot spinning
110 PROPERTIES Of MATTER
at a speed of 10 rev. per sec., and the wheels are then brought into mesh. What
is the resulting speed of each wheel ? (Cambridge Schorlaship Examination)
Ans. 89 and 178 rev. sec* 1 .
26. What do you understand by the term "precession 1 ? Show that if
the axis of the torque applied to a body be perpcnJicular to its axis of
rotation, the body precesses about an axis perpendicular to either of the first
two axes.
27. What is (/) a gyrostat and (//) a gyroscope ? Describe suitable experi
ments to illustrate their action.
What is meant by the term nutation ?
28. Explain the theory underlying a gyw static pendulum and obtain an
expression for its timeperiod.
29. Discuss in detail the case of a thin disc or hoop set rolling over a
plane horizontal surface and obtain expressions for (i) its critical velocity, and
(it) the radius of curvature of its path on the surface.
30. Write short notes on the following ;
(i) Gyrocompass, (ii) f'endulum Gyrocompass, (Hi) Rifling cf barrels of
funs and (jy) 'Precession of the Equinoxes.
CHAPTER IV
SIMPLE HARMONIC MOTION
47. Definitions. A simple harmonic motion is a particular case
of periodic motion, i.e., a motion which repeats itself over and over
again after regularly recurring intervals, called its timeperiod, and is
so called becauss of its association with musical instruments. Common
in nature, it is in fact th^ most fundamental type of periodic motion,
as all other periodic motions, (harmonic as well as non harmonic),
can be obtained by a suitable combination of two or more simple
harmonic motions.
If the acceleration of a body be proportional to its displacement
from its position of equilibrium, or any other fixed point in its path and
be always directed towards it, the body is said to execute a simple har
monic motion, (written, for short, as S.H.M.).
Now, a simple harmonic motion may be (/) linear, or (//) angular,
according as the body moves along a linear path, under the action of
a constraining force constantly acting upon it, or rotates about an
axis, under the action of a constant torque or couple.
The timeperiod of a body, executing a 5. H. M., is quite inde
pendent of the extent of its motion to either side of its mean posi
tion, (i.e., of its amplitude), and the motion is, therefore, said to be
isochronous.
Mathematically, a linear S.H.M. may be regarded as the projec
tion of a uniform circular motion, or of a rot at ing vector, on the dia
meter of the circle, or any other fixed line in the plane of the circle,
this circle being refer red to as the circle of reference, and may, in many
a case, be purely imaginary.
Thus, if a particle P (Fig. 66), moves with a uniform speed v
along a circle of radius a, and another particle M, along the diameter
YOY', such that when P is at X, M is at O,
and, as P starts along the circle in the anti
clockwise direction, M starts along OK, so that
when P reaches Y, M also reaches Y. As P
continues to travel further along YX', M
starts back towards O.
And, when Preaches X', M reaches O.
As Pnow traverses the lower half of the
circle along X'Y' 9 M proceeds downwards _
along OY', so that both Pand M reach Y' Y
together ; and, finally, when P travels further Fig 66.
on along Y'X, M starts back along Y'O, reaching its mean position
whe& P reaches X.
Thus, the particle M moves along the diameter YOY' from O to
F, from Y to Y' and back to O (i.e., completes one vibration), in
the same time in which P moves once round the circle, such that, at
111
112 PROPERTIES OP MATTE!
any given instant, the line joining the positions of? and M is perpendi
cular to the diameter TOY', or, the position of M on the diameter
YOY' corresponding to the position of P on the circle of reference, at
any instant, is given by the foot of the perpendicular from P on to YOY'.
This particle M is said to be performing a linear S.H.M. along
YOY', and is obviously, the projection of the particle P, moving uni
formly along the circle of reference XYX'Y', or is the projection of
the rotating vector OP on the diameter YO Y 1 of the circle.
If the motion of M be due, not to P, but to any other force
along its path, the circle of reference will, as indicated above, be purely
an imaginary one. Further, the path of M need not necessarily be
straight and may as well be curved.
Since a force acting on a body is proportional to the accele
ration it produces in it, it is obvious that the force acting on a body
executing a 8. H. M. must correspond to the changes in its accele
ration. In other words, it must also be proportional to the displace
ment of the body from its mean position and must always be directed
towards it.
Some familiar examples of simple harmonic motion.
(a) Linear.
(/) The up and down oscillations of the piston of a cylinder, con
taining a gas, when suddenly pressed down and released, (see solved
example 1).
(//) Tli 2 oscillations of mercury or water contained in a Utube,
when the column in one limb is depressed and released, (see solved
example 7).
(Hi) The vertical oscillations of an elastic string (or a spiral
spring) suspended from a rigid support and loaded at its lower end, (see
solved example 10).
(b) Angular.
(/) The oscillations of a pendulum, provided the amplitude be
small, (see Chapter VI).
(//) The oscillations o^ a magnet suspended in a magnetic field,
(see solved example 8).
(Hi) Torsional oscillations, in general, (see Chapter VIII).
48. Characteristics of a Linear S.H.M.
1. Amplitude. The maximum distance covered by the body
on either side of its mean or equilibrium position is called its ampli
tude. It is, obviously, equal to the radius of the circle of reference.
Thus, the amplitude of the particle M, in the case above, is OY
*=OY'=a, the radius of the circle XYX'T.
2. Displacement. The distance of a body from its mean posi
tion, at any given instant, measured along its path, gives its displace
ment at that instant.
Thus, the displacement of M, in the position shown, (Fig. 66), is
equal to OM, orj, such that
OP sin $. Or, y**a sin 0. (where
SIMPLE HAEMONIO MOTION
113
If a> be the angular velocity of P and t, the time taken by it in
traversing the distance OP along the circle, i.e., in describing the
angle 0, we have 0=tof ; so that, y=a sin wt.
This relation, giving the value of the displacement of a body,
executing a S.H.M., in terms of its amplitude and the angular velo
city of the rotating vector, (or of the particle in the circle of reference)
is referred to as its equation of motion. Thus, the equation of motion
of M along YOY' is represented by y= a sin cot.
If we consider the motion of a particle W alon? the diameter XOX',
such that both P and AT are together at X and as P goes round the circle in the
anticlockwise direction, N starts along XOX', so that when P reaches Y, N
reaches 0, and when P reaches X', N also reaches X', and when P goes along
the lower half of the circle, N starts back along X'O, reaching O when Preaches
F', and finally both arrive together at X, then the motion of ATalong XOX' is
also a S. H. M. And its displacement ON x is clearly given by
ON = OP cos o.
Or, x a cos 0. [ v ON = x and OP = a.
its equation of motion is x a cos cot. [v = o>f.
The position of the particle M executing a S.H.M. along the
diameter YO Y f may at any time, be found with the help of its dis
placement curve, which is a graph, showing the relation between the
time that elapses since the particle was at its mean position O, and its
displacement from O during this time.
DISPLACEMENT CURVE
Fig. 67.
Let time be represented along the horizontal axis AB and dis
placement along the vertical axis DC, (Fig. 67).
Let the circle XYX'Y' be divided up into an equal number of
parts, say 8, representing equal intervals of time T/8, where T is the
time taken by tho particle P to go once round the circle. Let these
intervals of time be also marked along the axis AB, taking A as the
origin or the starting point. Then, the perpendiculars drawn from
the points on the circle on to XOX 1 give the displacements of M
along YO Y', corresponding to the intervals of time represented by
them, as shown in tabular form below :
Time
J/4
772
3T/4
T
Displacement
(min.)
a
(max.)
(min.)
a
(max.)
(min.)
114
PROPEKTIES OF MATTER
Ordinates equal to these perpendiculars are then erected at the
corresponding points on AB. Thus, the ordinate/? represents the
displacement of M after a time T/8 of its starting from O; the ordi
nate a, after time T/4 ; the ordinate q after time 5 T/8, and so on.
The extremities of all these ordinates are then joined and a smooth
curve AJKLB is obtained, which is a harmonic or a sine curve, be
cause it is of the same form as would be obtained for the relation
between angles from to 360, and their sines, the maximum value,
viz., J, being at 90, and 1 at 270 and the least, i.e., zero at 0,
180 and 360. The displacement curve shows at a glance how
the displacement of the particle M changes along the diameter YOY'
and its value can be readily obtained from the curve at any given
instant.
3. Velocity. The velocity of the particle M is clearly given
by the component of the velocity of P, along the diameter YOY 1 ,
(Fig. 68).
Now, the velocity of P is v in a direc
tion tangential to the ^circle at P.
Representing it in magnitude and
direction by the straight line PK, we may
resolve it into two rectangular components
PN and PM along and perpendicular to
the diameter YOY 1 , as shown. The com
ponent PN represents the* velocity of M
Fig. 68. along YO Y'.
Now clearly, component PN = v cos <f> = v cos wt.
Since cos tf> = OQjOP
we have component PN = v
vV 2 >*/<* =
 V f y/*.
[v QP = OM
[v v
Or, velocity of M =
Alternatively the velocity of M may be obtained by differentiating its
displacement y with respect to t (because velocity is rate of change of displace
ment). Thus, since y = a sin <*t, we have
dyfdt = av.cos of. Or, dy\dt =*a<*.Va*y*la = oW/
O r velocity of M = wv/^ 2 ^ 2 
Thus, the velocity of M would be different at different points
alone its path, depending upon its displacement, or distance from its
mean position O, being a maximum when y is a minimum, and a
minimum, when y is a maximum.
So that (0 wheny =0, i.e., when M is at O, (or, its displacement
is zero), its velocity = wV^ 2 ~= w  a ^ v > ^ e same as ^ at f** ;
and '() wheny = a, i.e., when M is at Y or T, (i.e., its dis
placement is maximum), its velocity = wv/fl 2 fl 2 ^ >0 = 
Or, the velocity of the particle varies inversely as its displacement.
It will thus be readily seen that the velocity of M varies from a
maximum (v) at to a minimum (zero) at 7, then increases from zero
SIMPLE HABMONIO MOTIOtf
115
to v at 0, decreasing to zew again at 7', and again becoming v when
it comes back to O. In other words, at time 0, it is maximum ; at
tima 274, a minimum ; at 3T/2, a maximum ; at 32^/4, again a mini
mum, and finally at time T, again a maximum, as shown in the table
below, where the velocities of M are shown at different times.
Time
774
TI
37/4
T
Velocity
v
(max.)
(min.)
(max.)
(min.)
v
(max.) i
If, therefore, a graph be plotted botween time and velocity of
the particle M , we get the velocity curve of the particle, shown in
Fig. 69, where time is
shown along the hori
zontal axis and velocity
along the vertical axis.
The curve obtained is
a cosine curve, for it is
of the same form as the
curve plotted between
angles from to 360*
t
r#
T/ 2
and their cosines.
TIME *
Fig. 69.
It should be noted that the maximum value of the velocity of the
particle is aa> or v, and occurs when it passes /Ys mean position to either
side, or when its displacement is zero ; and its velocity is zero, when it
has attained its maximum displacement on either side.
4. Acceleration. As in the case of velocity, so also here, the
acceleration of M is the resolved part of the acceleration of P along
YOY', (Fig. 67). Now, the acceleration of P is v 2 /a or # 2 o> 2 /tf, or aa> 2 ,
and is directed towards O. Resolving it into two rectangular com
ponents, along PM and MO, we have the component along MO equal
to auP.sin $ or == aw 2 .yla = aj z y. And, as is clear from the figure,
it is directed towards 0, the mean or equilibrium position of M.
Thus, acceleration of M = &*y,
the negative sign being put to indicate that it is directed toward 0, in a
direction opposite to that of y, its displacement.
Alternatively, we may obtain the acceleration of M by differentiating its
velocity with respect to time, for acceleration is the rate of change of velocity.
Thus, since dyjdt = au.cos <of, [see page 114.
we have d*yldt 2 =* a^sin^t = ^.a sin cof = 2 >>,
Or, acceleration of M = o>V
Or, we may put it as acceleration of M = MJ>*,
where co 2 *= M, a constant of proportionality.
Or, acceleration ofM oc y.
*Sinrilarly, in the case of angular S.H.M , we shall have angular accekra*
tlon of the particle = /*,0, where is its angular displacement.
So that, here, angular acceleration oc Q.
116
PROPERTIES OF MATTER
Thus, we see that the acceleration *of the particle M is propor
tional to y t its displacement from O, and is always directed
towards it.
Obviously, this acceleration has its. maximum value, aof, the
same as that ofP, at the extreme positions Y and Y', where y ~ a, and
its minimum value, i.e., zero, at O, where y = 0.
It should be noted that the constant of proportionality /z is
equal to o> 2 , or the square of the angular velocity of the particle P in
the circle of reference.
Further, if y = 1, acceleration of M = p,. Thus,
defined as the acceleration per unit displacement of M.
Tabulating acce]eration of M against time, we have
may be
Time
A , ..
Acceleration
774 T/2
3r/4
o
(min) (max) i (m/f>) (nmx) (min)
If, therefore, a graph bo plotted between time and acceleration,
we get the accelera
tion curve of the
particle, as shown in
Fig. 70, which is of a
type similar to the
displacement curve ,
(Fig. 67), but is reci
procal in form, for
acceleration is directed
in the opposite sense
Fig. 70. to displacement.
5. TimePeriod and Frequency. The time taken by the particle
M, in completing ono vibration, (or one cycle), i.e., in going from O
to Y, Y to Y', and finally back to O, is called its limeperiod, period
of vibration or periodic time, usually denoted by the letter T.
Obviously, it is the same as the timeperiod of tho particle P, i.e.,
equal to the time taken by P in making one full round of the circle,
(from X back to X), or in describing an angle 2ir. Therefore,
27T 2?r 27T r v to = **, the acceleration
of M per unit displace
l~ mei
ment.
whence,
Or, timeperiod of M =
acceleration per unit displacement'
The number of vibrations made by the body per second is called
the frequency of vibration of the body, and is denoted by the
IIMPLE HARMONIC MOTION 117
letter n. Clearly, therefore,
frequency, n l/T = \/u /27T.
_ y acceleration per unit displacement
27T """
6. Phase. The term, 'phase' applied to a vibrating particle,
has a meaning similar to the ono associated with it when, we talk of
the 'phases 1 of the moo:i. Just as tho phase of the moon i.e., whether
it is a crescent, (or new moon), half moon or full moon tells us about
its position etc., so also th3 phase of a paroicle, executing a S.H.M.,
enables us to form an idea about its state of vibration.
Thus, the phase of a vibrating particle, at any given instant, may
be defined as its state or condition as regards its position and direction
of motion at that instant. It tells us in what stage of vibration the
particle is.
It is indicated either (i) in terms of the angle 0, described by
the rotating vector, measured as a fraction of the whole angle 2/T
that it describes in one full rotation, or (//') in terms of time t that
has elapsed since the particle last passed its mean position, in the
positive direction, measured as a fraction of its timeperiod T.
Thus, taking O as the starting position of the particle M
(Fig. 66), if its phase be zero, it indicates that the particle is at O,
tending to move towards Y. And, if the phase be Tr/2, or T/4, it
indicates that it is at Y, the position of the maximum positive dis
placement ; for, the radius of the circle of reference, or the rotating
vector, has, up to this instant, described onefourth of the total angle
27T, i.e., an angle 2?r/4 or Tr/2 ; or that onefourth of the timeperiod,
i.e., T/4, has elapsed sinco the particle last passed its mean position
O in the positive or upward direction, towards Y.
Hence, when we talk of a 'phase difference' between two simple
harmonic motions, we mean to indicate how much the two are out of
step with each other, or by how much angle, (measured as a fraction
of 2?r), or by how much time (measured as a fraction of T), one is
ahead of the other.
Now, because the phase of a vibrating particle merely indicates
its actual stage of vibration, it is clear that two vibrating particles, if
they happen to be in identical stages of their respective vibrations,
at any given instant, will be said to be in the same phase, at that
particular instant, irrespective of their amplitudes and velocities
being the same or different.
Thus, for example, they will be in the same phase, if they both
simultaneously attain their maximum displacements, positive or
negative ; or, when the two pass through their respective mean posi
tions at the same time and in the same direction. Similarly, if one
of the particles attains its maximum positive displacement simulta*
neously with the other particle attaining its maximum negative dis
placement, or when the two cross each other simultaneously in
opposite directions at their mean positions, they are said to be n
opposite phases,
US PROPERTIES OF MATTEE
7. Epoch or Initial Phase. We have deduced the reiatioii
y = a sin ojt for the displacement of the particle M , executing a
S.H.M., [ 48, (2)] on the assumption that the starting position of
the rotating vector is OX, or that the starting point of the particle
in the circle of reference is X, i.e., we start counting time when P
crosses the axis of x at X.
Sopaetimes, however, the starting position of the rotating
vector, or the position of the particle P in the circle of reference, is
fixed, not in some standard position, as on the axis of x or y, but
anywhere, in an arbitrary manner, such as at P', (Fig. 66), i.e., the
time is counted from the instant when P is at P' t such that the
angle XOP' = e.
Then, clearly, POP' = cot = 6+e. Or, = (wfe).
So that, y = a sin = a sin (a>te), where cot is the phase
angle of P. This angle e is called the 'epoch' or the 'initial phase 9
of the particle. It may also be measured, in terms of the time
taken by the particle P in describing this angle, i.e., by the time
It should not be confused with 'phase' ; for, whereas the 'phase*
of the particle continuously changes with time, its epoch or initial phase
remains the same all through.
49. Equation of Simple Harmonic Motion. Let y be the
displacement from its mean position of a particle, executing a S.H.M.
Then, if v be its velocity at that instant, we have
v = dy/dt.
n ,, , 1 . si i d 2 y dv dv dy dv
So that, acceleration of the particle = S  = , = . x *,  = V.T
Now, acceleration is also given by n.y, where /* is the cons
tant of proportionality and is equal to o> 2 , (o> being the angular
velocity of the rotating vector, or the particle in the circle of
reference).
Thus, d 2 yjdt 2 = aA.y. fthe negative sign indi
Or, v.dvldy = a>\y, { eating that acceleration
i' ' , a , ,/ i and displacement (y)
whence, v. dv =  a>*.y.dy. { are oppo ^ tely directed!
Integrating this expression, we have
v.dv = aP.y dy = co 2 ly.dy.
Or, }v* = jc
where C is the constant of integration, and has to be determined from
the condition of the particle at the instant considered.
Obviously, the velocity of the particle is zero, when it has its
maximum displacement a, or a, i.e., v = 0, when y = a, or a.
So that, we have, from relation (/), above,
 JwW+C. Or, C ** JciiV.
Heaoe, Jv* =* Jo^+JcoV ico 2 (a 2 ^). Or, v 1 = cu 8 (a*^).
And /. v = w^/o^y^. ,. () [Sec 48 (3).
SIMPLE HARMONIC MOtflOtt 119
llms, the velocity of the particle can be determined for any
value of its displacement y. Clearly, the value of v is the maximum,
i.e., = coa, when y = 0, i.e., when the displacement is zero, or" the
particle is in its mean or equilibrium position ; and, it is a minimum,
i.e., 0, when y = a, or, a, i.e., when the particle has its maximum
displacement, positive or negative.
Now, since v = dyjdt, we have, from (//) above,
dyjdt = oV** 1 }'*
n dy ,. fPut y = a sine ;
Ur > y~=.^ = ai.at. then, dyj~ a, cos 0.^0,
v a y I an d ^ a 2 y z =a cos Q,
Integrating this, we have ~^^==_dy =  w.dt. { so that, f ~r$^**
J^a*y* J j Jva 2 /"
Or, sin 1 y\a = ^r 4 C",  r __ . ^ y
whore C' is another constant of integration. I J 9^9 stn ^
Or, j = a sin (tt + C). ...(/)
(a) Now, if we start counting time when y = 0, /.<?., w/i^n //*
particle is in its mean position, moving in the positive direction, i.e.,
y = 0, when t 0, we have, from relation (in) above
= a sin (Q + Cy.
Or, a sin C = 0, whence, C' = 0.
.. substituting this value of C' in (Hi) above, we have y = a sin ojt.
(h) ]f, on the other hand, we start counting time when the particle
has its maximum displacement, i.e., t = 0, when y = a, we have
a = a sin (0+C), or, sinC'^a/a^l or C' = ?r/2 ;
so that, in this case, y = a sin (cot \7ij '2). Or, y = a cos a>t.
(r) Again, if we start counting time from an instant /' before the
particle has passed through its mean or equilibrium position, we have
t = t', when y = 0. Therefore, from (///) above,
= a sin (ut' + C), or, a>t' + C = 0.
Or, C = o>f = *,
where e is the epoch of the particle in the circle of reference.
Substituting this value of C' in (///'), we have y = a sin (wte). ...(iv)
(d) And, //" we start counting time from an instant t' after the
particle has passed through its mean position, we have
y = 0, when t f, or, = a sin (ut' + C),
whence, ut + C' = 0, or, C = cot 1 = e.
And, therefore, y = a sin (ut+e).
A mere glance at the above relations for y indicates that these
simple harmonic vibrations of the particle are a case of periodic
motion.
(e) Now, if the time t be increased by 2ir/co, we have, from
relation (I'v) above,
y = a sin [o>(f f 27r/o>) e] = a sin (o>ff 2ire),
120
PBOPERTJES OF MATTER
whence, y = a sin (wte), i.e., the same as be ore, in (z'v),
showing that the pos'tion and direction of motion of the particle is
the same as 2ir/a> seconds earlier, i.e., the particle repeat? its move*
ments after every 2ir/o> seconds. In other words, the timeperiod of the
particle is 2^/0). Further, since this value of the time period is quite
independent of a and e, it is clear that the vibrations or oscillations
of the particle are also isochronous.
The above results will also be true for angular S.H.M., if we
consider angular displacement, acceleration, velocity, etc., in place of
the linear ones.
Important Note. We have seen how the acceleration of a
particle executing a S<H.M. is given by
The general solution of this equation is of the form
y = a sin a> +b cos cut.
Thus, if the displacement of a vibrating particle be given by a
relation of the form y = ja sin cot+b cos cut, it is executing a Simple
Harmonic Motion. **
Clearly, as t takes up the values 0, 27r/o>, 47f/oj, 2n7r/co, etc., ajt
assumes the values 0, 2ir, 4?r, 2nrr, etc., with y assuming the same
value over and over again.
In other words, the timeperiod of the motion = 2irjaj t
Further, this equation can easily be reduced to the simple sine
form as follows :
Let
(Fig. 71).
Then, clearly, a = c cos Q = \/(a*+b*) cos
and b = c sin = i/(a*+K*).sinO.
Now, y = a sin cot+b cos wt.
So that, substituting the values of a and b,
obtained above, we have
s ; n w t cos Q+^(a^Wj.cos wt sin 0.
(sin wt cos 6 +cos cut sin 0).
sin (a>t+0),
which is the usual form of the displacement of a body executing
S.H.M.
Obviously, the displacement will have the maximum value
(a>t+0) = 90 and, therefore, sin (wt+0) 1.
And, since the maximum displacement of the particle is equal t
SIMPLE HARMONIC MOTIOJf
its amplitude, we Have
amplitude of the vibrating particle, here =
Again, the velocity of the particle is given by
V = .
so that, the maximum value o^v r= v /a 2 +fe 2 . v.
And, finally, acceleration of the particle = vy.
Hence, maximum value of acceleration of the particle
50. Composition of Two Simple Harmonic Motions. Just as a
particle may be subjected to two forces or two velocities simultane
ously, so also we may have a particle under the action of two simple
harmonic motions at the same time. Its final motion will then be the
resultant of the two simultaneous simple harmonic motions impressed
upon it. It does not, of course, mean that it will execute both the
motions simultaneously, any more than a particle, having two
velocities impressed upon it, w r ill move in both the directions at
the same time. All it moans is that its resulting motion would be
one as though it were simultaneously executing the two motions
together.
It should be clearly understood, however, that the simultaneous
execution of two rectilinear simple harmonic motions by a particle
is no guarantee that the resultant motion of the particle will neces
sarily be rectilinear or harmonic. Indeed, if their timeperiods be
incommensurable, it may not even answer to the definition of a
vibration.
We shall now take up first the simpler case of the composition
of a S.H.M. along one direction with a linear motion in a perpendi
cular direction and then pass on to the composition of simple harmonic
motions along the same straight line and at right angles to each other,
both graphically and analytically.
1. Graphical Method.
(/) Composition of a S.H.M. with a Uniform Linear Motion perpendicular
to it. The resultant motion will, in this case, be a sine curve. This may be
easily seen by attaching a small spike or style to the prong of a tuning fork (at
right angles to its length) and then drawing it* uniformly over a smoked plate of
glass, with the style just touching the plate, in a direction at right angles to that
of the vibrations of the fork. It will be found that a series of sine curves are
traced out on the plate, with the direction of motion of the fork as the hori
zontal or the timeaxis and the direction of vibration of the prong as the vertical
or the displacementaxis.
Composition of two linear simple harmonic motions in the same direction.
If two simple harmonic motions take place in the same direction, their resultant
is also a simple harmonic motion, defined by the resultant of the vectors which
define the two motions, this resultant vector being obtained by the ordinary law
of vector addition. This will be clear from the following :
*Qr, holding the fork in position and moving the platp. "
122
'KOFJEKTIES Olf MATTER
Let two simple harmonic motions, having the same timeperiod tiit
different amplitudes and phases, be represented by the projections of the vectors
,*.
J
Fig. 72.
respectively, on the axis of y, (Fig. 72), and let their equations of
equal to angles AOP
OP and O
motion be
y =* a sm cof and >> = b sin
where a and 6 are their amplitudes, and <o/ anJ
and AOQ respectively.
Then, if DEFG be the sinecurve for the first motion, and DffJK for the
second, we obtain the sinecurve DLMN for the resultant motion by adding up the
ordinates of the two curves at all points, because the displacements of the two arc
in the same direction and can be added up algebraically.
Now, the curve DLMTVis the same as would be obtained for the rotation
of the resultant vector OR, whose projection on the axis of y, therefore, gives
the resultant of the two motions.
For, RB = CB+RC PA+RC.
= OP. sin AOP+PR.sin CPR.
= a sin co//> sin (a>/f0),
because OP = a, PR = OQ = />, LAOP = w/ and LCPR = LAOQ ** w/4 $
Thus, the resultant motion is also a S H.M. and takes place along the same
line and, (since the rotating vectors OP, OQ and OR have the same velocities),
// has the same time*period as the two component motions.
The amplitude a' of the resultant motion is, clearly, equal to OR.
OR*~OP*+OQ*,20P.OQ.co S POQ.
0*}6 2 f 2ab.cos f,
Now,
Or,
whence,
Now, if ^ 0, i.e., if there be no phase difference between the two
motions, cos <f> = 1, and, therefore,
a
a'
algebraic sum of the amplitudes of the two component motions.
The phase angle of the resultant motion is, obviously, given by the angle
ROB, such that
JRj? CB+RC PA+RC
OB** OA+AB
tan ROB
SIMPLE SARMONIO MOTION
123
6r,
Now,
where LPOR = e , the phase angle by which the resultant motion is ahead of the
first motion,
tan ROB = tan

(at
sin of _j_b_sin cof cogjj f /> fas cof MM ^
a cos co/ f 6 cos to/ cos ^ b sin <*t sin <f>
Now, yiw/ <rt the start, t = 0, and .. wf = ; so that,
5/w w/ == 0, and c<?5 cuf = 1.
Hence
b sin <f>
tan e =  . r
a + bcos
~
Or, f =
_
b cos
Thus, the resultant motion is ahead of the first motion by a phase angle e,
. t b sin <f>
where e = tan 1 = ; . r*
a f b cos <f>
(//) Resolution of a S. H. M. into two components in the same direction.
The converse of the above is also tiue, viz. that a simple harmonic^ motion may
be resolved into two , by resolving its rotating vector into two vectors, in accordance
with the law of resolution of vectors, each vector defining a component simple
harmonic motion.
(Hi) Composition of two linear simple harmonic motions at right angles
to each other. The resultant of two, S.HM's, impressed simultaneously on a
particle, along directions at right angle* to each other, is a curve lying in the
plane containing the two motions and its character depends upon the ampli
tudes, timeperiods (or frequencies) and the phase difference of tee two compo
nent motions. Let us consider the different cases that arise.
(a) When the timeperiods (or frequencies) of the two motions and their
phases are the same, but their amplitudes are different. Let the two motions be
defined by the rotation of the
vectors in circles (/) and (//)
respectively, (Fig. 73), /<?., let
(i) and (//) be the circles of refe
rence of the two motions, with
radii equal to their amplitudes
respectively, say a and b.
Divide the two circles into a
number of equal parts in the
ratio of the frequencies of the
two motions, in this case, 1:1,
as shown, (each circle being
divided into eight equal parts,
for the sakg of convenience),
the starting point of the rotating
vector being marked zero. Then,
draw straight lines passing
through points bearing the
same numerals in the two circl
es, and parallel to the axes OX
and OY respectively, along
which the motions take place Fig. 73,
in the two cases*
124
PROPERTIES OF MATTER
Mark the points where these lines intersect and join all these points of
intersection. It will be found that, in this case, the straight line AB is obtained
as the path along which the
2 * resultant motion takes place,
the arrow heads indicating the
direction of motion about 0.
And, as will be readily seen,
this straight line is the diago
nal of the rectangle with sides
2a and 1b ; and the amplitude
of the resultant vibration of
the particle, i.e., OA or OB
is, therefore, clearly equal to
(b) When the timeperiods
or frequencies are the same,
plitudes are different and
phase difference is TT. Here, the
starting position ,(Fig. 74), of
the vector in circle 00 is at
the top of the circle, as shown, instead of at the bottom, (as in the first
case), the second motion being ahead of the first by a distance equal to half it*
path, the other numerals being shifted accordingly. Again, drawing straight
lines through the same numerals and parallel to the corresponding axes OA'and
Or, along which the two motions take place, and joining their points of inter
section, we get the straight line CD, inclined in the opposite direction, showing
that the resultant motion is again a straight line motion, about 0, but inclined the
other way, (i.e., the other diagonal of the rectangle of sides 2a and 2b), the direc
tion of motion of the particle being as indicated by the arrowheads.
74.
(c) When the timeperiods are the same, amplitudes different, and the phase
difference is 7t/4. We again proceed
exactly as above, with the only ;  ^^^~ ........ .
difference that, here, we shift the
zero, or the starting position of
the radius vector, by oneeighth of
its path in the case of the second
(lower) circle of reference, (Fig.
75), the second motion being
ahead of the first by r/4,
Joining smoothly the points
of intersection of the straight
lines through the same numerals,
parallel to the two axes respec
tively, we get an oblique ellipse as
the resultant path of motion of
the particle, the direction of mo
tion along it being indicated by
the arrowbead.
75.
S1MPLB HARMONIC MOTION
125
(d) When the timeperiods are the same, amplitudes different, andtha
phase difference is rr/2. In this
case, the starting point of the
radius vector in the second
circle of reference is taken a
quarter of its path ahead of its
original position, the phas
difference being rc/2.
Then, proceeding as
before, we get an ellipse as the
path of the resultant motion,
H ith its axes coincident with the
directions of the component mo
tions, the starting point being O
and the direction anticlockwise,
as shown in Fig. 76
76.
(e) When the I Ln > periods
or frequencies are the same, ampli
tudes different, and the phase diff
erence is 3*/2, Here, (Fig. 77),
0,8 the starting point of the radius
vcuo. in the second circle is
taken threefourths of its path
ahead of the original position.,
and we get, as in the last case,
an ellipse as the resultant path
the direction of motion being
clockwise, as shown, and the
starting point being 0.
Fig 77.
(/) When the timeperiods
or the frequencies are the same,
amplitudes equal, and the phase
difference is 3^/2. In this case,
(Fig. 78), we take both the
circles of reference of the same
radii and proceed as in case (c),
when a circle is obtained as the
path of the resultant motion, the
direction of travel along it
being anticlockwise and Jhe start
ing point being 0.
Fig. 78,
126
PBOPEETIBS OF MATTBB
(g) When the timeperiods
or frequencies are the same, am
plitudes equal and the phase diff
erence is 3rr/2 Here, again, we
take the radii of the two circles
to be the same, and proceed as in
case (d), when we obtain a circle,
as the path of resultant motion
the direction of travel along it
being clockwise, in this case, as
shown, (Fig. 79), and the starting
point being O.
Fig. 79.
(h) When the timeperiods
or frequencies are in the ratio of
2 : 1, amplitudes are different,
and the phase difference is zero.
In this case, (Fig. 80), we divide
the two circles into equal parts,
1/1 the ratij 2:1, (e.g., the first
one into 8 parts, and the second
one into four parts). Then, pro
ceeding as before, we get the
path of the resultant motion of
the form of the figure 8, as shown,
the direction of motion along
it being indicated by the airow
heads, and the starting point
being O.
Fig. 80.
0) When the timeperiods
or frequencies are f n the ratio 2:J,
amplitudes are different and there
is an initial phase difference
equal to a quarter of the smal
ler timepreiod. As in the case
above, we divide the two circles
here also into equal parts, in the
ratio 2: 1, but shift the zero of
the second circle, onefourth part
ahead. Then, proceeding as
before, we obtain, in this case, a
parabola as the resultant path
of motion of the particle, (Fig.
81), the direction of motion be
ing as indicated by the arrow
heads.
Ctf>
Fig 81.
N.B The epithet 'initial* has been deliberately used here with 'phase
difference* to emphasize that the timeperiods being different, the phase does not
remain constant, even though we start with the same phase originally. Inevi
tably, therefore, a phase difference comes in between the two motions,
ilotiOtf 127
Precisely in the same manner, we can obtain the path of the resultant
motion of a particle, subjected simultaneously to two simple harmonic motions,
perpendicular to each other, whatever their frequency ratio or the phase differ
ence between them.
The student may, as an exercise, try to determine the resultant path of
a particle, subjected to two simple harmonic motions, at right angles to each
PHASE DIFF
277
Fig. 82.
other, with the timeporiods and amplitudes equal, but with phase difference
changing from to 2n, when he will find that, as the phase difference changes
from to n, the resultant path changes from a straight line, inclined one way,
through an oblique ellipse inclined the same way, a circle, and, again, an ellipse,
inclined the other way, to finally, a straight line, inclined at right angles to the first
one, as shown in Fig. 82. And, as the phase difference changes from n to 2n,
*the same figures are repeated in the reverse order, as shown.
The superposition of such rectangular vibrations is of particular impor
tance in the subject of sound, since it serves as a test for the equality of the
periods of two vibrating bo Jies like tuning forks etc. The method was first
adopted by Lissajjus and aeace the various curves thus obtameJ, v/z., those in
Figs, 73 to 81 and others, are usually referred to as Lissajous' figures.
II. Analytical method.
(1) Composition of two linear simple harmonic motions along the
same line. Let two simple harmonic motions, having the same time
period^ be represented by the equations.
y l = a sin ait and y 2 = b sin (o>f+<),
where <f> is the phase angle by which the second motion is ahead of
the first.
The phase difference will throughout remain constant, because
the timeperiods of the two motions are the same.
Now, since the two displacements are along the same line, thd
resultant displacement y will, at any given instant, be equal to the
algebraic sum of the displacements of the two component vibrations.
Thus, y = Ji+JV
Or, y = a sin wt+b sin (wff <).
= a sin ajt^b sin a>t cos <f>\b cos a>t sin <f>.
Or, y = sin cot (a + b cos (f>)+cos wt.b sin $.
Putting (a+b cos </>) = a' cos e and b sin <f> = a' sin e, we have
y =a' sin ojt cos e+a' cos tot sin e.
Or, y = a' sin (a*t\~e)>
i.e., the resultant motion is also a S.H.M., along the same line > and
has the same timeperiod, its amplitude being a' , and its phase angte&e,
by which it is ahead of the first motion.
The values of a' and e may be deduced as follows :
We tove #' sin e b sin ^ and <?' cos e = (a^b cos ^),
128 PBOPBBTIES OF MATTER
So that, squaring and adding the two, we have
0' 2 sin* e+a' 2 cos 2 e b 2 sin 2 <f> + a 2 +b* cos 2 <f>+2 ab cos <f>.
Or, a' 2 (sin 2 e+cos 2 e = a 2 +b 2 (sin 2 (/>+cos 2 (f>)+2 ab cos <.
Or, a' 2 = a 2 + 2 +2ab cos <f>. ['.' sin*e+cos*e  1 ;
^ [_also,j/wV+c<wV = L
Or, a' = Vfl 8 +>T2flftciw"^
e <i' sw e b sin <f>
   1
. ,
And tan e =
 = ,, 
e a{b cos
Or, e = tan* ~*~ ^ . . [See page 123]
'
Now, (/) if < = 0, i.e., i/ fAe fwo motions be in the same phase, we
have e = ; so that, y = a' sin cut.
Or, the resultant motion will also be in phase with the two component
motions, with its amplitude given by
i.e., e^W(?/ to the sum of the amplitudes of the two component motions.
[See page 121.
And (ii) if < == TT, i e., i/ //ze ^v<? motions be in opposite phases,
we have again e 0, So that, again, y ~ a' sin cat.
Or, the resultant motion will be in phase with the first motion, with
its amplitude now given by
* = *** t v cos *  1
/ e., ##/ /o //ie difference between the amplitudes of the two motions.
Further, in this case, if a = &, '*.<?., //^Ae amplitudes of the two
component motions be also equal, we shall have a 1 ab = 0.
Or, the amplitude of the resultant motion will be zero. In other words,
there will be no resultant motion at all.
Note. In the above treatment, we have, for the sake of simplicity,
taken one motion a phase angle <f> ahead of the other.
The same result may be obtained, however, if we take the phase angles
of the two motions to be <f> l and <f> 2 respectively. For, in this case, we have
yi = a sin (u>t\$^ and >> a b sin
So that, y = y^y^ = a sin (wff ^) 6 sin
a sin w/ cos <f>ia cos at sin <^ t +b sin f cos fi+b cos wf sin ^ f .
= sin ut (a cos $ L + b cos <f>t)+cos cor (a sin $ L +b sin ^ a ).
Now, putting (a cos <f>i+b cos fi t ) = a' cos e and (a sin fa+b sin 8 ) = a' sin e $
we have y = a' sin co/ cos e f a' cos co/ sin e.
Or, y == a' .y/ ( w /f <?), as before ;
/ r., the resultant motion, is also simple harmonic, with the same timeperiod as
of the two component motions.
And,
_.
a' cos e a cos <f>i f b cos ^ g
Again, the values of tne amplitude a 1 of the resulting motion maybe
obtained in exactly the same manner, as before.
Thus, a' z sin*e+a' 2 cos z e = (a sin <f>i+b sin a ) 2 f (a cos fa+b cos &)*.
Or, w*(sin*e+cos*e) a 2 sin* h+b* sin 2 fa+lab sin fa sin fa.
ia* cos 2 <f> l + 6 2 cos 2 fa+2ab cos fa cos fa.
Or, <?' 2 * 2 (w 2 ^^ f co^Vt) f6 2 (w 2 ^ifco5 8 fa)
f 2a6(5/# ^ 5/w fa i cos fa cos fa),
129
Or, a' 1 = a*+b*+2ab cos (&0,), whence, a'  *Jf+b*+1ab cos (<f>if*\
which, when (<j>i~<f>i) = 0, gives the same result, as above, (page 80).
Proceeding again, as before, if, (/) ^ =* ^ a or (0i &) = 0, i.e., if the
two motions be in the same phase, we have
a'  (+&) ;
(//) if 01 & = w, j>., ;///ie /wo motions be in opposite phases, we have
a' = (fl ;
and, further, if a = b t i.e., if the amplitudes of the two motions be also equal,
we nave a' = ; i.e., we obtain the same results as above.
(2) Composition of two linear simple harmonic motions at right
angles to each other. Let the two simple harmonic motions be along
the axes of coordinates 'XOX' and YOY' t and let a and b be their
amplitudes respectively and <f>, the phase difference between them.*
Then, if their displacements at any instant t be x and y,
we have x = a sin cot y ...(/) and y = b sin (cut +</>)> (")
sin ojt = x/a ; [from (/), above.
and, since sin* cut + cos* a>t = 1, we have cos* a)t = 1 sin* wt.
/ x*\\
Or, cos* wt = lx*/a*, so that, cos cot == f 1  ^J '
Now, y = fc(.sm cu/ cos <f>+cos cot sin <f>), [from (//), above
Or, y\b = sin ojt cos $ f cos cot sin </>.
Or A c
So that, squaring both sides, we have
v a jc a 2jcv
Or ' i+^ COS * + ~ ab COS
^. r
Or,
ordinates, and may be inscribed in a rectatnglp of sides 2a and
Now, a number of special cases arise :
(a) When <f> = o, i.e. 9 when there is no phase difference between
the two motions. In this case, sin <f> = 0, andc05^=== 1. . ' ' 3. ^
So that, substituting these values in relation (///) above;
y* x* 2xy
'  .<Mh
This is the equation to an ellipse, inclined to th axes of tlj ~ ^"
130
FBOPtRTIE3 Of
0. Or, t
Or,
y_
b
X
a
Or,
0.
JL
a
This is the equation to a straight line, passing through the origin, such
that it meets the axis XX' at an angle
tan 1 b/a, [see case I, (ill), (a)].
The resultant motion, is, therefore,
along the straight line AB, (Fig. 83),
i.e., the particle describes a S.H.M.
along this line, with the same timeperiod
as that of the two component motions and
an amplitude equal to <\/a*+b 2  If the
amplitudes of the two motions be equal,
the straight line AB is inclined at an
angle of 45 to the axes of x and y.
(b) When <j> = TT, i.e., when the phase difference between the
two motions is JT. Here, sin < = 0, and cos <f> = 1. So that,
from (iii) above, we have
y + *.
b ^ a
This too is an equation to a straight line, passing through the
origin but inclined to the xaxis at an angle tan^ bja ; so that, the
resultant motion is again a S.H.M. , with the same t'meperiod but along
the straight line CD, (Fig 83), inclined the other way, [see case I,
(iff), (b)], the amplitude being again ^/ a *+b*.
(c) When <j> = ?r/2, i.e., when the phase difference between the
two motions is Tr/2.
Here, sin <f> = 1 and cos <j> = 0.
Substituting these values of sin </> and cos < in relation (ill), above,
=0. Or,
3
x
A
a
we have
1.
This is the equation to an ellipse*
whose major and minor axes coincide
with tht directions of the two given
motions, and whose semiaxes are
equal to b and a respectively. The
resultant path is, therefore , an ellipse,
(Fig, 84), which it describes once in
the timeperiod of each component
S.HM. t [see case I, (///), (d) 9 above].
The direction of motion of the
particle along the ellipse may be
determined aa follows :
Since ^ ir/2, and x = a sin <vt, Fig. *4.
we have y b sin (ut+j) bsin M+ir/2), whence, y *
b cos *t,
SIMPLE HARMOKIO MOTION 131
A differentiating y with respect to f, we have
velocity of the particle = dyjdt = u.bsin wt.
Now, x and, therefore, sin wt 9 is positive in the right half of the
figure. And, therefore, the negative sign of dy/dt means that the
velocity of y is negative, i.e., it is directed downwards in the right half
of the figure In other words, the direction of the particle along the
ellipse is clockwise.
v* x*
If, on the other hand, <f> = Tr/2, we have, again, , ^ f a = *
i.e., the resultant motion is again an ellipse.
But, since y=b sin (cot IT} 2)= b cos a>t 9 we have dy/dt = a).b sin cot,
i.e., the velocity of y is now positive in the right half of the figure, and
is, therefore, directed upwards. In other words, the direction of motion
of the particle along the ellipse is now anticlockwise.
(d) When <j> = 7r/2, and b = a, i.e , when the phase difference is
7T/2 and the amplitudes are equal. In this case, obviously,
sin <f> = sin ir/2 = 1 , and cos <f> = cos Tr/2 = 0. \f
/. substituting these values in relation (Hi)
above, we have
 4 ~ A = . \Jl' _ BC 1 ' v t )
Or OP fl* \
whence,
This is the equation to a circle, whose
radius is equal to the amplitude of either Fig. 85.
simple harmonic motion ; so that, in this case, the particle describes a
circle, (Fig. 85), once in the same time as that taken by any one of the
two component motions, [see easel, (///), (g), above] the direction of
motion along the circle being determined as explained above, in the
case of the ellfpse.
A uniform circular motion may thus be regarded as a combination
of two equal or similar simple harmonic motions, at right angles to each
other, and differing in phase by ?r/2.
(e) When <$> = Tr/4, i.e., when the phase difference between the
two motions is ?r/4.
Here, cos <f> = cos~r = ^ ; and so also, sin $ ** 5.
4 \/ v *
Hence, from relation (Hi) above, we have
j x 1 2xy 1 1 _ y 1 x 1 \/2xy 1
which Js the equation to an oblique ellipse. (See case I (///), (c) above.)
So that, the resultant motion, in this case, is an oblique ellipse.
Thus, we see ihat the two perpendicular linear simple harmonic
motions compound into a straight line motion, when they differ in phase
by or ir, and into an ellipse or a circle, when the phase difference
132 PBOPBKTIES OJT MATTJCB
For any other phase difference, the motion is still an
with its major and minor axes no longer coinciding with the direc
tions of the two component motions, but being inclined to them.
(/) When the amplitudes are different and the timeperiods or
frequencies are nearly equal: In the cases, dealt with above, where
the timeperiods of the two component vibrations are identical, the
elliptical paths of the resultant vibrations remain fixed. But if the
two timeperiods differ slightly from each other, there comes about a
gradual but progressive change in the relative phase (^) of the two
vibrations and the elliptical path consequently undergoes a corres
ponding cycle of changes, whose frequency is equal to the difference
between the frequencies of the two component vibrations.
Thus, (i) when < = 0, the ellipse coincides with one diagonal of
the rectangle of sides la and %b, within which the ellipse lies ; for, here,

a b
(ii) When <f> increases from to w/2, the ellipse opens out to the
x* v 8
form  2 + 73= 1, passing through intermediate obliq ue positions.,
And, if a = 6, the ellipse is reduced to the form of a circle.
(Hi) When </> increases from ir/2 to TT, the ellipse closes up again
and finally coincides with the other diagonal of the rectangle ; for now
And, the same changes take place, in the reverse order, when (f> in
creases from TT to 2?r, the ellipse ultimately coinciding with the same
first diagonal as in Case (/), all these changes being shown in Fig, 82,
above.
(g) When the frequencies are in the ratio 2 : I, or the periods
arc in the ratio, 1 : 2, and the amplitudes are different. In this case,
the angular velocity of the particle in the circle of reference of
one will be double of that of the particle in the circle of reference of
the other.
/. if the two motions be represented by
x = a sin wt and y = b sin (2wt\ </>),
where <f> is the phase angle by which the second motion is ahead of
the first, we have
xja sin wt, and .*. cos cut = \/lsin* wt ;
and y\b = sin (2^f <^) == sin 2a>t cos $+cos 2wt sin <f>.
Or, y\b s= 2 sin wt cos <*>t cos ^+(12 sin 2 a*t) sin <f>.
[v sin 2a>t = 2 sin a>t.cos wt and cos 2wt = (12 sin* wt)].
.'. substituting the values of sin wt and cos wt from above, we have
JC* \i / X^ \
I t jcos<f>+( 1 2~2Jsw ^.
I i .pj* cos f^.$in ^ ~ fin ^
SIMPLE HARMONIC MOTION 133
Or, */ t+^sin +  1* cos +.
Or,
Or, 
^ r> \ h ~~ s $ ) + r~ s * n * ^+2(r sin <f> Y j sin ^
= COS^ <f> : COS <f>.
(y . \' 4jc^ 4jc 4 4x^ / v \
y sin <b ) L  sin^ d>\~~ cos^ <A [ 7 sin }sin m
b J fl* " o fl \ b /
Or, ( r sin <b ) j ^(sin^ <b~\cos^ <f>)~\~ / ,5/w ^ >  .sin tp
\ t? / at) a
A y2
~ a 2 T*
Or,  ,m ^+ + 5/ # _(,/ #+W5 # = 0.
Or, (^,/ ^ + ^ 4 +^ (f rt, *_l) . 0.
Or, rih +   8 +rfi, *l = 0. . .(A)
This is the general equation for a curve having two loops, for any
values of phase difference and amplitude.
Let us now take some particular cases :
(/) If the phase difference, i.e., </>, be eqnal to o or ir.
i;2 4.^2 / x% \
Here, sin </> = 0, and, therefore, ~ + { 1 ) = 0,
r b 2 ' a 2 \ a* /
which is the equation to the figure ofS. [See case I (///), (h), above.
(//) If the phase angle </> == ir/2, i.e., sin <f> 1,
a 
134 FBOFJCBT1BS OF MATTER
This represents two coincident parabolas, each having the equation
o r , **._(_,) Ot , x > = ( y _ b
If, however, the frequencies differ slightly from the ratio 2:1,
(i.e., the timeperiods differ from 1 : 2), the variation in the resultant
path of the particle may be obtained by substituting the consequent
changes in the value of $ in the general relation above. The changes
occurring when < changes from to TT and then from IT to 2ir are
shown in Fig. 82, above.
Note. Alternatively, the student may, without deducing the general
equation (///), [50, (2), page 129], obtain the resultant motions in simple casei
as follows :
Taking displacements of the two S.H.M's, at right angles to each other,
as x = a sin cot, and y = b sin(ojt + fi),
where $ is the phase angle by which the second motion is ahead of the first we
have the following cases :
(/) When <f> o. We have x = a sin co/ and y b sin /.
So that, x\a = sin co/ = y/b, whence, yjx = b\a, [See 5 J, II, (2), (a).
which is the equation to a straight line, passing through the origin ; and inclined
to the *axis at an angle, tan 1 b/a (straight line AB, in Fig. 83>; /.*., the resultant
motion, here, is along the straight line AB.
(//) When # = n. We have x = a sin wf and y * b sin w/.
S . that : *1<* = sin co/ = y/b, whence, y/x = />/, [See ^ 50, II, (2), (fe).
which is, again, the equation to a straight line, inclined to the xaxis at an
angle, ta/riA/0, (/. e . f straight line CD, in Fig. 83), at right angles to that in
case (i). The re suit ant mot ion is thus along a straight line, at right angles to that
in the first case.
(Hi) When <f> = w/2. We have x a 5/11 co/ and ^ = 6 c<?5 w/.
So that, 5m co/ == x/a and cc?5 o/ yjb.
And " ^/o 2 + y*/b* (sm 1 co/ f ow* a>/) = 1,
Or ' Wit*) f (^ s /fl s ) = 1 , [See 50, II, (2), (c)
which is the equation to an ellipse, with its major and minor axes coinciding
with the directions of the two given perpendicular motions, and whose semiaxes
are equal to b and a (Fig. 84). The resultant motion is thus an ellipse here,' describ
ed once in the timeperiod of each component motion.
(iv) When ^ w/l and b = a, i.e., the amplitudes are also equal.
x a <j J/H co/ and >> 6 co5 to/ o coi w/.
So that, sin co/ and cos w/
a
x
And, ^r+~ fl ? * */n f <*t + cos* to/  1. Or, y*+x* o 2 , [See 50, II, (2), (d).
which is the equation to a circle, with a radius equal to the amplitude of either
of the two motions. The resultant motion, in this case, therefore^ is $ drch, d$$
in ffo timtywfad of each component motion,
S1MPLB HAKMOHIO
135
51. Composition of two equal circular motions in opposite direc
tions. Let twopwticles P, and P a move with equal velocities along
the same circle XYX'Y' of radus 2a, in
opposite directions, as shown, (Fig. 86),
such that when P l passes the point X,
P a passes X'.
Let the positions of P 3 and P 9 be
as phowi , at any given instant t y after
starting from X and X' respectively ; so
that, XOP^X'OP^t = <of, where o> is
the angular velocity of P! and P t .
Now, we know that a circular motion
is equivalent to two equal linear simple har
monic motions with a phas? difference ir/2
and along perpendicular directions to each
other, (sae pige 131).
The circular motions of both P, and P, may, therefore, be re
solved along perpendicular directions XX' and YY'. Then, the dis
placements x l and JC 2 , of P, and P.,, will, at the given instant, be
equal in magnitude but opposite in direction along XOX' and will,
therefore, cancel each other out, but their displacements y l and y t
along YOY' will be equal and in the same direction, so that the
resultant displacement is given by y^y^y^ along YOY'.
Since y l = y t a sin $ = a sin wt, we have
y = y l f y 9 = a sin <ot f a sin wt =* 2a sin wt.
Or, the resultant motion is a linear simple harmonic motion along the
diameter YOY', at the extremities of which the particles P l and P t
cross each other as they describe their circular motions.
And clearly, the amplitude of the resultant motion is 2a, and its
timeperiod the name as that of the two constituent circular motions.
52. Energy of a Particle in Simple Harmonic Motion. The
acceleration of a particle, executing a S.H.M , is. as we know, direc
ted towards its equilibrium position, or in a direction opposite to that
in which y, the displacement of the particle, increases. Hence, work
is done during its displacement, or the particle has potential energy.
Also, the particle possesses velocity and, therefore, has kinetic energy.
Thus, it has both potential as well as kinetic energy, or its energy
is partly potential and partly kinetic. And, if there be no dissipative
force at work, i e., if the energy is not dissipated away in any way,
the sum total of the two remains constant, although as the displace
ment increases and the velocity decreases, the potential energy in
creases and t!ie kinetic energy decreases.
Now, when the particle has its maximum displacement, positive or
negative, its velocity is zero and, therefore, its kinetic energy is then
eero ; so that, in this position, the whole of its energy is present in the
form of potential energy. And, when the particle is in the equilibrium
position, its displacement is zero and its velocity, maximum ; so that,
tfre who'e of its energy is MW present in the form of kinetic
If m be the ma?s of the particle ; a, its amplitude and 27r/w, its
timeperiod (T), i.e , if o> be the angular velocity of the rotating
vector, or that of the particle in the circle of reference, we have
velocity of the particle in its equilibrium posit ion =aa), a maximum.
And .% its kinetic energy =*\m. (aw) z =wa 2 w 2 ,
and its potential energy=Q, [ v its displacement ii zero.
Or, its total energy = J/w7 2 co 2 f 0=wa 2 oA
In other words, the whole of its energy, here, is present in the
kinetic form. Similarly, when the particle has its maximum dis
placement, the whole of its energy is in the potential form, which,
therefore, is also equal to J. tna 2 w* 9 in this position of the particle,
For any other position of the particle, its displacement is given
by y=asinwt.
A.nd /. its velocity is given by dy\dt =aaj.cos a>t.
Hence, its kinetic energy = \m.(aa) cos cot) 2 =%m.a 2 a)*.cos 2 wt.
4.nd, since its total energy = fynato 2 , we have
its potential energy s ^ma^^^ma
Alternatively, we may proceed as follows :
We have, acceleration of the particle, executing a S. H. A/., given by
<py[dt* =  coV
*. if m be the mass of the particle, ths force F required to maintain this dis
>lacement y is equal to m wV
Knd, therefore, work done by the force for a small displacement dy is equal to
Now, this work is also a measure of the potential energy of the particle
it this di8placement.
'. P. E. of the particle for a displacement dy ** F.dy = mu>*y.dy.
Hence, total work done for displacement y and, therefore, total P. E. of
^he particle for a displacement y is given by
f> fr
I mco'./.dfy moj 8 I
y.dy.
3r, P. E. of the particle for a displacement y Jwco 1 .^ 1 ,
r., Potential Energy oc jv f .
Thus the P. E. of a particle, executing a S.H.M. is, at any given instant,
lirectly proportional to the square of its displacement from its mean or equili
>rmm position, at that instant.
v, velocity of the particle at displacement y
,  (a sin to/) * oc.) cos w/.
at
Vnd, /. K. E. of the particle = \m(a<* cos tot)*bma*c**.cos*<t.
ience tolal energy of the particle=//j/?0te//0/ energy f its kinetic energy.
And, since to = 2?r/r = 2?r, where Tis the timeperiod of the
particle and w, the frequency of its vibration, we may also say that
total energy of the particle = \m ( } fl=
H1BMOHIO MOTIOW 137
Now, since in any conservative system the sum total of the
kinetic and potential energies of the system must be a const ant f it is
clear that the former can only increase at the expense of the latter,
and, therefore, attains its maximum value, when the latter is reduced
to its minimum value or zero, and vice versa. Thus, the maximum
value of any one of the two forms of energy measures the total
energy of the system, (see page 136).
53. Average Kinetic and Potential Energies of a Particle in
S.H.M. We have seen above that, at any given instant, the P.E. of
a particle is equal to }mu*w* sitfojt and its K.E. equal to m 2 o 2 cos z wt.
Now, the mean or average value of both sin'wt and cos^ojt for a whole
cycle, (from to 2ir), i.e., for a whole timeperiod is equal to i*, and,
therefore, the mean or average K.E. of the particle over the whole of
its period of vibration is equal to its mean or average P.E. over the
whole period, each being equal to J x i wa 2 co* = wtf 2 a> 2 . Thus,
average K.E of the particle = its average P.E. =*
And, .. total energy of the particle = 2 x Jwa 2 o> 2 = Jma 2 co 2 .
We may express this by saying that the energy of a particle,
executing a S.H M. 9 is. on the average, half kinetic and half potential
inform, the whole being present in the kinetic form at its mean or
equilibrium position, and in the potential form at its extreme posi
tion, on either side.
The results obtained above for linear S.H.M. are equally valid
for angular S.H.M. Only, the linear displacement x or y of the
particle or the body, and its mass in, are replaced by their rotational
analogues, viz., the angular displacement 6 and its moment of inertia
/ about its axis of rotation, respectively.
SOLVED EXAMPLES
1. A quantity of gas is enclosed in a cylinder, fitted with a smmooth heavy
piston. The axis of the cylinder is vertical. The piston is thrust downwards to
compress the gas, and then let go. Is the ensuing motion of the piston as S.H.M. ?
If so, what is its timeperiod ?
Let original volume of the gas be = V and its pressure = P.
Let a be the area of crosssection of the piston, (and cylinder). Then, if
*This is so, because the mean value of sin* / for a whole cycle from
"
sin*<*t.dt
*=
to 2 i. given by  _ P *=***
J
j
J
JO
r i 2 *
<*tl1sin 2o>//4
~ ^
dot.
" T
the pa$c with wV *9f * w^pjc timepcri0d f
138 PBOPBETIE3 OF MATTBB
the piston be displaced through a distance *, (Fig. 87), the change in volume
produced is given by x.a, the correspjnding change in pressure
being p. By Boyle's law, therefore,
PV  (P+p)x(Vx.a)  PVPx.a+pVp.x.a.
Or, = P.x.a+pVp.x.a.
Neglecting p x.a as the product of very small quanti
ties, compared with the other terms in the expression, we
have
p ft
P.x.a \pV. Or, pV P.x.a, whence, p y.x.
Now, the restoring force on the piston^ which is equal
to the disturbing force, is obviously equal to change in pres
Fig. 87. sure into area of crosssection of the piston p.a.
P.a Pa*
p.a. y .x.a  p.x.
Since, acceleration = force/mass, the acceleration of the piston * p.ajm.
__ ^ flf8 . ^ ^ fl2 f substituting the value of
Jr.
t
F /w  Vm ' ' L Pa> from above.
Or, accel<ra'ion of the piston * A*.x,
where P a*\Vm * t*> a constant of proportionality, which is equal to the acceleration
of the pistonper uwt displacement, (i.e., when x = 1).
Or, acceleration of the piston is proportional to x, its displacement.
Hence, the motion of the piston is a 5. H. M.
And .'. its timeperiod, T 2n \f 1
acceleration per unit displacement
r.
2. A body describing a simple harmonic motion executes 100 complete
vibrations per minute, and its speed at its mean position is 15 ft. per second. What
is the length of its path ? What is its velocity when is its half way between its mean
position and an extremity of its path ?
Here, timeperiod T of the body 1/100 mt. = 60/100 = '6 sees.
and velocity of body at is mean position = IS ft. I sec.
Since velocity of a body executing a S.H.M. = aca, at it* mean "* position,
where a is its amplitude and to, its angular velocity, we have
flo> = 15 Or, a.lr.lT 15, [ v <o 2rr/r.
Wh encc, ._._. or, .  1432 A
Now, /e/t#//j of path of the body = /we? 1/5 amplitude, because it goes the
same distance on either side of its mean position.
Hince, length of path = 2* 1'432 = 2'864/r.
Again, velocity of a body at a displacement y is given by
v = 6>\/a* >> 2 .
Here, displacement of the bodv its amplitude/2.
Or, >>  a/2  1432  716//.
So that, v  ~ V(f : 432) 2 ^716)^ = \/(T432 1 716)(i'432'71Q.
 V/2T481T71T. 1299 /r./^c.
Thus, the length of path of the body is 2 864 /Jr., and its velocity when it is
half way between its mean position and an extremity of its path, is 1299 ft. I sec.
3. If the earth were a homogeneous sphere, and a straight hole were
bored in it through its centre, show that a body dropped into the hole will execute
a S.H.M. , and calculate the timeperiod of its vibration. [Radius of the eartl*
4009 miles, aqd value of f op its surface  32 ft. per sec. per $ec.J
SIMPLE HABMONIC MOTION
139
We know that the force with which a body is attracted by the earth to
wards its centre is equal to the weight of the body, (m^), and also equal to
G.m.MIR*, where m is the mass of the body ; M, the mass of the earth ; JR, the
radius of the earth ; g, the acceleration due to gravity on the surface of the
earth and (7, the gravitational constant.
nig G.m. MIR 9 . Or, g  G.MJR*.
Since the earth is supposed to be a sphere, its volume 4* 8 /3, and, there
fore, if A be its density, we have
its mass, M * 4* R*.&/3. So that, * ~ . ^ ;~. G ** 4.7t/?A.C7/3. . .(0
If the value of acceleration due to gravity at a distance r below the surface
of the earth, (Fig. 88), be g' t we have, as above,
Dividing (//) by (i), we have
8'lg l.*.(Rr)&.Gl.*.R.&.G.
Or, g'Ig=(R~r)lR.
Or, # ' = g(Rr)IR  (Rr).glR.
Thus, the acceleration of a body at a distance
(# r) from the centre of the earth is equal to
(Rr).glR ; and since g/R is constant, /A/5 acceleration
is proportional to (/ r), //ie displacement of the body
from the centre O of the earth. The body, therefore,
executes a S.H.M., and its timeperiod is given by_ _
"
and
Now,
IT//?
R 4000 m//e5  4000 x 1 760 x 3 //.
^ =* 32 ft. I sec 1 .
substituting the values of /? and ^ in relation (111) above, we have
2 A/
V
32
5105 sec*.
. 2ff A / 12^176x107
V 32
8507 m/roife*
Thus, the timeperiod of vibration of the body would be 85'07 minutes.
4. If a body executes a simple harmonic vibration in time TI, under one
constraining force, and in time T 2 , under another, what will be its timeperiod under
both forces together ?
Let rrass of the body be m,
and let its acceleration under the first force F Y be a l$
,* second F, a t
and M ., f , both the foices Fi+F t bt a.
Then, clearly,
F,/m ; fl t F f /wi ; a
j '.'
L
arc.
Also, the ac:eleration of a body executing a S.H.M. is proportional to
its displacement x from the equilibrium position, i.e., acceleration a oc jc.
where /* t , ^, and /* are the constants of proportionality in the three caset
respectively.
Again^ because the timeperiod of a body executing a S.H.M. is given by
T 2n Vf/^ where A* is a constant of proportionality, we have
So that,
and
where Til the timeperiod of the body under both the forces acting together
140 PROPERTIES OF MATTEL
Since acceleration a, under both the forces acting together, will ob
viously be equal to the sum of the accelerations under the individual forces, we.
have a = fli+fl,
And, therefore, ( *.)'. , = (* )'. * + (  )'. x
Or, dividing by (2z)*.x throughout, we get
Or J ___ 1 , i _ JVIV . O 'r r r ' 2 ^l
or, t  fMT  ur '
or r = \ I ^ ^ 8  r, r. A / ~ '
u ' V TV+r,'" l ' V v+r,' '
Thus, the timeperiod of the body under both the forces together will be
5. A test tube of weight 6 gins and of external diameter 2 cms. is floated
vertically in water by placing 10 gins of mercury at the bottom of the tube. The
tube is depressed by a small amount and then released Find the time of oscillation.
(Oxford and Cambridge Higher School Certificate).
Here, mass of the tube and mercury = ' f 6 = 16 gms.
and external radius of the tube =2/2 = 1 cm.
area of crosssection of the tube =7rr 2 = rr.l 2 n sq. cms.
Let the tube be depressed through a distance x cms.
Then, volume of the water displaced = KXX = TCX c.cs. f Taking density of
and weight ,, ,, ,, ,, = :rx.l. = x.x gms.wt. {_ water = 1 gm.jc.c.
Therefore, upward thrust experienced by the tube is equal to the weight
of the water displaced, i e., TT.X g dynes.
Hence acceleration of the tube = ^f == ^ x. \'.'acc. =* '.
16 16 L mass
Since ^r^/16 is a constant, say, /*, we have
acceleration of the tube Px ; i.e., acceleration of the tube is proportional to its
displacement ; and, therefore, it executes a S.HM.
Hence, its timeperiod is given by / = 2?c \i
. 2, \/~T = 2rt A/I? = 2 A/ '
V ng\6 V * V ^
Or, //re //me of oscillation of the tube = '4527 sec.
6. A particle executing a S H.M. has a maximum displacement of 4 cms.
and its acceleration at a distance of 1 cm from its mean position is 3 cm /sec 2 .
What will its velocity be when it is at a distance of 2 cms from its mean position ?
Here, amplitude of the particle 4 cms.
and its acceleration, when its displacement is 1 cm., is equal to 3 cms. /sec*.
Now, acceleration  x./ 2 . Or, 3 = 1 to 8 ,
where x is the displacement and co, the angular velocity of the particle.
Or, <o* * 3, whence, co = V3 radians/sec.
.Now, the velocity of a particle executing a S.H.M. is given by
v =
where a is the amplitude and x, the displacement of the particle.
.*. when the displacement of the particle, i.e., x ** 2 cms., we have
v  1(3 . v/4^2 2 * ^3 Vlo^ = ^ 3 A/12.
*. the velocity of tbe particle at a distance of 2 pm^. frorn it mean positipr
Will
MOflOU
7. A vertical Utube of uniform crosssection contains water up to a height
of 3D cm?. Show that if the water on one side is depressed and then released, its
motion up and down the two sides of the tube is simple harmonic, and calculate its
tinuk nprinH \Uelnl. J"4/)
Let 'AA', (Fig. 89), represent the level of the water in the two limbs of
the Utube, to start with, and let the column on the left be depressed up to 0,
through a distance x cms. Then, the column on the right
will naturally go up, say to the level C, such that the diffe
rence of levels in the two limbs is now, B'C, where B' is at
the same level as B.
The weight of this column of length 2* will now act
on the mass of the water in the tube, as a result of which it
will oscillate up and d :>wn.
Now, obviously, the weight of this column of water
its massxg = its volume x its density xg.
= ax2xxl xg dynes.
Or, fora acting on the mass of water in the tube
= 2.x.a.g. dynes.
And, mass of water in the tube (in both its limbs)
=2x30xaxl = 60a gms.
.*. acceleration produced in the mass of water in the tube
six;
B
Fig. 89.
'30 ***
Where g P ""*' a constant
2jca.g
** 60a~ ~ 30
Or, acceleration is proportional to x, the displacement of the water column.
Hence, the motion of the water column ft simple harmonic, and its time
period is given by
t
'V?
T ' Or ' * 
The water in the Utube will thus oscillate with a
09S,
timeperiod equal to
8. Show that the timeperiod for the swing of a magnet in the earth's
field Is given by t = \/i /MH, where M is the magnetic moment of the magnet, I,
its moment of inertia about the axis of suspension and H, the earth's field.
Lei a magnet NS, of pole strength m, be suspended so as to make an
angle with the earth's field H, v Fig. 90 >.
Then, clearly, the forces acting on its two poles are
mH and mH, as shown. These two forces being equal,
opposite and parallel, constitute a couple, whose moment is
equal to the product of one of the forces and the perpendi
cular distance between them.
So that, couple C, acting on the magnet mHxST.
 mHx NS sin , [v ST  NS sin a.
Or, C*=MHsin*,
m x NS M , the magnetic moment of the magnet.
If a be small, we have sin a = (in radian measure).
So that, C =MH.*.
Since the magnet is in equilibrium, this must be
balanced by the restoring couple set up in the suspension
i.e , by I.d<*ldt, where dv>ldt is the angular acceleration of
the magnet and /, i is moment of inertia about the axis of
suspension So that, Ldu/dt *= A///.a,
Or, dtafdt <M#//M = /*., where /* MH\1, a constant,
Or, d<*\dt ccj*s\ .'. 
*H should be noted that the timeperiod is the same as tliat of a simple
pendulum of the same length * s the height of the water column* i.e., of length
equal to 30 cms.
142 PBOFJBBTIXS Of MATTE*
i.e., the acceleration of the magnet Is proportional to its displacement. The motioh
of the magnet is, therefore, simple harmonic and its timeperiod is given by
9. A particle is moving in a straight line with simple harmonic motion.
Its velocity has the values 5 ft./sec. and 4 ft/sec. $ when its distances from the
centrepoint of its motion are 2 ft. and 3 ft. respectively. Find the length of its
path, the frequency of its oscillation, and the phase of its motion, when it is at a
distance of 2 ft. from the centre. (London Higher School Certificate)
Length of path. We know that v o
So that, in theory/ case, when v=5//./s<?c , and x~2//., we have 5=yV 2*.
Or, 25  o'.(a 2 4). ...(/)
And, in the second case , when v = 4 //. ] 'sec. , and jc=*=3//., we have 4<o y/ fl aZ3.
Or, 16 8 V9). ..(//)
/. dividing (/) by (//), we have **  *. Or, 25(a a 9)~16(a 2 4).
Or, 25a 3 225  16a'64. Or, 25a a ~16a* ~ 22564.
Or. 9a'=161, Or, '  ^
And/.  <yii423/ir.
Or, the amplitude of the particle is 4'23 ft.
Sioce the length of the path traversed by the particle is twee the ampli
tude, (as it travels equal distances on either side), we have
length of path of the particle 2.a = 2x4'23 = 8'46//.
Frequency. The frequency of the particle, n I//, where t is its time
period ; and since / = 2*/w, we havo
"te/i^r'  ( ' v/)
The value of w may be obtained by substituting the value of a io either
(/) or (//). Thus, we have from (/),
t ^35 5 5
1 2)(4 232 \/6 23 x 2'23
And, therefore, substituting the value of w in (///) above* we haro
n *ir??~** ....... sy "* 0*2135.
2" \/6 23x2'23
Thus, the/rtfgtteflcj' of the particle is 2135.
Muue angl.. We h.vc the relation, , a .
Here, x 2/r. and fl 4'23/f. /w^ and r, thi
2  423 sin $. Or gin 6   '4729.
4729 * 28' 13'.
Hence, the phase of the motion of the particle, when its distance is 2ft.
from the centre, is 28 13'.
10. A light elastic string is suspended vertically from a point and carries A
heavy mass at its lower free end, which stretches it through distance / cms. Show
that the vertical oscillations of the system are simple harmonic in nature, and of a
timepwiod equal to that of a simple pendulum of length / cms.
SMPtB HARMONIC MOTION
143
Let original length of the string ,45, (Fig. 91), be  L cms , and lot the
mass attached to its lower end be mg dynes.
Then, the downward force acting on it =* mg dynes.
And, if T be the tension of the siring (upwards), we have
T mg, because the string is in equilibrium.
Now, Young's modulus for the string, i.e., Y . 
Or, stress Yx strain.
Obviously, stress = T[a t and strain =*//,
where a is the area of crosssection of the string.
So that, ~^ Y.4 . Or, T K. H
2 J j i*
Since m^ = T, we have m^ = ' ./,
JL
Ar m^ 7a i.*
Or. "/" *" (/)
If the string be pulled down a little through a distance x, the tension
in the string acting upwards will, clearly, ba
= . (/ix) ~.(l+x). [See (/) above. '
And, since the downward force acting on the string ~ mg t tht resultant upward
force acting on the string will now be
_ , , f mg(l+x)mgl mg.x
Or, retultant upward force =*  .  * =  
Now, acceleration  = ~~ = ^.x, where ^// = /*, a constant.
Or, acceleration oc displacement.
.*, the oscillations are simple harmonic in nature ; and since the time period of a
body executing a . //. M. is given by / = 2^^^^ we have
MJ/ ^/ a simple pendulum of length 1 cms.
11. A particle moving in a straight line with simple harmonic motion, of
period IT/CO, about a fUed point O, has a velocity ^3 6o>, when at a distance b from
0. Show that its amplitude is 26 and that it will cover the rest of its distance in
time ?t/3c>>.
We know that the velocity of a particle executing a S.H.M., at a dis
placement y> is given by <*\/a*^y* 9 where a is Us amplitude.
Here, displacement of the particle is b, and its velocity at this displace*
ment is \3.o>.
i3.ta>  CD / a *31 Or,
Or, 36' o 1 ^, whence, a 1 * 46*.
Or, * * 26, i.e., the amplitude of the panicle is 26.
Now, we havt y =* a sin o>r.
Hore, y =6, and a = 26. And, therefore, 6 ** 26 sin *f.
Or/ */n l  6/26 i, Or, wf i//!" 1 J * n/6.
Or,, ^ ^/6ca.
. ":\x * Hence tho time taken by the particle in covering the distance 6 from O it
liual t
144 PROPEETIBS OF MATTBK
Now, since it takes time >/co to complete one vibration, it will take i of
2*/co or, time 7t/2o> to complete Jr/i of its vibration, i.e., its amplitude, 2b. And,
time already taken by it is :t/6<o.
.*. the time it will take to cover the rest of its distance is clearly equal to
7t 7t STC TC 2n n
2o> 6o> 6o> 6o> 3o>
EXERCISE IV
1. Deduce the equation for the simple harmonic motion of a particle.
Two simpl e harmonic motions, having the same period but differing in
phase and amplitude, are acting in the same direction on a particle Show that
the resultant motion is simple harmonic, and deduce the expression for the
resulting amplitude and phase. (Calcutta 1940)
2 Find the resultant of two mutually perpendicular S. H. motions
which agree in period but differ in phase. Consider the important cases for
phase difference varying from to 2n (Punjab, 1953)
3. A particle executes a S. H Af of period 10 seconds and amplitude 5
r t. Calculate its maximum acceleration and velocity.
Ans. 1974 ft Isec*. ; 3'l42ft.jsec.
4. The path of a b^dy executing a S. H. M. along a straight line is 4
cms. long : and irs velocity, when passing through the centre of its path, is 16
cms. /sec Calculate its timeperiod. Ans. '7854 sec.
5. The maximum velocity of a particle undergoing a 5. H M. is 8 //./sec.
and its acceleration at 4ft. from the mean position is Ibft.fsec*. What is (/) its
amplitude and (ii) its period of vibration. Ans. (i) 4ft. ; (//) 3'142 sees.
6. Explain the characteristics of a simple harmonic motion and show
how to find the velocity at any phase of the motion.
A particle executes simple harmonic motion of period 16 sees. Two
seconds after it passes the centre of oscillation, its velocity is found to be 4ft.
per second. Find the amplitude. (Madras, 1949)
Ans. 1441 //.
t 7. Define simple harmonic motion and show that if the displacement of
t moving point at any time is given by an equation of the form
x = a cos o>f + sin /,
the motion is simple harmonic. If a = 3, b = 4, nnd 2, determine the
period, amplitude, maximum velocity and maximum acceleration of the motion.
(Madras, 1949)
Ans. (/)3142, (i7)5, (//) 10, (iv)20.
8. Find the velocity aad acceleration of a point executing simple har
monic motion. M ^
A point describes simple harmonic motion in a line 4 cms. long. "Ita
velocity, when passing through the centre of the ljre, is 12 cm. per second.
Find the period. (Calcutta, I94<r)
Ans. 1*047 sees.
9. Define a S.H. motion, explaining the meanings of the terms, period,
amplitude and phase.
A particle is subjected simultaneously to two S.H. vibrations of the
lame period but of different amplitudes and phases, m perpendicular directions.
Find an expression for ttie resultant motion and show that the path traced by
the particle is an ellipse.
For what conditions may the path be a circle and a straight line ?
(Calcutta,
10. A mass of 15 Ibs. is suspended from a fixed point by a light spring.
In the equilibrium position; the spring is extended by 15 inches. The mass isttien
pulled down by 4 inches and released from rest. Show that it executes a S.H.M.
ana calculate its timeperiod Also calculate the energy of its mass.
c. ', 4Q ft.pound*l*.
SIMPLtt iUtttfOtflO MOflOH
11. Show that a compound pendulum would swing most rapidly when
the distance of its e.g., from the axis of oscillation equals its radius of gyration.
12. A thin and square metal plate, of aside 2/, is suspended from one
corner so as to swing in a vertical plane. Calculate the length of the equivalent
simple pendulum. Ans. 4 A/2// 3.
13. Calculate the timeperiod of a circular disc of radius r, oscillating
about an axis through a point, distant r/2 from its centre and perpendicular to
its plane. " Ans. 2n\/3rl2f.
14. Find the velocity, acceleration and the periodic time of a point exe
cuting Simple Harmonic Motion.
A particle is moving with simple harmonic motion in a straight line.
When the distance of the particle from the equilibrium position has the values
x l and * g , the corresponding values of the velocity are u and a . Show that the
period is
Find also (i) the maximum velocity and (11) the amplitude.
(Madras, 1949)
15. A particle moves with uniform speed in a circle. Show that the
motion may be resolved into two simple harmonic motions at right angles to
each other. How do they differ in phase and amplitude ? Show how the
potential and kinetic energies ot a particle executing siniple harmonic motion
vary. (Calcutta)
16. Show that the total energy of a particle executing simple harmonic
motion is proportional to (a) the square of us amplitude, (b) the square of its
frequency.
Show how, on an average, its energy is half kinetic and half potential in
form.
17. In ths HCl molecule the force required to alter the distance between
the atoms from its equilibrium value is 5 '4 x I O 5 dynes per cm. What is the
fundamental frequency of the vibration of the molecule, assuming the vibration
to be simple harmonic, and the mass of the Cl atom to be infinite compared to
that of the H atom which is 166 x 10~ 24 gm. ?
(Cambridge Scholarship Certificate)
Ans. 9'1 x 10".
18. Find graphically the resultant of two simple harmonic motions at
right angles to one another (a) when the amplitudes and periods are equal and
one vibration differs in phase by */2 from the other, (b) when the amplitudes are
equal, the period of one is twice that of the other and the slower vibration is
w/2 ahead of the other.
19. The total energy of a particle executing a S.H.M. of period 2rc sec. is
10,240 ergs rc/4 sec, after the particle passes the midpoint of the swing its dis
placement is 8^ 2 cm. Calculate the amplitude of the motion and the mass of
the particle. (Oxford and Cambridge Higher School Certificate)
Ans. 1 6 cms. ; 80 gms.
20. Show that the motion of the piston of a steam engine is approxi
mately simple harmonic if the connecting rod is long compared with the crank.
CHAPTER V
MEASUREMENT OF MASS THE BALANCE
54. Mass and Weight. The mass of a body is the quantity o*
matter contained in it and is an inherent, invariable and fundamental
property of it, quite independent of the presence or absence of any
other neighbouring bodies or of the place where the body happens to
be situated. Thus the mass of a given body will be the same at the
equator, at the poles of the earth, or, for that matter, anywhere else
in the whole of the universe.
The weight of a body, on the other hand, is the force with
which it is attracted by the earth towards its centre, and is equal to
the product of its mass and the acceleration due to gravity.
Thus, if m be the mass of a body, and g, the acceleration due to
gravity, its weight is given by w = m.g.
Since the value of g changes from place to place, being inversely propor
tional to the square ol the distance from the centre of the earth*, the weight
of the same body differs from one place to another, being about half a per cent
greater at the poles than at the equator, twentyeight times its weight on earth,
on the sun and about onesixth its weight on earth, on the moon.
It will thus be seen that the weight of a body in a variable property of it,
depending not only upon its own mass but also on its distance from the centre
of the earth, i.e., on its position, relative to the earth.
Then, again, since the mass of a body endows it with the property of
nertia or of reluctance to chin^e of both rest and motion, we may also define it as
the digree of resistance of matter to changes of motion.. As against this, the weight
of a bady, being a force, directed towards the centre of the earth, tends to accele
rate it 1 ! own mjtion in that direction. Thus, whereas the one resists, the other
tends to produce, motion.
Nevertheless, at a place, since g is constantf, at any rate, within a small
space, the weights of two bodies are directly proportional to their masses. For, if
w and w' be their weights and m and m', their masses, we have
w mg and w' = m'g.
So that w\w' = mg/tn'g^ m/m'.
If follows, therefore, that the common physical balance may be
used to compare masses. For, although, strictly speaking, it really
compares weights indicating a measure of their equality or want of
equality, but since the value of g, for the body as well as the
standard weights, placed in its two pans respectively, is the same,
the forces exerted at the two ends of tho beam, in its equilibrium
*SecChapt er VI . ~
fThis was first shown by Galileo in 159), by his famous experiment of
dropping simultaneously two unequal masses from the top of tne Leaning Fower
of Pisa, wnen they reached the ground together. The same fact was confirmed
by Newton and Uter by Bes.se I, by using pendulums with hollow bobs, filled with
materials of different densities and, observing no variations in the value of g
beyond those within experimental error. And finally, it has been shown con
clusively by Eotvos by his experiments with an ingenious modification of the
Torsion Balance.
JHURA5UHBM&NT OJf MASS TUB BALJLHUJi **'
position, are evidently equal, thus indirectly establishing the equality
of the two masses, irrespective of the value of g.
If it be desired, however, to determine the weight of a body,
we make use of a spring balance, the stretch of the spiral spring of
which, if riot unduly large, is proportional to the force applied to it
by the weight of the body suspended from it, and this, as can be
readily seen, will be different for the same body at different places,
depending on the value of #.
55 The Common Balance. It is, in essentials, an equi arm
lever of the first order and depends, for its action, on the principle of
moments.
The essential feature of its construction is a symmetrical rigid beam
usually in the form of a triangular lattice girder, as shown in Fig 93, (to ensure
lightness vviih strength), pivoteJ centrally, so a* to be free to rotate in the ver
tical plane about thehorizonUl axis provided by a knifeedge of steel or agate,
resting on an agate plane carried by a stout vertical pillar. A long and light
pointer, hxed at tight angles to it moves over a small ivory scale below, whose
central division marks/its normal position, when the beam is in equilibrium or
at rest. A screw, worked upwards and downwards, at the top of the pointer,
enables the e.g. of the beam (together with the pointer; to be rahed or lowered,
as desired*.
Two other knife edg:s, similar to, and equidistant from, the central one,
are carried by tlie beam itself on either side, with two identical scale pans, of
equal mass, suspended Irorn the agate planes resting on them.
The whole instrument is enclosed in a glass case, with sidewindows and
a sliding front, to safeguard against disturb ince due to air draughts or tempera
ture variations, all weighings being earned out with the glass case propeily
closed on all sides.
The bodyf to be weighed is placed in the lefthand pan and standard
weights from the weight box, in the right hand pan*, starting with the seemingly
heavier ones, until the pointer swings evently on either side of the central mark
on the wofy scale If the ////$ (/ <*., the two halves of the beam, on either side of
the central knife edge) be of the same length and the scale pans be of the same
weight , the beam will come to ic^t in the horizontal position, but if the weights
of the scale pans clilfer even slightly, it will be tilted towards the side of the
heavier weight, with the pointer moving correspondingly over the scale below.
The use of the Rider. Since the weight boxes are not provided with
weights smallei than 1 milligram, the final adjustment for the equilibrium of
the beam is made with the help of what is called a *rider\ which
is just a piece of wire, weighing 1 centigram, and bent into the
form shown, (Fig. 92), and can be moved over the right half of
the beam by a levei device, manipulated from outside the case,
this arm of the balance being graduated into 100 equal divisios
from the central to the end knifeedge. With the rider at the
100th division, the effect is equivalent to placing a centigram
weight in the righthand pan ; so that, when it is, say, at the nth
division, ihc etfect is equivalent to adding a weight of w/100 _^
centigram or lOrt/100 or w/10 milligram to the pan. *
56. Essentials or Requisites of a Good Balance. There are three
essentials of a good physical balance, viz., (/) Truth, (//) Sensitiveness
(or Sensitivity), and (///) Stability or Quickness.
*If the beam were to be pivoted exactly at its e.g., it would be in neutral
equilibrium, and will remain at rest at any angle with the horizontal. Its e.g.
is, therefore, arranged to be below ihe central knifeedge, because as it tilts
one way or the other, the c g. rises upward, and the beam is thus, in stable
equilibrium.
tToo heavy bodies, likely to break or bend the beam, should be avoided.
Jit is purely a matter of convenience, with no principle involved io it.
148
BOFEBTI1S OF MATTltt
1. Truth. A balance is said to be true, when, with its scale pans
unloaded, or equally loaded, the beam remains horizontal.
S
Fig. 93.
Let a and b be
the lengths of the two
arms of the balance,
(Fig. 93), and S and
8', the weights of its
two scale pans.
Then, with the
scale pans unloaded,
the beam will remain
horizontal, when the
moments on either
side of the central knifeedge C balance each obher, i.e., when
Sxa=S'xb. ..(/)
Now, let the scale pans be loaded with equal masses m and m.
Then, for equilibrium, we have (S+m).a=(S t +m).b. . .(')
Subtracting relation (/) from (//), we have m.a~m.b.
whence, a=b. ("'0
Substituting this in relation (/) we have
S=S'. (iv)
Thus, a true balance must have (/) arms of equal lengths and (//) pans
of equal weights.
2. Sensitiveness. A balance is said to be sensitive when, for a
small difference ofhads in the two scale pans, the beam (and, therefore,
the pointer) swings through an appreciable angle, it beinj assumed that
the balance is true.
Thus, the ratio between the deflection of the beam or the
pointer and the difference of load, (usually 1 m.gm.) causing it,
measures the sensitiveness or the sensitivity of the balance. So that,
the greater the displacement of the pointer for a given difference of
load, or, conversely, the smaller the difference of load required to
produce a given displacement of the pointer, the greater the sensi
tiveness of the balance. Usually, a balance is regarded to be quite
sensitive, when a difference of 1 m.gm. in load causes the pointer to
be displaced through 1 division on the scale.
(/) Case of a Balance with the three knifeedges in one plane.
Let Fig. 94 represent a vertical section of the balance through the
centre of the beam,
passing through the three
knifeedges at A, B and
C, all in one plane.
Let a be the length
of each arm, d, the depth
of the e.g. (0) of the
beam from the central
knifeedge C and Jf, its
mass. Further, let S and
(8im) be the masses
of the two scale pans
MEASUREMENT OF MASS THE BALANCE 149
together with their loads, the difference of load m between them
being small.
Then, if 6 be the deflection of the beam from its initial horizon
tal position, so that it takes up the position A'B' t with its e.g. shifted
to (?', we have, taking moments about C,
(8 + m)g.a cos = Sg.a cos + Mg.d sin 0.
Or, (S + m).a cos 6 = S.a cos + M.daia 0.
Or, m.a cos = M.d sin 0.
~ sin m.a ^ m.a ,..
Or, =TT> Or > tawfl^iTj (')
cos Md Md v '
And, if be small, tan = ; so that, in that case,
m.a ,... ~ a ....
6 = m <"> <* sf  ira ' <">
where Q\m measures the sensitiveness of the balance,
It is thus clear that to increase the sensitiveness of the balance.
(/) a must be large, i e., the arms (or the beam) must be long,
(ii) M must be small, i.e., the beam must be light, and
(///) d must be small, i.e., the e.g. of the beam must be close to the
central knifeedge.
Now, a cannot be increased beyond reasonable limits. For, as
Blrge correctly pointed out, the bending of a beam being propor
tional directly to the cube of its length and inversely to the cube of
its thickness or depth, its thickness will also have to be increased in
the same proportion with its length if its original stiffness is to be
maintained, and this will inevitably increase its mass in a much
greater proportion, thereby seriously impairing sensitiveness.
Nor can the beam be made light beyond a limit, or else it will
break or bend permanently ; so that, the only workable alternative is
to decrease d. This may be done with the help of the vertical move
ment of the screw, provided at the top of the pointer, (Fig. 93),
though, carried to an excess, this too has its own drawbacks, v/z.,
(/) loss of stability, and (//) a longer period of swing of the beam. We
have, therefore, to content ourselves with a judicious compromise
between all these factors.
Further, if / be the length of the pointer, its displacement s on
the scale will obviously be 10. And, therefore, if be small, so that
tan 00, we have from relation (//) above,
, m.a
5 = l 'M~d'
To determine the displacement of the pointer, it is by no means
necessary to wait until it actually comes to rest ; it can be easily
estimated from its swings to the right and the left.
Thus, suppose we have a scale with its zero at one end and the
successive turning points of the pointer occur on it at the Wth, the
2nd and the Sth divisions, (Fig. 95). Then, clearly, if S be its resting
point, i.e., the division where it will eventually come to rest, it is
clear that the successive displacements of the pointer from this
division are s l = (10 S), s^ = (S~~2) and $ t = (8 .S).
150
PROPERTIES OF MATTBB
Now, although, theoretically, as we shall soon see, (page 152),
the oscillations of the pointer muss be simple harmonic, it really
seldom happ3ns that the oscillations of any vibrating system remain
truly so. The oscillations always die down and their amplitude goes
on progressively decreasing due to airresistance and other causes,
but the ratio between the successive swings to the left and the right,
very aptly called 'decrement', is found to remain constant.
Thus, with 5 l9 s z and S 3 as the successive swings of the pointer,
we have
P ri
 1 ^= 2 ~, and so on. So that,
J
3
"AV,
/
\ N
/
* \
/
\
*
}
1
\\
1
. V
V
1
\ \
I
\ \
\ \
\ \
I V
\ *
__ 1 \
\
\
J^Li
\ \Jr
< 6 S 6 7 8 
10
~s
(8
Fig, 95.
S ^ 5~2
Or, (^2) 2 = (105)(85)
O r> 5 2 45+4 = 5 2 185'+80.
Or, 145 = 78,
whence, S = 76/14.
= 543.
Thus, the pointer will
ultimately come to rest at
the 5*43r</ division on the
scale.
(//) Case of a Balance with the end knifeedges in a different plane
from the central one The balance, diseased above, with all the three
knifeedges lying in the same plane, is re illy only an ideal one,
this condition being hardly ever attainable in ordinary balances.
For, the beam does yield, however so little, to the forces acting
at its two ends, so that the end knifeedg^s do get depressed a
little below the central one, and no longer remain in the same plane
with it.
Let us see how the sensitiveness is affected when the three
knifeedges are not co planar.
Let h be the
height of the end knife
edges A and B, above
the central one, C,
(Fig. 96), and let the
beam be deflected
through an angle 0, fas <i/"
before, for an extra . *_ ^i j ^
mass m in the right ^ /
hand pan. Then, for
equilibrium, we have
here,
(S4 m).g.(a cos 0+h sin 6)
= S.g.(a cos 0h sfn Q)
+Mg.d sin 0.
Or, (S+m)(a cos 0+h sin 0)
Fig. 96.
S(a cos 0h sin 0)+M.d sin 9,
MBASTTBEMlMfT OF MASS THE BALAffOJ 151
Or, S.a cos g+S.h sin 0+m a cos 0+m.h sin = S.a cos 0S,h sin
+M.d sin 9.
Or, 2 S h sin 9+m.a cos ft +m.h sin = M.d sin g.
Now, if be small, so that sin 9=0, and cos d = 1, we have
2S.h.0+ma+mh0 =* M.d.O.
Neglecting mh0 as the product of very small quantities, we have
2Sh.e+ma = Md.Q. Or, ma = M.d.62S h.$ = 0(Md2Sh) t
n.
.
whence, sensitiveness, = ^ ~ ... (iv)
Now, three possible cases arise :
(/) When h = o, i.e., when the three knifeedges are coplanar.
Here, ti\m = a/Md, [See relation (Hi) above], and the sensitiveness "is
quite independent of the total load (2S).
(it) When h is positive, i.e., when the end knifeedges are higher
than the central knifeedge. In this case, obviously, the sensitiveness
increases with the total load. But, as will be readily seen from Fig, 96,
the effective length of the arm, in the tilted position of the beam,
becomes greater on the side of the heavier, and smaller on the side of
the lighter, pan than its true length ; so that, for a given value of the
excess load m, and for a given deflection tf, the difference of moments
due to the pans on either side of C is greater for a heavier than for a
lighter load, witL the result that the greater the load in the 'pans, the
longer the beam takes to attain equilibrium.
(Hi) When h is negative, i e., when the end knifeedges are lower
than the central one. In this case, clearly, the sensitiveness decreases
with increase in the total load.
N.B. We have seen above how, in the ideal balance, with its three knife
edges in the same straight line, the end knifeedges get depressed with the beam
a little below the central one when the pans are loaded. This results in a
decrease in the sensitiveness with increasing load. If, therefore, the end knife
edges could be arranged at the Correct height above the central knifeedge, the
decrease in sensitiveness due to flexure could be just offset by its increase due to
the latter, and the balance thus nvde equally sensitive for all loa'ds. The method
has actually been used with the success in building balances whose sensitiveness
is quite independent of the total load placed in their pans.
3. Stability or Quickness A balance is #aid to be stable (or
quick), if, with the pans unloaded or equally loaded, the beam be dis
turbed, its time of swing is small and it comes back to rest quickly,
thus making for convenience in weighing.
Now, as we have seen above, with the three knifeedges in the
same horizontal plane, the condition for equilibrium is that
(S+m)g.a cos $ = S.g a cos 6+Mg d sin 0. [See page 149.
Therefore, if the two pans be equally loaded, i.e., if m = 0, or there
be no extra load in the right hand pan, the only restoring moment
about C, tending to br<ng the beam baek to its original position, is
Mg.d sin 0, or Mgd 0, if be small. This, obviously, tends to accele
rate the motion or swing of the beam ; so that, if a be the angular
acceleration it produces in the beam and /, the moment of inertia of
the moving system about the central knifeedge C, we
Mg.d0
Qt BBS ~,
152 PROPERTIES OF MATTER
Attd, clearly, / =* moment of inertia of the beam about C
+
moment of inertia of the two scale pans about C.
Or, /= Mk*+2S.a*,
where k is the radius of gyration of the beam about C.
Thus,  0. Or, = M. Totting
, . . . a constant for a given
where ^ is a constant. L balance.
Or, a oc 6,
i.e., the angular acceleration of the beam is proportional to its angular
displacement. The swing of the beam is thus a simple harmonic
motion, and its time period t is, therefore, given by the relation,
0, / _ > A I , ' M S d 
r ' ' = ** V + M 8 .d
In order, therefore, that / be small, i.e., the balance be stable, k, S
and a should be small and M and d should be large. We thus see that
a balance would be stable when
(/) its arms are short ;
(i7) Us beam is heavy, with its e.g. far below the central knife
edge ',
(iii) the radius of gyration of the beam about the central knifeedge
is small
and that the stability diminishes with increasing load. It will be seen
at once that almost all these conditions are opposed to those for
sensitiveness. So that, sensitiveness and stability of a balance are, to
a great extent, mutually exclusive, and we have, therefore, to^trike a
working balance between the two.
57. Faults in a Balance Determination of True Weights.
1. Arms unequal in lengths and pans unequal in weights. The
co'mmonest fault in a balance is that it may appear to be true, i.e.,
the beam may swing evenly on either side of the central knifeedge,
with the moments on either side balancing each other, and yet the
arms may have unequal lengths and the pans, unequal weights.
Thus, if $! and S% be the weights of the two scale pans and a
and b t the lengths of the two arms respectively, we have
The true weight w of a body may be determined with such a
balance by the method of double weighing, i.e., by weighing the body
first in one pan and then in the other.
Let the counterpoising weight in the right hand pan be w lt
when the body is placed in the left hand pan. Then, clearly,
... ()
MBASUBBMENT OF MASS THE BALANCE 153
And, .. subtracting relation (f) from (//) we have
w.a = wfi. ...(///)
Now, let the body be placed in the right hand pan and let the
counterpoising weight required in the left hand pan be vv a . So that,
Again, subtracting relation (/) from (/v), we have
w.b = w^.a. ( v )
Multiplying relations (///) and (v), we, therefore, have
w 2 .ab =. jv r w 2 .0&, whence, w 2 = w lt w 2 .
Or, w = vX^V
i.e., the true weight of the body is the geometric mean of its apparent
Heights in the two pans.
The same will be true if the pans be equal in weight and the
arms slightly different in lengths.
Note. If we multiply relation (///) and (v) above, crosswise, we have
Or,
Or,  Or,
b 2 w, b M , 2
And, since from relation (/) above, a\b 5 2 /5i, we have
= A /
V "
Thus, we can determine the ratio between the lengths of the two arms
or that between the weights of the two scale pans.
2. Scale pans unequal in weights. Another common fault in a
balance is that whereas the arms may be equal in length, the scale
pans may not be truly equal in weights, so that the beam does not
remain perfectly horizontal.
To determine the true weight of a body with such a balance, we
again resort to double weighing, i.e., to weighing the body first in one
pan and then, in the other.
Let its apparent weights in the two scale pans respectively be
w l and H' 2 . Then, if the length of each arm be a and the weights of
the two scale pans, S l and $ 2 , we have
in the first case, (S l +w).a = (# 2 fu\),0, or S^+w = S^+w^.^i)
And, in the second case, (S t \w).a = (S i \w 2 ).a, or 82+^=^ + ^2. ..("')
Adding relations (i) and (//), we have
S L +Si+2w = S^+^+H^+HV Or, 2w = Wj+w,,
. W t 4 Wo
whence, w = ~> =,
2t
i.e., the true weight of the body is now the arithmetic mean of its
apparent weights in the two scale pans.
3. Inaccuracy of the Brass Weights. A possible source of error may
also be the inaccuracy of the brass or 'standard' weights, supplied in the weight
box, due to their getting worn out by use or getting slightly rusted by discuse.
The probability of error due to such causes is presumably the least in the case
of the larger weights and the greatest in the case of the smaller ones. So that,
assuming the largest among them to be accurate, others, of smaller denomina
tions, are counterpoised against it ; these smaller ones are then counterpoised
against others smaller than them, the process being continued up to the very
smallest ones, and, in this manner, the errors in the smaller weights are easily
154 PROPEBTIES OF MATTER
detected. Thus, for example, a weight of 100 gms. is first counterpoised againsl
50+20+20+ 10 ; then, the weight of 50 8 ms  against 20 + 10 4 10 r 5+2 i 2 M, and
agam the weight of 20 gms, against 10^5 + 242+1 and so on. "lo make
sure, the weights assumed to tx correct must also be tested against a weight
known accurately in terms of the, Inter 'national Standard.
4. Blunting of the KnifeEdges. Due to constant use, the knife edges
get blunted or rounded off, in course of time; so that, with the tilt of the
beam, the point of contact with the plane of support may shift slightly. This
tantamounts to a slight change in the lengths of the arms and must also be
corrected for. 4
58. Correction for Buoyancy. Ordinarily, we make all our weighings
in air. But air, in common with all other gases and liquids, exerts an upward
thrust on a body immersed m it, in strict obedience to the Principle of
Archimedes. So that, the body to be weighed, as well as the brass weights,
agamst which it is weighed, are subject to this upthrust or buoyancy due to the
air displaced by them, which is equal to the weight of the displaced air, in
either case,
If the body weighed happens to hnve the same density as that of the
material of which the standard weights are made their volumes too would
obviously be the same, when they are counterpoised against each other, and the
volume of air displaced by both, and hence the buoyancy or upthrust due
tD it, would just be counterbalanced and the standard weights used would
straightaway give the true weight of the body in vacuum. This is, how
ever, rarely the case. But, it makes it clear why, in realby true and accurate
balances, we insist upon the arms and the pans being identical in length, volume
and mass.
More often than not, the density of the body is quite different from that
of the material of the standard weights, and, therefore, even when they counter
poise each other, their volumes, and hence also the weights of the air displaced by
them are altogether different. Let us, therefore, deduce the necessary correction
in this commonly occurring case.
Let the true mass of the body we'ghed be M and its density p. And, let
the mass of the weights required to counterpoise it be M', an J the density of their
material p'. And, finally, let the density of dir be .
Then, clearly, volume of the body and hence the volume of air displaced
by it = M/p'.
And, so the weight of this displaced air and, therefore, the upthrust due
to it = M.&.gl?.
So that, their apparent (or observed) weight of the body
Similarly, the volume of the standard weights and, therefore, of the air
displaced by them Af'/p' ;
and the upward thrust due to this displaced air = M'.S.gfc'.
(Tkf/ "^
M'g , .8g \
Since the body and the standard weights counterpoise each other in air,
their apparent weights must be equal. And, therefore,
Wh cnce,
Or,
r / S M fNeglectng the prouct o
' 1 + (  , ) . { S/P and 6/p', compared with
L VP P /J $ othe terms.
MEASUREMENT OF MASS THE BALANCE 155
Prom the above it follows at once that
M > = < M ' according as p < = >p',
i.e., the true weight of a body (ie. t its weight in vacuum) is greater than, equal
to or less lhan, its observed or apparent weigtit in air, according as its density
is Ijss titan, equal to or greater than that of the material of the standard
weights used.
SOLVED EXAMPLES
1. The arms of a balance are unequal in length but, without the scale pans,
the beam and the scalepan holders are correctly balanced. The scale pans A and
B are of weights 2w t and 2w 2 respectively. A body placed in pan A has an apparent
weight Wj and placed in pan B has an apparent weight W 2 . Show that the true
weight of the body is
1/[W X W a +2(Wi Wi+w, W,) +0"i +*>*] (*!+*> f ^
(London Higher School Certificate)
Let the true weight of the body be W and the lengths of the lefthand
and right hand arms be a and b respectively.
Then, since equilibrium is attained with the body in the lefthand pan
and a weight W in tho righthand pan, the moments on either side of the
central knifeedge must be equal ; so that, neglecting moments due to the pan
holders, which already balance each other, we have
(2vv 1 + PK).a = (2w 9 +W l U.  ..... (0
Again, equilibrium is attained with the body in the righthand pan and a weight
W % in the lefthand pan ; so that,
(2w z + W).b == (2^+Wj.a. ...... (")
Multiply the corresponding sides of relation (/) and (), we have
Or, (2w^ W}(2
Or, 4w l H> a r2H' 1 W'h2H' a Wr W l = 4w l w^2wJV n i2w l lV l f
Or, ^ 2 h2^(>Vt 4 w 2 ) *
Adding OvjHv;) 2 to both sides, we have
The left hand expression is clearly the complete square of
so that,
Or,
And, .. W
Or, the true weight of the body
2. The arm? of a balance are similar and of equal length, a. The scale
pans are similar and of equal weight, P. When the beam of the balance is
horizontal the central knifeedge is a distance x vertically above the middle of the
line joining the knifeedges of the scale pans, and the e.g. of the balance is a dis
tance y vertically below the same point Assuming that the weight of the moving
system of the balance is W, derive an expression for the angle of deflection of
the beam when weights w x and w, are placed on the scale pans. fw t > w 2 ].
(Joint Matriculation Board High School Certificate)
Let AB, (Fig. 97) be the position of the beam, when the pans are yet un
loaded, C, that of the central knifeedge, P and P, of the scale pans and p,
that of the pointer, with G, as the e.g. of the beam.
Let the heavier weight H'i be now placed in the righthand pan and the
lighter weight w,, in thr lefthand pan and let the beam, and, therefore, also
the pointer, deflect through an angle 9, into the positions A'B' and p' 9 with the
eg of the beam at G', where OG'=~y (O being the midpoint of the line joining
the knifeedges of the two scale pans). Let the central knifeedge be now at C',
where OC'~x (given).
Then, the different forces, all acting vertically downwards,on the
beam are
156
PBOPBBFIBS Of MATTEB
(/) (P+Wj) at B' t (//) (P+w> 2 ) at A' and (///) W, the weight of the moving
system, at G'.
Since the beam is in equilibrium in this position, the moments about the
knifeedge C', on either side, must be equal.
Fig. 97.
(P+\v*).DEW.G'L=(P{ wJ.DF.
DE^OE+OD=OA' cos Q + OC' sin Q=a cos Q+x sin $,
Or,
But
y sin 9+ x sin (xf y) sin 9,
and DF^OFOD^OB' cos 9 OC sin = (a cos 0x sin 0).
So that, CPHw> 2 ). f a cos 0+x sin 0)f W.(x+y).sin Q iP+wJ.ia cos Q~x sin 0).
Or, P.a cos 0+P.x sin Q + w 2 .a cos 0f w 2 .x sin + W.(x+y).sin Q.
=P.a cos Q~P.x sin Q\w v a cos w^x sin Q.
Or, 2P.x sin Q+w^x sin 9 f w 2 .x sin 9 f W.(x+y).sin 9
= H' 1 .a cos 9 w 2 .ci cos 6.
Or,
whence,
cos 9
6*
@==
and
This, then, is the angle of deflection of the beam, when weights
w are placed on the two scale pans.
3. With a balance of which the arms were 10 cm. long, it was found that
0*010 gm. extraload on one pan deflected the beam of mass 20 gms. through 1 and
that this deflection was independent of the loads placed on the pans. What can you
deduce from these measurements ? (Oxford Local Higher School Certificate)
Since the deflection of the beam for the given extra load of '01 gm. is
quite independent of the loads placed on the pans, it is clear that the three edges
are coplanar.
And, since, in the question, every other factor is given except the depth
of the e.g. of the beam below the central knifeedge or trte centre of the beam,
we are obviously expected to determine its value. Let it be h.
Since the beam is in equilibrium at angle 9 1 from the initial horizontal
position, it is evident that the moment about the central knifeedge due to its
158 PBOPERT1BS Ot MATTEB
We know that sensitivity of a balance is given by the relation,
= TT>J r* ~ r where 2S is the total load. [Page 151.
m Md2S.h
Now, in the first case, $ = 3'0, m  1 mg. *001 gm. and 25 0.
/ /&? second case, 6 = 2'70, m = 1 m.?. => '091 gm. and 2S  100 gms.
Sothat  )Y = AfJlOOA .......... (//)
dividing relation (/) by (//), we have
3 MJ100/i 0.7 *,/ > 7 AA
* =  , . Or, 3 Afa 2 7 Ma 270/r.
Or, 270/1 = 3M, whence, A = '3 A/r// 270.
The negative value of /z thus cbarly indicates that the end knifeedges are
below the central knifeedge.
Now, let thi sensitivity of the balance be x divis ons per mg. for a load
of 200 gms. Then, we have
x_ __ <* _ <* r Substituting
001 ~ Md2i)0h ~~ MJ~200/~'3 \fd\' the value of
r ustt
~'3 \fd\' the val
270 L// from a
\ 270 / L// from above.
x _ _a ___ 270a
001
.*. dividing relation (/// by (/), we have
x 001 270 Md_ o x _ 270
001 X ^3 ^ 330V/J X '"* f) 3 " 330'
270x3 27
whence, "330 ^ Ti =3 2455.
Thus, the sensitivity of tUs balance for a load of 200 gms. is 2455 di\i
sions per mg.
6. A piece of metal weighs 300 gms in air If the densities of the metal and
the brass weights used by 19 gms. c.c , and 8 gms. /c.c., respectively, and that of air
00123 gm./c.c., calculate the true mass of the piece in vacuum and the cor
rection due to buoyancy.
We know that the true mass of a body in vacuum is giv^n by the relation,
M  M 1 [l +8 ( *  p, )J, [Sec page 154.
where Mis the true mass of the body, p, its density, M', its apparent mass, p',
the density of the weights used and 3, the density of the air.
Here, M' = WQgms., p = 19 gms lex., p' = 8 gms./c.c. and 8 '00123
gms.lc.c. Substituting these values in the relation for M above, we have
 300 100123 x  300
300(1 '00008902)  300026706  299'91 3294 gms.
Thus, the true mass of the body in vacuum will be 299973294 gms., the
buoyancy correction being obviously '026706 gm.
EXERCISE V
1. What are the essentials of a good balance and how are they secured
in actual practice ?
Show that the sensitiveness of a balance is independent of the load in tho
two scale pans, if the three knifeedges be coplanar. How does the position of
the centre of gravity of the beam affect the working of the balance ?
MEASUREMENT Off MASS THE BAUUSCJB
* 2 " ^ e arms of a Dalance ar ? eacn 7 ww. long, the length of the pointer
is 12 cms. and the mass of the beam is 50 ms. If the knifeedges are in a plane
and the centre of gravity ot the beam is 0*02 cm. below the centre knifeedge,
now much will the end of the pointer be deflected when the difference in load in
tne pans is 1 milligram ? (Cambridge Local Higher School Certificate)
Ans 0*84 cm.
f 3. A certain balance has a beam weighing 200 gms., with knifedges
carrying the pans 15 cms from the central knifeedge. What is the depth of
the centre of gravity of the beam bslow this knifeedge, if a weight of 1 mg.
placed on one of the pans displaces the end of the pointer through a distance of
U'5 cm., the pointer being 15 cms. long ? Br Inter )
Ans. 0*0225 mm.
4. Two balances, made of the same material, are alike in all respects
except that the linear dimersn ns of one are n times those of the other. Compare
the angular deflections of the beams for H given difference in load
(Cambridge Local Higher School Certificate)
Ans. 1/fl 3 : 1.
5. A body is weighed first in the left and 'then in the right hand pan of
a balance, the respective weights being 9'842 gms. and 9'833 gms. Find the true
weight of the body and the ratio ot the lengths of arms of the balance.
Ans. True weight 9 837 gms ; Ratio of arms 1 '0005 : 1.
6. Discuss the points to be taken into consideration in the design of an
accurate, sensitive and convenient balance. If the arms are of unequal lengths,
show how the error on this account can be avoided. How would you except the
sensitiveness to vary with the load ? (Bombay, 1933)
1. What are the requisites of a balance ? Obtain the general expression
used for determining the conditions for these requisites and show that the con
ditions for two of these are mutually contradictory. (Punjab, 1933)
8. Sketch ihe essential parts of a balance in which the two end knife
edges arc h cms. below the centra) kiufeedge and discuss the conditions of itt
sensitiveness.
9. Obtain an expression for the true mass of a body in vacuum, when its
apparent mass in air is M ' gm , its density p, the density of the standard weights
used P and the density of air 8, Would tne same treatment be applicable, if the
body be weighed in a liquid ?
10. Given the apparent weights of a body in two different liquids, of
Known densities, similar standard weights being used in both the cases, how will
you proceed 10 calculate the density of the body ?
* Jc 1 ' A gla&s st PP er Density 2'5 gms Ic c.) is first weighed in water of den
sity ws gm. fc.c and then in oil. Its apparent weights in the two liquids are
loand to be 1 8'6^mv. and 2') 4 #wv. respectively, the brass weights used being
similar in cither case. Calculate the density of oil. Ans. '8489 gm /c.c.
CHAPTER VI
ACCELERATION DUE TO GRAVITY
59. Acceleration due to Gravity. Galileo was the first person
to have performed in 1590, the then bold and spectacular experiment
of dropping a cannon and a musket ball from the Leaning Tower of
Pisa, which, contrary to the teachings of Aristotle 9 reached the ground
simultaneously. He thus clearly showed that, at any given place,
all bodies, big or small, when dropped so as to fall freely, do so at
the same unijorm rate, neglecting, of course, the resistance to their
motion due to air. That is to say/ all bodies, irrespective of their
mass or nature, falling freely in vacuwn, will have the same acceleration
at a given place. This acceleration is called the acceleration due to
gravityJ&s it is due to the gravitational attraction of the body by
the earth, towards its centre, (see Chapter VIT). It is denoted by
the better g, and is numerically equal to the force with which a unit
mass is attracted by the earth towards its centre, i.e., equal to the
weight of unit mass*.
The value of g differs from place to place, being the greatest at
the poles and the least at the equator. Its value,, for all practical
purposes, is however, taken to be 981 _cms. jsec*., in the C.G.S.
system, and 32^ ft. I sec*., in theT F.P.S. system. Due to this
comparatively large value of g, bodies fall much too quickly to the
surface of the earth, when dropped freely, and hence it becomes
difficult to measure it directly with any great accuracy. It is,
therefore, determined indirectly with tho help of a simple or a
compound pendulum, or by other methods. We shall now proceed
to consider some of these in proper detail.
60. The Simple Pendulum. A simple ( or, a mathematical )
pendulum is just a heavy particle, (ideally, only a pointmass),
suspended from one end of an inextensible, weightless string, whose
other end is fixed to a rigid support, the point where the string is
fixed to the support bJng known as the point of suspension of the
penduluii. In practice, we usually take a small and a heavy metallic
spherical bob, tied to a fine silk thread.
/ The motion of ths pendulum is simple harmonic and isochronous, f
i.e., its timepsriod is quite independent for its amplitude of small
swings, and is given by the relation, t = 27T\/ Ijg^ where / is the length
of the pendulum, (or the distance between its point of suspension and
*This is so, because the force with which a b>dy of mass m is attracted
by the earth, towards its centre, is equal to its weight mg\ and, therefore, if,
m = 1. i.e., ii' the body be of unit mass, this force of attraction on it, or its weight,
is equal to g.
For the discovery of this property of the pendulum, we are again
indebted to Galileo, who noticed a swinging lamp in the cathedral at Pisa
and timed its oscillations against his own pulse beats. The time taken for
each swing was found to be the same and, as far as Galileo could judge, quiU
independent of the size of the swing.
160
ACCELERATION DUB TO GRAVITY 161
the centre of gravity of the bob), and, g, the acceleration due to gravity
at the place. This will be clear from the following :
Let S be the point of suspension and O, the mean or equilibrium
position of the bob, (Fig. 99). Then t if the bob be given a small
angular displacement 6, in the vertical plane, (or the plane of the
.pendulum itself), so as to occupy the position A, it is clear that it
will be under the action of two forces, viz., (/) its weight mg, acting
vertically downwards, (m being the mass of the bob), and ( ii ) tJie
tension T of the string acting, along the string,
towards its point of suspension S.
Resolving the weight (mg) of the bob,
into two rectangular components, we have ~o\
(a) component mg cos 0, acting along
the string, as shown, and
(b) component mg sin 0, at right angles
to it, /.e., along the tangent to the arc OA,
in the direction AO.
Obviously, the former component (mg cos 0)
is just balanced by the tension (T) of the
string, there being no motion along it either
way; so that, the only force left acting on the
bob is mg sin 0, towards its mean or equilibrium
position O. ' m 9
Now, acceleration = forcefmass. Fig. 99.
And, therefore, acceleration of the bob = mg sin 6/m = g sin in the
direction AO, or towards the mean or equilibrium position of the
bob. And, if 9 be small, we have sin Q = Q,
Hence, acceleration of the bob = gO, directed towards O.
Again = OAJSO = x/l, [v angle = arc /radius.
where, OA = x, the displacement of the bob, and /, the length of the
pendulum.
acceleration of the bob = g. = ^ .x.
Or, putting gjl = /^, a constant, for a given pendulum at a given
place, we have
acceleration of the bob = ^ Jt.
Or, acceleration of the bob oc x, and is directed towards O,
i.e , the acceleration of the bob is proportional to its displacement
from its mean or equilibrium position, and is always directed
towards that position. The bob thus executes a simple harmonic
motion, and its timeperiod I is given by the relation,
t =
Alternative Methods.
Method (/). Let the bob be given a small angular displacement
oc in the vertical plane, so as to occupy the position A, (Fig.
In other words, let a be the angular amplitude of the bob.
f 02 PROPERTIES OF MATTER
Then, clearly, the e.g. of the bob has been raised up through
A vertical distance OC, whon AC is the perpendiculat dropped from
A on to SO.
\ .*. potential energy of the bob at A~mg. OC.
As the bob is released at A, it starts moving to
\ wards 0, thus acquiring kinetic energy, (due to ita
I 9 \\ motion), at the expense of its potential energy.
i \ \ Consider the bqb to be at B, on its way to
! \ * wards O, and let its angular displacement here be
{ \ ^l\ 6. Then, clearly,
I \ \ potential energy of the bob at B=mg.OD,
i \ \ where BD is the perpendicular dropped from B on
j \ \ to SO.
J \ \ .'. loss in P.E. of the bob in moving from position 1
/; < ?+'''A A to posit ion B=mg.OCmg.OD=mg (OC OD).
o """"""" ^ =mg[(SO8C)(SOSD)].
Fig. 100. Now, SO = SA=SB=l, the length of the pendulum;
and /. SC=^SA cos <x=/ cos a and SD=^SB cos 0=1 cos 6.
So that, loss in P. E. of the bob=mg[(ll cos a)(// cos G)]
mgl (cos 9 cos a).
This must, therefore, be equal to the gain in K.E. of the bob
at B y Or equal to the K.E. of the bob at B, (since its K E at A was
equal to zero). If / be the moment of ineitia of the bob about the
axis of suspension through S, and oj, its angular velocity at B,
we have
K E. of the bob at j? = /co 2 .
And /. lja)*=mgl(cos 6cos a). .. .. (/>
Differentiating this expression with respect to time (/), we have
aco i aQ
i it}  ^^ ~ /77I?/ Sin v t ~~
' dt dt
the angular amplitude (a) of the bob being a constant quantity.
Now, dBldt=o) 9 the angular velocity of the bob at B.
And /. dajjdt is its angular acceleration, here.
Thus, Ia>.du)ldt*= mgl sin Q.J.
Or, f.da)./dt= mgl. \in 6,
whence, dwjdt = mgl. sin 6/1.
knd since 6 is small, sin Q = Q(radians), very nearly.
Hence, acceleration of the bob = f  *0, . . (//>
the ve sign merely indicating that it is directed towards O, opposite
ro that in which the angular displacement (0) increases.
Or, neglecting the ve sign, and putting wg.///=^, a constant
For a given pendulum at a given place, we have
acceleration of the bob = v.$,
ttid, therefore, proportional to , its angular displacement from it*
mean or equilibrium position.
ACCELERATION DUB TO GRAVITY 163
The bob thus executes a S.H.M., and its timeppriocUs given by
fiT =2* V ~mg7,r = 2 " V mir*
Now, if r be the radius of the bob, we have, by the principle
of parallel axes, 1= 2/wr/5 +m/ J .
/Twr*/5 + w/ r
And, therefore, /=27rA/ f .
Since the bob is a small one, its radius r is negligible compared
with /, the length of the pendulum; so that, we have
/ = 27ri/nil'lmgl. Or, t=2iri/ljg~~
Method (//). See Borda's pendulum, (next article).
Calculation for g. Squaring the expression for the timeperiod
of the pendulum, we have
r 2 = 47r 2 //^, whence, = 47r 2 /// 2 .
Thus, knowing /, the length of the pendulum, and f, its time
period, we can easily calculate out the value of g at the given place.*
Drawbacks of Simple Pendulum. Though simple in theory, and
easy to perform, this method of determining the value of g is n >t
quite an accurate one, due to its numerous drawbacks, the more
important ones of which are the following :
1. A simple pendulum is just an ideal conception, not realizable
in actual practice, for we can neither have a point mass, nor a weight
less string: so that, the string too has a moment of inertia about the
suspension ax is.
2. The resistance and the buoyancy of air appreciably affect tlte
motion of the bob.
3. The relation for the timeperiod (/), obtained above, is trite
only for oscillations having an infinitely small amplitude.
4. The motion of the bob is not, strictly speaking, a motion, of
translation, for it also has a rotatory motion about the axis of suspen
sion pass ing through the point of suspension.
5. The bob has also a relative motion with respect to the
suspensionthread at the extremities of its amplitude on either side.
61. Borda's Pendulum. In this pendulum, the bob is a sphere
of large radius and, assuming thit it is rigidly fixed to the string
and oscillates only about the axis of suspension, (there being no
relative motion between the bob and the string), its timeperiod
//+ V 2 //
is given by / = ^TT A/ >
where / is the length of the pendulum, and r, the radius of the bob.
This relation for t may be deduced as foil >ws :
*The earliest determinations of the value of g w^re all made oy means
of simple pendulums Thus, Picard, in 1669, used a pencul im in which a
copper bob of diameter 1 inch was sispen^ed by an ah? fib>e (\vhich remains
unaffected by moisture) and, in the year 1792, Borda and Castini used one, with
a platinum ball of diameter 1*5 inches, suspended by an iron wire about
J 275 ft. long.
164
PROPERTIES OF MATTER
Let SO be a Borda's pendulum, (Fig. 101), suspended from the
point of suspension S and let m be the mass of the bob, of radius r.
Imagine the bob to oscillate in the plane of the paper, and let it
be displaced from its original position A to the position B, through an
angle 0, at any given instant.
The only restoring force on the
S bob in this position is its weight mg,
acting vertically downwards, which
has, obviously, a moment about the
axis of suspension through 8 and
perpendicular to the plane of the
\l paper,
\ = mg x O'D = mg.l sin 6,
\ where / is the length of the pendu
lum, and .'. O'D = I sin 0.
This, then, is the restoring
couple, aoting on the bob in the
position B, tending to bring it back
to its original position.
]f dwjdt be the angular accele
ration produced in the bob, and I,
its moment of inertia about the axis
of suspension through 8, the couple is also equal to I.dwjdt.
I.dco/dt = mg I sin 6 = mgJd. \
Or, dojjdt = 0.wg//7 = jutf,
where mgl/I = u. t a constant.
Or, daj/dt oc 0,
i.e., the angular acceleration ofth3 bob is proportional to 0, its angular
displacement ; so that, its motion is simple harmonic ; and, therefore,
the timeperiod of the pendulum is given by
TTIQ
Fig. 101.
[V if be small, sin 9 = 0.
v the couple due to mg is
clockwise and '. negative.
27T
VT 
I ""
lngl/1
Or,
t =
/
~mgl
Now, / =: the M. I. of the bob about an axis through its e.g.
and parallel to the axis of suspension] w/ 2 . [Principle of II axes.
= M. I. of the bob about its diameter +ml 2 .
Or, / = mr 2 +/7t/ 2 , (v M. I. of a sphere of mass m, about its
diameter is equal to % mr 2 ).
Or,
27T
+ t
"gl
ACCELERATION DUE TO GRAVITY
165
Thus, as will readily be seen, the time period of this pendulum
is the same as that of a simple pendulum of length (/+jjr 2 .//) 'which,
for this reason, is called the length of an equivalent simple pendulum,
or the reduced length of the pendulum.
Further, if r be equal to zero, i.e., if the bob be just a point, or
the pendulum be a simple pendulum, we have, substituting r =s in
the relation for t, above,
. Or, t =
which is the expression for the timeperiod of a simple pendulum,
given above, ( 60).
This pendulum too cannot give an accurate value of g, as, in
the first place, the string has also a moment of inertia about the axis of
suspension and secondly, there is relative motion between the bob and
the stnng, the bob oscillating about it at each extreme end.
62, Compound Pendulum. A compound, rigid (or, a physical)
pendulum is just a rigid body, capable of oscillating freely about a hori
zontal axis passing through it. Its vibrations are also simple harmonic
and its timeperiod is given by the relation,
*
"/
mg. I '
where / is its moment of inertia about the axis of suspension ; m, its
mass and /, its length (or, the distance between its axis of suspension
and its centre of gravity).
This may be seen from the following :
Let S be the point of suspension of the body, (or the pendulum),
through which passes a horizontal axis, perpendicular to the plane of
the paper, about which the body oscil
lates, its e.g., G, \\ill obviously he verti
cally below /S>, in its normal position of
rest, (Fig. 102).
Let the body be displaced through
an angle #, into the dotted position
shown ; so that, its e.g. is now at G'.
Then, the couple acting on the
body due to its weight mg will, obvious
ly, be mg.l sin 6, tending to bring it
back into its original position, (SG'
being equal to /, the length of the pen
dulum).
If the angular acceleration pro
duced in the body by this couple be
da)jdt f thq couple will also be equal to
I.dwjdt, where / is the moment of
inertia of the body about an axis
Fig. 102.
through the point of suspension S, and perpendicular to the plane of
the paper.
166 PROPERTIES OF MATTBK
So that, I.!~ = mg.l sin 6 = mg.l0...(A) [v if be small, sin e ~ 0
~ dct) mg.l
Or,  = _~.0 = M#. [where /w^./// = A*, a constant.
* y
Thus, the angular acceleration, (dc^/dt), of the body is propor*
tional to its angular displacement, (0). The body, therefore, execute*
a S.H.M., and its timeperiod is given by
t == 2?r A / l = 2v\/ \. r = 27T /V/^r ... " (B)
\ u, V mgl I \ mgl
If / be the M. /. of the body criow/ 0H flx/^ through G., /Yj? e.g.,
and parallel to the axis of suspension through 5, we have, by the
principle of parallel axes,
I = /,+/!!/*.
And, if k be the rad>'us of gyration of the body about the axis
through G 9 obviously I g = mk : so that,
/ = mk* +w/ 2 .
.*. substituting the value of / in relation (B) above, we have
Or, t  2
i.e., the period of viberation is the same as that of a simple pendulum
of length . f/, or ~t~ , which is, therefore, the length of an equi
valent simple pendulum, or the reduced length of the pendulum. It
is sometimes denoted by the letter L.
Since k 2 is always greater than zero, this length of an equivalent
simple pendulum is always greater than 1.
Centre of Oscillation. A point 0, on the other side of 0, at a
distance fc 2 // from (7, is called the centre of oscillation, and a horizon
$ tal axis through it, parallel to the axis of suspension,
is known as the axis of oscillation of the pendulum.
Thus, GO = & 2 //, (Fig. 103). Putting it equal to /',
2
K
2 i /2 2
we have SO = J  = / + * = /+/'.
And, .. t = 2;r
L
~g
/ 63. Interchangeability of the Centres of Suspension
' and Oscillation. If the pendulum is inverted and sus
pended about the axis of os< illation through 0, its time
period of vibration will obviously be given by
ACCELERATION DtJl TO GBAYITY
167
And, since fc 2 /' = /', we have fe2 = w/ ?
that, the expression for the timeperiod t becomes
V
: is the same as about the axis of suspension through S.
Or, the centres of suspension and oscillation are interchangeable,
i.e., are reciprocal to each other, a property of the pendulum, first
discovered by Huyghens.
Thus, we get the same values for the timeperiod and the length
of the equivalent simple pendulum whether the pendulum be suspended
at S or at O, i.e., at a distance I from the e.g., (G), or at a distance
Jc 2 /lfrom it.
If, therefore, we draw two circles or arcs, with G as centre, and
radii equal to / and k z /l respectively, they will cut SG produced at S
and Q above, and at O and P below, G.
Then, clearly, SG = GP = I and GQ =G0 = W\l.
And, therefore, QP = GP \ GQ = /+ 2 // = /+/' = SO.
Thus we have four points, S, Q, O and P, collinear withG, the
e.g. of the pendulum, about which the timeperiod is the same.
If. therefore, we can determine, by experiment, these four points,
we can easily find out th* length L or (/+/') of the equivalent
simple pendulum, and hence the y^He of g at the given place, with
the help of the relation, t = 2 Try^/*, where ' is the timeperiod of
the pendulum.
64. Centre of Percussion, Fig. 104 shows a section of a rigid
body, of mass m, by a vertical plane passing through its e.g., G
with S, as its poii.t of suspension, the axis of ~
suspension through which is perpendicular to
the plane of the paper.
Let a force F be applied at O, in the
direction shown, so as to be perpendicular to
both the line SCO and the axis of suspension
through S. Then, this force is equivalent to
(/) an equal, and a like parallel force F at G,
and (ii) a clockwise couple, formed by the force
F at O and an equal and opposite force F at G,
the moment of which is clearly equal to Fxl'
where /' = distance GO.
Now, this force Fat G tends to produce
linear acceleration a, say, in all th^ particles of
the body, including the centre of suspension
S ; so that, a = F/m,
in the direction of the force, i.e., from right to
left.
This, then, is the acceleration produced at S by the force F at
G. The couple, on the other hand, tends to produce an angular
acceleration a, say, in the body, about a parallel axis through G.
[f / be the moment of inertia of the body about this axis, we have
104.
168 PBOPBBTIES OF MATTER
La = Fx I', whence, a^Fx /'// = Fxl'/mk*,
where / = m& 2 , (fc being the radius of gyration of the body about
this axis).
Now, linear acceleration = angular acceleration x distance from
the axis.
Hence, linear acceleration produced by this couple is given by
a' = Ixa =^ Fxl'xllmk*, in the direction left to right, i.e., opposite
to that of a.
In order, therefore, that the force F applied at O may produce
no effect at S, the linear acceleration a due to force F at G must bo
equal and opposite to the linear acceleration a' due to this couple,
i.e., a = a'. In other words, Fjm = F x I' X Ijmk*,
whfence, /'X// 2  1. Or, /'  *//.
This, therefore, is the distance of the point O from the e.g. of
the body, and the point O is thus the centre of oscillation, (see page
166), and is here called the centre of percussion, with respect to S.
It is thus clear that if a body be struck at the centre ofpercus
'sion, or the centre of oscillation, in a direction perpendicular to its axis
of suspension, it does not move bodily, as a whole, at its point of suspen
sion, but simply turns about the axis passing through it.
This explains why when a ball strikes against a bat such that
the point where it strikes the latter is the centre of oscillation, or the
centre of percussion, corresponding to the point where it is held in
the hand as the point of suspension, no sting or shock of any kind is
felt. Similarly, a good hammer should be so constructed that its
centre of percussion lies in a line with the driving force.
65. Other points, collinear with the e.g., about which the time
period is the same. Squaring the expression, f =27ry'/ 2 f/c 2 //g"> for the
timeperiod of a compound pendulum, where / is its length and k, its
radius of gyration about the e.g., we have
. Or,
Or,
gt*
Dividing both sides by 47T 2 , we have / 2 4& 2 = ,,/
Or, ^ 2 ~"/ a ./+&* = 0, which is clearly a quadratic equation in L
Thus, / has two values, viz.,
1  + V e * 1 
" I and
Therefore, there are two values of / at distances (af fe) and
(a b) from the e.g., for which the time period is the same.
ACCELERATION DUE TO GRAVITY 173
even less than onethousandth of that of the bar, and yet maintaining its strength
and rigidity. The values of g and k may then be determined in the usual manner.
68. Eater's Reversible Pendulum. Devised and first used by
Captain Kater, in the year 1817, to make the celebrated determination
of the value of g in London, it is a compound pendulum,
consisting of a brass or steel bar with a fixed heavy bob
B and fitted with two adjustable and mutually facing
knifeedg^s F l and F 2 * near its two ends, so that the pen
dulum may be suspended from either. (Fig. 111). Two
weights, Wi and W^ can be made to slide along the length
of the bar and clamped in the position desired, the smaller
weight W 2 having a micrometer screw arrangement for
finer adjustment of its position. The position of the e.g. of
the pendulum can be altered by changing the relative
positions of the two ,weights, their positions being, how
ever, so chosen that the e.g. always lies inbetween the two
knifeedges f
The pendulum is first suspended from the knife
edge F,, and its timeperiod determined. It is then sus
T
pended from the knifeedge F 2 , and its timeperiod deter K X'/ B
mined again. If there be a divergence in the two values
of the timeperiod, the heavier weigtit W l is moved up
or down and a proper position found for it so that the
timeperiod is very nearly the same, whether the pendu
lum be suspsnded from F i or F 2 . The smaller weight IV 2
is then adjusted by means of the micrometer screw M, until the
time periods, in the two cases, are as nearly equal as possible (say,
differing only by 01 sec. or less, i.e., until the number of oscillations
made by the pendulum in 24 hours, in the two cases, differs by just a
fraction of one full oscillation). When this is so, we have, obviously,
one knifeedge at the centre of oscillation of the other. The distance
between the two knifeedges is measured carefully**. This gives the
length L of the equivalent simple pe,idulum$ and the value of g is
then calculated from the relation g = 4ir 2 Lit*, (see page 171), where
/ is the mean of the timeperiods about the two knifeedges, which
Kater determined by the method of coincidences, (see 69).
The values of L and t may, however, be determined more easily
and accurately as follows :
*It is gratifying to observe that in Kater's own pendulum, these were
made in India, from a special variety of steel, called *wootz\
tin fact, the heavier weight W l is there to ensure that this is so.
**This is done by means of a travelling microscope. The pendulum is
laid horizontally on a table with Us knifeedges lying alongside, and on a level
with, a standard steel scale, and the positions of the knifeedges read on the
scale with the help of the microscope. Or, a better method is to use a vertical
comparator (an instrument carrying two microscopes fitted on two massiye stone
slabs), when the cross wires of the microscopes are first focused on the two
knifeedges of the suspended pendulum, and then on a standard steel scale, fixed
vertically in plape of the pendulum. The distance between the positions of the
ttwo crosswires then gives the distance between the two knifeedges.
jThis deduction from the reciprocal nature of the points of suspension
and oscillation was first pointed out by Bohnenberger, in the year 1811.
174
PROPERTIES OF MATTER
d(BOB DOWN)
C(BOB UP)
Fig. 112.
First a near equility in the timeperiods of the pendulum Is
obtained about either knifeedge, by adjusting W l and W % , as ex
plained above, the timeperiod
being slightly greater, say about
F, than about F 2 , i.e., when the
bob B i* down than when it is up.
The weights are now kept fixed in
their positions and one of the
knifeedges, say F,, moved up or
down a bit to further narrow down
the discrepancy in the two time
periods until a position is attained
when a little more displacement
of F, makes the time period
greater about F 2 (with the bob up)
than about F,, (with the bob
down), i.e., a reversal in the relative magnitudes of the timeperiods
about the two knifeedges takes place. The timeperiods of the
pendulum for two slightly different distances between F, and F 2 are
noted, just before this happens, and similarly for two slightly different
distances between them, after tins happens. These four distances
between F t and F 2 are measured accurately and the time periods
corresponding to them plotted on an exag/eratod scale, as shown in
Fig. 112, where a and b represent the distances and timeperiods for
oscillations about F, in the first two cases and c and d in the second
two cases. The coordinates of the point of intersection O of ab and 1
cdthen give the true length L of the equivalent simple pendulum and
the true time period / corresponding to it.
69. Eater's Method of Coincidences. Kater determined
the timeperiod of his pendulum by what is known as the method
of coincidences, in which the oscillations of the experimental pendu
lum, (Kater's, or any other), are compared with those of a standard
second's pendulum, (i.e., a pendulum of time period two seconds), which
may be a simple pendulum or a clock pendulum. This gives better
results than those obtained by simply timing the oscillations against
B
A
T
Fig. 113.
a stop watch or clock, the accuracy of which can hardly be expectedf
to go beyond *5 sec.*
This means that the time taken for several thousand of swings will have
ACCELERATION DUB TO GRAVITY 175
The experimental pendulum A (Fig 113) here is suspended in
front of the second's pendulum B, with its knifeedge resting on a
rigid support, care being taken to see that the lower ends of the two
pendulums lie in the same level and exactly coincide with each other
when viewed from in front, in their mean or equilibrium position.
A suitable marking device* M is arranged behind the peri*
dulum 5, such that when the two pendulums are in their mean posi
tion, this mark M is just covered by their lower ends and is thus not
visible to the observer viewing them through a telescope T some
distance away.
The whole idea is to enable the observer, watching the oscilla
tions, to judge as accurately as possible as to when exactly do the
two pendulums come into coincidence, i.e., as to when exactly do they
simultaneously pass a particular reference point, in the same direction.
And this is perhaps'best done by using a cross wire in the eyepiece
of the telescope itself, (in which case the marking device M becomes
quite unnecessary).
The two pendulums are set oscillating, and, if they start
together, and have identical timeperiods, thpy continue to oscillate
'in step', i e., they appear to the observer b&just 'one' pendulum,
pendulum B being just hidden behind pendulum A, all the time. But
if their timeperiods differ, ever so slightly, they soon get 'out of step 9 ,
and their oscillations are watched carefully until they both simultane
ously pass the reference point fixed upon, (say, their lowest positions),
in the same direction. When this happens, a 'coincidence' is said to
occur, and the mark M behind the pendulums is just not visible to
the observer. After this, the pendulums again get out of step and
the next coincidence occurs when one of them gains or loses a whole
swing or oscillation over the other. The oscillations made by both
between two successive coincidences are carefully counted.
Let n and (n+l) be the oscillations made by the seconds and*
the experimental pendulums respectively.
Then, if t' and t be their respsctive timeperiods, we have
=,'( 1 n
a f ' f i ^ ver^ nearly,
neglecting the terms involving the second and the higher powers of
n ; for, with t and t' nearly the same, n is sufficiently large, (about
500 or more), and these terms become negligibly small.
/ 1 . r pendu
Now, f'=2 sees., so that, ?=2 f 1  J, B being a second's
^ n ' L pendulum
pendulum
B being a s
pendulum
whence f, the time* period of the experimental pendulum can be easily
*A white pafcer pointer or just a black (iron) stand, with a white chalk
mark on it, would do
176 PROPERTIES OF MATTER
calculated out. This value of / is then used in the expression for g,
above.
It will be easily seen that with all the care taken, it is not
really possible to determine with absolute certainty the particular
oscillations at which the exact coincidence occurs. Luckily; however,
even a difference of a couple of oscillations, this way or that, hardly
matters, ifn be fairly large, as it usually is. For, supposing there ia
an error of two whole oscillations made in judging the point of
coincidence. Then, clearly,
,,^2") = t' [ 1   { y^y J
1
The error introduced is thus 2/ 2 , which, if fi=500, worka
out to be +'0008% and is, therefore, not of much consequence.
Note. In Kater's own determination, made at Portland place, the
pendulum was set in front of a standard clock and the swings of the two pendu
lums observed by means of a telescope from a distance of 9 feet.
The standard clock was checked every 24 hours by stellar measurements,
io that the rate of swing of the pendulum was, in effect, compared with the rate
)f rotation of the earth itself.
Successive coincidences occurred every 530 sees., during which time the
reversible pendulum completed 528 swings or half oscillations. The error thus
vorked out to 1 part in 1,00,000 and the length of seconds pendulum at sea
level, in the latitude of London, cams to 39*13829, inches.
70. Computed Time Bessel's Contribution. Kater, in his
reversible pondulurn, made the timeperiod about the two axes
exactly equal, which, as we have seen, is an extremely tedious process.
But Bessel showed that it was by no means necessary to make
them exactly equal and that it was enough to make them only nearly
equal.
Thus, suppose the timeperiods about the two axes are ?, and
f 2 , respectively, (both being very nearly equal), and that / and I' are
their respective distances from the e.g. of the pendulum.
Then, we have
V&2lL/
~~~ fa~ ' (*) rwhere k is the radius of
5 gyration of the pendului
^ , o A / 2 + /' 2 ,.., l<*but its e.g.
and f 2 =27r A / ., ... (n)
Squaring and rearranging (/) and (//), we have
/ 1 2 /g=47r 2 (/: 2 +/ 2 ), . . (///) and
3o that, subtracting (iv) from (Hi), we have
g
(//') + (*.***) (/+/')
2(1 1'\
ACCELERATION DUE TO GRAVITY 177
8
This quantity, *.'+'' +*'''' . l L = T , [> ~ =< 2 '
^  ^~~^ L simple pendulum
where T is called the computed time of the pendulum.
This distance (/+/'), between the two knifeedges, can be
determined accurately, but (//'), the difference between the
distances of the two axes from the c.g cannot bo determined to any
high degrea of accuracy, because of the difficulty of locating the
position of the c.g. of the pendulum correctly. Smce, however,
(/ 2 / 2 2 ) is much too small a quantity compared wirh (/ 1 2 +/, 2 ),
this can only introduce an inappreciable error which does not^
matter.
71. Errors in the Compound Pendulum and their Remedies.
Besides the difficulty of adjusting the time periods to be exactly
the sain? about either knifeedge, and of correctly measuring the
distance between them, there are a number of other sources of error
in a Kater's pendulum (or the compound pendulum, in general) for
which proper corrections must be applied to obtain an accurate result,
the chief among them being the following :
(/) The Finite Amplitude of the Fenfluhim. The expression for
the timeperiod has been deduced on the assumption that the
amplitude of swing of the pendulum is vanishingly small ; for, then
alone, will its motion be truly simple harmonic in nature. In actual
practice, however, it has always a finite value. This reduces its
acceleration and thus increases its timepoiiod The observed
time period may be corrected for this error by multiplying it with
(1 ^.^g/IH), where a, and 2 are halfswings (in radians) at the
beginning and at the end of the experiment, respectively, as can be
seen from the following :
Proceeding in a manner similar to that discussed in connection with a
simple pendulum (Ahernative method (/), page 162), we have
i/w 2 = mgl (cos 0cos a). [Relation (/) page 162.
Or, /G>* = 2mgl (cos 9 cos a),
where a is the angular amplitude of the pendulum and Q, its angular displace
ment at time /.
Now f / *= mk*+ml 2 = m (A: 2 !/ 2 ), and <o  ^  .
So that, m(k*+ / 2 ) f j 1  ^ 2mgl(cos Q  cos a).
Or,
Or
178
OF MATTJfiA
Integrating this expression for the limits to f/4 and to , where f la
the observed timeperiod of the pendulum, we have

O

o (cos Qcos*Y
V2]7" fa ^
~~*i Jo TT^7
f a
J o
* \ **\ )*
Or
Or
Or,
Putting j/ y= 5//i * J/V? 5^, we have J d$ cos ^ "* ^ 5/w 2
cos
Substituting the value of dQ in the above expression, we have
Now,
sin' ~ s
^ 5W !  /i s <f> )*
* = sin 
)* =
J//Z 2 cos
Or,
  y ^TT . y 1^4  (i
we have
ACCELERATION DTTB TO GRAVITY 179
Or, ,  2 */y *^[l+i sin' ~ + .
Now, lTt\l +L would be the timeperiod of the pendulum, for an infinitely
imall amplitude. Denoting it by /, we have
If a be small, sin = ~ , and we have
Since the amplitude (or half swing) does not remain constant but de
creases from a, in the beginning to a, at the end, both being small, we replace a 8
by a^. So that
<.
(//) Aireffects. There are three distinct ways in which the pre
sence of air affects the timeperiod of a pendulum :
(a) Buoyancy of (he air. This tends to reduce the restoring
couple acting on the pendulum, due to slight decrease in its weight,
similar to tho one produced in a body immersed in a liquid. For, if
m be tho mass of the pendulum and m', that of the air displaced by
it*, the restoring couple is reduced from the value my./ 5/72 to
(mlm'h).g sin 0, where /?f is the dibtance between the e.g. of the
displaced air and the axis of rotation of the pendulum, and can be
obtained sufficiently accurately from its physical dimensions.
The equation of motion of the pendulum thus becomes
m(tf+l*Y ^ g sin e(mlm'h).
This is obviously the equation of a simple harmonic motion, of time
period t t given by
/ :
(mlm'hg)
which is clearly greater than/ 2w \/(* +/*)//*, the expression for the time
period in vacuo.
The timeperiod of the pendulum is thus slightly increased due to buoy?
ancyoftheair.
This was the only correction taken account of by Newtw, followed by
Kater, and it was left to Bessel to show that other corrections due to airefFectf
were also called fox. ____,
*Th>s can easily be obtained from the volume of tfre pendulum and the
density of the air, at the time.
t The value of /j may not be the same as that of /, im]es> the .pendulum
hat a uniform density.
180 PROPERTIES OP MATTER
(b) Some air being dragged along with the pendulum, during it$
motion, (Du Buat's Correction). The p3ndulum during its 'to and
fro* motion, ^carries air with it" and this increases it effective mass,
and hence its moment of inertia, making the obssrved timeperiod
greater than the true one, as will be clear from the following :
That the pendulum does 'carry' ^ormair with it as it oscillate , can bo
shown by a simple experiment, viz , by attaching a feather to its bob, in a direc
tion at right angles to its direction of motion. It will be found that the feather
tilts in a direction opposite to that of the mon'on of the bob, showing that the
air surrounding it is at rest. If, however, the feather be sufficiently close to ihe hob t
it is not found to tilt at all, clea^lv indicating that the air in immediate contact
with the bob moves along with it, or that it "carries air with it*.
Let the mass of this air 'carried' by the pendulum be m" and let the dis
tance of its centre of mass from the point of suspension of the pendulum be d.
Then, clearly, the effective moment ot inertia of the pendulum, i.e , the moment
of inertia of ths pendulum and the adherent mass of air with it, is equal to
w( 8 f /*)f/wV a . And, therefore, the equation of motion of the pendulum now
becomes
Q Tt " " _.. _ ^1 / r.'!LZiLC___. e .
The timeperiod of the pendulum in thus given by
t  2n^~J t ^I^j^
It follows at once from the above that
n ** _A^4/' w^l PM 2 tn^h f Neglecting second
Ur> 4**  / + ml f 7 * ml ' L order terms '
assuming the timeperiod here to bs already corrected for the finite arc of its
swing.
Now, if / t and /., be the distances of the two points of suspension from
the centre of gravity, on cither side, such that the timeperiods in the two cases
are nearly the same, then, if h l3 h 2 and */,, d* be the re*pective distances from the
point of suspension of th3 centres of buoyancy and the centres of mass of the air
adhering to the bob, in the two positions, we have
., ....
'.
So that, subtracting relation (//) from (/), we have
'.'  W W 4 (**) 4
Since t l is very nearly equal to / a , we have k 2
And, therefore,
Here, / l _/ a is the sc l uare of the computed time, (see 70). Denot
ing it by T 2 , we have
AOCELBBATIOH DTJE TO GSAVIfir 181
Thus, the obvious method to eliminate this correction is to make ^Aand
hi h 9t (i.e., to make the psndulum symmetrical in shape), which will reduce
the two expressions on the righthand side of relation (///) to zero. This is
precisely what has been done in Repsold's reversible pendulum, (see S 72,
page 188).
Both effects (a) and (6) due to air can, however, be made almost negligible
by arranging to swing the pendulum in reduced pressure, a procedure now being
increasingly adopted for the residual effect in low pressures is found to be a
linear function of the pressure. The required correction can thus be directly
obtained by plotting a graph between pressure and timeperiod and obtaining the
value of the latter by extrapolating the graph to zero pressure.
(c) Viscosity of the air. The viscous drag due to air produces
a damping effect on the pendulum and tends to reduce its amplitude,
thereby increasing its time period.
For, taking the viscositydrag for small velocities to be proportional to*
velocity, the equation of motion v ould be of the type
Let th solution of this equation be QAe** . Then, clearly,
0. Or, o> 2 f wr+/* = 0,
which is a quadratic equation in eo.
c ... r\ / 7 I ^4i*  r 4./A/~I r r f" where 7
So that,  2  y J V ~4~ * Land
Hence the general solution is
 +7 V/^*/4 1*  I j V~^r 2 /4 I/.
f Be L 2 J
/* J) /+;M "" B) J/n V( ^"r
which is a simple harmonic motion of decaying amplitude, of a timeperiod
'
_
Now,*2rr/v ^ r , the timepsnod of the pendulum in the absence of
any viscous drag.
So that, I  *.[ 1 + 
And, if we make use of ths approximate relation f = 2n/^^r for the time
period of the pendulum, we have /* = 4w 2 /f /a . So that, substituting this value
of f* in the expression for / above, we have
1  jjT^J' very nearly.
This correction due to viscosity is however much too small, being of
the order of 10 ~ 9 and is, therefore, usually neglected.
(///) Nonrigidity of the Support. Due to the yielding of the
support, the timeperiod tends to be greater than, the correct value,
as explained below ;
182 PROPERTIES OF MATTER
It might, at first sight, appear improbable that the stipport should yield
by the mere swinging of a pendulum suspended from it. This is, however, not
so, for the simple reason that no support is perfectly rigid In fact, any ordinary
support, we consider to be rigid, would yield under a weight of 100 k.gms. or so.
True, a pendulum is seldom as heavy as that, but in view of the fact that we
can measure lengths and timeperiods to an accuracy of one in several thousand,
it 1$ only in the fitness of things that we must take into account even this
slight yielding of the support, ii we really aim at a high degiee of precision in
Our work.
Again, it is also true that we can adopt ways and means of eliminating
this error altogether (as well as that due to the presence of air) in so far as the
pendulum is concerned, as explained in 72, below, we should, nevertheless,
acquaint ourselves with the method of deducing a proper correction for it should
it become necessary in other similar cases, wfoere its outright elimination is not
feasible or possible.
Now, then, let as pass on to a brief consideration of it.
We know that a vibrating body tends to set into vibration any other body
in contact with it the degree of response of the latter depending upon how nearly
its natural timeperiod agrees with that of the vibrating body, the closer this
agreement between the two, the greater the response and vice versa.
In the case of the pendulum, therefore, the support carrying it also yields
a little to its vibrations and is forced to oscillate copenodicaily with it.
This oscillation of the support may be resolved into two rectangular com
ponents, (i) along the vertical and (//) along the horizontal, the latter having a
more pronounced effect on the timeperiod of the pendulum
than the former.
Thus, if OQ be the mean or equilibrium position of the
pendulum, (hig 114), of length /, uith the axis of suspension
passing through 0, then, as it swings through an angle into
the position OG' 9 there are two forces acting upon it. (j) along
the arc of its swing, to which its motion is due and (//') the
other at right angles to it, i.e., along G'O, (the centiipetal
force). So that,
acceleration of the pendulum along the arc
<Ps d ds ^ . d 2 Q J" where
and acceleration along the length of the pendulum
v a 1 / ds "* ' J " "*
U4.
*' v = ds l dt
& /j \
[ \ dt J
These accelerations, obviously, act at the support 0, with the component
in the horizontal plane
and the component in the vertical plane
So that, If 9 be small we have
horizontal component of the acceleration
add vertical component of the acceleration
</e
(/l)
1OOULLJEHAT10M DOE TO QRAVttt 153
,/S/j
Now, from the equation of the pendulum, ^j /a * tng't*0, we have
And, from ths energy equation of the pendulum, viz.,
mgl (cos 8  cos a) jm (&H/ 2 ) ( ~
<* ^/ (c^ ^cos )
we have ^^  ^ p
whence, expanding cos and cos a and retaining only the first two terms, in view
of the small values ot and a, we have
/ de V gi ( 2 ~e 2 ) r . i 2 e 4 e 9 .
I ir } = j*2~r>2 " *" cos e ~ *"~~'T"i + ~A :r ~^~i
\ dt / k z +r L 2 ! 4 ! 6 I
Substituting these values of  , 2 and ( rJ in relations (/) and (//)
above, we therefore have
horizontal acceleration /. (,5, ,5, ). 6f /^. /TTTa
and vertical acceleration =* ^ ( T* /a ) ^ 8 + ^ ~^T7T >
Or, neglecting the second term as being extremely small, we have
(/ \
&* + /2 )'^
/ #/ 2 \
and vertical acceleration ( ~jj*Tjr )'&*
And, therefore, horizontal force on the support **(  ^ ^ j.Q
and vertical force on the support =f ~~fciZ/r ) "
/.., horizontal force on the support a 9
and vertical ,, ,, ,, a 6 8 .
Since, 6 is small, O 2 is comparatively very much smaller. In other words,
the horizontal foice on the support is very much greater than
the vertical force on it and the latter may, therefore, be easily
ignored.
Now, as the pendulum moves from O to G' t the point
of suspension moves from O to 0', say, so that its displace
ment is equivalent to shifting its axis of suspension to Q,
(Fig 115).
Then, if P be the displacement of the support per unit
force, in the horizontal direction, when displaced through an angle
6, its displacement due to a horizontal jorce (w#/ a /&*t/ i )0 vvil
clearly be equal to (A>/'/ 2 p/PF/ 2 )G.
Since Q is now the effective axis of suspension, ~^GQG'
and hence the displacement of the point of suspension
00' = OQ.Q.  8.0. [Putting OQ = S.
We, therefore, have .e r A whence, * .
It follows, therefore, that the effective length of the pendulum duo to
this yielding of the support is
184 PttOPEBTIfiS OF MATTER
And, therefore, the timfr}eriod of the pendulum is now given by
c A / , / ^(/f ) ^ ^ A jk**liW [W
V A^w ~ V ^^ ' L l
fWriting (/+*)
for /.
From this it follows at once that
So that, if /! and / 2 be the lengths of the pendulum for which the timeperiods
fi and / a are v^ry nearly the sams, we have
s t 2 / C 2 / x f wrnre 5 t and 5 2 are
4 7 = Ci+ s i) f y( 1 r ) \ the corresponding
2 'J v 7l y ^ additions to the
and L ( / a+ s 2 ) + ^ f i _ i ^ two lrigths of the
4^:^ /a V ^a / L pendulum.
Since /! is very nearly equal to r a , we have 2 = / A / 2 . And, therefore,
(/,+/,, + V.* ' j.fcV..
C.T7, +,,>/r ) v *
So that, putting (/ l 9 / 1 f,*/ 1 )/(/ 1 / 1 ) = T. (where T is the computed time),
we have
P For, /!^/ 2 L, the length of the
' I e Q u ^ va ^ ent simple pendulum.
s\' f\ Clearly, here, mg is the weight of the pendulum ;
 S; so that, //** correction factor wep 75 f/itf displacement of
( +" m gp * j/j^, SU ppori due to a horizontal force equal to the weight
of the pendulum and can b^ determined directly by sus
pending the pendulum from a string passing over a
pulley and attached horizontally to the support at O, as
shown, (Fig. 116), when ihc displacement OO'~mg$
(THE PENDULUM) can bc read accuratelv bv mcans of a microscope.
^ Vening Meinesz suggested a method by which
this correcion could bo considerably reduced, viz.,
that of using two pendulums, swinging from the same
support but in opposite p ases with each other. This
involves however, the d fficulfy of having to adjust
Uvir timeperiods to very near equality. The correction
>y is thus bcsi eliminated as explained in 72, (page 187).
Fig. 116.
('V) The knifeedges not being perfectly sharp, (hut more or less
rounded). Due to this also the effective length of the pendulum i$
altered.
ACCBLEttATIOtf DUlfi TO
186
For, the axis of suspension is not a mathematical line we have so far
tacitly assumed it to be In actuil practice, it has a definite shape, generally
symmetrical with a finite radius of curvature.
Let us, as a first approximation, assume the edge
to be the pai t of a cylinder, a cylinderical cone, as shown
in Fig. 117. Then, if be the centre of nirvaiure of the
edge, a line perpendicular to the plane of the paper and
passing through O represents the axis of the c>iindrical
edge.
As the pendulum (of length /) swings, the edge
also moves along with it about thi\ ax s through O, so
that the axis of suspension is. in effect, shitted fiom S to
O, i.e., through the distance SO r, the radius of curva
ture of the edge, and the effective length of the pendulum
thus becomes (/ 1r).
Since, 'however, the instantaneous axis of rotation
of the pendulum stiH passes through the bottom of the
knifeedge (S), the moment of inertia (/) is still to be
taken about this axis In other uords, we still have
/ = m.(k*+l 2 ) t where m is the mass of the pendulum.
The
becomes
equation of motion of the pendulum thus Fig. 117.
mg(l+r}B. [ Q being small.
And, therefore, its timeperiod is given by t
whence.
. =
/4r
Or.
If we find two lengths of the pndulum, say / t and /, on the two sides of
its c.g , such that the timeperiods ( 1 and r 2 for tnein are nearly the same, (with,
of course ^ not equal to J 2 ), then, if r x and r a be the radii of the two knifeedges
respectively, we have
So that, subtracting the second expression from the first, we have
Since t^ is very nearly equal to / 3 , we have k*
Or,
And, therefore,
iW [ (/iM.)  (Ijzjl)^!^) ]
Or,
have
Again, putting J ~] ~ f  = T 1 , where T is the computed timeperiod, we
*i 'a
18ft FttOPteBTlfis OF
Hcfe, clearly, the correction term ( ! (Ota 'i) tectfffiei zero, (Wily if
\ /i*i y
fi=r t , /.*., only if the two knifeedges have the same radius of curvature.
Since it is difficult to make the two knifeedges of exactly the same radius
of curvature, the suggestion at once comes to the mind that the same knifeedge
may be used at both the two points of suspension. But this may affect the
position of the e.g.. which might be different for the two positions of the knife
edge. And, then, it would disturb the symmetry of the pendulum, necessitating
the troublesome aircorrections. This difficulty may be tided over by ananging
two knifeedges of the same shape and mass, and by using only one of them for
suspension, i.e.. by interchanging them when we change the sHe of the pendulum.
Here too, however, an error may creep in if we do not succeed in replacing one
knifeedge with the other exactly in its true or original position. This difficulty
too may be got over, however, by performing the experiment fourtimes, first
taking the two observations for / x and / 8 with one position of the knifeedges and!
then two similar observations with the knifeedges inter changed . Thus,
if T! and TJ be the respective computed timeperiods in the two cases,
we have
So that, adding the two, we have
of the correc
2(/H/,),
47i* v x
whence, ^ r ^J 1  = (/!+/).
{*. * the sum of the corre
tion terms (involving
andr^O.
But, even this correction does not help much. For, a* the pendulum
swings to and fro about its mean position, the edges invariably get chipped off,
resulting in the loss of weight of the pendulum.
The one and only way of eliminating this correction, now being increas
ingly ustd, is to replace the two knifeedges in the pendulum by just plane
bearings, / <?., by flat plates, and to provide a fixed knife edge on the support,
the latttr being carefully ground to a sharp edge and the foimer being accu
rately plane or flat and always placed in the same position on the knifeedge.
N.B. In a bid for an extremely high degree of accuracy, the effect of
the elasticity of the pendulum was also investigated at Potsdam under the
supervision of Helmert, viz., its periodic extension under the varying longi
tudinal strain and its flexure under the changing bending moment to which
it is subjected as it swings the latter being the more important of the two and!
resulting in a reduction in the effective length of the pendulum.
Thanks to the work of Clark t Heyl and Cook, an accuracy of one in ai
million is now more or less easily attainable
(v) Change of Temperature during the Experiment. This'
results in a corresponding change in the length and hence the time
period of the pendulum.
A correction for it can, however, be readily applied, if we know
the coefficient of expansion of the material of the pendulum. Or,
the error may be eliminated altogether by using what are called
invariable pendulums, (see 76).
(vi) Other Errors. We have considered above the errors and
corrections in wo far as they relate to the pendulum itself. To obtain
an accurate value of g at a place, however, certain other corrections
must also be applied, v/z., the corrections (a) for rot at ion of the earth,
Ib) for latitude, (c) for altitude, (d) for elevated masses and (e) fop
ACCELERATION DUB TO (iRAVIT* 1
the terrain or the topography of the place, all of which are discussed in
the succ^edin^ chapter.
Nevertheless, as a method for determining the value of g, a
compound pendulum, (e.g., the Kater's pendulum), is distinctly
superior to a simple pendulum. For,
(i) whereas a simple pendulum is just an ideal conception, not
realizable in actual practice, the length of an equivalent simple pendu
lum, and hence the value of g, can be easily and accurately determined
with its help ;
(//)* it vibrate? as a whole, there being no lag between the bob
and the string, as in the case of a simple or a Borda's dendulum ;
(Hi) the length to be measured here is clearly defined, viz., ttie
distance between the two knife edges, and can thus be easily and
accurately measured whereas, in thn case of a simple pendulum, the
point of suspension and the e.g. of the bob are both more or less
indefinite points, and hence its true length can hardly be expected to
be determined correctly ;
(iv) due to its large mass, t\e compound pendulum keeps on
oscillating for a fairly long time, thus enabling its timeperiod to be
determined with accuracy. In a simple pendulum, on thn other hand,
the oscillations die down much too soon due to the comparatively
small mass of the bob, and it becomes difficult to determine its time
period to an equivalent degree of accuracy.
The one obvious disadvantage in the case of a compound
pendulum, however, is that during its vibrations to and fro, about
its mean position, some air is dragged aloni* with it, as mentioned
above, thus increasing its effective mass and hence its moment of
inertia. But it has been clearly shown by B^ssel that if it be of a
form, symmetrical about the centre of its geometrical shape (which is
not the same thing as its centre of gravity), this error is automatically
eliminated. This explains the symmetrical shapes of various types
of compound pendulum* we use, though, theoretically, a rigid body
of any shape whatever would do.
72. Other Improvements due to, Bcssel. Not only ha
Bessel done away with the trouble and the tedium of having to make
the time periods about the two axes identical, but he has also
succeeded in removing quite a few other important errors. Thus,
for example :
(i) The error due to some air being dragged along with the
pendulum is removed by the symmetrical physical form of the instru
ment, as suggested and shown by him.
(ii) The error due to the knifeedges not being perfectly sharp,
for which a correction, proportional to their radii of curvature, would
be necessary, (unless they be of the same radii of curvature), [see 71,
(/v), page 184], has also been eliminated by him. For, he has shown
that this error would automatically vanish if the two knifeedge,
at the two ends of the pendulum, could be made interchangeable.
Tben f if f, and f, be the computed times, before and after the in^er
188 fBOPERflfcS OF MATtfitt
change of knifeedges, the true timeperiod / is given by the relation
R
^/"ff^ Bessel, unfortunately, died before he could put his theory
into actual practice but, later, Rep>old did actually
construct in the year 186), a reversibh pendulum of this
type and used it with success.
Repsold's Pendulum is more or less a Katertype
pendulum but is svmmetrical in geometrical form^about its
midpoint, (Fig 118). Here, we have a rod 7? fixed on to
two rings R L and R z at its two ends, which, in their turn,
have two short rods screwed into them, terminating in
knifeedges E l and 2 inside the rings and carrying two
bobs B l and B 2) one solid and the other hollow*.
The timeperiod of the pendulum can be made nearly
equal about either knifeedge by moving the bobs up and
down and screwing them into the desired position.
With the symmetrical form of the peivlulurn, the
error due to air effects is automatically eliminated, as
Fig 118. explained in 71, (ii) above.
And, (iii) finally, ths error, due to the yielding of the support
has boon eliminated by D^ffarges, by using two reversible Rep sold
t)p3 p3nlulurm, of the same inns but different lengthy the sime
ratio of I to I' and h.ivin ; a common pair of knife edges, (to be used
with either of them). He has shown that if L l and L> be the reduced
lengths of tha two reversible pendulums (i.e., the lengths between
the knifeedges) and TJ and T 2 , their computed times, then \/(T 1 2 ~T 2 2 )
gives the correct timeperiod of a pendulum of length (L L L 2 ), as
can be seen from the following :
We have
and
g*i
4?r a
(See pages
LI 84 and 186.
where / t and / 2 are the two lengths on the two sides of the c g. in the
case of one pendulum and t t ' and // in that of the other.
So that, subtracting th3 83con<l expression from the first, we have
Clearly, ths correcting term, (i.e., the second term) in this expression
can be made zero if /,:/,:: V : /,', and this is easily done by adjust
ing the positions of the bobs of the two pendulums. With this adjust
ment made, we have
*This is to ensure that the lengths / t and / t of the pendulum on the two
sides of the e.g. are not very nearly equal, or else the correcting terms for the
c/ror due to air effects, (page 180), will not be small. This is the reason why in
a Kater's pendulum one bob is made smaller than the other.
ACCELERATION DUB TO GRAVITY 189
Or, if we use only one knifeedge, (fixed on the support, as
explained in 71 (///), page 181), we have
^ o = Li+mg.fi and J^ 2 = L +mg S, Seepage 184.
47T 47T a L
where Z^ and L 2 are the reduced lengths and r l and T 2 , the computed
timeperiods of the two pendulums respectively.
Thus, subtracting the second expression from the first, we have,
straightaway,
** ' \
VT 2 2 )=L i L z .
This removes at one stroke the errors due to yielding of tke
support and curvature of the knifeedges, as also those due to air
effects, and we have
*. a
This is about the most accurate method of determining the
\alue of g at a given place.
73. Conical Pendulum. A simple conical pendulum is just
a simple pendulum, (ie., a srnill heavy bob attached to a light, inex
tensible string), which is given such a
mot ; on thit the bob describes a horizontal S
circle and the strirg traces out a cone. The < *
length of the pendulum is the distance
between the point of suspension and the /
e.g. of the bob. /
\r
r,
Let m be the mass of the bob J?;
v, its velocity and r, the radius of the
circle it describes, (Fig. 119). Then, its
centripetal acceleration towards 0, the
centre of the circle, is equal to v 2 /r, and */. " nL. "."." I ".13
the centripetal force on it is, therefore, '" T>^*'
mv*/r in that direction, ma
Let SO be equal to h. pig, 119^
Clearly, the forces acting on the
bob are (i) its weight, wg, vertically downwards, and (U) the tension of
the string T, in the direction BS.
The weight mg is balanced by the vertical component T cos 0,
of the tension T of the string, and its horizontal component T sin
provides the centripetal force wv 2 /r towards O, where o is the semi
vertical angle of the cone
Thus, T sin = mv 2 /r and T cos = mg.
n T sin mv*lr v 2
Or, = tan = L ^ Jl_ t
Or, v*/rg=~tan 0. Or, v*=r.g.tan 0.
Since v 2 ==r 2 .co a , where w is the angular velocity of the bob, we
have r*.w 2 =r.g. tan ^ r.g,r//j=r 2 g/A, [\ tan 0=r/A.
whence, uP^glh, and ,*. o>
? 1
PEOPlfiBTIBa OF MATTEB
Now, the timeperiod of the pendulum is given by
2* 27T _
= JrTTA /
g
t =27rA > ..(it) [v ft /we.
where / is the length of the pendulum.
It will thus be noted that the timeperiod is the same as that, of a
simple pendulum of length h. the axial height of the cone.
If he very small, cos is nearly equal to 1 ; so that h = /,
i.e., the time period is almost independent of 0. /H ctf/zer won/5, the
timeperiod remains the same whether the bob moves along a circular or
a linear path.
74. Steam Engine Governor. It will be seen from relation (/)
above, (73), that the angular velocity (co) of the bob of a conical
pendulum varies inversely as the square root of the depth (//) of its
e.g. from the point of suspension ; or, conversely, that the depth of
the e.g. of the bob, below its point of suspension , varies inversely as the
square of its angular velocity,
This is made use of in the construction of what is called the
"governor* is a steam engine, which is just a device to maintain the
speed of the engine constant by regulating, or 'governing', the supply
of steam from the boiler to the steam chest,
In essentials, it is just a combination c*f two similar conical
pendulums, mounted on either side of the vertical shaft (with a com
mon point of suspension), rotated by the engine,
and cons, sts of two rods OP and OQ hinged to
gether at their upper end O to the shaft OS, and
carrying two spherical metallic bobs P and Q at
their lower ends, (Fig. 120). Two other smaller
rods connect O/^and OQ to a metallic collar C,
which slides freely along the shaft, thus operat
ing a lever which controls the throttle valve, or
the steam valve, opening it partially or fully,
according as the collar moves up or down the
shaft.
Now, when due to a greater supply of steam
to the cylinder, the shaft, and, therefore, the
bobs rotate faster, i.e., & increases, h propor
tionately decreases, or the bobs rise up, thus
partially closing the steam valve, thereby partially cutting off the
supply of steam to the cylinder. This automatically results in a
falling off of the speed of rotation (<o) of the shaft or the bobs, and
when this happens, h increases, i.e. 9 the collar slides down with the
bobs, thus opening the steam valve more fully, allowing more steam
into the cylinder, which then, increases the rate of rotation of the
shaft. So that, by proper adjustment, the rate of supply to the
steam chest or cylinder, and, consequently, the rate of rotation
the shaft, can be maintained at any constant v
DUB TO GRAlfttt 19)
The sensitiveness of the device, however, decreases with the
increasing speed of the engine. For, we have the relation
co 2 g//f, ...(/)
differentiating which, we have 2cu.rfco =* ~~~dh. (' v )
And, therefore, dividing relation (fv) by (///), we have
Zw.da) g.dh h ~ , . ,,
2 as TT X   Or, 2rfa>/a>= aft/A,
^
whence, dft =s  ~
0}
Or, substituting the value of ft, from relation (Hi) above, we have
Thus, it is clear that dh oc l/o> 3 , i.e., dh decreases as w increases.
In other words, the change in the position of the e.g. decreases with
increasing angular velocity of the bobs or the shaft, thus slowing
down the 'up and down motion' of the collar along the shaft or
decreasing the sensitiveness of the device.
75. Other methods for the determination of <g'. The following
are a few other methods that m iy bs used to determine the value of
g at a place. Although they do not compare favourably with the
pendulum methods in point of accuracy or ease of performance, they
are, nevertheless, valuable laboratory exercises, affording good illus
trations of tbe various principles employed for the purpose. Here,
then, are these different methods :
(1) The Inclined Plane. We have seen before, in 39, (page
88), bow the acceleration a of a body, rolling down an inclined plane,
(without slipping), is given by the expression,
a=*[r*t(k*+r*)]g sin a,
where r is the radius of the body ; jfc, its radius of gyration about its
axis of rotation ; a, the angle of inclination of the plane, and g, the
acceleration due to gravity at the place.
It follows, therefore, thatg= *
* *
sin a
So that, knowing r, (by means of a vernier calliper) k, (from the geo
metrical shape of the body) sin a,, from the height and length of the
plane) and a, (by direct experiment, as explained below), we can easily
calculate out the value of g at the place.
The value of a can be easily and accurately obtained by noting
the distances covered by the body, down along the plane, in succes
sive equal intervals of time*, and plotting the distancetime curve
for it. The equation of the curve being 5 = af 2 , (u being zero, be
cause the body starts from rest), it will, obviously, be parabolic in
*This may be easily done if the angle of inclination (a) of the plane be
small ; for, then the acceleration of the body will also be small and the time
taken by it in rolling down the place will be fairly accurately measured by means
of a stop watch.
192
PROPERTIES OP MATTER
form, with its axis coinciding with the distanceaxis. So that, sub
stitu ing tha coordinates of any snitabh p'jint on th 3 curve in the
relation S \at 2 , the value of a can be easily determined.
(2) The Dynamical Spherometer. Let S be the centre of
curvature of a spherical surface, arranged horizontally with its con
cavity upwards, and, /?, its radius, (Fig.
121) ; and let a steel ball, of mass m and
radius r 9 be allowed to roll to and fro
on it, wilhout slipping. Then clearly,
the ball oscillates on the inside of the
surface as though it were a compound
pendulum, wiih its centre of suspension
at S and its centre of oscillation at O t
the centre of the ball ; so that, SO
(*')
Let the angular amplitude of the ball be 0', (i.e., the angle that
it makes at S, when in its extreme positions). Let it be in the posi
tion B 9 at any given instant, such that the angle it now makes at
s is 0.
Then, clearly, work done on the ball by the force of gravity in
bringing it down from A to B is equal to weight of the ball x the verti
cal distance through which the ball has fallen down.
This must, therefore, be clearly equal to the loss in its poten
tial energy, i e., equal to mg x PQ where mg is the weight of the ball.
Or, loss in P.E. = mg(SQSP).
Now, SQ = (Rr).cos and SP = (Rr).cos 0'.
And .. SQSP = (Rr).(cos 6cos 0').
So that, work done by the force of gravity in moving the ball
from A to B is equal to mg.(Rr) (cos cos 0') and is equal to the
loss in the potential energy of the ball.
This must, clearly, be equal to tha gain in the kinetic energy of
the ball i.e., equal to /.o/ 2 , where / is the moment of inertia of the
ball and a/, its angular velocity about the line of contact.
So that, mg (Rr).(cos B cos 0') = J/o/ 2 .
If co be the angular velocity of the ball about a horizontal axis
(jy .
_H_ Va,.
.. mg.(Rr).(cos 0cos 0')
whence, differentiating with respect to time, we have
, since dd/dt = w, we have
0.<o
(J?r) da,
ACCELERATION DXTB TO GRAVIT7 193
Or, mg.(Rr)jin 6 = 7 . . ~ .
^ wS'r*(R~ r ) sin 9
mg.r 2 .sm w#.r 2 . p.
* I(RrT ~~ /.(.R^rp' L
. T .
Now, /.
/n 0=6, if
besmal1 
_ 5g_
" f/ ' '
Here, clearly, dw\dt is the angular acceleration of the ball ;
5?
so that, angular acceleration of the ball = T^^V*^*
/.., the angular acceleration of the ball is proportional to 0.
[v 5g/7(Rr) is a constant].
The ball thus executes a simple harmonic motion, and its time
period is, therefore, given by
o . ^. i,t .~
Squaring this expression, we have /* r = r 
287r 2 (/?r)
whence, g = ~
o/~
Thus, knowing the radii of the concave surface and the ball
(with the help of a spherometer and a vernier calliper, respectively),
and noting the time period of oscillation of the ball, we can easily
calculate the value of g at the given place.
N.B. Rearranging the expression for t 2 , obtained above, we
have 5f 2 = 28w 2 jR287T 2 r. Or, 287TIR = 5g/ 2 +287T 2 r.
Or R
ur, K
So that knowing r, the radius of the ball, t, its period of
oscillation and the value of g, we can easily calculate the radius of cur
vature (R) of the given spherical suriace.
(3) The Atwood's Machine. In the ribbontype of machine,
[Fig. 122 (a)], a strip or ribbon R is passed round the flat rim of a
light and frictionless pulley, (running on ballbearings), with two
equal masses M and M at its two ends, one of which is initially kept
resting on a platform P.
An identical ribbon is attached to the lower ends of the two
8 shown, 90 that when th ayvtem iff set into motion, ap
194
PROPERTIES OF
additional length of the first ribbon, passing on to the right side of
the pulley, is exactly balanced by an equal length of the second, pass
ing on to its left side, thus ensuring thai
no extra mass is transferred from one siJt.
of i he pulley to the other.
A steel strip or vibrator F, of a
known timeper.od T, is clamped hori
zontally at one end, and carries a light
style, or an inked brush B, which just
touches the paper ribbon going round
the pulley.
A small rider r, of mass m t is placed
on the mass M, rest ng on the platform,
to make it slight'y heavier than tho other.
Then, with the rna c ses not yet in
motion, the bnibh is moved across the
paper ribbon to mark a horizontal line on
it, indicating the starting point.
The platform P is now suddenly made to fall, (by means of
('trigger releases') and the vibrator simultaneously set vibrating.
Naturally, the mass loaded with tho rider, moves down and the
other up with a common acceleration, say a. And, as the ribbon
runs, past the brush, a \\avy curve, duo to the transverse vibrations
of K, gets traced OH it, and goes on gradually lengthening out,
[Fig. 122 (/?)], on account of the accelerated motion of the masses,
and hence that of the ribbon.
Since one wave is traced out on the ribbon during one vibration
ofV, the distances occupied by successive waves represent the dis
tances covered by the masses during successive time periods of it.
Thus, if S M So, S 3 etc., be the distances covered by the masses
in the first, second and third etc., timeperiods of K, we have
[/ u = 0, the masses starting from test.
here u = aT, the velocity
after time T.
Fig. 122.
and S. 2 = aT.T+laT* = aT*+laT*,
S 2 = T0P/2
5, = 2aT. T.
c
V now u 2'T, the velocity
after time 2T.
i.e.,
And, similarly,
So that, S 2 5 l = 5 3 S^ = aT 2 = x, say. Or, a^x/T*.
Thus, 7 being known, the va'ue of a can be easily calculated out.
Now, if v be tho velocity acquired by the masses, when they
have covered a distance, h we have
gain in K.E. of the pulley and rhe masses ('ogether with (he rider)
= /0ss in P.E. of the masses and the rider,
ACCELERATION DUB TO GRAVITY 195
where /is the moment of inertia of the pulley about its axis of rota
tion and w, its anguLir velocity at the time.
Now, if R \ e the radius of the pulley, v = Rw, or w = vjR.
Hence, \LviR 4 \(*M\ m).v 2 = wg/7.
Or, Iv
Or,
whence, , _v'(//JP + 2M + m).
* 2mA
But v 2 = 2ah. ['.' u = and S  A.
__ 2flA(//JP + 2 A/ + w) g(//JP + 23f f w)
g ~~ 2mh  ~ m~ ' W
whence, the value of g can be easily calculated.
It H, however, 'lesir.ible to eliminate /from this expression, by
repeating the experiment with the same masses but a different ridtr,
of mass /'. If a' bo no\v the accaleration of the masses, (determined
as before), we hav r e
Rearranging relations (/) and (//), we have
mgla = (///? + 2Jf + m),
and m'gja' = (7//J 2 + 2A/ + W).
So that, fliibtiMctin^ relation (/V) from relation (///), we have
mgla  m'%[a' = (mrn f ). Or, g(Aw/a  m//') == (mm'), ^
, , ~ ,'/ m m' \*
whence, g = (/HTH ) ^ _ J .. ( v )
Thus knowing w?, w', ^ and a', the value of g can be easily obtained.
A possible source of error, here, is the fi'idional force encoun
tered by the pulley as it rotates about Jts axle, which, obviously,
tends to lower its angular velocity. This may be easily remedied by
placing another auxiliary rider on the loaded mass,,, such that, with
the main rjder (r) removed from it, if an initial velocity be given to
it, to sot the svstem in motion, it continues t"> move with the same
uniform velocity, i e., (its motion, is neither accelerated nor retaded).
Obviously, tlnn, the weight of this auxiliary rider exactly counter
acts the retarding force due *r friction. If, therefore, kept on the
miss throughout the expe ent, it completely eliminates the error
due to friction, and, clearl, neither its weight nor the fractional
force need enter into our ca' at ions.
(4) The Dropping Plate. A plate of glass, P, smoked by holding
it over burning <urnph'>r, is suspended with its plane vertical, by
means of a thread, as shown in Fig 123 (a), and a tuning fork f,
of a known frequency \ f is mounted close to it, so that a light alu
minium tyle, (or better still, a hog's b*ii>tle. such as may be obtained
from a discarded hair b ush), attached to one ot its prongs, just touches
the surface of the plate.
*Or, we could u*e different ma^^c* M a.id M' but the *amc rider, in vhich
case w: shill tuvs g = 2(AfAf' )/mU/ !/') where a' is the acceleration of
tjie masses in the second case,
196
PROPERTIES OF MATTER
The fork is set vibrating by lightly drawing a bow across it, or
by simply pinching it strongly, and the thread, supporting the plate,
burnt or cut simultane
ously. Thus released,
the plate starts fall
ing, with an accelera
tion, equal to the value
of g at the place, and
the style traces out a
wavy line on it, of the
form shown in Figs. 123
(b) and (c), the waves
being smaller and closer
together at first, but
gradually lengthening
out and getting further
apart, due to the accele
rated motion of the
plate, though the time
taken to trace each
wave remains the same,
viz., equal to 1/N, the
timeperiod of the fork.
(a)
JL.
Y Three points D, E
* and F, [Fig. 123 (b)\
' are then marked on this
wavy line, such that DE
and EF contain the same number of waves, say, n, each Let distances
Fig. 123.
'and 7'" be 8 l and S 2 respectively, as measured by means of a
travelling microscope, both being covered by the plate in the same
interval of time t njN, taken by the fork to complete n vibrations.
Then, clearly, 5, = ut+\Qt*. Or, 2S t = 2w/fgJ 2 , ... (/)
where u is the velocity of the plate at D.
And, " (S,+S ? ) = 2w/+g.(2f) a .
because, here, distance = (8 {S 9 ) 9 and time = (2f).
80 that, subtracting equation (/) from (//), we have
Or >
whence, g = ( iT~LtL ...(m)
Or, substituting the value ,w/JV for /, in relation (///), we have
g 5. . ~  y as ^ ... (JV)
Thus, knowing N, n and (S 2 Sj), we can easily calculate out the
value of gat the place.
N.B. It will be readily seen ttfat the mass of the style (or the 'hog's
bristle 9 ), attached to the prong of the foik, together with the friction it encounters
at the plate, will slightly lower its frequency, so that it will actually be somewhat
less than N. For greater accuracy, therefore, the frequency of the fork (with the
style attached to it) must be determined by the method of 'beau\ by sounding it
to
with another fork of an accurately known frequency. The frequency, thus deter
mined, should then replace N in relation (iv) for g, above.
Alternative Calculation. The following is a comparatively more
accurate method of calculating the value of g, because, here, the
possible error in correctly counting n is eliminated.
Three points A, B and C are marked on a portion of the wavy
line, [Fig. 123 (c)], where the waves are clearly visible and can be
distinctly counted. Let there be n l waves (and, therefore, x vibra
tions) made by the fork inbetween A and B, and n 2 waves (or
W 2 vibrations) in between B and C, and let the total distance AC be S.
Then, clearly, (tf 1 +w 2 ) waves or vibrations are made by the
fork in time ( 1 +fl 2 )/jV. So that,
S z=~
^
. Or,
A /"F "1
= V 4 ~
Y 2 ^
and
Plotting
against
n i9 therefore,
we obtain a straight line, (Fig. 124),
of slope A /? / N 9 from which the
value of g can at once be calculated,
without knowing n.
Incidentally, the dropping plate
method also shows that a freely fall
ing body is subjected to a constant
acceleration due to gravity,* a fact,
not easy to demonstrate otherwise.
(5) Vertical Oscillations of a
Flat Spiral Spring. A spiral spring is
Fig. 124.
just a uniform wire or ribbon, designed to have, in its normal, un
strained condition, the form of a regular helix, such as may be
obtained by winding the wire closely and uniformly round a cylinder,
of a diameter much greater than its own.
If the plane of each coil of the spiral is perpendicular to the
axis of the cylinder, it is called a flat spiral, but if it be
inclined at a small angle to this axis, it is spoken
of as an inclined spiral. We shall concern ourselves
here only with the flat spiral of a wire of circular
crosssection.
If a small force be applied to a flat spiral, along
its axis, (i.e., along the straight line passing through the
centre of each coil of it), and perpendicular to its plane,
it increases in length a little, but still preserves its
helical form, as will be clear from Figs. 125 (a)
and (b).
Let us consider a flat spiral, of length L and
radius R, (where R is much greater than its pitch),
suspended from a rigid support, with its upper and
lower ends (A and B), bent as shown in Fig. U6 (a) 9
ao as to lie along its
198
PEOPEBTIBS OS MATTBB
Then, if a mass m b3 suspond^d from its lower free end, & forcd
equal to mg (the weight of the mass) acts vertically downwards
along its axis, producing a statical
extension/ n its length, (/e., an
extension, with the mass m at
rest).
The effect of this force mg,
acting along the axis of the spiral
is to produce a turning monent,
equal to mg R. at every section of
it, [Fig. 12(5 (/?)]. And, this, in its
turn, subjects the wire to a uni
form twist 0, say, per unit length
of it.
Now, the twisting or tor
sional couple* per unit length
of the wire is equal to rar 4 tf/2,
where n is the coejfif ient of rigidity
of the material of the wire ; r, its
(a; radius and #, its angle (>ftwi>t.
Fig. 126.
This, therefore is the torsional resistance, opposing the turning
moment mg.R due to the weight nig. For equilibrium, therefore, we
have mg.R = n.ir^d/2. ... (/)
Now, a twist per unit length of the wiro corresponds to a
displacement or extension R.6 per unit length of it. [Fjg 12t> (b)].
Charly, therefore, the extension produced in the \\hole length of the
wire is equal to L.R.6.
Or, / = L.R.O, whence, 6  IjLR. \ v ' is the to ' al '
' . L tension produced.
Substituting this value of 6 in relation (/) alove, we have
Tiw 4 / Trnr 4 /
mg.R a= r~ . jg = ..  p , whence,
The expression Tnr*.l/2LR' thus represents the force of elastic
reaction for an increase / in the length of the spiral . and. therefore,
the elastic reaction per unit incievse in the length of the spiral
as 7tnr*/2LR*, (because /I). Denoting this by K, we have
mg = K /, whence, K = mg/l.
If the suspended mass (m) Le now displaced or pulled vertically
downwards through a distance x and then released, so as to produce
vertical oscillations in the spiral, the restoring force F. acting on the
mass may for small oscillations, be taken to be dirjctly proportional
to its displacement. So that, F = K.x.
And, if d*x/dt 2 be the acceleration of the mass at the given
See chapter VII, where it is shown that the twisting couple on a cylinder
(or wire) is equal to OTtr 4 G/2/ f where Q i^ the angle of twist and /, its length*
ACCELERATION DUE TO GRAVTl^ l&S
instant, when its displacement is x, the inertial reaction of t\e mass
is clearly equal to m.d'x/dt*.
Hence, by Newton's third law of motion, we have
m.d 2 x/dt 2 =Kx, whence, d x/rf/ 2 = x.K[m.
Now, since K and m are both constant quantities for the given
spiral, we have Kjm = a constant, p,, say.
So that, d*xldt* *=  n. x. Or, d z x[dt* oc x ;
i.e., the acceleration of the mass is directly proportional to its displace
ment x ; and it, therejore, executes a simple harmonic motion, its time
period being given by the expression
/ = 2irvT/ ~ 27TA / * , Or, = t*^lR = 2* A / ...(//)
V Kim V nig 1 1
Or, t = 2ir v '/7F ;
i e., ///e time period is ihe same as that of a simple pendulum of length /,
the extension produced in the spiral.
Squaring and rearranging this expression for /, we have g = 47T ? .//f 2 ,
whence the value of g, at the given place, can be easily calculated out.
In the above treatment, we have not tak'm into consideration
the mass of the spring, assuming it to be negligible, compared with
the suspended mass m. For gre itcr accuracy, however, it must also
be taken into account. So that, if the effective mass of the spring be
m s , the total mass acting downwards along the axis of the spring
becomes m+ 3 and the expression (//) above, for the timeperiod of
the spiral, becomes / = 27t\/(m\w s )/K^.
It is, however, best to eliminate m s altogether. This is done
by performing the experiment with two different suspended masses,
m 1 and w 2 .
Then, if t l and t% be the respective timeperiods of oscillation
in the two cases, we have
t l = 27rv\w 1 Fwj7A' and f 2
So that; squaring and subtracting the second from the first, we have
.....
...(in)
Now, if /, and / 2 be the statical extensions corresponding to the two
mabses, w r e have
m v g = AT/j and m^g = K1 2 ,
whence, m r g^m 2 .g^Kl } Kl 2 . Or, (w, m 2 ).g = K(l L l^.
A g
Substituting this value of (m l m. 2 )IK in expression (i/i) above,
have (^/a 1 ) = 47r^~ i? , whence, g
Thus, observing f, and ^ 2 directly, and measuring / t and /,, by
noting the positions of a light pointer, attached to the spiral, on *
2UO
PBOFBRTIBJS OB MATtlA
vertical centimetre scale fixed alongside it, the value of g can be eatfi
ly calculated.
(6) The Bifilar Suspension. If a heavy and uniform bar or
cylinder, (or, in fact, any rigid body), be suspended horizontally by
means of two equal, vertical, flexible and inelastic threads, equidistant
from its centre of gravity, the arrangement constitutes what i called
a bifilar suspension. On being displaced a little in its own plane, i.e.,
in the horizontal plane), and then released, the bar or cylinder exe
. cutes a simple harmonic motion about the vertical axis through its
centre of gravity.
Now, two cases, arise, (/) when the two suspension threads are
parallel, and (//') when they are not. Let us consider both.
(/) Bifilar Suspension, with Parallel Threads. Let AB [Fig. 127,
(a)], represent the original or equilibrium position of a cylinder, of
mass m, and with its e.g. at 0, where its weight nig acts vertically
downwards. Let the two suspension threads PA and QB b*
parallel to each other, and distance 2J apart ; and let the length of
each be I.
Now, if the cylinder be displaced a little into the position A'B',
through a small angle 6, about the vertical axis through (9, the sus
pension threads take up the position PA' and QB' at an angle </> with
their original positions, where <f> is small.
Let T be the tension in in each thread, acting upwards along it.
Then, resolving it into its two rectangular components, we have
() (W (c)
Fig. 127.
(/) the component T cos <f>, acting vertically upwards ;
nd (//) the component Tsin ^, acting horizontally along B'Band A' A
[Fig. 127 (b)].
Obviously, the vertical components support th weight of the
cylinder. Hence,
?r cos J> mg. Or, T cos i
fO GfcAVlTt 201
And, since <f> is small, cos <f> = 1, very nearly.
So that, T = wg/2.
The components, T 5fw <, (acting at ^4' and B')< on the other
hand being equal, opposite and parallel, constitute a couple, tending
to bring the cylinder hack into its original position. And, since A' A
and B B are practically at right angles to A'B', we have
moment of this restoring couple = T. sin </>.2d = T.<f>.2d. ' S mall, C
L .y/fl = 0.
Now, = BB'IOB = #'/rf ; so that, BB' = e.d. [Fig. 127 (c).
And, therefore, ^ = BB'\l = 0.<///. [Fig 127 ().
Hence, r^ormg C*HJ>& = T.'4'2d = 2 .** . 2L^ f . tf . But
/ & l i
the restoring couple is' also =? Ld 2 &ldt* 9
where Us the moment of inertia of the cylinder about the vertical
axis through 0, (its e.g.), and d 2 $/dt*, its angular acceleration.
T d*8 mg.d* n
' 
* V* jf J.
Now, mg.d 2 jll is a constant quantity, in a given case, and, there
fore, putting it equal to n, we have,
d^Qldt 1 = n.Q. Or, d*0/dt* oc 0,
/.., f/t angular acceleration of the cylinder /? proportional to its angu
lar displacement, and is clearly directed towards its mean position. The
cylinder, therefore, executes a simple harmonic motion and its time
period is given by
T
mg~d*Jll
Or,
O  o
Or, if we put / = mfc 2 , where k is the radius of gyration of the cylifi^
about the vertical axis through <9, we have
ri>__ x
s ^7f __^
M ^
k ,
whence, T = 27r.~7 / V  s ^ (")
d * g
Now, squaring relation, (i) or (//), and rearranging, we have,
from relation (/), g = 47T 2 /.//m.rf 2 .r a ... ... ...(/)
and, from relation (), g = 4:r a .fc 2 .//d 2 .r a . ... ... ... (/v)
And, thus, the value of g at the given place may be easily cal
culated out.
(ii) Bifilar Suspension with NonParallel Threads. Let the rod or
cylinder AB, [Fig. 128 (#)], be suspended symmetrically by two equal
but nonparallel threads*, each of length /, and let the distance bet
ween the threads at the top and at the bottom be 2d lt and 2d t res
pectively, where (d % dj = X
*The threads, in this position of the rod, are not shown, to avoid COJOQ
cHicatini the Figure
202
FROFERTIES OF MATTER
If the cylinder be displaced through an angle 0, in its owri
plane, into the position A' B' t the suspension threads take up positions
PA' and QB'. Then, as befora, tension T acts upwards along each
thread, and may be resolved into two rectangular components, v/z.,
(/) T cos <, acti'g vertically upwards, as shown, and (ii) T sin <,
acting horizontally, along A'K and B'L [Fig. 128 (0)J. where <f> is
the angle that each thread makes with the vertical, or the perpendiculars
PK and QL, from P and Q on to AB.
Id,
Ts>n
tfl ff '7TT  * "> ^ '' z
CP^'>
(a)
T S,r>
Fig. 128.
The vertical components T cos <f> support the weight mg of the
cylinder, acting vertically downwards at its e.g., O ; and, therefore,
2T cos j> = mg. ' ...( v )
Now, clearly, LB' = ^^d^^d^d^sJ. [Fig 123 (c)
L 2 2,J^d L , approximately.
B'R
Or, LB' = (did^) = #, approximately.
And, c0$ $ ss
Now, T
r 6 b;i
L
cing small, and
cos o=l, nearly.
2
Zv/ 2 *'// "
[Fig 128(6).
From relation
(v) above.
And, resolving forces T sin <f>, acting at A' and B' along A'K
and 5X respectively, into their rectangular components along and at
right angles to A'B', we have
the components at right angles to A'B' = T sin <f>. sh a, [ Fig. 128 (c).
Since these two components act in opposite directions at A'
and B\ they constitute a couple, tending to rotate the cylinder back
into its original position AB ; and, clearly,
moment of this restoring couple = T sin <f> . sin a. A'B'.
Or, restoring torque on the cylinder =* Tsln <f>.sin oi.2d t .
ft , mg.l x C . r
*= 2d r A= r. sin a.
from Fig. 128 (b).
Now, the sides in a triangle, be ; ng proportional to the sine* of
the angles opposite to them, we have, from Fig, 128 (c),
djsin * LB'Isin 9,
ACCELERATION DTTE TO GRAVITY
^ ,,,,,. , A , . 1 n
So that, rf,/5/n a = x/0. And .. sin a = .0.
Hence, restoring torque on the cylinder
X 
,] Q
fiut, restoring torque is also = ^.r 2
where /is th3 mvnent of inertia of the cylinder about the vortical axis
through its e.g., and d^^jdt 2 , its angular acceleration.
^
^ Or,
where, ' ? ' = a constant (*.
7. v / *'
Tims, cc 0.
Or, ///e angular acceleration of the cylinder is proportional to its angular
displace'mit. It. therefore, executes a simple haimonic motion and its
time period is given by
Or, r= 2; r^. ...(v)
v rf, ^/ 2 .mg v ;
And, if we put / mk* 9 where k is the radium of gyration of
the cylinder about the vertical axis through its e.g. we have
Now, if 7 be tin vertical dista y ic3 bstwean the two ends of each
suspension thread, we have y = B'R ~ LQ = y/^ #. [Fig. 128 (6).
So, that, " r = 27r.A = A / J!_. ... (v/ii)
i ^ V S
Again, squaring and roarran.jiiig relations (v//) and (v//7), we
have
/ i j / ..v 47T*.A.\/* ^^ . v
from relation (v//) g= r/ r r" ....... (^)
MI a**m*j. ~*
47T* fc 2 V
and, from Nation (v///), g = ,'1.^. ^ ...... (x)
a^.u c .7
The value of g, at the place, can thus be calculated from either
of thee relations
It will be readily seen that if </, = d. =* d, and y = /, so that
x = 0, i.e., vr/zgfl //ie /wo threads are parallel and vertical in the
original equilibrium position of the c^ Under, we have
(tf) relation (v//) reduced to 2* *= STT ^ A/  1  , the same as
r elation (/), for parallel threads,
204
kHOPKBTIBS
fflATTKB
(b) relation (v/ff) reduced to T = 27r.~rA/ ,the same ad
relation (//) for parallel threads, (see page 201).
Note. It will be clear from the above that the bifilar suspension may
also be used to determine the value of /for the suspended cylinder etc. For,
relation (///) above, when rearranged, gives / =  . n 2/
and relation (ix) 9 when rearranged, gives / = '
4n \r x"
In fact, this method is more suitable for determining th
moment of inertia of a body than for determining the value of g.
76. Variation of the value of f g'. The value of g at a given
place is affected by a number of factors, viz.,
(i) latitude of the place ; (// altitude aiid (Hi) depth. We shall
DOW proceed to study the effuct due to each of these factors a little
in detail.
(/) Effect due to Latitude. The effect of latitude on the value
of g may be considered under two headings, v>z., (a) the effect of
the rotation of the earth, and (b) the effect due to the bulge at the
equator.
Let us consider each separately.
(a) Effect of Rotation. We know that the earth is rotating
about its axis from west to east. If it were at rest, and were a
homogeneous sphere, the acceleration due to grav'ty would be the
same for a body at all points on its surface and would be directed
towards its centre. Due to its rotation, however, part of the force
of gravity on the body is used up in overcoming the centripetal force
acting on it, and thus the resultant acceleration on it is different,
both in magnitude and direction, at different places, i.e., the
apparent value of '#' is different in different latitudes, as will be clear
from the following :
Let NWSE, (Fig. 129), be a section of the earth, (supposed to
yy be a perfect sphere), through its polar
diameter NS, and let its radius be r. Then,
if a* be the angular velocity of the earth
about the axis of rotation NS, all points on
its surface rotate about this axis with the
angular velocity o>. The linear velocity of
ach point will depend, however, on its dis
tance from the axis. Thus, the linear velocity
of a particle at the points E, N, W and S will
be r.co, and that at a point P, distant
PM = x from the axis, will be x.eo, where
x is the radius of the circle that the point P
describes as it rotates with the earth.
Let <f> b the latitude in which the point P is situated. Then,
since the radius PM, of the circle described by P, is r cos 4>, the
linear velocity of P = r cos <f>.aj ; so that, the centrifugal force acting
QB P, oway from the centre (M) of the circle it describes, and acting
ACCELERATION DUE TO GRAVITY 205
along MP, is clearly given by m.r.cos <f>.aP. Let it be represented in
magnitude as well as direction by the straight lino Pp.
The force of gravity mg, which would act on the boly if the
earth were at rest, (g being the acceleration due to gravity, with the
earth at rest), would obviously act towards the centre of the earth O.
Let it be represented in magnitude as well as direction by the
straight line PO.
Thus, there are two forces acting simultaneously at the point
P, viz., (a) the centrifugal force m.r.cos <.eo 2 , ahng PF, and (b) the
force mg due to gravity along PO. Completing the triangle of forces
POQ, where PO represents the gravitational force mg, and OQ, the
centrifugal force m.r.cos <.o> 2 , we have the resultant force at P
represented by the third side PQ of the triangle, both in magnitude
and direction, where
PQ = ^POOQ*~2PO~OQcos>OQ. [See Appendix 1 "7 (2),
r.r* cos 2 <j> to 4 2m*. g r cos^.a)*.
Now, the value of r.w* comes to be about 339 cms. I sec*., or
about 1/288 of the value of g; for r ^ 6378xl0 8 cms., and
<o = 27T/86164, (where 8(5164 is the number of mean solar seconds in
one day),
Thus, the expression jn*.r*.cos*<f>a>* is negligible, compared with
the other terms involving g, and, therefore,
PQ = y/mg* 2m*.~g.rcos*i~^>* = \/m*(g*^i.f~c(^^)'
9 * i V frt
mg 1  
Or, PQ = mg( 1  x~ ' w + some other negligible terms )
\ + g ^ /
Or, the resultant force on P = mg( 1 fOJ ' c \
.'. if acceleration of the point P, in latitude <f>, be g ,, we have
T
" '""' a
This value is obviously smaller than g, and is directed towards
, and wo/ towards 0, the centre of the earth ; the angle OPQ, or
e change in direction of the gravitational force is, however, very
small.
For pointy on the equator, since <f> = 0, and, therefore, cos<j>= 1,
the value of the centrifugal acceleration = r.<o a , i.e., a maximum.
And, for points on the poles, because ^ = 90, and, therefore,
cos <^ = 0, the value of the centrifugal acceleration is zero, i e., a
minimum.
It follows, therefore, that the apparent acceleration of a body is
the least at the equator, and the $reat$st qt the poles, with \n
206 PROPERTIES OP MATTER
Substituting tho valua of r.aj* ! g = 3'39'978 = 1/288 in the
expression for g above, (the value of g being 978'03 cms.jsec*. at the
equator), we havy
,*( '!?) _
This is a result, not quite in agreement with the experimental
value.
The discrepancy may, however, be ascribed (/) to the elliptic ity
of the earth, its radius increasing as wo proceed from the poles towards
the equator, so that points in the higher altitudes arj nsaer to its
centre than those near th^ equator ; (//) to the nonhomogeneity of its
comyosilio'ii tho density of its different layers b3inz different, with
the i m,r layers compirjiivelv much denser ih:m (about more than
twice as dense as) the outer ones.
(b) Effect of Bulge at the Equator. It was RVier, whose ex
periments in 1672 first showed a variation in the value of 'g' at
two different places.
Determining the length of a seconds pendulum at Cavenne
(in French Guiana) and at Paris, he found its length at Paris to be
just over onetenth of an inch greater than at Cayenne, clearly
showing the value of g to be greater at Paris.
Newton soon explained this variation on the assumption that
the earth behaved as though it wera a 'uniformly gravitating fluid
globe' ; so that, by virtue of its very rotation, it was bound to have
a spheroidal shape, with a bulge or a protuberance at tho equator,
and comparative flattening oh the poles under the influence of the cen
trifugal force acting on it, tho valua of which varies from zero at tho
poles to a maximum at the equator. In fact, even if the earth \\ero
perfectly rigid, it should have assumed this shape before it actually
cooled down. As a consequence, the equatorial radius is about 13
miles greater than its polar radius. Hence, all bodies in the equatorial
regions are farther from the centre than those in the polar regions,
and the force of attraction due to gravity on the latter is, therefore,
greater than that on the former.
It can be shown that the true value of g at a place in latitude A
is given by the relation,
g = (98O61 *025 cos 2 A) cms. /sec*.
These changes in the value of g due to latitude are of great
help in determining the figure or the shape of the earth.
(//) Effect of Altitude. The correction for altitude we really
owe to Laplace and Stokes, particularly to the latter.
Let g be the value of acceleration due to gravity on the surface
of the earth and g'. its value at a height h above the surface. Then,
if the earth be considered to be a sphere of homogeneous composition,
the acceleration duo to gravity at any point above its nurfaco will
vary inversely as the square of the distance of that point from its
centre ; so that
, ^5 == : .f ==: f= * + + pf [^radius oftheeartfy
AOCBLBEATION D0B TO GBAVTTY 207
If h b3 small, compared with r the quantity W\r* will ba negli
gibly small, and we shall, therefore, have
g/g' = 1+iVi/r. Or, g'lg = l/(l+?/i/r) = l~2/i/r appro*.
Or, g' g(lvA/r),
/.., the greater the value of A, the smaller the value of g'.
Or, the value ofg decreases with altitude.
The general expression for the acceleration due to gravity at
altitude h and in latitude \. thus becomes
g' = (l^/r)(9SO61~025 cos 2A) cmvJsec*.
(ili) Effect of Elevated Masses The correction term fl 2/i/r)
for altitude wo ill only be valid when there is mth'ng but spice
between the surface of tlie earth and the point* /i above, e.g., f<>r
an observer in an aeroplane at height h But if we consider the point
to Ii3 o i the top of a rrn intain, of height A, a complication comes in
due to the effect of the attraction by the mountain.
/ I?// 3 h p \
Boug uer suggested the correction ( I " _j_ 'lj } known
as Bxigier's Rde, where A i 3 the mean density of the earth and p,
that of the mountain.
It is now found, however, that Bouguor somewhat overestimated
the effect of the mountain and his correction*, therefore, gives the
upper limit, as it were, of its effect, th^ lower limit being that in
which its attraction is neglected altogether. The Board of Trade
have, therefore, adopted the following relation for the combined
effect of latitude and altitude :
g' = (9SO6l0:55 cos 2x)(l5/i/4r) cms.lsec*.
(/v) Effect of Depth. Again, imagining the earth to be a homo
g^neous sphere, let us take a body of mass m, inside the earth, at a
depth h below the surface, so that its distance from
the centre of the earth is (r /;), where r is the
radius of the earth. Imagine a sphere of this
radius (r h) to be drawn concentric with the earth,
(Fig. 1 0).
Then, clearly, the body lies on the surface of
this inner sphere, and inside the outer hollow sphe
rical shell, of thickness h. Let g and g' be the
accelerations due to gravity at the surface of the Fig. 130.
earth and at a depth h below it, respectively. And since the force of
attraction on a body inside a hollow shell is zero, the only force of
attraction on the body is that due to the inner solid sphere, of radius
(rh), and is directed towards its centre, its magnitude being clearly
given by
' mass of the spheres mass of the body
m g _ ^^
where G is the gravitational constant.
*Thi<? co f rection by Bouguor was prompted by the same idea which ins
pired his Momt.ti'i experiment for the determination of the Gravhationd
Constant G, (Sse pge 231), v/z., that the attnction on a mass due to the
mountain cpvrtd sfmply be added up ty tl^at 4ue Q U*e fajth, (taken to be a,
208 FKOPEKTIBS OF
Now, mass of the sphere = its volume x its density,
= J.7r.(r/0 3 xA,
where A is the density of the earth, supposed uniform.
.'. force of attraction on the body at a depth h inside the surface of
the earth is equal to
whence, g' = .TT. &.G.(rh). ..II
And, if the body were kept on the surface of the earth, the
force of attraction towards the centre of the earth would be given by
mg =   .'~ 9 O 9 whence, g = ~.7r./\.G.r. ...Ill
Dividing relation II by relation IIT, we have
Or, g'= g (i^\ ...IV
i.e., the value of g decreases with depth from the surface of the earth.
And it follows at once, from relation IV 7 above, that at the centre of
the earth, where h~ /*, the value of g will be zero ; i.e., the accelera
tion due to gravity and, therefore, the weight of a body at the centre of
the earth will be zero.
(v) Effect of Terrain (Topographical Correction). This correc
tion consists in reducing the result at any given station to that we
would obtain if the laud or the terrain in which it is situated were
just a horizontal plane, instead of its actual form.
Obviously, some parts of this terrain would be above and
others below the horizontal plane, so that the former would exert an
upward attraction, thus decreasing the value of g and the latter, a
downward attractive force, thereby increasing tko value of g.
It so turns out, however, that this correction is always a posi
tive one.
77. Determination of the value of g at Sea. Until comparatively recently,
the value of g at sea was determined indirectly, because it was not considered
possible to use a pendulum on board a ship. The method, suggested by Hecker
and Duffield, and usually adopted, was to determine the atmospheric pressure in
two different ways, one of which involved g and the other did not, so that, by
equating the two, the value of g could be easily calculated out.
Thus, for example, the atmospheric pressure P could be obtained (/) from
a barometer which involved g, because P H ?.g., where // is the height of the
mercury column and p, its density, and (//) from the boiling point of water,
which did not involve g, because it could be calculated from the Tables, giving
the relation between temperature and the saturated vapour pressure of water
vapour ; or directly from aa aneroid b irometer, (again, without involving g).
Then, equating H.p.g, against P,as obtained from method (), we have
P
f . .
The results obtained by this method give us an error of about '01 cm. /sec*.,
which is considerably greater than that given by pendulum methods on land,
the chief source of error being the oscillations or 'bumpings' of the mercury
columns in the barometer, caused by the movement of the vessel, the ship or the
AUUBliBKATION DUB TO GRAVITY
Vening Meinesz has shown, however, that pendulums can be used for the
purpose with far greater accuracy, particularly in a submerged submarine. Hii
argument is as follows :
A pendulum is subject to four types of disturbances on board a ship, viz.,
(/) the point of suspension having a horizontal acceleration, (//) vertical acceleration
of the support, (Hi) angular movement of the support or 'rocking 9 of the plane of
oscillation, and (iv) slipping or sliding of the knifeedges on their agate planes.
Of these, the first disturbance is the most marked, but it can be complete
ly eliminated by simultaneously oscillating two identical halfsecond pendulums*,
suspended from the same support, oscillating in the same vertical plane, but with
different phases, and noting their angular displacements Q l and 2 . Then, it can
be easily shown that (Oi 2 ) gives the angular displacement of a pendulum,
altogether unaffected by this disturbance.
The vertical acceleration of the support can, however, not be eliminated,
without eliminating g itself, but the disturbance due to this can be greatly mini
mised by taking the mean of a large nu Tiber of observations. For, the value of
# seems to be affected only by the nmn value of the vertical acceleration during*
the whole period of observation. And, sinee the vertical motion is alternately
up and down the zero position, the mean value of this acceleration becomes
almost inappreciable.
The error due to 'rocking* can be easily corrected for, by noting different
values of the rocking angle and computing the necessary correction, which is
usually quite small.
And, with all these errors eliminated, or minimised, to an extent thai
the total angular deviation due to them does not exceed 1, the fourth error, viz.,
the slipping of the knifeedges gsts automatically eliminated.
What is done, in actual practice is tint three halfsecond pendulums are
suspended from the same support and set oscillating, and continuous photogra
phic records of (0 the difference (9 t 2 ) between the angular displacements of
the first and the second, and (//) the difference (9283), between those of the
second and the third pendulums obtained on a sensitized paper, by means of a
suitable optical arrangement. Ths value of g is then calculated from each of
these two sets of observations and their mean taken.
The whole system is suspended in gimbals, to avoid external disturbances
due to small angular movements of the ship or the submarine ; and, further, to
avoid any possible errors due to any slight change in temperature, the whole
apparatus is kept properly thermally insulated and any small correction, still
necessary, applied. And, finally, to make sure that no magnetic disturbances
affect the result, the pendulums are made, not of invarsteel^ ', (which would be
so helpful in minimising any temperature corrections), but of brass.
The probable error in the value of g thus obtained is claimed to lie with
in *0018 cm./.rec 2 ., obviously, a marked improvement over Duffield's earlier
indirect method.
A recent and comparatively much more accurate method consists in
measuring the change in the frequency of transverse vibrations of a wire under
tension, due to a weight suspended from it For, whereas, any variations in the valua
of ? produce next to no effeet o i tru d snsity of the wire (density being the ratio
of mass to volume), they dp naturally affect the pull of the earth on the suspended
weight and hence the tension in the wire, resulting in corresponding changes in
its frequency of vibration These can be easily detected to just a fraction of a
vibration in a frequency of several thousands, by comparison with the vibrations
of a quartzcrystal oscillator by the methods of beats. This explains the high
accuracy of the method, which is obviously equally applicable to the measurement
of the value of g on land, particularly at places where it is difficult or impracti
cable to use the usual method, as for example, at the bottoms of boreholes etc.
78. Local and Temporal Changes in the value of g. The value
of g at a point is also affected, to some alight extent, by local causes,
"" */.., pendulums, whose full timeperiod is one second] ~"~
fit is an alloy of nickel and steel, whose coefficient of thermal expansion
is exceptionally low, and which is, therefore, used in the construction of what
are called invert able pendulums, i.e., pendulums whose lengths remain practically
unaffected by temperature variations, the name 'invar' for the all% being
suggested by the word 'invariable'.
210 PROPERTIES OF MATTER
like smalt geological deposits near about, the topography of the region?
or even by masses like buildings etc., in the neighbourhood. It is alsa
affected by time, because deformations in the earth's surface take.
place periodically, thus bringing about a change in the equipotoatiall
surface, and hence in the direction of the force of gravity, which is
always perpendicular to this surface. These changes are, however,
much too small to be measured by ordinary pandulum methods
whose accuracy is limited to within 10~ 4 cms. /sec 2 ., or 10 1 milligals,
(where i cm./sec 2 . =1 gal = 1000 milligals*). More sensitive methods
have, therefore, to be used for the purpose. A detailed study of these
is beyond the scope of this volume, and we shall, therefore, deal with
them only briefly here.
(/) Local Changes. Small changes in the value of g due to
local causes are measured with the help of (/) what are called
invariable pendulums and (//) gravity meters or balances. The former
are suitable only for the measurement of placetoplace variations
in the value of g in regions, free of all marked local abnormalities,
and the latter, for changes due to abnormal conditions like irregu
larities in the density of surface constituents and such other causes.
For the most accurate determination of small variations in the value
of g, however, a still more sensitive instrument viz,, the Eotvos gravity
balance must be used.
(l) The Invariable Pendulums. These pendulums are so called, because
of their being standardised to such an extent that their time periods (/) vary
$olely due to variations in the value of g and to no other factor.
They are usually rigid pendulums of invarsteel, suspended from a massive
tripod in a partially evacuated chamber, with a specified air pressure inside it, to
make all air corrections constant. And the variations due to temperature already
small on account of the use of invarsteel, (with its negligible coefficient of ex
pansion), are further corrected for by a direct determination of the change in
time period with temperature.
The timeperiod of such a pendulum is first determined at a chosen base
station, i.e., at a place where the value of g is known and then at the field station,,
i.e., at thf place where it is to be determined. Then, clearly, the gravity ratio,
or the ratio between the values of g at the two stations, will be given by the
inverse ratio of the squares of its time periods there, since
The only error possible, after all this standardisation, is that in noting
the timeperiods of the pendulum at the two stations, or in the 'timing opera
tion\ as it may be called, and the utmost accuracy is attempted to be secured
here by arranging to have precise timesignals broadcast at frequent intervals.
In the ultimate analysis, however, the results obtained will be restricted to the
same order of accuracy to which the timeperiod of the pendulum and the other
constants involved have been determined at the base station.
The use of the timesignals at the field station may be obviated by the
technique used by Bullard in his determination of the value of g in East Africa
(in 1933), v/z., that of using two pendulums one at the base station, (in his
case, Cambridge) and the other at the field station and recording an agreed
Morse signal, alongside the oscillations of the pendulum, at each station, on a
photographic film, repeating the same an hour or so later. The timeperiods of
the pendulums can then be compared with the equal timeintervals given by the
Morse signals, and a high degree of accuracy thus attained in their measure
ment.
(2) Gravity Meters. Next in sensitivity come the gravity meters, various
ftorms of which are now in commercial use as prospecting instruments and other
wise. We shall consider here only a few of them. ___
*The milligal is a new unit, now increasingly being used to express small
changes in the value of #.
ACCELERATION DUE TO GRAVITY
211
TheThrelfall and Pollock Gravity Meter. Used first by Threlfall and
Pollock, in 1899, it is perhaps the earliest gravity meter and consists of a tine
quartz thread AB* (Fig.
131), stretched horizon
tally, with the end A
fixed or 'anchored* and
the end B attached to an
axle which c<m be rotat
ed, in line with the
A
'/?
p;
rig *
thread (the latter being thus twisted) by means of a pointer (or a vernier) wljich
moves over a circular scale S A small metal rod R is fused athwart the thread/
near about its midpoint and is so weighted (by a bob or weight w) that its e.g. ,
lies on one side of the thread.
The end B is twisted by means of the pointer, until the rod becomes
horizontal about three full turns of the thread being necessary for the purpose
in which position it is just stable under the balancing forces due to the tension
of the thread and the gravitational pull on itself, / e., when the torsional couple
due to thread just balances that due to the pull of the earth. The position of the
ponter is now read off on the scale, the slightest further movement of it
making the rod lose its precarious balance and turn right over. This is safe
guarded against by a suitable stop or arrester, but the veiy fact of this tending
to occur enables its position of approaching instability to be readily determined.*
Thus, with a change in the value of g, the rod will no longer remain horizontal
and the end B of the thread will have to be twisted to restore it to that position.
The angular twist thus given to the thread can be read on the scale from the
position of the pointer, and is a measure of the variation in the value of g, the
pressure being kept constant and proper correction for temperature effects (i.e.,
for expansion and change in the rigidity of the thread etc.) being made.
The instrument is made directreading by first noting the positions of
the pointer at two stations, where the value g is accurately known, its variation
with temperature being determined at one of them. So that, if now the instru
ment be carried from place to place, the various positions of the pointer indi
cate the values of g on the scale straightaway.
With proper precautions taken, this simple appliance can yield results of
a fairly high degree of accuracy.
(//) The Boliden Gravity Meter. A later form of gravitymeter, shown inr
Fig. 132 is due to Boliden (1938), m which two pieces of spring S, S, support a
mass M which ends in two flat plates D and E
above and below, each forming one plate of
, the parallel plate condensers AD and BE r
whose other plates A and B are properly insu
lated from the framework of the instrument
by means of insulating slabs FandG. The
condenser AD above forms part of an oscil
latory (or LC) circuit, whose frequency (N) i
compared with a standard oscillator.
A change Bg in the value of^ brings
about a change in the flexure of the springs
and hence a proportionate change Bx in the
air gap (d) between the condenser plates A and
D. This, in its turn, results in a change BO
in the capacity (C) of the condenser, such that
CB/C = Bx/d. And, finally, the change m the
capacity of the condenser is then responsible
for a corresponding change BN in the fre
quency (AT) of the oscillatory circuit.f So that,,
Bg oc Bx oc SAT,
Fig. 1 32. whence Bg can be easily calculated out.
" ~~*For, with the approach of the position of instability, the net couple
acting on the thread varies only slowly with the change in its inclination anfd
hence the timeperiod of the torsional vibrations of the thread about its equi
librium position goes on increasing.
tBecause the frequency of an oscillatory circuit depends upon the capaci
tance (C) and the inductance (L) included in it.
212
PROPERTIES OF MATTER
The instrument is calibrated by applying known potential differences to
the plates and of ch2 lovvar coadeaser, calculating the force of attraction
between them (and hence on D) and ihs attendant frequsncy change of tho
oscillatory circuit and plotting a graph between the latter two.
The sensitivity of the instrument is rather low, being only about 1 X 10**
cmjr./iec 1 ., or just 1 milligal.
(Hi) The Gulf Gravity Meter. This is a more recent (1941) and sensitive
type of gravity meter and depends uponths same principle as a spring balance,
v/2., that the weight of a bidy is proportional to the acceleration due to gravity,
so that a mass suspended from a spring will exert a different pull on it for diffe
rent values of g, the stretch of the spring thus indicating the variations in g,
The method fails in the case of the ordinary spring balance purely for
want of requisite sensitiveness. In the case of the present instrument, however,
this sensitiveness is well assured, as much by the choice of a suitable type of
spring as by the accuracy of the means of observation.
We u^e here a flat, metallic ribbonspiral spring, fastened to a torsion head,
at the top, and carrying a load at its free end below, including a mirror m, (Fig.
* 133), which untwists the spring by about 8 full revolutions.
jl Any change in the value of g will bring about a
$ proportionate change in the weight of the suspended mass and
D the consequent pull on the spring, resulting in a correspond
ing rotation of the mirror, which can be measured by the devia
tion of a beam of light from an illuminated slit, reflected
from it. Th? angle of deviation is magnified by making the
beam travel four time* between m and a fixed reflector and the
image of the slit finally observed by means of a microscope,
fitted with a micrometer eyepiece. The slight changes in
the value of g corresponding to thsse deviations can thus bo
easily determined.
The sensitivity of the instrument is found to be about
5 X 10~ f cms. I sec*., or 5 x 10 ~ 2 milligals.
(iv) Eotvos Balance. None of the above appli
ances possess the necessary sensitivity to be able to
measure the small change in the value of g due to
neighbouring buildings or small geological deposits
etc. Instruments far more responsive to small varia
Fig. 133. tions in the value of g must be employed for these
delicate measurements and the gravity balance, devised by Baron
Eotros, admirably answers this requirement. It is not only used
for a comparative or an absolute determination
of g, but also for the measurement of other
important quantities connected with the earth's
gravitational field and for purposes of gravi
tational survey, the accuracy claimed for the
instrument being 10~* cms. I sec*., or 1CT*
milligals.
In essentials, the Eotvos Balance consists
of a rectangular torsion beam B, (Fig. 134),
of aluminium about 40 cms. in length, and
between 3 X 10~ 4 to 4 x 10~ 4 mms. in diameter,
suspended from a torsion head T, by means of
a fine suspension S, about 60 cms. long, of the
alloy platinumindium, through an aluminium
rbd R, fixed on to the beam at its c.g , O. The
rod carries a small concave mirror C, to enable
the deflections of the beam to be read by the Fig. 134.
ACCELERATION DUE TO GRAVITY 213
lamp and the scale method, with the help of a telescope. A small
cylindrical weight P, of platinum, gold or silver, of mass about 30
gms. is suspended from one end of the beam, by means of a fine wire
(H>) of platinum, and a counterpoise weight M, of mass about 25 gms ,
is slid on to, or suspended from, the other end of the beara, as shown.
If the instrument be taken to a place, where the value of g
varies from point to point, its suspended system experiences a couple,
producing a twist in the wire and deflecting the beara from the
position, (not known), that it would occupy if the value of g were
constant.
Let the beam, in its equilibrium position, make an angle with.
the xaxis, (lying along the northsouth direction), i.e, let 6 be the
'azimuth angle', as it is called, and let S l be the reading on the scale
in this position. Then, if 5 be the scale reading in the (unknown)*
position, in which there would be no gravitational torque on the
beam, it can be shown that
where A and C are the constants of the instrument, 17, the gravitational
potential, and 9C//3*, a*7/aj and 3t7/3z, the values of the gravitational
attraction along the North, the East, and the vertical directions
respectively, (this last one being the value of g). The origin of the
three axes ," along these three directions, is taken to be the midpoint O
of the beam,
we have, from relation (1) above,
SiSo = A' sin 20+ B' cos 20+C sind+D' cos 0. . .(2>
Now, taking 6 = 0, 60, 120, 180, 240 and 300, in equation
(2) above, and taking the corresponding values of S L to be S 19 S 2 , S 9 >
$i> S & , S*> we have
 S 2 + S 4 +S 6 = 3S ,
A' =
2V3C' =
and 2D' = S.S^
Thus, all the constants of relation (1) being known, the rate of
change of g northwards, (given by C'/C), as also that in the eastern
direction, (given by D'fC), can be easily determined.
On account of its high sensitivity, the balance is used for
geophysical prospecting (see 80). And, Shaw and Lancaster Jones
have successfully mapped out with its help the local gravitational field
in a laboratory.
214
PROPERTIES OF MATTER
(ii) Temporal Changes The Horizontal Pendulum. As we hare
seen above, a deformation of the earth's surface and its gravitational
equipotential surface results in a change in the direction of the force
of gravity and hence in that of g. Since a plumb line always sets
itself normally to the gravitational equipoential surface of the earth,
it is clear that measuring a change in the direction of g at a point, is
tantamount to measuring a change in the direction of the plumb line
at that point. These changes, however, are much too small seldom
exceeding 1", to permit of their accurate measurement by means
of a plumb line. The most commonly used device to measure
these is what is called the horizontal pendulum, devised by Hengler, in
the year 1832.
This horizontal pendulum essentially consists of a rod AB, (Fig.
135), carrying a knob or bob at #, with G as its e.g. It is supported
in an inclined position, by moans of two pieces of a light string, AP
and Cg, attached to a rigid support at P and C respectively, such that
the straight line CP, joining the two, meets AB in O, and makes an
angjle <j> with the direction of the forco of gravity. The pendulum
thus takes up a position in a plane parallel to the force of gravity.
On giving the bob a slight lateral displacement (towards or away
from the observer), it begins to oscillate slowly, with a small
amplitude, along an arc with O as its centre and OG as radius. Its
period of vibration is deduced as follows :
B
I
/r\
Fig. 135. Fig. 136.
Let the bob oscillate along an arc GG', (Fig. 136), which lies in
a plane, making an angle < with the normal to the plane of the force
of gravity.
If the bob be displaced through an angle in this inclined
plane, into the position shown, its weight mg acts at its e.g., G' (v G
is now at G') in the direction of the force of gravity. Resolving it
into its two rectangular components, (/) in the inclined plane of its
rotation, and (ii) perpendicular to //, we have the former component
s= mg sin <f>, and the latter = mg cos $.
Further, resolving the component mg sin < into two rectangular
components, along and perpendicular to 06?', we have
the component along OG' = mg sin <f> cos Q.
and the component perpendicular to OG' = mg sin <f> sin 0.
This latter component (mg sin <f> sin $ ) has, clearly, a restoring
ACCELERATION DUE TO GRAVITY 215
about the inclined axis = mg sin ^ sin Q.I, (where OG* = OG
=/), tending to bring the pendulum back to its original position.
If be small, sin = 9, very nearly. And, therefore,
the restoring moment = mg sin <.#./.
And, if the angular acceleration of the pendulum be da>ldt, and
its moment of inertia about O be 7, we have
restoring moment (or torque) also = I.dto/dt.
So that, neglecting frictional and viscous forces, we have, for
equilibrium,
Lda>ldt = mg Lsin +.0, whence, dw/dt = mgl * in *.e.
Or, putting mg.l.sin </>/! = /i, a constant, we have
daj/dt = n.S. Or, </oj/<# oc 0,
.i.e., the angular acceleration of the pendulum is proportional to its
angular displacement.
It is, therefore, a case of simple harmonic motion; and its time
period Tis given by the relation,
Or, T = 2*
But / = 7w/c 2 , where fc is the radius of gyration of the pendulum
about O.
Hence T = 27rA/ "!^ , = 77 A/ , k * 
V ing I sin $ n 'V ./.*<
in practice, to make Tlargo, ^ is made as small as possible.
Now, if <f> = 90, j/71 < = 1, and, therefore, the timeperiod !F,
in this case, is given by the relation T' =
Hence r/r = ^/T^inf = (l/ 5 f ^). And /. T^/T'* = l!sinJ>,
whence, sin <f> = T' 2 /T 2 .
Thus, knowing T' and T t we can easily calculate sin <f>, and
lience ^, which represents the change in the direction of the force oi
gravity, and, therefore, that of g, in the equilibrium plane of the
pendulum.
79 Gravity Survey. The purpose of a gravity survey is
twofold, viz., (i) the main one being to determine the value of the
force of gravity and its direction at various points of the sealevel
surface of the earth, or the l geoid' 9 as it is called, and (if) a secondary
one being to deduce from it the possible distribution of matter in the
earth, and thus to form an idea about its structure and internal
condition.
Now, in any gravity survey, it is found necessary to introduce
two new quantities, connected with the earth's gravitational field,
These are (1) the gravity gradient, denoted by the letter O, and (2
the horizontal directive tendency (written as H. D. T., for brevity)
denoted by the letter R. Let us try to understand their meaning.
216 PROPERTIES OF MATTER
(1) The Gravity Gradient (G). It is a convenient abbreviation for
"maximum gradient of g in a horizontal direction, near a point", where ^ is the*
vertical gravitational intensity at the point, i.e., g = QU/dz The gravity gradient
may, therefore, he denoted by tig Ids, the rate of variation of g per unit distance, in*
the direction of the maximum rate of change in its value, and it is thus obviously a
vector quantity.
Now, if the gravity gradient G, (=&e/0s) makes an angle <j> with the axis
of x for the north south direction), and if its components along the axes of x and*
y be Qgld* and dg/Qy respectively, then, clearly,
dgld* = G cos <f> and dgfty = G sin <f>.
And, since g  QUIdz, we have fa fix = 2 //0Jc0z U xs ,
So that, t/^ = G cos $ and C/^ = G sin <f>,
.'. squaring and adding the two, we have
G* cos 2 f+ G 2 sin* f = V\ Z +V\*. Or, G* (w 2 ^M/i 2 #) = U*
Or, G a = U* xz +U\ zt whence, G = (C/ 2 az f l/^)*.
Now, {/as and U v * can be easily determined by means of an Eotvof
Balance, (see page 212), and thus the value of G can be calculated out from the
above expression.
(2) The Horizontal Directive Tendency (R). It is what is called a 'cur
vature vector', i.e., a directed quantity, though not a true vector. Its value at a.
point is given by the relation,
where r and r, stand for the maximum and the minimum radii of curvature of
the level surface, or the gravitational equipotential surface at the point. Its
direction, according to an agreed convention, is taken to be the direction i/r
which the level surface has the least downward curvature and, therefore, the maxi
mum radius of curvature.
If the direction of H.D.T. makes an angle with the axis of x, or the
northsouth direction, it can be shown that
R sin 28  2U XV , r where, U xv = fU/dxdy,
and R cos 2o * U* xx U* vv . Uy^ = cW/0^* and
The dimensions of both G and R are [T]~* and they are generally ex
pressed in what are called Eotvos units, where one Eotvos unit is equal*
to 10 9 /sec*.
In survey maps, the gravity gradient at a point is
represented, in magnitude and direction, by an arrow
head drawn from the point, whereas the horizontal direc
tive tendency is just represented in magnitude and direc
tion, by a straight line, passing through that point, wim
out any arrowhead or feathered tail, as shown in Fig. 137,
where O is the point in question.
Further, points, where the value of g is the fame,,
are joined by curves, which are called isogaras, G being
Fig. 137. always directed along the normals to these.
80. Geophysical Prospecting. We have seen above how, due
to the presence of local geological deposits, (i.e., minerals etc.), inside
the earth, small variations are produced in the value of g. Similar
changes are produced, by their presence in the normal values of the
other quantities, magnetic, electrical, seismic* etc., associated with it ;
so that, by measuring these variations, with the help of specially
designed instruments, we can detect their presence. This is technically
called 'geophysical prospecting. 9 We are concerned here only with
the gravitational methods adopted for the purpose, the principle
*See article on, 'Earthquakes' in the next chapter*
ACCELERATION DUE TO GRAVITY 2 IT
underlying which is to measure the gravity gradient G t and the hori
zontal directive tendency R, at various points in the region under
survey, with the help of a sensitive instrument, like an Eowos gravity
balance, as explained above in 78, (iv) the instrument, when so used,
being called a 'gradiometer'.
Thus, if & be the angle that R makes with the axis of x, or the
northsouth direction at a point, we have
Ssin 28 = 2U^, and R cos 26 = U* xx [/%,. [ 79, (2), above
And, therefore, tan 26 = 2U x y/U* xx U 2 yVy
the two solutions of which, BI and tt t , differ by 2/7T and give the
directions of the two principal axes of curvature of the equipotential
surface, (i.e., the level surface), at the point. The values of G and R
at various points are then plotted, the direction of a being such that
secant 2$ and (V 2 XX U 2 VV ) are of opposite signs.
A graphical representation of the variations of g over the
region, under examination, is thus obtained, and closed curves or
isogams, [see 79, (2), above], are then clearly marked out on it,
(which, as we know, are at every place perpendicular to G), so that
we have an isogam chart of the region in question.
Interestingly enough, the physical form of these isogams almost
faithfully represents the physical form of the subterranean deposits.
Thus, for example, a uniformly monoclwic type of region uould give
isogams t which are all parallel and equally spaced, whereas if the sub
terranean deposits form a domelike structure, the isogams obtained
also resemble the outline or the contour of a surface dome, as it were.
This method can, however, succeed only in the hands of those
welltrained in the use of the delicate instruments employed and in
the proper interpretation of the results obtained from them.
SOLVED EXAMPLES
1. A metal disc oscillates in its own plane about an axis passing througft
a point on its edge. What is the length of the equivalent simple pendulum ?
Let the disc of radius r oscillate about an axis through the point Pon ita
edge, (Fig. 138).
Then, clearly, the timeperiod of the disc is given
by the relation, / = 2n \/ iiMg /,
where / is its A/./, about the axis through P ; M, its
mass and /, its length.
Or, t = 27c\/ /,A/#.r, for / = r, the distance between
the point of suspension (P) and the e.g. (O) of the disc.
Now, /=/0+A/r*.
where I g is the ML of the disc about a parallel axis
through O, U. 9 l g = Mr*\ 2.
So that, /  (Mr 2 /2)+Mr a _3A/r f /2. Fig. 138.
And, therefore, t  2* A / l
'
2* A /
'y
Mg.r
I.*., the same as that of a simple pendulum of length / = 3r/2.
Or, the length of the equivalent simple pendulum is 3/2 times the radius of the disc.
2. Find the period of small oscillations of a rig>'d body, free to turn abou
t fixed horizontal axis, and also find a formula for the length of the equivalent
simple pendulum.
218 PROFEfcftES OF MATTER
Three particles of the same mass m are fixed to a uniform circular hoop of
mass M and radius a at the corners of an equilateral triangle. The hoop is free to
move in a vertical plane about the point on the circumference opposite to one of the
masses m. Prove that the equivalent simple pendulum is equal in length to the dia
meter of the circle. (London Higher School Certificate ; Patna 9 1948)
For first part ; see 62 (page 165).
Let the three equal masses, m, m and m, be fixed to the hoop, of radius
a, as shown, (Fig. 139), so as to lie at the corners of an equilateral triangle.
Since they are all equidistant from the centre, the e.g.
of the triangle is at O, the centre of the circle. The
whole arrangement is thus equivalent to a hoop of
mass (Aff 3m), with its centre of gravity at its cen
tre O.
Clearly, then, the moment of inertia of this
loaded hoop about O, (i.e., about an axis through O
and perpendicular to its plane) == (M>3w)a a . And,
therefore, its moment of inertia about a parallel axis
through the point of suspension P is (by the principle
of parallel axes), given by
I  (M+3m) a z +(M+3m)a* = 2(M+3m)a*.
Fig. 139. Now, the timeperiod of the hoop about P is given by
= 2 _ ~ A m .
2 * '
Or, / = 2rr
te., the same as that of a simple pendulum of length la, the diameter of the hoop.
Or, the length of the equivalent simple pendulum is equal to the diameter of
the circular hoop.
3. How much faster than its present rate should the earth revolve about
its axis in order that the weight of a body on the equator may be zero, and how long
would it take to make one revolution then ? What would happen if (/) the rotation
became faster still, (//) the rotation were stopped altogether ? (g = 978 cms./sec 2 .)
We have seen (page 203), that the value of g is different in different lati
tudes, due to the rotation of the earth, and that, assuming the earth to be a per
fect sphere,
r.c05 2 ^.o> 2 \ Twhere g . is the value of V
8 ' Lin latitude <f>, (see page 205)
Now, at the equator, ^ 0, and .*. cos 2 ^ * 1 ; so that,
where, g is the value of 'g 9 at the equator.
With the actual value of g, the value of r.w 2 comes out to be 3'39 and,
therefore, we have ^ . 3 ' 39 = JL.
g 978 288
Thus, in order that the weight of a body may be zero, the value of F O
should be zero, i e., r.^/g should be equal to 1, or the value of r<o 2 should be
288 times greater than its present value, r being a constant. It follows, there
fore, that <o should be \/288 times, /.<?., 16'97 times greater than its present value.
When this is so, the outward centrifugal force on the body will, obviously, be just
balanced by the inward force due to gravity.
//, therefore the earth rotates, 16 97 times, or 17 times, faster than at pre
sent, the weight of the body at the equator will be zero.
Now, the earth makes one complete revolution in 24 hours, i.e., dis
cnbes an angle of 2* m 24 hours. But, in the case considered, viz., when the
weight of the body at the equator is zero, it rotates 17 times faster, and will,
therefore, describe an angle 17x2* i n 24 hours, or an angle 2* in 24/17 hours
or 1412 hours.
ACCELERATION DUE TO GRAVITY 219
.*. the earth will then make one rotation in 1*42 hours.
If the rotation became faster still, i.e., faster than 16*97 or 17 times its
normal rate, obviously, all objects kept loose on the equator will start leaving
fthe surface of the earth ; for, the increased centrifugal acceleration on them will
be greater than that due to gravity, and, therefore, a resultant force will be act'
ing on them outwards, away from the centre of the earth.
If, on the other hand, the rotation of the earth about its axis were stop
ped altogether, we shall have = ; so that, substituting this value of to in the
relation,
o = g( 1  \ for the value of V at the equator, we have
*o=*U0)=*.
i.e., the value of g increases by (g g Q ) !'<**/ times g.
Or, = (3'39/978) times g = 1/288 times g, or = /288. [ For r.o> 2 = 3'39.
Thus, // the motion of the earth were stopped altogether, the value of g
would increase by 1/288 of its normal value.
4. Assuming that the whole variation of the weight of a body with its posi
tion OH the earth's surface is due to the rotation of the earth, find the difference in
fthe weight of a gram as measured at the equator and at the poles. (Radius of the
earth = 6 '378 x 10 8 cms. )
We have the relation,
g, g ( 1 r  r ' os ? \ for the value of V in latitude tf.
r \ g x
Since r.o 2  6378x 10* x( J^")^ 3 ' 39 > and ' r  to V* = 97^ = 1/288,
we have gj (lo?s 2 0/288).
Now, at the equator, ^ = 0, so that, cos <f> and .'. cos 2 $ 1.
Hence ^ = #(11/288),
where gp is the value, of '#' at the equator.
And, at the poles, ^ = 90, so that, cos $ and .*. cos 2 $ = 0.
Hence gp = #(10) = g t
where g# is the value of 'g 9 at the poles.
Since the weight of body is mg, where m is its mass, we have
weight of I grn. at the equator w = 1 x # .
= 1x^(11/288) = ^(11/288), (/)
and its weight at the poles w' = 1 xgp = 1 *xg == g. (//)
Hence, the difference in the weights of this mass at the poles and at the
equator w'w.
= g{l 1/288) = gg #/288 ^/288 = 978/288 == 3'395 dynes.
Or, the difference in weights of a gram at the poles and at the equator is
3'395 dynes.
5. The mass of a railway train is 100 tons. What will be its weight when
(a) stationary, (/?) travelling due east, (c) travelling due west, along the equator at
60 miles per hour ? Radius of the earth is 4000 miles. (Punjab)
(a) When the train is stationary. When the train is at rest, its apparent
weight is 100 tons wt. 100 x 2240=224 x 10 3 Ibs. wt. t (because 1 ton = 2240 Ibs.).
<(b) When the train is moving East. When the train travels due east, its
angular velocity about the axis of rotation of the earth increases, because the
earth itself is rotating about the axis from west to east. The centrifugal force
on the train, therefore, increases, (being proportional to r.w 8 ), and hence the
.apparent force of gravity on it and, therefore, the apparent acceleration towards
the centre of earth, i.e., apparent acceleration due to gravity, decreases. And,
since weight = mas sx acceleration due to gravity, the apparent weight of the train
decreases.. Let us see by how much.
The radius of the earth, r = 4000 miles = 4000x1 760x3 //., and, there
fore, linear velocity (v) of a point .on the earth = 2^/24 x 60x60 ft.lsec., since
&fce earth makes one complete rotation in 24 hours.
220 PROPERTIES OF MATTER
A point on the equator will also thus describe a distance 2*r in 24 hours ;
its velocity is, therefore, given by
2*x 4000x1760x3
==  24x60x60 
.*. centrifugal acceleration of the point, when at rest relative to the earth;.
is given by
v 1536x1536
T = Too6TT760x~3 =
Since velocity of the train = 60 m./hr. =  = 88//./sec.,
the resultant velocity of the train, say, v'
 1536488 = 1624 ft./ sec.
And /. centrifugal acceleration on the tram moving at 60 m/hr. v"/r.
1624x1624
.'. increase in centrifugal acceleration, or decrease in acceleration towards
the centre of the earth, i.e.. in the acceh ration due to gravity
= O'1248O 1116  00132//./WC*.
.*. decrease in the weight of the train = mass x decrease in acceleration due to
gravity.
= 100x2240x00132 poundah  100x2240x00132/32 Ibs. wt.
 100x2240x00132/32x2240 = 1 00 xO'Ol 32/32 ton wt.  0'0412 ton wt.
.'. apparent weight of the train = 10000412 = 99'9588 tons wt.
(c) When the train is moving West. In this case, since the train is moving
from east to west, opposite to the ditection oj rotation of the earth, its angular
velocity about the axis of rotation of the earth decreases and, therefore, the
centrifugal acceleration on it also decreases, with the result that the acceleration
towards the centre of the earth, i e , the acceleration due to gravity increases. The
apparent weight of the train on the equator, therefore, increases. Let us calcu
late this apparent increase.
As before, velocity of a point on the equator, i.e.,
2nr 2nx4000x 1760x3
centrifugal acceleration of the point, when at rest relative to the earth
v 2 1536x1536
  40UOX17605T
Hence, "the resultant velocity of the train, say, v" =* 153688 = 1448 ft. I sec.
And .*. the centritugal acceleration on the train moving at 60 m./hr. is
clearly given by
v' /a 1448x1448
" T * 4000 x 176071
.'. decrease in cenrtifugal acceleration or increase in acceleration towards the
centre of the earth, i.e., increase in acceleration due to gravity
 0111600993 = 00 123 //./sec 2 .
And .*. increase in apparent weight of the train
100x2240x00123^^^/5 = 1 00 x 2240 xO'Ol 23/32 Ws. wt^
100x2240x00123/32x2240 = lOOx '0123/32 ton wt.
C '03 844 ton wt.
Thus, the apparent weight of the train
 100+003844  10003884 100'04 tons weight.
EXERCISE VI
1. What is a simple pendulum ? Is it obtainable in actual practice 7
Deduce an expression for its timeperiod and show how the value of g maybe
determined with its help. What are the drawbacks of this pendulum ?
ACCELERATION DUE TO GRAVITY 221
2. Deduce the formula for the time of vibration of a compound pendu
lum and show that this is a minimum when the length of the compound
pendulum equals its radius of gyration about a horizontal axis through the centre
of gravity of the compound pendulum. (Punjab, 1951)
3. Distinguish between a simple and a compound pendulum.
For a given compound pendulum, show that the centres of oscillation and
suspension are interchangeable.
How is the value of ** determined with the help of a compound pendu
lum ? (Agra, 1948)
4. Give the theory of Kater's pendulum and find an expression for the
acceleration due to gravity in terms of two nearly equal periods of oscillation
about the two parallelknifeedges.
Indicate the sources of error in an experimental determination of ^.
(Bombay, 194041 ; Punjab, 1948)
5. A Borda's pendulum his a bob of radius 12 rwv., which i suspended
by a fine wire, 94 cms. long. Calculate the length of the equivalent simple pendu
lum. Ans. IGO'144 cms.
6. Tf a pendulum beats seconds at a olace where # = 32*2 ft. /see*., how
much would it gain or lose per day at a place where g 32*1 8 ft./*ec z .
Ans. Gains 3 min. 36 sees.
7. Explain the Dropping Plate method for the determination of tho
value of g. If there be an enor of 1% in m ^asurin r the distance covered by the
plate as also in measuring the frequency of the fork, how would it affect the
result ? Ans. The remit will b* wrong by 3%.
8. The length between the knifeedjes of a Kater's pendulum is 89*28
cms., while the times of oscillation abrjt the txvo edjes ire 1'920 sec*, and 1*933
sees., respectively. The e.g. of the pendulum is about 54*4 cwy. from one edge
What is the value of g 1 Ans. 979 cms. /sec 9 
9. An Atwood's machine has a pulley of radius a and moment of inertia
/ ; the masses attached to the ends of the string are each Mand the rider is of
mass m.
Prove that the acceleration /of the masses is given by
assuming that the string does not slip on the pulley, and neglecting axle friction.
(Madron, 1949}
10. A uniform rod of length 100 cms can rotate about a horizontal axis
through one end. Find the angular velocity which will enable the rod just to
make a complete rotation. (Madras, 1947)
Ans. 3*83 radians I sec.
11. A solid cylinder, of radius 4 cm?, and mass 250 gms.* rolls down an
inclined plane, with a slope of 1 in 10. Find the acceleration and the total energy
of the cylinder after 5 sees. (Bombav, 1944)
Ans. 65'4 cms./sec^. ; 4*799 Joules.
12. A cylinder, of mass 100 /6s*. and diameter 12 inches, rolls from rest
down a smooth inclined plane of 1 in 8 and 20 feet long. Calculate the total
kinetic energy and its energy due to rotation, when it reaches the bottom.
(Madras, 1949)
Ans. (/) 8*0 x 1 0* ft. poundah.
(it) 2'6xlQ*ft.poundals.
13. Define 'centre of suspension* and 'centre of oscillation'. Show that
in a compound pendulum they are interchangeable.
What is the distance between the centre of suspension and the centre of
oscillation on a uniform cylindrical metal bar used as seconds pendulum ?
(Diameter of the bar=l cm., to density, 8 gms./c.c. and #=978 cmi.lsec*.)
(Allahabad, 1949)
Ans. 99'1 9 cms.
14. Obtain an expression for the timeperiod of a compound pendulum,
and show that
222 PROPERTIES OF MATTER
(0 there are four points, collinear with its e.g., about which its time
period is the same.
(//) its timeperiod remains unaffected by the fixing of a small addi
tional mass to it at its centre of suspension.
15. Obtain an expression for the period of vibration of a compound pen
dulum and show that the centre of suspension and the centre of oscillation art*
interchangeable.
A thin uniform bar of length 120 cms. is made to oscillate about an axis
through Us end. Find the period of oscillation and other points about which
it can oscillate with the same period. (Punjab, 7953>
Ans. 1*795 sees. ; at 40 cms , 80 cms.
and 120 cms. from the top.
16. Derive an expression for the period of oscillation of a circular disc,.
supported on a horizontal rail passing through a narrow hole, which is bored
through the disc halfway between the centre and the periphery. (Bombay, 1946)
Ans. T = 2v x 3r/2.
17. A uniform rectangular sheet of metal is supported by frictionless
hinges, attached to one edge which is horizontal. Determine the period of
oscillation of the sheet if / denotes the length of the side of the rectangle which
hangs downwards. (Patna, 1951)
Ans. T== 2nVT3//L2*.
18. A metal bar is suspended in turn from two parallel axes on the same
side of its c.^., and its timeperiods are four.d to be 1 42 sees in each case. If
the distance of the two axes be 108 cms. and 39'2 cms. respectively from the
e.g., calculate the value of g and the radius of gyration of the bar about a
parallel axis through the e.g.
Ans. g = 979*2 cms.lsec*., and K  20*58 cms.
19. A narrow uniform bar of mass 1000 gms. oscillates about an axis, 40
cms. from the centre, with a period of 7 '48 sees., and about a parallel axis, 10
cms. from the centre, with a period of 1*67 sees. Find the value of g, the
moment of inertia of the bar about its e.g. and the length of the bar.
Ans. 990 cms. jsec*., ; 6'02 x JO 6 gm.cm z . ; 85 cms.
20. What is meant by a simple equivalent pendulum ? If the periods of
a Rater's pendulum in the erect and inverted positions are equal, prove that
the distance between the knifeedges is equal to the length of the simple equi
valent pendulum.
A uniform circular rod, with a radius of 2 cms. oscillates when suspended
from a point on its axis at a distance of 4 cms. from one end. It the length of
the rod is one metre, find the point or points from which, if suspended, the
periodic time would remain unaltered. (Bombay, 1942)
Ans. At 31'87 cms. and 68*13 cms., also at 96*0 cms. from the same end.
21. Define a conical pendulum, and show that, for a small amplitude, its
period equals thU of a "plane" pendulum of the same length. Do simple pen
dulums exist ? What are the nearest approximations to them ? Why are they
discarded in favour of compound pendulums and what are the main applications
of pendulums ? (Bombay, 1941}
22. Describe a conical pendulum and derive an expression for its frequency ^
Explain how it is used to regulate the speed of steam engines. Show that the
sensitiveness of the pendulum used as a governor increases with diminishing
speed. (Bombay, 1937)
Ans. n = IjInV big,
(where h is axial height of the cone described by it, and equal to / cos 0, where /'
is the length of the pendulum and 0, its angular displacement ; see 74).
23. What is a steam engine governor ? Explain clearly the principle
underlying its action, and discuss its limitations.
24. If the earth were to cease rotating about its axis, what will be the
change in the value of g at a place of latitude 45, assuming the earth to be a*
sphere of radius 6'38 x 10 8 cms. ? (Madras, 1947)
Ans. 1 6895 cwj. /,*"
ACCELERATION DUE TO GBAVITY 223
25. Find an expression for the period of swing of a compound pendulum.
A disc of metal, of radius R, with its plane vertical, can be made to swing
about a horizontal axis passing through any one of a series of holes, bored along
a diameter. Show that the minimum period of oscillation is given by
T = 2nv/l414 "Rig. (Saugar, 1948)
26. Give the theory of the compound pendulum and show that the centres
of suspension and oscillation are reversible.
In a reversible pendulum, the periods about the two knifeedges are t and
(ff!T), where T is a smaJl quantity. The knifeedges are distant / and /' from
the centre of gravity of the pendulum. Prove that
/+/' . *Lt+L T . (Madras, 1949)
27. A heavy uniform rod, 30 cms. long, oscillates in a vertical plane, about a
horizontal axis passing through one end. When a concentrated mass is fixed on to
it at a distance x from its point of suspension, its timeperiod remains unaffected.
Calculate the value of AT. Ans 20 cms*
28. Explain how the length of the simple pendulum which has the same
period as a given compound pendulum may be found experimentally.
A uniform cube is free to tuin about one edge which is horizontal. Find
in terms of a seconds pendulum, the length of the edge, so that it may execute
a complete oscillation in 2 sees. (Central Welsh Board higher School Certificate]
Ans. 3A/2/.
29. A body of mass 200 gms. oscillates about a horizontal axis at a dis
tance of 20 cms. from its centre of gravity. If the length of the equivalent simpk
pendulum be 35 cms., find its moment of inertia about the axis of suspension.
(Patna, 1954}
Ans. 1 4 x 1 6 gms.cm*.
30. A pendulum, whose period slightly exceeds 2 sees , is compared with a
standard seconds pendulum by the method of coincidences. Successive coin
cidences occurred at times min., 2 nun? 58 sees., 5 wins. 48 sees., 8 mms
48 sees. Find the exact period of the pendulum. Ans. 2'0224 sees,
31. A thin rod is suspended bv means of two threads parallel to each
other and tied to its two ends. Compare the timeperiod of the rod when it
oscillates thus in its own plane with that when it oscillates as a compound
pendulum about a horizontal axis, passing through one of its ends.
Ans. 1 : 1 414.
32. Give the theory of the compound pendulum and show that the
centres of suspension and oscillation are interchangeable.
A uniform thin rod AB, of mass 100 gms. and length 120 cms., can swing
in a vertical plane about A, as a pendulum. A particle of mass 200 gms. is
attached to the rod at a distance x from A. Find x such that the period oi
vibration is a minimum. (Madras , 1951}
Ans 2 748 cms,
33. How does g (acceleration due to gravity) vary with latitude anc
height ? Obtain a general relation, assuming the earth to be a Homogeneous
sphere. Does the relation agree with observed values ? Give reasons.
(Punjab, Sept , 1955}
34. Give the theory of Kater's pendulum and mention the errors to
which pendulum experiments are liable. How is the value of g compared at
different places ? (Punjab, Sept., 1956]
35. What are gravity meters and balances ? Describe the constructioc
and working of one you consider to be the best.
36. Write short explanatory notes on :
(/) Gravity survey, and (') Geophysical prospecting.
CHAPTER VII
GRAVITATION
81. Historical. The celestial bodies have been an object of
interest to scientists all through the ages, and the first astronomical
observations, of which we have any definite knowledge, were perhaps
made by the Chinese, as far back as 2,000 B.C , though the Baby
lonian astronomers are credited with having mapped out the constel
lations even earlier, near about 2700 B.C. The first authoritative
treatise on the subject, however, was due to Ptolemy, working in
Alexanderia, about 100 A.D. who formulated his theory on the basis
of the catalogue showing the nightly positions of planets and some
1000 stars, prepared earlier by the Greek astronomer Hipparchus.
Ptolemy's book, the Almagest, enjoyed the authority of the Bible
and reignsd supreme for 1400 years. According to him, the whole of
the heavens, carrying the stars, revolved round the earth, supposed
stationary. The forward and retrograde motion of the planets*
among the stars was explained by postulating that the planets
revolved in circles, with their centres revolving in larger circles round
iihe earth, the former circles bsing termed epicycle* and the latter
ones, deferents. And, it stands to his credit that, with a suitable
choice of radii and velocities, he could explain quite accurately the
/observed facts of the day.
The Ptolemaic theory was first challenged in 1543, by the
famous Polish monk, Nicolaus Copernicus, in his book, 'Concerning the
Revolutions of the Heavenly bodies^, his geometrical solution being
much neater than that of Ptolemy, involving only thirty four epicycles
as against the eighty of the latter. In it he propounded his helio
centric theory,! according to which the planets moved in perfect
circles round the Sun, which was supposed to be fixed. The theory
was, however, received with reserve and scepticism, being objected
to on the ground that (/) the rotation of the earth should result in
bodies being hurled from its surface, and (//) with greater justifica
tion that, no parallax (or relative motion) could be noticed between
stars as was always observed between objects at different distances
from a moving ship. This parallax has since been shown to actually
exist, and was first measured by Bessel, in 1838. It is, however,
*From the Greek word, meaning 'wanderer', because a planet moves
forwards and backwards or 'wanders' about among the stars.
tHe hesitated and deferred publishing bis book until he was dying,
dedicating it to the Pope, who, not taking it seriously, expressed himself
pleased with it. And Martin Luther was positively contemptuous towards it
'Did not Joshua, (m the Bible) command the Sun to stand still and not the
Dearth?', he asked.
JTnis was4eally a revival of the theory, first propounded by the Greek
Astronomer Aristarchus, that the earth was not the centre of the universe but
revolved round the Sun, as also did the other planets.
224
GRAVITATIOK 25
extremely email, on account of the enormous distances of the stars
from the earth.
As is so well known, Galileo was compelled to recant his
belief in it, a century later, and legend has it that blind and helplpss,
he was thrown back into prison for murmuring 'And still it moves',
{E pwr si muove), until he died nine years later, and that Giordano
Bruno was actually mercilessly burnt at the stake for refusing to
do so.
Then, appeared on the scene, twenty five years later, hi 1569,
Tycho Brahe, an imperious nobleman and a brilliant astronomer*,
who rejected the Copernician theory and made careful observations of
the motions of heavenly bodies, on every clear night for thirty long
years, particularly of the motion of the planet Mars, from v his
observatory in Denmark, with his celebrated wooden quadrant,
(about 10 ft. in radius), carrying a brass scale. In view of the fact
that the telescope was yet to come, soms forty years later, we cannot
but marvel at the unprecedented accuracy of his observations. No
wonder, they were usad by navigators for centuries together, much in
the manner of the Nautical Almanack tolay. With all his great
mathematical and experimental skill and his 'infinite capacity for
taking pains', however, Tycho Brahe could not somehow piece his
results together into a proper theory. But later, Keplerf, his
assistant at the Royal Observatory at Pragus, an impocunious but a
gifted mathematician into whose hinds passed all his data on the
subject, carried on tho work and, accepting the Copernician theory J,
which his chief had rejected, worked on the latter 's figures and finally
succeeded, after twenty two years of caaaeless work, in evolving the
famous three laws, known after him, the first two in the year 1009
and the third, ten years later, in 1619.
82. Kepler's Laws. The following are the three laws, formu
lated by Kepler.
1 . The path of a planet is an elliptical orbit, with the Sw at one
of its foci.
2. The radius vector, drawn from the Sun to a planet sweeps out
*He was reputed to be 'an unsurpassed practical astronomer' and made
his own instruments for his wellequipped laboratory at Uraniborg, built for
him by Frederick II, King of Denmark. He had, however, a violent temper and
lost part of his nose in a duel, while still young, going about for the rest of his
life with this lost part replaced by an artificial one of aa alloy of silver. On , the
death of Frederick, he had to flee and seek asylum at Prague, under the patron
age of Rudolph //, .Emperor of Bohemia. It was here that Kepler joined him as
his assistant.
fHe actually succeeded Tycho Brahe, who died after a little over one year
of his migration to Prague, under the impressive designation of 'Imperial
Mathematician', at a high salary which was, however, seldom paid, tte Sad,
therefore, to supplement his income by practising astrology, 'the foolish "and
disreputable daughter of astronomy, without which the wise old mother would
starve'. He was also the fou tder of Gsomstrical Optics.
{And, for this he had 10 migrate to a Protestant country to save himself
from persecution.
He was so filled with ecstasy at his success in enunciating his third law
that he declared *I will indulge in my sacred 1 fury ; I will 1 triumph over mankind
by the honest confession that I have stoleil the golden vases of the Egyptians ta
build up a tabernacle for my God.'
226 PROPERTIES OF MATTER
equal areas in equal time, i.e., its area! velocity, (or the area swept on
by it per unit time), is a constant.
3. The square of a planet's year, (i.e., its timeperiod, or its tim*
of revolution round the Sun), is proportional to the cube of the majoi
axis of its orbit.
Unfortunately, Kepler was not aware of the property of inertia
and so could not proceed any further. For him, it was necessarj
to suppose a power acting continuously on a body, in order to make
it move. Most of the fellows of the Ro} 7 al Society*, which included
among others, men like Robert Boyle, Edmund Hailey and Somuei
Papys, were convinced, by the year 1 685, that a planet could move
in an elliptical orbit, only if it were attracted by the Sun with a
force, varying inversely as the square of its distance from the Sun,
but they couid not prove it mathematically. Newton, who was also
a member, was at this time / ucesian Professor of Mathematics at
Cambridge and seldom attended the meetings of the society, mostly
held at London and Oxford. Edmund Hailey, therefore, went all the
way to Cambridge to ask him if he could furnish the required prooi
and was simply astonished to learn that he had already done so years
earlier, but had somehow lost his papers.
Realising that no other member of the Royal Society could
hope to provide the required proof and also that Newton hid really
already achieved something much more than this, Hailey pleaded
with him to reproduce his papers in book form and, though not a
rich man himself, offered to bear all the cost of publication of the
same, which ultimately resulted in the appearance of the celebrated
Principia, in the year 1687.
Newton knew that both rest and uniform motion along a
straight line were equally natural and, after a careful study of
Kepler's laws, he showed (i) that it follows from his second law that
only a central force acts on the planet and is directed towards the
Sun, it alone being responsible for keeping the planet in its orbital
path,t (//') that it can be deduced from his first and third laws that
this^force between the planet and the Sun is inversely proportional
to the square of the distance between them f, and (Hi) that it is an*
easy further deduction from the above that this force of attraction.
between the two is also directly proportional to the product of their
masses.
He further proceeded to verify these deductions from Kepler's
laws by comparing the value of acceleration of the Moon towards
the Earth, calculated on their basis, with its value obtained experi
mentally, the two values showing a close agreement with^each other,
a$ will be seen from the following :
If g m be the acceleration of the Moon towards the Earth, v, its
Royal Society, the fellowship of Mhich today is consider* d to be
a very high honour, really grew out of informal group meetings of men interested
in natural philosophy, about the year 1645, and received its Royal Charter,
in 1662 from Charles II, who, according to Samuel Papys, 'mightily laughed at
them for spending time only in weighing air*.
fScc 'Note on Newton's deductions from Kepler's laws 9 on page 228.
GRAVITATION 227
linear velocity in ita orbit about the Earth and R, the distance
between the centres of the Earth and the Moon, we have
gm = v*/R = (wR)*IR, = <JR,
where <o is the angular velocity of the Moon.
Since to = 27T/7 1 , where T is the time taken by the Moon in
going once round the Sun, we have
Now, T = 273 days = 273x24x60x60 W.T., and
R = 60 ///Her r he radius of the Earth
= 60x40uO/w/feJ. rv radius of the
= 60 X 4000 X 1760 X 3/r.L to 4000m?les.
?r x4000x1760x3 nAQAA , , .
Hence g m = ____ r _ 00899/r./^.
Again, if the acceleration due to gravity be g on the surface of
the Earth, its value at the distance of the Moon from it would, in
accordance with deduction (it) above, be equal to g/60 2 ,
i.e., g m = /60 2 .
So that, taking the value of g to be 32'2ft./sec 2 . t on the surface
of the Earth, we have
g m = 322/60* = 00084 ft. I sec*.,
which is practically the same as the one deduced above, thus fully
vindicating the deductions made by Newton, and convincing him of
the existence of a universal and mutual force of attraction between
any two masses.
Not only this, but Newton also put to test his assumption that
in so far as the attraction at external points is concerned, both the
Earth and the Moon behave as though their masses were concentrated
at their respective centres. He actually showed that the force of
attraction, exerted at an external point, by a uniform sphere, or by a
sphere consisting of a number of concentric uniform shells, one inside
the other, is the same as that exerted by an equal pointmass, occupy
ing the same position as its centre. In other words, the sphere behaves
as though the whole of its mass were concentrated at its centre.
Thus fortified with a clear and complete confirmation of his
deductions and assumptions, Newton announced to the world, in the
year 1687, his celebrated Law of Gravitation in his monumental work,
the Principia*, which the entire scientific world later hailed, in the
words of Langrange, as 'the greatest production of the human
mind'.
'Having lost his papers, as mentioned already, Newton had to iccreate
the whole, step by step, all over again and accomplished the almost superhuman
feat in only 18 months. He used geometrical methods, partly due to his
admiration for ancient geometers and partly to avoid being baited by 'Mttte
imatterers in mathematics*. Perhaps he had Robert Hooke in mind, who had
claimed priority in the discovery of the Inverse Square Law ; the Royal Society
had, iiowever, sided with Newton.
228 PROPERTIES OF MATTER
83. Note on Newton's deductions from Kepler's laws.
I. Let A be the position of a planet, (Fig 140), at a given instant t in
its elliptical path round the sun S, situated at one of its foci. Then, if tho
planet moves on to B in a small interval of time
dt, the area swept out by the radius vector SA, in
this interval of time, is equal to the area of
the triangle SA B.
i.e., equal to i SA.AB = } R.R dQ,
because SA = R and AB R d$.
.'. area! velocity of the planet J/?*.</6/df.
But, this, according to Kepler's second
Fig. HO. law, must be a constant. Putting it equal to A/2,
therefore, we have R*.d$ldt = h.
Now, the fact that the planet moves in a curved path and thus conti
nually changes its direction, means, in accordance with Newton's first law of
motion, that it must be under the action of a force, and must consequently be
possessing an accelerai ion in the direction of the force Resolving thif accele
ration into its two rectangular components, along and at right angles to the
radius vector, we have
(i) component a ly along the radius vector, i.e., the radial acceleration of the
planet, given by
s i pa result obtained from
simple dynamical con
Lsideration.
and (if) component a t , at ri%ht angles to the radius vector, i.e., the transverse
acceleration of the planet, given by
1 . d
a t _
elf)
But, we have seen above that R*. ~  is a constant (h), and hence its
differential coefficient must be equil to zero. It follows, therefore, that a 2 = 0.
In other words, the planet has no transverse acceleration ; so that, the only
acceleration it has is the radial one t and, therefore, the only Jorce acting on it is
towards the Sun.
2. Now, since 7? 2 . 4Q Idt = h, it is clear that dQjdt = hjR 1 . Or, putting
l/R = w, (or R = I /), we have dQ/dt = hu .
It follows, therefore, that
d * = A ( l ^ _ 1 du ^ _ * **L d *  _/ dl L
~df dt \Ju )** V ' ~di ~u* ' <to ' dt '~dij '
. 1 d n ^ c/0
because ? . . = R 2 . j h.
u* dt dt
Differentiating this again with respect to f, we have
d*R , d'tt do lf , d'u f Je h . ,
*  * ^ dt = ""' dv ' L v </, = ^  Aw
Substituting the values of do/dt and rf'/f/A' in the expression for a,
above, we have
**
Now, let the equation of the elliptical orbit of the planet be
I] R = l + e cos 0. Or, lu = 1 f e cos 0,
where / is its latus rectum and e, its eccentricity.
Differentiating this expression twice, with resp^ect to Q, we have
' T~
GBAVITATION 229
And, adding relations (//) and (///), we have
d?u
lu + l.f = 1 +e cos Qe cos 6 = 1.
whence
Substituting this value of fw+^a ) in relation (0, above, we have
ai  *V//  . . [Putting 1 for *.
Or, denoting the constant A 2 // by K, we have n x KIR 2 .
Or, a 4 oc I//? 2 . ... (/v)
i.e., the acceleration, and hence / he force acting on the planet is inversely propor
tional to the square of it* distance jrom the Sun, (the ve sign merely indicating
that the force in question is one of attraction).
Now, the limeperiod (T) of the planet (i.e , the time taken by it to com
plete its one full revolution round the Sun) is given by
r g, area f tne e [liP s e ___ ___ *^L_
areal velocity oj the radius vector ~~ . D2 d$ "
** *
where a and 6 are the semimajor and semiminor axes of the elliptical orbit of
the planet.
Or, r=*f And .*. T* ^. a z b 2 lh\ [vi^ 1 .^ 4
n\2. L at 2
Now, clearly, 2 /a = /, the latus rectum of the ellipse, and, therefore,
al ; so that, ^
. * ^
But, since, in accordance with Kepler's third law, T ? oc a 3 for every
planet, it follows Um 4x*,K is a constant, or ihai K is a constant for every
planet. In other words, K is quite independent of (he nature o\ a planet.
3. Fin^llv, if AH and M be the nspcciivc masses of the planet and the
Sun, and F and F', trie force of attraction, exerted by the Sun on the planet,
and the reaction of the planet on the Sun respectively, we have, from relation
(iv) above,
F  hnlR* and F' = KMIR\
where k and K are constants.
And, since by Newton's third law of motion, action and reaction are
equal and opposite, we have F F' ; so that,
k.m K.M. Or, [k/M  Kim ~ a constant, say, C.
So that, k = M.G.
Substituting this value of A ia the expresssion for Fabove, we have
P m M
F p.G,
showing that the force of attraction between the planet and the Sun is directly
proportional to the product of their ma**es.
84. Newton's Law of Gravitatiaar This law states that every
particle of~~fnatter in the universe attracts every other pa> tide with
a force which is directly proportional to the product of their masses, and
inversely proportional to the square of the distance between them.
Thus, if m and m' be the masses of two particles, distance r
apart, and F, the force of attraction between them, we have
F oc m m'lr*. Or, F = G.w.m'/r 1 ,
where C? is a universal constant, called the Gravitational Constant.
Obviously, if m = m' 1 gm., and r = 1 cm., then, F = G.
230
PBOPERTIES OF MATTER
Thus, the Gravitational Constant is equal to the Jorce of attraction
between two unit masses of matter, unit distance apart. Its dimensions
are A/ 1 ! 8 !* 2 , and its latest, accurate value, (as determined by Heyl
in 1930), is taken to be 6669 x 10~ 8 C.G.S. units. The gravitational
constant is also sometimes referred to as the astronomical unit of
force.
The law is universal in the sense that it holds good, right from
huge interplanetary distances to the smallest terrestrial ones. The
minimum distance up to which it is valid is probably not yet known
with absolute certainty, but it seems to break down at molecular
distances, which are as small as 10~ 7 cm. We shall discuss latter, in
this chapter, some of the overwhelming evidence in favour of this
law, as well as the small deviations from it and the proper explana
tion for then, on the basis of the new ideas put forth by Einstein.
85. Determination of the Gravitational Constant. The methods
for the determination of the gravitational constant, (and, therefore,
also those for the determination of the density and the mass of the
earth), may be divided into two categories, viz ,
(/) Mountains and Mine Methods. which involve the measure
ment of the force of attraction exerted by a large natural
mass, like a mountain or the earth's crust*, on a plumb
line suspended on one side of it, which is then compared
with the force of attraction on it due to the earth, as a
whole.
(ii) Laboratory Methods, wh'ch involve the more delicate
measurement of attraction between small masses.
We shall deal here only briefly with the former, more for their
historical interest than othorwiss ; for ; the results obtained were not,
indeed, they could not be, very accurate. The latter, i.e., the
laboratory methods, we shall however study in proper detail.
(0 Mountain Methods.
1, Bouguer was the first to have attempted a determination of the
value of G. Wnilc engaged in geographical measurements in the Andes (Peru),
in the year 1740, he suspected a
deflection of his plumb line due to
large mountainmasses. He decid
ed to verify this, and selected a
mountain, Chtmhorazo, 20.000 //.
high, (in the Andes) for the pur
pose. Shorn of experimental
details, his method was the follow
ing :
He chose two stations A
and B[Fig. 141. </) and O/)], the
former due south of the summi t of
the mountain and ch*e to it, and
the latter, in the same latitude, and
at about the same altitude some
distance to its we*t, away from its
A r
Fig. 141 (/)
influence. At stat ion B, he observed
a star passing the meridian directly
overhead, so that the plumb line
*The word is prob tbly a relic of the times when the earth was supposed
to be a globe of water, bounded by a solid shell or crust. It. is now used, how
ever, to signify the rigid surface layer of the earth, which is heterogeneous and,
more or less, in a state of permanent stress and strain.
GRAVITATION
231
hung exactly vertically parallel to the telescope. But at station A he
observed that it wa* attracted by the huge mountainmass, (ou td
its nearness). He measured this A ,
deflection of the plumb line at A ^
and thus compared the horizon
tal pull of the mountain with the
vertical pull of the earth. For,
if F and F' be the forces of gravi
tational attraction acting on the
plumb line due to the mountain
and the earth respectively, and
4, its deflection from the vertical,
(Fig. 141, (*')] we have
tan f = F(F'.
Now, dearly, F' = nig,
where m is the mass of the plumb
iine and , the acceleration due to
gravity. So that,
F
And, if V be the volume and p, the density of the mountain, and r, the
distance of its c g. from the plumb line, we have
V mass of the moun
tain  K.p.
.G.
(
Hence m.K.p.G/r 2 = mg tan $. Or, G g.r* tan
Thus, a knowledge of the volume, density and shape of the mountain,
and hence its cen're of gravity, wa* needed to determine the value of G. Bouguer,
therefore, proceeded 10 do so, but did not quite succeed ; for, due to the most
adverse conditions of snow and storm under which he had to work, he could not
properly survey the mountain, anJ ihe results hs obtained were very much
wide of the mark. Thus, for example, he found that his plumb line was drawn
aside by about 8*, and his calculation showed that if the mountain were as
dense as the earth itself, the deflection of the plumb line would have been
twelve times as gr*at, indicating that the earth was about twelve times at
dense as the mountain. And this, as we know, is very much beyond the truth*
Nevertheless, he had the satisfaction of showing that the attraction due to the
mountain masses did actually exist and thai the method was, therefore, possible.
Not only that, but he also deserves the fullest credit for proving conclusively
that the earth was not just a globe of water or a hollow shell, as was fairly widely
supposed at the time.
2. Maskeiyne, later in the year 1774* repeated, at the request of the
Royal Society, Bousuer's experiment on the mountain Schiehallion, in Perth
shire (Scotland). 3547 feet hi eh, an elaborate survey of \vhich *as first made to
determine as accurately as possible, its volume and density (and hence iti mass)
and centre of gravity.
Two stations were then chosen at fqual distances from the c g. of the
mountain, on the northsouth line (Fig. 142), and the tarn* star was observed,
(as in Bou^uer's experiment), by means of a special type of telescope, called the
Zenith Sector^, first at the Sduth Station acid, a rmmh later, at the North
StaHon At the former Station, the star which, in the absence of the mountain,
would be directly overlmd, appeared to shift slightly to the north, because the
plumb lins was pulled by the mountain towards the north, (and the zeniih
*He was the Astronomer Royal at the time.
fThe instrument could rotate about a horizontal (Fast ard West) uxfa
at its objectglass end. pointing upwards, and was provided with a pit rob
line, suspended from this axis, over a scale, graduated in deuces, so that the
angular distance of the telescope from the v crtical could be directly read on it*
232
PROPERTIES OF MATTER
thus shifted to the south). At the other Station, on the other hand, the exact
opposite was the case, (the plumb lioe being pulled towards the south, the zenith
thus shifting to the north) and ihe
star, therefore, appearing to shift
eqally to the south Thus, the
total shift of the star was double
of the deflection of the plumb line
at either station due to its attrac
tion by the mountain. This wa
carefully measured and was found
to be 55". Out of this, a shift of
43* was calculated lo be due to the
curvature of the earth's surface
so tnat the net shin or deflection
of the plumb hne, due to the gravi
tational pull of the mountain, \\as
(55"43") = 12*. In other words,
the plumb line* at each of the two
stations, was deflected by 6" due to
the mountainmass The valoe of G
was then calculated, as explained
above <in Bouguer's experiment),
and was found to be 7*4xlO~
C.G.S units.
Further, it was estimated
that if the mountain had the same
density as that ol the earth, the
deflection of the plumb line, due
to its attraction, wouki have been
9/5 times the observed deflection,
showing that the earth was 9/5 time*
Fig. 142.
denser than it. And, since the density of the mountain, determined from
pieces of rocks composing it, was fouiit to be 2 5 gms jc c., the density of the
earth came to be 9x2'5/5 or 4 5 gms. Ice. This was corrected and increased to
50 gms.lc c. after a careful resurvey of the moumain, some thirty years later,
i result nut very much wide of the mark.
Since, however, it is almost impossible to determine correctly the mas*
md position of the e.g. of a huge natural iruss like a mountain, the value of G,
>b tamed by the above methods, is far from reliable, and is at best only ao
approximation.
(//) Mine Methods. In these, the timeperiods of a pendulum, (say, a
seconds pendulum), arc compared on the surtace of the earth and at the bottom
>f a mine. It is obviously greater in the latter case, the value of g being less
,here than on tne surface of the earth, {see page 206]. The change iti its time
>eriod enables a comparison to be made between the values ol acceleration due
o gravity, and hence between the density of the layers immediately above the
wndulumbob and that of the rest of the earth, which, in its turn, Teads to a
letermination of Af, the mass of the earth and thus to a calculation of the value
>f G, as will be clear from the following :
We know that the weight of a body at a place is the force with which it
5 attracted by the earth towards its centre, and is numerically equal to the pro
iuct of its mass (m) and the acceleration due to gravity at the place.
Thus, if g be the acceleration due to gravity on the surface of the earth
nd g', that at the i bottom of a mine of dcptn h, we have its weights at the two
laces given respectively by
+ mM  , , w(A/ '
mg~  .G ~
,
and
mg
'here M is the mass of the earth, R its radius, and m' t the mass of the outer
tie 11 of the earth, of thickness //.
i> , M  , (Afm'> ,
t>that, *Br.0 and g'  ="' G>
hence,
g' Mm' f_R_ V
T~ ST' \RitS '
GBATITATION 233
And, clearly, if p be the density of the outer shell of the earth, its mass m' i
obviously given by the relation,
m' = tolPGRWp. ...(//)
Now, the average value of p was obtained by determining the densities
of the samples of rock at different levels, down to the bottom of the mine, and
thus the mass (m') of the outer shell evaluated. Substituting this value of m' i
relation (/) above, the mass (A/) of the earth was easily determined. And, then,
putting the value of M in the expression for g above, the value of G could be
calculated out straightaway.
Airy was the first person to have made a successful attempt of this nature
in the Harton coal pit in Sunderland, in the year 1854, two earlier attempts*
made by him, in a Cornish coppermine, as early as 1826 and 1828, having come
to naught, due to unfortunate accidents in the mine. Airy's value of G came to
5'7xlO~ 8 C.G S. units, and that of the density of the earth's surlace, to 6*5
gms./c.c.
Like the earlier Mountain experiments, these experiments by Airy too
gave far from satisfactory results, due mainly to the difficulty of determining;
accurately the density of the outer shell of the earth. His methods, however r
wth improved modifications, now find a wide and useful scope in the branch?
of Geophysics 9 [ 80, (page 216)].
(///) Laboratory Methods. In these methods, the attraction
between the masses is inevitably feebler clue to their small ness. But
this is more than compensated for by the high degree of accuracy
with which the masses and their sizes can be determined. The first
successful attempt at an accurate method of this type, for the deter*
mination of the Gravitational Constant (G) was made bv Cavendish r
in the year 1798, in which he made use of the Torsion Balance.
It will be of some interest to recall that Cavendish was prob
ably also associated with Maskelyne in his Mountain experiment,
performed some twenty five years earl it r. He, however, took his^
cue from Rev. John Michell, who had devised an apparatus almost
similar to Cavendish's own, but was not destined to use it, due to*
his sudden death. His apparatus fell into the hands of Prof.
Wollaston, who passed it on to Cavendish.
(a) Cavendish's Method. The apparatus ussd by Cavendish,
and installed in an outhouse in his garden on Clapham Common, was
as shown in Fig. 143. A long cross bar PQ, about 6 ft. (or about
180 cms.) long, was suspended from the ceiling of a room and was
free to turn about a vortical axis by means of an arrangement, mani
pulated from outside. It carried two large and equal lead spheres C
and >, about 8 to 10 inches (or 20 to 25 CMS.) in diameter and weigh
ing about 350 Ibs. each, at the ends of two metal rods attached to
its two ends.
Immediately below the midpoint of the crossbar was a torsion
head M which could also be worked from outside, from which was
suspended a dealrod RS. (slightly bigger than PQ), by means of a
fine torsion wire W, of silvered copper. Two wires (w. vv), fastened
the ends of the dealrod to a vertical rod r in the middle, which was
attached to the suspension wire. This increased the strength of the
rod, without increasing its moment of inertia. Two small lead balls, 2
inches (or 5 cms ) in diameter and weighing about a pound and a
half (or 680 gms.) each, were suspended from the two ends of th&
dealrod RS such that the centres of the four balls lay in the same
horizontal plane, roughly in a horizontal circle of about 3 ft. (or 90
534
PROPERTIES OF MATTER
cms.) radius. The arrangement was such that, when the line joining
the centres of the large lead spheres was at right angles to the tor
sion rod, there w&* no twist or torsion in the suspension wire W *
Each erd of the torsion rod carried a vernier, (of five divisions),
which moved over a fine ivory scale, fixed to vertical stands, and with
each division equal to 05".
To guard against any changes of temperature, and consequent
airdraughts, which would otherwise mask the gravitational effect,
the room was closed and observations were taken with the help of
telescopes T and T, fixed into the walls of the room, as shown. And,
Further, to avoid the effect of any outside electric charges, the whole
apparatus was enclosed in a gilded glass case, supported on four
levelling screws.
The method of procedure was the following :
The rod PQ was rotated until the line joining the centres of the
arge spheres was at riidit angles to the torsion rod, i.e., in the post'
Fig. 143.
ion in w}iich there was no twist in the suspension wire, and the read
ing on the verniers, attached to the torsion rod at either end, taken.
The large spheres
were then rotated un
til they lay on oppo
site sides of the tor
sion rod and near to
the small balls at
either end, i.e., in the
positions C and D, as
shown in Fig. 144,
such that thQ lines
* *  &
F~~IIJ" joining the centres of
g * I44< each pair of near
balls were equal in length and perpendicular to the torsion rod. Obvi
ously, then, the forces exerted by the big spheres on the correspond
ing near small balls were equal and opposite, thus constituting a
couple, tend ing to rotate the torsion rod. This was resisted by the
torsional couple set up in the suspension wire, and equilibrium was
attained when the deflecting couple, due to the forces between the two
GEAVITATION 23f
pain of balls, was just balanced by the restoring torsional couple, set
up in the suspension wire. The position of the verniers was again
noted on the scales by the method of oscillation, as in the case of an
ordinary physical balance In Cavendish's own experiment, this
distance between each pair of balls was 8" (or 20 cms.), and the
small balls were displace 1 through '7681" (or 1*915 cms.). The rod
PQ was then rotated again about its vertical axis, until the large
spheres now occupied the positions C r and D' respectively, and the
same adjustment was mad a as before?, viz., that the lines joining the
centres of the two near balls were of the same equal lengths as before
and perpendicular to the torsion rod. The positions of the verniers
were r*?ad on the scales, as before, and their mean taken as the deflec
tion of the torsion rod.
The value of G was then calculated as follows :
Let M be the mass of each large sphere, m, that of each small
ball an' I d tho distance batween each pair of near balls. Then, the
force of attraction between each pair of balls is, clearly, equal to
G.Kf.m/d 1 ; and, therefore, if/ be the length of the torsion rod, the
deflecting couple formed by this pair of equal, opposite and parallel
r . i , M.m .
forces, is equal to , 2 G.l. [/ IcosQt&l.]
Any], if ba tho twist (in radians) in the susp3nsion wire, and
C, tho Factoring couple p^r unit (raJia.i) t\yist set up in it, the restor
ing co'iplo (Itu to torsion is eqpiil to C.Q. Since the torsion rod is in
equilibrium under these two couples, we have
 /  .0./ == C.0. whence, = ./ , .0.
d 2 ' M.m I
In fwl T to determine the value of (7, the torsion rod alone was
set into torsional vibrations abo'it the suspsnsion wire, and its time
period was measured. Cavendish found it to be 28 minutes in his
apparatus.
Then, if / b3 tli3 munint of inertia of the torsion rod (together
with the small balls, about tho wire as axis, and /, the timeperiod of
the rod, we have
t = 27r A/ p , whence, C = J 2  .
Substituting this value of C in the relation for G above, we have
~~ "M m.7.r 2 ''
Or, if we ignore the mass of the torsion rod, we can put
/ = 2/w(//2) 2 = iw/a/2.
So that, substituting this valua of/ in the above expression, we have
Corrections and Sources of Error in the Experiment. Correc
tions were applied for the following :
( I ) force of attraction between each large sphere and the distant
imall ball ;
236 PROPERTIES OF MATTER
(2) force ofafyrqption between the two large spheres and the tor
sion rod ; and
(3) forces exerted by the rods carrying the large balls.
Tho following are the sources of error in the experiment :
(1) The gravitational forces between a pair of balls being small,
the torsion rod had to be made long to increase the deflecting couple.
This also minimised the force of attraction between a large sphere
and the distant small ball, but required a large chamber, and thus
convection currents could not be avoided. In addition to this, the
torsion rod was also disturbed by heavy traffic on a nearby road.
(2) For a given deflecting couple, the deflection of the torsion rod
was small, because the suspension wire required a JJarge torque per
unit deflection.
(3) The torsion wire, being not perfectly elastic, did not return
to its normal position when the applied forces were removed, and
thus the torque was not strictly proportional to the angle of deflection.
(4) The distant large spheres decreased the angle of deflection
while the rods carrying them increased it.
(5) The method of measuring the angle of deflection was not sen
five enough.
( (6) 1'he time period of the torsion rod, with the attached small
rolls was much too big, i.e., its swings wcro much too sluggish and
impaired, rather than improved, the accuracy of measurement.
The value of G obtained by Cavendish, as the> in9an of twenty
nine observations, was 6*754 x lir 8 C.G.S. units *
Many other attempts wera made since Cavendish's time, by
Jolly and Pjynting, amon^ others, to obtain the volue of G more
accurate ly but the method adopted by Boys, with his newly invented
quartz fibre, used as suspension for the torsion rod, was by far the*
best of these. We shall, therefore, discuss that first.
(b) Boys* method. Sir Charles Veroon Buys removed, almost a
century later, in the year 1895, all the defects of Cavendish's experi
ment by
(/) reducing greatly the size of the chamber, thus considerably
minimising convection currents and making it easier to con
trol its temperature ;
(11) arranging the pairs of balls at different levels, thus making
the attraction between the distant large and small balls
almost negligible ;
(ill) using, for the suspension wire, a quartzfibre, which required
a comparatively very small torque per unit deflection ; and
(iv) which being almost perfectly elastic, besides being fine and
strongf. the angular deflection produced was appreciably
large and also proportional to the torque ; and
(v) measuring the angular deflection by the telescope and scale
method, which greatly enhanced the accuracy of measure
ment.
*A musingly enough, Cavendish made a slip in calculating this mean and
It was later pointed out by Baity.
fA quartzfibre is found to be much stronger than a steel wire of the
wane dimensions. Boys was thus able to use fibres having a diameter as small
li '0125 mm.
GRAVITATION
237
Thus, Boys greatly reduced the size of the apparatus and yet in
creased its sensitiveness. This ra%y, at first sight, appear to be a
contradiction in terms ; for, it is commonly b3lieved thit the larger
a piece of apparatus, the greater the degree of accuracy obtained
from it. Boys clearly showed, however, that this was not so, that
the sensitivity of Cavendish's apparatus was quite indep3ndent of its
dimensions and that there was no point, therefore, in attempting a
larger version of it. He argued as follows :
,,,, . ' . r i , Deflecting couvie
The sensitivisy of the apparatus =7 
K/f m I
Clearly, deflecting couple oc * '
i.e. t
Restoring couple
(mass > 2 x length of torsion rod
Now,
So that,
d* ' '* distance he t ween near balls ,
mass oc volume oc 4* (radius)* 13 oc r 8 .
deflecting couple oc ' t 9
T 5
i>
And,
. I moment of inertia
restoring couple oc  T oc  r  y *
* ^ H (time of swing?
Now, there is a practical limit to the time of swing which should not
exceed 5 minutes, whatever the size of the apparatus, or else the swings become
very sluggish, thus impairing the accuracy of measurement. This being so, we
have
3v Jl. P being the same
1 Lfor any apparatus.
. MK 2
restoring couple oc   <
Or,
restoring couple oc L 5 . And .*. sensitivity
L 6
In other words, the seniitivity is independent of the size (L) of the
apparatus.
Thus, if, for example, we double the dimensions of Cavend'sh's appara
<O> keeping the time of swing (/) the same, we find that M is increased by (2)',
because it is proportional to (radius)*, d is increased by (2) 8 and / by 2.
The net result is that the valuo of d ^MV 2 .G/2*r 2 d 2 / remains the same, i.e.,
no advantage is derived by doubling the size of the apparatus, i.e., by increasing
sthg dimensions of all its parts in the same ratio.
What Boys, therefore did was to reduce the
dimensions of the different parts of the apparatus
in different ratios, as will be clear from the
following :
A small mirror strip S t (Fig. 145), about 25
cms. long which acts as the torsion bar, is suspend
ed by means of a fine quartzfibre, from a torsion
head T inside a glass tube, about 4 cms. in dia
meter. From the two ends of the strip are sus
pended, by similar quartzfibres, two small gold*
balls A and B, about 0'5 cm. in diameter and 265
fins., in mass each, one ball being about 15 cms.
ibove the level of the other. In an outer coaxial
iube, which can be rotated about the common axis,
jwo large lead balls C and D each about 11*0 cms.
n diameter, (and about 74 kfgm. in mass), are
Fig. 145.
*On account of the higher density of gold (19 3 gms /c.c.) compared with
hat of lead (l\*3 gms lc.c.) t the spheres of gold for the same mass are smaller
ban spheres of lead and thus enable the distance d between the centres of the
arge and small balls to be reduced.
238 PROPERTIES OF MATTER
suspended such that the centre of C is in level with that of A, and
that of D, in level with that of B, (to .ensure greater precision in the
measurement of the distance between each pair), the distances between
the centres of the pairs A and C, and B and D, being exactly equal.
The deflection is measured by the telescope and scale method, a half
millimetre scale being placed at a distance of about seven metres
from the strip.
The experiment is performed by rotating the outer cylindrical
tube until the large (lead) balls he on the opposite sides of the two
gold balls (but not in a line with the mirror strip*), so as to exert the
maximum moment on the suspended system, i.e., when the angle of
deflection is the largest.f The tube is then rotated, so that the lead*
balls now lie on the other sides of the gold balls in a similar position,
again exerting the maximum moment, or producing the greatest
deflection. The mean of the two is then taken. Let it be Q.
The calculation for the value of G arj then made as follows :
Let A, B, C and D, (Fig. 146), be the four balls, when they are
in equilibrium, in the position of maximum deflection 0 To visualise
the balls in these positions, we must remem
ber that to start with, the centres of all the
four balls lie in the same vertical plane, there
being no twist in the suspension fibre and
hence no couple acting on the suspended system
If we now rotate the larger balls C and Z>
through a certain angle, the plane containing
their centres will also rotate through the same
angle ; with the result that a gravitational
couple now comes into play on the suspended
system, tending to rotate it into a position of
equilibrium in which, once again, the centres
of the small balls come to lie in the same
vertical plane with the centres of the large
Fig. 146. balls. this being equally true when the sus
pended system suffers its maximum deflection.
Here, the small balls A and B are shown in their initial posi
tions, corresponding to 0=0 and the large balls in their final posi
tions when they have been rotated into a position BO as to exert the
maximum couple on the suspended system, tending to make the
latter suffer its maximum deflection 0. Obviously, equilibrium will
be attained again only when the centres of the small balls come to
lie in the vertical plane of the centres of the large balls. To bring
the small balls back to their original positions, (shown in the Figure),
, therefore, the torsion head will have to be rotated in a direction
opposite to that in which they have been deflected by this couple.
In other words, the deflecting gravitational couple exerted by the larg<
*For, in this position, the gravitational forces due to large balls on the
tmall balls near to them will act in opposite directions along the same straight
line and will thus neutralise each other.
tThis position of the lead balls is chosen because when the couple on the
impended system is the maximum, the rate of variation of the couple is small
and the relative positions of the balls need not be known with any greal
accuracy.
GBAVITATION 239
balls on the suspended system, in the position shown, is just balanced
by the restoring torsiona! couple set up in the quartz suspension.
Now, let O be the midpoint of the mirror strip, and let / be its
halflength (i.e., let OA^OB=1). Let OC=OD=b, AC^BD^a
and let AOC~BOD~a. Let OE be the perpendicular drawn
from O on to DB produced.
Then, clearly, in the triangle OBD, we have
BD = \/O~D*+OB 2 ^OD.Utf cos a. [See Appendix 1, 7, (11);
Or, rf= ^b*+l* k 2blcosa ~= (b*+l*2bl cosa)}.
sin a BD d p.* in a triangle, the sides are proportional
Also,  n =r* = 7; D = " / to the sines of the angles opposite to them*
sin EDO OB / L [See Appendix 1, 7, (/;.]
and, therefore, sin BDO = ^ a  ...... (/)
Now, in the rightangled triangle OED, we have
OE = OD.sin EDO =b.sin BDO.
Or OE = b '' S * [From (/) above,
' d
Obviously, the attraction between the t\\o balls of each of the pairs,
A, C and B, D is equal to M m.GjBD* = M.m.G/d 2 , where M and m
are the masses of each large ball and small ball respectively.
These two forces, being equal, opposite and parallel, constitute a
couple, tending to rotate the mirror strip ; and, quite clearly, the
moment of the couple = G ' ' EF = G  ,, 2.OE,
where EF = 2.OE is the perpendicular distance between the lines of
action of the two forces.
^ , . a . . M.m b.l sin a r Substituting ther
Or, the deflecting couple = G .,. z.   va j ue fO, fron?
^ a L above.
_ r 2M.m.b.I sin a __ 2M.m b.l sin a
Now, this deflection of the mirror strip is resisted by the torsion
or twist, set up in the suspensionfibre, and the mirror comes to rest,
when the deflecting couple due to the attraction between the two
pairs of the balls is just balanced by the restoring torsional couple
set up in the suspension fibre. If C be the torsional couple per unit
deflection, set up in the suspensionfibre*, the restoring torsiona)
couple is equal to C. 6. and, therefore,
2G. M m b I sm a n
w ' u= 2M.mbJ*ina ~* w
whence the value of Gjmy be easily calculated out.
' * ifhis^js determined, as in Cavendish's experiment, by oscillating the mov
ing system, in the absence of the lead balls, and noting its timeperiod.
t The value of a is clearly the angle through which the torsion head T
must be rotated to bring the small balls back into their originalpositions and can
be easily read on a circular scale attached to it. Since a quartzfibre is nearly
perfectly elastic, the value of may be taken to be the same as that of 6.
240
PROPERTIES OF MATTER
The value of G obtained by Boys was equal to 66576 X 10~ 8
C.G.S. units.
Alternative Calculation. The value of G may alternatively be calculated
out as follows, using the same symbols as above.
Gravitational pull between each pair of near lar^e and small balls = ^ *G 9
in the directions A to C and B to D
respectively, (Fig. 147). Resolving these
into their two rectangular components
each, a'ong and perpendicular to AB t
we have the latter components of each
equal to
F ^ G ^ cos ^^ G Mmp
Fig. 147.
.Now, AC*
So that,
BD* = d 2 = /? 2 fa; 2 , where p =^ b sin a and x
d*~'
where CK DL ~ p are the perpendicu
lars drawn from Cand D respectively on AB
produced.
Or,
And, therefore,
Thus,
= 6*
F ** G
*/).
1 af^cs5 t ah/ 2 ~26/ cos*.
(OK OA)
(b cos a/).
[/ siri*
M.m.b sin a
Hence, the deflecting couple on AB dse to these forces F and F t
= F.AB = F.21 G M^jA 5 i w __ a : 2/ _
Now, if C be the torsional couple set up in the suspension fibre per unit
twist of it, (/e., for unit deflection of the suspended system), the total restoring
storsional couple for a twist a in it = C.Q.
Since, for equilibrium, the two couples must balance each other, we have
G M.m.b sin a.2/ r
. 3 1^
Or,
M.m.b, 21 sin a
the same as expression I in the cas^ above, whence the value of G can be easily
obtained.
N B. In case the centres of the neir large and small balls do not lie in the
same horizontal plane, but a verticil distance h apart, as shown in Fig. 148, then,
if d' be the actual distance between them, we have
gravitational couple on the suspended system == ' ' ^
But, clearly, d'* * (</ 2 4/r), where c/is the horizontal (or
the perpendicular) distance between the balls, or rather between
their centres. So that,
gravitational couple on the suspended sys'em =
And, therefore, proceeding as above, we have
2bl cosx+h*)?
M.m.b.U sin a
.(ID
GRAVITATION
241
an expression which, when h ; when the centres of the near large and small
balls lie in the same horizontal plane, is reduced to relation (I), above.
(iii) Balance Methods. These methods do not compare in accu
racy with Cavendish's, Boys' or Heyl's methods, but are given here
only for their historical importance.
(a) Jolly's Method. As early as 1881, Von Jolly had suggested that a
common balance could be used to measure directly the gravitational force
of attraction exerted on a mass, placed in one pan, by a large lead mass, placed
immediately below it.
He actually performed this experiment in Munich, where he had a common
balance fixed at the top of a tower, 21 metres high, and suspended two long wires
from its two scale pans, carrying two other pans at their lower ends.
Two equal masses were then placed in the two upper scale pans and
balanced against each other. One of the masses was then moved down into the
lower scale pan, on the same side, so that, being now comparatively (about 20
metres) nearer to the centre of the earth than the other mass, its weight increased
a little, this increase (due to the earth's attraction) being equal to the extra
weights needed in the other scale pan to balance the beam.
A large lead sphere (of known mass) was then placed immediately below the
lower pan carrying the mass, so that due to the additional attraction of it by the
lead sphere, its weight again increased a little. This increase was also determined,
as before, by putting some more weights in the other scale pan. The attraction of
the mass by the lead sphere could thus be compared with its attraction by the
earth. And, since the distance between the centres of gravity of the lead sphere
and the earth was known, the masses of the two could also be compared. Then,
the mass of the lead sphere being known, the mass of earth could be easily calcu
lated out. And, once the mass of the earth was obtained, the value of G could be
deduced as in g 85 (*/), page 232.
(h) Poyn ting's Method. The balance method has perhaps been used to the
best advantage by Prof. J.H. Poynting, whose arrangement was much more elabo
rate and susceptible of a much higher degree of accuracy. He performed his
experiment in the year 1891, in the basement of the University of Birmingham.
The apparatus used by him, (shown diagrammatically in Fig. 149), con
sisted of a strong and sensitive bullion type of balance, with a gunmetal beam,
provided with steel
knifeedges and planes.
The whole apparatus was
fully enclosed in the room
and all necessary manipu
lations were made from
the outside.
Two equal spheri
cal balls, A and #, of an
alloy of lead and anti
mony, weighing about
50 Ibsi each, were sus
pended from the two
ends of the beam. A
large sphere S, of the
same alloy and weighing
about 350 Ibs. was
arranged below, on a /""
turntable, which could fcea===
be turned about its .  AQ
vertical pivot P, so that ' g. 149.
the sphere could be brought to lie immediately under the ball A or /?, as desired.
To guard against the tilting of the turntable due to the weight of the sphere S, a
smaller sphere 5', of half the mass of S, was placed on the other side of the pivot
at double the distance of S from it, so that, in accordance with the principle of
moments, the turntable was kept in equilibrium.
242
PROPERTIES OF MATTER
To start with, the sphere S was brought to He under the ball A, so that A
was attracted downwards with a force equal to G.M.to/r*, where AT and m ar the
respective masses of 5 and A, and r, the distance between their centres, (which was
about one foot).
The turntable was next rotated about its pivot until the sphere 5 came
from under A to under B, and the balancing sphere 5' moved on to the other side,
(into the dotted positions shown), so that 5 now exerted a pull on B instead of on
A, resulting in the beam being tilted in the opposite direction to that in the first
case, the angle of tilt being now obviously twice* that due to S on A or B. Let
this deflection or tilt of the beam be 0.
Then, if a be the length of each arm of the balance, (i.e., if 2a be the length
of the beam), we have
change in torque or couple due to the shifting of S from the first position
(under A) to the second position (under B)
And, if C be the torque or couple required per unit deflection of the beam,
the torque for deflection 9 of the beam isalso equal to C$.
Hence G m .2a*=C.O So that, G ^/wIL* '" ^
Thus, knowing M, m, a, r, C and 0, the value of G could be easily
calculated.
To determine the value of C, a centigram rider was moved along the arm
of the balance and the deflection a of the beam, for a shift / ems', of it along the
arm, was noted. Then, clearly,
01
Ca, whence, C
B
B
1
Substituting this value of C in relation (/) above, we have
r*j9 ;OU./ _OUr 2 ./
M .tn 2a * a "" 2Mm a ' a '
whence, G can be easily evaluated.
The effect of 5 and S" on the beam was eliminated by repeating the experi
ment with A and B, a foot higher up m the dotted positions shown, and proper
corrections were also applied for the cross effect of 5 and S' on A and B.
Both the angles, 0, and a, being very small, (0 being only about one second)
were measured by Kelvin's double suspension mirror method, as illustrated in
Fig. 150, where M is a small mirror, suspended
by means of a bi filar suspension (w and >v')
from two horizontal brackets B and B', in level
with each other, with a small adjustable gap
between them, the former being a movable
one, attached to the pointer (p) of the balance,
and the latter, a fixed one. Thus, when the
beam turned, the wire w also turned with it,
making the mirror M turn about the stationary
wire w f . This, with the gap BB' suitably adjust
ed, magnified the deflection of the beam about
150 times. The scale, graduated in halfmilli
metres, was arranged about 5 metres in front of
the mirror, and its image in the latter viewed
from the room above by means of a vertical
telescope, fixed up in the ceiling.
The effect of the air draughts or currents
was eliminated by using what arc called damp
ing vanes V, suspended from the mirror and
kept immersed in oil.
*""" Poynting obtained the value of G to be
66984 x 10~ 8 C.G.S. units.
bSni firsfcbnies to its'original position from its
first tilt and then gets tilted equally in tfce opposite direction, due tp the attrac
tion pf S on #.
GRAVITATION
243
(c) Heyl's Method. The value of G obtained by P.R. Heyl, in
year 1930, is taken to be the most accurate one so far. His
is a modification of Braurfs Torsion Balance experiment,
which, in its turn, was a revised version of Boys' earlier experiment,
referred to above, (page 236).
Heyl performed his experiment in a constant temperature
enclosure*, with the pressure inside reduced to about 2 mms. of
mercury column, in order to minimise
convection currents. The attracting large
masses, used by him, were massive steel
cylinders, each of mass about 66 3 k.gms.
suspended from a system, free to rotate
about a vertical axis midway between the
two. The smaller masses, each weighing
244 gms., were balls of gold, platinum
and optical glass, in three different sets
of experiments respectively, and were
suspended from the two ends of a light
aluminium torsion rod R, 28*6 cms. long,
(Fig. 151), supported by a tungsten
thread, T.W., (1 metre long and 025 mm.
in diameter), and two inclined copper
wires, (u\ vr), so that almost the whole of
the moment of inertia remains in the balls
themselves. He chose as suspension a
tungsten thread in preference to a quartz
fibre, because the latter is sometimes found to break quite unexpect
edly and for no apparent reason .
This suspension system (of the torsion rod and the two small
masses) was made to oscillate in the gravitational field of the two
large masses, which
were arranged once
^^ ^^ . with the centres of
(_) O O v yd : all the four masses
t Fig. 151.
j
o
d
(CL) NEAR POSITION
(b) DISTANT POSITION
lying along the same
horizontal line, and
then, with the hori
zontal line joining
their centres, along
the right bisector of
Fig. 152.
the torsion rod, the two positions being referred to as the 'near' and
the 'distant' positions respectively, [Figs. 152 (a) and (b)] the gravi
tational attraction accelerating the oscillations in the first case and
retarding them, in the second. Or, as a variation of this, the time
period of the suspension system was first determined, with no other
masses in its neighbourhood and then with the large masses brought
in the near position, shown in Fig. 152 (a).
To set the system oscillating, bottles of mercury were brought
near the smaller masses for a short while and then removed, when, for
*It was, in fact, the constanttemperature room of the American Bureau
of Standards, 35 />, below ground level.
244
OF MATTER
an angular displacement of 4*, the system continued oscillating for
about 20 hours.
The usual telescope, lamp and scale method, was employed to
observe the oscillations, and the passages of the lines on the image of
a scale across a vertical cross wire of the telescope were recorded
automatically by a pen on a chronograph, another pen marking down
on it the 'second' signals from a standard clock.
From the timeperiods of the suspension system, in the two
cases, the value of was then calculated out as indicated in brief
outline below :
Let T l be th timeperiod of the suspension system, when the large
masses are not yet brought in its neighbourhood. Then, if G be the torsional
couple per unit twist of the suspension wire and /, the moment of inertia of the
system, we have 7\ =2n\///C.
The large masses are then brought into the near position, shown in Fig.
152 (a), such that the distance between the centres of the neighbouring large and
small balls is the same on either side, say, equal to d. This will obviously result
in a gravitational pull Fby each large mass over the corresponding small one,
towards itself, given by F = w G,
whepc M and m are ths values of each large and small mass respectively.
Considering the gravitational pull between the neighbouring large and
small masses to remain unaltered by any small displacements of the small
. spheres from their initial or equili
" brium positions, the gravitational
^ n , pull of each large mass over the
C<cc rf& small one, when the small masses
* *** 1 A and B are deflected a little
through an angle 0, into the posi
tions A' and B' ', (shown in Pig. 153)
will also be equal to
p = M.m G
towards itself, along the line joining
the centres of the two masses.
And, clearly, resolving this gravita
tional pull on both sides into two
rectangular components, along and perpendicular to A'B', we have the compo
nents perpendicular to A'B' (the line joining the centres of the two small
masses) equal to Fcos a, represented by A' C and B'E respectively, where
LOA'C = LO'B'E = a.
So that, we have F cos a. '!? cos a = ^ *sin $,
d 1 d*
Fig. 153.
where = LOA'D
Now, since
And, therefore,
LO'B'J  (90a).
LA'OP = LB'O'P =
y, we have
(0+y).
F coi a =
M.m.G . ,. . ^
, r .sin (0fy).
Or,
F cos
(I)
r$Id*
M.m.G
d* ""'
If d and y be small, as they are in actual practice.
Now, clearly, A' A = B'B = Jy = r$, whence, y
Substituting this value of y in relation I above, we have
F cos a 's ' ( f j~ ) *= ji'
d 2 \ d / a* ^
These two forces acting at A 1 and B' obviously form a couple, tending
to bring the small balls back into their original positions A and B ; and, clearly,
GRAVITATION 245
moment of this couple  M '' G .( 1 +  \0 x 2r. [where A 'B'  2r.
Since C is the torsional couple per unit twist of the suspension wire, the torsional
couple set up in the suspension wire, also tending to bring the small balls back
into their original positions A and B is equal to CQ.
Thus, the total couple acting on the suspended system of the small balls
If, therefore, T a be the timeperiod of the suspension system now, we have
... /i I /~ i 2, Mi m \j ( ret \ f ) I >^ . <~VA
And, thus, T l j ~ ^ C+  d , J C  1 4 crfj
Or,
2MmG(rdr*) T,* T.r.
Or
Ur '
whence the value of G can be easily calculated.
The mean of Heyl's results (for the different small masses, mentioned
above) gave the value of G to be 6*670 x 10~ a C.G S. units.
Birge estimated the probable error in this result to be '005. So that, the
best value of G obtained so far is 6 67 005) x 10* dynes cm 2 . gm~*.
86. Density of the Earth. We know that the weight of (or, the
force actiiTDfbh) a body of mass m t on the surface of the earth, is equal
to mg, where g is the acceleration due to gravity at the place.
Also, if M be the mass of the earth and R, its radius, the force
acting on the mass is, by Newton's Law of Gravitation, equal to
M t
n . i ivi.ni /^ /\ ^^^ /^
So that, wg = ^ 2 &. Or, g = ^ 2 .O.
Taking the Earth to be a homogeneous sphere, its volume F =
u
Therefore, if A be its density, its mass M = 4.7r/? 3 .A/3.
47r/? 3 A ., n 4
Hence g = *G. Or, g = .itR.&.G
6 K o
whence, the value of A mav bo easily obtained.
This gives the value of A t be 55270 gms./c.c., takings the
value ofCto be 6'b'576x 10~ 8 C.G.S. units, (the value obtained by
Boys). And, with Poyntinij's and Heyl's values of G, the values of
A come respactively to 5 4934^/;iy./c.c. and (55150'004)^m5./c.c.
The most probable value of A is however, taken to be 5'5247
gms.jc.c. ; and, since the density of the upper layers of the earth is
found to be only 2*7 gms./c.c., it follows that the density of its inner
layers must be very much greater than 552 gms./c.c.
It is interesting to observe how Newton intuitively made a lucky
guess at the probable density of the earth, placing it so aptly between
246 PKOPEETIES Of MATTES
5 and 6, a truly inspired 'Newtonian' guess, which stands so amply
justified today ! As to the reasoning that led him to it, we can do
no better than to listen reverently to as he puts it himself in his cele
brated Principia :
'But that our globe of earth is of greater density than it would be if the
whole consisted of water only, 1 thus make out. If the whole consisted of water
only, whatever was of less density than water, because of its less specific gravity,
would emerge and float above. And upon this account, if a globe of terrestrial
matter, covered on all sides with water, was less dense than water, it would
emerge somewhere ; and the subsiding water falling back would be gathered to
the opposite side. And such is the condition of our earth, which, in great
measure, is covered with seas. The earth, if it were not for its greater density,
would emerge from the seas, and, according to its degree of levity would be
raised more or less above their surface, the water and the seas flowing backwards
to the opposite side. By the same argument, the spots of the sun which float upon
the lurid matter thereof, are lighter than that matter. And however the planets
have been formed, while they were yet fluid masses all the heavier matter subsided
to the centre. Since, therefore, the common matter of our earth on the surface
thereof is about twice as heavy as watei, and a little lower, in mines, is found to
be three or four or even five times more heavy, it is probable that the quantity of
the whole matter of the earth may be five or six times greater than if it consisted
all of water, especially since I have before showed that the earth is about four
times more dense than Jupiter.'
The attempts made by different workers to determine the values
of G and A are tabulated below in chronological order, for the con
venience of the student.
Year Name of Types of Value of G Value of A
Experimenter Experiment
1775 Maskelyne Mountain method 7*4 x 10~* dynes crn z .gm~* 5'0 gms jc.c.
1898 Cavendish Torsion Balance 6'754xlO 5'448
1854 Airy Mine method 5'7xlO 6'5
1881 Von Jolly Sensitive Common
Balance 6'465xlO~ B 5'692
1891 Poynting 6'6984xl()' 5493
1895 Boys Torsion Balance 6'6576xlO 8 55270,,
1896 Eotvos 6*66xlO 8 5'53
1901 Burgess 664xlO~ 8 5'55
1930 Heyl Oscillation method 6'670x 10~ 8 5'517
87. Qualities of Gravitation. We shall now proceed to see
whether the gravitational attraction between t\u> Icdii s is in any
way affected by the nature of the intervening medium between them,
by the nature of the masses themselves or by physical conditions, like
temperature etc.
1. Permeability. From the similarity of the formula (F =
for gravitational attraction between two bodies, with these for magnetic and
electrostatic attraction, it might appear that, like the constants M and A there,
the value of G might also depend upon the nature of the intervening medium
between them.
That this is net so has been clearly shown by Austin and Thwig, who
performed a direct experiment with a modified form of Boys* apparatus, in which
they placed slabs of different materials in between the two attracting masses and
could detect no change whatever in the value of G, within the limits of theii
experiment.
This stands further confirmed by (/) the fact that whereas in the expcri
ments for the determination of G, discussed above, air is the intervening medium
in the case of planets, the intervening medium is just free spoce, and yet thi
astronomical predictions, deduced on the baiii of the same law, come out so sur
GRAVITATION 247
prisingly true, showing clearly that the value of G cannot possibly be very diffe
rent in the two cases ; and (/*) the very close agreement between the values of G,
obtained by different pendulum experiments, with their bases of different mate
rials, so that different materials lie between the pendulum and the A earth. We,
therefore, conclude that little or no effect is produced in the gravitational attraction
between the masses by the nature of the medium interposed inbetween them.
2. Selectivity. The law simply states that the force of attraction between
two masses depends only upon their magnitude, having nothing to do with their
nature, or their chemical combination, etc. This is amply borne out by the large
volume of experimental evidence in its favour. For, it has been shown by Eotvos
and others, by their experiments with Boys" apparatus, using a laige variety of
materials as the attracting masses that the values of G obtained in the different
cases all agree admirably, even in the case of radio active substances, thus showing
clearly that gravitational attraction is by no means a selective phenomenon.
3. Directivity. We know that in the case of amsotropic* substances, their
physical properties, like refractive index, conductivity for heat and electricity etc.
depend upon their orientation, i.e., upon the direction of their crystallographic
axes. Or. Mackenzie and a host of otherworkers, therefore, tried to investigate
as to whether it had any effect upon the gravitational attraction also, and they all
obtained negative results. An additional obvious proof of the independence of the
value of G of the orientation of crystals is the fact that their weight (which is just
another name for the force of gravitational attraction between them and the earth)
is exactly the same whatever the orientation, showing that the phenomenon of gravi
tational attraction is far from directive.
Poynting and Gray confirmed this fact in an ingenious experiment, in
which two quartz crystals or spheres were suspended close to each other, one
enclosed in a case, whereas the other, outside it, being free to rotate. If there
were even a trace of a directive influence in gravitational attraction, the rota
tion of one crystal would sjt up forced vibrations in the other ; so that, if their
timeperiods agreed, the enclosed crystal or sphere would be set into appreciably
large resonant or sympathetic oscillations with the outer one. Nothing of the
kind, however, was found to occur.
4. Temperature. Poynting and Phillips, together with a whole lot of
other workers, tried to investigate the effect of temperature on the value of G,
and, once again, the results obtained were absolutely negative. Only Shaw,
experimenting with a BoysCavendish type of Torsion Balance, observed that the
value of G, increased slightly with the temperature of the attracting bodiesf, the
value of the coefficient of increase (a) being negligibly small, being only about
l6xlO* between (TCand 250C.
All the abovo mass of evidence thus goes to suggest that gravi
tational attraction is purely a function of the masses of the attracting
bodies and of t lie distance between them, being quite independent of all
other factors.
No wonder, then, that Newton's theory of gravitation held such
an unquestioned sway over the minds of scientists all over the world.
And, for the non scientific people in general, it had equally spectacular
predictions which, when found true, could not but impress them
deeply as to its unerring truth.
Thus, for example, Adams, in 1845, predicted, from his calcula
tions, based on the disturbance of the orbit of Uranus, the presence
of a hitherto unknown planet. Unfortunately, Airy, the then Astro
nomer Royal, did not as much as care to look for this planet, per
haps from sheer scepticism ; and Challis, the Director of the Cam
bridge Observatory, though he actually saw a new star, looking like
a, disc did not care to verify whether it was the one predicted by
*Aniso tropic substances are those whose properties, are different in different
directions,*.?., crystals, in general.
t According to Shaw, G  G. (l+o/), where G and G 9 stand for the values
of the Gravitational Constant at tC and O'C respectively.
PROPERTIES OF MATTEB
Adams. A year later, Leverrier, a French mathematician, made
similar calculations to those of Adams and communicated his results
to the Berlin Observatory and, lo and behold, the planet we call
Neptune, was there for all to see at the very spot predicted !
And, once again, in 1930, the disturbance of the orbit of Nep
tune itself led to the discovery, by American astronomers, of the
planet Pluto.
These two profound discoveries put Newton's theory beyond
the pale of any doubt or scepticism and it came to be looked upon as
infallible. Indeed, it continued to enjoy its 'infallible' status until
the arrival on the scene of that genius of modern times, Albert
Einstein, who showed it to be no better than a close approximation
to the actual law of gravitation propounded by him, as we shall see
in the next article.
88. Law of Gravitation and the theory of Relativity. Although, as
we have seen above, the Newtonian Law of Gravitation is found to be
valid over a wide range and is supported by a large mass of experi
mental evidence, there are certain small divergences, not quite in
conformity with it. But such, indeed, has been the general faith in
the infallibility of the law that any divergences from it were ascribed
to some hitherto undiscovered disturbing influences rather than to
any possible discrepancy or flaw in the law itself. It was only after
Einstein put forward his "Theory of Relativity" that it camo to be
realised that Newton's Law was only an approximation although
an extremely close one to the true or thfc fundamental Law of
Gravitation.
A detailed discussion of this theory is beyond our present scope,
and we shall, therefore, consider here only one or two salient points
of it, to show how Newt6n's formula lays itself opon to criticism in
the light of these :
(i) One consequence of the theory, fully confirmed experiment
ally, is what may be called the 'inertia of energy 1 , viz., that wherever
a change in the energy of a body is brought about, a corresponding
change takes place in its mass. In other words, energy and mass are
mutually convertible, one into the other, the relation between the two
being the following :
Change in mass (in grams) = Change of energy (in ergs)jc*,
where c is 'the velocity (in cms. [sec.) of light in vacuo, (equal to
3 X 1C 10 cms. /sec.). Further, according to this theory, we have
m = m Q l^i~Ti*Jc*,
where m is the mass of a body, moving with a velocity w, (called its
moving mass, and m , its mass when at rest, (called its rest mass).
Thus, the mass of a body is different when in motion from that
when at rest, i.e., it changes with its velocity, and Newton has not
specified which one of these is to be used in his formula for gravita
tional attraction, a noticeable omission.
(ii) Another consequence of the theory is that the numerical
value of the distance between two points varies according to the
system of spaoetimo coordinatea chosen, BO that the distance bet
GBAVITATIOB 249
Ween them changes with the circumstances of the observer, making
the measurement.
The effect of these two discrepancies is, of course, only slight,
but it is there, nevertheless. Einstein took both these factors into
account and was thus led to the formulation of his famous Theory,
which explains satisfactorily the deviations from the Newtonian law,
among which may be mentioned the following :
(a) The precessional motion of the perihelion of the orbit of the
planet Mercury, with a period 0/ 3 x 10 6 years.
On the Newtonian theory of gravitation, it could be explained
only as being due to the influence of another planet, for which the
name Vulcan was chosen but which has never been located.
(b) The deflection of a ray of light in a gravitational field A ray
of light, due to its great velocity, behaves like a material body,
possessing mass and momentum, etc,, [see (/) above], and it must,
therefore, suffer deflection in a gravitational field. Thus, a ray of
light from a star would get deflected near the edge of the Sun, due to
its high gravitational field, resulting in an apparent shift in the
position of the star. Calculated on the basis of both these laws, tho
value of this shift, by Einstein's theory, comos out to be twice that by
Newton's law, and actual expoximental observation* fully supports
the former result.
(c) The shift in the spectral lines in the solar spectrum. Due to
the Sun's gravitational field, the spectral lines in the solar spectrum
must have different positions, and, therefore, different ^\ a ve lengths,
from those they would have, when emitted by some terrestrial
source. This has been fully verified in the year 1924, the shift
being only very slight, about onehundredth of an Angstrom Unit
(or = f^th of 10~ 8 cm., i.e., = 10 10 cw.)
This has been further confirmed by Dr. Adams, who measured
a shift of as much as half an Angstrom Unit in the spectrum of the
dwarf companions of Sirius, due to the greater gravitational field at
the surface of these stars than at that of the Sun.
It is thus clear that whereas the correct Law of Gravitation is
that due to Einstein, the Newtonian Law is a sufficiently close ap
proximation to it for our ordinary experimental purposes, except in
a few rare cases, here and there.
89. Gravitational FieldIntensity of the Field The area
round about a body, whithin which its gravitational force of attraction
*This was made on the Island of Principe (on the African coast) aad at
Sobral, in Brazil, on May 29th, 1919, during a solar eclipse, (the stars being not
visible otherwise). Two wellequipped expeditions at these two places obtained
photographs of the portions of the sky, near the Sun, just before the eclipse and
again after the eclipse, when the Sun had shifted away from its earlier position.
The stars were found to have been displaced, with respect to the Sun, the respec
tive values of this displacement at the edge of the Sun obtained by the two
expeditions being 1*61" and 1*98*. Their mean, (1*795"), agreed admirably with
the value predicted by Einstein.
A similar observation was made again at another solar eclipse, three
years later, in 1922, this time at Wallal in West Australia, when the shifts of
as many as eighty stars were observed The mean shift was found to be 1*74'
which wai only about *01' short of the calculated value.
PROPERTIES OF MATTER
is perceptible, (no other body being near about it), is called its
gravitational field.
The intensity or strength of a gravitational field at a point is
defined as the force experienced by a unit mass, placed at that point in
the field. I^may also be defined as the rate of change of gravitational
potential or the potential gradient , at the point, (see 91). Thus, if /
be the intensity of a gravitational field at a point, we have
/ dV >
1 ~ ~ dx
where dV is a small change of potential for a small distance dx.
N B. The strength or intensity of a field at a point is often spoken of
merely as the field zl that point,
90. Gravitational Potential Potential Energy. Consider a body A t
with its gravitational field around it. It will naturally attract any
other body B, placed at any point in its field, in accordance with the
Law of Gravitation, and this force of attraction will decrease with the
increase in the distance of B from A, so that at an infinite distance
from A it will be zero. But, as B is moved away from A, work has
to be done against this force of mutual attraction and, therefore, the
potential energy of B increases, its value depending upon the masses
of A and B arid their distance apart. The work done in moving a
unit mass from infinity to any point in the gravitational field of body A
is called the gravitational potential of that point due to the body A,
and is an important gravitational property of that point. It is
usually denoted by the letter K. Obviously, it will also be the poten
tial energy of the unit mass at that point, with its sign reversed, (for
whereas the potential decreases^ the energy increases, with the in
creases in the distance from A). If we, therefore, replace the unit
mass by the body B, the potential energy of B will, clearly, be equal
to the product of its mass and the gravitational potential (with its
sign reversed) at that point. Thus, the potential energy of a body
at a point in a gravitational field is equal to the product of the mass
of the body and the gravitational potential (with the sign reversed) at
that point.
91. Gravitational Potential at a Point distant r from a Body of
mass m. Let a mass m be situated at 0, (Fig. 154), and let a unit
mass be situated at P. Then,
^ ....... x ...... ^ ^ the force of attraction on the
Q^ ____ / r ____ ,_^ ' unit mass due to m is clearly
* ......... ^ ..... ~~ ,. mxl ~ m
equal to , G = , (7,
Fig. 154. d ** * 2
where x is the distance of P from O, the force being directed towards
O. Therefore, work done when the unit mass m>ves through a
small distance dx, towards O, is equal to m.G.dx/x*.
And, therefore, work done when it moves from B to A
[Am. [A 1
= f.G.dx = G.m fdx.
GRAVITATION
251
wtere t arid r l are the distances of A and B from 0. This, obviously,
is the potential difference between the points A and B.
If B be at infinity, i.e., if r l = oo, we have
potential difference between A and oo = G.m f  _ ^ = m .G.
But potential difference between A and oo is equal to the poten
tial at A, because the gravitational force at oo, duo to m, is equal to
zero and, therefore, the work done in moving a mass about, at oo, is
also zero. In other words, the potential at infinity is zero. Therefore,
the gravitational potential at A duo to the mass m is equal to
G.m/r.
Or, denoting the gravitational potential at a point, distant r from
a body of mass m by K, we have F  <7.
It will be noticed that whereas the value of gravitational poten
tial at an infinite distance from a mass is zero, it goes on decreasing
as we approach that attracting mass, i.e., it is an essentially negative
quantity*, its maximum value being zero at infinity, where, at all
points, therefore, the potential will be the same.
92. Velocity of Escape. We know that, ordinarily, a body say, a rifle
bullet, projected upwards comes down to the earth due to the gravitational pull
of the earth on it. Let us see it it is possible to project it with a velocity such
that it will never come back. Obviously, it will be so if it can be given a velo
city that will take it beyond the gravitational field of the earth This velocity
of the bullet is called the velocity of escape.
Thus, if m be the mass of the bullet and Af, that of the earth, the force
acting on trie bullet at a distance x from the centre of the earth is clearly
= m.M.G./x 2 ,
.*. work done by the bullet against the gravitational field, when the body moves
,. , . rn.M.G ,
a distance dx upwards == . dx.
* x*
,'. total work done by the bullet escaping
r
J/
m M.G fa = m M G ("where R = radius
" x R [_of the earth.
This, as will readily be seen, is m times (i.e., the mass of the bullet times) the
gravitational potential on the surface of the earth.
If v be the initial velocity of the bullet , (i.e., its velocity of escape), its
initial kinetic energy (t c , at the time it starts) will be ^wv 2 . And this must,
therefore, be equal to the work done by the bullet during escape. So that,
9 or, v*
which, on substituting the value of A/ f G and R t woiks out to
v= \\'l9xW cms. /sec.
Thus, velocity of escape 11 1 9 x 1 0* cms. I we.
93. Equipotential Surface. A surface, at points of which
the gravitational potential is the same, ia called an equipotential
surface.
Thus, if we imagine a hollow sphere, of radius r, with a particle
of mass m at its centre, the potential at each point on it will be the
*Thc negative sign is often omitted in writing, but it must always bt
understood to be there.
252
PROPERTIES OF MATTBB
same, vfe., G,w/r. The surface of the sphere is thus an equipoten
tial surface.
Now, since the difference of potential between any two points
on an equipotential surface is zero, no work is done against the
gravitational force in moving a unit (or any other) mass along it.
In other words, in moving a mass along an equipotential surface, we
must be moving it in a direction perpendicular to the gravitational
field at every point on it. Or, the direction of the gravitational field
is, at every point, p8rp3ndicular to an equipotential surface, being
directed towards the nearest equipotential surface, having a poten
tial lower than it.
94. Potential at a Point Outside and Inside a Spherical Shell.
(a) At a point outside the shell. Let P be a point, distant d from the
centre O of a spherical shell of
radius a, (Fig. 155), and surface
density, (i.e., mass per unit area of
the surface), p.
Join OP and cut out a slice
CEFD, in the form of a ring, by
two planes close to each other
and perpendicular to the radius
OA, meeting the shell in C and Z),
and in E and F respectively. Let /_EOP = 6, and let the small
^COE = do.
Clearly, the radius of the ring is EK = OE sin = o sin ; BO
that, its circumference = 2na. sin Q, and its width = CE = a.dQ.
.. area of the ring or slice = its circumference X its width.
= 2ira. sin Qxa.dQ
And .. its mass == 2xa. sin OXa.ddXP = 2x0*. sin O.dO.p.
If EP = r, every point of the slice is at a distance r from /*',
and, therefore, the potential at Pdue to this small slice is given by
_ mass of slice Q = _^*^?P >Cj . . (/) [See 91.
r r
Now, in the triangle OEP,
EP 2 = OE* + OP*20E.OP.cos 0. [Appendix I, 7 (2)
/^ , n j /i i~V OE= a, the radius
Or, r 8 = a*+d*2 a.d. cos 0. of lhe she , K
Differentiating the above expression, we have
2r.dr = Q+Q+2a.d.sin d.de = 2a,d.sin
Hence
r =
a and d being
constants.
2a.d.sin 0.d6 __ a.d.sin B.d6
2.dr ~~ dr
Substituting this value of r in expression (/) above, we have
27ta 2 .sin B.dQ^ G , __ 2?r a.p.G dr
Integrating this between the limits, r = AP = (d a), and
r=s DP =s (d+a), we get V, the potential due to the whole shell at the
point P.
GRAVITATION
Thus,
Tf ,.. I * _ * * _ , T. _ * _ I
](da) d d ](da)
</r.
Now, 47T0 2 is the surface area of the whole shell, and, therefore,
4ir.a 2 p is equal to its mass M.
We thus, have
.
a
Or, the potential at the point P due to the whole shell is equal to
M.Gjd, i.e., the same as it would be due to a mass M at O.
The mass of the whole shell thus behaves as though it were con
centrated at its centre.
In the above case, if we imagine the point P to be at A, i.e., on
the circumference of the shell itself, we get the potential there by in
tegrating the expression for dV ', between the limits r = and r = 20.
So that, in this case,
f* 27r  a P G
Jo "~d
Or,
Hence
' d
M.O
. 
d
\_ T J
M.O
a
. ,
[ . here, d a.
(b) At a point inside the shell. Imagine now the point P to lie
inside the spherical shell, (Fig. 156).
Proceeding as above, we have
potential at P due to the slice, or ring
CEFDJ.e.,
In this case, the limits of r are (ad)
and (af d). So that, we havo
2ir.a.p.(?
Fig. 156.
Now,
.  . .
d " a
47f.a 2 f p = A/, the mass of the shell,
dividing by a.
UJf
Hence V s= .(?,
a
i.e., the same as at a point on the shell. [See case (a) above]
Since the above value for V has been obtained for a point P,
anywhere inside the shell, it follows that the potential at all points
inside a spherical shell is the same, and is numerically equal to the value
of the potential on the surface of the shell itself.
95. Gravitational Field inside a Spherical Shell or a Hollow
Sphere. We have seen above that the gravitational potential at all
points inside a spherical shell is the same.
Now, the field at a point is given by the potential gradient (i.e.,
the rate of change of potential with distance), at that point.
Or, 7 = dl'ldx.
Since V is constant for all points inside the shell, dVjdx = 0,
the field in the interior of the shell, due to the shell, is zero ; in
other words, there is no gravitational field inside
a spherical shell.
Alternative proof. Let P be any point
inside a spherical shell, or a hollow sphere, of
surface density, (i.e , mass per unit area), p.
Through P, draw straight lines, so as to form
two small cones, with their apices at P, and
intercept? ne: small areas 8 and 5' on the shell,
as shown, (Fig. 157), on the opposite sides of
the plane XY 9 drawn perpendicular to the
diameter passing through P. Let S and S' be
at distances r and r' from P, and lot the solid
angles* at P be equal to o>, each.
i.e.
S'
Fig, 157.
Then, clearly, the area of the right section of the cone S is equal
to r z .a> and of that at 5', equal to r' 2 .o>.
If, therefore, a be the angle that the right sections of the cones
\ respectively, we have
S and S' make with S and
S cos a = r 2 .co and S' cos a = r
'2 ,
And
r'.o,
C OS a
and S' =
COS a
So that, mass of area S *='' , and mass of area S' =
cos 9. cos a
intensity at P due to S =
.
cos a r*
cas a
in the direction PS.
'
in the direction PS 1
And intensity at P due to S' = ^,.0
cos a.r f cos a
These two intensities at P, being equal and opposite, their resul
tant is zero. Similar is the case for all other pairs of cones on oppo
site sides of AT, into which the shell may be divided ; so that, the
resultant intensity or field at P due to the whole shell is zero. And,
the same is true for any other point inside the shell. In other words,
there is no gravitational field inside a spherical shell.
_______ , , _ t
QBAVITATION 255
96. Potential and Field Intensity due to a Solid Sphere at a
point, (/) Inside the Sphere and (//) Outside the Sphere.
(a) Potential at a point P, inside the solid sphere. Let the point
P lie inside a solid sphere of radius a, and of volume density er, at a
distance d from the centre O of the sphere,
(Fig. 158),
The solid sphere may be imagined
to be made up of an inner solid sphere
of radius d surrounded by a number of
hollow spheres, concentric with it, and with
their radii ranging from c/to a. The potential
at P due to the solid sphere is, therefore,
equal to the sum of the potentials at P due
to the inner solid sphere and all such
spherical shells outside it.
Clearly, P lies on the surface of the solid sphere of radius d > and
inside all the shells of radii greater than d.
.*. potential at P due to the sphere of radius d
= _mas 8 of the sphere
a
4 4 . fv mass of the sphere
=  3 ir.d*a.GId =  j .TrJ*o.G ...(/) I = *</.<,.
To determine the potential at P due to the outer shells, imagine
a shell of radius x, and thickness dx.
Clearly, its volume = area x thickness = 4irx*.dx,
and .. its mass = <lirx*.dx.G.
Now, the potential at any point within a shell is the same as at
any other point on its surface.
So that, potential at P   47f  x ? dx * G ^ ___ 47r x dx.a.G.
Integrating this for the limits, x = d and x = a, we get the
potential at P due to all the shells.
Thus, potential at P due to all the shells
= I 4ir.a.G.x.dx = 47r.a.G x.dx.
}d Jd
Now, the total potential at P duo to the whole solid sphere is
equal to the potential at P due to the inner sphere of radius d> plus
potential at P due to all the outer shells.
*This expression is equal to In a.G (cPd*) ; so that, the potential a
Fdue to all the outer shell = 2n.Q.G(a z <**), and, the potential at P due to
the whole solid sphere is, therefore, also
256
PROPERTIES Off MATTER
total potential at P
_ 4
*= 3 w
4
~~ 3 7r *'
4
 3  (
7T
2tf 3
But .7T.fl 3 .a is the mass of the sphere = M.
["Multiplying and
[^dividing by a 2 .
potential at P due to the sphere =

(/?) Potential at a point P, outside a solid sphere. Imagine the
sphere to be broken up into a number of thin spherical shells, con
centric with the sphere, and of masses
iw lt w 2> m 3 , etc., (Fig. 159). Then, as
we have seen before, the potential at P
due to each spherical shell will be equal
to its mas X G/d, where d is the distance
of the point P from the centre of the
sphere i.e , the same as though the
mass of each shell were concentrated at
its centre O. So that, the potential at
Fig. 159 P duo to the different shells will bo
~m v Gjd, m^G\d, m 3 .Gjd f and so on.
Therefore, the potential at P due to all such shells, /.., due to
the whole solid sphere
= _["
/Hi
m
A/
<T
where
h . . . . = M, the mass of whole solid sphere.
M
Hence potential at P due to the solid sphere i G.
(c) Gravitational field due to a solid sphere at a point inside the
sphere. We have seen that the potential at a point inside a solid
sphere, distant x from the centre of the sphere, is given by
y jtf.G^" [See above.]
Now, intensity, or gravitational field, at a point is equal to the
potential gradient, (or the rate of change of potential with distance),
at the point.
Therefore, intensity or gravitational field at a point distant x
*u dV d
from the centre a = 5
dx dx
Af.G.
M.G.X*
a* 
* Taking the value of V at distance x from to be 2w<j(7/ a* ^j
(see footnote, page 255), we have intensity or field at distance x from O
aRAVITATION
257
This shows that the force decreases as x decreases, being zero at
the centre of the sphere. The negative sign only indicates the attrac
tive nature of the field.
(d) Gravitational field due to a solid sphere at a point outside the
sphere. We know that the potential at a point outside a sphere,
distant x from the centre of the sphere is = G, as explained
above in 98 (6), (page 256).
And, since intensity or gravitational field at a distance x is
equal to the potential gradient at x, we have
gravitational field at distance* = . = . M.G. (x~ l )
 M.Q. X (lxx) =  M ; Or, =  *<?.
Alternatively, we may get the same result by applying, in this
case also, the assumption, found valid in tli3 case of potential outside
a sphere, viz., that the potential is the same as though the whole mass
of the sphere were concentrated at the centre of the sphere ; so that
gravitation il intensity or field at a point, distant x from the centre,
and outside the sphere = M.G/x 2 , the negative sign merely indi
cating, as before, the nature of the field.
N B. This result is of great historical importance in that it enabled
Newton to apply his law of gravitation to the motion of the moon. For, the
radius of the earth not being negligible compared with that of the moon's orbit
around it, there would have been no means, in the absence of the above result,
of determining what correction terms, if any, would be necessary in the equa
tion of the moon's motion, in view of the distribution of matter* inside the earth
and the finite value of its radius.
97. Intensity and Potential of the Gravitational Field at a Point
due to a Circular Disc. Let MN, (Fig. 160), represent a circular disc
of radius R, with its plane per
pendicular to the plane of the
paper and let be a point on
its axis, distant x from its centre
C, where the intensity and
potential due to its gravitational
field are to be determined.
Imagine the disc to consist
of an infiuito number of concen
tric rings, with C as their com
mon centre.
Consider one such ring PQ
of radius r and thickness dr. Pig. 150,
Join O to the extremity A of the radius CA of the ring. Then,
if angle AOC be equal to 6, We have
r = CA ^ x tan 0.
And, therefore, differentiating ifc with respect to 8 9 we have
j
Or, dr x.sec*6,d9.
258 PBOPEBTIE3 OF MATTBB
And, AO = x.sec 0.
Now, area of the ring = circumferences thickness =*2irr,dr t
And, therefore, mass of the ring = 2irr.dr.9,
where p is the mass per unit area of the disc.
Considering a very small element A of the ring, we have inten
sity at O due to this element = mass of the element x G/AO 2 , along
the direction OA.
This can be resolved into two rectangular components, (/) along
OC and (//) at right angles to it. So that,
,, r ^ mass of the element . . *
the former component = AQ* & cos Q> along OC,
, tl , AA , mass of tie element n * * n
and, the latter component =  AQ* ~ ~~~~ ' sm ' v e rtica "y
upwards.
Similarly, for an equal element at 5, diametrically opposite to A, we
have
intensity at O, resolved into the same two components, v/z.,
,., mass of the element , ,
(i)  J ^p  .G cos $, along OC ; and
.... mass of the element . .. , ,
(j/) Am ' sm 0> v ertica "y downwards, as shown.
A\J
Thus, the two vertical 'components, being equal and opposite,
cancel out, and the components along OC alone being effective, are
added up, both acting in the same direction OC ; the same will be the
case with other elements into which the ring may be supposed to be
broken up ; and, therefore, the intensity at O due to the whole ring
is equal to the sum of the components along OC due to th different
elements, i.e.,
mass of the ring

Or, intensity at O due to the ring= yUg 'Gcos 0.
Or, substituting the values of r, dr, and AO, we have
^ ^ j x ^  x tan 6.x sec 2 tf.dQ _
intensity at O due to the rmg=  2 2 cosS.O.
x . sec u
= 2irp.G.sin 8.d6 t along OC.
Therefore, intensity at O due to all the rings into which the
disc is supposed to be divided up, i.e., intensity at O due to the
whole disCj is obtained by integrating the above expression for the
intensity due to the ring, between the limits, r = and r=/?, or 6 =
and = a, where a is the angle between OC and the line joining
with the extremity M of the radius CM of the disc.
Thus, intensity at O due to the disc is given by the expression,
I torf.Qsin Q.dQ = 2ir.p.<?  sin
Jo Jo
OBAVITATIOB
259
o 27r.p.G cos  = 27r.f>.G. cos a (cos 0) J
Or, intensity at 0, </we f
= 27r.p.0.( cos a+1)
disc = 27r.p.G.(l coy a)
(0
A i
And, since cos a =
CO
~ MQ
,
9 we have
ity at 0, due to the disc = 27rp.#/l ~/^Tj?a )'
intensity
Or, again, because 2?r (1 cos a) is the solid angle, co, say, (see
on solid angle below), subtended by the disc at the point O, we have,
from relation (/'), above,
intensity at O, due to the disc = p.G.co.
(KI)
Now, potential at P due to the ring of radius r is equal to the
intensity at P due to the ringxx = ZitQ.Gsin O.dQxx, because field
at P is equal to the potential gradient at P.
Hence, potential at P due to the whole disc is
= P* 27T.P.G.X sin Q.dB = 27T.p.G.x sind.dQ,
JO JO
Or, potential at P due to the disc
...... (iv)
...... (v)
Or, =
Or, =
to give it its proper negative sign.
Note on Solid Angle. Suppose, we have an area PS, (Fig. 161), as the
base of a cone, with its apex at point O. Then, if we draw a sphere, with centre
O and any radius R, so that a sur
face of area pq of it is cut off by the
cone, then pq is proportional to r*,
where r is the radius of the spheri
cal surface pq ; and, therefore, area
pqlr* is constant for any given cone.
This quantity, area pqjr*, is called
the solid angle of the cone, or the
solid angle subtended by the area pq
at O, and is usually denoted by the
letter co. Obviously, it is also equal
to area PS/OP*, or the solid angle
subtended by area PS at O, and its numerical value is equal to area pq, if
r 1 cm.
Now, suppose the given area be PQ and not PS. Then, if AN be the
normal to it at its centre A, we have area PS  area PQ cos a, where a is the
angle between AO and the normal AN to the surface PQ at A.
Thu,.
260 PROPERTIES OF MATTER
Now, to determine the solid angle o>, subtended by a circular shell or disc MN* t
of radius R, at a point O, distant x from its centre C, (Fig. 162), we draw a
sphere, with O as its centre and (R 2 + x 2 ) as its radius, such
*X that its circular face lies on this sphere. Then, we have,
\^ (from above),
\ surface area of slice MFN
/ Since the area* of a sphere, lying between two parallel
' planes, is equal to the area of the circumscribing cylinder
in between these planes, with its axis perpendicular to them,
we have
area MFN  2rr(RH* 2 )*xFC = 2n (tff x 1 )* X (FOCO).
X(MOx). [v FO  MO, and CO x,
Hence .  * . 2ir r I _
Or, o * 2n(l cos a),
where a is the semivertical angle, subtended by the shell or disc at O.
98, Intensity and Potential of the Gravitational Field at a Point
due to an Infinite Plane. In tho caso of the disc, above, ( 97),
if its radius R becomes i a finite, the disc becomes an infinite
plane. In this case, obviously, a becomes 7T/2, so that cos oc = 0,
and a> = 2?r.
Thus, if we put R = oo in expression (//), or cos a =0, in
expression (/), or a> = 2?r in expression (///) above, we have
intensity at O due to an infinite plane = 2n?G.
which is, clearly, quite independent of the distance x from it.
Similarly, putting these values of R, cos a and o> in relations,
(iv), (v) and (vi) above, we have
potential at due to an infinite plane =2^p.(/ f .x.
99. Inertia! and Gravitational Mass. We ordinarily define the mass
of a body by the acceleration produced in it by a known force. This is known as
its inert ial mass.
But since, as we have seen, the gravitational field due to body is propor
tional to its mass, it is also possible to define the mass of a body as proportional
to the gravitational force of attraction it exerts on a standard test body at unit
distance away from it. Thus defined, the mass of the body is called its gravita
tional mass.
Now, Galileo showed that the acceleration of a falling body was quite in
dependent of its mass and the same is found to be true in the case of pendulums
used for the determination of the value of g t showing that the gravitational
force between a given mass and the earth is proportional to the inertia of the
mass.
There appears to be no a priori reason, however, why this should be so ;
for, in the case of an inclined plane, for example, we have seen how the accelera
tion of a body loliirg cc\\n the plane depends not only on the mass of the
body but also on the distribution of its mass, ( 39, page 87). The above may
thus be regarded to be only an experimental law.
*MN 9 here, represents the sideview of the shell or disc.
OEATITATION 281
It is confidently claimed by some, but equally hotly disputed by others,
that the equality of the inertial and gravitational mass can be 'predicted' from
the general theory of relativity, so that nothing very definite can yet be said
on the point. Their eqiulity*, however, is of great consequence in astronomy.
For example, dus to the proportionality of gravitational force to inertial mass,
the orbit of a satellite round a planet / quite independent of its mass, and we
can thus 'wv/VV, the planet from a mere observation of the orbit of its satellite.
And, agiin, we can determine the mass of one component of a double star, by
observing; the orbit of ths othsr ronj thiir CD mm centre of gravity, the
required value of G being obtained from terrestrial experiments, dealt with
above.
100. Earthquakes Seismic Waves and Seismograph. An earth
quake is ciusod by a portion of the ri*id crustf of the earth giving
way or getting: fractured, soim distanco balow its surface and the
consequent sudden slipping of the resulting portion, or due to 'fault
slipping*, as it technically called. 80 to speak, it is just a landslide
on a largo scale, or a readjustmont of the earth's crust, in response
to a change of forces, or more precisely, to changes of pressure deep
in the earth's crust, down to a distance of 100 w/ev or so, brought
about by a variety of causes like erosion, deposition, tidal forces,
cantrifurral forces, etc etc. An earthquake thus represents the energy
released by this 'relative motion of portions of the earth's crust*.
The place whore the actual fracture occurs is called the focus of
the earthquake, and it not a geometrical point, but an extended
region. The point nearest to the foaus, on the surface o the earth,
is called the epicentre.
From the focus, (which we may, for our purposes here, regard
as just a point), originate a number of different types of waves,
collectively called LQN6 WAVS
seismic waves, which , P SM L \
spread on to different "t/l^^^lj
points on the surface
of the earth and which
we fed as 'earthquake Fig. 163.
tremors'. The general pattern of these seismic waves is as shown in
Fig. 16J, and thoy consist of the following different types of waves :
(a) The Primary or P Waves. The first to arrive at the
Observing Station, these are longitudinal waves, in which the particles
of the earth vibrate about their mean position, along the direction of
the waves themselves.
If the earth be regarded to be a homogeneous sphere, these
waves, starting from the focus, travel along the chord of a huge
circle of the earth, with a velocity equal to \/ jl~&, vherQ j is what
is called the 'elongational elasticity' $ of the earth and A. its density.
These waves arc also variously called as ' condensational', '/> 'rotational'
and 'push' waves and their velocity is found to be about 5 miles per
second.
*Since, as we have seen, they are proportional to each other, a proper
choice of units can make them equal.
fSee foot note on page 230.
JThe elongational elasticity j = y (1 <*)/(! f<*) (12<J), where rand a
stand for Young's modulus and Poison's ratio respectively.
262 PROPERTIES OF MATT1R
(b) The Secondary or S Waves, These are transverse waves,
in which the particles of the earth vibrate at right angles to the
direction of propagation of the waves, thus having no component
along this direction. Starting from the focus, these waves also
travel along a chord of a huge circle of the earth and are the next to
arrive at the Observing Station, with a velocity equal to \/w/A>
where n and A represent the modulus of rigidity and the density of
the earth respectively. The other names given to these waves are
'distortional', 'equivoluminal' and 'shake' waves, their velocity being
about 3 miles per second.
(c) Rayleigh Waves. Discovered by Lord Rayleigh, these waves
are found to remain confined to a comparatively thin layer in the
close vicinity of the earth's surface. Unlike the P and S waves,
they start from the epicentre and arrive at the Observing Station,
along a huge circle of the earth, the displacement of the particles
at any point on the earth's surface, due to them, being in the vertical
plane containing their direction of propagation. Resolving this
displacement, we have (/) a vertical component and (it) a horizontal
component, along the direction of propagation, there being no horizontal
component at right angles to it. These waves thus persist over long
distances along the surface of the earth, and are almost unique in
this respect. If the earth were a homogeneous sphere, these waves
also would travel with a constant velocity, but, due to its hetero
geneous character, each single wave, starting from the epicentre,
gets split up into a number of different sets of waves, each set
having a different wavelength, velocity etc. ; so that, what we
receive at Observing Station is a series of oscillations, instead of
one single 'kick' or 'throw' as would be the case if there were no such
splitting up of the original wave, i.e., if the earth were really homo
geneous in composition.
(d) Love Waves. The heterogeneity of the layers of the earth
is responsible for yet another type of surface waves, known as Love
Waves, in which the displacement of the earth is horizontal, but
transverse to the direction of their pro^apation. The velocity of
these waves is less in the earth's crust than in the matter below.
Immediately after an earthquake, oscillations, corresponding to these
waves, can be detected at almost any place on the surface of the
earth.
Unlike P and S waves, which are separately and distinctly
received and recorded at the Observing Station, these waves get
intermingled with Rayleigh waves to form a somewhat complicated
system of waves, (not yet properly understood), called long or L
waves, or the main shock, registering themselves as a long series of
oscillations.
101. Seismology. The study of the seismic waves constitutes
what is called the science of Seismology, and it owes a great deal
to Prof. John Milne, who did almost the whole of the initial pioneering
work on the subject. As early as the year 1883, when he was
residing in Japan, he predicted that * 'every large earthquake might
ORAV1TAT1UB 263
be, with proper appliances, recorded at any point on the land surface
of the globe". And, then, in the year 1889, a curious incident
confirmed his prophetic words. For, a delicate horizontal pendulum,
set up for the measurement of the gravitation action of the moon,
gave recordings, which turned out to be due to an earthquake, with
its origin somewhere in Japan. This started a new era of intensive
researches on the subject, with Prof. Milne in the very forefront ;
and, in 1895, ho set up his own observatory at Shide, in the Isle of
Wight, which became the centre of a worldwide seismic survey.
By the year 1901, the main facts as to how the tremors tra
velled through and round the earth were fully established, again, due
in main, to the labours of the eminent Professor himself.
His reports to the British Association on Earthquake Pheno
mena in Japan from 1881 to 1895, together with those on Seismolo
gical Investigation from 1895 to 1913, (the year of his death), form a
fascinating and a detailed study of the growth and development_of
the presentday science of Seismology.
102. Seismographs. A seismograph (or a seismometer), is an
instrument used to record the earth tremors or the seismic waves,
to some dynamical function of which, (like displacement, velocity,
acceleration, etc.), they respond or react. The record of the vibra
tions so obtained is called the seismogram. The instruments, res
ponding to displacement, are of the mechanical type and we are,
therefore, concerned here only with those. The following is, in
brief, the theory underlying the mechanical type of seismographs.
AH vibrations of the earth may ultimately be resolved into (/)
vertical and (//') horizontal components*. The problem thus reduces
itself to merely recording these vertical and horizontal vibrations. We
shall confine our attention here only to the measurement of the
horizontal displacements, accompanying these latter vibrations. There
are 'two types of instruments in use for the purpose, viz., (a) the
vertical pendulum and (b) the horizontal pendulum type.
(a) The Vertical Pendulum Seismographs. A vertical pendulum
is just a rigid body, suspended from a stand resting firmly on the
ground ; so that, with the horizontal displace
merit of the ground and the stand with it, the 
point of support of the pendulum also gets
displaced horizontally.
Thus, if the point of support S of the
vertical pendulum, (Fig. 164), is displaced
horizontally to 5", due to the horizontal
displacement of the ground, it can be shown
that a style or pen, attached to its lower end,
reproduces faithfully the movements of the
support, with precisely the same frequency, Fig. 164.
(though on a different scale), it being assumed that the support moves
with a definite frequency and amplitude.
*These components may be along East and West or along North and
South but will be horizontal, nevertheless. These can also be used to measure
the horizontal velocity and acceleration of the earth, or rather of the earth's
crust.
264 PROPERTIES OF MATTSK
Those vertical pendulum seismographs, however, suffer from
two defects, viz., (/) they have to be very heavy, as much as 20 tons
or more, if a good magnification of the vibrations be desired, and
(//) their period of vibration is rather small.
(b) The Horizantal Pendulum Seismographs. We are already
familiar with the horizontal pendulum, [see 78 (//), page 214]*.
Only some slight additions to it convert it into a sensitive and
a reliable seismograph. With the horizontal movement of the earth,
the supports of the pendulum, which are firmly fixed on to it,
also share its movement, thus setting its stem or 'boom' into motion,
which can then be magnified mechanically or electrically by various
devices.
The best known seismograph of its class is that due to Prince
Boris Galitzin, in which the greatest care has been taken to see that its
indications correspond exactly to the actual movements or vibrations
of the earth. We shall, therefore, discuss in some detail only this
one instrument here.
103. Galitzin's Seismograph. This seismograph measures the
horizontal velocity of the earth's crust, and consists of a horizontal
pendulum, having a boom or stom, 28 cms. lon^. carrying a cylindrical
brass bob, weighing 7 k.gms* ani having its centre at a distance of
14 cms. from the inner end of the boom. The suspension of the
pendulum is of the Zollner type, (as shown in Fig. 135, page 214),
with a very small inclination of the axes, so that the period of
oscillation of the pendulum is about 24 seconds. The whole pendulum
is built up on a rigid framework, firmly secured to the ground, and
consisting of four m3tal pillars, braced together, and arranged
rectangularly on four points on a inetal base or plate, provided with
levelling screws.
The recording of the vibrations or tremors at the Observing
Station is done eloctromagnetically, and, for this purp>se, a flat
copper coil is wrapped round a portion of the stern or boom of the
pendulum, extending beyond its cylindrical bob, and connected to a
sensitive movingcoil mirror galvanometer.
With the motion of the stem, (caused by the motion of the
ground), the coil moves in the strong magnetic field of a pair of per
manent horsashoe magnets of tungsten steel. A current, which is
proportional to the angular velocity ofihe stem, is thus induced in the
coil and produces a deflection in the galvanometer.
A beam of light, reflected from the mirror fixed on to the sus
pension of the galvanometer coil, is passed through a semicylindrical
lens and allowed to fall on a sensitized (i.e., photographic) paper,
wrapped round a rotating drum, worked by a clockwork arrange
ment and moving uniformly along its axis, with a peripheral speed
of 3 cms. per minute. Timesignals are also similarly recorded on
the paper by cutting off, by means of an accurately timed shutter,
the beam of light for two seconds at the beginning of each successive
minute. A permanent record of a series of curves, (i.e., the seismo
The student would do well to refresh his memory by going over this
article once again before proceeding further.
GRAVITATION 265
gram), is thus obtained on the sensitized paper, from which the time
of occurrence of any seismic phenomenon can be determined aoou*
rately to within one second.
In order that the horizontal displacement of the earth may be
correctly calculated from the seismogram thus obtained, it must
faithfully correspond to the movements of the earth. To achieve this,
d imping of both the pendulum and the galvanom3t3r is necessary.
Or, else, if the period of oscillation of the penduluni'agrees, ornearly
agrees, with that of tho saismic wave, resonance will occur, producing
largo deflections, which would give an utterly deceptive picture
of the actual movement of the ground. And, if the damping be
made critical, (i.e., deadbeat), the calculations become greatly
simplified.
This damping is produced by attaching to the outer end of the
boom, a horizontal brass plate, which moves in another strong mag
netic field, duo to a separate pair of horseshoe magnets, arranged
above and below it. The eddy currents, thus induced in the plate,
then produce, with proper adjustments, the desired damping effect on
the pendulum.
This seismograph has the additional advantage of great magni
fication*, as also of enabling the recording apparatus to be arranged
in a separate compartment, away from the pendulum.
N.B. It will be readily understood that for a large or severe earth
quake, less sensitive seismographs are more suitable, while, for smaller, local or
nearby earthquakes, the nure sensitive ones or the short period ones, are the
more desirable.
104. Determination of the Epicentre and the Focus.
The Epicentre. To determine the epicentre of an earthquake,
we determine what are called the epicentral distances of it from a net
work of Observing Stations or Observatories, the epicentral distance
of an earthquake from a given station bei/ig the shortest distance of its
epicentre from the station, measured along the surface of the earth, in
terms of the angle it subtends at the centre of the earth. This is done
with the help of the Tables, compiled by Zoppritz, Turner and others,
which give the relation between the epicentral distances of past
earthquakes and the interval between the first arrivals of the Primary
(P) and the Secondary (S) waves at a station, i.e., which express the
epicentral distances as functions of the corresponding timeintervals
S P.
Thus, from the seismogram of an earthquake, obtained at an
Observing Station, we can determine the timeinterval S P for it at
that station, and the Tables then give the epicentral distance of the
earthquake from it. This is done at as many stations as possible.
Circles are then drawn on a globe, with these different stations as
their respective centres and their epicentral distances as the radii.
The point of intersection of these circles then gives the most probable
position of the epicentre of the earthquake in question. Or, the same
may be obtained from the method of least squares.
*Out of a set of 8 seismographs at the Fordham University in New York
city, there are three which magnify the motion of the ground about 2,000 times.
266 FEOPEBTIBS O* MATTttft
Usually, three seismographs are used for the purpose, one res
ponding to motions of the ground along the East West direction, the
other along the NorthSouth direction, and the third, having a pen
dulum suspended by a coiled spring, to respond to the vertical dis
placements of the ground. Tha information supplied by the three,
when pieced together, enables not only the epicentre, but also the
character, of the earthquake to be determined fairly accurately.
(ii) The focus. To determine the position of the focus, imme
diately below the epicentre, we use what is known as Seebach's
method, explained below :
Let F, (Fig. 165), be the focus of an earthquake, (taken to be
a point here), a distance h vertically below the epicentre , and let
O be the position of the Observing
Station, a horizontal distance d from the
epicentre.
Then, assuming the homogeneity
of the medium inbetween the earth's sur
face and the surface of a sphere, concen
tric with it and passing through F, the
time t taken by the P waves to travel
from F to O is clearly given by
t =
* whence, (d 2 +A 2 ) = v 2 ./ 2
' .. '. .V ( where v is the velocity of P waves in the
medium and can be determined inde
Fig. 165. pendently by other methods.
Now, if the time at which the earthquake occurs be T O *, and
the time at which the first P waves arrive at O be T, we have / =
(T T O ), where T naturally varies with distance d.
Thus, relation (i) above may be put in the form,
(</ a +/l 2 ) = V^TTO)*.
Obtaining the corresponding values of d and t from a number of
different Observing Stations, we plot a graph between d and v/, which
gives a hyperbola, from which h can bo easily calculated out, and
hence the position of the focus determined.
A better method, however, is to calculate, by the method of
least squares, the most probable values of h and T O .
N.B. It may be of interest to know that the severest earthquakes have
their foci about a hundred kilometres below the earth's surface.
105. Modern Applications of Seismology. The development of
the modern science of seismology has led to its application in four
important fields, viz., (/) investigation of the nature of the interior of
the earth, (ii) prospecting for oils and minerals, (Hi) construction of
quakeproof buildings, and (iv) forecasting of the occurrence of
earthquakes.
(i) It is now almost fully established (according to Jeffreys)
that the earth consists of a dense core of a molten mass, mostly of
*Tbis, though not known to us, is certainly constant.
GBAV1TATIOS
267
iron, together with some nickel, of a density of about 12*0 gms.jc.c.,
(at the pressure existing there), surrounded by a solid outer shell or
crust, about 3,000 kilometres thick, the density of which decreases
from about 5'Qgms./c.c. at its innermost layers to about 27 gms.jc.c.
at the outermost layers, or at the surface of the earth.
The existence of the dense core is deduced from the observed
refraction of the seismic waves, as they pass through the earth, and
is further confirmed by the production and propagation of the secon
dary or shake waves (5) through the core. These waves, as we know,
are transverse in nature and, as such, can only be produced and pro
pagated in media, possessing elasticity of shape or rigidity, viz., in
solids.
(/i) Prospecting for, oil, coal and other minerals is now being
increasingly done with the help of seismic waves*, the process being
technically known as 'seismic prospecting'.
Artificial earthquakes as set up in the groundregion to be
surveyed for the purpose, by detonating an explosive, like guncotton
Fig. 166.
or gelignite, at a point O on the earth's surface, (Fig. 106), and the
time of explosion noted. The time of arrival of the first low fre
quency longitudinal waves, or the primary waves, thus produced,
is noted, with the help of seismographs, at different stations P, Q, R,
S, etc., all lying in the same plane. The distances from 0, covered
along the chords OP, OQ, OR, OS etc. of the earth are carefully
measured and the mean velocities of the waves calculated along these
different paths or chords.
If one of the paths or chords, say, OS, happens to pass through
a mineral deposit, like a salt dome, the value of the mean velocity
along this particular chord will be different from that along the other
chords. The experiment is then repeated along a direction, perpendi
cular to the first, by exploding a fresh charge of explosives. And,
if this confirms the results of the first experiment, a more elabo
rate survey determines the positions of the top and the sidesf of the
salt dome.
(Hi) It has now been found possible to erect 'quakeproof
buildings in California, Japan and other places, frequently visited by
earthquakes, at a surprisingly low additional cost of just 15%. For,
it has been shown by Prof. Suyehiro that the severest earthquakes of
*We have already studied the gravitational methods of prospecting, by
means of the Eotvos balance etc, (sees 80, on page 216).
t The locating of the sides of the dome is equally important, because some
mineral oil is almost always found to be there.
268 PROPERTIES OF MATTER
Japan can do but little damage to buildings, designed to resist a hori
zontal force, equal to onetenth of their total weight. The day is thus
not far off when damage to D lildings due to earthquakes will just
become a memory of a dreadful past.
(/v) And, finally, the prediction of the occurrence of an earth
quake, a good timo in advanos, i<* also fast coming into the realm of
practical possibility. For, it has now boen established that the region,
where an earthquake occurs, exhibits, for quite a few years before, a
'////', or a gradual rise, very much like the rubber tube of a pneumatic
tyre or a football bladder swelling up before it actually bursts.
There seems to be but little doubt that much sooner than we
can imagine at the moment, an earthquake forecast will become as
general and universal an affair as the weather forecast is today.
But even as it is, the loss in buildings etc., due to the severest
earthquakes, seldom exceeds 5%, due to their being confined to a
very small area, and, quite often, an uninhabited one. The disas
trous effects of earthquakes have thus been unduly magnified ; and,
for all we know, they may be for our own good, designed by a benign
Providence, by way of safety devicos to save us from being blown up,
all in a heap.
SOLVED EXAMPLES
1, Given G =* 6*7xlO~ 8 c.g s. units, the radius of the earth = 64xlO
cms. and its mean density, 55 gms./c c., calculate the acceleration due to gravity
at the earth's surface.
Imagining the earth to be a perfect sphere, we have
volume of the earth = ..n(6 > 4x 10V c cs.,
And .*. mass = *.TM6*4xl0 8 )x5'5 gms.
Consider a mass m gms. on the surface of ths earth. Obviously, the
force with which it is bsing attracted by ths earth towards its centre is, according
to the Law of Gravitation,
^^6^10)2il5 <y = *m.n.(64xl0 8 )x55x6*7xlO 1 .
(6'4xl0 8 )" * '
^rn.it x 6*4 x 5 '5 x 6' 1 dynes.
/. this must be equal to the weight of ths mass, i.e., mg.
Thus, nig = *.7r.mx 6*4x5*5x6*7.
whence, g ~ .*.mx6'4x5'5x6*7 = 988'3 cm Iscc*.
Or, the acceleration dus to gravity at the earth's surface is 988*3 cms. /sec*.
2. Two lead spheres of 20 cms. and 2 cms. diameter respectively are placed
with their centres 100 cms. apart. Calculate the force of attraction between the
spheres, given the radius of the earth as 6*67 x I0 8 cms. and its mean density as
533gms,/c.c. ; (Sp. gr. of lead = 11*5).
If the lead spheres be replaced by brass spheres of the same radii, would the
force of attraction be the same ?
. . . . product of the masses ~
Clearly, force of attraction between the masses  ^/ 5 / fl ^ e \t ~ "
 *.*.(10)*x* n.(l)B X (ll*5) 3 .G/100 8 16n'xlO'x(ll'5) a .G/9xlO<.
 16"*x(ll5>*xG/90 8n a x(ll'5) 2 .G/45.
Now, force on a mass of one gram on the eajth's surface
a&AVTTATtOH 269
Or, G.M/R* 980, taking g  980 cm*. /we*.
Or, G~~X 533 980. Or, G.4.7c.*x5'33  960x3.
Or, G9$Ox3/4.nJ?x5'33.
force of attraction between the lead spheres is given by
980x3
* " 45 x 4.n. J Rx5'33*
Since the force of attraction between the spheres depends upon their
masses, it will naturally be different in 'he case of brass spheres whose mass wili
be much less than that of the lead spheres, (the density of brass being much less
than that of lead).
3. Calculate the mass of the earth from the following data :
Radius of the earth 6x10* cms. ; Acceleration due to gravity = 980 cm./sec*.
and Gravitational Constant =6 6x 10~ 8 cm.* gm.~ 1 .sec.~ JI
We know that the force with which the earth attracts a unit mass towards
itself, (/'*., towards its centre) = 1 xg 980 dynes.
Also, the force of attraction between the mass and the earth is given by
AfxlxG/K*. where M is the mass of the earth, R, it * radius and G, the Gravita
tional Constant. Clearly, therefore, M.GJR*^ g.
Or, M~g.R*IG. ... ... (/)
.*. substituting the values of g, R and G in relation (/), we have
A/~ 980x(6xl0 8 ) 2 /66xlO  53'47xl0 26 gms.
Or, the mass of the earth is equal to 53*47 x 10 26 gms.
4. Calculate the mass of the Sun, given that the distance between the Sun,
and the Earth is 1*49 x 10 13 cms., and G * 6 66x 10~ 8 c.g.s. units. Take the year to
consist of 365 days. (Punjab, 1942}
Let the mass of the Sun = M gms. and that of the Earth = m gms.
Distance between the two, or the radius of Earth's orbit round the Sun, i.e.,
r = 149 x 10 18 cms.
Time of ons revolution of the Earth round the Sun = 365 days.
* 365 x 24 x 60 x 60 sees.
Clearly, fone of attraction between the Sun and the Earth
 G.M.mlr* = xfrWx 10 dynes.
Now, centripetal force acting on the Earth in its orbit =mv a /r.
And, the distance covered by the Earth in 1 revolution, i.e., in 365 days,
clearly=2nr = 2x l'49x 10 18 cms.
.. , ,u i 2TTX149X10 1 '
distance covered by it in 1 sec., or, v
Hence, centripetal force on the Earth = mv*/r.
/2nxl49xlO l8 \*
= "\365 x 24 x 3600 /
This must, clearly, be equal to the force of attraction between the Sun and
the Earth ; and, therefore,
wx MX 6*66 xlQ' 4**xl 49x1 0"
m '
_
J49X10 18 )* ~~ '(365x24x3600) 1 *
.  _
' (365x24 X3600) 8 666xlQ " (365x24 x 3600; f x6'66xlO"**
Or, M 19'72xlO M gms.
Or, the mats of the Sun  19*72 x 10 M gms.
270 PttOWBBTiBS 0* MATTBB
Assuming that a sphere of mass 40 kilograms is attracted by a second
sphere of mass 80 kilograms, when their centres are 30 cms. apart with a force
equal to the weight of J mg. ; calculate the Constant of Gravitation.
Here, force of attraction between the two masses
40x1000x80 x 1000 _ 32 x 10 8 32 x 10*
30* G * 9xTo r * G  9  G '
But this is equal to J mg. wt. i.e.,
4 l "4 X981
r * G * 4600 9 x32x 10* = * = 6 ' 898x
Or, the value of G ~ 6*898 x 10~ 8 C.G.S. units.
6. Two small balls of mass m each, are suspended side by side by two
equal threads of length /. If the distance between the upper ends of the threads
be a, find through what angle the threads are pulled out of the vertical by the attrac
tion of the balls.
Let the upper ends of the threads be at A and B, (Fig. 167), such that the
distance AB a. D "
Due to mutual attraction, the balls are drawn
B towards each other, say, through a distance x each,
~ 71 from their original positions.
Considering the forces acting on the ball 2,
which keep it in equilibrium, we have
jy (/) the weight of ball = mg, acting vertically
N
downwards,
(fi) the force of attraction,
m x m
* """" f \* V ~" f v m
(ax) 1 (axY
j F ' ** (111) the tension of the thread T.
* Since the ball is in equilibrium, the three
m 9 forces can be represented by the three sides of a
Fig. 167. triangle, taken in order.
If the dotted lines show the positions of the threads when the balls are
in equilibrium, the three forces can be represented in magnitude as well as direc
tion by the sides of the triangle BQN y the side BQ, representing the weight mg,
the side QN, representing F and the side NB, representing the tension T of the
string, i/i cylic order.
So that, clearly,  gj  tan 6. Or, ( _  tan 6.
whence, tan 9   Or, 9  /*/
Thus, the threads will be pulled out of the vertical through an angle
$ = tan 1 mGI(ax)*s.
7. The radius of the Moon's orbit, r, is 240,000 miles, and the period of
revolution is 27 days ; the diameter of the Earth is 8,000 miles and the value of
gravity on its surface is 32 ft./sec*. Verify the statement that the gravitational
force varies inversely as the square of the distance.
Here, distance covered by the Moon in 27 days 2rr x 240000 miles.
f . ^ 2* x 240000x1 760x3 A .
.. velocity of the Moon, v = 27x24x60x60 ~ f*l* ee '
Now, centripetal acceleration of a body moving in a circle = v'/r. Hence
the centripetal acceleration of the Moon towards the centre of the Earth, is giveo
by
i
* m "" [ 27x24x60x60 J 240000 x 1760x3*
AVITATIO 271
4frx 24x1 76x3x10^ 4frj<I76x3x 10.
"" ~(27x24x36)xlO ~T27x36)x24
 0009189 ft.lsec*.
Then, denoting the acceleration due to gravity on the'surface of the Earth
by #, (= 32 //. per sec*.), and supposing it to be inversely proportional to the n ih
power of the distance, we have
where R g is the radius of the earth, and K m9 the distance of the Moon from the
Earth.
Or C'009189 ^ / 4000 x760_x 3 V 1 / 1 V
32 * V2406o6xl760xV " V 60 / '
Or, taking logarithms, we have
4 4582  n(22218). Or, n
2' 221 o
Thus, g varies inversely as the second power of the distance and hence the
gravitational force varies inversely as the square of the distance.
8. The radius of the earth is 6'37 x 10~ 8 cms., its mean density, 5*5
gms./c.c. and the gravitational constant, 6*66 xlO~ 8 c g s. units. Calculate the
earth's surface potential.
We know that potential, V GM/x. [Taking the earth to be a perfect sphere.
Now, mass of the earth, M volume x density = *.n(6'37x 10 8 ) 3 x 5'5,
distance, x r, in this case, = 6 37 x 10 8 cms.
G = 666 x 10~ 8 C.G.S. units (given).
6 66 x JO 8 x 4^(6 37 x 10 8 )x55 6'66x 10 8 x 471(6 37) 8 x5'5
3x637xl0 8 " ** 3^
2'22x 10 8 x4rr(6 37)'x5'5 = 62'27x 10" ergs Igm.
9. Calculate the intensity at a point due to an infinitely long straight wire
of line density p.
Let AC be a portion of the wire,
(Fig. 168), of line density p, and consider
an element AB of the wire, of length dl.
Let AO be a length / of the wire ; and let
P be a point at a distance x from O.
Join PA, and let IAPO = 0.
Then, clearly, tan $ = IJx. Or, / == x.tan 0.
Differentiating it with respect to 0,
we have
dl  x.sec*Q.dQ. Fig. 168.
Therefore, mass of this element AB = ?.<#= x. sec'Q.dQ.?.
And, .'. intensity at P due to the element  x ' sec *j 9 '?.G, in the direction PA.
/. intensity at Fdue to the element
x.sec*e.de.? G _ P.C.* along
x.jec 8 x
Resolving it into two rectangular components, along PO and perpendicular
to it, we have (/; the component along PO, equal to p.G.</0.co$ 0/x, and (ii) the
component at right angles to it, equal to p.G.dQ.sin Qjx.
Similarly, the intensity at P due to an equal element dl and the other end,
at a distance / from O will be p G.dQIx, in the direction PC. Resolving it into
, two rectangular components, the component p.G.dQ.sin 0/x, being equal and
opposite to that due to the element AB will cancel out and the component
272 FHOPKBTilS OF MAtflfifl
ot $lx will act along PO, as before, and the tW6 will, therefore, be added
up. The same is true for any two similarly situated equal elements of the wirt.
Therefore, considering the whole wire, we have
intensity at P due to the whole wire = 2 I p> ' &  cos ** . sin
J x L x JO
Or, intensity at a distance x due to an infinitely long straight wire is 2p.G/x.
EXERCISES VII
1. Mention different methods for determining the Constant of universal
gravitation, and describe one which you consider to be the most accurate.
(Punjab, 1940 and 1944)
2. What is meant by 'gravitation constant' ? What are its dimensions ?
Give an account of the experiments of Cavendish and Boys to determine this
constant. (Banaras, 1945}
3 If G = 666 xlO~ 8 c.g s. units, what is the force between two small
spheres weighing 2 k.gms, placed 30 cms. apart. Ans. 2*931 x 10 4 dynes.
4. State and explain Newton's law of gravitation and describe an accu
rate method of measuring the gravitation constant. What celestial evidence led
to the formulation of the law ? Is this law universally correct ? Explain your
statement. (Calcutta, 1945)
5. If the earth were a solid sphere of iron, of radius 6 37 million metres,
and of density 7'&6gms /cms 3 ., what would be the value of gravity at its surface,
taking the gravitational constant to be 6 658 x 10~ 8 c.g.s. units ?
Ans. 1396 cms./ sec*.
6. Give the theory of Cavendish experiment, explaining how the density
of the earth is determined. Explain why and how Boys modified the Cavendish
method. (Madras, 1950)
7. Explain how Cavendish determined the value of gravitation constant.
Indicate how, from the knowledge of the value of the gravitation constant,
it is a possible to calculate the mass of the earth. (Saugar, 1948)
8. If G 6*66 xlO~ 8 c g.s units, and the radius of the earth equal to
637 x 10 8 cms., what is the density of the earth ? Ans. 5 62 gms.lcms*.
9. The earth moves round the Sun in a circle of radius 9*288 x 10 7 miles,
and completes a revolution in 365 days A satellite of Jupiter moves about the
Jupiter in a circle of radius 1*161 x 10* miles, completing one revolution in 16*6
days. Calculate the mass of Jupiter in terms of the mass of the Sun.
Ans. 945xlO~ f .
10. Assuming the law of gravitation, find an expression for the period of
revolution of a planet.
The moon describes a circular orbit of radius 3*8 x 10 5 km. about the earth
in 27 days and the earth describes a circular orbit of radius l*5x 10" km. round
the Sun in 365 days. Determine the mass of the Sun in terms of that of the earth.
(Bombay, 1935)
Ans. 3*366x10*.
11. Define the gravitational constant and describe a laboratory method
for measuring it accurately.
A small satellite revolves round a planet of mean density 10 gmsJc.c., the
radius of its orbit being slightly greater than the radius of the planet. Calculate
the time of revolution of the satellite. (G = 6*66 x 10~ 8 c g.s. unin)
(Bombay, 1940)
Ans. 1044 hours.
12. Define 'Potential' and 'Potential Energy* of a gravitational field.
Derive an expression for the potential due to a sphere of uniform density at an
external point.
The radius of the earth is 637 x 10* cms., its mean density 5*5 gmsjcm*.
and the gravitation constant, 6 66x 10*. Calculate the earth's surface potential.
(Agra, 1940)
Ans.
GRAVITATION 27?
13. What is meant by the gravitational potential? How does it vary
with the distance from^tbe centre of the earth ? What initial velocity would be
required to project a body be>ond the attractive force of the earth ? (Radius of
earth is 6*4 x 10 8 cms.) (Cambridge Scholarship)
Ans. 1*12 xlO cms.jsec.
14. Explain what you mean by gravitational potential at a point. How
does it differ from other kinds of potential with which you are familiar ?
Find an expression for gravitational potential due to a thin hollow sphere
of uniform density at a point outside it. (Calcutta, 1947)
15. Two balls, each weighing 10 gms , are hung side by side by threads, 10
metres long. If the threads are I cm. apart at the upper ends, by how much is the
distance between the centres of the balls less than 1 cm.
Ans. l5xlO e cmi.
16. Describe one of the most accurate methods of measuring the constant
of gravitation.
The star Sirius has a mass of 6*9 x 10 3S gms. and its distance is 8x 10 18 km.
The mass of the earth is 6,x 10 27 gms. The tensile strength* of steel is about
20,000 kg./cm* Calculate the crosssection of a steel bar which could just with
stand the gravitational pull between Sirius and the earth. (G = 6'67xlO~ 8
dynecm*. I gm~*. (Bombay, 1951)
Ans. 2*169x W sq. cm.
1 7. Prove that the least velocity with which a particle must be projected
from the surface ofji planet of radius R and density p in order that it may escape
completely is /?\/8rcO>/3, where G is the gravitational constant. Calculate the
velocity in the case of the moon from the following data :
mean density of earth * 5*52 gms /c.c, ; mean density of moon 3'36 gms.jc.c. ;
mean radius of earth 638 km. ; mean radius of moon = 1740 km. ;
Acceleration of gravity at earth's surface = 980 cms. per sec. per sec.
(Oxford Scholarship)
Ans. 2'38 xlO 5 cm. see' 1 .
18. Describe an accurate 'balance method* for the determination of the
value of G, and write a short note on the 'qualities of gravitation\
19. What are seismic waves ? Give a brief description of their charac*
teristics. How may they be detected ? Also mention some of the applications of
tcismology.
20. What is an earthquake ? How is it caused ? Describe in brief the
principle underlying seismographs. Why are they so called ?
21. Describe in detail Galitzin's seismograph and explain how the epicentre
and thefccus of an earthquake may be determined with its help.
22. What is geophysical prospecting ? Write a short descriptive note oa
(/) the gravitational and (11) the seismic methods used for the purpose.
*See next chapter.
CHAPTER VIII
ELASTICITY
106. Introductory. All bodies can, more or less, be deformed
by suitably applied forces. The simplest cases of deformation are
those (/) in which a wire, fixed at its upper end, is pulled down by
a weight at its lower end, bringing about a change in its length and
(//) in which an equal compression is applied in all directions, so that
there is a change of volume but no change in shape, or (///) in which a
system of forces may be applied to a body such that, although there
is no motion of the body as a whole, there is relative displacement
of its continuous layers causing a change in the shape or 'form' of
the body with no change in its volume. In all these cases, the body is
said to be strained or deformed.
When the deforming forces are removed, the body tends to
recover its original condition For example, the wire, in the case
above, tends to come back to its original length when the force due
to the suspended weight is romoved from4t, or, a compressed Volume
of air or gas throws back the piston when it is released, in an attempt
to recover its original volume. This property of a material body to
regain its original condition, on th a removal of the deforming forces,
is called elasticity. Bodies, which can recover completely their
original condition, on the removal of the deforming forces, are said
to be perfectly elastic. On the other hand, bodies, which do not show
any tendency to recover their original condition, are said to be plastic.
There are, however, no perfectly elastic or plastic bodies. The nearest
approacli to a porfectly elastic body is a quartz fibre and, to a per
fectly plastic body, is putty. But even the former yields to large
deforming forces and, similarly, the latter recovers from small defor
mations. Thus, there are only differences of degree, and a body is
more elastic or plastic when compared to another.
We shall consider here only bodies or substances, which are (/)
homogeneous and (//) isotropic, i.e., which have the same properties at
all points and in all directions. For, these alone have similar elastic
properties in every direction, (together with other physical properties
like linear expansion, conductivity for heat and electricity, refractive
index etc.). Fluids (i.e., liquids and gases), as a rule, belong to this
class, but not necessarily all solids, some of which may exhibit
different properties at different points and in different directions, i.e.,
may be heterogeneous (or nonhomogeneous) and anisotropic (or non
isotropic). Examples of this class of solids are wood, and crystals in
general, including those metals, which are crystalline in structure.
As a class, however, metals, particularly in the form of rods and
wires, may b3 regarded to be more or less wholly isotropic, in so far
as their elastic behaviour is concerned.
107. Stress and Strain. As a result of the deforming forces
applied to a body, forces of reaction come into play internally in it,
. 274
ELASTICITY 275
due to the relative displacement of its molecules, tending to restore
it to its original Condition. The restoring or recovering force per unit
area set up inside the body is called stress, and is measured by the
deforming force applied per unit area of the body, being equal in
magnitude but opposite in direction to it, until a permanent change
has been brought about in the body, i.e., until its elastic limit has
been reached, (see 108, below). If the force be inclined to the sur
face, its component, perpendicular to the surface, measured per unit
area, is called normal stress* an'l the component acting alon^ the
surface, per unit area, is called tangential or shearing stress. Further,
the former may be compressive or expansive (i.e., tensile) according as
a decrease or increase in volume is involved, Obviously, being force
per unit area, the units and dimensions of stress are the same as
those of pressure, viz., ML~ J T~ Z , (see page 5).
The change produced in the dimensions of a body under a system
of forces or couples, in equilibrium, is called strain, and is measured by
the change per unit length (linear strain), per unit volume, (volume
strain), or the angular deformation, (shear strain, or simply, shear)"\
according as the change takes place in length, volume or shape of the
body. Thus, being just a ratio, (or an angle) it is a dimensionless
quantity, having no units,
It will be readily seen that for a perfectly elastic body (/) the
strain is always the same for a given stress ; (//) the strain vanishes
completely when the deforming force is removed and (Hi) for maintaining
the strain, the stress is constant.
108. Hooke's Law. Hooke's law is the fundamental law of
elasticity and states that, provided the strain is small, the stress is
proportional to the strain ; so that, in such a case, the ratio stress/strain
is a constant , called the modulus of elasticity, (a term first introduced
by Thomas Young), or the coefficient of elasticity.
Since stress is just pressure, (or tension per unit area), and strain
is just a ratio, the units and dimensions of the modulus of elasticity
are the same as those of stress or pressure.
When the stress is continually increased in the case of a solid, a
point is reached at which the strain increases more rapidly than is
warranted by Hooke's Law. This point is called the elastic limit,
and if the b'xly happens to be a wire under stretch, it will not regain
its original length on being unloaded, if the elastic limit be passed, as it
acquires what is called a 'permanent set'. On loading it further, a
point is reached when the extension begins to increase still more
rapidly and the wire begins to 'flow down* in spite of the same constant
load. This point is called the 'yield point' ; and, after a large ex
tension, it reaches the 'breaking point 9 and the wire snaps. In the
case of plastic substances, like lead, there is a long range between the
yield point and the breaking point.
*The stress is always normal in the case of a change in the length of the
wire, or in the case of a change in the volume of a body, but is tangential in the
case of a change in the shape of a body.
tThis will be dealt with more fully later in 109 (3).
276 PROPERTIES OF MATTER
Thus, if we were to plot a graph between the load suspended
from a wire, fixed to a rigid support at its upper end, and the
extension produced thereby, w*
obtain, in general, a curve of
the form shown in Fig 169,
the straight part OA of the curve
showing that the extension pro
duced is directly proportional to
the load applied, or that Hooke's
law is obeyed perfectly up to A,
and that, therefore, on being
unloaded at any point between
and A, the wire will come
back to its original condition,
(represented by O). In other
words, the wire is perfectly elastic
up to A, which thus measures
the elastic limit* of the specimen
in question, the extension here being of the order of 10~ 8 of the
original length.
On loading the wire beyond the elastic limit, say, up to B, the
curve takes a bend almost vertically upwards, as shown, and, on being
unloaded at any point here, (at B, say), it does not come back to its
original condition but takes the dotted path BC, thus acquiring a 'per
manent set' OC.
On increasing the load still further, a point D is reached, where
the extension is much greater even for a small increase in the load,
i.e., Hooke's law is obeyed no longer ; and, beyond D, the extension
increases continuously, with no addition to the load, the wire starting
'flowing down', as it were. For, due to its thinning down, the stresS
(or the load per unit area) increases considerably and it cannot
support the same load as before ; and, if the wire is to be pre
vented from 'snapping', the load applied to it must be decreased.
That is why the curve starts turning towards the extensionaxi
beyond this point D, which thus represents the yield point of the
wire. And, once the yield point is crossed, the thinning of the
wire no longer remains uniform or even, its crosssection decreasing
more rapidly at some points than at others, resulting in its develop
ing small 'necks 9 or 'waists' at the former points, so that the stress
is greater there than at the latter points ; and the wire ultimately
'snaps' at one of these. This point on the curve, at which the
snapping or the breaking of the wire actually occurs, is called ita
breaking point, the corresponding stress and strain there being
referred to as the breaking stress (or tensile strength) and the breaking
strain, respectively.
Note. The tlastic limit of a material is also sometimes defined as the
force producing the maximum reversible or recoverable deformation in it, and may,
*Jn quite a few cases, Hooke's law is obeyed only up to a point a little
below the elastic limit, represented by A. The portion of the curve from O
to this point (below A), is then**aid to indicate the limit of proportionality, to
distinguish it from the elastic limit. The two are thus not always identical,
though they are generally regarded to be so, in view of the very small difference
between them.
ELASTICITY 277
for a given specimen, be determined by loading and unloading it with a number
of different loads and measuring its length afterVacA unloading, until it acquires
a permanent set. The latter is then plotted against the load, and from the curve
thus obtained, the particular load at which the permanent set just starts, can be
easily estimated.
Even within the elastic limit, however, few solids come back to
their original condition, directly the deforming force is removed.
Almost all of them onfy 'creep' back to it, (i.e., take some time to do
so), though they all do so, ultimately. This delay in recovering back
the original condition, on the cessation of the deforming force, is called
clasticafter effect. Glass exhibits this effect to a marked degree, the
few exceptions to this almost general rule being quartz, phosphor
bronze, silver and gold, which regain their original condition as soon
as the deforming force ceases to operate. Hence their use in Caven
dish's and Boys' experiments for the determination of G, in quadrant
electrometers and movingcoil galvanometers etc. etc.
As a natural consequence of the elastic aftereffect the strain in
a material, (in glass, for example), tends to persist or lag behind the
stress to which it is subjected, with the
result that during a rapidly changing
stress, the strain is greater for the same
value of stress, when it is decreasing than
when it is increasing, as is clear from the
curve in Fig. 170. This lag between stress
and strain is called elastic hysterisis, (the
term 'hysterisis\ meaning 'lagging be
hind 1 ). The phenomenon is similar in its
implications to the familiar magnetic hys
terisis, where the magnetic effects tend
to persist or lag behind even after the
magnetising influence is removed, the XTAWOAf > X
curve referred to above may thus be called p gt 170.
the elastic hysterisis loop. And, exactly
in the same manner the energy, dissipated as heat, during a cycle of
loading and unloading is given by the area enclosed by the loop.
There is, however, very little hysterisis in the case of metals or of
quartz.
Further, it was shown by Lord Kelvin, during his investigation
of the rate of decay of torsional vibrations of wires, that the vibra
tions died away much faster in the case of a wire kept vibrating con
tinuously for some time than in that of a fresh wire. The same
happens to any elastic body, subjected to an alternating strain. The
continuously vibrating wire got 'tired* or 'fatigued', as it were, and
found it difficult to continue vibrating. Lord Kelvin fittingly express
ed this by the term 'elastic fatigue'.
A body, thus subjected to repeated strains beyond its elastic
limit, has its elastic properties greatly impaired, and may break under
A stress, less than its normal breaking stress even within its elastic
limit. This phenomenon is, obviously, of great importance in cases
like those of the piston and the connecting rods in a locomotive,
which, as we know, are subjected to repeated tensions and compres
sions during each revolution of the crank shaft.
278 PROPERTIES OF MATTER
It may be mentioned here that all these elastic properties of a material are
linked up with the fine mass of its structure. It is now finally established by care
ful microscopic examination, that metals are just an aggregation of a large
number of fine crystals, in most cases, arranged in a random or a chaotic
fashion^ i.e., their cleavage planes (or the planes along which their constituent atoms
can easily slide over each other), being distributed haphazardly, in all possible
directions. Now, single crystals, when subjected to deformation, show a
remarkable increase in their hardness. Thus, for example, a single crystal of
silver, on being stretched to a little more than twice its length, is known to
increase to as much as ninetytwo times its original strength or stiffness. So
that, operations like hammering and rolling, which help this sort of distribution,
i.e., which break up the crystal grains into smaller units, result in an increase or
extension of their elastic properties ; whereas, operations like annealing (or heat
ing and then cooling gradually) etc., which tend to produce a uniform pattern of
orientation of the constituent crystals, by orienting them all in one particular
direction and thus forming larger crystal grains, result in a decrease in their clas
tic properties or an increase in the softness or plasticity of the material.
This is because in the latter case, slipping (or sliding between cleavage
planes), starting at a weak spot proceeds all through the crystal and, in the
former, the slipping is confined to one crystal grain and stops at its boundary
with the adjoining crystal. Indeed, the former may be compared to a small cut,
developing into a regular tear all along a fabric and the latter to the tear stopping
as it reaches a seam in the fabric. Thus, 'paradoxically', as Sir Lawrence Bragg
puts it, */ order to be strong, a metal must be weak,* meaning thereby that metals
with smaller grains are stronger than those with larger ones.
A change in the temperature also affects the elastic properties of a
material, a rise in temperature usually decreasing its elasticity and vice versa,
except in certain rare cases, like that of invar steel, whose elasticity remains prac
tically unaffected by any changes in temperature. Thus, for example, lead becomes
quite elastic and rings like steel when struck by a wooden mallet, if it be cooled
in liquid air. And, again, a carbon filament, which is highly elastic at the ordinary
temperature, becomes plastic when heated by the current through it, so much so
that it can be easily distorted by a magnet brought near to it.
109. Three Types of Elasticity. Corresponding to the three*
types of strain, we have three types of elasticity, v/z.,
(/) linear elasticity, or elasticity of length, called Young's Modulus,
corresponding to linear (or tensile) strain ;
(i7) elasticity of volume or Bulk Modulus, corresponding to volume
strain ; and
(Hi) elasticity of shape, shear modulus, or Modulus of Rigidity,
corresponding to shear strain.
(1) Young's Modulus. When the deforming force is applied to
the body only along a particular direction, the change per unit length
in that direction is called longitudinal, linear or elongation strain, and
the force applied per unit area of crosssection is called longitudinal
or linear stress. The ratio of longitudinal stress to linear *trnin, within
the elastic limit, is called Young's Modulus, and is usually denoted by
the letter Y.
Thus, if F be the force applied normally to a crosssectional area
a, the stress is F/a. And, if there be change / produced in the origi
nal length L, the strain is given by //L. So that,
Young's Modulus, Y = J = . .
ijju a, i
Now, if L 1, a = 1 and / = 1, we have Y F.
In other words, if a material of unit length and unit area of cross
ELASTICITY 279
section could be pulled so as to increase in length by unity, i.e., to
double its length, the force applied would measure tbe value of
Young's Modulus for it.
Since, however, the elastic limit is exceeded when the extension
produced is 10~ 8 cm./cm., the material will snap before this much
extension is produced.
In cases, where, elongation produced is not proportional to the
force applied, we can still determine Young's Modulus from the ratio
L.dF/a.dL, where dF/a is the infinitesimal increase in the longitudinal
stress and dL/L, the corresponding increase in strain.
Or, *'%'
a dL
N.B. The particular case of rubber may, with advantage, be mentioned
here, which the beginner finds so confusing, when, in ordinary conversational
language, we refer to it as being 'elastic*. For, he knows well enough that it
requires a much smaller force than steel to stretch it, (and that, therefore, its elas
ticity is much less than that of steel). In fact, the value of Young's Modulus for
rubber is about onefiftieth of that of steel. What we mean when we say that it is
elastic, therefore, is just that it has a very large range of elasticity, for, whereas a
crystalline body can be stretched to less than even one per cent of its original
length before reaching its elastic limit, rubber can be stretched to about eight
times (or 80%) of its original length.
This high extensibility of rubber is due to its molecule containing, on an
average, some 4,000 molecules of isoprene (C 6 # 8 ), whose 20,OCO carbon atoms,
spreading out in a chain, make it very long and thin, about 1/4000 mm. in length.
Rubber, in bulk, has thus been rightly compared to an intertwined mass
of long, wriggling snakes, its molecules, like the snakes, tending to uncoil when
stretched and getting coiled up again when the stretching force is removed.
(2) Bulk Modulus. Here, the force is applied normally and uni
formly to the whole surface of the body ; so that, while there is a
change of volume, there is no change of shape. Geometrically speaking,
therefore, we have hero a change in the scale of the coordinates of the
system or the body. The force applied per unit area, (or pressure),
gives the Stress, and the change per unit volume, the Strain, their ratio
giving the Bulk Modulus for the body. It is usually denoted by the
letter K.
Thus, if F be the force applied uniformly and normally on a sur
face area a, the stress, or pressure, is F/a or P ; and, if v be the
change in volume produced in an original volume K, the strain is v/K.
and, therefore,
Bulk Modulus, K= F !* = F ' y  rv Fla /.
v/V a.v v l '
If, however, the change in volume be not proportional to the
stress or the pressure applied, we consider the infinitesimal change in
volume dV, for the corresponding change in pressure dP ; so that,
we have K = d
The Bulk Modulus is sometimes referred to as incompressibility
and hence its reciprocal is called compressibility ; so that, compressi
bility of a body is equal to l/#, where K is its Bulk Modulus. It must
thus be quite clear that whereas bulk modulus is stress per unit
strain, compressibility represents strain per unit stress.
280
PBOPHBTIBS OF MATTER
Since fluids (i.e., liquids and gases) can permanently withstand
or sustain only a hydrostatic pressure, the only elasticity they possess
is Bulk Modulus (K), which is, therefore, all that is meant when we
refer to their elasticity. This, however, is of two types : isothermal
and adiabatic.
For, when a fluid is compressed, there is always some heat pro
duced. If this heat be removed as fast as it is produced, the tempera
ture of the fluid remains constant and the change is said to be
isothermal ; but if the heat be allowed to remain in the fluid, its
temperature naturally rises ard the change is then said to be
adiabatic.
It can be easily shown that the isothermal elasticity of a gas (i.e.,
when its temperature remains constant) is equal to its pressure P, and
its adiabatic elasticity equal to yP, where y is the ratio between C>*
and C y * for the gas in question, its value being 1*41 for air, [see
solved Example 1 (b) at the end of the Chapter.]
It will thus be readily seen that the Bulk Modulus of a gas
fwhether isothermal or adiabatic) is not a constant quantity, unlike
that of a solid or a liquid.
(3) Modulus of Rigidity. In this case, while there is a change
in the shape of the body, there is no change in its volume. As indi
cated already, it takes place by the movement of contiguous layers of
the body, one over the other, very much in the manner that the cards
would do when a pack of them, placed on the table, is pressed with
the hand and pushed horizontally. Again, speaking geometrically, we
have, in this case, a change in the inclinations of the coordinate axes
of the system or the body.
Consider a rectangular solid cube, whose lower face aDCc, (Fig.
171), is fixed, and to whose upper face a tangential force Fis applied
, in the direction shown. The couple
so produced by this force and an
equal and opposite force coming
into play on the lower fixed face,
makes the layers, parallel to the
two faces, move over one another,
such that the point A shifts to A' t
B to B', rf to d' and b to 6', i.e.,
the lines joining the two faces turn
through an angle 0f.
F '3 171. The face A BCD is then said to
be sheared through an angle 8. This angle (in radians), through which
a line originally perpendicular to the fixed face is turned, gives the strain
or the shear strain, or the angle of shear, as it is often called. As will
*The symbols Cp and^C*, stand for the specific heats of a gas at constant
pressure and at constant volume respectively, their ratio r = C>/C, being the
highest (1*67) for a monoatomic gas, like helium, goes on decreasing with in
creasing atomicity of the gas but is always greater than 1.
fAs a matter of fact, if this were the only couple acting on the body, it
would result in the rotation of the body. This is prevented by another equal and
opposite couple, formed by the weight of the body (plus any vertical force applied)
and the reaction of the surface on which the body rests.
^.,
''] >/' /S.
''i
i
/
i
'
/
/
/A'
1
1
1
i
1^
I
?'
/
B IB
i
t
i
1
4 ^
1^
ELASTICITY
281
be readily seen, & = A A' /DA = II L, where /is the displacement AA'
and X, the length of the side AD or the height of the cube ; or 9 =
relative displacement of plane ABbajdistance from the fixed plane
aDCc. So that, if the distance from the fixed plane, i.e., L = 1, we
have 9 = / = relative displacement of plane ^4Z?6a.
Thus, shear strain (or shear) may also be defined as the relative
displacement between two planes unit distance apart.
And, stress or tangential stress is clearly equal to the force F
divided by the area of the face ABbd, i.e., equal to Fja. The ratio of
the tangential stress to the shear strain gives the coefficient of rigidity
of the material of the body, denoted by n.
Thus, tangential stress = Fla, and shear strain = 6 = //L.
And, therefore, Coefficient of Rigidity, or Modulus of Rigidity of
the material of the cube is given by
n = F/a  Fl  a = ~~ L (i)
to ~~ I/L ~~ a 1
This is a relation exactly similar to the one for Young's Modu
lus, with the only difference that, here, F is the tangential stress, not
a linear one, and I, a displacement at right angles to L, instead of along
it.
Again, if the shearing strain, or shear, be not proportional to
the shearing stress applied, we have
*L fl
where d0ia the increase in the angle of shear for an infinitesimal in
crease dF/a in the shearing stress.
Further, it is clear from relation (/) above, that if a = 1, and
Q = 1 radian (or 57 18'), we have n = F.
We may thus define modulus of rigidity of a material as the shear
ing stress per unit shear, i.e., a shear of I radian, taking Hooke's law
to be valid even for such a large strain*.
110. Equivalence of a shear to a compression and an extension
at right angles to each other. Consider a cube A BCD, (Fig. 172),
with the face DC fixed, and let the face A j R n ^ ^,
A BCD be sheared by a force, applied
in the direction shown, through an
angle 0, into the position A'B'CD.
Then, clearly, the diagonal DB is in
creased in length to DB', and the dia
gonal AC is shortened to A'C.
 The shear is really very small in
actual practice, and, therefore, triangles
AFA' and BEE' are isosceles rightangled * F j g i 72 ~
triangles, (i.e., rightangled 45 triangles).
*In the case of metals, however, Hooke's law no longer holds even if the
shear exceeds 11/200 radian, or '33.
V"
282
PROPERTIES OF MATTER
And, therefore, EB 1 = BB 1 . cos BB'E =
v Z.#fi' = 45 and cos 45 =
If AB = /, then, clearly, DB = = /y/2,
., extension strain along diagonal D2?
__ "5' BB' I BB' e
DB y/2 /\/2 2/ 2
Similarly, the compression strain along the diagonal
is given
'. cosA'AF
45
AC
[v
Thus, we see that a simple shear B is equivalent to two equal
strains, an extension and a compression, at right angles to each other.
Corollary. The converse of the
above follows as a corottary^viz., that
simultaneous equal, "compression and
extension at right angles to each other
are equivalent to a shear, as will be
seen from the following :
Let the cube ABCD, of side /,
be 'compressed along the diagonal AC,
so that the new diagonals become A'C'
and B'D', (Fig. 173).
Let AA' = BB' = a.
And since OA = AB cos BAO
Fig. 173. = AB. cos 46 = AB/\/2
we have
and
Clearly, ..
OA'OAAA'^a).
'  OB+BB'  ( ' + a).
OB
(A'B'f = (OA')*+(pB')*
I 2
In practice, 2a* is very small as compared with / a , and may,
therefore, be neglected.
So that, (A'B')* = /*. Or, AB' = / = AB.
Thus, A'B'C'D' may be rotated through the angle DGD' =
angle ^l 7 ^', so that D'C coincides with DC. Then, it is obvious that
A'D' would make an angle 2^4F^4' with AD, so that the angle of shear
is equal to twice the angle AFA ', i.e., is equal to 2 LAFA '.
Or, flflgfe of shear = 2^'/F, (/ the angle is
ELASTICITY
283
where A'E is the perpendicular from A' t to AF.
Now, A'E = *Av/2 and EF = 7/2.
2 ~ V2 f ~ /
Denoting this angle of shear by 0, we have 6 = 2fl\/2//*
Now, compression strain along the diagonal ^4C is
/L4' ^ fl\/2 __
Ad  //V2 ~ 7 ~~ 2 *
Or, the compression strain is half the angle of shear, i.e., the
angle of shear is twice the angle of compression.
Similarly, it can be shown that the extension strain is also half the
angle of shear.
Thus, we see that simultaneous and equal compression and exten
sion at right angles to each other are equivalent to a shear, the direction
of each strain being at an angle of 45 to the direction of shear.
111. Shearing stress equivalent to an equal linear tensile stress
and an equal compression stress at right angles to each other. In the
case of the cube above, if Fwere the
UtVDO VJL tilt? UULJC itUUVU, 11 JT VrClU l/ilt? J?jC ' \
only force acting on its upper face it ^ p f " /p\
would move bodily in the direction of
this force. Since, however, the cube is
fixed at its lower face DC, an equal and
opposite force comes into play in the
plane of this face, giving rise to a couple
F./.*, tending to rotate the cube in the
clockwise direction, (Fig. 174).
Again, since the cube does not
rotate, it is obvious that the plane of
DC applies an equal and opposite couple
B
FF
^
F'./, say., by exerting forces F' and F' "" Fig. 174.
along the faces AD and CB, tending to
rotate it in the anticlockwise direction, as shown. Thus, because the
cube is in equilibrium under the two couples, we have
F.I = F'.l Or, F = F',
i.e., a tangential force F applied to the face AB results in an equal tan
gential force acting along all the other faces of the cube in the directions
shown.
Clearly the resultant of the two forces F and F' or F and F
along AB and CB respectively is F\/ 2 along OB t and of those acting
along AD and CD is also F\/2 along OD. And, thus, an outward pull
acts on the diagonal DB of the cube at B and D resulting in its
extension, as we have just seen above, (110). Precisely similarly,
an inward pull acts on the diagonal AC at A and C, thereby bringing
about its compression.
*/ being the length of each edge of the cube and hence the perpendicular
distance between the two forces Fand F.
284 PROPERTIES OP MATTER
Thus, a tangential force F applied to one face of a cube gives
rise to a force F\/2 outward along one diagonal (BD, in the case
fihown) and an equal force F\/2 inward along the other diagonal (AC)
of the cube, resulting in an extension of the former and a compression
of the latter.
Now, if the cube be cut up into two halves, by a plane passing
through AC and perpendicular to the plane of the paper, each face,
parallel to the plane, will have an area lxl\/2 / 2 \/^ an< * dearly,
the outward force F\/2 along BD will be acting perpendicularly to it.
So that, we have
tensile stress along BD = F\/'2/l* \/2 = F// 2 .
Similarly, if we cut the cube into its two halves by a plane
passing through BD and perpendicular to the plane of the paper,
we shall have an inward force Fi/2 along AC acting perpendicularly
to a face on an area / \/2. So that, we have
compression stress along AC = F\/2jl *i/2 = F// 2 .
Obviously, F// 2 is the shearing stress over the face AB of the cube,
which produces the shear $ in it, (see page 281).
Thus, it is clear that a shearing stress is equivalent to an equal
tensile stress and an equal compression stress at right angles to each
other.
112. Work done per unit volume in a strain. In order to deform
a body, work must be done by the applied force. The energy so
spent is stored up in the body and is called the energy strain. When
the applied forces are removed, the stress disappears and the energy
of strain appears as heat.
Let us consider the work done during the three cases of strain.
(i) Elongation Strain (stretch of a wire). Let F be the force
applied to a wire, fixed at the upper end. Then, clearly, for a small
increase in length dl of the wire, the work done will be equal to F.dl.
And, therefore, during the whole stretch of the wire from to /.
work done =
Now, Young's modulus for the material of the wire, i.e.,
Y = F.L/a.1.,
where L is the original length, /, the increase in length, a, the cross 
ectional area of the wire, and F, the force applied.
And /. F = Y.a.ljL.
Therefore, work done during the stretch of the wire from to /
is given by
Y.a t* 1 Y.a.l ,
L' 2 " 2""/r
ELASTICITY
285
But Y.i.llL = F, the force applied.
Henoe W = ^ F.I =  x stretching force x stretch.
. , .
work done per unit volume =
I F I 1
= U ' T ' IT = 2
1 _ f 1
^ Fix y~
Lt, a
fV v
w
L
volume of the
w i re _. Lxa
//L=strain .
Alternatively, the same result may also be obtained graphically
as follows :
Let a graph OP be plotted between the streitching force applied
to the wire and the extension produced in it, within the elastic limit ,
as shown in Fig. 175.
Consider a small exten
sion pq of the wire and erect
ordinates at p and q to meet the
graph in p' and q 1 respectively,
where pp' is very nearly equal
to <?#'> (the extension pq being
really small).
Then, clearly, work done
upon the wire or energy stored
up in it
s=stretching force pp' ^extension
pq.
=pp'Xpq=area of strip pp'q'q.
So that, imagining the
whole extension OB = /, of the
wire, to be broken up into small
bits like/N? and erecting ordinates at their extremities, we have
total work done upon the wire or total energy stored up in it
= sum of the areas of all such strips formed
= area of the triangle OBP = \OBxBP = $/x/s
where the total extension OB = / and the stretching force corres
ponding to it is BP = F.
Now, if L be the original length of the wire and a, its area of
crosssection, clearly, volume of the wire = L x a.
.. work done, or strain energy, per unit volume of the wire
1 F I 1
= \lFIL.a *= v~  != o st r ess x strain.
A a ju
(ii) Volume Strain. Let p be the stress applied. Then, over
an area a the force applied is p.a, and, therefore, the work done for a
small movement dx t in the direction of p, is equal to p.a.dx. Now,
a.dx is equal to Jv, the small change produced in volume. Thus,
work done for a change dv is equal to p.dv.
And, therefore, total work done for the whole change in volume, from
to v, is given by
EXTENSION
Fig. 175.
Now,
K
W = T p.dv.
p.Yjv ; so that, p == K.vjV,
286
PROPERTIES OF MATTER
where V is the original volume, and K, the Bulk Modulus.
And .. W  f^ V . dv = f v.rfv = J 4v.
r K J K Z
== ~ v^ v = /?.v = 9 stress X change in volume.
Or,
work done per i/mV volume = % p*vjV = \ st r ess x strain.
(Hi) Shearing Strain. Consider a cube (Fig. 176), with its
lower face DC fixed ; and let F be the tangential force applied to its
J $ 3 B' v u PP er f ace in the plane of ^4, so that
the face A BCD is distorted into the
position A'B'CD, 6r sheared through an
angle 0. Let tike distance AA' be
equal to BB'= x\ Then, work done
during a small displacement dx is equal
to F.dx. And, thereiS^, work done for
the whole of the displacement, from to
x. is given by
W = I F.dx.
oo i
1
t
i
1
1
t
t
1
I L
oc *
i
i
i
i
t
f
D C
Fig. 176.
^ow, n = FjaQ, or /
1 = n.a.Q, and a = L 2 ; also # = x/L,
where L is the length of each edge of the cube.
So that, F = n.L^.xjL = /*..*.
.*. work done during the ivo/^ stretch from to x, i.e.,
= r
Jo
work done per unit volume =
n.L.x.dx =
2 "
n.L.x*
of
F'. 4 volume
[_the cube
11 V 7" V I C* >*
n.x.JL x L r x
= 2 ' // X L ^T' a " % L
Thus, we see that, in any kind of strain, work done per unit
volume is equal to J stress x strain.
113. Deformation of a Cube Bulk Modulus. Let A BDCOHEFA
be a unit cube and let forces T x , T v and T e act perpendicularly to the
faces BEHD and AFGC, ABDC
and EFGH, and ,4Fand DHGC
respectively, as shown, (Fig. 177).
Then, if a be the increase per unit
length per unit tension along the
direction of the force and (3, the
contraction produced per unit
length per unit tension, in a direc
tion perpendicular to the force , the
elongation produced in the edges
AB, BE and BD, will, obviously,
be T x .<x., T^.OL and T B .a, respective
ly, and the contractions produced
perpendicular to them will be T K .$,
T v .$, and T^. The lengths of the Fig. 177.
edges thus become the following :
ELASTICITY
287
AB = i+arva2vpr..p.
BE =
BD =
Hence the volume of the cube now becomes
2P)
neglecting squares and products of a and (3, which are very small
compared with the other quantities involved.
Tf fn /TT ___  /TT /77
the volume of the cube becomes 1+ (a2p).3T.
And, therefore, increase in the volume of the cube
= l+'3!T(a2p)l == 3T(a2(3).
If, instead of the tension 2 r outwards, we apply a pressure P,
compressing the cube, the reduction in its volume will similarly be
3P(a 2(3), and, therefore, volume strain is equal to 3P(a 2(3)/l, or
equal to 3F(a 2(3). ["" original volume of the cube = 1,
Hence Z?w/A: Modului, K =
Or,
volume strain 3P(a 2(3)
[
3(a2p)' "* (/)
And, Compressibility, which is the reciprocal of Bulk Modulus,
is, therefore, equal to 3(a 2(3).
114. Modulus of Rigidity. Let the top face ABHQ, of a cube
(Fig. 178), be 'sheared* by a shearing force F, relative to the bottom
face, such that A takes up the position A'
and B y the position B\ the angle ADA' being
equal to the angle BCB' = 6. Then,
, rr ^ _ _ /L_
~~ area of the face ABHG
= 2 = r, say,
where L is the length of each edge of the
cube.
Let the displacement A A' = BB' <= /.
Then, 5/z^^fr .y/r^/n = // = ^.
And .. coefficient of rigidity, n =T/0.
Now, extension of the diagonal DB, due to extension along AB
is DB.T.a, and that due to contraction* along fA is DB.T,$.
Therefore, /o/a/ extension Jj[^^2fA^22^^ 5 now becomes
^^~^^^^3<>^1^
= L^.Tfa+p). [/ jDJ5 =
Drop a perpendicular BE from 5 on to DB'.
* See 117, page 288.
Fig. 178.
288 PBOPBBTIBS OF MATTER
Then, increase in length of DB is practically equal to EB'.
And, clearly, EB' = BB'. cos BB'E.
= / cos 45 = //<v/2 [v <BB'E = 45, very nearly, and cos 45* ~ i/v2.
o ^T _ JL n T
' ~~' '
* T =2,% tv '/i 
And, since TjQ = n, the coefficient of rigidity of the material of
the cube, we have
115. Young's Modulus. If we now imagine a cube of unit
edge, acted upon by unit tension along one edge, the extension
produced is a. Then, clearly,
stress 1, and linear strain = a/1 = a.
Therefore, Young's Modulus, Y = I/a. ...(///>
116. Relation connecting the Elastic Constants. We have from
relation (/), above,
a2p = 1/3AT. ...(/)
And, from relation (77), af fj == Ij2n. .. (//)
.*. subtracting (/) from (H), we have
30  1 l **"
p ~ 2/j
whence, p =
Again, multiplying (//) by 2 and adding to (/), we have
Q 11 ZK+n ZK+n
~ *
Or
Y r ~ r>
[/ l/y from (III), above.
O  *K+ n _ ?5 "
f 7 ~ A>  JOi + Rn *
, 931
whence, =  + ...(6)
This, then, is the relation connecting the three elastic constants.
117. Poisson's Ratio. It is a commonly observed fact that
when we stretch a string or a wire, it becomes longer but thinner, i.e.,
the increase in its length is always accompanied by a decrease in its
crosssection (though not sufficient enough to prevent a slight
increase in its volume). In other words, a longitudinal or tangential
strain produced in the wire is accompanied by a transverse or a lateral
strain in it. And, of course, what is a true of a wire, is true of all
other bodies under strain. Thus, for example, when a cube is subject
ELASTICITY 289
ed to an outward force perpendicular to one pair of its faces, there is
elongation produced along this face, but a contraction in a direction
perpendicular to it, (as we have seen already in 113).
The ratio between lateral strain (fc) to the tangential strain (a) is
constant* for a body of a given material and is called the Poisson's
ratio for that material\ It is usually denoted by the letter a.
Thus, Poisson's ratio = lateral stramjtangential strain ; or, a = p/a.
It follows, therefore, that if a body under tension suffers no
lateral contraction, the Poisson's ratio (a) for it is zero ; and, because
its volume increases, is density decreases. '
The relations for K and n above may now be put in terms of
Poisson's ratio, asjollows :
We have, from relation /, above,
K = __L_ _ a _ I _  . l~v ^ Y [see (III) above.
2 3ai2a ~" 3l*o L <C
a
whence, Y = 3^(l2j),f ...... (iv)
Similarly, from relation (II) above, we have
i i y
n =
L'a(l+a) 2(1
whence, y=2w(l + a)f (v)
Now, from relations (/V) and (v) we have
3A"(ll>a) = 2/7(1 + a),
, :iA:~2Ai
whence, a = ^^
which gives the value of Poisson's ratio in terms of K and n.
Similarly, if we eliminate c from (iv) and (v), we have
y _ <*, , [Sime as relation (a), above.
931
whence, =  u  [Same as relation (b), above.
1 J 71 A
Limiting values of a. We have seen above how
where A^ aiifl n arc essentially positive quantities. Therefore,
(/) if a be a positive quantity, the right hand expression, and
hence also the left hand expression, must bt> positive, and for this to
be so, 2a<l, or a< or *5. And,
*i>., the lateral strain is proportional to the longitudinal strain. Xnis
is, however, so only when tue latter is small.
J These relations would not b^ foand ta apply in ths cass of wire speci
matenals for the simple reason that ths process of wiredrawing brings
about at least a partial alignment of the minute crystals of the substance, which
thus no longer remain oriented at random, with the result that the substance.. loses
its isotropic character,
290
f BOFKRTIEB Of MATTBB
(//) if Q be a negative quantity, the left hand expression, and
hence also the right hand expression, must be positive, and this is
possible only when a be not less than 1.
Thus, the limiting values of o are 1 and '5. Or, else, as will be
readily seen from relations (iv) and (v) above, either the bulk modulus
or the modulus of rigidity would become infinite. Further, a negative
value of a would mean that, on being extended, a body should also
expand laterally, and one can hardly expect this to happen, ordinarily.
At least, we know of no such substance so far. Similarly, a value of
a = Q5 would mean that the substance is perfectly incompressible,
and, frankly, we do not know of any such substance either.
In actual practice, the value of a is found to lie between *2 and
4, although Poisson had a theory that the value of a for all elastic
bodies should be *25, but this is not borne out by any experimental
facts.
118. Determination of Young's Modulus. Young's modulus, as
we know, is the ratio between tensil stress (or tangential force applied
per unit area) and elongation strain (or extension per unit length).
The extension produced is rather small and it is difficult to measure
it with any great degree of accuracy. The different methods used are
thus merely attempts at measuring this extension accurately. We
shall consider here only two methods, viz., one for a wire, and the
other for a thick bar.
(/) For a Wire Searle's Method. Two wires, A and B, of the
same material, length and area of cross sect ion, are suspended from
a rigid support and carry, at their
lower ends, two metal frames, C and
Z>, as shown in Fig. 178, one carry
ing a constant weight W to keep
the wire stretched or taut ' and the
other, a hanger //, to whic'h slotted
weights can be slipped on, as and
when desired.
A spiritlevel L rests horizon
tally at a point P in frame C, and
on the tip of a micrometer screw (or
spherometei) 5, working through a
nut in frame D.
The screw is worked up or
down, until the air bubble in the
spiritlevel is just in the centre.
Weights are now slippedo n to the
hanger, so that the frame D moves
down a little due to the extension
of wire B, and the air bubble shifts
towards P. The screw is now
worked up to restore the bubble
back to its central position. The
distance through which the screw
is moved up is read on the vertical
174 acale. ar&duated in half millimetres.
ELASTICITY
and fixed alongside the disc of the screw. This gives the increase in
length of wire B. A number of observations are taken by increas
ing the weight in the hanger by the same equal steps and making the
adjustment for the level for each additional weight. The mean of all
these readings of the screw gives the mean increase in the length of
the wire, for the stretching force due to the given weight. Thus, if
I cms. be the increase in the length of wire B, and L cms., its original
length, we have
elongation strain = //L.
And if W k.gms. be the weight added each time to the hanger,
the stretching force is equal to IfXlOOO gms. H'/. WxluOOx981
dynes, or equal to F dynes, say.
So that, if a sq. cms. be the area of crosssection (irr 2 ) of the
wire, we have tensile stress =* F/ct.
And, /. Young Modulus for the material of the wire, i.e.,
y~ J 7 L.1 ^ FxL .
a ' L axl
The other wire A merely acts as a reference wire, its length
remaining constant throughout, due to the constant weight suspend
ed from it (which need not be known). Any yielding of the support
or change in temperature during the experiment affects both the
wires equally, and the relative increase in the length of B (with respect
to A) thus remains unaffected by either change.
If a graph be now plotted between the load suspended and the
extension produced, it would be found to be a straight line (just like
OA in Fig. 100), passing through the origin, showing that the exten
sion produced is directly proportional to the load. Hooke's law also
can thus be easily verified.
(n) For a thick BarSwing's Extensometer Method Ewing's
Extensometer is raorely a device to magnify the small extension of
the bar under test and con
sists of two metal arms. APS
and CQD, (Fig. 179), pivot
ed at P and Q, by means of
pointed screws, on the verti
cal bar B itself, (the Young 1 s
modulus for the material of
which is to be determined),
so that they are free to rotate
about P and Q. The arm
APS is bent at right angles,
as shown, and carries a mic
rometer screw S at its lower
end, and a microscope M,
fitted with a micrometer
scale, at the end of an arm,
pivoted at its uppe? end 4 Fig. 179.
292
PBOPEBTIES OF MATTER
180.
The other horizontal arm CQD> has a Fshaped groove at D for
the micrometer sore v to rest in, and a fine horizontal line marked on
the end C.
The bar B is fixed at its upper end, the two metal arms are
adjusted to be horizontal, by means of the micrometer screw S 9 and
f, the microscope focused on the
S Q JL horizontal line on C. The bar
is now stretched downwards
(by means of a testing
machine), so that the horizon
tal arm CQD gets tilted a little
about D as its fulcrum, the
end C, with the fine mark on
it, moving down to C", (Fig. 180), and the point Q to Q' . The micro
scope^is again focused on the mark and the distance CC' through
which it has shifted downwards is measured accurately on the micro
meter scale of the eye piece. Let it be equal to h.
Now, obviously, the increase in the length PQ of the rod is
QQ' = /, say.
Thon ; clearly, in the two similar triangles SQQ' and SCC\
we have QQ'/CC' = SQjSO. Or, ///i = SQjSC,
whence, / = SQ.h.jSC.
Thus, knowing SQ, SO and A, we can determine / to quite a
high degree of accuracy.
Then, from the length PQ of the bar, its area of crosssection
and the stratching; force applied to it, we can easily calculate the value
of Young's Modulus for its material.
N.B. A modification of Ewing's Extensometer, as shown in Fig. 181, is
called the Cambridge Extensometer, in
which there is a vibrating reed R arrang
ed, as shown, the arrangement being such
that as the bar B is stretched by the test
ing machine, that part of the reed which
touches the micrometer screw M, moves
downwards through a distance five times
the extension of the rod. Thus, by noting
the micrometer screw readings, when the
vibrating reed just touches the micro
meter screwpoint both before and
after the rod B has been stretched, we
can directly obtain the increase / in the
length of the rod.
Fig. 181.
119. Determination of Poisson's Ratio for Rubber. To deter
mine the value of cr for rubber, we take about a metrelong tube AB
of it, (Pig. 182), such, for example, as the tube of an ordinary cycle
tyre, and suspend it vertically, as shown, with its two ends properly
stoppered with rubber bungs and seccotine*. A glass tube C open
at both ends, about half a metre long and about 1 cm. in diameter,
graduated in cubic centimetres, is fitted vertically into it through
*atvpeof liquid glue.
BLASTtCITY
293
a suitable hole in the stopper at the upper end A, so that a major
part of it projects out.
The rubber tube is completely
filled with water until the water rises up
in the glass tube to a height of about C
30 cms. from A. A suitable weight W
is now suspended from the lower end
B of the tube. This naturally increases \
the length as well as the internal
volume of the tube. The increase in ~~(~TF
length is read conveniently on a vertical . 1 1:
metre scale M, with the help of a
pointer JP, attached to the suspension
of W , and the increase in volume, from
the change in the position of the water
column in C.
Let the original length, diameter
and volume of the rubber tube be L, D
and K respectively.
Then, its area of crosssection,
A = ir(/)/2) 2 = TrD 2 l4, . . . . (/)
differentiating which, we have
JA = *dD,
whence, dA = 2A.dD/D. ... (//)
[From (/) above,
[_by eliminating TT.
Now, if corresponding to a small i7i
crease dV in the volume of the rubber
tube, the increase hi its length be dL,
and the decrease in its area of cross 
section be dA, we have
V + dV = (AdA)(L+dL). [ v "S  x '\^ h
= AL+A.dLdA.LdA.dL
Or, V+dV = V+A dLdA.L, f where ^.L V Unoriginal
' [_ volume of the tube.
neglecting dA.dL, as a very small quantity, compared with tho other
terms in the expression.
Substituting the value
from ('), above.
dV
'dL
Fig. 182.
So that, dV = A.dL  dA.L = A.dL^ dD.
Or, dividing both sides by dL, we have
d?L = A 2AL dD Or *~ d
dL D dL D dL
dD f . dV\ /2AL AD
whence ,
dL
D
2AL
dY
'dL ' 2AL
^
2L
dV D
' 2/il'
294 P&OPERTiES OF MATTER
...(in)
 =_.
' ~dL 2L A dL
lateral strain dDjD dD L
Now, Poissons ratio, a =  ~
L dD
Or, <, = . dr
Or, substituting the value of dD/dL from relation (///) above, we
have
n L D f. 1 </K\ I/, 1 rfK\
* 'D '2L I X" 'dZV ~ 2 ^ 1 X dLj*
Thus, knowing the area of crosssection (A) of the tube, the change in
its volume (dV) and the change in its length (dL), we can easily calcu
late the value of o for its material.
N.B.An identical method may be used for the determination of the
value of o for glass, but since the change in its volume is comparatively much
too small, we have to use a capillary tube, instead of an ordinary glass tube, to
measure it to an adequate degree of accuracy.
120. Resilience. By the resilience of an elastic body we understand
its capacity for resisting a blow or a mechanical shock, without acquir
ing a permanent set* and we measure it by the amount of work done
in straining the body up to the elastic limit. Let us consider it for
a uniform bar of length L and area of cross sect ion a.
We know that when the bar is subjected to a stretching force
W, so that it increases in length by /, we have
Young's modulus for the material of the bar, Y = 
WJa W_ L ^ F_
l/L a I strain '
where F denotes the stress Wja.
Now, work done per unit volume in elongation strain = stress x strain.
.. work done in producing extension I = \ (stress x strains) x volume.
F IF 2 KF* f
* i F.yXvolume (V) ~ ~2~~ ~T ' V ^ 2Y I '*' Strai " FlY '
Thus, work done, or resilience of the bar,
_ _ _
** 2Y ~~~ 2 xlfoung's modulus '
And /. resilience per unit volume of the bar
F* (stress)*
~~ 2Y "" 2x Young } s~modulus
Height from which the bar can be dropped without acquiring a per
manent set. Since resilience is a measure of the power to resist a
*The meaning attached to the word 'resilience* in our common everyday
parlance is different, viz., that the body comes back to its normal condition wheo
(be applied forces are removed.
ULASTIOITY
295
blow or shock without acquiring a permanent set, let us see from
what height the bar can be dropped without taking on a permanent
set.
This height must obviously be one in falling through which
the bar acquires energy equal to its own resilience. Let it be h. Then,
if w be the weight per unit volume of the bar, cJearly,
energy acquired by the bar in falling through height h = Vw.h.
Equating this against the resilience of the bar, therefore, we
have
Vw.h. = VF*I2Y, whence, h = F 2 l2wY.
Thus, the bar can absorb a blow or a shock due to fall from this much
height.
Proof Resilience. The maximum amount of energy per unit
volume that can be stored in a body or a piece of material, without
its acquiring a permanent set, i.e., without its undergoing a perma
nent strain, is called its proof resilience. Thus, if F m be the maximum
stress to which a material, in the form of a wire, can be subjected
i.e., if F m be its elastic limit, we have
proof resilience of the material = PffiY.
121. Effect of a suddenly applied load. Suppose we have a uniform
bar of length L and area of cross section a, suspended vertically
from one end with a collar C provided at the other
and with a weight W ', in the form of a ring, thread
ed on to it at a height h from the collar, as shown
in Fig. 183.
If we now allow the weight to fall freely so
as to hit the collar, so that the length of the bar
is increased by a small amount /, with the collar
taking up the position C", clearly, the total height
through which the weight has fallen is (h+l).
.. potential energy lost by the weight = W(h+l).
This has obviously been utilised in stretching
the bar through / and must, therefore, be equal
to the work done in so stretching it.
If F m be the maximum stress in the bar, the
resistance offered by the bar 9 or the restoring force
set up in it == F m .a.
And, since work done during stretch = \ stretch
ing force X stretch, we have [see page 285.
work done in stretching the bar = \.F m .a.l
And, therefore, W(h+l) = J F m .a.l
tensile stress F m
 = 1JL
W
CL:
Now, as we know, Y
tensile strain
Substituting this value of / in expression (j) above, we
1 F m .L
Fig. 183.
and .. / = F m .LlY.
296
OF MATTER
Or,
Solving this quadratic equation, we have
Wh
W
a
2aLIY
aL
So that, if h = 0, we have F m = or 2W/a.
Since the zero value of F m has no physical significance, we have
F m = 2W/a.
This clearly shows that when the full load W is applied to a bar
all at once, the maximum stress is 2Wja, which is clearly twice the
value of the maximum stress W\a* which is set up in the bar when
the load (W) is applied gradually to it, as for example, when the bar
is stretched in a testing machine.
In other words, the effect of a suddenly applied load is to produce
a stress double that produced by a gradually applied one.
122. Twisting Couple on a Cylinder (or Wire). If we have
a cylinder or a wire, clamped at one end, and twist it through an angle
about its axis, it is said to be under tension. Due to the elasticity
of the material of the cylinder or the wire, a restoring couple is set
up in it, equal and opposite to the twisting couple.
Consider a cylindrical rod of length / and radius r, of a material
of coefficient of rigidity n.
Let its upper end be fixed and let a couple be applied, in a plane
perpendicular to its length (with its axis coinciding with that of the
cylinder) twisting it through an angle 6 (radians).
This, incidentally, is an example of what is called a 'pure' shear,
A 2.TTX 1
i
i
i
\
\
\
\
v\
t
\
\
\
\
\
\
\
\
\
\
\
\
\
\
\
\
\
1
\
1
\
\
i
J8 7
() (6) (c)
Fig. 184.
for the twist produces a change neither in the length nor the radius
of the cylinder, the value of the twist for any crosssection of the
ELASTICITY
291
cylinder being quite independent of the direction of the couple, a re
versal of which also reverses the direction of twist.
Now, in the position of equilibrium, the twisting couple is equal
and opposite to the restoring couple. Let us calculate the value of this
couple.
Imagine the cylinder to consist of a large number of coaxial,
hollow cylinders, and consider one such hollow cylinder of radius X,
and radial thickness dx, [Fig. 184 (a)]. Each radius of the lower end
is turned through the same angle Q, but the displacement is the greatest
at the rim, decreasing as the centre is approached, where it is reduced to
zero.
Let AB, [Fig. 184 (6)], be a line, parallel to the axis, before the
cylinder is twisted. On twisting, since the point B shifts to B' ', the
line AB takes up the position AB', such that, before twisting, if this
hollow cylinder were to be cub along AB and flattened out, it will
form the rectangular plate, A BCD, but, after twisting, it takes the
shape of a parallelogram, AB'C'D, [Fig. 184 (c)]. The angle through
which this hollow cylinder is sheared is, therefore, BAB' = ^, say.
Then, clearly,
BB' = 14. Also BB' == x.e. .. </> = x.d/i [See Fig. 184 (a).
Obviously, <f> will have the maximum value where x is the
greatest, ie., the maximum strain is on the outermost part of the
cylinder, and the least, on the innermost. . In other words, the shearing
stress is not uniform all through.
Thus, although the angle of shear is the same for any one hollow
cylinder, it is different for different cylinders, being the greatest for
the outermost and the least for the innermost cylinder.
. shearing stress F
Since n ~ . .  = ~~ ,
strain or angle oj shear <f>
we have F = n.(f> = H.x.0/7.
Now, face area of this hollow cylinder = 2i:x.dx.
And /, total shearing force on this area
, n.xjf B 9 .
= 27tx.dx x , = 2irn.j~.x*.dx.
Therefore, moment of this force about the axis OO' : [Fig. 184 (&)},
of the cylinder is equal to 2vn.0.x*.dx.xll = 2irn.Q x*.dx/L
Intergrating this expression between the limits, x = and x=r,
we have
ff B
total twisting couple on the cylinder = 2nn. ~..x 8 .dx.
Znn.O [r
~ f ~l
/Jo
298
OP MATfftft
If 9 SB 1 radian, we have
twisting couple per unit twist of the cylinder (or wire) = 7fr 4 /2/.
This twisting couple, per unit twist of the wire, is also called
the torsional rigidity of the cylinder or wire.
Note. If the cylinder be a hollow one, of inner and outer radii, equal to
fa and r t respectively, we have
twisting couple on the cylinder
2rr/i. y.,
123. Variation of stress in a twisted cylinder (or wire). Let us
again imagine a cylinder or wire, of length / and radius n, to consist of a
large number of coaxial, hollow cylinders and
consider both a cylinder of radius x and the outer
most cylinder of radius r, (Fig. 185), in which the
lines AB and CD respectively are parallel to
the axis OO', before the cylinder is twisted, and
shift into the positions AB' and CD' after it
hag been twisted, as explained above. Then,
clearly,
LBOE' = DOD' = 0,
and the angles through which the two cylinders are
sheared are BAB' = <f> and DCD' = <j> m respective
ly, where <f> = BB'/l and <j> m = DD'/l, this latter
being the strain on the surface of the cylinder and,
therefore, the maximum on it. [v DD' = r&.
Since OB = OB' = x and
Fig. 185.
So that,
whence,
OD = OD' = r, we have =
DD' ^ r$ r_ BB' _ r
/""""/^/'x^x
x
9 ms  r ?
57* = Z)PVr.
J?^ 7 r
Or, 5/ra/n fn the cylinder or wire at distance xfrom the axis
x
X maximum strain.
Now,
n =s=
radius of cylinder or wire
and, therefore, F = n$ == n,
where F is the shearing stress at distance jc from the axis.
And, since c/> m is the maximum strain, we have n.^ w = F w , the maxl*
mum shearing stress on the wire t i.e. , on its surface.
We, therefore, have F= .F m ,
Or, shearing stress at distance xfrom the axis
~ maximum stress.
299
In other words, both shearing strain and stress, go on increasing
as we proceed away from the axis of the cylinder or wire and acquire
their maximum values on its surface.
Let us now see whether there is any variation of shearing strain
and stress along the length of the cylinder or wire also.
Let us, therefore, consider the stress in the plane EO"E' of the
cylinder at a distance a/ from its upper fixed end, where a>0 and
<1. Here, clearly, ^EO"E f between the radii O"E and 0"'=a0.
= EE'jr and .. EE' = r.ad. [v O"E = O"E'=r.
.*. shear strain on the surface of the wire, in this plane
EE'lCE = r.aO/al = rOll.
Now,
= ~.<f> m > as we have seen above.
So that, shear strain on the surface of the wire in plane
r I , ,
i.e., the same as in plane DOD', as discussed above. Clearly, therefore,
shearing stress in this plane is also the same as in plane DOD',
namely, n<t> m = F m .
Thus, we see that the shearing stress at a point in a cylinder, or
a wire, depends only on the distance of the point from the axis, and not
its vertical distance from either end, of the cylinder or the wire.
124. Strain energy in a twisted cylinder (or wire). Let C be the
couple applied to the lower end of a cylinder of length / and radius r,
with its upper end fixed and y
let B be the angle of twist pro
duced at the former (i.e.,
lower) end. Then, if the limit
of elasticity is not exceeded,
the relation between C and Q
is a linear one and we obtain a
straight line graph OP between
the two, as shown in Fig 186.
So that, for a small in
crease bC in the value of the
couple, the increase in the
angle of twist is dft, and the
work done on the cylinder, or the
energy stored up in it, is, there
fore, C.dQ, where C is the
average value of the couple.
o
dO B m j
ANUE OF TWIST >
Fig. 186.
This is represented by the area of the shaded strip in the Figure.
And, therefore, the total work done on the wire, or the total energy
stored up in it for the maximum twist O m (represented by CM), to
which its lower end is subjected, is represented by the whole area
OAP.
r*i
This strain energy is obviously equal to E = I C.d0.
Now, for a twist 6 in the wire is, as we kow f equal to m8r*fllt
300 p&ottSKTifls of
And, therefore, E  JJ" ". 9 .d 9 
_ _. ___^_._, ^ mm ,
where itnr*0 m /2l is tha value of the maximum couple C m , correspond
ing to the maximum twist m ; i.e., C m ~ nnr*0 m l2l.
Or, substituting from this the value of Q m in the expression for E
above, we have
strain energy in the twisted cylinder, E = C m O m =
As will be readily seen, this is half the energy (C m .Q m ) that
would be stored up in the cylinder if the stress in it were to have
the same value throughout, equal to its maximum value on the
surface of the wire, which, as we know, is not actually the case,
the stress increasing from zero at the axis to a maximum on its surface,
(see 122 above).
125. Alternative expression for strain energy in terms of stress.
We know that strain energy per unit volume
Tp IT*
= } stressx strain = \.F.<f> = J F. n =  . ... (/)
Now, if we consider an element of the cylinder or wire, defined by
radii x and x\dx y the stress will, as we have seen in 123, be constant
at all points in it and its value will be x.F m fr, where F m is the maxi
mum value of the stress in the wire on its surface ; i.e., F = x.F m /r.
Since the volume of the cylindrical element we are considering
is 2nx.l.dx, we have, from relation (/) above, energy of the cylindrical
element
_ x f /
~ rfaa * I*"' V
Or, dE = x*.dx.
* r< 7T,ljT tri 0r '*'* ' '*'**?
And /. E = ^ I x*.dx = ;* * r = v.r.F,
 . ~
nr* Jr nr* 4 4rc
.
m
If <f> m bs tho maximum shear strain corresponding to the maxi
mum stress F m , we have F m = n<f> /n == n.r.Bjl, (see Fig. 123), where
O m is the angle of twist for the maximum value of the couple.
1 that, E = ^ TJ SHE j
i.e., the strain energy is again half the value it would have if all the
elements of the cylinder (or wire) were subjected to the same
maximum stress F m .
126. Torsional Pendulum. A heavy cylindrical rod or disc,
suspended from one end of a fine wire, (attached to its centre), whose
ELASTICITY
301
upper end is fixed, constitutes what is called a torsional pendulum ,
(Fig. 187). The rod or disc is turned in its own (i.e., in the horizontal)
plane to twist the wire, so that, on being released,
it executes torsional vibrations about the wire as axis.
Let 6 be the angle through which the wire is
twisted. Then, the restoring couple set up in it is equal
tO 7T.W 4 .0/2/ = C,0,
where ?ntr*/2/ is the twisting couple per unit (radian) twist
of the wire, usually denoted by the letter C.
This produces an angular acceleration dco/df, in
the rod or the disc.
,\ If 7 be the moment of inertia of the rod
about the wire, we have
Ldw/dt = C.O Or, da>!dt == C.0/7,
i.e., the angular acceleration (da)[dt) t of the disc or the rod is proportional
to its angular displacement (0), and, therefore, its motion is simple
harmonic, Hence, its timeperiod is given by
Fig. 187.
Or,
V
c '
__
moment of inertia of the disc or rod about the wire
restoring couple per unit twist of the wire
ill. Determination of the Coefficient of Rigidity (n) for a
Wire.
(1) Statical Method. This method is based on a direct appli
cation of the expression for the twisting couple on a wire deduced
in 122. There are two different types of apparatus used for the
purpose, according as the specimen under test is a rod or a wire. We
shall now consider these in detail.
(a) Horizontal Twisting apparatus for a Rod. Here, a couple^
which can be measured directly, is applied to a horizontal rod and
equated against the
expression for the
torsional or twisting
couple, 7Tr 4 0/2/,
whence the value of
n for the rod can be
easily calculated.
The arrange
ment of the appara
tus is as shown in
Fig. 188, where one
end of the rod, under
_^ test, about 50 cms.
^^^^^^^ \]r i* 1 length and of
^^Z^^* radius about '25 cm. t
is secured firmly
to a block B lt with
its other end atta
ched to a steel axle
of a large pulley J? ? .
Fig. 188.
302
PROPERTIES OT MA1TER
A cord is wound round the pulley and has a mass M suspended
from its lower free end. Thus, a couple acts on the rod, tending to
twist it about its own axis.
Two pointers p l and /? 2 , are clamped on to the rod, at two
points, a known distance / apart, so as to move freely over the circu
lar scales Sj and S 2 , graduated in degrees, on which the twist produced
in the rod at those two points can be read directly.
Now, if R be the radius of the pulley, the couple acting on the
rod, due to the suspended mass (M) is, clearly, equal to Mg.R.
This couple is balanced by the couple due to the torsional re
action of the rod, equal to mrr* (0 a 0j)/2/, where r is the radius of
the rod and 1 and 2 , the angles of twist (in radians*) produced at the
two chosen points, as indicated by the two pointers.
So that, /mr 4 ( 2 0,)/2/ = Mg.R. Or, n = ~JL
whence, the value of n for the material of the rod can be easily deter
mined.
The apparatus, though quite simple in manipulation, suffers
from two serious drawbacks, viz.,
(i) there being one single pointer moving over the circular scale,
an error is caused due to eccentricity of the axis of the rod with respect
to it ,'
(ii) there being just one pulley, only one single force is applied to
the end of the rod, attached to it, thus exerting a sidepull on it. This
results in friction between the rod and the bearings, thus appreciably
hindering the rod from twisting freely.
(b) Vertical Twisting apparatus for a Wire. This was designed
by Barton, and here also a couple, which is measured directly, is
applied to the lower end of the vertically
suspended wire, and the twist produced in
it is noted. Then, equating this couple
against the expression H7rr 4 0/2/ for it, the
value of n for the wire can be easily calcu
lated.
The wire W, whose coefficient of rigi
dity is to bo determined, is clamped at its
upper end T, (Fig. 189), and has a heavy
cylinder C attached to it, at its lower end.
Two pieces of cords are wound round
the cylinder and, leaving it tangentially at
either end, pass over two frictionless pulleys,
as shown, with equal masses M and M,
suspended from their free ends.
The couple* formed by two masses^
rotates the cylinder about the wire as axis,
and thus twists the wire through an angle
M
Fig. 189.
*To convert degrees
multiply by */}80 f be^apc J80
ELASTICITY 303
(radians), say, which is read directly on the horizontal circular
scale S by the movement over it of the pointer, attached to the wire,
or by the more sensitive lamp and scale method, with the help of a
telescope.
Then, if d be the diameter of the cylinder, we have
twisting couple applied to the wire = Mg.d.
But, the twisting couple for a twist of radians is also
where r is the radius of the wire, /, its length and n, the coefficient
of rigidity for its material.
Clearly, therefore, 7rr^/2/ = Mg.d, whence, n
from which n for the material of the wire can be easily obtained.
The two sources of error present in the first method are elimi
nated here. For, (/) due to the very nature of the arrangement of
the pointer and the scale, the error due to eccentricity of the axis of
the wire does not arise and (//) due to the use of two pulleys, the
sidepull on the wire is also avoided.
N B. It will be noted that the weak point in the above two methods is
the radius r of the rod or or wire, the fourth power of which occurs in the expres
sion for n. It must, therefore, be measured most carefully.
(2) Dynamical Method Maxwell's Vibrating Needle. The
dynamical method of determining n for the material of a wire
consists in determining, by direct observation, the time period t of
a body, like a disc or a rod, suspended from the wire and executing
torsional vibrations about the wire as axis, i.e., of a torsional
pendulum.
Then, since t = 27T\///C', where 7 is the moment of inertia of
the body about the wire, and (7, the couple per unit (radian) twist of
the wire, we can easily obtain from it the value of C. Equating
this against the expression ?jw 4 /2/ for it, the coefficient of rigidity (n)
for the wire can be easily calculated.
It is not, however, easy to determine the moment of inertia (/)
of the body accurately. Maxwell, therefore, devised a method in
which the necessity of determining it was altogether obviated. Let
us study his method in detail.
Maxwell's Vibrating Needle Method. A hollow tube or cylinder,
open at both ends, is rigidly fastened in the middle to the wire,
the coefficient of rigidity of the material of which is to be determined,
and w}iich is suspended vertically from a support, and has a
small piece of mirror attached to it, as shown, (Fig. 190), to enable
the vibrations of the tube to be observed by the telescope and scale
method.
Two hollow and two solid metal cylinders, of equal lengths and
diameters, can be fitted into the tube such that, put ei4 to eucj,
just fid it completely.
304
PROPERTIES MATTER
The solid cylinders are first put into the inner positions and
the hollow ones in the outer positions, as shown in Fig. 190 (0), and the
tube, so loaded, is then
given a torsional vibra
tion, and its time
period determined. Let
it be t v Then,
fr
)Af
H
/v*.
V/s
H
H
H
where C is the twisting
couple per unit deflection
or twist of the wire and
is equal to ni:r 4 l'2l, and
(tv w I lf the moment of inertia
Fi s 19  of the loaded tube about
the suspension wire as axis.
The solid and hollow cylinders are then interchanged in posi
tion, i.e., the hollow cylinders are now put in the inner positions and
the solid ones in the outer positions, as shown in Fig. 19 ) (ft), and the
timeperioi of the torsional vibration of the loaded tube determined
again. Let it be t 2 .
Then,
t t = 2:: y 7/C,
... (ii)
where 7 2 is now the M.I. of the loaded tube about the suspension
wire.
Squaring and subtracting (/) from (//), we have
C ' {
.. (Hi)
Now, let the mass of each hollow cylinder be m { and that of
each solid cylinder, w 2 , and let the length of the hollow tube be
20, so that the length of each solid oj hollow cylinder is 20/4, or
a/2.
Clearly, then, the centres of mass of the inner and outer cylin
ders are at distances 0/4 and 30/4 respectively from the axis of
oscillation.
Therefore, the change from the first adjustment, when the solid
cylinders occupy the inner positions to the second adjustment, when they
occupy the outer positions, consists in transferring an extra or excess
mass (m^m^ from a distance 0/4 to a distance 30/4 from the axis
of oscillation, on either side of it. The moment of inertia of the
loaded tube, therefore, increases, and, by the application of the
principle of parallel axes, we have
Here, we multiply the mass by 2, because the change takes
place oj} both t^e
ELASTICITY 305
Or, /,  Ma^md x  
/.<?., 7 2 = / 1 4( W2 wj.fl 1 And .. (7 2 /,) = (w 2 w^.a 1 .*
Substituting this value of (/a/!) in relation (///) above, we
have
47T 3
(tfV) =  c .(mim^a*.
Further, substituting the value of C, we have
47T 2 , % . 47
^ , , ox ?r..a , % , 7r... 2 ,
Or, U 2 f 'i 2 ) = n r *  faaflM, whence, n = _* i_i/.
Thus, knowing /, a, Wj, w t , f 2 , fj and r, the value of n for the
material of the given wire can be easily determined.
NB The vHue of n obtained by the dynamical method is slightly
higher than that obtained by the statical method, because, in most cases, the
twist produced by a torsional couple depends, to some slight extent, upon the
time for which the couple is applied and so, in the dynamical method, where
the time of vibration is rather short, the twist (o) is smaller for the same value
of the couple than in the statical method.
Further, since wires are made by squeezing the molten metal through
holes, (as in a sieve), their outer layers are invariably tougher than the inner
ones, and hence the value of n for a thinner wire needs must be higher than for
a thicker wire of the same material.
128. (a) Determination of Moment of Inertia with the help of
a Torsional Pendulum. The moment of inertia of a body of a regular
geometrical shape can be easily calculated from its mass and dimensions.
But, if it be of an irregular shape, it is not possible to do so. In
either case, however, it may be determined by using a torsional
pendulum with a disc or a rod of known moment of inertia / about
the suspension wire and noting the time period (/) for its torsional
vibration. Then, mounting on it the given body, such that the axis
of the moment of inertia of the, two together is again the snme wire,
the timeperiod (t^) for the torsional vibration of the combination is
determined. Then, if / t be the moment of inertia of the body about
the wire as axis, the moment of inertia of the combination, in the
second case, is clearly equal to /+/,. So that, if C be the torsional
couple per unit twist of the suspension wire, we have
/ =* 27TV///C ... (/) and t l = Zic^T+FJC. ... (ff)
.. squaring and dividing relation (//) by (/), we have
,2
. *Or, this may easily be deduced as follows :
If / be the moment of inertia of the hollow tube about the suspension
wire, and h and /*, those of the solid and the hollow cylinders about the vertical
axes through their respective centres of mass, we have, by the principle of paralle
axes,
A  /+2[/,fm 8 .(fl/4) 2 ]+2[/ A fm 1 .(3a/4n,
and /i * /f 2[/* +/M0/4)'] + 2[/ f + nt r (3a/4) 2 ].
So that, (/i/i) (/Wt
30ft PBOFEBTIBS OF AlATTKB
, t fSubtracting the denomi
So that, f = ^ * nator from the numera
Mor on either side.
Or,  f whence, / a
Thus, knowing /, t and f,, we can easily calculate I lf the
moment of inertia of the given body.
(b) Comparison of Moments of Inertia. If, however, it is simply
desired to compare the moments of inertia of two bodies, we first
use one and then the other, as the disc or rod of the torsional pendulum,
and determine the timeperiods t i and t 2 respectively for their
torsional vibration about the wire as axis. Then, if /j and /g, be
their respective moments of inertia about this axis and C, the torsional
couple per unit twist of the wire, we have
r, =2irVA/C and f 2 = Zn^TjC
So that, squaring and dividing one by the other, we have
f
and thus, knowing t l and t z> the moments of inertia of the two bodies
may be easily compared.
Note. In the above cases, the amplitude of vibration need not be small,
because it is found that the restoring couple continues to be proportional to the
twist B in the wire, up to fairly large values of 0. The assumption made, however,
that even with different bodies suspended from the wire, resulting in a change
in its longitudinal tension, the value of C (or the twisting couple per unit twist of
the wire) remains the same is found to be only approximately true.
129. Bending of Beams Bending Moment. We must first be
clear about the terms, beam and bending moment.
Beam. A beam is a rod of uniform crosssection, circular or
rectangular, whose length is very great compared with its thickness,
so that the shearing stresses over any section are small and may be
neglected.
Bending Moment. When a beam is fixed at one end and loaded
at the other, it bends due to the moment of the load, the plane of
bending* being the same as that of the couple applied. Restoring forces
are called into play by this deformation of the beam and, in the equi
librium state, the restoring or resisting couple is equal and opposite to the
bending couple, both being in the plane of bending.
Irrespective of the manner in which the beam is bent by the
couple applied, its filaments on the inner or the concave side get
shortened or compressed, and those
on the outer or the convex side get
lengthened or extended, as shown
in Fig. 191. Along a section, in
between these two portions, there
is a layer or surface in which the
191. filaments are neither compressed
*In the case of uniform bending, the longitudinal filaments all get bent
into circular arcs in planes parallel to the plane of symmetry, which is then
known as the plane of bending. And, the straight line, perpendicular to this plane
on which lie the centres of curvature of all these bent filaments, is called the a*ii
of bending.
ELASTICITY
307
nor extended. This surface is called the neutral surface and its ste
tion (EF) by the plane of bending which is perpendicular to it U
called the neutral axis.
In the unstrained condition of the beam, the neutral surface
becomes a plane surface, and the filament of this unstrained or un
st retched layer or surface, lying in
the plane of symmetry of the bent
beam, is referred to as the neutral
filament. It passes through the e.g.
(or the centroid) of every transverse
section of the beam.
The change in length of any
filament is proportional to its distance
from the neutral surface.
Let a small part of the beam be
bent, as shown in Fig. 192, in the
form of a circular arc, subtending an
angle at the centre of curvature O.
Let R be the radius of curvature of
this part of the neutral axis, and let
a'b' be an clement at a distance z
from the neutral axis.
Then, a'b' = (R+z).0,
and its original length db == RQ.
.*. increase in length of the filament = a'b' ab.
= (R+z).g R.O = z.e.
And, since the original length of the filament = R.Q, we have
strain = z.e j R.O = z/R,
i.e., the strain is proportional to the distance from the neutral axis.
Since there are no shearing stresses, nor any change of volume,
the contractions and extensions oj the filaments are purely due to forces
acting along the length of the filaments.
If PQRS (Fig. 193), be a section of the beam* at right angles
to its length and the plane of bending, then, clearly, the forces acting
p f * on the filaments 'are perpendicular to
t kj g section, and the line AfW lies on the
neutral surface.
Let the breadth of the section be
PQ = 6, and its depth, QR = d.
The forces producing elongations
and contractions in filaments act perpen
dicularly to the upper and the lower
halves, PQNM and MNRS respectively,
of the rectangular section PQRS, their directions being opposite to
each other.
l
Af
JV
S
a,
Fig. 193.
R
*The section is shown rectangular purely for the sake of convenience.
308 PROPERTIES Off MATTEH
Consider a small area Sa about a point A, distant 2 from the neu
tral surface. The strain produced in a filament passing through this
area will be z/R, (see above).
Now, F = stress j strain and /. stress = Yx strain.
Therefore, stress about the point A = Y xz/R* where Y is the
value of Young's Modulus for the material of the beam.
And, there fore, force on the area Sa = Sa.Y.z/R
and, moment of I his force about the line MN = Y.zx$axz/R.
= Y.Sa.z*/R.
Since the moments of the forces acting on both the upper and
the lower halves of the section aro in the same direction, the total
moment of the forces acting on the filaments in the section PQRS is
given by
Now, a.z* is the geometrical moment of inertia (I )* of the sec
tion about MN, arid, therefore, equal to ak 2 , where a is the whole
area of the surface PQRS and k, its radius of gyration about MN.
Y YI
Hence, the moment of the forces about MN = D >ak* = * .
J\ i\
This, then, balances the couple of mo nent M, say, called the
bending moment, acting on the beam due to the load, when the beam
is in equilibrium ; for, there is no resultant force acting on the area
PQRS, and the resultant moment about EF, perpendicular to MN,
is also zero. In other words, it is the moment of the stress set up in
the beam or the moment of resistance to bending, as it is usually called
in engineering practice, and is also of the nature of a couple, for
only a couple can balance a couple. Obviously, it acts in the plane of
bending and is equal to the bending moment at the section due to the
load, though, quite frequently, (but, not strictly correctly) it is itself
referred to as the bending moment. This forms tho very basis of the
theory f regarding the bending of beams and is, therefore, a relation
of fundamental importance.
*lt is so called because it is proportional to the mechanical moment of
inertia of a plane lamina of the same shape as the crosssection. It is denoted,
here, by the symbol / 7 , so that, the student may not confuse it with the ordinary
mechanical moment of inertia, denoted by /.
fine theory is subject to the limitations mentioned in 131, (page 313),
which the student would do well to keep in mind.
Imagine the section as a rectangular plate of unit mass per unit area,
(Fig. 194).
Then, area of the strip AB, of length b and breadth dz> is equal to b.dz.
And, therefore, its mass = b.dz.l b.dz.
p A Q Hence, geometric moment of inertia of
the strip about MN b.dz z*, and, therefore,
moment of inertia of the whole plate or sec
tion about MN
I*
m
26 ^
S R "124J 12
Fig. 194.
ELASTICITY
The quantity YJ = Y.ak* is called the flexural rigidity of th
beam.
/. bending moment = (Y/R)x geometric moment of inertia of the
section.
= flexural rigidity / R,
whatever the shape of the crosssection of the beam.
For a rectangular crosssection, a = bxd, and fc 2 = d 2 /12.
Hence, bending moment for a rectangular crosssection = Y.b.d*ll2R.
For a circular section, a = 7rr a and A 2 = r 2 /4.
I 9 = 0A2 = 7rr 4 /4,
/.., the same as the moment of inertia of a disc about a diameter.
.*. bending moment for a circular crosssection = 7.irr 4 /4/?.
Note. We have seen above how strain in a beam is proportional to the
distance z from Us neutral axis, and is equal to z\R, where R is the radius of
curvature of the poition of the ncutial axis under consideration. So that, if F
be the stress cor res ponding to the strain z//?, we have
F _ F Y
>*/*' r ' z ~R '
If, therefore, F lf F 2 e,c. be the values of stress at distances z lt z 2 . from the
neutral axis, we have
And .*. bending moment M = y / ff //?.
ra ^L / ^ 2 / ^tr Jfr F f "~ F etc
. /(/  ./ 7 etc. ./i r a CIL.
^1 2 2 I 2 2
 Zi F, * Z 2 F 2 etc.,
where Zt /^/^ and Z a =/ ff /z 2 are called the moduli of the section under consi
deration.
~, u , .  Ai geometrical moment of inertia
Thus, modulus of a section = ^. .  r  ,. . ~i ;
distance from the neutral axis
Now, in the case of a flat bar or beam, of rectangular cross
section, if the bending be small, there is brought about a change in
the shape of th3 section, such that all lines in it, originally perpendi
cular to the plane of bending, get bent into arcs, which are all con
centric and convex to the axis of bending. In other words, the layer of
the beam, which was originally plane and perpendicular to the plane
of bending, and which contained the neutral filament, now gets changed
into what is called an anticlastic surface (Fig. 195), of radius / in
the plane of bending (which, here, coincides with the plane of the
paper), ^nd, of radius R f in the plane perpendicular to it. the two
centres of curvature lying on either side of the beam. This is uhat
is to be expected, because a transverse bending must, of necessity,
be associated with a longitudinal bending of the beam, with the cur
vature of the former opposite to that of the latter. For, the filaments
above the neutral axis, which get extended, must obviously suffer a
lateral contraction a times as great and, similarly, the filaments below
the neutral axis, which get compressed, must suffer a lateral extension.
310
Thus, by way of illustration, if a rectangular piece of India
rubber Le bent longitudinally in the form of an arc, it takes up the
form shown in Fig. 196, with its longitudinal
fibres bent so as to be concave with respect to
\A a P^ below, and the transverse fibres, so
as to be concave with respect to a point
above, the rubber piece, in the case shown. It is
this bending, which occurs in a plane normal to
the longitudinal plane, that gives the rubber
piece (or the beam) an anticlastic curvature.
Fig. 195.
And, therefore, as we have seen before, (page 307), the longi
tudinal and lateral strains in a filament, distant z from the neutral axis
will be given by zjR and zjR' res
pectively. So that, Poisson's ratio a, for
the material of the beam, is given by
the expression
lateral strain __ zJR^ _. R Fig. 196.
~~ longitudinal strain zjR R''
This, then, gives us a method for the determination of cr for the
material of a given boam or bar, the two radii being determined
directly by attaching suitable pointers to the rod and noting the
distances and angles traversed by them, when a known couple is
applied to the beam.
130. The Cantilever. A cantilever is a beam fixed horizontally
at one end and loaded at the other.
(i) Cantilever loaded at the free end. Here, two cases arise, viz.,
(a) when the weight of the beam itself produces no bending, and (bj
when it does so. Let us consider both the cases.
(a) When the weight of the beam is ineffective. Let AB, (Fig. 197)
represent the neutral axis of a cantilever, of length L fixed at the end
A, and loaded at B with a
weight W, such that the end B is
deflected or depressed into the
position B' and the neutral axis
takes up the position AB, it
being assumed that the weight
of the beam itself produces no
bending.
Consider a section / of the
beam at a distance x from the
fixed end A.
The moment of the exter
nal couple at this section, due to
W
Fig. 197.
the load W 9 or the bending moment acting on it
311
Since the beam is in equilibrium, this must be equal to
YI^R = Y.ak*IR, where R is the radius of curvature of the neutral
axis at P.
Therefore, W.(Lx) = F./,/^ = 7.^ 2 /^ (0
Since the moment of the load increases as we proceed towards
the fixed end A, the radius of curvature is different at different points
and decreases as we approach the point A. For a point Q, however,
at a small distance dx from P, it is practically the same as at P.
So that,
PQ = R.d6. Or, dx = R.d&, iwhere do is the L POQ
whence, R = dxjdO*
Substituting the value of R in (/) above, we have
* 
Draw tangents to the neutral axis at P and Q, meeting the
vertical line through BE' in G and D respectively. Then, the angle
subtended by them is also equal to d6 t the radii at P and Q being
perpendicular to the tangents there.
Now, clearly, the depression of Q below P is equal to CD, equal
to dy, say.
rru i IT ^ so (LX)W.(LX).dX
Then, dy = (Lx).dO = ~  jTak*~~ * fFrom ( ' 7) above>
JV(Lx)*.dx
" Y.ak* ....... "' V ' ;
Therefore, the depression y = /?' of the loaded end B below
the fixed end yl, is obtained by integrating the expression for dy
between the limits, x = and x = /.
"377"* [Putting back I g for oA: 1 .
Thus, the free end of the cantilever is depressed by
_
ZY.ak* "" "377, '
(b) When the weight of the beam is elective. In this case, in
addition to the weight W at B, the weight of the portion (Lx) of
the beam is also acting at the midpoint or the e.g. of this portion ; so
that, if w be weight per unit length of the beam, a weight w(Lx)
is acting at a distance (L x)/2 from the section PQ. And, therefore,
*See solved example 4, page 332, where it is shown that l/R d*y/dx*, the
rate of change of slope. A mathematical minded student will find the solution
given there with this value R. much neater and also perhaps a trifle easier.
312 PBOPEBTIES OF MATTEB
<
the total bending moment on the beam
w
Since the beam is in equilibrium, this must be equal to YLJR or
And,... rffl== _
Then, dy = (Lx).dd ==
And ... y=
Now >v.L = W^j, say, the weight of the beam.
WL* WJ2
so that, y+ 8 ; 7/
Or, ' "' ' 3
i.e., the beam now behaves as though it is loaded at the end B with
a weight W plus 3!8ths of the weight of the beam.
(ii) Cantilever loaded uniformly. Let the uniform load on the
cantilever be w per unit length. Then, the weight of the portion
of the beam (Lx), ie., \v(L x) alone produces a bending moment
about the section PQ, there being no weight suspended from the end
B. And, since this weight w(Lx) acts at a distance (L x)/2 from
the section PQ, we have
ding moment due to it = w(Lx).(Lx)l%.
For equilibrium of the beam, therefore,
J*
whence, d0 = _..
Substituting this value of dd, in the relation dy=z(Lx)d9 t
?e have
Clearly, w.L = FF, the total load on the beam
So that,
BLASTIOITB 818
WL*
'817,'
It will be seen that this expression could be obtained directly
from the result in case (/) 6, above, by putting W ', the load at the end
B equal to zero.
131. Limitations of the simple theory of bending. In discussing
the above simple cases of bending, we have tacitly made the following
assumptions :
(i) That the crosssection of the beam remains unaltered during
bending. This, as we have seen, is not strictly true ; for, the exten
sion of the filaments above the neutral surface brings about their
lateral contraction, a times as great, and the contraction of the
filaments below the neutral surface brings about their lateral exten
sion. So that, the crosssection of a rectangular beam, bent so as to be
concave downwards along its length is convex downwards across its
length. Similarly, a circular crosssection may change into an oval
form. This change in the shape of cross section of the beam, due to
bending, results in a change in the value of the crosssection and
hence in that of I g for it. Usually, however, it is much too small to be
of any practical consequence and may safoly be ignored.
(/"/') That the radius of curvature of the bent beam, or rather that
of its neutral surface, is large compared with its thickness. This is
almost always true for all cases of elastic bending.
(///) That the minimum deflection of the beam is small compared
with its length. This, while more or less true for ordinary engineering
problems, is not strictly so in quite a number of cases. Thus, for
example, in the case of a clock spring, the deflection produced is very
large evon within the elastic limit. We shall, therefore, do well to
discuss this particular case here, as representing the more general case
of strongly bent beams.
132. Strongly bent beams When a beam bonds very strongly, its
inclination to its original, unbent or unstrained position, and hence
the tangent of this inclination, is no longer small. Consequently, its
curvature (l/R) can no longer be taken to be equal to the rate of
change of slope, d*yl<Jx 2 , as is done in the ordinary cases of small
curvatures (see solved example 4, page 332) but it is now given by
The problem thus bee >mes quite complicated in many a case, with
the differential equations obtained not being amenable to easy
solution.
However, there are also some cases which can be investigated in a
much simpler manner and we shall here consider only one of these, v/3.,
that of & flexible cantilever, like a clockspring, clamped at one point
and loaded at its free end. As we pass or 'pay out' more and more of
the spring through the clamp, keeping the load constant, its free end
drops further and further down, as a result of the large amount of
bending, until finally it becomes quite vertical. The horizontal distance
between the clamp and the loadod eud of the spring is now the
maximum and any more of the spring 'paid out' through the clamp
merely hangs vertically.
314
*BO* BRIMS Of MATTBB
Thus, let OAB, (Fig.198), be the bent position of the spring,
changed at O and loaded with a weight W at the end B 9 such that
^ ne tangent at A, (coordinate x = x),
makes an angle #, and that at B
(coordinate x = tf), an angle <, with
the axis of x. Then, the bending
moment at A is clearly equal to
W(ax). So that,
x) = YJ g . 1 R ,
where ^ is the radius of curvature of
portion OA of the spring.
'Now,  =
d6
Fig. 198.
And
r~Sec solved
Kxample 4,
Lpage 332.
de
ds'
YI 
YJ '
*'
dx
dx
ds
Y.I a . cos 6 ~.
['. dxjds = cos 6.
k
Or, W(ax).dx== Y.I g .cos Q.dB.
.*. integrating this expression between the limits x and
X = a, the horizontal distance of the loaded end from 0, we have
[
W(ax).dx =
cos 6.dQ.
Or,
Or,
~ = Y.I ff .sin6
Now, when the loaded end becomes vertical, <f> = 90 and the
horizontal distance a becomes the maximum, say, a m . So that, sub
stituting a = a m and <j> = 90 in expression / above, therefore, we
have
H>fl w 2 /2 5= Y.* g> whence, a m * =
Or, a m = ^/ZtQW. ...(II)
It will be seen that the value of Y for the material of the
flexible beam can easily be determined from either of the relations /
or // above, if we know the angle of inclination $ of the loaded end
of the beam with the horizontal, or the maximum horizontal dis
tance, (i.e., a or a m ) of the loaded end from the clamp.
133. Transverse Vibrations of a Loaded Cantilever. If the
loaded free end of a cantilever be depressed a little and then released,
it starts moving up and down its original position, i.e., executes
transverse vibrations. Let us calculate the timeperiod of these
vibrations.
Sift
We have seen above, ( 130, pages 311), how
WL* , TT7
y = srrv , whence, W =
which gives the load or the force W \ required to maintain the canti
lever in equilibrium, with its free end depressed or displaced through
y. This, thus, also measures the elastic reaction of the cantilever,
which is oppositely directed to it.
Now, if a = d z yjdt* be the acceleration of the mass M suspended
from the free end of the cantilever, (i.e., if M be the mass of the load
W), the force of inertial reaction on it is equal to M .a.
Hence, since the cantilever is in equilibrium and there is np
other external force acting on it, we have
.. 3.YyI g , 3.37
M.a =  i' 9 whence, a=
Or, a = u.y,
where 3 YI g [ML* = fi 9 a constant for the given cantilever, with the
given load.
Thus, a oc y,
i.e., the acceleration of the mass (or of the free end of the cantilever) is
directly proportional to its displacement. It is thus a case of simple
harmonic motion, and its timeperiod is, therefore, given by
t = 27T A / JL = 27T A / _ X
V ft yZYIJML
As can be seen at once, this relation for t gives a good dynami
cal method for the determination of the value of Young's modulus
( Y) for the material of a given beam or rod. It is particularly suit
able for beams like a metre stick etc.
134 Depression of a Beam supported at the ends.
(/) \* hen the beam is loaded at the centre. Let a beam be
supported on two knife edges at its two ends A and B, as
shown in Fig. 199, and let it be \y/ w/
loaded in the middle at C with a A 2 " * 2
weight W.
The reaction at each knife
edge will clearly be W/2, in the up'
ward direction. Pig. 199.
Since the middle part of the beam is horizontal, the beam may
be considered as equivalent to two inverted cantilevers, fixed at C, the
bending being produced by thu loads Wfi, acting upwards, at A
and B.
If, therefore, L be the length of the beam AB, the length of
each cantilever (AC and BC) is L/2 f and the elevation of A or B
316
PROPERTIES F MATTER
above C or, what IB the same thing, the depression of C below A and
B is given by
.
whence,
y 
4
.
[v
.
V
If the beam be of a circular crosssection, we have ak z = 7rr 4 /4,
where r is the radius of the crosssection ; so that, for such a beam,
WL* 4 H/L 8
......... CO
And, if the beam be of a rectangular cross section, of breadth
b and depth d y we have ak 2 = bd*/l2, and, therefore, for such a beam,
(//) When the heara is loaded uniformly. Here, let w be the load
per unit length of the beam, so that the total u eight < r icting down
** wards at tho c g, is M'L = IV,
wliere L i,s the length of the beam.
The reaction at each knifeedge is
thus obviously J ivL, acting up*
wards* (Fig 2i)(>) with the beam be
having as a s} T stem of two canti
s, fixed at"C.
( 1 tnsidoi'ing anain, a section
PQ distance .v from the midpoint
W(lX)
W*urL
pi 200
C of the beam and taking halflength of the beam equal to /, the
weight of the portion (/Jc) of the beam, /., a weight w(lx) acts
downwards at a distance (/ ,v)/J from tho section PQ.
Thus, the bending momjnt about the section
For equilibrium, therefore, this must be equal to the moment
of the resistance to bending viz.. Y I g iR, where R is the radius of
curvature of the neutral axis at PQ.
i.e.,
wl.ence,
Y IJR = y./ f . d ,
C/ J*
/. /
[_ _
'JO. page
j 1 U.
A ,,
And hence
f/
y =
Jo
f' Fr f. C a 
ELASTICITY
317

2 3 _
wL
' YL
wL
12
1 w
.,2YI..
Now, substituting L/2 for /, we have
w (L* L*\ M'
y YL \ 48 '1287
J*
6
877,,
_
Yf "384
_
YT 384
But wL=W t the total weight on the beam.
Hence
Determination of F by bending of a beam. It will be easily seen
that if we measure the depression (y) of a beam of known dimen
sions, supported at the ends and loaded at the centre, as in case (/)
above, we can easily determine the value of Y for its material, by
applying relation (/) or (//), as the case may be. In practice, it IB
convenient to use a beam of rectangular cross section ; so that, know
ing W, L, b } d and y t Y can be easily calculated from relation (//)
above.
The arrangement is as shown in Figs. 201 (a) and (b). The beam
is supported horizontally and symmetrically, on two parallel knife
Fig. 201.
edges, a known distance L apart and the load is applied by placing
weights in a scale pan, also supported on a knifeedge, midway bet
ween them, as shown. The depi ession y of the midpoint, thus pro
duced, is noted directly with the help of a micrometer screw, [Fig.
201 (0)], or, more accurately, with the help of a microscope, the eye
. piece of which is fitted with crossfires, [Fig. 201 (b)].
Readings are taken, first with the load increasing, in equal steps,
and then with the load decreasing, in the same equal steps, and their
mean taken. This gives y. Then, if the load were increased (or
decreased) in regular steps of IV each, we have, as explained above,
WL*
y " A v i% wk*
318 FBOPBBTIES OF MATTE*
where b and d are the breadth and the depth of the beam, and Y, the
Young's modulus for its material.
Hence Y = rrji
4y &.d 8
Now, since the depression of the beam is given by the relation
y = WL*/4Y.b.d 3 , it is clear that, for a given load, the depression of the
beam is
(i) directly proportional to the cube of its length,
(ii) inversely proportional to its breadth,
(Hi) inversely proportional to the cube of its depth, and
(iv) inversely proportional to the Young's modulus for its material.
It follows, therefore, that in order that the depression of a beam
may be small for a given load, its length should be small, i.e , its span
should be small, its breadth and depth should be large and the
Young's modulus for its material should also be large.
When a girder is supported at its two ends, its middle part is
depressed, and the surfaces above and below the neutral surface are
respectively compressed and extended, the compression being the
utmost at the upper face, and the extension, the maximum at the
lower face, the stresses being the maximum there and decreasing as
we proceed towards the neutral surface from either side. It follows,
therefore, that the upper and the lower faces of the beam should be
much stronger than its middle portions In other wor Is, the middle
portions may be made of a much smaller breadth than the upper and
the lower faces, thus affecting a good deal of saving in the material.
It is for this reason that girders are usually manufactured with their
crosssection in the form of the letter I.
Stiffness of a beam. The ratio between the maximum deflection
of a beam and its span measures what is called the stiffness of the beam.
It is usually denoted by the symbol 1//7. For steel girders of large
span, n should lie between 10UU and 2000 and for those of shorter
spans, between 500 and 700. And for beams of timber, the value of
n should in no case be less than 360.
135. Searle's Method for the Comparison of Young's Modulus
and Coefficient of Rigidity for a given material. A short length of the
wire, the values of Y and n for the material of \*hich are to be com
pared, is fastened to the middle points of two similar and equal metal
bars AB and CD, (Fig. 202), of circular or rectangular crosssection.
The bars are then suspended from a rigid support by means of two
small vertical lengths of threads, so that, when the wire is straight,
the bars are parallel to each other, as shown.
On slightly pulling together the ends A and C of the two bars
symmetrically and through equal distances, the wire is bent into a
circular arc, (Fig. 203). On releasing the bars, they begin to vibrate
in a horiz >ntal plane from a circular arc on one side to a similar arc
on the other, due to the torque exerted on them by the wire, the mid
point of the bars remaining almost at rest.
If /be the length of the wire and 0, the angle of deflection of
each bar from its normal position, the angle subtended by the wire at
ELASTICITY
319
the centre of curvature of the circular arc into which it is bent; is
clearly 20, such that
/ as R.2Q, where R is the radius of the arc.
Or, R = 112$.
Fig. 202. Fig. 203.
Now, the bending moment of the wire and the couple exerted
by it on each bar is, as we know,
YL
Yirr*
R
Lfbr the wire.
This couple produces an angular acceleration dwfdt in each bar,
and, therefore,
r // is* Y.TT.r'.O
I.datjdt =  2[  ,
where 7 is the M./. of each bar about an axis through its midpoint
and perpendicular to its length, i.e., about the thread from which it is
suspended.
7 Trr 4
And /.
i.e., the acceleration is proportional to the (angular) displacement.
Therefore, the motion is a simple harmonic one, and hence the
time period of each bar is given by
whence, Y = ? . ... (I)
The suspension threads are then removed, and one of the bars
is clamped horizontally, so that the other bar hangs vertically below
it at the other end of the wire. The suspended bar is then turned
about the wire in the horizontal plane, so as to twist the wire when,
on being released, it begins to vibrate torsionally. Its timeperiod f
is noted.
Now, t l a 2wy^/C where C is the twisting couple set up in the
wire per unit deflection or twist, and is equal to nnr^fiL ( 122)
whence, n =* ysyr*
320 PROPERTIES OF MATTE*
Dividing relation (I) by relation (II), we hav
Z  SnJJ r   JL1
n ~~ r*.tS "87T./.7 "" //*
This gives the ratio of Young's modulus and the coefficient of
rigidity of the material of the wire in terras of t l and f ,.
Y
Now, Poissorfs ratio, a = ~ 1.
>Yl
So that, substituting the value of Y[2n, we have
/ 2 / 2 o/ 2
* * ^*
Thus, Poisson's ratio for the material can also be easily deter
mined.
N.B. As will be readily seen, the radius (r) of the wire, the measurement
of which is the chief source of error ;see page 303) has been eliminated altogether.
136. Strain energy in a bent beam We have seen before, (page 307), how when a
small portion AB of a beam is b^nt into ths form of a circular arc (Fig. 204), sub
tending an angle 89 at its centre of curvature,
the strain produced in an element of it at a
distance z from the neutral axis is given by
z/R, where R is the radius of curvature of the
neutral axis.
Now, energy per unit volume in any
type of strain i stress* strain.
So that, energy associated with, the element in
question = J stress X strain X volume of the
g lenient.
Then, if the stress be F, the strain, e,
the area of cro^s section of the element
normal to the plane of the diagram equal to
<M, and ths length of the neutral axis, Sx,
clearly,
Fig. 204. energy in the element =
Now, Y = stress I strain = Fje 1 and .'. F = Ye = Y.zjr.
z 2
So that, energy in the element = f Y. ^$A$x.
Integrating this expression over the whole cross section, we have
strain energy in the entire portion AB of the beam
Since, as we know, fz*.dA =/<,, the geometrical moment of inertia of the
section considered.
1 f Y1 1 f 1
Or, strain energy in portion AB of the beam ^ ~ . ^rp.&x.
* L K J X'ff
But YlglR A/, the bending moment of the section.
Hence strain energy in portion AB of the beam = ~*r * y^ &*.
And .*. strain energy E in the whole beam of length L = I v>y . dx>
JO y/ ^
^fjf
ELASTICITY
321
So that, substituting the relevant values of the bending moment, we can easily
determine the strain energy of the beam in different cases.
Thus, for example, in the case of a light cantilever with a load W suspended
from its free end, we have M = W(L x), (see page 310), and, therefore,
1 f 1
strain energy, in this case  2 Y[ W*(Lx)*.dx l .W Z L*.
Jo g
137. Resilience of bent beams. The work done in deflecting a horizontal
beam of whatever type, loaded with a weight W is equal to \W x (deflection at the
loaded point). Since this work do.ie measures the resilience of the beam, as ex
plained in 120, we have
resilience of the beam = i W.y,
where y is the deflection of the loaded point.
Thus, in the case of a light beam, of length L y breadth b, and depth d, sup^
ported at the two ends and loaded with a weight W at the centre, we have
y ~ WL*/48YI g , (page 316) and
So that, we have
resilience of such a beam
I g , in this case, is given by bd 3 /\2, (page
308).
96Y
12
' bd*
138. Colums, Pillars and Struts. A long beam of an isotropic
material used for supporting loads is called a column, a pillar, or a strut.
Now, whereas a column or a pillar must always be vertical and
generally fixed rigidly at its ends, a strut may be vertical, horizontal
or inclined and may either have bath its ends fixed rigidly or both con
nected to the surrounding structure through flexible joints, or it may
have one end rigidly fixed and the other connected to a joint, The
theory underlying the two is, however, the sama, the commonest case
being that in which the load applied is a compressive one, i.e., acts at
one end of the column or strut, along its axis, tending to compress
r shorten it lengthwise, though, in some cases, there may as well be
a lateral load, in addition.
Let us take the case of a column or a pillar first,
139. Critical load for a long column, Let us take a long and straight
strip AR, [Fig. 205 (a)] t of wood or metal, arranged in a vertical
position, representing a column, with
both its ends rounded and fitted into
metal sockets, as shown magnified in Fig.
205 (#), so as to allow it freedom to bend
all along its length, and let a load be
applied to it at the top in the form of .a
metal cylinder, containing lead, shot or
mercury, so that its magnitude may be
varied at will, with the cylinder moving
between two parallel guides GO to ensure
its vertical descent.
Now, first, with the load insuffi
cient to bend the strip or the column, we
apply a lateral force / t at its midpoint
0, 0,3 shown, to make it bend a little and
we find that, on removing the lateral force,
the column straightens itself out. We
increase the load at the top, apply the
lateral force, as before, and then remove it. Perhaps, the column
(a)
Fig. 205.
322
FROPEBTIES OP MATTER
again bends and then straightens itself. In this manner, we go out
repeating the experiment with successively increasing loads at the
top until we find that when the lateral force is applied and then*
removed, the column remains bent. At this stage, we find that
whatever the deflection we produce in the column by the lateral force,
the column continues to retain the same on the removal of this force,
provided, of course, that the elastic limit has not been exceeded. This
load which just keeps the column bent, but does not bend it further, z>
called the critical load/br it.
If we increase the load beyond the critical value and give a
slight bend to the column, as before, we find that the load now
increases the bending further and the column either acquires a perma
nent set or collapses due to buckling,*
Let us see how we may account for this critical load. Let us,
therefore, consider the equilibrium of the column AB, under a verti
cal load P l and a lateral force
\(P+Q) /! at its midpoint 0, with the
deflection of the column equal
to y v as indicated in Fig. 206
(a).
Since, for equilibrium, the
lateral force f v must bo balanced
by two horizontal forces, each
equal to fJ2, acting at A and
B, in the opposite direction to
that of/j, we have
total bending moment about O
If now we decrease the lateral force /i to and increase load P l
to \P (the critical load) so that the deflection of the column remains
the same y lt clearly, the condition for equilibrium demands that
P. yi = ^Z, whence, P = F l Z/y 1 [v/i is now 0.
 Again, if the column be in equilibrium when subjected to a ver
tical load P 2 and a lateral force / 2 , with its deflection now equal to j a
and we reduce / 2 to and increase P 2 to P', with the deflection
(a) (b) = A>
And, if jpj be the maximum
Fig. 206. stress due to bending and Z,
the modulus of crosssection at
O (see note on page 309), the moment about due to stress or the mo
ment of resistance to bending = F X Z. 4 So that, for equilibrium,
remaining unaltered at J 2 , we have, proceeding as before, P f =
where F a is now the maximum stress due to bending.
Since the bending is proportional to stress, we have F 1 /y 1 =F 2 ly^
And, therefore, P' = P (the critical load), thus clearly showing that
the column will remain^ equilibrium under the same critical load P for
/.. bv bending or bulging out.
1LASTICITY 323
any value of the deflection we choose to give it within the elastic limit,
as we have seen already in the experiment discussed above.
If, however, wo increase the vertical load to a value beyond its
critical value P, say, to (P+Q), as shown in Fig. 206 (b), the bend
ing moment will clearly increase to (P\Q) y^ or to P^i+Qy^ And,
since the moment of resistance to bending, v/z., F r Z balances only the
portion P l y l of it (as we have just seen above), the portion Qy l
remains unbalanced, resulting in an increase in the bending or deflec
tion of the beam beyond < y 1 . In order to keep the deflection at the
same value y v therefore, we shall have to apply a force/, say, at the
midpoint 6 of the column, in the opposite direction this time, so as
to balance the portion Qy l of the bending moment. Since/ is sup
posed to be balanced by two equal forces //2 and//2 acting at A and
B, as shown, its moment about O =s (//2).(i/2) = /.L/4. So that,
to prevent the column from bending further (beyond y\) we must
have/L/4 = Q.y r
Now, within the elastic limit, the moment of resistance to
bending is proportional to the stress, i.e., FZ oc F, and hence also to
the deflection of the column (because, then, y oc F). But, once the
elastic limit is exceeded, the column acquires a permanent set, though
it is also possible that, due to the moment of resistance due to bend
ing now increasing more rapidly (as it always does beyond the elastic
limit), the column may acquire a new position of equilibrium under
the additional load Q. But, if this does not happen, the column will
continue to bend further and further and finally collapse.
140. Filler's Theory of Long Columns.
(/) When the two ends of the column are rounded or hinged. Let
AB represent a long and initially straight column of an isotropic
material, of length L and of a uniform crosssection
and uniform elasticity, with rounded or hinged
ends so as to be free to bend throughout its length.
Further, let the critical load P act axially upon it, f ~
i.e., in a line with its axis in its straight unloaded ^
position, and let it be given a slight bend by the ;
application of a lateral force for an instant, (Fig. i 2
207). !
Now, consider a point C in the column, at a * ~
distance x from its midpoint O. If the deflection
here be y, clearly, the bending moment here due to
P = P.y t But, if the radius of curvature of the
bend at C be R, the moment of the resistance to
bending there is YJ gi IR. And, clearly, therefore,
Y.I /R = P.y, whence, IjR = P.ylY.I g .
But, as we know, l/R = d 2 yldx 2 , (the ve sign Fig. 207.
being given to make jR positive, for dyjdx decreases
as y increases). We thus have the differential equation
Yl g
324
PROPERTIES OF MATTER
The solution of this equation gives
y = A sin A x+ B cos A x l9 (!)
where A = <\/ P\Y.I g and A and B are constants to be determined.
Differentiating this, we have
dy/dx = A\ cos \x~~ B\ sin pc.
(a) Now, when x = 0, rfy/rfx = and, obviously, sin Ax = ;
so that, A = 0. From equation I, therefore, we have 7 = B cos \x.
But when x = 0, we have j> = ; J 1? Ax = and .*. 05 AX = 1
and sin Ax = 0. So that, from equation I, we have y = B.
(b) When x == L/2, (i.e., at 4), >> = 0. And .. B cos A^/2=0.
[From (a) above.
This means that either B = or cos AL/2 = 0. But, as we
have seen [in (a) above], B = y\. It follows, therefore, that
cos 7\ L/2 = 0, or. that A /2 = ir/2,
L/2 =" w / 2 [for> A =
whence,
9 87 YI g
L*
This is, therefore, the value of the critical load, or the load which
can just keep the column bent at the initial curvature given to it An
addition to this makes the column collapse.
It will be clear from this expression for the critical load that for
the same values of Y and /,, the smaller the length of the column, the
greater the critical load for it.
(ii) When the two ends of the column are fixed. Let the ends A
and B of the column bo now fixed, as shown in Fig. 208 (a), so that
when it gets bent or deflected, the tan
gents to it at points A, O and B are all
vertical, with the line of action of the
resultant load now no longer passing
through the centres of its endpoints. It
passes, instead, between the initial un
bent position AB of the column, and its
midpoint O in the bent position, cutting
the bent column in points C and D. At
these points, therefore, there is no bend
ing moment, RO that they are points of
opposite flexure.
Now, considering the portions CA
and CO of the bent column, we observe
that the deflections at certain points in
the two curves (as measured from the
vertical line through C ) are equal. At
all such points in the two curves, there
fore, the bending moments must bo
equal, and hence also the radii of curva
ture there must be the same, in view of
(I))
Fig. 208.
ELASTICITY
325
the fact that the column is of a uniform cross section. Further, the
two curves (GA and CO) have clearly the same slope at C and also
at A and O, the tangents at all these points being vertical. Obviously f
therefore, the two curves are equal and similar. The same is also*
true of the curves DB and DO. The points (7, O and D thus divide
the whole column into four equal parts and the length of the portion
COD of the column is, therefore, equal to half its total length, i.e.,
COD = L/2.
Clearly, this portion COD of the column, the whole of which i
bent, behaves exactly as the cases considered above, i.e., like a
column of length L/2, with its ends rounded or hinged and carrying
an axial load P at C So that, proceeding as in case (/) above, we
have
P 
L*
i.e., the critical load (P) for the column, in this case, is four times
that in case (/).
Thus, a column with its ends fixed, has four times the strength (to
resist thrust) that it will have with its ends rounded or hinged. Or,
putting it differently, ci column, with its ends fixed, can support,
without bonding, the same load as one of half the length, with its
ends rounded or hinged, would do.
(Hi) When one end of the column is fixed and the other loaded,
This is an easy deduction from case (i) above. For, suppose we have
a column AS, with rounded or hing
ed ends, and of length L 1 ', as shown in
Fig, 209 (a), with P as the critical
load on it. Then, the tangent to
it at its midpoint O is ver
tical. If, therefore, we clamp it
tightly at O, without disturbing the
direction of the tangent at that point,
the lower half OB of the column might
as well be removed, without in any
way affecting the upper half OA.
So that, the upper half then behaves
as an independent column, of length
L s= L'fi, fixed at its lower ends and
loaded at the top, as indicated in
Fig. 209(&).
Fig. 209.
(b)
All that we have to do to calculate the critical load, in this
case, is to consider the column of length L and fixed at one end, as
equivalent to a column of length L' = 2L, with both its ends round
ed or hinged. Therefore, proceeding in the same manner as in case
(i), we have the critical load given by
P =
which is clearly onefourth of the critical load for a column of length
L, with its ends rounded or hinged.
326
PROPERTIES OF MATTER
Thus, we find that a column, hinged at one end and loaded at the
other, has only onefourth the strength of the same column when hinged
at both ends.
N.B. Exactly a similar treatment to that in cases (i) and (//) applies to
the corresponding cases of a strut, arranged horizontally, so long as the strut is
Fig. 210.
toaded axially, or along its axis, like the columns incases (/) and (//). Thus, if
the two ends of the strut be rounded or hinged, so that the whole of it can bend,
we can represent its behaviour as in Fig. 210 (a), and, when its two ends are fixed,
as in Fig. 210 (b). It will be noted that these are essentially the same figures as in
cases (i) and (//), respectively, but are rotated, as it were, through an angle of 90,
so that instead of a vertical load we now have a horizontal load. The method of
calculation for the critical load, therefore, remains the same.
141. Elastic Waves. When a system of stresses, to which a body or a me
dium is being subjected, is suddenly altered, we have (/) a corresponding motion
of the body or the medium itself and (11) propagation through it of the changes in
stress, i\\t two occurring simultaneously and constituting what is called the propa
gation of an elastic or a stress wave.
Now, as we know, even in the case of an iso tropic medium, a deformation
in one direction is invariably accompanied by deformations in two other direc
tions, at right angles to the first, (the familiar case of the deformation of a cube),
so that the theory of elastic waves is really quite a complicated one ; and this
complication is further aggravated in the case of bodies like the earth, for
example, where the elastic properties vary with depth, which explains at once
the complicated pattern of the seismic waves ( 100).
In general, however, we have three types of elementary elastic waves in the
case of a uniform, isotropic medium, viz., (i) compressional, (//) shear and (///)
fltxural waves. We shall only briefly touch upon them here.
(/) Gompressional Waves. These waves are produced when we give an
axial blow to a long bar, i.e., strike it along its axis, and, assuming the sides of the
bar Jo have freedom of movement, their velocity is given by the relation v
V Y l9, where Y is the value of Young's modulus for the material of the bar and p,
its density. But, in case the sides of the bar too are fixed, Y is to be replaced by
Y(l a)/(l + a)(l2(j), where a is the value of Poisson's ratio for the material
of the bar. This expression takes many forms, the simplest among them being
K+4nfi, where K and n are the coefficients of bulk and rigidity modulii for
the material of the bar. Thus, we have different types of compressional waves, all
of which, however, have the common features that (i) the vibrations occur along the
direction of propagation of the wave, i.e., the wave is of the longitudinal type, and
<//) the velocity of the wave is given by
modulus of elasticity
~~ aentity
Thus, in the case of a liquid or a gas, n = and we, therefore, have the
HYDBOSTATICS
.Pascal discussed this result in his 'Treatise on the equilibrium oj
liquids', published in the year 1663, with reference to vessels of different shapet r
known as Pascal's vases, (Fig. 213), all having equal bases and containing wafer
upto the same vertical height h, so that the pressure on the base of each vessel was*
equal to hgms. w/./rmV, and, therefore, the thrust on it was h.a. gms* wt., where^
a is its area.
He was perhaps the first person to have pointed out the paradoxical truth
that even if vessel (i) contains 100 Ibs of water and vessel (v) only 1 oz. of it r
the thrusts on the bases of
both is the same. Aptly,
therefore, it is called the
hydrostatic paradox.
ELB
Strange as it may
seem, but if the water in
vessel (v>) be frozen into ice
and detached from its
sides, the thrust exerted by
this ice on its base will be only 1 oz. >v/., but once this ice is melted back into
water, the thrust again increases to 100 Ibs. wt. The explanation of this seem
ing paradox is, however, simple. The ice does not exert any upward thtttittyl
the part of the vessel opposite to the base and the latter, therefore, exerts tify
squal and opposite thrust on it. But the water does exert an upward thrust O$
it and hence receives back an equal and opposite downward thrust from itt
In case of vessel (/), the thrust on the base is equal to the entire weight
of the water on it.
In vessel (//), the upward component of the thrust due to the left side of
the vessel supports the weight of water in it, between the left side and the dotted
line A, while the downward component of the thrust due to the right side of the
vessel exerts a downward thrust on it, equal to the weight of the water in
between the right side of the vessel and the dotted line B ; so that, the thrust or*
the base is the same as due to a vertical column h of water.
In vessel (///), the upward components of the thrusts due to both the left
and the right side* of the vessel support the extra weight of the water, in
between the two sides and the dotted lines C and D, and, again, therefore, the
thrust on the base is equal to that due to the cylindrical column h of water inbet
ween the dot ted lines C and D.
And, similarly, in vessel (/v), the downward components of the thrust
due to the two sides of the vessel exert an extra thrust on the base, equal to the
weights of the water contained between either side and the dotted lines and F;
so that, once again, the total thrust on the base is the same as that due to a cylindri
cal water column of
Pascal ex t
a separate stand,
.verified the above fact by supporting, by means of
vases of the above shapes, one by one, on a large
disc, D (Fig. 214), suspended from
the shorter pan of a hydrostatic
balance and kept pressed against
their bases by placing a heavy weight
in the longer pan, and pouring water
into the vessel. The disc just got de
tached from its base as the water
reached the same level in each case r
thus clearly demonstrating the equi
valence of the thrust on the disc in*
each rase and fully vindicating his
deductions.
Fig. 214.
328 PEOPBBTIES OF MATTER
mum values being respectively 1, in the case of perfectly elastic bodies,
and zero, in the case of perfectly plastic ones.
Thus, if M t and u 2 be the velocities of two bodies before the im
pact and v l and v 2 , after the impact, we have
( v v ) ("neglecting the ve sign of the
a/ = e , relative velocity after the im
(U l u t ) L pact.
Or, ( Vl v a ) = <?Kw 2 ), ... (i>
where (U L w 2 ) and (v l v 2 ) are their relative velocities, before and
after impact.
It will thus be seen that if e = 1, i.e., if the bodies be perfectly
elastic, (Vj v 2 ) = (i^ w 2 ), i.e., //*e relative velocities of the bodies are
the same before and after the impact, (suffering only a reversal of
direction, in the latter case).
But if e = 0, i.e., if the bodies be perfectly plastic, (v^v^) 0,
or, v l = v 2 , i.e., the two bodies move with the same common velocity r
after the impact.
What happens is that when one elastic body, of mass m v moving
with a velocity u v collides against another elastic body, of mass m zr
moving with velocity u^ (where u l > w 2 ), the surface between then*
gets compressed and when this compression or pressure reaches
its maximum value, their relative velocity becomes zero. Thereafter,
the elastic stress between them makes them recede from each other,
the compression is released, and the two bodies move away with
different velocities, say v 1 and v 2 . v
Clearly, then, in accordance with the law of conservation of
momentum, we have
sum total of momenta after the impact
= sum total of momenta before the impact.
Now, sum total of momenta after the impact = /W 1 v 1 +/w a v
and sum total of momenta before the impact = m^+m^.
So that, m jVi+ AW 2 v 2 = m 1 u 1 {m 2 u 2 . ... (//')
From relations (/) and ('), we can easily calculate the values of
v t and v 2 For, multiplying relation (i) by m 2 , we have
"Vi w 2 v a = w^K ,). ...(/>
And, adding relations (//) and (///), we have
'WiVi+/w a v 1 +m a v 1 m 2 v 2 = m^ im 2 u 2 +m 2 e.(u l u 2 ).
Or, v^
,  ll99 ^ t 9
whence, v l = x 1 ^ , 2 2 , 2 . v l a/ . ,..(/v
v
Similarly, multiplying relation (i) by m lt and subtracting from
relation (ff), we have
(m 1+ m 2 )
Further, it can also be shown that
_ Impulse during restitution
""" Impulse during compression
ELASTICITY 329
The value of e, nowever, is found to diminish with increase in the
velocities of the approaching bodies, and vice versa, and it has been
shown by Sir C.V. Raman, that its value is very nearly equal to 1, if
the collision be very weak.
143. Loss of Kinetic Energy on Impact. It can be shown from
the above relations, that
i * , i > i * , i 9 ^ (1 e^.ttiiWVfW! w 2 ) a
where J^v^+Jw^Vj 1 is the total kinetic energy E 2 of the colliding
bodies, after the impact, and Jw 1 w 1 2 +m 2 w 2 2 , their total kinetic
energy E 1 before the impact.
Thus E 2 = , 1 . 2
2
Or E E ~ l ( 1 r
1 2
/ f ^'
i.e., toss o/ energy on impact = >
J
an expression, with always a positive value, showing that there is
always a loss of energy on impact between two bodies.
Now, the following special cases arise :
(i) When e = 1, i.e., when the colliding bodies are perfectly
elastic. In this case, (E L E 2 ) = 0, i.e., there is no loss of energy on'
impact of perfectly elastic bodies.
(ii) When e = 0, i.e., when the colliding bodies *are perfectly
plastic. Here (E 1 E 2 ) has the maximum value, viz.,
1 (/W 1 m 2 .(w 1 w 2 ) 2
2 * (Wi+wtaj
i.e., there is maximum loss of energy on impact of plastic bodies.
(Hi) When u^ = w 2 , i.e., when the bodies have the same velocity,
(in magnitude as well as direction). In this case, the relative velocity
of one body, witli respect to the other, is zero, so that no impact takes
place at all between the two bodies, and, therefore, (E l rE 2 ) = 0, or
again, there is no loss of energy.
The question now arises as to what happens to this loss oi
energy on impact ? Until very recently, it was supposed that the
energy lost during impact was converted into (i) sound, (ii) heat, or
(7) vibration or rotation of the colliding bodies.
Sir C. V. Raman's experiments have shown, however, that the
production of sound is in no way related to the energy of impact,
being solely due to the impulse set up in the air during the reversal
of the motion of the colliding bodies, after impact.
The change in temperature too is almost always very small and
hence it appears that an appreciable portion of this energy lost during
impact is used up in bringing about a redistribution of the molecules
in the surface layers of the colliding bodies. Indeed, it has been shown
by Hertz that impact produces a definite flattening of the point of
contact of the colliding bodies, with a finite common area between
33U PROPERTIES OF MATTER
them, each body being compressed in its neighbourhood, the com
pression increasing at first to a maximum, (which is proportional to
the twofifth power of the velocity), then diminishing and finally
vanishing altogether, when the bodies get separated from each other.
Prof. Deodhar has also verified this molecular displacement in
the surface layers of the impinging bodies, by making different bodies
('similar and dissimilar') impinge upon each other, with 'extremely low
velocities' and measuring their velocities before and after the impact.
Prom his experiments, he has come to the following conclusions :
(/) With great velocities of the colliding or impinging bodies, the changes
in their surface are 'vivid*, and a greater portion of their energy of impact is used
jp in producing these deformations.
(//) With very small velocities of the colliding bodies, on the other hand,
he value of e increases, in the limit, to unity the increase of e with the 'minimal
velocities' being independent of the nature of their material.
(HI ) The rate of change of e is qui te independent of the medium in which
the impact takes place.
(iv) The duration of impact is observed to be greater in water than in air,
i.e., it depends upon the density of the medium.
(v) A distinct change in the structure of the impinging bodies is noticeable
under the microscope, though no trace of it is visible to the naked eye.
He estimates from this that energy, of the order of 1000
Jc.gms.lcm*. is used up 'in displacing the molecular aggregates'. Further,
bodies, when strained, take time to recover their original condition,
and a rapid rise and fall in the stress may result in the dissipation of
some energy, provided the elastic limit of the bodies, for gradually
varying forces, is not exceeded.*
144. Relative masses of colliding bodies. If, in the above
example of two colliding bodies or balls, the second one be at rest, so
as to have no kinetic energy, we have
total kinetic energy before impact = Jm 1 .w 1 a .
A i if i  , ! (
And, loss of energy during impact =  v
So that. !?"f"*W
total energy
Or, loss of energy = ^ m 1
I "h
Clearly, therefore, the loss of energy will be small if mjm 3 be large,
and vice versa.
Thus, in order to minimise loss of energy, the ratio mj/n, must be
made large, i.e., the mass of the striking body must be much greater
than that of the body struck. Hence it is that a slowmoving heavy
hammer is more suitable for imparting momentum to a body than a
*This should not, however, be understood to mean that an exceeding OF
'overlapping* of the elastic limit is necessary for a loss of energy to occur.
ELASTICITY 331
quick moving lighter one, even though the two may possess the same
'momentum.
On the other hand, if the loss o energy is to be converted inta
useful work, the ratio m 1 /m 2 must be small, i.e., w 2 must be much
greater than w r That is why while forging instruments etc., we
(must have heavy anvils underneath them.
SOLVED EXAMPLES
1. Show that (a) a small and uniform strain v is equivalent to three linear
strains v/3, in any three perpendicular directions;
(h) the bulk modulus for a gas (/) at constant temperature (i.e., under
Isothermal conditions) is equal to its pressure and (/*) when the temperature is not
constant, (i.e , when the conditions are adiabatic), it is equal to r times its pressure,
where y is the ratio c p /c v for it.
(a) Imagine a unit cube to be compressed equally and uniformly from all
ides, so that the length of each edge is decreased by a length /, i.e.. becomes
Then, clearly, decrease in volume of the cube, i.e.,
v . l(l/) = ii + 3/_3/4/ = 3/, i.e.. I = v/3,
neglecting / 2 and /*, the value of/ being small.
Thus, a small uniform volume strain is equal to three linear strains, each
equal to v/3, in three perpendicular directions.
(b) (/) Let P be the pressure and K, the volams of a gas, and let it be
compressed isothsrmally* by increasing the pressure to (P+dp), so that the
volume is reduced by dv and becomes (K dv).
Then, clearly, stress = force per unit area = pressure applied = dp, and volume
strain = change in volume/ original volume = d\\V.
.. Bulk modulus for the gas, i.e.. K = j?. y = * .K.
Since the temperature of the gas remains constant, Boyle's law holds
good, and we, therefore, have
PV = (pdp}x(Vdv] = PVP.dvdp.Vdp.dv.
Or, PV = PYP.dv+dp.V. Or, P.dv  dp.V. [neglecting dp.dv.
whence, V.dpjdv = P.
Since V.dp/dv = K, we have K = P.
Or, the Bulk Modulus for a gas, at constant temperature, i.e., its isothermal elas
ticity, is equal to its pressure.
(//) If, on the other hand, the change in the volume is brought about
we have
= a constant.
Diffrentiating this, we have
t fa (" The ve sign merely indi
PyK r Vvf V r dp = 0. Or, V~r = rP j cates that dv and dp arc
Or, JTyr. V  of opposite signs.
thus, the adiabatic elasticity of a gas is equal to y times its pressure, i.e., is y
times its isothermal elasticity.
This may be done by using a cylinder and a piston of a perfectly con
ducting material, so that the heat H conducted out into the surrounding air as
soon as it is generated and the temperature of the gas remains the same as
before.
fin this case, the cylinder and piston are of a perfectly nonconducting
material or the cylinder is placed on a perfect insulator, so that the heat gene
rated on compression of the gas cannot escape out but remains inside the gas
itself, thus raising its temperature a little.
332 PROPERTIES OF MATTER
2. Show that the Bulk modulus K, Young's modulus Y and the Poisson'ff
ratio a are connected together by the relation, K = Y/3(l 2<*).
(Punjab, 1940 ; Delhi. 1947}
We have A'  ^ = ^^ . [See page 28 7 .
Now, l/a=y, and p/a = a. [See pages 2888^.
Therefore, K  7/3(12(7).
3. Show that l he rigidity n, and Young's modulus Y are connected by the
relation, n = r/2(l fo), where <r is Poisson's ratio. (Punjab, 1938)
We have n I/2(a+ = * . [See page 289.
zaiitp/a)
But I/a  Y, and p/a or.
4. Obtain an expression for the radius of curvature of a flat curve in term*
of the slope of the curve, and use the result to find the value of deflection in the case
of a bar fixed horizontally at one end and loaded at the other. (Bombay, 1928)*
Let APQ be a flat curve (Fig. 21 1), and let P and Q be two points on it
small distance $x apart. Draw tangents to the curve at Pand Q, and let O be
the centre and /?, the radius of curva
ture of the portion PQ of the curve.
Then, if LPOQ = 9, we have PQ R.Q.
Or, Sx  R.e. (')
Now, 6 = difference in slope of the tan
gents at P and Q.
And since slope of the tangents at a
point is measured by dy\dx at the point,
we have
Now, the rate of change of slope is given
Fig. 211. by the second differential coefficient r
f change in slope from P to Q = *x d*y/dx*.
Or, = $v.d*yldx\ And, /. 8x = '*x.R.<Pyldx\ [From (/) above.
t.e. 9 R.d*yldx z = 1.
Or, l/R d*y/dx* = rate of change of slope at P.
Since in the case of bent rods, or beams, the curve of the neutral axis is
very slight, the relation \/R d*yjdx* gives the radius of curvature of the axis
at any given point.
Now, for a bar fixed horizontally at one end and loaded at the other,.
(/.*., in the case of a cantilever), we have
W.(Lx) = Y.I g !R, [See page 310.
*he axis of x being taken along the horizontal and the axis of y, vertically down
vards.
Here, L is the length of the bar from the fixed to the free end, x, the dis
tance of the section PQ from^the fixed end, and W 9 the weight applied at the
free end.
Therefore, substituting the value of 1 /R 9 from above, we have
A ~ Y J d ' y Or YJff . d * y (T y\
x) r./flr.. ur, F'* =* (Lx).
Integrating this, we have
Y.I
f
J 
Or, j'
ELASTICITY 333
where Q is a constant of integration, to be determined from the conditions of the
experiment.
Clearly, dy/dx is zero at A, i.e., dy/dx = 0, when x0.
Substituting these values of J and x in (//), we have Q =0.
dy r x 2 ,... v
W d* = LX ~ 2 '  (l ">
Again, integrating this expression (//), we have
y/ T Y a a; 8
Or. TT>" 2 6 + C "  (/V)
where C t is another constant of integration.
To determine this constant, we observe that the depression y of the rod
us zero at the end A ; so that, j=0, when x=0.
Putting these values of y and x in (iv), we have C a = 0.
r./^ v Lx z x 9
Hence ^ .y  2 ~ 6 '
Now, to obtain the deflection of the loaded end, let us put x = L.
TU . , y./r LJT' 8 3
Tben, clearly, ^ * .;> = 2  6 = 3 '
Or, y =* W L*l3Y.I g .
if the rod be of rectangular crosssection, k bd 3 /l2, where ^ and </ arc
its breadth and depth respectively ; so that,
f i j
for a rectangular rod, y
And, if the rod be of a circular crosssection, (i.e., if it be cylindrical),
i a rcr/4, where r is its radius ; so that,
W / 3 4H^ / 3
for a cylindrical rod, ^ =  '
5. A brass bar 1 cm. square in crosssection is supported on two knife
edges 100 cms. apart. A load of 1 k.gm. at the centre of the bar depresses that
point by 2*51 mm. What is Young's modulus for brass ?
We know that the depression of the midpoint of the bar is given by
y = jF/ 3 /48 Yf ff . [See page 3 1 6.
Now, for a bar of rectangular crosssection,
j g = b.cl s l\2. [See footnote page 308.
Here, b d = 1 cm., becauss the bar is 1 cm. square in cross* section.
b.d* = 1 x 1  1 ; W = 1 k. gm. wt.  1000x981 dynes.
/ = 100 cms. and y = 2*51 mm. "25 i cm.
Wl* \2Wl* Wl*
TMrcfort. ,    , Or, 1  
r . x _ . , 77x , ou
Or, the value of Young's Modulus for brass is 9 77 x 10 11 dynes/cm 9 .
6. Establish an expression for the work done in stretching a wire through
J cms. assuming Hooke's law to hold.
Find the work done in Joules in stretching a wire of crosssection 1 sq. mm.
and length 2 metres through 01 mm., if Young's modulus for the material of the
^vire is.2x 10 18 dynes /cm 1 . (London Inter. Science)
For answer to first part, see 112 (i), where it is shown that work done
In stretching the wire *= J stretching force* the stretch.
334 PROPERTIES OF MATTER
r, work done =  F.l=^ '' ' ., wnereFis the stretching forces
Numerical. Here, Y = 2x 10 l> dynes /c/w 2 ., a = 1 sq. mm. = 1/100 = '01 sq. cm.,
/ '1 TW/W. = 01 cm. ; and L =2 metres ~ 200 aws
.  . . 1 Y.al . \ 2xl0 12 x01x Olx01
Therefore, work done = ^  ^ ./ = . 
10 18 x01 3 10 10* c ._ n 5000 5
c ,A .
200 SOO 2 ^5 10?  ro4  5x10
Thus, work done in stretching the wire is 5 x 10~ 4 Joules.
1. Show that for a homogeneous isotropic substance, YIN= 2(ahl),
where Y is the Young's modulus, A r , the simple rigidity, and a, the PoissonV
ratio.
A gold wire 32 mm. in diameter, elongates by 1 mm., when stretched by
a force of 330 gm. wt., and twists through 1 radian, when equal and opposite torques*
of 145 dynecm, are applied at its ends. Find the value of Poisson's ratio for
gold.
An isotropic substance is such that two equal, similar portions cut from it,
with any orientation, arc exactly like and indistinguishable.
For proof of the relation, Y/N = 2(crHl), see solved example 3, above,
(page 332), where it is shown that N = Y/2(crH), whence, Y/N = 2(crh 1).
Now, Y = FyL r
a <l
Here, F=330x 981 dynes; 1=1 ww.=l cm. ; and a=rcx('016) 2 s<?. cms.,
[because radius = 032/2 = *016 cm., and a = nr 2 ].
v 330x981xL
Y ~ WX 016 xl
Since couple acting on the wire ~ 145 dynescm., and angle of twist = 1 radian ^
we have, couple per unit twist = 145/1 *= 145 dynescm.
This must be equal to N.nr*l2L, where r is the radius of the wire.
1
Thus, Mir/2I  145, whence, N 
7 __ 3^0x98 lxl
Ur> ^ ""
_
l 145X2L*
330x981x(016) 2 330x911 x( 016) 2
"* 2x : lxl45 29 ^^.
Since ^ = 2(afl), we have 2(<j + l)=2'858.
Or, (afl) = 2858/2  1'429. And, .. a = r429l = '429.
Hence, Poisson's ratio for gold = 0'429.
8. A square metal bar of 2*51 cms. side, 37*95 cms long, and weighing 826*
gms. is suspended by a wire 37 85 cms long and 0501 cm. radius. It is observed
to make 50 complete swings in 335 7 sees. What is the rigidity coefficient of the
wire ?
Here, timeperiod of the bar, i.e., t  335'7/50 = 6'714 sees.
Now, timeperiod of a body executing a torsional vibration is given by
f=2TrV i\Ct where / is its moment of inertia about the suspension wire as axis an(T
C, the twisting couple per twit deflection or twist of the wire.
Here, 7  mass (^"'*' n "'" ) = 826. ^
63
99540 gm. cm 2 . .'. 6714 = 2*W ^
826. ^ 1440 ^ 6301 ) . 826 x ^J 3  826x1205.
99540
ELASTICITY 33 1
squaring which, we have
^TUM A 2 9954 ^ ^ 4Tr*x99540
(6714)' 4*'xc. Or, C
Now, C is also = .wr 4 /2/, where the symbols have their usual meanings.
77Xnx(0501) 4 = 4Tt*x 99540
2x3785 (6'714) a '
87CX 99540x37'85
whence,/i 
Or, the rigidity coefficient of the wire is 3'357x 10 11 dynesjcm*.
9. A disc of 10 cms. radius and mass 1 k. gm. is suspended in a horizontal
plane by a vertical wire attached to its centre. If the diameter of the wire is 1 mm.
and its length is 1*5 metres and the period of torsiona? vibration of the disc i:
5 sees,, find the rigidity of the material of the wire. Prove the equation you employ.
(Bombay, 1931}
For proof of the formula, see 126, (pages 300301), which gives the
relation,
Here, in this case, t 5 sec., and /, for the disc, about the axis of sus
pension, is given by A/> 2 /2, where M is the mass of the disc, and r, its radius,
Since M = 1 k.gm. = 1000 gms : and r = 10 cms , we have
/  100 2 X 102 = 500 x 100 = 5x 10* gms. cw a .
Therefore, r  5 = 2*^/5 x~10 4 /C. Or, 25  
, _ 4Ti 2 x5xl0 4 4n3 X 10 4
whence, C ~~ = 
WTTA** 4?^ X 10 4
Since C is also equal to nnr*!2l, we have   =
/. substituting the values of r and /, we have
nxnx('05)*_ ^ 4* 2 xlO^
2x150 " 5 """'
(V r == '5 771/w. *05 r/77. and / ~ 1*5 metres = 150 0777.5.)
4^ x 1 4 x 2 x 1 50 __ _4_xj
whence, /i    X7TX ( 5) 4 "~ Ifo
.'. the rigidity of the material of the wire is l*206x 10 ia dynes/ cm 2 .
10. An elastic string has a mass M suspended at its lower end, the uppei
being fixed to a support. The mass is pulled down over a short distance and let go
Explain the motion that ensues and find an expression for the time of oscillation.
If a mass 777 is added to the mass M, the time is altered in the ratio of 5 : 4,
Compare the masses 777 and M. (Bombay, 1936]
For first part, see solved example 10, Chapter IV, (page 143), where it i<
shown that if / be the extension produced in the string in the equilibrium positior
due to the mass, the string executes a S.H.M., its time period being given by
Let the timeperiod in the first case, when the mass M is suspended frorr
the end of the string be r 2 , and, in the second case, when the mass (77* hM) is sus
pended, be t t . Then, if /! and l t be the respective extensions produced by the twc
masses in the equilibrium position, we have
t l = 2*V / i/ r and / 2"VQg
So that, /i 8 //, 2  /i//,.
if /! 4 sees, and f t = 5 sees., we have 16/25 /V/V
Now, /! and / are "directly proportional to the stretching force applied
v/i., Mg and (Mf m)g respectively ; for,
336 PROPERTIES OF MATTER
in the 1st case, Y *** and in the 2nd case Y 
a.li fl./i
whence, /,. = Mg.L/ya and / f = (M+m)g.LIY.a,
where L is the original length of the string, a, its area 0/ crosssection and y, the
Young's modulus for its material.
M+m 25 . MhmM 2516 . m 9
Hence ' M = 16' Whence ' M "16" ''" M = 16 '
Thus, the two masses, m, and A/, are in the ratio 9:16.
11. The breaking stress of Aluminium is 7*5 X 10 8 dynes cm."* and of Cop
per, 22 xlO 8 dynes/cm. 2 . Find the greatest lengths of the two wires that could
hang vertically without breaking. Density of Aluminium = 2*7 gms./c.c. and of
Copper = 8 '9 gms /c.c.
Let a sq. cms. be the aica of crosssection of the wires and / x and l t cms.,
their lengths respectively that could hang vertically without breaking.
(i) Case of the Aluminium wire.
Here, volume of aluminium wire = lxac cs., and .'. its mass /j x ax 2 7 gms.,
and its weight = ^ x a x 2 7 gms. wt. = /j x a x 2 '7 x 981 d>wi .
This must be equal to the maximum force the wire can withstand.
Now, breaking stress = 7 9 5 x 10 s dynes /cm a .
Therefore, total force that can be applied to the wire, without breaking it,
is equal to 7 5 x 10 8 x a dynes.
Or, / 1 XflX2 7x981 75 x'lO'xo,
7<; v 10 B
whence, I, = ~r^t = 283,100 cms.* 2'831 kilometres.
( ii ) Ci ^ of coppe r wire .
Proceeding as above, we have, in this case,
/,Xflx8'9x981 = 22xl0 8 Xfl,
22 y 10 8
whence, /, = Q * ^ ,  252,000 cms. = 252 kilometres.
Thus the required lengths of the aluminium and copper wires are 2*831 and
2 52 kilometres respectively.
12. A copper wire 3 metres long for which Young's modulus is 12*5 x 10 11
dynes per. sq. cm., has a diameter of I mm. If a weight of 10 k. gms. is attached
to one end, what extension is produced ? If Poisson's ratio is 0*26, what lateral
compression is produced ?
Here, original length of the wire (L) = 3 metres = 300 cms.,
Young's modulus for the wire (Y) = 12*5x 10 U dynes. cm.~ a ,
radius of the wire (r) ** J mm. '5 mm. = '05 cm.
and .". its area of crosssection wr* 7ix('05) 1 5^. cms.
And force applied (F) = 10 k. gm. wt. = 10 x 1000 1 gms. wt t
= 10 x 100 x 981 dynes ** 981 X 10* dynes.
Now, we have the relation,
v FxL . , F.L
 whence, /
98 1 x 10* x 300
'5xlO n
981_x3 __
n x (05) xT2 r 5 x W*
ELASTICITY 337
Or, extension produced '2997 cm.
. . , _ . , lateral strain
Again, we have Poisson s ratio, a  , ong if u( , inal strain
JA
26 l,L '
Or, lateral strain '26 x //L '26 x 2997/300= 2'598 x 10*.
This, therefore, gives the value of lateral strain, i.e., d\D, where d is the
decrease in diameter and D, the original diameter of the wire.
Hence d\D ==2598x10*.
Since D = 1 mm. = 1 cm, we have
dl'l  2598 x 10*. Or, d  2*598 x 10* x '1 2'598 x 10~ 5 .
Hence, lateral compression produced 2'598 x 10~ 6 cms
13. A uniform glass tube is hung from a support and stretched by a weight.
It is found that one metre of the tube stretches by '08 cm , but that a column of
water 1 metre long contained within the tube lengthens by only 0'4 cm. Find POH
son's ratio for glass.
We know that Poissorfs ratio, a = &/.
Now, let the stress be ~ P dynesjcm*.
and internal radius of the tube = r cms.
Then, increase per unit length of the tube =P cms.
And, decrease per unit^radius of the tube = P.p cms.
s. increase in 1 metre or 100 cms length of the tube = 100.P cms.
and decrease in the radius of the tube P.p r. cms.
So that, the radius of the tube is now (r~P.p.r ) cms. = r(lP p) cms.
Now, increase in length of the tube = 06 cm.
lOO.P.oc = 06, whence, a = 06/100 P.
And clearly, initial crosssection of the tube = *r* sq. cms.,
and, final crosssection of the tube = "[/(I P )] a ^. rmv.
= nr 2 X 1 1 2P ft 4 (P )*] ^. cms.
7rr a x[l 2P./S] ^. c/W5.,
neglecting (PjS) 2 as a very small quantity.
Therefore, volume of water column initially = 100 rcr 2 c cs.
And volume of water column finally = 103 4x *r\\ 2P./3) c c^.,
v length of the water column is now 100I04 cms.
Since volume of the water column remains the same, we have
Or, 100410004x2P=100. Or, 100 04 x2Pj8=100'04 10000 ='04,
whence, ft = 2Px TdOW
04 100P 04x100
==  X
a == 2Px 10004 *06 2xlOO04x06
4 ______ 1
~ 2 x 1 00 : 04 x '06 ~~ 5002 x '06 =
Hence, Poisson's ratio for glass 0*3332.
14. A steel wire* 2 mm. in diameter is just stretched between two fixed
points at a temperature of 20C. Determine its tension when the temperature falls
to 10C. [Coefficient of linear expansion of steel is 000011 and Young's modulus
for steel is 2'lx 10 12 dynes per sq. cm.]
Let the length of the wire be / cms.
Then, on a fall in its temperature, from 20C to 10C, its length will de
crease by an amount =/x '000011 x 10 cms.
PROPERTIES OF MATTER
And .% strain produced in it = /x 000011 x 10// = 000011 x 10=11 x 10~ 5 .
Let T dynes be the tension in the wire. Then,
stress=Tlnr***TjT*('\)*=TlKX '01 sq. cm.
V radius of the wire, r=\ mm. '1 cm., and /. its area of crosssection*
=irr 8 = n x(*l) 2 sq. cms.
Now, Young's modulus, (Y) stress/strain. So that,
r ~ X T Or ' ic
whence, r= 2'lxl0 12 X7rxll xlO~ B =2 Ix 10 7 xnx'll=7 257x10' dynes.
Therefore, tension of the wire = 7'257x 10 dynes.
15. If one body impinges on another which is at rest, find the relation
between (a) momenta, (b) the kinetic energies of the system before and after
impact.
A steel ball is let fall through a height of 64 cms. on a plate of steel. The
height through which it rebounds is 36 cms. Calculate the coefficient of restitution.
,,, . .. . relative velocity after impact
We know that e = 77  r t'    /
lelative velocity before impact
Here, relative velocity after impact, say, v= \/2..36, ['' n ~ 36 c ms 
and before impact, say, u = \/2.g.64. [v here, /i= 64 cms.
Therefore, e = V^36/2 g 64= V36/64  v/9; 1^3/4 '75.
Thus, the coefficient of restitution =75.
EXERCISE VIII
1. Define Young's modulus. Show that a shear is equivalent to a com
pression and an extension.
Find an expression for the work done in stretching a wire and hence de
duce an expression for the energy per unit volume of the wire.
(Madras B.A., 1947)
2. A wire 300 cms. long and 0*625 sq. cm. in crosssection is found to
stretch 0*3 cm. under a tension of 1200 kilogrammes. What is the Young's
modulus of the material of the wire ? (A.M I.E., 1961)
Ans. 2'3x 10 1 * dynes I sq cm.
3. Explain the terms : stress, strain, Young's modulus, Poisson's ratio,
bulk and rigidity moduli. Show that the value of Poisson's ratio must lie bet
ween  1 and +1/2. (Calcutta)
4. Define Young's modulus, Bulk modulus and modulus of Rigidity. If
, IT and n represent these moduli respectively, prove the relation E=9nKj3K+n.
(Allahabad, 1943)
5. A solid ball 330 cms. in diameter is submerged in a lake at such a
depth that the pressure exerted by water is 100 k. gm. wt Icm 2 . Find the change
in volume of the ball. (K for the material = 1 '00 x 10 7 dynes/cm*.)
(Bombay, 1959)
Ans. 1 386 c.cs,
6. While at 0C., a square steel bar of 1 cm. side is rigidly fixed at both
ends so that it cannot expand. Its temperature is then raised to 20C. What
force does it exert on the clamps ? (Young's modulus for steel = 2x 10 lf
dynes/ sq. cm. and coefficient of expansion of steel = 000011).
Ans. 448 k. g m.
7. Find the formula for the work done in stretching a wire, and apply it
to find the elastic energy stored up in a wire, originally 5 metres long and 1 mm.
in diameter, which has been stretched by 3/10 mm. due to a load of 10 k. gm. Take
g  300 w. (Bombay)'
Ans. 45Tcxl0 4 er,s,
8. A bar of iron, 0'4 sq. /. in crosssection is heated to 100C. It is then
Hxed at both ends and cooled to 15C Calculate the force exerted by the bar on the
ELASTICITY 339
fixings. Young's modulus for iron is 30,00,000 Ib./sq. in. The coefficient of
linear expansion of iron is 00000121C.
(Institute of Civil and Electrical Engineering)
Ans. 5464 tons.
9. A wire of length 50 cms., and diameter 9 mms. was fixed at the upper
end while a wheel of 10 cms diameter was fastened to the lower end. Two threads
were wrapped round the wheel and passed horizontally over pulleys ; each
thread supported a scale pan. On placing a weight of 230 gms. on each pan
the lower end of the wire was twisted through 45C'. What is the rigidity coeffi
cient of the material of the wire ? Ans. 796 x 10 11 dynes tmr*
[Hint. Convert degrees into radians n radians = 180.]
10. Explain what you understand by 'shearing strain*. What are its
dimensions ? Deduce an expression for the moment of the couple required to
twist the lower end of a rod of circular crosssection by 90, the upper end being
clamped. (Agra, 1945)
[Hint, e ~90 ~ Tt/2 radians.] Ans. Couple w 2 .nr 4 /4/.
11. What couple must be applied to a wire, 1 metre long, 1 mm. diameter,
in order to twist one end of it through 90, the other end remaining fixed ? The
rigidity modulus is 2 8x 10 11 dynes cm" 2
[Hint. 90 7t/2 radians.] Ans. 4'3x 10 6 dynes cm.~*
12. Explain what is meant by 'modulus of rigidity' and find out its dimen
sions. Describe one method of finding experimentally the modulus of rigidity of
a v/ire and give the theory oi the method. Find the force necessary to stretch by
1 mm. a rod of iron 1 metie long and 2 mms in diameter. Also calculate the
energy stored in the stretched rod, [Young's Modulus for Iron =2x 12 U C.G.S.
units ] (Patna, 1949)
Ans. (/) 64 k gm wt (//) TCX 10 6 ergs.
13. Find the relation between the bending moment and the curvature of
the neutral axis at any point in a bar.
A vertical rod of circular section of radius 1 cm is rigidly fixed in the
earth and its upper end is 3 metres from the ground level. A thick string which
can stand a maximum tension of 2 A gm. is tied at the upper end of the rod and
pulled horizontally. Find how much will the top be deflected before the string
snaps. (Y for steel = 2x 10 U> C.G 5. units, g = 1000 C.G.S. units).
(Saugar. 1948)
Ans. J 1 '47 cms.
14. If a brass bar, 1 cm. square in section, is clamped fiimly in a horizon
tal position at a point, 100 <v//5. from one end, and a weight of one k. em. is
applied at the end, what depression would be produced ? (Y for bra^s 978 x
IQ 11 dynes cm.*). Ans. 401 cms.
15. A uniform beam is clamped horizontally at one end and loaded at
the othei. Obtain the relation between the load and the depression at the
loaded end.
Compare loads required to produce equal depression for two beams,
made of the same material and having the same length and weight, with the only
difference that while one has a circular crosssection, the crosssection of the
other is a square. (Saugar, 1950)
Ans. 3 : TT.
16. A weight is suspended from the free end of a uniform cantilever.
Find the equation of the curve into which the cantilever is bent. The weight of
the cantilever may be neglected.
A uniform rigid rod 120 cms. long is clamped horizontally at one end.
A weight of 100 gms. is attached to the free end. Calculate the depression of a
point 90 cms distant from the clamped end. The diameter of the rod is 2 cms.
Young's modulus of the mateiial of the rod is l'013x 10 11 dynes per sq. cm. and
g =='980 cm.lsec*. (Bombay, 1940}
Ans. 2*834 mmx.
17. A light beam of circular cross section is clamped horizontally at one
end and a heavy mass is attached at the other end. Find the depression at the
loaded end.
If the mass is pressed down a little and then released, show that it will
form simple harmonic motion. Explain how from a knowlege of the period
340 PROPERTIES OF MATTER
of oscillation, the mass and the dimensions of the bar, the value of Young's
modulus for the material of the bar may be determined. (Madras)
18. A vertical wire is loaded (within the limits of Hooke's Law) by
weights, which, produce a total extension of 3 mms. and 5 mms. respectively.
Compare the amounts of work necessary to produce these extensions.
Ans. 9 : 25.
19. A sphere of mass 800 gins, and radius 3 cms. is suspended from a
wire of length 100 cms and radius 0*5 mm. If the period of torsional vibration
is T23 sees , calculate the'rigidity of the material of the wire. (Bombay)
Ans. 7'654 x 10 11 dynes cm~*
20. A bar, one metre long, 5 mmi. square in section, supported hori
zontally at its ends and loaded at the middle, is depressed T96 mm. by a load of
100 zms. Calculate Young's modulus for the material of the bar.
(Take g 980 cm.} sec*.)
Ans. 1999 x 10 U dynes cm.~*
21. Calculate the time of vertical oscillation of amass of 1 k. gm. hang
ing by a steel wire 3 metres long and *5 mm. in diameter. (Y for steel = 2x 10 11
C.G.S. units). Ans. '05 sec.
[Hint. Find extension I produced Then, t 2K^ljg. (See solved example
W,page 143J.
22. Prove that Young's modulus Y, the bulk modulus K, the modulus of
rigidity n and Poisson's ratio cr satisfy the relations :
(/) 2n " 1+ff ; ( "> IK = 1  2ff ; 3nd <" V > 4 K = I '
23. Define Poissoa's ratio, and show mathematically, from first princi
ples, that it must be ,* than 0'5 and cannot be less than 1. Calculate
Poisson's ratio for, anc. :.'w rigidity of silver from the following data :
Young's modulus for silver wire 725 x 10 11 dynes cw.~ a
Bulk modulus for silver wire = 1 1 x 10 U dynes cm.~*
Ans. n = 2 607 x 10 11 dynes cm.*, and cr = 0'39.
[Hint, (i) From (/) and (ii) above (Ex. 22>, we have 3K (/2or) = 2n(l i <*).
Since K and n are both iive, G cannot be more than *5 and less than !.
(ii) See 116, page 288, whence, it can be shown that
3 FAT nY A v
n ~~ (9K~ Y) 9 ^ (9/i"3 K) and Y ~~
AI 2 1 K 3Y~2n
Also 3^= 2n  ^, whence, p 
A 1 ^ v 1 9AT/I
and a  ~ v . Or, F = 
Or, directly from (/) (Ex. 22), a 2 1.
24. A wire 1 metre in length and 1 mm. in diameter is stretched by 0'
mm. by a load of 10 k. gms M/. and is twisted through 70 by a force of 5 gms.
wt. applied to each end of a 20 cm. rod soldered at its midpoint to the end of
the wire. Calculate (1) Young's modulus, (2) Shear modulus, and (3) Bulk
modulus of the wire.
Ans Y = 2081 x 10 11 dynes cmr* ; n = 8268 X 10 11 dynes cm~ 2
and K = 14 35 = 10 11 dynes cm.
[Hint. (/) Twisting couple = mtQr*!2l = 5 x 981 x 20 dynecm.,
and 9 = 70 x TT/ 1 80 radians.
(ii) K nYI(9n3Y). [see Ex. 23, Hint (//))
25. A block of soft rubber, 5" square, has one face fixed, while the oppo
site face is sheared through a distance *5* parallel to the fixed face by a tangen
tial force of 39 Ibs. wt. How much work is done per unit volume of the cube to
do this? Ans. 8'64/r. Ibs.
ELASTICITY 341
26. Calculate the depression at the free end of a thin light beam, clamp
ed horizontally at one end and loaded at the other.
For the same mass per unit length, show that a beam of square section
is stiffer than one of circular section, the deflections being in the ratio 3/w.
(Bombay, 1949)
27. A rectangular bar of iron is supported at its two ends on knifeedges
.and a load is applied at the middle point. Calculate the depression of the
middle point.
How can this be utilized to determine Young's modulus of iron ?
(Allahabad, 1947)
28. Find the value of Young's modulus for copper. In an experiment,
the diameter of the rod was 126 cms. and the distance between the knifeedges
70 cms. On putting a load of 900 gms. at the middle point, the depression was
0*025 cm. Calculate the Young's modulus of the substance. (Agra, 1948)
Ans. 2042 X 10 U dynes [cm*.
29. Define Poisson's ratio and describe a method for its determination
Derive the formula used. (Agra, 1947)
30. Derive the expression for the bending of a tube supported at the two
ends and loaded in the middle. (Banaras, 1947)
31. How do you differentiate between a column and a strut ? Obtain an
expression for the critical load for a long column with its ends rounded or
hinged.
32. Discuss Eulefs theory of Ions columns for the case (/) when both ends
of a column are rounded or hinged, (ii) when both ends of the column are fixed.
33. Show that (/) a column, with its ends fixed, has four limes the strength
to resist a thrust than a similar column, with its ends rounded or hinged and
(11) a column, hinged at one end and loaded at the other has only onefourth the
strength of the same column when hinged at both ends.
34. Two steel balls of masses 1 and 10 k.g. respectively are moving
each towards the other with a relative velocity of 4 metres per second. Find the
loss of energy after impact and state the reason thereof. (Bombay, 1932)
Ans. 50290 ergs.
35. A sphere of mass 3 Ibs., moving with a velocity of 7 ft.jsec., impin
ges directly on another sphere, of mass 5 Ihs., at rest ; after the impact, the velo
cities of the spheres are in the ratio of 2 : 3. Find the velocities after impact
and the loss of kinetic energy. (London University)
Ans. (i) 2ft.jsec. and 3ft.jscc. (ii) 45 ft. poundals*
36. Explain briefly the terms : resilience and stiffness of a beam. What
is proof resilience ?
37. Write a brief note on clastic waves.
CHAPTER IX
HYDROSTATICS
145. Fluids Liquids and Gases. Hydrostatics deals witn tne
mechanics of fluids in equilibrium and our first step, therefore, is to
understand clearly as to what exactly do we mean by a 'fluid 9 .
Unlike a solid, in which, as we have seen, the strain set up
under a shearing stress lasts throughout the p3riod of application of
the stress, a fluid may be defined as that state of matter which cannot
indefinitely or permanently oppose or resist a shearing stress. In fact,
it constantly and continuously yields to it, though the yield may be
rapid in some cases and slow in others. In the former case, the
liquid is said to be mobile, (like water, alcohol etc.) and in the latter,
viscous, (like honey, treacle etc.). In either case, however, & fluid has
no definite shap 3 of its own and assumes ultimately the shape of the
containing vessel.
And yet, with all this seemingly clearcut distinction between
a solid and a fluid, it is not quite so easy to distinguish between the
two in many a borderline case. Thus, for example, pitch, which
looks so much like a solid that it has to be hammered in order to be
broken, is essentially a fluid ; for, when subjected to the shearing
stress of its own weight, by putting a piece of it in a funnel, or by
putting a barrel of it on its side, it does begin to yield or flow,
although intinitely slowly. On the other hand, metal wires, which
Are obviously solids, when subjected to an excessive tension, begin to
flow in the manner of fluids, and, indeed, may be considered to be
so, for the duration of the yield Once the yield is over, however,
they behave like solids they in fact are.
Then, again, we have, on the one hand, highly elastic solids,
like quartz, in which no change of shapo is discernible even in
millions of years, as is evidenced by the sharpness of its crystals
which look as though they head just been formed, and, on the other,
fluids, like water, the rapid flow of which almost instantaneously
does away with any sharpness of its edges and which, in small
quantities, assumes a spherical form, with no sharp edges or corners
whatever.
The fundamental distinction between the two nevertheless
remains, and we declare a substance to be a fluid or a solid according
as it does or does not yield to a shearing stress applied to it over a
long enough period.
Now, fluids too are further divided into two classes, viz., (i)
liquids and (ii) gases.
A liquid is a fluid which, although it has no shape of its own,
occupies a definite volume, which cannot be altered, however great the
342
HYDROSTATICS 343
force applied to it*. In other words, a liquid is a fluid which is quite
incompressible and has a free surface of its own ; as, for example,
water, alcohol, ether, honey, treacle etc.
<4* A gas, on the other hand, is a fluid, which cannot only be easily
compressed when subjected to pressure but, which, with a progressive
reduction of the pressure on it, can also be made to expand indefinitely,
occupying all the space made available to it. Thus, the whole of the
gas will escape out from a vessel, if there be the tiniest aperture in
it somewhere.
Summarizing then, a gas is a fluid which has neither a shape nor
a free surface of its own ; as, for example, oxygen, hydrogen, carbon
dioxide, air (a mixture of gases) etc. We shall consider first the case
of liquids.
146. Hydrostatic Pressure. Since a liquid possesses weight, it
exerts force on all bodies in contact with it, e.g., on the bottom and
the walls of the vessel containing it, the force duo to it being always
spread over an areaf . And, if this force be uniformly distributed
over tho whole area, i.e , be the same on each small equal element of
the surface, its value per unit area is called pressure or hydrostatic
pressure of the liquid, meaning pressure due to the liquid at rest%.
And, if the force be not uniform, the ratio between the small force
SFand the area BA on which it acts gives the pressure.
Thus, pressure = 8F/BA.
So that, when S A is progressively diminishing, we have
, . Limit force
pressure at the point  ^ 
Or, denoting pressure at the point by p, we have, in mathematical
notation,
p =
Tho total force exerted by a liquid column on the whole of the area
in contact with it is called thrust.
Thus, thrust = pressure X area.
That a liquid, at rest, always exerts a thrust normally to the surface
in contact with it is obvious. For, if it were not so, there would be
a component of the thrust along, or parallel to, tho bounding surface,
and an equal and opposite thrust on it due to the reaction of the
surface would cause it to flow, since it must, by its very nature,
yield to a tangential force. It follows, therefore, that since the
liquid us at rest, the thrust due to it must be perpendicular to the
bounding surface at every point.
*Strictly speaking, all liquids do get compressed a little, when subjected
to very high pressures. The compression is, however, almost negligible. Thus,
water, when subjected to a pressure of about 200 atmospheres, undergoes a reduc
tion of only a hundredth part of its original volume. I
fThe same is true of the force exerted on a liquid, as, for example, when.
we press the piston down in a cylinder containing the liquid.
}It is also sometimes called the pressur* in a liquid due to gravity.
344
PBOPEaTIBS OF MATTER
In other words, the free surface of a liquid at rest must, be at
right angles to the forces acting on it. Thus, when the only, force
acting on it is duo to gravity, its surface remains horizontal*, being
perpendicular to the force of gravity, but, with a steady wind blow
ing, it is slightly inclined, again, however, at right angles to the
resultant of the forces due to gravity and the wind.
Further, since every layer of a liquid at rest is in equilibrium,
it follows that the downward thrust on it, due to the liquid column
above it, is just balanced by an equal upward thrust due to the
liquid column below it. In other words, at any given level, in a liquid
at rest, tfte downward thrust due the liquid column is equal to jhe
upward thrust on it.
147. Hydrostatic Pressure due to a Liquid Column. Let us now
calculate the hydrostatic pressure due to a liquid column A. Imagine
a narrow metal cylinder, of area of cross 
section a and fitted with a frictionless pis
^ on> ^ ne gl*e*k]e weight, to be supported
in a liquid of density p, (Fig. 212).
Then, if the upthrust on the piston
due to the water below it be F, obviously
an equal and opposite force F has to be
exerted on the pistor^to keep it in position,
Hence, if the piston be moved down
through a distance x, work done on it i&
clearly equal to F.x.
This downward motion of the piston*
will obviously expel a volume x.a of the liquid out of the tube, its
mass being x.a.9 and its weight, equal to x.fl.p.g.
Since the level of the liquid in the containing vessel is thus
slightly raised, it is tantamount to this weight of the liquid x.fl.p.g.
rising up through a vertical distance h up to the liquid surface.
In other words, increase, in potential energy of the liquid
x.a.p.g.h. And, this gain in potential energy will obviously be
equal to the work done on the piston. So that. we have
F.x  x.a.p.g.h. Or, f\a h^jf '" ^
t depth h from
pressure).
2l2 
i.e., the hydrostatic pressure due to a liquid ronfigi
the surface is equal to h.p.g., (v Fja = force /area
N.B. The argument remains the same even if the metal tube is inclined
and not vertical, so long as the vertical depth of the piston remains the same. It
will thus be seen that the pressure due to a liquid column depends only upon its
depth and density, and not to any other factor like the surface area of the contain
ing vessel etc.
148. The Hydrostatic Paradox. A remarkable fact follows from the
above, viz,, that so long as the vertical height of the column of a liquid remains
the same, the pressure exerted by it remains the same, 'irrespective of its actual
mass or weight.
*In the case of a large expanse of water, the surface is spherical and
thus again perpendicular to the direction of gravity at every point.
HYDBOSTATICS
.Pascal discussed this result in his 'Trefttise on the equilibrium oj
liquids', published in the year 1663, with reference to vessels of different shapet r
known as Pascal's vases, (Fig. 213), all having equal bases and containing water
upto the same vertical height h, so that the pressure on the base of each vessel was
equal to hgms. w/./rmV, and, therefore, the thrust on it was h.a. gms* wt. t where*
a is its area.
He was perhaps the first person to have pointed out the paradoxical truth
that even if vessel (0 contains 100 Ibs. of water and vessel (v) only 1 oz. of \\ r
the thrusts on the bases of
both is the same. Aptly,
therefore, it is called the
hydrostatic paradox.
EO3
W
Strange as it may
seem, t>ut if the water in
vessel (v) be frozen into ice
and detached from its
sides, the thrust exerted by
this ice on its base will be only 1 oz. \vt., but once this ice is melted back into
water, the thrust again increases to 100 Ibs. wt. The explanation of this seem
ing paradox is, however, simple. The ice does not exert any upward thwft$fl
the part of the vessel opposite to the base and the latter, therefore, exerts mjj
squal and opposite thrust on it. But the water does exert an upward thrtist Oil*
it and hence receives back an equal and opposite downward thrust from it.
In case of vessel (/), the thrust on the base is equal to the entire weight
of the water on it.
In vessel (//), the upward component of the thrust due to the left side of
the vessel supports the weight of water in it, between the left side and the dotted
line A, while the downward component of the thrust due to the right side of the
vessel exerts a downward thrust on it, equal to the weight of the water in
between the right side of the vessel and the dotted line B ; so that, the thrust or*
the base is the same as due to a vertical column h of water.
In vessel (///), the upward components of the thrusts due to both the left
and the right side* of the vessel support the extra weight of the water, in
between the two sides and the dotted lines C and D, .and, again, therefore, the
thrust on the base is equal to that due to the cylindrical column h of water in~bet
ween the dot ted lines CandD.
And, similarly, in vessel (/v), the downward components of the thrust
due to the two sides of the vessel exert an extra thrust on the base, equal to the
weights of the water contained between either side and the dotted lines E and F;
so that, once again, the total thrust on the base is the same as that due. to a cylindri*
cal water column of htigtefa
Pascal ex^lm^ti, t verified the above fact by supporting, by means of
a separate stand, bottoMS* vases of the above shapes, one by one, on a large
disc, D (Fig. 214), suspended from
the shorter pan of a hydrostatic
balance and kept pressed against
their bases by placing a heavy weight
in the longer pan, and pouring water
. into the vessel. The disc just got de
tached from its base as the water
reached the same level in each case*
thus clearly demonstrating the equi
valence of the thrust on the disc ir>
ach case and fully vindicating his
deductions.
Fig. 214.
346
PROPERTIES OF MATTER
149. A liquid transmits pressure equally in all directions
Pascal's Law. Since we do not have any boundary demarcated in
the interior x of a liquid,
we may define pressure
there as the force exerted
per unit area across any
plane in it, and it can
be easily shown that this
pressure is exerted
equally in all directions
in the liquid.
Thus, let us consi
der a portion of the
liquid, in the form of a
triangular prism, [Fig. 215 (/)] with its faces ABC and A'B'C' verti
cal and its edges AA' , BB' and CC' horizontal.
This triangular liquid prism is obviously in equilibrium under
the action of the forces acting on its different faces. Let us study
the interrelation of all these forces.
It is clear that due to its small size, every part of this prism
can be taken to be at the same depth from the liquid surface and also
the pressure on each face of it to be uniform*.
Now, the forces on the two end faces are equal and opposite,
thus neutralising each other's effect and may, therefore, be ignored in
our discussion. Hence, if P lt P 2 and P 3 be the pressures on the faces
BCC'B', CAA'C' and ABB' A! respectively, and /, the length of the
prism, we have
force F l on face BCC'B' = /^xarea BCC'B' = Prl.BC,
force F 2 CAA'C' = P 2 xarea CAA'C' = P 2 .lCA,
and force F, ,,
ABB' A' = P 3 xarea ABB' A' = P^.lAB.
Since these three forces keep the liquidprism in equilibrium,
they can be represented by the three sides of a triangle, taken in
order. Let PQR, [Fig. 215 (/7)], be this triangle of forces, with its
sides PQ, QR and RP representing F v F 2 and F 3 respectively. Then,
clearly,
Or,
Or,
sin a
P r l.BC
sin a
P V BC
sin a
sin
sin 7
P 2 .l.CA
2
sin
sin B
sin y
sin
[Lame's theorem.
..(0
Now, since angles A, B and C, of the triangle ABC, are respec
tively equal to angles a, p and y f (the sides PQ, QR and RP being
perpendicular to BC y GA and AB respectively, we have
*This is so, because of the small dimensions of its faces.
" HYDROSTATICS
EC CA BA ("The sides of a triangle being
*L ^ _^L. = _f ~ . . (ft) proportion to the sines of the
sin a sin fi sin y Bangles opposite to them.
From relations (/) and (f/), therefore, we have
= P 2 =
v.e.> the pressures on the three faces are equal
Further, since the same relation holds good in whatever position
the prism may be rotated, it follows that a liquid exerts pressure
equally in all directions within itself.
N.B. It also follows from the above that if we replace the liquid prism
by a solid ons, of the sans size and weight, the forces acting on the latter would
.also be the same and hence it would also be in equilibrium.
150. Thrust on an Immersed Plane. If we have a plane hori
zontal surface of area A, immersed in a liquid of density P, the pressure
jPon it is uniform, since ail points on it are at the same depth h, say,
from the liquid surface, which, as we know, is also horizontal and,
therefore, parallel to the immersed surface.
So .that, P = h.p.g.
Now, force or thrust, i.e., F = pressure xarea = P. A = h.p.g.A.
If, on the other hand, the plane of the immersed surface be
inclined' at an an^le 6 to the liquid surface, (Fig. 216), we must first
determine the thrust on a
small area dA of the surface
and integrate its value over
the entire surface.
Let h be the depth of
this element of area dA.
Then, the thrust dF on this
area is clearly equal to
Ji p.g.d.A.
Now, if x be the dis Fig. 216.
tance of this element from
the line OF, in which the plane of the immersed surface meets the
liquid surface, we have
sin 6 = h\x. Or, h =x sin Q.
So. that, dF = p.g.sin b.x.dA.
And, therefore, the thrust on entire area A of the surface is given by
F =s fp.g.sin e.x.d A = p.g.sin d J x.dA. . .(/)
Now, x.dA is clearly the moment of the element dA of the sur
face about the lino OF, and, therefore, J" x.dA is the moment of the
whole area A about CF, i.e.,
$x.dA = A.X,
where ^.is the distance of the centroid* G of the area from CF.
 "The term 'Centroid' or 'Centre of mass* is ordinarily used synony
mously with "Centre of gravity'* and, in a uniform gravitational field, the two are
one and the sams point. Bat, in a nonuniform field, the weights of the particles
are not proportional to their misses. In such a case, therefore, the weights may
not form a system of parallel forces, reducing to a single resultant force, but
may form a couple, instead, varying with the different orientations of the body,
whereas the centre of mass is quite independent of the gravitational field.
PROPERTIES OF MATTER
So that, substituting the value of fx.dA in relation (i) for F above,
we have F = p.g.sin 0.A.X.
Again, if H be the depth of the centroid G of the immersed
surface from the liquid surface, we have sin 9 = H/X.
And, therefore, F = *.g.~.A.X.= H.P.g.A. . . (fl)
A.
Clearly, H.p.g is equal to the pressure at the centroid or the centre of
area. So that, we have
resultant thrust on the immersed plane = pressure at centroid or centre
of area x area of the plane.
It should be carefully noted that the thrust on the immersed plane
is quite independent of its angle of inclination (0) with the liquid
surface.
151. Centre of Pressure. Having obtained the value of the
resultant thrust on the immersed plane, our next step obviously is to
determine the point of the plane through which this resultant acts,
this point be ing known as the centre of pressure.
We know that the liquid pressure acts normally at every point
of the immersed plane. So that, the forces h.p.g.d.A acting on elemen
tary areas of the plane, (like dA), are so many like parallel forces. We
ma y, therefore, determine the centre of these parallel forces (i.e., the
point through which their resultant acts) b}^ an application of the
principle of moments, viz., that the algebraic sum of the moments of a
system of parallel forces, about a given axis, is equal to the moment of
the resultant about the same axis.
Now, clearly, moments of the thrust (or force) h.p.g.dA acting on area dA
about CF t (Fig. 216) = h.p.g.dA.x = p.g.x sin 0.dA.x. r ... .
. n x sin cr
= p.g f sin0.x*.dA. L
Therefore, total moment of the forces on such like elementary
areas dA of the plane about CF = J p.g.sin 9.x 2 .dA,
= p.g. sin 6 J x 2 .dA, ("')
where the integration extends over the entire surface of the plane.
Now, J x 2 .dA is the geometrical moment of inertia I g of the area
A of the plane about OF. So that,
total moment about CF = p.g. sin 8.1 ff .
And, since I g = Ak 2 , where k is the radius of gyration of the
area A about CF, we have
total moment about CF = p.g.sin Q.Ak 2 .
Again, if A" be the distance of the point P through which thfr
resultant thrust F acts on the plane, i.e., the distance of the centre of
pressure from CF, we have
moment of the resultant thrust about CF = F.X .
And, therefore, by the principle of moments, we have
= p.g. sin
whence, X,
H.Q.g.
HYDROSTATICS
349
Or,
..(iv)
B= XsinB.
If fc be the rqdius of gyration of the area about a parallel axis
through its centroid G, we hav, by the principle of parallel axes
f where 7 is its moment
7 = IQ^A.X*,  of inertia about f to axis
L through G.
Or, * Ak 2 = Ak*+A X\ whence, k*  k
And, therefore,
(v)
whence, Z may be easily determined.
Alternatively, equating F.X Q against expression (Hi) above, for
total moment about GF due to tho thrusts on elementary area dA, wo
have F.X Q = p.g.sw J x*.dA . . (vi)
And, clearly, if H Q be tho distance of the centre of pressure from
the liquid surface, we have 7/ /A" = s/Vz ; and, therefore,
A" ffjsin 0.
Putting this value of A^ in relation (v/) above, we have
F'Hn ^ ,
jw / ^ Pig ' 5w * ^ %
Or,
Or,
sin
.dA. \ '\ h
L and.'.
R.g.j//i 2  , dA
j oin (7
TT*
h ~ x sin 0.
/f
J//I
depend
the value of the integral J A 2 .<i4, like ths expression J
ing upon the shape of the immersed plane.
We thus see that whereas the distance of the centre of pressure
from the liquid surface is quite independent of the density of the liquid, it
depends upon the shape of the plane.
152. Particular Cases of Centre of Pressure. Let us now con
sider some simple cases of centre of pressure on surface of a definite
geometrical shape.
(/) Centre of Pressure on a Rectangular
Lamina. Suppose we have a rectangular
lamina of length I and breadth b, immersed
vertically in a liquid to a depth h below its
free surface, (Fig. 217).
Then, since for a rectangular lamina
7 = A.k Q * = /.6 3 /12, (see page 308),
where 7 is its geometrical moment of inertia
about an axis through its centre of area (or
centroid) and parallel to its length, and k ,
its radius of gyration about this axis, we have
Fig. 217.
350 PROPERTIES OF MATTER
A.k Q *=l.b.b*II2=Ab*l\2 t p.. ixb=A, the area of the
lamina.
And, clearly, X == h + ~ .
2i
b 2 , / i , ft
V2  + (//+
So that, * = V == " 12 "& ~
2
whence, the position of the centre of pressure for the rectangular
lamina may be easily calculated for any depth h.
Clearly, therefore, if the lamina be just submerged in the liquid,
i.e., with its upper edge just lying in the liquid surface, A 0, and we
therefore, have X$ = lb*/b = fr,
i.e., the centre of pressure, in this particular case, lies twothirds below
the top of the lamina.
(ii) Centre of Pressure on a Circular Lamina. Let the centroid
of a circular lamina, ot radius r, lie at a depth X from the free sur
face of the liquid.
Then, since & 2 for a circular lamina is r 2 /4, we have
r * +x*
X = X = X = 4X
r 2
Or, the centre of pressure lies at  ^ +/Y below the liquid surface.
//*, however, the lamina be submerged in the liquid, with its edge
just touching the liquid surface, we have X = r, and, therefore,
X  r2 +r = r +r  5 r
^~ 4r + 4 +r ~ ^~ r '
Or, the centre of pressure, in this case, lies at 5r/4 from the liquid
surface.
(iii) Centre of Pressure on a Triangular Lamina. Hero, two cases
arise viz., (a) when the lamina is immersed upright into the liquid, i e.,
with its vertex up and base down and (b) when it is immersed upside
down, with its base up and vertex down.
(a) When the Lamina is immersed Upright. Let h be the height
of the lamina and let its apex^be at a depth d below the free surface
of the liquid. Then, clearly,
k* = /i 2 /18 and X = d+\h. [ Its centroid lying 2/J/3
[below the apex,
* 8 
And, therefore, A>= 
_, =
"
HYDBOSTATICS 3511
If, however, the vertex of the lamina just touches the free surface'
of the liquid, we have d = 0. In this case, therefore,
X Q = 3A 2 /4/z = f h.
i.e., the centre of pressure now lies $ths down the height of the lamina.
(b) When the Lamina is immersed Upside Down. Again, let the
depth of the base of the lamina be d from the free surface of the*
liquid.
Then, we have Jr 2 = /z 2 /18 and X = (
And, therefore, X =
And, i/ Me lamina be just submerged in the liquid, with its base in the
surface of the liquid, we lave, d = 0. In that case, therefore,
/.., the centre of pressure lies a distance \h below the liquid surface.
153. Change of Depth of Centre of Pressure. Let a plane
lamina of area A be immersed in a liquid such that its centroid is*
at a depth X from the liquid surface CD,
(Fig. 218),
y
Then, if k Q be the radius of gyration X >
of the lamina about the axis AB passing ^J^v _ ./_ N V.
through G, in the pJane of the lamina, snd ^^j
parallel to the liquid surface CD, itBradiu8^..^i
of g} ration about a parallel axis, lying in ^f^E 7 ^^!^
the liquid surface, will clearJy be k l9 ^uc
that kf = k *+X*. r
Now, let the level of the liquid sur
face be raised through a distance h by Fig.
adding some more liquid to that already present. Then, clearly,
radius of gyration k z of the lamina about a parallel axis to AB, and'
lying in this elevated liquid surface CD 1 , is given by kJ = /:</
Subtracting one from the other, we have
k*k* = [k
Or, & a 2 &j 2 =
If X Q be the depth of the centre of pressure of the original
amount of the liquid from its free surface, we have
X Q = kf\X, [ See relation (/v), Page 349v
So that, after the addition of a liquid layer of thickness h, its depth*
becomes
jrt y i L _
352
PROPERTIES OF MATTER
whence, k,*+Xh  X.X Q ~. Or, k* = X.XJXh = X(X Q 'h).
And, if X " be the depth of the new centre of pressure from the
raised surface of the liquid, after the addition of the liquid layer, we
jfaave
*'(*  *'*<*+>
80 that, k^W = X '(X+h)X(X 9 'h). ...()
From relations (/) and (//), therefore, we have
XJ(X+h)X(XJh) = h(2X+h).
Or, X H (X+h) X.X a ' + Xh = 2Xh+h 2 .
Or, X Q "(X+h) 
Or,
X "(X+h) =
*.(* +fc)
Dividing throughout by (X+h). we have X Q " h +
&nd, therefore, the shift in the position of the centre of pressure is
liven by
Or, A O '~AY' 
It will also be easily seen that the distance between the depths
of the new centre of pressure and the centroid of the lamina is given by
Or, X Q "
which approaches zero as h approaches infinity.
Thus, the greater the depth of the liquid, the nearer does the
centre of pressure come to the centroid of the lamina ; so that, at an
infinite distance, the two must just coincide with each other.
154. Principle of Archimedes. Imagine a body ABCD to be
immersed in a liquid, (Fig. 219), and a vertical line GA to travel
round it, touching it along the line
AECF and meeting the liquid surface
in the curve GHKL.
Then, clearly, the resultant up
ward thrust F l on the surface ABCEA
and the resultant downward thrust F 2
on the surface ADCEA are given by
the weights of the liquid that would
occupy the spaces ABCKG and ADOKO
respectively, acting through their res
pective centres of gravity. So that,
the resultant thrust on the whole body
Fig. 219. is given by
HYDROSTATICS 363
F l F a = 'weight of liquid occupying space ABCKG minus weight of
liquid occupying space ADCKO.
= weight of liquid occupying space ABCD.
= weight of liquid equal in volume to that of the immersed body,
= weight of liquid displaced by the body.
In other words, when a body is immersed in a liquid, it experi
ences an upward thrust equal to the weight of the liquid displaced by
it. It can easily be shown that the same is also true for/ a body
which is only partially immersed in the liquid, the upward thrust oa
it being equal to the weight of the liquid displaced by its immersed
part. We may, therefore, generalise and state that
when a body is wholly or partly immersed in a liquid, it experi
ences an upthrust equal to the weight of the liquid displaced by it, (i.e.,
by its immersed part}.
This is known as the Principle of Archimedes*, for it was he
who first enunciated it.
The point where this upthrust acts is obviously the e.g. of the
displaced liquid, which is called the centre of buoyancy, the upthrust
being referred to as the force of buoyancy.
N. B. The applications of Archimedes Principle are many and. various
It gives us the method of determining ipecific gravities or densities of liquids
as well as the instruments, kn^wn as Hydrometers, with which the Degree
students are no doubt already familiar.
155. Equilibrium of Floating bodies. A body, immersed wholly
or partly in a liquid, is subject to two forces, viz., (i) its own weight
W> acting vertically downwards at its , v
e.g., G, and (il) the upihrust W* ', act
ing vertically upwards at its centre of
buoyancy B, (Fig. 220).
If these two points of applica
tion, (G and B), of the two forces : ~
respectively, coincide or lie in the
game vertical line, called the centre
line, the body sinks, just remains sus _}~\
pended (or float ing), or rises up, accord i ji~rJ I _ .!*'_"
ing as W is greater than, equal to, or
less than W. Fig. 220.
For, obviously, if W>W' t the body sinks further down, dis
placing more and more of the liquid and thereby increasing the
upthrust until the two balance each other, and the body just stays
there, i.e., is in equilibrium.
If W = W, then obviously, the two forces on it are equal and
opjtosite and their line of action being the same, they just neutralise
* Archimedes, (287212 B.C.), was a Greek philosopher. He was asked
by King Heiro, at Syracuse, to test the goldcontent of a crown. Engaged on
this problem, he suddenly discovered the law of upthtust, while taking a bath,
w hich enabled bun to determine the specific gravity, and hence the quantity of
gold in the crown, without in any way damaging it Overjoyed at his success,
he ran home, with the triumphant cry 'Eureka', 'Eureka'. 'I have found it, 1
have found it.*
354
PROPERTIES OF MATTER
each other and the body remains suspended or floating in the liquid.
And finally, if W <W ', the body rises up, so that a lesser volume of
it is under the liquid, i.e., it displaces a smaller volume of the liquid,
and the upthrust on it is now less. This rise of the body continues
until the upthrust is just equal to the weight of the body, and the
body then continues to float in that very position.
Thus, two conditions are necessary for the equilibrium of a float
ing body, viz., (/) its weight must be equal to the weight of the dis
placed liquid*, and (//*) the e.g. of the body and the centre of buoyan
cy of the displaced liquid must either coincide with each otherf or lie in
the same vertical line.
156. Stability of Equilibrium. If the floating body be tilted a
little to one side or the other from its original equilibrium position,
through a small angle 6, (Fig. 221),
so that the weight of the displaced
liquid, or the upthrust on it, remains
pj the same, then, since the shape of the
jz displaced liquid changes, its centre
of buoyancy shifts a little, say, in
to the position B' ; so that, the ver
tical line drawn through the new
centre of buoyancy B', meets the
old centre line in M. This point M
is called the metacentre of the body
and the distance MG', where G' is
its centre of gravity, is called its
metacentric height.
Now, whether the metacentre (M) coincides with, lies above, or
below, the shifted position of the centre of gravity (G') of the body,
depends upon the shapo of the body and determines whether the body
will be in neutral, stable or unstable equilibrium.
(/) Thus, in the case of a sphere, [Fig. 222 (/')], a tilt this way
or that brings about no change in the shape of the displaced liquid,
Fig. 221.
(/) Fig 222. (//)
the metaoentre coinciding with the e.g. of the body all the time.
*lt is for ihis reason that the weight of a ship or boat is often referred to
is {^displacement^ the weight of the water displace J by it being equal to its
own weight
 As happens in the case of a spherical body.
HYDROSTATICS
355
Fig. 222. (///)
It is, therefore, in neutral equilibrium and continues to float in ail
positions.
(ft") In the case of a rectangular body, floating in a liquid, as
shown in Fig. 222 (ft*), M lies above G' ', the position of the e.g. of
the body in its tilted position, and a couple is formed by W and W\
the two equal, opposite, parallel and noncollinear forces acting at G'
and B', which represent the shifted posi
tions of the e.g. of the body and the centre
of buoyancy of the displaced liquid respec
tively. This couple tends to rotate the body
back into its original position, thus making
its equilibrium a stable one.
Obviously, the moment of this restor
ing couple = W.G'M sin 0. It has been
appropriately called the 'righting moment',
(particularly in the case of ships and other
floating vessels), because it tends to bring
the body, or to 'right it' into its original
position.
(//'/) la the case of a rectangular body, floating in the manner,
shown in Fig. 222 (///'), M lies below G' ', and the body is, therefore,
in unstable equilibrium. For, the couple formed by W and W tends
here to rotate the body further in the same direction in which it has
been tilted already. There is thus no prospect of its coming back (or
being righted) into its original position.
Let us now discuss the problem in a little more detail, with
particular reference to a floating vessel or ship.
Identical consideration to the above applies also to a floating
Tihip ; so that, when the ship is 'on an even keel\ its centre of gravity
</ md the centre of buoyancy B, of tho displaced water, he in the
same vertical line, i e. 9 its plane of symmetry (W'} is vertical
(Fig. 220).
If, however, the ship rolls or gots tilted through an angle 0,
(Fig. 221), its plane of symmetry (VV') is no longer vertical, and, al
though this roiling or tilting does not alter its. e.g. with respect to the
ship, the centra of buoyancy shifts to B\ giving rise to the righting
moment W.G'M sin or W.h sin 0, where h denotes the metacentric
height G'M. If $ be small, so that sin 9 = 8, this righting moment is
equal to W.h 9.
It will thus be clear that the greater the value of h, the metacen
tric height of a ship (or a floating body, in general), the greater the sta
bility of its equilibrium. It is for tiis reason that heavy cargo is stow
ed as low as possible in a ship or that it is provided with a leaden
keel, to lower its e.g. or to increase h*.
" N.B The lowering of the c g. is, however, not quite so desirable beyond a
certain point. For, due to the waves ia the sea, the ship is subject to lateral for
ces in different directions and the moment of thsse forces can be quite consider
able if the e.g. of the ship is very low down, resulting in its being tossed about
*The~stabiHty may also to increased by making the ship quite broad at the
line, i.e.. ths lins along which it is just touched by the watersurface.
856 PROPERTIES^ OF MATTER
this way and that, which is obviously most unpleasant and annoying to those on
board. Judicious care must, therefore, be taken to lower the e.g. of the ship with
in reasonable limits.
157. Rolling and Pitching of a Ship The righting moment W h.O, acting
on the ship when it is tilted through a small angle results in its oscillation (or
rolling), as we have seen above; and if T be its natural period of oscillation, we
have
T = 2 V W [See pages 300301 ,
where /is the moment of inertia of the ship and M, the turning moment on it per
unit (radian) deflection.
Obviously, the turning moment per unit deflection is also equal to W.h ;
for, if = 1, the value of the righting or turning moment becomes W.h. So that,
substituting W.h for M, we have
It is thus clear that the period of rolling (T) of a ship is inversely proportional to the
square root of its metacentric height (h). A ship, with a small metacentnc height,
is, therefore, less liable to rolling. It is for this reason that large ocean liners are
designed to have a comparatively small metacentric height of just a few metres, for
small displacements, which obviously makes them much steadier. At the same
time, however, to avoid the danger of the ship turning over or capsizing, if the
deflections be large, the designing is such that the metacentric height increases for
large deflections*.
Similarly, to avoid 'pitching', or tilting of a ship in the direction of its
length, its metacentric height in this direction also is suitably adjusted.
158. Determination of Metacentric Height. The displacement
of the ship through an angle B causes a wedge shaped portion of the
ship, (shown shaded in Fig. 221), to be immersed on the right hand
gide and an equal wedgeshaped portion of it to rise out of the water,
on the left hand side. Let these wedgeshaped portions be divided
into a number of elementary vertical prisms, by planes perpendicular
to the water surface, on either side, and consider one such prif m, of
height //, at a dt&tance x from O, where the plane of symmetry meet&
the water surface.
Since is supposed
small.
Then, clearly, H *= x tan Q =3 x.0.
If the base area of the prism be dA, we have
volume of the prism = x.Q.dA, an its mass = x.O.dA.p,
where P is the density of water.
Clearly, therefore, weight of the prism, or the weight of the water dis
placed, i.e.,
the upthrust on the prism = x.Q.dA.p.g.,
and its moment about O =x0 dA.p.g.x = p.g.0.x*.dA.
Similarly, considering the equal wedejeshaped portion on the left
hand side, we find that there is a loss of upthrust due to its rising out
of the water, whose moment about O is, obviously, also equal to
fi.g.Q.x z .dA> the direction being anticlockwise in either case. Hence,
the moment of the couple acting on the ship (or a floating body, in
general, due to its displacement 0, is given by
J P.g.0.x 2 .J4 = P.g.0. jx*.dA = p g.OAk* ....... (/)
The rolling motion of a ship can be greately minimised by the use of a
Gyrocompass (See page 98).
HYDROSTATICS 367
where / x*.dA = I g =.y4fc 2 , the geometrical moment of inertia of the
surfaceplane of the liquid about the axig through O, k being its
radius of gyration about this axis.
This displacement (0) of the ship being small, the volume V of
water displaced remains unaffected by it, and the upthrust p g.F, due
to this displaced water, acts through its new centre of buoyancy after
the displacement 0. The floating body or ship is thus acted upon by
a couple equal to
p.g.V.BM sin 6 = p.g.V.BM.O. ...(ii) [Q being small.
Equating the two values of the couple, we have
p.g.V.BM.O = 9 .g.0.Ak*.
Or, V.BM = Ak*. Or, EM =
So that, the metacentric height h of the ship or the vessel is given by
Ak 2
<}'M=BMBG' = BG' and may thus be easily determined.
Alternatively, in the case of a ship, its metacentric height
may be easily determined by moving a known weight w from
point A to another point B across
the deck, say through a distance */,
i.e., AB = d.
Now, this shift of weight w
from A to B is equivalent to an
,upward force w at B and a down
ward force w at A, (Fig. 224), thus
constituting a couple, of moment
w.d cos 6. For equilibrium, there
fore, this must be equal to the
couple due to the weight W of the
ship and an equivalent upward ^^"^^^^3p~^^^ T ?j^ r r
thrust at the new centre of buoy ~ ~r_:r _rjr~ JT
ancy B', i.e.. equal to couple of mo Fig. 224.
ment W.G'M sin 9. [See above]
Or, W.G'M sin = w.d cos 8.
s\ rntr W.d COS 6 W.d n W.d 1
**> GM = W sine W ' C te ^tanO
Or G'M = Wt(i T $ being smaU '
W.6 L rfl/l ~ 0.
Thus knowing w, W, d and 0, we can easily calculate the metacentrio
height of the ship.
159. Pressure due to a Compressible Fluid or a Gas. A gas
differs from a liquid in that, unlike the letter, it is highly compres
sible* and, therefore, also highly expansible, tending to expand
perpetually and indefinitely.
*An idea of the high compressibility of a gas, compared with that of a
uquid, can be had from the fact that whereas the density of sea water at a deptb
3f 5 miles is about the same as that of the surface layer, the density of the
atmosphere at the same height above sea level is reduced to one quarter of that at 1
the latter.
PROPERTJ&S OF MATTER
The pressure exerted by a gas, is thus fundamentally different in
the nature of its cause from that of a liquid and cannot be taken to be
proportional to the height of the gaseous column, for the simple
reason that, except in the case of a small volume of a gas, the
density goes on progressively increasing as we go further down the
column, due to the layers above pressing down upon the layers
below, thus giving riso to a well defined pressure gradient ail along the
column.
So that, whereas a liquid everts pressure only under the action
of gravity, i.e., due to its weight, or due to an external force applied
to it, as, for example, when it is pressed down by a piston, the pres
sure due to a gas is entirely a consequence of the incessant mobility and
the kinetic energy of its molecules, or due to what Boyle called the 'spring
of the gas.
The mad and random motion of the gaseous molecules results
in their colliding not only agairst eech other but also against the
walls of the containing vessel, and it is this bombardment of the
walls by the fast and haphazardly moving molecules that causes the
pressure. (See Chapter XV).
We are here concerned mainly with the pressure exerted by the
gaseous mantle or envelope, surrounding us over land and sea alike
and in all latitudes, which we call 'air' or 'atmosphtre' and which, as
we know, is a mixture of a number of gases and vapours, In
pursuance of its inherent property of indefinite expansion, this air or
atmosphere should expand to an infinite distance above the earth,
but the earth's gravitational attraction on a huge mass like it sets a
limit to its expansion. Even so, it has been known to exist up to a
height of 300 miles from the surface of the earth, although even at
25 miles or so, its density and pressure start falling oif so rapidly
that at altitudes above 300 miles, it may be said to be as good as
non existent, \*ith just a void or a vacuum beyond.
Now the atmosphere can be divided into two very distinct
regions, viz., (i) a lower region, called the troposphere or the convective
zone, and (//) an upper region, called the stratosphere or the advective
zone, the surface ol separation of the t\*o being known as the tropo
pause, which varies with the latitude and falls from a height of about
14 kilometres at the equator to about 8 or 10 kilometres at the poles,
and is found to be higher in summer than in winter.
(0 The Troposphere This extends to a height of about 6 miles at the
poles and about 10 miles at the equator, with a vertical distribution of temperature
as its chief characteristic, the temperature falling off rapidly with altitude, there
being a vertical temperature gradient or a lapse rate* of 1C per 500 feet rise in
altitude.
This temperature gradient is probably due to a variety of causes. Lord
Kelvin attributes it to the atmosphere being in a state of C3nvective equilibrium,,
which is brought about, on the one hand, by the earth getting hcaied by the
solar radiation Jailing on Uf, and then warming up the layers above it, by direct
*UsuaUy, a vertical temperature gradient is taken 10 bd the fall in tem
perature per 100 metres rise in altitude and the lapse rate, as the fall in tem
perature per one kilometre rise in altitude.
t Little or no heat is absorbed by the air during the passage of the solar
radiation to the earth through it, and whatever little is, is distributed over too
large a mass to be able to produce any appreciable rise in its temperature, this
absorption being the same at all altitudes,
HYDROSTATICS 359
contact and by emitting out radiations which are absorbed by them, and, on
the other, by the lower atmosphere getting cooled by radiation due to its
emitting out more energy than it absorbs at the ordinary temperature. The
two processes, going on side by side, produce changes in the density of the air,
conducive to the setting ip of vertical convection currents, the lower warmer
air rising up and getting cooled by adiabatic expansion and the upper coldei
air coming down and getting heated up by adiabatic compression. A vertical
temperature gradient is thus established and maintained throughout this region
of vertical convection. Hence the name, 'convective zone* also given to it.
This seems to be amply borne out by the fact that the lapse rate for
dry air, calculated on this assumption, comes out to be 3C per 1000 ft., which,,
though appreciably higher than the observed value, is quite understandable,
considering that the air is really never 'dry' and the moisture present in it
inevitably tends to lower the lapse rate.
(//) The Stratosphere. Also known as the a&dctive zone, it is the regidb
above the troposphere, where the vertical convection, relerred to above,
becomes much too feeble, with the temperature falling to such an extent that
the heat radiated out is equal to the heat absorbed from radiations from the
earth and the solar radiation parsing through it, there being set up a radiative
equilibrium in the region, the temperature remaining constant at about 55C,
hence the name, 'isothermal layer* , also given to it.
It will thus be readily seen that the stratosphere is a direct consequence
of, and is characterised by, the cessation of vertical convection and the setting
up of a radiative equilibrium, with the temperature constant at 55 c Cuptoa
height of 300 miles or so, after vihich it probably shoots up to 700C or there
abouts.
160. Measurement of Atmospheric Pressure. The instruments
used to measure the atmospheric pressure are known as barometers,
one of the bast forms of which is the cisterntype Foriin's baro
meter. Another hand} 7 and portable type of barometer is the
Aneroid barometer, (from 'a 9 without, and 'neros' liquid) s no
mercury or any other liquid is used in it. We have already studied
these in good detail in the junior classes and shall not, therefore,
repeat them here. Instead, we shall pass on to a consideration of the
corrections that must be applied to the readings obtained from them.
161. Correction of Barometric Reading. Although the Fortirfs
barometer is quite an efficient instrument, a few corrections Lave to
be applied to its readings for greater accuracy. We shall consider
here only t\vo important ones of them, v/z.,
(/) correction for the expansion of the brass scale, on which the
reading is taken, and which is usually calibrated at 0C ;
(//) correction for expansion of mercury, and consequent lowering
of its density.
(/) Let the temperature, at which the reading is taken, be /C,
and let H t be the observed reading at this temperature. Then, H t is
in fact just the value of the divisions of the scale, correct only at 0C.
If, therefore, a be the coefficient of linear expansion of brass, the
correct length at tC, is given by H = H t (l+at) cms.
(ii) Again if v and P O be the volume and density of a certain
mass m of mercury at 0C, and v t and P/, its volume and density res
pectively, at tC, we have m = v .P = v t?f
^ P n V (l+70 _ JL* r whe r e '>' is the cofficient of
Or, ^ = . ur,  ^  ^cubical expanssion of mercury
Or, po/p, = 1+yt, whence, Po = p,(l+70
360 PROPERTIES OF MATTER
Now, clearly, H Q .? Q .g = tf.p,.g,
where H is the true barometric height at 0<7.
Or, # . Po = H.? t . Or, // O .p/(l+70 = H.? t = //,(l+a/). P< ,
B
whence, #
neglecting squares and higher powers of a and J.
Or > # = #,!l(y<*)']
For mercury, y = 00018 and for brass, a =000019.
#o = fli[l  (00018 000019X]. Or, #,=jf7,(l'0001610,
whence, the barometric height at 0C can be easily calculated.
Other errors, due to pressure of mercury vapour and capillarity
etc., are much too small for tubes of reasonably wide bores, and are,
therefore, usually neglected.
162. Change of Pressure with Altitude. Consider two points
A and B 9 distance dx apart, vertically below each other, in
air, (Fig. 225). If A be at a height jc above the ground,
, the height of B from the ground is obviously (x+dx).
x * Since the density of air and, therefore, its pressure, de
creases with altitude, for a pressure p at A, that at B will
T be pdp, say. If, therefore, p be the density of air bet
j ween A and B, and g, the acceleration due to gravity, we
x have dp = p.g.dx, ...(/)
I the ve sign being used, because the pressure decreases
i _ with height.
Fig. 225. If the temperature of the air be constant, P oc p.
~ r n T v l/^ A (Boyle's law)
Or > P = K 'P> [ and l\V oc p.
where K is a constant, equal to p//?.
Substituting this value of p in relation (/) above, we have
dp  K.p.g.dx. Or,  2 = K.g.dx.
Or, dptp+K.g.dx = 0.
Integrating this, we have
log e p+K.g.x s= a constant C. ...(#)
Kfow, if the pressure at heights h and I/be p and P respectively,
we have, from (ii) above,
log, p+K.g.h. = C, ...(/ii) and log, P+K.g.H = C ...(/v)
.. subtracting (iv) from (in). w h av
log, plog, P = K.g.HK.g.h = K.g.(Hh).
Or, log,(^) ^K.g.(Hh). ...(v)
Thus, (///( ) = ~  ...(v)
HYDROSTATICS
361
Or, substituting the value of K 9 i.e., p//>, in (v/), we have
(Hh)
.P
p.g P
Thus, knowing p 9 P, p and g, the altitude (Hh) can be easily
determined.
In the above treatment, it has been assumed that the tempera
ture of the air, or the atmosphere, remains the same throughout.
This, as we know, is far from being the case. Nevertheless, the result
is accurate enough for the determination of small heights.
If now we have a number of heights, A 1? A 3 , A 3 etc., in arithme
tical progression then (A a A,) = (63^2) and s6 on 
And, if p L9 /> 2r A etc., be the pressures at these heights, we have
from (v) above,
log, (pjp 2 ) = K.g.fahJ, and log, (pM = K.g.fahJ.
Since (A 2 Aj = (A, A a ), we have
log, (ft/A) == lo S (A/ft)
Or, A /A = A/A
f ^> Pv A A etc  are in S eometrlcat P r Sression.
Thus, we see that as the height or altitude increases in arithmetical
progression, the pressure decreases in geometrical progression.
Note. To convert logarithm* to the base e into common logarithms, (i.e.,
o the base 10), multiply the former by 2'302.
SOLVED EXAMPLES
 A . . . , r f t long and 5 ft. wide is filled with water
A !u ^ r J f ct l ng l u . lar clster V Af water to weigh 62'5 Ibs., find the magnitude
a a depth of 3 ft. Taking one cu. ft. of water w B
nd position of the resultant fluid thrust on
ach side.
2ft
(a) Here, clearly, (Fig 226),
depth of water 3 ft.
. centre of area for each side of the cistern
= 3/2 = 15 ft.
*. pressure at centre of area = h.p.g.
= I'5x62'5x32poundals.
= 15X625 lb. Wl . Fig 7 226
Now, area of each longer side in contact with water = 3 x 3 = 9 sq. ft.
and area of each smaller side in contact with water = 2x3=6 sq.ft.
Since thrust pressure at centroid or centre of area X area,
we have pressure on each longer side = 1'5 x 62 5 x 9 = 843*7 Ibs. wt.
and, pressure on each smaller side = 1'5 x 62 5 X 6 = 562*6 Ibs. wt.
And, centre of pressure = ~  x depth =   x3 == 2/r.
2. Find the position of the centre of pressure of a triangular plate immers
ed in a liquid with its plane vertical and one side in the surface.
ABC is a vertical triangular door in the side of a ship, AB is horizontal,
C below AB, and the triangle equilateral of side 5 ft. The door is hinged along
AB, and kept shut against the pressure of the water by a fastening at C. If the
362
PBOPEETIES OF MATTER
water rises to the level of AB, find the force on the fastening. (One cu. ft. of water
weighs 625 Ibs.) (Liv. Inter.)
The centre of pressure of the triangular plate, with one of its sides in the
plane of the liquid surface will be at a depth /z/2, from the liquid surface,
where h is the height of the plate, (see page 351).
Let ABC be the triangular door hinged
along AB and having a fastening at C wheie
ABlies in the plane of the water surface, (Fig.
227).
Obviously, height h of the triangular door
I 75 = 4 329 ft.
Fig. 227.
Therefore, pressure at the ccntroid=
Since the centre of a triangular lamina is
\rds of its height below the vertex, its depth be
low AB, or the water surface, is $rd ot its height,.
,e., equal to Jx4329=r443/r.
Ibs wt. = 1443x625 Ibs. wt.
And /. thrust on the door = 1 '443x62 5 Ibs. X area of the door.
= 1 443 x 625 x i base x altitude = 1 "443 x 62 5 x i x 5 x 4329 Ibs. wt.
= 1443 x 625 x 25 x 4 329 Ibs. wt.
Now, centre of pressure of the triangular door lies at i h, i.e., at
ix4329 = 2 164ft. [See above
/. moment of the thrust about AB
=~t'nrustx depth of centre of pressure.
= 1443x62 5 x2 5 X 4329x2 J 64 /6s wt.
And, if Fbe the magnitude of the force on the hinge, its moment about
Ibs. wt.
Clearly, therefore, Fx4 329=1443x62 5 x2'5x 4 329x2 164.
Or, F=l 443x625x25x21644875 Ibs. wt.
Thus, the force on the fastening at C = 487 5 Ibs wt.
3, Find the centre of pressure of a rectangular sheet 'a' in, long and 'b' in.
wide, of uniform thickness, immersed in a liquid of unifoim density, with one side
of length V?' in. in the surface, the plane of the rectangle being inclined at a
angle to the vertical.
If the rectangular sheet remains in the same position with respect to the
\essel containing the liquid, and the depth of the liquid be increased by h in , find
the new position of the centre of pressure. (London Higher School Certificate)
Let^BCDbe the rectangle, immersed in the liquid, of density p, with
its side AB = b in the liquid suiface F, and its plane loclincd to the vertical
a tan angle 0, (Fig. 228)
Then, since the vertical depth of the rect
angle is clearly BK^BC cos = a cos 9, its cen
troid
lies at a vertical depth r cos 0, from the
liquid surface.
And, hence, proceeding as in 152 [case
(i), page 349], we have
depth X . ef the centre of pressure P from
the surface of the liquid, clearly given by
X =  x vertical depth = a cos Q.
Fig. 228.
Now, let a column of liquid EE'F'F, h in. thick, be added on to the top
of the liquid surface to increase its depth by h in., and let P' be the new centre
of pressure of the rectangle, whose position is otherwise unchanged with respect
to the vessel; at a distance XJ from the new surface E'F'.
HYDROSTATICS 363
Then, clearly, thrust on the rectangle due to the original liquid column
area of the rectangle X depth ofcentroid G x density oj the liquid* g
=ab^ . cos e x p xg = r a*b cos 6 x p xg ;
and increase in thrust on the rectangle due to the new layer h of the liquid added'
of rectangle x depth of new liquid column added x density of liquid* g.
So that, the total thrust on the rectangle = \a*b cos 9 X P x g} ab X h x p X g.
~ab$g(\a cos 0f/z).
Clearly, therefore, the moment of this thrust about the new liquid surface E'F'
~ab9g(\a<osQ + h).X Q '. ... ..... (i)
Again, the distances of the new centre of pressure P' and the centroid of the
rectangle from the new liquid surface E'F' = (X ^fi) = ( a cos h h) and
(J a cos 0f/z) respectively.
And, therefore, the moment about the new liquid surface E'F' is also equal to
ka*b>cos 6>pg(% a cos f h) + ab.h pg. ($ a cos 6 +h),
=abpg[$ a cos B ( a cos e+A)+Mi a cosQ + h)]. (//)
Equating (/) and (//), therefore, we have
cos e + /0 Xo'=a.b.p.g[k a cos 0(*. a cos 6h/iH /*(i a cos Q+h)]
Or, ( a  cos  * '= K cos* o + lah cos <j + $ah cos
Or, A'o', or the depth of the new centre of pressure of the rectangle from
the new liquid surface
_ .
( ro5^ + 2/z\
 2 )
4. Neglecting atmospheric pressure, find the depth of the centre of pressure
of a circular lamina just completely immersed with its plane vertical in an incom
pressible liquid.
A circular door in the vertical side of a tank is 'hinged' at the top and
opens inwards, and the tank contains water to a height just sufficient to cover
the door. If the diameter of the door is 2 ft , find the magnitude of the force that
must be applied normally to the centre of the door in order just to open the door.
Find also the reaction at the hinge when this force is being applied.
(Cambridge Higher School Certificate)
The centre of pressure of the vertical circular lamina, just immersed in an
incompressible liquid, is equal to 5r4, as explained in 152 (//;, (page 350).
Here, obviously, the centroid of the circular lamina is at its centre, at a
depth equal to its radius from the water surface, / e. 9 at a depth 2 ft. 12 or 1 ft.
below it, (the lamina bemgyw^/ immersed in water).
pressure at the lamina = // p.# = 1 x 62 Sxgpoundals = 62'5 Ibs. wt.
and area of the circular lamina = nr 2 = TT x 1 = TT sq. ft.
So that, thrust on the lamina = pressure on the centroid x area of the lamina
= 62*5 XTT /fo. wt. = 1963 Ibs. wt.
This thrust acts at the centre of pressure of the lamina, whose depth from
the water surface, as we know,
= 5r/4 = 5x1/4= 5/4 ft.
.'. moment of this thrust about the liquid surface, or about the hinge
= 625 XTTX 5/4 = 2455 Ibs. wt.
If F Ib. wt. be the force applied to the centre of the door, just to open it,
we have
moment of F above the hinge = Fx r = Fx 1 Ib, ft.
And /. Fx 1  245*5. Or, F  245*5 Ibs. wt.
364
PBOPERTIBS OF MATTER
The force required to be applied to the centre of the door to just open it is thus
equal to 245*5 Ibs. wt.
Thus, the forces acting on the two sides of the door are respectively 245 '5
Ibs. wt. and 196*3 Ibs. wt. ; and, therefore, reaction at the hinge
= 245 51963 = 49 '2 Ibs. wt.
5. If a load of 20 tons, moved 50 ft. across the deck of a ship of 20,000
Ions displacement, causes the ship to tilt through i, vhat is its metacentric
height ?
Let a load w 20 tons be moved across the deck of a ship from A to B
through a distance of 50//. (Fig. 229;, and let the ship be tilted through an angle
i into the position shown. This, as we
_r _~^ .^__~ ______ _i  know, is equivalent to the application
of a couple, of moment 20x50 tonfeet,
tending to turn the ship clockwise, [see
158 (alternative treatment), page 357}.
And, the couple, formed by the
weight of the ship acting vertically at
/X2 its c g., G' and the equal weight of dis
JjJE placed water at its centre of buoyancy B 1 ,
^= tends to restore the ship back into its
original position, the moment of the
couple being WxMP. where MP is the
perpendicular from the metacentre M on
to the vertical line through G 1 .
Since the ship is in equilibrium
under the action of these two opposite
couples, they must obviously be balanc
ing each other. So that,
Or, 20,000 x MP = 20 x 50.
= 1/20 /V. = 05 ft.
'= AfG'X'0087,
Fig. 229.
= 20x50.
Or,
Now,
20x50/20,000
MP^ MG' sinV
whence,
MG'
0087
= = 5748 ft.
Thus, the metacentric height of the ship = 5 '748 ft.
6. State the theorem of Archimedes, and explain what you understand by
the terms "force of buoyancy", "centre of buoyancy". A cylinder of radius 1 cm.
and length 4 cms., made of material of specific gravity 0*75 is floated in water with
its axis vertical. It is then pushed vertically downwards so as to be just immersed.
Find (a) the work done, (b) the reduction in the force on the bottom of the contain
ing vessel when the cylinder is subsequently taken out of the water,
(Oxford and Cambridge Higher School Certificate}
Here, the volume of the cylinder is clearly equal to m*l = TT. 1. 4 ccs.
And .'. its weight = n X 4 x 0*75 = 3rc gms. wt.
Since it floats vertically in the water, we have
weight of displaced water also = 3 n gms. wt.
Let length of the cylinder inside water be = x cms.
Then, the volume of the immersed part of the cylinder = IT. 1.x =* *x c.cs.
i.e., volume of water displaced by the cylinder = *x c.cs.
And /. the weight of this displaced volume of water = it.x.l gms wt.
Hence, *x = 3^. Or, x 3 cms.
i.e., length of the cylinder inside water = 3 cms.
And, therefore, length of the cylinder outside water = 43 = 1 cm.
Thus, to immerse the whole of the cylinder just inside the water, we have
to simply push it down through 1 cm.
Obviously, the volume of the displaced water or upthrust on the cylinder
will be TT x 1 x4x 1 = 4* gms. wt.
Hence, increase in the upthrust on the cylinder will be = 4rc 3* = n gms. wt.
HYDROSTATICS 365
Since this increase in the upthrust takes place gradually from to w, a*
the cylinder is pushed steadily down into the water from its initial position, we
may take the average value of the upthrust against which we work in pushing the
cylinder down through 1 cm. to be (0j7i)/2 or ir/2 gms wt.
And, therefoie, work done in pushing the cylinder down through 1 cm ,
against this average increase in the upward thrust, will clearly be equal to
average thrust x the distance through which the cylinder is pushed down.
i.e , work done = (^12) x 1 ?i/2 = 1571 gms. wt. cm.
Now, with the whole of the cylinder immersed in the water, the weight of
water displaced
= 4n gm. wt. = 1257 gm. wt.
This must also, therefore, be the downward thrust ori the bottom of the
containing vessel. So that, when the cylinder is removed out of the water, the
reduction in the thrust on the bottom of the containing vessel will also be the
same, viz., 1257 gm. wt. '
7. Calculate the metacentric height and determine the necessary condition
for the stable equilibrium of a cylinder of length /, radius r, and density p, floating
vertically in water.
Let a portion x of the cylinder be inside water.
Then, volume of water displaced by the cylinder, i.e., v n r z .x c.cs.
And .*. weight of water displaced or upthrust on the cylinder nr*.x.l m z x.
This must, for equilibrium, be equal to the weight of the cylinder, i.e.,
equal to w.r 2 . logins wt.
Or, *r 2 .x = 7tr 2 ./.p gms. wt. Therefore, x / p.
The centre of buoyancy of the displaced liquid must, therefore, be at a
height xj2 = /p/2 from the bottom of the cylinder.
Now, as we know, the distance between the centre of buoyancy of the
displaced liquid and the metacentre is Ak 2 jv t (see page 357), where Ak z is th(
moment of inertia of the surfaceplane of the cylinder about its diameter. Sc
that, /c 2 =/ 2 /4,
k being the radius of gyration of the plane about the surfaceline or the diamete
of the cylinder.
Substituting the value of v = iir 2 x, therefore, we have
distance between the centres of buoyancy and metacentre
rtr* x
Now, distance of the e.g. of the cylinder from the centre of buoyancy
1 _ *  JL JP .JLLiP)
""2 2" ~~ 2 ~ 2""~ 2
And, therefore, distance between the e.g. of the body and the metacentre
or the metacentric height, h, of the cylinder = distance between the centre o
buoyancy and the metacentre minus distance between the e.g. of the body an<
the centre of buoyancy.
r a /(1p) _ r 2 ~2/p/(lp) _r a 2/ 2 p(l~p)
Or, n  ^ 2   ^ p   4/p
Now, for stable equilibrium of the cylinder, the metacentre should be
above the e.g. of the body, i.e., h should have a positive value.
And, obviously, this is possib^ only when r 2 >2/ 2 p<l p). This is, therefore,
the necessary condition requirea,
EXERCISE IX
1. Define pressure at a point in a fluid. Find the total thrust on the
sides and vertical ends of a Vshaped trough, 1 ft. deep, 2 ft. wide at the top
and 4ft. long, when nlled with water, density 62*5 Ib.jcu. ft.
(Oxford and Cambridge Higher School Certificate)
Ans m Ibs. wt.;2Q'Zlbs wt.
2. Define clearly the term 'Centre of pressure 9 . Determine its position
for (/) a circular lamina of radius r t just immersed vertically, and (11) a triangular
366 PROPEREIES OF MATTER
lamina of height h, immersed vertically with (a) its apex, and (b) its base in the
mrface. Ans. (/) 5r/4 ; () (a) 3H/4 ; (b) h/2.
3. A square lamina with its sides 3ft. long is just immersed vertically
a water with an edge in the surface and is then lowered 10 ft. Find the dis
,ance of the centre of pressure in the new position from the centre of the square
;Neglect the pressure of the atmosphere in each case).
(Joint Matriculation Board and HS. Certificate)
Ans. 00303 //.
4. A circular area of radius a ft is immeised in water, with its plane
vertical. The surface of water rises from 2a ft above the centre of the circle to
[aft above it. Neglecting atmospheric pressure, prove that the centre of prts
,ure rises through a distance a\ 16 ft (London Higher School Certificate)
5 State the Principle of Archimedes and define clearly the terms (i)
Centre of buoyancy, (it) Metacentre, and (///) Me tacentric Height. Discuss in
^detail the conditions for the stable equilibrium of a floating body, with particu
lar reference to a floating ship.
6. Show that if a floating body be given a small rotational displacement
in its plane of symmetry, the distance between the centre of buoyancy of the
displaced liquid and the metacentre is Ak^\V \ where A is the area of the surface
plane of the body, k, the radius of gyration about the surfaceline and V, the
volume of the displaced liquid
Discuss the conditions necessary for a hollow cylinder of height h, and
density p, open at both ends, with i\ and r a as its internal and external radii, to
float vertically in a liquid in stable equilibrium. Ans. r^f r 2 2 >2/z a .p(l p)
7. Discuss how the atmospheric pressure changes with altitude above
the surface of the earth, the temperature remaining constant, and show how if
the altitude increases in arithmetical progression, the pressure decreases in
geometrical progress ion.
8. A mercury barometer is known to be defective and to contain a small
quantity of air in the space above the mercury. When an accurate barometer
reads 770 mm , the defective one reacts 760 mm and when the accurate one
reads 750 mm., the defective one reads 742 mm What is the true atmospheric
pressure, when the defective barometer reads 750 mm. ?
(Cambridge Scholarship} Ans. 758'8 mm.
9. A simple barometer has the glass tube attached to a spring balance.
What weight does the balance record when the open end of the tube is just dip
ping under the surface of the mercury in the reservoir, and what changes occur
when the tube is lowered so that m:>re of it dtps under the mercury ?
(Oxford Higher School Certificate)
Ans. (/) The balance records the weight of the tub , and the mercury column.
(//) A progressive decrease in the weights, due to buoyancy of the tube, un
til the fatter is full, when finally it decreases to zero.
10. A sealed spherical cellophane balloon has a diameter of 5 metres and
weighs, with the apparatus it carries, 1 k.gm* It contains onetenth of the
volume of hydrogen required to fill it at atmospheric pressure. The balJoon is
illowcd to ascend ; if the cellophane does not expand and if the temperature of
,he atmosphere is assumed to be constant at 0<7 at all heights, calculate at
vhat height the envelope becomes full and the height to which the balloon rises.
The pressure p at height h (km.) is related to that at the ground (p Q ) by the re
ation/i = 20 log lo ip n lp). (Densities of air and hydrogen atO'O and atmos
)heric pressure are 1'29 and 09 gm per litre respectively).
(Oxford and Cambridge Higher School Certificate)
Ans (/) 20 km. and () 34'5 km.
CHAPTER X
FLYING MACHINES
Jet Planes, Rockets and Satellites
163 Flying Machines. There are two types of flyingmachines,
mz., (/) light erthanair machines, or Air ships, (//) heavierthan*air
machines, or Airplanes.
The Airship. An airship is based on the principle of Archimedes. The
weight of the air displaced by it is greater than its own weight, i.e., the upward
thrust on it, due to the displaced air, is greater than the 1 downward thrust, (due"
to its weight), and hence it rises up.
An airship is "in fact a big cigarshaped balloon of a light material, like
aluminium or its alloy, covered with a specially treated water proof linen or silk
and divided up into a number of compartments contain iag bags filled with a
'light g is like hydrogen or helium (preferably the latter, due to its noninflamma
ble nature) from which it derives its buoyancy, i.e., which makes the total weight
of the airship less than the weight of the air displaced by it, or the upward thrust
on it greater than its weight. This excess of upward thrust that it possesses over
jts weight is called its liftm ; power, and gives the maximum extra load it can be
made to carry.
For steering purposes, it is fitted with rudders or other suitable devices and
for horizontal motion, it is provided with propsllers, worked by light and power
ful engines.
164. The Kite. Before dealing with the airplane, it will be help
ful to study the principle underlying the ordinary kite. This will be
understood from the following :
Let AB, (Fig. 230) represent the midline of the kite. Then,
the different forces ou it are (/) its weight W, acting vertically down
wards at its e.g., G. (//) the tension
T of the string, acting along the
string, as shown, (///) the pressure
due to the wind, acting along the
direction of the wind, all along the
undersurface of the kite.
Now, this pressure of the
wind may be resolved into t\\o
rectangular components at all
points, (/) perpendicular to the
plane of the kite, and (b) along the
plane of the kite. These latter com
ponents play no part in supporting
the kite and may thus be ignored ;
and the former components p, p...
being so many like parallel forces,
have a resultant P, equal to their sum, called the effective pressure of
the wind, acting at the point C. Thus, the three forces acting on the
kite are
(/) W> acting vertically downwards at G, the e.g. of the kite.
367
368 PROPERTIES OF MATTER
(//) T, acting along the string at E, and
(Hi) P, acting perpendicularly to the plane of the kite at C.
Condition for the Kite to be in Equilibrium. In order that the
kite may be in equilibrium, these three forces acting on it must be*
represented by the three sides of a triangle, taken in order. And in
order that this may t>e so, they must all meet in a point, say, at O.
It will easily be seen that this can be possible only when G, the e.g.
of the kite, lies below the point C, where the effective pressure (P) of
the wind acts. In other words, the e.g. of the kite must be pretty low
down for it to be in equilibrium. It is for this reason that the lower
part of the kite is made slightly heavier, and that a small paper tail
is sometimes attached to it, which, in addition to bringing its e.g.
down, also makes it look more attractive.
Condition for the Kite to rise up. The effective pressure P
of the wind may also bo resolved into two rectangular components,
viz.,
(/) D, along the direction of the wind, called the drift, or the
drag.
(ii) L, upwards, perpendicularly to //, called the lift.
If be the angle that the kite makes with the direction of the;
wind, we have
Drift (D) = P cos (90 0) = P sin 0,
and Lift (L) = P sin (90 0) = P cos 6.
Similarly the tension (T) of the string may be resolved into two
rectangular components, (/) along the horizontal and (ii) downwards,*
along the vertical, (shown dotted).
Now, clearly, the only force tending to make the kite rise up
wards is the lift L = P cos 0, and the forces tending to make it fall
downwards are (/) its weight W arid (//') the downward component of
the tension of the string. The moment, therefore, that the lift (L) is
greater than W 4 the downward component of T, the kite rises upwards.
Thus, to make the kito rise up, we must increase the lift, i.e^
P cos 6. This can be done by increasing P, i e. t by running against
the wind, and by decreasing 0, by giving small jerks, (Tanka) to th&
string. If, however, the drift (P sin 0) be greater than the lift
(P cos 0), the kite drifts along in the direction ol the wind.
165. The Airplane. An airplane is a heavierthanair machine
and the principle underlying it is in main the same as that of
the kite. Obviously, however, there is no tension of the string,
here, pulling it downwards, so that the only force ttnding to
take it up is the lift and the force tending to take it down is its
weight.
Farther, as the propeller blades rotate rapidly, they throw the
air backwards from in front of the plane, and its reaction is a thrust
R, forwards.
Let us now consider the relation between these different forces'
on the plane in the different phases of its flight, viz., (i) when itflie&
level, (ii) when it climbs, (Hi) when it dives and (/>) when it glides.
FLYING MACHINES
36
(/) Level Flight. An aeroplane flying along the horizontal is
said to be flying level.
HORIZONTAL
ATTITUDE
LINE OF FLIGHT
I
P<
W
w
W (c)
Fig. 231.
Fig. 231 (a)* shows an aeroplane in level flight, from right to left,
with a constant speed K. This is tantamount to wind blowing from
left to right with velocity V and striking the undersurface of the
plane ; so that, proceeding, as explained above, we have
Reaction or Thrust R, forwards = the drift or drag D y backwards,
both acting practically along the horizontal ;
and, weight W of the plane downwards = the lift if, upwards, both
acting along the vertical, [Fig. 231 (b)]
Or, L=W, (/)
and, R = D, ... ... ...(//)
which, represented vectorially, form a closed polygon, [Fig. 231 (c)],
the plane being in equilibrium.
It will at once be clear from relation (//) that, for level flight, the
forward thrust R must just be balanced by (i.e., must be equal to) the
backward drag D, at that particular speed of flight.
Further, if the speed falls, the lift decreases and the 'plane
starts losing altitude ; so that, a minimum speed (about 50 m./hr.) is
essential to keep the plane at a certain height.
(ii) Climbing. If an aeroplane flies obliquely upwards, it is
said to be doing a <climb\
HORIZONTAL _ * * ^
ANGLE OF AT JACK
(WOW*) \f
W
*&,
%
W (W (c)
Fig. 232.
We shall, for the sake of simplicity, take the line of flight dur
*For simplicity, the student may simply show these forces acting on the wing
or the aerofoil, instead of sketching the whole plane.
fine lift is not necessarily vertical. It is just the component perpendicular
io the current of air.
370
PBOPEETIBS Of MATTEB
ing the climb to coincide with the direction of thrust R due to the
propeller, or with the attitude of the plane.*
Since the relative velocity of the wind now makes an angle a
with the horizontal, the lift (L) no longer acts in a vertical line with
the weight (W) of the plane and is, therefore, balanced by the com
ponent W cos a of the weight [Fig 232 (a) and (b)] ; and, similarly,
the thrust R of the propeller, by the drift or drag (D) along the line
of flight plus the component W sin a of the weight, i.e., now
L = W cos a ...(/)
and R = D+W sin a. (*>)
It will thus be clear that for a climb, the thrust R must be greater than
the drag (D) by the factor W sin a and that it increases with the angle
& or the steepness of the climb.
It follows, therefore, that if a = 0, sin a = and cos fl=l ; so
that, equations (///) and (/v) reduce to (/) and (//) respectively. In othei*
words, the plane then flies level with a constant velocity, without a
climb.
The forces in equilibrium, during the climb of the 'plane, repre
sented vectorially, give a closed polygon [Fig. 232 (c)], which, in the
case of level flight, reduces to a rectangle, [Fig. 231 (r )], with R and
D equal and horizontal and L and W, equal and vertical.
(Hi) Diving. When a plane flies obliquely downwards, it is said
to be making a dive.
Again, taking the speed of the plane to be constant and its line
of flight coincident with its attitude, the different forces on the 'plane
are as shown in Fig. 233 (a) and (fe).
VERTICAL
COMPONENTS
Of
L&D ~
L/
(a) (b) (c)
Fig. 233.
Since the relative wind velocity (V) makes an angle a with the
horizontal, the lift here also does not act along the vertical line with
W\ and, since it makes an acute angle with the downward vertical line,
the vertical components of both the lift and the drag act upwards,
thus opposing W. A& will be readily seen from Fig. 233 (a) and (b).
the lift is balanced here by the component W cos a of the weight (W)
of the 'plane, the other component W sin a of which acts along the
same direction as the thrust (R) ; so that, for equilibrium, we now
have
*lt is by no means necessary that it should always be so. The line of flight
can & fact it often does make an angle with the thrust (R), which, of course .
jicts along the attitude of the 'plane.
FLYING MACHINES
37)
R = D  W sin a, . . .(v/)
showing that,/0r a dive, the thrust R must be less than the drag (D) by
the factor W sin a and, therefore, it decreases with a or the steepness
of the dive, necessitating the throttling down of the engine.
And, once again, if a=0, sin a =0 and cos a = 1, so that re
iations (v) and (v/) reduce to (i) and (//) respectively, the 'plane flying
level with a constant velocity, without a dive.
Representing the different forces vectorially, we again obtain *
closed polygon. [Fig. 233 (c)], which with a = 0, becomes a rectangle,
with R and D equal and opposite, and acting along the horizontal,
and L and W ', equal and opposite, and acting alohg the vertical.
(iv) Gliding. With the engine not functioning, i.e., with R = 0,
a,8 the 'plane descends down, it is said to be gliding.
In this case, obviously, the lift and the drag are balanced by
the components W cos a and W sin a of the weight (W) of the 'plant
respectively, and we have, for equilibrium,
L = Wcosa, ...(vff)
and D = W sin a ; ...(viff)
so that, with the increase of a, the drag increases and the lift
decreases.
166. Different parts of an Airplane and their functions.
The following are the different parts of an airplane and their
respective functions.
(i) The Wings. The wings or the aerofoils, as they are techni
cally called, are, appropriately, the most important part of an
airplane (a flying machine) and much research has gone into perfect
ing their design, in order to obtain the maximum lift for the 'plane.
In fact, the lift due to them accounts for as much as about two thirds
of the total available lift.
To minimise the fractional force to its motion and to ensure a
smooth airflow along its surfaces, the wing is shaped to the stream*,
lines of the air through which
ft passes, with a gradual taper
ing ff of its thickness from its
front or leading edge to its rear
or trailing edge, with the upper
surface more curved than the
lower, as shown in Fig. 234.
The axis of wing (shown dotted)
is called the chord and the angle
that the chord makes with the
Fig. 234.
direction of the wind is called the angle of attack.
The air, moving more rapidly over the upper than the lower
surface, brings about a difference of pressure on the two surfaces,
In accordance with Bernoulli's principle, (see Chapter XII) and
this gives the wing an upward lift. This lift on the wing really
consists of (/) an upward thrust on its lower surface and (it) a suction
effect on its upper surface. For, as the leading edge of the wing
372 PROPEBTIES OF MATTER
moves through air, it parts the air stream into two parts, which tend
to flow as close to the two surfaces as possible. The upper stream
is, however, deflected upwards by the curved shoulder of its upper
surface and its return back to that surface is retarded due to its
inertia, resulting in an area of partial vacuum above it and a conse
quent upward pull on it due to suction.
For a given wind speed, the lift increases with the angle of
attack up to a certain limit, beyond which it begins to decrease and
the 'plane begins to sink. This limiting value of the angle of attack is
called the stalling angle and its value varies from about 15 to 20.
The ratio lift/drag is, however, the maximum when the angle of attack
is about 4. Hence we have the maximum efficiency in flight at this
angle of attack.
An important consideration in the structure of the wing is to combine
lightness with strength, and it, therefore usually consists of two main spars of
wood or metal, running all along its length, with light girders of the same
material, set perpendicularly to them at suitable intervals, the whole frame
work being covered with a 'skin' of sheet metal or thin plywood, having a tightly
stretched fabric over it, well coated with a liquid solution, called 'dope', which
not only shrinks the fabric and makes it taut like the skin of a drum, but also
serves to increase its strength and to make it water and airproof.
It is found that the force or effective pressure (P) due to the
wind, as it is called, depends (/) directly upon the area A of the
aerofoil, (ii) directly upon the square of the velocity (V) of the wind,
and (i/f) directly upon the density (p) of the air at the height of the
'plane. Thus,
P oc A.?.V*. Or, P=K.A. 9 .V 2 ,
where K is a constant, depending upon the shape of the aerofoil and
the angle of attack.
Since the lift, L = P cos 0, and the drag D =* P sin 0,
we have
L = K.A. ? V*.cos 9 and D = K.A.p.V 2 . sin 6.
Or, multiplying and dividing each expression by 2, we have
L = 2K cos 6 \A$V* and D = 2K sin
The factor 2K cos is called the Lift coefficient and the factor
2K sin 0, the Drag coefficient, usually denoted by the symbols
CL and C^ respectively. So that,
L = C L . } ApV* and D = C D .\ A?V\
where L and D are expressed in Ib. wt. ; A, in sq. ft. ; p, in slugs* per
c.ft. and V, in ft. per sec.
The Lift and Drag coefficients increase with the angle of attack,
the former having its maximum value 12 at about 16, when the
value of C D is about 20. The ratio of the two coefficients i.e., C L \C D
or the ratio Lift /Drag also varies with the angle of attack, and has
its maximum value (12) at about 4, at which value of the angle of
attack, therefore, we have the maximum efficiency in flight. Nor
mally, the angle of attack is arranged to lie between 3 to 6.
Further, it will be clear from the expression for L above that a
certain minimum wind speed is essential for the lift to be large enough
to make the 'plane rise up against the force of gravity. It is for this
*Mass in slugs is equal to weight in pounds weight, divided by 32.
FLYING MACHINES
373
reason that the plane must first be made to run on the ground before
it can take off.
() The Propeller or the AirScrew. It is a large fan like
structure, carried right in front* of the plane and rotated rapidly
about a horizontal axis by an internal combustion engine. Its tw<?
(or more) blades are set at an angle in a central hub, a shown in
Fig. 235, and may be made of
wood or metal, consisting, in
the former case, of a number
of layers firmly glued to Fig 235
gether, with their edges tipped with metal, and their surface provided
with a suitable protective covering of fabric or cellulose.
A propeller blade is in fact a small wing and functions precisely
as such. For, just as a moving wing, meeting the air at an angle,
experiences an upward thrust in a direction almost at right angles to
that of its motion, so also does a revolving propeller blade experience
a thrust at right angles to its direction of motion, i.e., along the
horizontal, for the very air which it sweeps from in front of it and
throws backwards, pushes it forwards.
Thus, because the propeller cuts its way through air, much in
the manner of a screw cutting its way through wood or metal, it is,
on that analogy, also referred to as the airscrew.
Apart from the two most important parts of an airplane, discussed
above, there are others which make for its stability and easy manageability
In any desired position and direction in the air. These together constitute what
are called the surface controls of the airplane and we shall now deal briefly with
these.
(Hi) The Tail Unit. Carried at the rear end of the airplane, it consists
of two sets of surfaces, (/) vertical and (//) horizontal, each being made up of two
parts, one fixed and the other movable, viz., the/z/j and the rudder ; and the tail
plane (or stabilizer) and the elevator respectively.
(a) The Fin. It is the fixed or the front part of the vertical surface of
the tail unit and takes the form of a vertical plate, arranged at a small angle with
RUQDEP
ELEVATOR
LEFT
WING
Fig. 236.
the central line of the fuselage or the body of the 'plane, (Fig. 236). Its function is
*This is the most usual position of the propeller in most 'planes, such
planes belonging to the tractor type, because of their being pulled through ail
by the action of the propeller. In what are called the pusher type of 'planes, the
propeller is carried behind the line of the wings, so that it exerts a pushing action
OB them.
474 PROPERTIES OF MATTER
to give directional stability lo the 'plane, (very much in the manner of the featheis
tipping a dart), making for its straightline flight in the horizontal plane and
tending to bring it back to its original course, should a disturbance cause it to
turn to one side or the other.
Thus, if the 'plane be thrown to the left, the fin will deflect the air to the
right, which would then push it back to the left, to resume its original straight
course and vice versa.
The student may perhaps wonder as to why the/w, with its avowed purpose
of keeping the 'plane along its straightline course, should be offset a few
degrees from the central line of the fuselage. The reason is that the air stream,
blown back by the propeller, (and called the slip stream), shares with the latter
its corkscrew motion and would strike the fin at an angle, were it set along the
central line, producing precisely the opposite of the desired effect ; for, it would
result in turning the plane rather than keeping it along its straight course. The
small inclination of the fin to the central line just counteracts this turning effect due
to the slip stream.
(b) The Rudder. It is the rear portion of the vertical surface, (Fig. 236),
hinged on to the front portion or the fin, and has freedom of lateral movement in
the vertical plane. Its function is very much similar to the rudder on a boat and
it enables the 'plane, in level flight, to be steered to the right or to the left ID
the horizontal plane.
Connected by means of cables to the rudder bar, pivoted horizontally OD
a central vertical pin in the cockpit*, it is operated by the pressure of the pilot's
feet, (see Fig. 236) a pressure with the right foot (i.e., on the righthand end ol
the bar) makes it swing out of the central line and turns the plane to the right,
and a pressure with the left foot similarly turns the plane to the left.
(c) The Tail Plane or the Stabilizer. This is the fixed pat t of the horl
lontal surface of the tail unit, (Fig. 236), and its function is identical with that
of the fin, but in the up and down direction, i.e., it serves to give the airplam
Mobility in the vertical plane, or the 'fore and aft* stability, as it is called.
(d) The Elevator. It is the movable part of the horizontal surface
of the tail unit and controls the vertical motion of the 'plane, i.e.,
its climbing and gliding movements.
Lying normally in level with the tail plane or the stabilizer, it
up and down movement is controlled by the central column, or the
central stick, (or, simply the stick, as it is usually called), which is
connected to it by cables and is arranged conveniently in front of the
pilot'i seat, (see Fig. 236, above). A backward or inward pull on the
stick raises the elevator up above the level of the tail plane and the
air, rushing over the surface of the plane, strikes against it, tending
to blow it down to its original position, in level with the tail plane,
thus exerting a downward pressure on the tail of the 'plane, as a
whole, with the result that its nose is pushed upwards and it climbs
up.
Similarly, a forward or outward push on the stick lowers the
elevator below the level of the tail plan and the air thrust on it now
pushes the tail up, which is the same thing as pushing the nose down,
and the 'plane, therefore, now glides down.
(/v) The Ailerons. These are hinged flaps, free to move up and down ai
the rear or the trailing edges of the two wings, extending from the tip of eacl
wing to almost its midpoint, (Big. 236), their up and down movement being con
trolled by the sideways pull on the stickf, to which they are connected by meani
*The Cockpit is a closed or open well, in the front portion of the aer*
plane in which the pilot takes his seat, (Fig. 236), with different controls and 10
it rumen ts arranged in front of him .
fin the larger type of aircraft, the aileron is controlled not by the sticl
but by what looks like an incomplete steering wheel of a motor ear, fitted on t<
4he top of the stick.
FLYING MACHINES 375
of cables. The arrangement is such that as the stick is pulled to one side, it
simultaneously causes one aileron to be raised above, and the other to be lower*
ed below, the undersurface of the corresponding wing, with the result that the
lift on one wing increases and that on the other decreases, making the plane
'bank' or heel over to one side, a pull on the stick towards the left making the
'plane bank to the left and a pu)l towards the right, making it bank to the right.
It will thus be seen that the stick and the rudder bar, between them
selves either singly or in combination with each other enable the 'plane to bf
manoeuvred into any desired position and to perform all sorts of aerobatics.
(v) The Tail Trim. If an airplane continues to fly level, even when the
pilot releases his hold on the stick for a while, it is said to be 'flying trimmed 9 *
This ideal state of affairs may however be easily disturbed by the entry or exit
of a passenger or two, the plane becoming 'nose heavy' or 'tail heavy' and thui
starting to fall down or to rise up. This puts an undue strain on tbe pilot, al
ways alert to exert an inward or an outward pull on th;s stick.
The tail trim is just the device to prevent all this and to enable the 'plane
to fly trimmed even with different loads in it, by automatically adjusting the in
ward or outward pressure on the stick, to suit the load. Of immense help to the
pilot during 'take offs' and 'landings', it just consists of a lever on one side of the
cock pit which, working on a quadrant, suitably alters the tension of a spring
Attached to the lower end of the stick*, always exerting the requisite pull, com
mensurate with the load in the 'plane.
(vi) The Undercarriage or the Chassis. It is that part of the airplane
behind the engine and at the base of the fuselage, which serves as a carriage for
the 'plane to run on the ground and includes the wheels and a shockabsorbing
mechanism (the oleomechanism) to take up the unavoidable impact on landing or
the bumps on uneven ground, which may otherwise severely strain the fuselage
even to the extent of damaging it.
To minimise the air resistance to the flight of the airplane, the under
carriage is now almost universally made retractable (except perhaps in the case
of very small aircraft) ; so that, it can be drawn up into the fuselage once the
'plane is up in the air, and lowered again when about to land, there being a
suitable device to warn the pilot in time, when preparing to land, m case his
undercarriage remains retracted.
(v//) The Wheels. The undercarriage is supported on twof wheels (ex
cluding the one at the tail end), fitted with widetrack pneumatic rubber tyres, in
flated at low pressure. These, besides enabling the 'plane to run on the ground
before a take off also absorb part of the shock of impact, on landing, passing on
the rest to the oleomechanism.
In modern aircraft, we have also wheel brakes fitted more or less in the
manner of our motor car brakes, which (a) keep the plane stationary during the
running of the engine on the ground and (b) also shorten its run on landing. In
addition, they enable more pressure to be applied to one wheel than to the
ther, thereby greatly facilitating the steering and the manoeuvring of the plane,
while still on the ground.
(v///) The Tail Skid. The rear of an airplane is supported either on a
small wheel or a sparlike structure, called the (ail skid When the two front
wheels and this spar, or small wheel, touch the ground simultaneously on land
ing, the plane is supposed to have made a perfect 'three point landing'.
a (tx) The Slot. Oftentimes, when an aeroplane climbs too steeply, or
when it is about to land, and in fact, when for any reason, the speed of the
'plane falls below a certain minimum, the lift on the wings becomes insufficient
to keep the 'plane flying and there 15 every possibility of its 'stalling'. Not only
that, but with an insufficient airfbw, the other controls, and particularly the
ailerons cease to function properly and the 'plane starts dropping in a dive.
^Sometimes, the lever is replaced by a wheel, whose movement suitably
adjusts the position of the tail plane instead of acting on the stick.
tin some cases, we have a threewheeled or a 'tricycle' undercarriage, the
third wheel being arranged well ahead of the other two. This not only prevents
the 'plane tipping on its nose, thus greatly reducing the possibility of accidents
on landing or manoeuvring the plane on the ground, but also greatly simplifies
both takeoffs and landings.
"376
PROPERTIES OF MATTER
An ingenious saftcy device, known as the Handley Page Slot, or, simply
the Slot, is, therefore, used to avert this danger of a 'stair. It is just a small gap
between the upper surface of the wing and another miniature winglike structure
the 'slat 9 arranged over its leading edge*.
Without the slat, if the airplane were to stall, the airflow ceases along
the upper surface of the wing and breaks up to form a series of eddies, as shown
THE SLOT
(a) (b)
Fig. 237.
In Fig. 237 a), thus depriving the 'plane of about 60% of its lifting power
When, however, the slat is fitted to the 'plane, it opens up as shown in Fig. 237
(6), and forms t small passage or slot between itself and the wings and the ait
stream is directed through it on to the wing surface, instead of breaking up into
eddies. The lift on the wing is thus avoided and the danger of a stall averted.
If, however, the wing be tilted too steeply, a stall may eventually occur,
but the 'plane recovers from it much sooner than would be possible without the
slat.
(x) Engine Controls and Other Instruments. Among these, the main 01
the important ones are the following :
The Throttle. This corresponds to the accelerator of the car and controls
the speed of the 'plane. Operated by the throttle lever on one side of the cock
pit, it differs from the car accelerator in that it stays in the position in which it
is set, without spring ng back when the pressure is released on it, thus enabling
'plane to fly at the desired constant speed. There is no gear changing or slowing
down for negotiating corners, for which, indeed, it must fly a little faster.
(xi) The Altimeter. As its very name indicates, it is an instrument to
indicate the altitude of the airplane. It is, in fact, a modification of the aneroid
barometer and is calibrated to indicate height or altitude in terms of 'thousands'
of feet. Thus, if the pointer be at 5, it indicates a height of 5000 ft and so on.
Since, ho ^ever, the altimeter really measures variations of pressure at ground
level, which can occur due to changes of weather, it may indicate different height
even at one fixed point on the ground, and its readings may thus be highly mis
leading and may prove dangerous. To obviate this risk, therefore, it is so
arranged that the pilot sets it at zero altitude before taking off, so that its read
ings later indicate the heights above this starting point, and not the absolute height
above the ground at any given moment. Thus, even if it indicates a height of 5000
//., it may well be within a couple of hundred feet from a mountain top.
Improved instruments to indicate the absolute height of the 'plane above the
ground at a given moment (instead of from the starting point) are however well
in the offing and would greatly reduce the hazard of an airplane flight in fogg>
weather.
(*//) The 'Engine' Revolution Counter This enables the pilot to feel the
pulse of the engine, as it were, telling him all about the condition of the engine,
Including its undue vibrations and uneven running etc. Further, should there be
an unexpected or unaccountable drop in the revolutions of the engine, it is a
warning to the pilot that trouble is jmminent. The revolutions are measured in
terms of hundreds per minute.
(xiii) The Oil Pressure Gauge. It is a small but vitally important instru
ment and indicates the pressure (in pounds) under which the oil is pumped round
to the different parts of the aeroengine, an operation about just as essential to
*Sometimes the slot is also arranged close to the aileron flap, when it
helps to maintain the requisite air flow over the aileron surface, thus enabling it
to function effectively even at low speeds of the 'plane.
FLYINQ MACHINES
377
Us life as the blood supply to ths various parts of our body. A sudden drop in
this pressure forewarns the pilot of a coming serious trouble and alerts him to
take remedial measures in time.
I.
2.
It is a
machine.
Airship
'lighterthanair 1 flying
It is based on the principle of
floatation and its lifting power is
provided by the buoyancy of the air
displaced by it.
It rises vertically upwards, directly !* 3.
from the ground. I
1.
It is a
machine.
Airplane
'heavierthanair' flying
Here, the lifting power is due
to the thrust produced by a
strong artificially created wind
and the characteristic shape of its
wings.
It must first be made to run
on the ground before it can 'take
off*.
4.
It is very much bigger in size than
an airplane.
4.
It is
size.
comparatively smaller ID
Thus, though the airship arid the airplane are based on entirely
different principles, they have in common (/) an upward motion against
the action of gravity and (//) propulsion through air.
167. Jet Propulsion, We are all familiar with the meaning of the
word 'jet* which is just the term applied to a high velocity stream oj
fluid (liquid or gas) issuing out of a nozzle, as for example, a 'jet of
water' or a 'jet of steam' etc. And, therefore, jetpropulsion is
obviously the method of driving or propelling a body or a machine
forwards through the agency of a jet, the body or the machine thus
driven being said to be jet propelled.
That a jet possesses such a motive or tractive force can be
easily seen from a number of facts of every day life, if only we care
to stop a while and analyse them. Thus, for example, when a bullet
is forced out of the barrel of a rifle by the exploding mixture of gases
inside it, the rifle suddenly moves or 'kicks' back in a direction oppo
site to that of the bullet and the exploded gases.
So that, if we continuously fire a rifle fastened to the rear of a
boat, with its barrel facing outwards, we shall find that the boat
continues to move forwards with a jerky motion so long as the firing
continues, each bullet fired producing a push forwards. We, therefore,
have here a jetpropelled boat ! In fact, even when we ply the oais,
the action is similar. For, what we do is simply to push some volume
of water backwards and the boat, as a consequence, moves forwards,
Indeed, if we did nothing else but simply sit quietly in the boat
and throw stones into the water, with our face towards the stern of
the boat, the boat will still move forwards (i.e., opposite to the
direction in which the stones are thrown).
All these examples are, as the student is no doubt already
aware, a consequence of the well known Newton's third law of motion,
according to which action and reaction are equal and opposite, or what
follows from it, viz., the Jaw of conservation of momentum, which
378 PEOPERTIES OF MATTBB
demands the equality of the momentum lost with the momentum
gained. So that, the boat, in the cases above, moves oppositely as a
result of the reaction to the motion of the bullet (and the gases) 01
the stones, or because the momentum lost by the bullets or the stones
is equal to the momentum gained by the boat. A force such as the one
experienced by the boat is called the reactive force ; and, in the case
of a jet, sometimes alo the jetforce.
Now, does it surprise the student when he is told that even the
usual type of airplane, in which we use the ordinary reciprocating
(i.e., the piston type) engine makes use of a jet for its propulsion for
wards* ? For, the propeller, as it whirls round at a high speed,
throws a jet of air (or in the case of a ship, a jet of water) backwards,
as a reaction to which the plane (or the ship) is pushed forwards
against the viscous resistance of tiie air (or water). The question,
therefore, naturally arises as to why then do we not call them jet
propelled planes. The answer is that, technically speaking, the
narrower the crosssection of the high velocity fluid stream, the more
nearly does it come up to the definition of a jet, and the term jet
propelled planes is, therefore, reserved for planes in which the jet is a
narrow one, about one foot in diameter, as compared with ten feet of
more in the ca.se of the ordinary airplane.
Again, it must not be inferred from wheat has been said above
that a jet must necessarily consist of hot gases. No, it may just as
well be of cold air, as in the case of what are called the ductedfan
type of planesf, or as was the case with perhaps the earliest jet
propelled plane, constructed by CampM in Italy, the jet in this
latter case, being produced by a compressor, driven by the ordinary
reciprocating type of engine.
168. Thrust supplied by the jet. Let us now calculate the thrust
supplied to an aircraft by the jet produced by the power unit
inside it.
Suppose we have an aircraft travelling with a speed V and fitted
with a powerunit which produces a jet of fluid, of velocity w, relative
to the aircraft, where u is higher than F, the velocity u of the jet
being measured at a point in it a little away from the nozzle, where
the static pressure is the same as that in the surrounding air. Then,
if we impose a velocity V on the aircraft in the opposite direction to
its own, the aircraft comes to rest, with the air streaming past it with
velocity V. So that, if a be the area of crosssection of the jet at the
point where its velocity is u, the volume of the fluid flowing per second
In the jet is clearly a.u. If, therefore, p be the density of the fluid, we
have
massflow of the fluid per second in the jet =a.u.p = m, say
And, therefore, momentum of this fluid in the jet = m.u.
If this mass (m) of the fluid finally emerges out from the aircraft
*The same being the case with a ship.
tin these planes, air is sucked in through two holes or ducts, by two fans,
rotating in opposite directions, the latter thrusting the air away with consider
able force, propelling the plane forward. This sucking action also helps to buoy
the plane up. Further, the gyroscopic effect, produced by the oppositely rotatini
fans greatly helps in enhancing and assuring the stability of the plane.
PLYING MAOHINBS . 3713
with a velocity F, its momentum is clearly reduced to m.V. It thus
suffers a loss of momentum, equal to (mu mV) or m(uV)pcr second t
I.e., its rate of change of momentum = m(uV).
And this, therefore, in accordance with Newton's second law oj
motion, must be the force, or the thrust, F supplied to the aircraft bj
the jet in the direction opposite. So that,
F = m(uV) = 0wp (ti V)
169. Efficiency of the jet. If we consider the exact state oj
affairs in the case above, viz., that the aircraft is really not at rest, as
fre had imagined, but is moving with velocity V, then, the velocity ot
the final jet, moving in the opposite direction clearly becomes (w V)
and, therefore,
K.E. of the final jet = \ m (uV)* = \F(uV) [. m(u~V)P}
Also, the aircraft does FV amount of work per second against
ihe air resistance'as it moves forwards. So that,
total energy that must be supplied by the jet propulsion unit,
(i.e., by the powerunit) = FV+\F(u V).
Of this, obviously, the portion usefully employed is only FV, the
rest being simply a waste, creating a disturbance behind the aircraft.
So that,
efficiency of the jet, or the Froude efficiency, as it is commonly
. _ energy converted into useful work _ FV _
~~ total energy supplied ~~ FV~+\F(u^V)'
2V
~ u+V '
Note. Clearly, the efficiency will have the maximum value 1, when
a V y /.., when the initial jet velocity is equal to the flight velocity of the air
craft, for, then, the energy wasted in the form of K.E. of the final jet [\F(u V)]
will also become zero. But, then, the thrust on the aircraft [m(uV)} will also
become zero. .This condition of maximum efficiency is, therefore, not a practi
cable proposition, just as it is not in any other type of machine also.
170. Effect of smaller crosssection of the jet. As indicated earlier,
the crosssection of the jet in a jetpropelled plane should be narrow,
Let us see what advantage is to bs gained by it.
Apparently, from the relation F = m(u V) for the thrust
supplied to the aircraft by the jet, we find that a reduction in its
crosssection will mean a diminution in the value of the mass flow of
the fluid, m, so that, to obtain the same thrust F, as before, (u V)
will have to be correspondingly greater. This will naturally mean a
higher value of \(u K)*. the K.E. of the final jet, which, as we have
seen, is a mere waste of energy. Not only that, but, as a natural
oonsequence, the efficiency of the jet FF/FK+F(wF), will also fall
below its previous value. It would thus appear that a decrease in the
crosssection of the jet, far from improving matters, does just
the reverse, viz., increases the loss of energy and decreases the
fficiency for propulsion, In what manner, then, is jetpropulsion a
better mode of propulsion ?
The answer is manifold :
(i) Initially, when jetpropulsion was just introduced round
bout the year 1940, it was intended to render auxiliary support to
380 PROPERTIES OF MATTER
the then prevalent gas turbine engine. The materials of the gas
turbine could not function satisfactorily at the temperatures obtaining
in the earlier reciprocating type of engines and the products of com
bustion required to be diluted with a large excess of air. This seeming
difficulty was actually turned into an advantage by the engine
designers, who used this necessary excess of air as a narrow jet
to supply the entire thrust required to be given to the aircraft, thus
eliminating the necessity of the propeller and quite a few other
accessories. The jet was made to escape through a small turbine
which then supplied the necessary power to the generator, the fuel
pumps and the compressor etc. Thus, although the introduction of the
jet inevitably entailed a loss in efficiency, with the fuel consumption
rate rising higher, it gave the distinct advantage of reducing the weight
of the whole unit for the same value of power. In view of this smaller
weight but higher rate of fuelconsumption, the turbojet engineg, ae
these engines were aptly christened, came to be considered more
suitable for flights of shorter durations, say, of less than 2 hours in
those early days when the highest speed was only 400 miles per hour.
(//) It was found that although the efficiency of a narrow jet is
rather low at moderate flight speeds, it increases rapidly with the
flight speed. In fact, if we take into consideration also the other
advantages that go with higli speed, (e.g., assistance given to
the compress ion process in the engine, etc.), the overall result is that
the po^er output (FF) increases directly with flight speed with only a
comparatively very small increase in fuel consumption, i.e., FV oc V.
Clearly, therefore, F remains practically constant for varying flight
epeeds.*
This linear increase in power (FK) with speed (K), with practi
cally a constant fuelcon sumption rate, necessarily implies that if the
flight speed bo high, the turbojet unit will also be about as economi
cal as the ordinary propellerengine and will, in addition, possess the
advantage of (a) having less weight and (b) capacity of packing large
power in a smaller space.
In fact, both the turbojet and the propeller engine will have
the same efficiency, i.e., their power output for the same fuel
consumption will be the same, at a speed of 700 m.p.h., provided the
propeller engine had a constant poweroutput upto this speed. And this
is the point where the jetunit scores over the propeller unit. For, the
power output of the propeller engine does not really remain constant
with speed but falls steeply as the flight speed approaches the speed
of sound, v/j., 762 m.p.h. at ground level and 660 m.p.h. at altitudes
above 3600 ft. This is so, because a propeller may be regarded
essentially as a wing, with the difference that whereas the latter pro
vides a lifting force to the aircraft against the force of gravity,
the former supplies a similar force in the form of a thrust in the direc
tion of its motion, for which purpose it is rotated in a plane perpendi
cular to the direction of flight, the lifting force in the case of
the wing and the forward thrust in the case of the propeller being
always roughly perpendicular to the direction of their respective
motions through air, both experiencing an airresistance or 'drag*
"That is why, in the case of a turbojet unit, only its thrust (F) is indicated
and not its power (FV).
FLYING MACHINES
381
opposing their motion, with some power used up in overcoming the
same.* So that, despite all improvements made in the designs
of propellers (such as making their sections near the tips very thin,
etc.) their propulsive power falls greatly at high flight speeds, whereaa
that of a jetunit rises equally greatly. This comes about because the
actual velocity of the propeller blade is the resultant of its velocity of
rotation and translation and, as such, is higher than the flight speed
of the aircraft itself, even a nonrotating wing experiencing a large
increase in the drag on it much before the speed of sound is attained.
Thus, from the point of view of overall efficiency, a jet unit is
certainly superior to a propeller unit at speeds of 600 tn.p.h. and
above.
Then, again, another advantage that a jet unit possesses over
the propeller unit is that there being no accessories and profcubrancea
like radiators, oilcookers etc., the drag is comparatively less. And
the absence of the propeller which makes for a smoother flow over the
entire surface of the plane, cuts down the drag over these surfaces by
as much as 20 to 30 per cent.
171. Rocket Planes. The small firework rockets, rising pretty
high up in the air, to the amusement of onlookers, are a common enough
sight everywhere and they
are obviously jetpropelled
on their own small scale, A
rocket plane is merely a large
scale version of the same
phenomenon. It possesses a
higher speed and can rise to
a much greater height than
even a turbojet plane. In
fact, it is a turbojet plane,
OXIDI r
SER foil
PROPELLED,
NOZZLE
'(HOTGASES)
<s
(COMBUSTION JTmtm .
? CHAMBER! J X
t/F
(0
in which the
jet propulsion
stage further,
PROPELLER
NOZZLE
(HOT6ASZS)
technique of
is carried a
with the jet
still narrower in crosssection
and its velocity higher.
Is this difference in
degree, then, the only factor
that distinguishes it from a
F . 2 , 8 jetplane? Of course not;
s ' for, the essential difference
between the two lie* in the method of production of the jet. In a jet
propulsion unit, the fuel alone is carried on the aircraft, with the oxygen
necessary for its combustion being drawn from the surrounding air, [Fig,
238 (/)] only a fraction of which is usually consumed, the rest, together
with the considerable larger quantity of nitrogen 'swallowed', merely
serving to keep the temperature down throughout the jetpropulsion
_____ ^ ^ ^ ~~ comprised iairki front of the aircraft is of little
consequence at speeds below that of sound, for it simply moves away with the
loeed of sound. But when the speed of the aircraft is higher than that of sound,
the condensed air in front can oniy move sideways but not forward with the result
that the nose of the aircraft has to carry along a bulk of compressed air, witn a
consequent large increase in the drag on it.
382 PROPERTIES OF MATTEB
tinit. In the case of a rocketpropulsion unit, on the other hand,
the fuel as well as the oxidising agent, required for its combustion, are
together carried on the aircraft and no air has to be drawn in from tht
surroundings. It is thus a selfcontained unit in itself, ['Fig. 238 (//)].
This is a point of great importance in that it makes the working
of the motor quite independent of the presence or absence of any
surrounding air. Thus, whereas a jet plane can only attain a height
of 50 to 80 thousand feet up to which it can have its supply of
air from the surroundings, there is no such limit to the height
of a rocketplane, which alone is capable of rising up to higher alti
tudes beyond the earth's atmosphere, where there is obviously no air
to be drawn in. So that, its being selfcontained, with its own supply
of the oxidising agent, while it may be a comparative disadvantage at
lower altitudes, is clearly a tremendous advantage at higher altitudes
Calculating the thrust (F) given to the aircraft by the issuing
jet in the same manner as in the case of the jet propulsion unit
( 168, page 378), we have Fin.u, where m is the massflow of the flui<3
per second in the jet and u, its final velocity, its initial velocity hert
being zero, since all the constituents are carried on the aircraft itself.
Its efficiency thus works out to 2Fw/(w 2 +F 2 ), which again, as in the
ase of jet propulsion, will have the maximum value 1, when u = V.
Further, the powerunit, in the case of the rocket plane, is also
much lighter. Thus, while, for supplying a thrust of 1 Ib. at ground
level, or of 02 Ib. at an altitude of 50,000 ft. a turbojet unit weighs
about 0*3 Ib., a rocketunit, weighing only 01 Ib. can supply the
same thrust of 1 Ib. at all altitudes. It follows, therefore, that a
rocket plane is the more suitable for use only for flights of short
duration or at very high speeds.
172. Rocket Fuel. The rocketunit, in which both the fuel and the
required amount of oxygen for its consumption are carried on the
aircraft itself, is much simpler than that other one which requires the
compression of large quantities of air. For in this case, the only
problems with which we are concerned are those of the combustion
chamber and the propelling jet.
The oxygen may be carried either in the liquid form, or in the
form of oxidisers rich in oxygen, like hydrogen peroxide, (H 2 O t ) or
nitric acid (HNO Z ). In the latter case, the remaining part of the oxidi
ser, going into the propelling jet, merely serves to cool the jet and the
combustion chamber.
Now, if the fuel and the oxidiser are carried in separate con
tainers, the system is known as a bipropellant rocket ', but if the fuel
contains its own oxidiser with it and is carried in a simple container,
It is known as a monopropellant and its decomposition is broughl
about either by the application of heat or through the agency of
catalyst. Obviously, a monopropellant must be some sort of an ex
plosive and, therefore, requires careful handling. Quite a commoi!
one being hydrogen peroxide, which decomposes as shown by thi
equation
2H 2 O a 2H i O+O a +69Q C.H.U. Ib*
*1 C.H.U. (Centigrade heat unit) is the heat required to raise the tempera
tore of 1 Ib. of water through re.
FLYING MACHINES 383
The products of decomposition of a monopropellant substance
are sometimes themselves rich in oxygen, as we can see in this case
of /JgOj, and the substance can, therefore, also be used as one of the
components of a bipropellant. Thus, for example, H^O^ used with
methyl alcohol (CH$OH) would react as shown :
releasing heat at a much higher rate and hence resulting in a much
higher exit velocity of the gases through the exhaust nozzle or the
*enturi, as it is called, and consequently a much higher thrust*
The propellant may be injected into the combustion chamber in
one of the two ways : (/) by exerting pressure by a compressed gas, like
nitrogen or air on the propellant tanks or (') by means of a pump
ing mechanism, usually a turbine. The former method admits of no
variation or control of thrust and is, therefore, suitable only for
shortduration flights or remotely controlled missiles, and the latter
Is the one commonly used for rocketpropelled aircrafts.
173. Specific Impulse. The performance of a rocket motor is
measured in terms of what is called the specific impulse or the specific
pull, /, which is the thrust generated by unit rate of fuelconsumption, i.e.,
thrust (Ibs,) _ F
~~ rate of fuel consumption (Ibs. /sec.) ~~ nig'
So that, the dimensions of / are the same as those of time. Phy
sically, therefore, it is the time for which a unit thrust can be generated
by a unit weight of fuel.
Now, as we have seen, the thrust in the case of a rocket is
equal to mu, where m is the massflow through the nozzle and u, the
exhaust velocity of the gases. So that,
/ = mujmg = ujg.
And, therefore, the higher the jetvelocity, the higher the specific im
pulse and the smaller the fuelconsumption for a given thrust.
Besides fuelconsumption and thrust, there are quite or few
other factors which determine the suitability of various fuels, e.g., the
weight of the engine, the temperature in the combustion chamber etc.,
etc. In the modern rocket motors, the total weight of the pump,
control and installation etc. must be about onetenth of the maxi
mum thrust developed. In short, the performance of a rocket depends
chiefly upon three factors, (/) jet velocity, (ii) density of the propellant
and (Hi) weight of the power plant, which includes that of the propel
iant tanks and the fuelsupply system etc., into details of which we
need not enter in an elementary discussion of the type we are con
cerned with here.
174. Shape of the Rocket. During an upward flight, particularly,
through the denser layers of the atmosphere, the components of the
rocket are subjected to intense air pressure, and also a lot of heat
is produced due to viscous friction of the air. Both these factors
are taken into account while designing a rocket. Its frame is accord
ingly made of a heatresisting material and its velocity during the
first part of its flight, through the denser layers of the air, kept suffi
eiently low. Further, it is so designed as to reduce the air pressure
384
PBOPBETIBS OF MATTBB
on its each individual part to the very minimum, its overall shape
being more or less like that of a cigar.
175. The Multistage Rocket. If a rocket is hurled into space be
yond the earth's gravitational field then, supposing that its acceleration
takes place in tho latter region, where the value of g is 32 ft. /sec*.,
the velocity F that it must acquire to escape from the earth's gravi
tational field or the 'escape velocity 7 , as it is called, is given by the
relation V 2 = 2MG/P, (see 92, page 251) from which the value of
K works out to about lM9x 10 s cms. [sec. or about 36000 ft.jsec.
Now, at the present stage of rocket development, no single roc
ket can achieve this velocity. To tide over the difficulty, therefore,
we make use of what is called a ww///
stage rocket, which is just a combination
of rockets, either (i) joined consecutively
or in series, as it were, or (//') one inside
the other or (///) with the rear port oj
one inside the nozzle of the other, as in
dicated diagrammatically in Fig. 239. In
all these three types, the first stage
rocket is the largest in both dimen
sions and weight, and the last stage
one, the smallest.
Naturally, the first stage rocket is
used first and when it has done its job,
it gets detached and IB discarded, with
the second stage rocket taking over the
task of producing further acceleration.
Then, this too is discarded and the
third stage rocket takes over and so on.
The velocity thus goes on increasing at
each stage by the same amount as it
does in a single stage rocket and each
stage has its own propulsion and con
trol system. Obviously enough, the
fuelconsumption and the thrust for
the first stage rocket are the highest
of all, say about a hundred times the
corresponding values for the third stage
rocket ; and the fuelstock too in the
first stage is about sixty times that
in the third stage, the same being the
ratio of the total weights carried by
the former to that by the latter.
Considerations of both weight and cost demand that the number
of stages should not be large and that, therefore, the payload of
each stage (which includes, in addition to the useful load of the
final stage, the weights of the intervening stage rockets to be discard
ed later) be limited to about 20% of its own weight. Clearly, the
useful pay load of the final stage thus works out to be a very small
fraction of the initial overall weight. Thus, for example, if there be
n stages in all, this fraction is just l/5 n of the initial total weight. Or,
3RD STAGE
ENGINE Of
3RD. STAGE
ROCKET
2ND. STAGE
ENGINE Of
2ND. STAGE
ROCKET
1ST. STAGE
ENGINE OF
1ST. STAGE
XOCKET
Fig. 239.
FLYING MACHINES 385
to give a more concrete example, a space ship of the size of the well
known V2. designed by Dr. Verner Von Braun, would be about just
sufficient to land a match box or a packet of cigarettes, by means of
a purachute, on the planet Mars.
Each individual rocket of the multistage rocket, has its own
independent design and basic characteristics, with its function
correlated with those of the others. These characteristics include the
following :
(i) Net weight. The net weight for a single stage rocket includes also
the weight of the instruments and appliances or the weight of ammunition,
if any, etc. And, in the case of a multistage rocket, obviously, the total weight
of the second stage is the net weight of the first stage and the total weight
of the third stage, the net weight of the second and so on, the ratio between the
two being usually for 3 : 1 for each stage.
(ii) Steering Equipment. This is necessary to steer the course of the
rocket during its flight during the other stages except the first which only serves
as a sort of runway for the rocket, as it were.
(///) Design. This includes the frame of the rocket or of the indivi
dual rockets, m the case of a multistage rocket, with its fortifications and fasten
ings etc.
(iV) Rocketlength. This obviously means the height of the rocket or
that of the individual rockets of the multistage one. This is an important fac
tor in as much as the very stability of the rocket in its trajectory depends upon the
ratio between its length and in mean diameter (ie. y the mean diameter of the
whole rocket or of each one of the stage rockets)
(r) Number of Motors. Each stage rocket has its own separate motors.
The first stage rocket, naturally, in view of the highest total weight it has to
carry and the greatest resistance of the lower denser layers of air it has to
overcome, has more thin one motor and the last stage rocket, because of its
lightest load and the least resistance to be overcome, is provided with only one
motor.
Apar* from these, there are also other characteristics of a rocket, like
\\.% fuelconsumption, thrust, specific pull or impulse, time of combustion (in seconds),
acceleration, lift or range etc.
176. Take off of the rocket. This is perhaps the most important
part in the flight of a rocket and must be fully ensured to be correct.
The slightest error in the timing or the accuracy of firing makes all
the difference between the rocket returning back in this generation or
the next or perhaps not at all.
177 Salvaging the various stage rockets. Let us wind up our
elementary study of a rocket flight with a word about salvaging the
various stage rockets which are discarded after they have performed
their respective functions. This problem cannot yet be said to have
been satisfactorily solved. Experiments are. however, being made with
various systems of parachutes and other devices and, if they succeed,
it will mean a tremendous economy in cost. And for all one knows,
the ideal solution may turn out to be the utilisation of the material
of the stage used up as fuel for the next stage.
178. Satellites. Among celestial bodies, a satellite is what may
be called a minor or a junior member of the solar system revolving
round one of the major planets in its own prescribed orbit. Till recently,
it was not thought possible that anything manmade could also be so
placed round the earth or any other major planet to revolve in a given
orbit. But, then, with the development of jetpropulsion (in the year
1940), followed by that of high speed rockets, man began to dream of
386 PBOPBBTIES OF MATTER
flight into space an'd of interplanetary travel, when, all of a sudden,
on October 4, 1957, the Russian scientists made the whole world gasp
with wonder and surprise by launching their first 'sputnik' or artificial
satellite. This Sputnik /, of the form of a ball, 58 cms. in diameter
and weighing 836 kilogrammes (roughly 185 Ibs.) was placed into an
elliptical trajectory round the earth exactly like a celestial satellite
making, in its initial phase, one full revolution in 96 2 minutes and
attaining a speed of 8 km. or nearly 5 miles/' sec. at a distance of 950
km. from the earth.
The progress of this latest wonder was watched with dumb ad
miration by scientists all over the globe and the radio signals sent out
by it listened to attentively as long as its source of power lasted. It
existed as a satellite for full 58 days, during which it made 1400
revolutions of the earth, thus covering a distance of 39 million kilo
metres. Its existence, however, continued for 92 days and the entire
distance covered by it totalled up to the enormous figure of 60 million
kilometres, when, finally, on January 4, 1958, it entered the denser
layers of the atmosphere and got burnt out due to the intense heat
produced by friction.
This artificial satellite was obviously an automatic rocket, hurled
into its predeterminod and wellcalculated orbit by a multistage
rocket. Indeed, the rocket carrier too continued to revolve round the
earth at about the same height as the sputnik but at a distance of
about a thousand kilometres from it ; and, then, while descending
through the denser layers of the atmosphere, it also began to burn
out, with fragments from it falling somewhere in Alaska and North
America.
After almost exactly a month, on November 3, 1957, the
Russians put their socond artificial satellite 'Sputnik IT into orbit
round the earth, containing scientific equipment for exploratory pur
poses, as well as the first space traveller, the dog 'Laika 9 , in a sealed
cabin, which they successfully retrieved back, safe and sound. The
total weight of the Sputnik was this time much greater, being 508'3
kgms. or 1126 Ibs. (including the dog). Its distance from the earth
was alo greater, 1700 kms. f its period of revolution, 102 sees.,
with the angle of tilt of its orbit roughly 65 from the equatorial
plane.
This wis followed by the first American artificial satellite, 'The
Explorer', on January 31, 1958, though of a comparatively much
smaller weight and size.
These sensational events brought still more sensational and
breathtaking ones in thoir wake, with the Russians putting the first
cosmonaut of the immortalised name, Major Yuri Gagarin, into space
in a much larger spacevehicle or spaceship and retrieving him back,
with the Americans later repeating the performance. The race in still
on in right earnest and who knows what greater wonders yet are in
store for us.
Let us try to understand the basic principles underlying this
phenomenon.
179. Conditions for a satellite to be placed in orbit. It is obvious
that an artificial satellite goes round the earth exactly as a celestial
SATELLITES 387
satellite goes round a planet, as the moon, which, for all practical pur
poses, is a satellite of the earth, goes round it, or as the earth and the
other planets go round the sun, i.e., in accordance with the laws, first
enunciated by Kepler, leading to Newton's celebrated Law of Gravita
tion, which forms the basis of the entire celestial mechanics.
The student is quite familiar with the whirling motion of
a stone, tied to one end of a string, the other end of which is held in
the hand. Precisely similar is the case with a planet going round the
sun or an artificial satellite going round the earfch, with the force of
gravitational attraction replacing the tension in the string. There is,
however, one fundamental difference between the two, viz., that where
as the tension in the string is, within limits, a variable quanti
ty, permitting a lower or a higher velocity of the stone, the attractive
force of the earth onihe satellite in a specific quantity and thus per
mils only a specific velocity for the satellite, if it is to remain in orbit,
this velocity for a satellite close to the earth being, as mention
ed already, about 8 kms. or 5 miles per second. Since, however, the
gravitational force decreases with increase of distance from the centre
of the earth, a satellite further away from the earth will need
a smaller velocity to remain in its orbit than the one nearer to the
earth, though up to about a 1000 kms. above the earth's surface, this
reduction in velocity is only nominal. This is clear from the fact that
the moon, which is roughly 38000 kms. away from the earth and,
therefore, moves in a much larger orbit, has only a velocity of about
1 km. /sec., which is about oneeighth of a satellite close to tho earth ,
so that, whereas the moon makes only one revolution of the earth in
one month, the satellite makes as many as 15 revolutions in one day.
Now, the question is how to have the satellite with such a high
velocity away from the earth, to enable it to go into arbit around it.
As can be seen, not only has the opposing gravitational force to be
overcome but also the very considerable air resistance, particularly
in the lower denser part of it. As we have seen above, the least velo
city for the purpose is about 8 km. or 5 miles j 'sec. , called the first
cosmic velocity. But, if the velocity rises to about 112 km. /sec.,
called the second cosmic velocity or the velocity of escape, the satellite
passes right out of the earth's gravitational field and flies away into
the cosmos, within the range of the solar system.
This formidable problem, can, as mentioned earlier, be easily
solved by carrying the satellite on a multistage rocket, for no single
rocket can possibly (at any rate, not yet) achieve the requisite velo
city all by itself alone. We have already discussed the essential
features of such a rocket in 175. Let us now see how exactly to
launch the rocket, carrying the satellite, into the required orbit.
180. Launching of the Satellite. Apparently, the shortest route
for the satellite to take from the launching base to its assigned orbit
would be the vertical one. This, however, is not feasible in actual
practice, for the simple reason that the gravitational pull of the earth
will then be in the directly 'opposite direction to its motion and coun
teract the pull of the engines. So that, before it can gather the
necessary speed, its limited fuel stock may get exhausted, resulting in
its first coming to a stop and then starting falling down. Vertical
launching of the satellite is, therefore, not a practicable propositian.
388 PROPERTIES OF MATTEB
To ensure that the satellite does not fall back to the earth, it is
essential to give it a sufficient horizontal velocity. Its upward flight
is, therefore, so arranged that it is brought into its orbit in the
shortest possible time, acquiring meanwhile the requisite horizontal
velocity. It is thus clear that the particular trajectory that will take
the satellite to its assigned orbit has first to be most carefully
calculated.
It is usual to arrange the first portion of the flight of the rocket
to be vertical, so that it may pass through the first 20 kms. of the den
ser portion of the atmosphere the earliest. Thereafter, as it enters the
rarefied portions of the atmosphere, it is given a gradual tilt by means
of a mechanical pilot, so that it emerges into its orbit with a horizontal
velocity large enough for the centrifugal force coming into play on it,
(on account of its circular motion), to just balance the force due to the
gravitational pull. And, the trajectory of its path is so chosen that the
loss of velocity entailed, due to airresistance and the earth's pull, is
a small percentage of its required or characteristic velocity. In fact,
to make up for this loss, the actual velocity given to it is a little
higher than the computed value of its characteristic velocity. When
launched laterally to the earth's rotation, however, an increase in its
velocity is automatically obtained at the expense of the velocity of
the earth's rotation, depending upon the latitude of the launching
site. Thus, for example, this increase is the maximum at the equator,
being as much as 400 met res /sec., which is higher than that of the
fastest fighter planes of the day.
If it be desired to give the satellite an elliptical orbit, instead
of a circular one, the rocket carrying it must either be given a higher
velocity than the perepheral on$ or its velocity, immediately alter
completion of the motor's performance, must not be directed along the
tangent to the circular orbit. In the elliptical orbit, the point near
est to the earth is called 'per the' and the farthest from it, the 'acme'.
And it is quite possible that the satellite at the former point may be
nearer to, and at the latter, farther from, the earth than at any
point in its circular path.
In any case, the accuracy demanded in the firing of the rocket
into its correct orbital path is really exacting. For, even an error of
1% in the direction of velocity may produce a height variation of the
perihe and the acme which may be as much as 120 kms. or more. This
firing accuracy is secured by means of proper steering devices, direct
ing the course of the rocket at every stage of its flight. And, clearly,
rudders of the type used in the ordinary jet aircraft, are hardly
suitable for the purpose, since they cannot possibly function equally
effectively both in the denser and the rarefied regions of the atmos
phere.
The necessary steering control can, however, be effected in a
number of ways but the one usually resorted to is to so design the
rocket as to enable it to change the direction of the escaping jet by
a mere tilt of the longitudinal axis of its motor with respect to its
.own. This is actually the device adopted in most of the presentday
longrange rockets.
The manner in which the angle of inclination of the longitudi
SATELLITES
axis of the rocket with the horizon is varied, will be clear from
Fig. 240. As will be readily seen, the trajectory of the rocket from
BURN OUT OF
2nd STAGE
IGNITION
Jrtf STAGf
(10 MINUTES
Af TR L A UNCH/H6)
OffBH OF SATELUT6
2SOOOft /Sec)
(2OO TO 400 MILES)
SEPARATION oft SEPARATISTS
2nd STAGE \OF 3rd STA6f
BURN OUT
AND
SEPARATION OF
STAGE ROCKET i
Fig. 240.
its very start until its longitudinal axis takes up the horizontal posi
tion, (/.., until its outward motion towards its orbit) is split up into
a number of stages, indicated by A 1 , A 2 , h 3 etc., depending upon the
height of the orbit. The angles that its horizontal axis makes
with the horizon at each stage is carefully, calculated before hand and
the control instruments set accordingly, to ensure that the rocket
takes its assigned trajectory. And, this very setting of the instru
ments also regulates the fuel sup ply in keeping with the pre
determined requirement at the lime.
Now, it will be easily understood that, while going round in it
allotted orbit, the Scitellite passes over different parts of the globe in
its successive rounds. For, by the time it has completed one round,
ihe earth has also rotated about its axis and hence, in its next
round, it naturally passes over other parts that now fall below its
orbit. This will always be so except when the satellite goes round
an orbit coinciding with the equatorial plane, in which case, obviously,
it will always pass over the same parts or countries situated at the
equator. It docs not mean that we can launch the satellite in
Any orbit we choose. For, the orbit must be one such that its plane
passes through the centre of the earth and it will, therefore, clearly
depend upon the site of launching.
Not only that, but even the time of the day and the season at
'the time of launching matter a great deal. For, a satellite receives
energy direct from the sun through special type of solar batteries fit
ted into it, a particular side of which must all along be illuminated
by the sun. The satellite must, therefore, be launched in an orbit, the
plane of which is perpendicular to the rays of the sun, and this
is possible only at a particular hour of the day, v/z., when
the radius of the earth connecting the starting point of the satellite
with its centre is perpendicular to the sun's rays. And, the season is
important because, with the satellite launched in its orbit, as
explained, the earth which also moves round the sun, comes inbetween
it and the sun at a particular time, thus preventing the rays of the sun
from reaching it. Account has, therefore, to be taken of this occur
rence and the season of launching chosen such that the satellite
390 PBOPBBTI1S OF MATTER
can get the maximum time to store up enough energy from the
sun to suffice for the period when the sun will remain hidden from it
later during its flight.
And, finally, it must also be clearly understood that in view
of the uneven distribution of the mass of the earth and, there
fore, with its e.g. some 500 km. away from its geometric centre,
the satellite in its orbit is subjected to varying forces of attraction at
different intervals, with the result that its real course is neither
circular nor elliptical. It does not even lie in either of the two planes
and is, in fact, a curve of a complicated pattern. For the same
reason, there are variations in the velocity of the satellite at different
points along its path.
181. Stability of the rocket during flight. It is imperative that
throughout its flight, along its allotted trajectory, the rocket should
not get tilted. This is achieved by means of an autopilot (see 46)
and a suitable gyroscopic arrangement.
182. Form of the Satellite. In designing a satellite, attention is
naturally paid to the geometrical shape it should be given to ensure
its smooth motion in its orbit. The present view is that this form
should be spherical, for, then, it will always have the same area
of resistance and thus help calculation of the air resistance to its
motion at higher altitudes and hence in the assessment of the density
of air at those altitudes. Further, with a spherical shape, there
will be less chances of its getting overturned than if it were cylindri
cal or of any other shape. At the samo time, a spherical shape
is also a drawback, since it doss not make for an easy setting of
the various instruments and other equipment inside it For, as will
be easily realised, the instruments must be sot, not haphazardly
but in a definite order so as to ensure both an equitable distribution
of the total weight inside the satellite and a specific position of
its e.g. This 'balancing* of the satellite, as it is called, is obviously
important and must be done with great precision.
183. Weight and size of the Satellite. The weight of a satellite
clearly depends essentially on the potentialities of the rocket carrier,
and its dimensions, upon those of the last stage rocket, which is
usually the third stage one.
The satellite which gets detached from the last stage rocket
may not necessarily be included as part of the rocket itself and may
simply be arranged to lie inside a cavity in the nosepart of it. In
such a case, it is possible to give the satellite a bigger diameter
than the mean diameter of the rocket, as a whole, but only slightly
so, or else it will mean a change in the ballistic characteristics of the
rocket as also an increase in the airresistance encountered. The
satellite in the cavity is sometimes covered by a protective streamline
cone, during the course of the flight of the rocket, which is later
discarded and the satellite pushed out by means of a spring or a
compressed gas, when the rocket has actually reached the orbit irk
which the satellite is intended to move. This was exactly the case
with Russian Sputnik /, whereas Sputnik /f formed part of the third
stage rocket itself and did not get detached from it.
184. Material of the frame of the satellite. Obviously, the material
SATELLITES 391
of the satellite frame must be both light and strong, the former from
considerations of its weight and the latter, to make sure that the
instruments etc. inside it are securely attached and that it 'can
withstand the onslaught of micrometeorites to which it is subjected
during its orbital motion in space. Then, the material must also
be less sensitive to changes of temperature and must be able to
properly reflect radiowaves. It must, therefore, be either aluminium,
magnesium or one of their alloys, with, in some cases, a suitable outer
covering,
If, however, it is desired to study the electric currents in the
ionosphere, the frame of the satellite should neither be a conductor
of electricity nor should it possess any magnetic properties. So that,
in this case, a metallic frame is clearly ruled out in favour of one of
a plastic material, some of the modern varieties of which are just as
tough and durable as steel.
185. Duration of satellite's existence. It is only natural to enquire
as to how long can a satellite be expected to stay in its orbit. Well, if
the space in which it moves along its orbital path were completely
devoid of air, there would be nothing to stop it and it could go on
perpetually, like the moon, for instance. But there being air even
at a height of 1000 kms. and above, it has to encounter resistance
due to it, however small, this resistance being greater for orbital
paths nearer the earth than further away from it. So that, when
its velocity is thus sufficiently retarded, it cannot possibly remain
in its orbit and starts falling down along a spiral path. In doing
so, it either gets burnt up due to the heat produced by friction in
the denser atmosphere or drops down to the earth with the help of
parachutes.
The actual calculation for its 'life' is rather a complicated one,
but it basically depends upon the density of the upper regions of the
atmosphere, i.e., on the height of its orbit from the earth.
186. Other Essentials. In case a man is to be placed in* the arti
ficial satellite or the sputnik, there are quite a few other problems
to be tackled, as, for example, provision of an hermetically sealed
ca,bin, with requisite conditions for the sustenance of life, and with
windows fitted with the type of glass that absorbs ultraviolet and
Xrays, a prolonged exposure to which is harmful in its effects. It
is, however, almost impossible to afford any protection to the cosmo
naut inside the cabin against cosmic rays which, as we know, can
penetrate even through a block of lead, one metre thick. Luckily,
although their effect on human or animal life is yet not quite clear,
they do not appear to produce any baneful effects. Then, there are
other problems, like those of weightlessness etc. All four have now
been more or less overcome, as is evidenced by four Astronauts,
two Russian and two American having made orbital flights and
come safely back to the earth.
Another very essential item is the special type of dress that
an astronaut must wear during his voyage in the cosmos. This is
fittingly called the Astrosuit and must at once be airtight and
loosefit to allow free respiration. In fact, the astronaut needs one
type of dress during the tifceoff of the rocket from the earth, which
392 PROPERTIES OF MATTES
must be so designed as to free him from the feeling of overload, ex
perienced during a vertical ascent. Then, he should have another
lighter dress for free locomotion inside the cabin. A small cylinder,
provided in the girdle of this dress, creates an artificial pressure
on him to increase his blood pressure (which falls appreciably at
greater heights) as also to counteract weightlessness. The dress should
have provision to ensure normal respiration and the requisite body
temperature and must not restrict movement.
187 Return of Artificial Satellite. For the return of the satellite
back to the earth, the main problem is of sufficiently slowing down
its motion or braking it. There are two devices for it, v/z., (i)
utilising air as the resisting medium and (ii) using rockets.
If airresistance is to be used for slowing down the motion
of the satellite on its journey back, it must be given the form of a
rocket. For, then, as it enters the denser layers of the air, its
velocity falls but it rebounds back into the cosmos ; it then reenters
the air with a reduced velocity and goes a little deeper than before
and there is a further reduction in its velocity. This process is
repeated a few times and the velocity of the satellite is thus suffi
ciently reduced to enable it to continue falling on specially provided
slide wings and sliding planes.
On the other hand, if a rocket is to be used for its downward
journey, an automaticallycontrolled rocketmotor is necessary, the
reaction of which is in the opposite direction to that of the motion
of the satellite, i.e., it produces an effect opposite to that of the
rocket carrier during upward flight. It is, therefore, called a retro
rocket, (i.e., a rocket, taking the satellite back). The velocity
of the satellite is thus reduced and can be controlled by regulating
the fuelsupply to the rocket motor, the distance it thus has to cover
up to the landing strip being carefully estimated with the help of
a radar or other similar appliances. And, an automatic guidance
system is provided to control and manoeuvre the downward descent of
the satellite.
Now, the first method is certainly the simpler of the two, from
the technical point of view, but its great handicap is that it is
extremely difficult to design a landing strip to receive the landing
satellite. The second method, although more complicated technically,
ensures a smooth and an accurate landing on a properly constructed
landing strip.
188. Uses of an Artificial Satellite. Ignoring the military uses to
which a satellite may be put, we shall concern ourselves here only with
its uses for strictly scientific purposes, among the more important
ones of which may be mentioned the following :
(/) Proper study of the upper regions of the atmosphere. Despite the fact
that the atmosphere is being studied for a long enough time, our present
knowledge of it is still much too meagre and superficial, particularly about
the region, called the Ionosphere, as also about cosmic rays. The artificial satellites
will, it is hoped, help to improve this.
(//) Weather forecasting. This can be made much more accurate and
dependable with a number of satellites around the earth in various orbits.
Meteorological observations over various countries could then be made simul
taneously, thereby greatly improving the reliabilr of weather forecasts.
SATELLITES 393
(iii) Determination of the exact shape and dimensions of the earth. This
4s the task that scientists all over the earth have set for themselves during the
third International Gcophysical year.
(iv) Detailed study of the solar radiation.
(v) Study of meteorites.
(vi) Experimental verification of the theory of relativity.
(vii) Use of a system of three artificial satellites for universal telecasting.
(viii) Study of propagationcharacteristics of radio waves in the upper regions
*f the atmosphere.
(ix) Astronomical observations, without atmospheric and other disturbances
etc., etc.,
EXERCISE X
1. Explain clearly the principle underlying an airplane. How does it
remain in equilibrium in air and how does it rise up ?
2. Differentiate between climbing, diving and gliding of an airplane and
explain the co relation of forces in each case.
3. Name the principal parts of an airplane and mention concisely
but clearly their respective functions.
4. What do you understand by the term 'jetpropulsion' ? Give, in brief,
in account of jetpropelled planes.
5. What is a rocket 1 How do rocketplanes differ from etplanes 1
Explain the principle underlying a multistage rocket.
6. What is an artificial satellite 1 Explain as clearly as you can how a
latellite may be placed in its orbit around the earth.
7. Mention the essential prerequisites and conditions for a satellite to
be placed in its orbit and its return back to the earth.
Also mention some scientific uses of an artificial satellite.
CHAPTER XI
FRICTION AND LUBRICATION PRINCIPLE OF VIRTUAL
WORK AND ITS SIMPLE APPLICATIONS
189. Static Friction Laws of Friction. In all cases of motion
of material bodies, counter forces come into play inbetween their
surfaces which tend to nullify or neutralise the effect of the driving
force applied. These counter forces are called resistances, the most
important among which is friction. As we have already learnt in our
junior classes, when one solid body is sought to be moved over the
surface of another on which it rests, an opposing force, called the
'force of friction', comes into play inbetween the two surfaces, tend
ing to destroy the relative motion between them,* and which is
usually measured by the force required to produce uniform relative
motion between the two surfaces. It is this force which always acts
in a direction opposite to that in which the motion is desired and
which is called the force of friction or rather static friction^ . Experi
ment shows that friction roughly obeys the following laws, called
the 'laws of friction', discovered by Amontons (1699) and Coulomb
(1779) and hence sometimes referred to as Coulomb's laws :
(/) The f fictional force is a selfadjusting force and increases
with the applied force, so as to be equal and opposite to it,
until motion is just about to ensue ; this maximum
fnctional force is called the 'limiting friction' and its
value is different for different pairs of surfaces.
Before this limiting value of friction is reached, its magnitude
is just enough to preserve equilibrium. J
(ii) The limiting friction between the surfaces of two bodies is
directly proportional to the normal reaction of the support
ing surface,
Thus, if R be the normal reaction of the supporting surface and
F, the limiting friction set up between the two surfaces, we have
F oc R, Or, FIR .= M ,
where p is a constant, called the 'Static Coefficient of Friction' or
simply the coefficient of friction for the given pair of surfaces.
... ^ . , rr . . limiting friction
i.e., static coefficient of friction = , .
" JJ normal reaction
*Strictly speaking, this is not the only force that opposes the relative
motion between the two surfaces. There is also another force, called the l force of
adhesion', (from the Latin word 'adhaerere' to stick or to cling, which is mole
cular in origin and which tends to make the bodies cling together.
t'Sffl/Jc', because the two surfaces are initially at rest with respect to each
other.
tit will be readily ssen that this really follows from Newton's third law oj
motion.
394
FRICTION
395
Thus, obviously, F = nR ; so that, if R = 1, we have F = p,
i.e., the coefficient of friction for a given pair of surfaces may be denned
as the limiting friction coming into play inbetween them, for unit
normal force applied to them, or, as the fraction of the normal force that
is required to keep the two surfaces in uniform relative motion.
(Hi) The frictional force is independent of the surface areas, in
contact with each other* and of their relative velocities.
(iv) The frictional force is independent of the relative* velocities
of the two surfaces.
It may be pointed out here that these laws apply only in the
case of clean and smooth or wellpolished surfaces.
190. Sliding Friction. We have already seen how the force of
friction between the surfaces of two bodies, one resting over the
other, continues to increase with the applied force and is always equal
and opposite to it until its maximum or limiting value is reached for
that particular pair of surfaces.
If at this stage, v/z., when the friction is about to attain its
limiting value, we apply a force in the form of a gentle push to the
body resting over the other, such that it is maintained in uniform
motion over the latter, then, this force measures what is called the
sliding friction between their surfaces, i.e., the frictional force inbet
ween them when motion ensues It is also spoken of as kinetic or
dynamic friction, to distinguish it from static friction (that comes into
play before motion actually takes place) and is found to be some
what less than the limiting friction for the same pair of surfaces. That
is why we find it easier to maintain a body iii uniform motion over the
surface of another than to start it moving.
The ratio between this sliding friction and the normal reaction
then gives the coefficient of sliding friction for the given pair of
surfaces and is also obviously less than the coefficient of static fric
tion for them. The difference between the two is however quit
small and we usually assume them to be the same for all practical
purposes.
191. Angle of Friction Cone of Friction. If we place a body
Fig. 241.
*This is no longer so, if what are called lubricants, like grease, graphite,
talc etc., are introduced inbetween the two surfaces. For, the normal force
applied is more likely to squeeze out the lubricant from inbetween the two
surfaces, when applied to a small area than when applied to a larger area.
396 PROPERTIES OF MATTER
on an inclined plane, [Fig. 241 (a)], it is clear that it is under the
simultaneous action of three forces, viz.,
(/) its weight W, acting vertically downwards at its e.g., G ;
(//) the normal reaction R of the plane, acting perpendicularly to
the plane ; and
(in) the tlimiting friction F, acting upwards along the plane.
If the angle of inclination of the plane be so adjusted that the
body is just on the point of sliding down, it is clear that the three
forces are in equilibrium and, therefore, concurrent, so as to be re
presented by the three sides of a triangle, taken in order. This angle
of inclination of the plane, ft, at which the body is just on the verge
of sliding down, is called the angle of friction.
Resolving W into its two rectangular components, we have (/)
somponent W sin \, along the plane, tending to move the body down
bhe plane and component W cos A, at right angles to the plane.
Since the body is in equilibrium, we have
F = W sin X and .R = W cos \.
So that, ., 5 = * = tan >.
Or, the coefficient of friction for a given pair of surfaces is equal to
the tangent of the angle of friction for them.
From the above relation for /*, we have F = R tan ft, from
which it follows that the resultant of F and R must lie along the sur
face of a cone, with \ as its s^mivertical angle, and the direction of
the normal reaction, as its axis.
The same is true in the case of two horizontal surfaces, where
the frictional force F acts along the supporting surface and the nor
mal reaction R, perpendicular to it, [Fig. 241 (b}]. Their resultant
P then makes with the latter the angle of friction ft and, again, there
fore, the resultant (P) lies along a cone of semi vertical angle X, such
that tan A = F/R.
This cone is called the cone of friction, and it is obvious that
no force, whatever its magnitude, with its line of action lying within
the cone, can possibly produce motion in the body, its component
along the surface of contact being less than the limiting friction (F)
between them.
192. Acceleration down an Inclined Plane. We have just seen
in 191, above, that a body placed on an inclined plane will not start
sliding down along "it until the angle of inclination of the plane is
equal to the angle of friction ft for the surfaces of the body and the
plane ; for, at a smaller angle of inclination than this, tan 0<F/J? or
/*, where F is the limiting friction and ^, the coefficient of friction for
the two surfaces in question. And, when 0=\, clearly tan 9= tan A
= F/R and sliding just commences.
But when > ft, so that tan Q > tan ft and hence greater than
F/R, the body slides down the plane with an accelerated motion. Let
us calculate this acceleration of the body.
As before, resolving the weight of the body W=mg (where m is
FRICTION 39T
the mass of the body) into two rectangular components, along and
perpendicular to the plane, we have
component along the plane == mg sin 0.
and component perpendicular to the plane = mg cos 6.
Since there is no motion perpendicular to the plane, the normal'
reaction R of the plane is equal to mg cos 6 and the two, being equal
and opposite, neutralise each other. And, thus, the only two forces
effective on the body are (/) mg sin 0, downwards along the plane,
and (//) the sliding fractional force F upwards along it. So that, the
resultant force acting on the body downwards along the plane is equal*
to mg sin 6 F.
Now, if n be, the coefficient of sliding friction for the given pair
of surfaces, we clearly have
F = /ijR = n mg cos Q.
So that, the resultant force on the body downwards along the plane
= mg sin dnmg cos 6 = mg (sin QLL cos 6).
And since acceleration = force/mass, we have
acceleration of the body downwards along the plane
= mg (sin Qn cos 0)jm = g (sin 0v> cos 0).
If the plane be perfectly smooth , the acceleration of the body
sliding down the plane would be g sin 6. Clearly, therefore, the acceler
ation of the body down the plane is reduced by n g cos 6 due to
the frictional force between them.
193. Rolling Friction. The frictional forces between two sur
faces when one roils over the other is called rolling friction and is
found to be much less than when sliding occurs between the same
two surfaces. That is why vehicles are provided with wheels and
their axles, with ball bearings*, the latter converting the chief fric
tional loss of the wheel that occurs at the axle or the journal in the
form of sliding friction, here, called journal friction into rolling
friction.
It was shown by Osborne Reynolds that in rolling an appreci
able amount of slipping or sliding of one surface over the other
occurs and that the frictional resistance to this slipping, or sliding,
really constitutes rolling friction. As extreme cases of this slipping
between two rolling surfaces may be mentioned (?) an iron cylinder
rolling over a plane rubber surface or (ii) a rubber cylinder rolling
over a plane iron surface. In the former case, the cylinder covers
a distance equal to only nine tenths of its circumference in its one
full turn and, in the latter case, a distance equal to eleven tenths of
its circumference ; eo that, in either case, there is a slip of onetenth
of its circumference In ordinary cases too, some slip always occurs
between two rolling surfaces, even when the two surfaces are of the
same material*
It follows as a natural consequence that rolling friction between
two surfaces would be zero, (a) T\hen either tie sliding friction
between them is zero, i.e , when v. for them is zero, (b) or when no
*Here, there is a ring of small balls between the wheel, (pulley or disc
etc.) and the axle, so that when the former rotates, the balls all roll also.
398
PROPERTIES OF MATTER
slipping occurs between them during rolling, i.e., when v> for them is
infinite. For all other values of \i (between and oo) there must be
friction and, for a particular value of M, it must have its maximum
value.
It may as well be mentioned here that while lubrication of
the surfaces always reduces the value of the coefficient of sliding friction
(M) for them, it may or may not reduce the rolling friction between
them. Thus, as is so well and so generally known, lubricating ball
bearings only results in increasing friction*.
194. Friction and Stability. When a body, say, a block of wood,
rests on a plane horizontal surface, it does so because the weight
D A
D
<>
Fig. 242.
of the block W, acting vertically downwards at its e.g., is just
neutralised by the equal and opposite normal reaction R acting there.
And, when a horizontal force Fis applied to the block to move it for
wards on the plane, it does not move or slide along it so long as
F<uR, where u is the coefficient of friction between the surfaces of
the block and the plane, (see page ,196). The possibility is, however,
there that the block may topple over ; for, the moment the horizon
tal force F is applied, at a point P, say, a frictional force F', equal
and opposite to F, comes into play inbetween the surfaces of the
block and the plane, [Fig. 242(0)], thus constituting a couple =Fx PB,
tending to rotate the block (in the clockwise direction, in the
case shown), and thus making it topple over.
Now, as Fis gradually increased, this couple formed by Fand
F' makes the centre of reaction of the plane shift from //towards J5f,
with the force at C progressively decreasing and that at B, increasing,
until, in the limiting case, the whole reaction R acts at B (that at C
being zero). We thus have another couple, formed by W and 7?,
equal to WxHB, tending to rotate the block in the opposite direction
to that due to the first couple (in this case, anticlockwise) which thus
tends to restore the block back to its original position. So lon#,
therefore, as this restoring couple W XHB, is greater than the couple
*This might raise a question in the mind of the student as to why then
are they lubricated at all ? The simple answer is that it is done only with a view
to reducing wear.
tFor, with no force F acting on the block, its weight is uniformly distri
buted over its base B,
FBICTION 399
FxPB, due to F and F', the block remains at rest and upright,
i.e., in stable equilibrium, but if WxHB<FxPB, it topples over.
Let us now investigate the conditions under which the block will
remain upright but slide along the plane. For this, let us first deter
mine the resultant of F and W by the ordinary application of
the parallelogram law of fordes and then take the moments of this re
sultant R' and the frictional force F 1 about B. Let R' be represented
by the diagonal EL of the parallelogram EMLN, with its adja
cent sides EM and EN representing Fand W respectively, and let it
cut the plane in K, [Fig. 242(6)].
Now, clearly, moment of F' about B is zero, since its line gf
action passes through B. So that, so long as HK<HB, i.e., so long as
the resultant R' passes through the contour of the ba^e of the block,
it will have a restoring moment about B and the block will remain
upright. But, if the angle made by R' with the vertical, or the
direction of the normal reaction R, be greater than A, the angle
of friction for the surfaces of the block and the plane, (see 191), it
will slide along the surface. If however, HK>HB, i.e., if the
resultant R r , of F and W, passes outside the contour of the base
of the block, the block will topple over.
Another important case of stability due to friction is that of a
fast moving vehicle on a curved track, discussed already in 18 (4).
195. Friction, a Necessity. Taking most of our daily activities
in life as a matter of course, we seldom care to pause and think as to
bow much they are dependent on the existence of friction. Thus, for
example, in the absence of friction, we would find it impossible
to walk or to drive on a road, and if we just start moving, we shall
not be able to stop ; again, it would be impossible to climb a tree, tie
a knot or even fix a nail in the wall. Brick would not stand on brick
and buildings would tumble down like a house of cards and so
on. Indeed, we find it so much of a necessity that we deliberately in
crease it for many of our purposes, as, for instance, when \vo
.apply brakes to our bicycles or cars.
In many other cases, on the other hand, we find friction
so irksome ; as, for example, in the various parts of our machines,
making their speed slower and their output lower and bringing about
a greater wear and tear in them. And, yet, we know that friction is
necessary even for thorn. What we do, therefore, is just to adopt
ways and means of minimising it in such cases by means of oils
and other lubricants, and ballbearings etc. etc., (see page 404).
196. Simple Practical Applications of Friction Rope Machines. Apart
from the ordinary uses of friction, some of which have been mentioned above,
there are various types of useful machinci based on it. We shall consider here
a couple of them by way of illustration of the principle underlying them.
1. The Prony Brake. This is a simple appliance to measure the power
of machines, which we owe to Baron G. C. F, Prony, a French Mathematician,
(1750 1839). It is in fact a broke dynamometer and consists of two wooden arms
or, *cheeks\ A and B (Fig. 243) in between which can be clamped the shaft of
the machine whose power (i.e., rate of working) it is desired to measure. The
frictional force between the shaft and the cheeks is regulated by tightening or
loosening the screws S and 5. provided on the uoper cheek A % to which is also
attached a small rod /?, about *5 to 1*0 metre in length, carrying a scale pan at
its other end.
400 PEOPEBTIBS OF MATTEB
Suppose the shaft of the machine turns in the anticlockwise direction, at
shown. Then, if there be no weights placed in the scale pan, i.e., if the brake
Fig. 243.
be unloaded, it will tend to be carried around with the shaft due to friction
between itself and the shaft. But, if the brake be sufficiently loaded, before the
shaft starts rotating, i.e., before starting the machine, the moment due to
the load may be enough to overcome the nioment due to friction between the
brake and the shaft and the brake may turn in the clockwise direction ; so that,
it is quite possible to so adjust the weights W in the scale pan, i ?., to so load
the brake, that the rod R remains quite horizontal inbetween and equidistant
from the stops s t and J 2 placed on its two sides, a little distance away from it.
When this is so, obviously, the frictional resistance between the brake and the
shaft is equal to the force F exerted by the machine on the periphery of the shaft
and is exactly balanced by the \\eights or the load W placed in the scale pan. So
that, we have
moment of force F(due to machine) on the shaft = moment of weight W on it.
If r be the radius of the crosssection of the shaft and /, the length of the;
rod R, we clearly have
moment of F on the shaft = JFx r,
and moment of W ,, ,, = W xl.
Thus, Fxr = Wxl, whence, F  W.lfr.
If the shaft makes one full rotation in time r, we have
work done by the shaft, i.e., by the machine, in time T
= Fx circumference of the shaft = Fx2nr.
And .'. work done by the machine per unit time, i.e., the power of the machine
Fx2nr
~ T '
Or, substituting the value of F, obtained above, we have
power of the machine = ^x =  W.I = _ x moment of the load.
Thus, knowing W, I and T t we can easily determine the power of th<
machine, which, as we can see, is clearly proportional to the moment of the
load.
N.B. If W be taken in dynes, I in cms. and Tin sees., we get the power ol
the machine in ergs per second.
2. The Rope Brake. Before discussing any ropemachines, we
must be clear in our minds as to what exactly is meant by a rope and
what, if any, are the peculiar properties possessed by it that make
for its usefulness.
A rope, then, is any flexible body or combination of flexible bodies,
capable of transmitting tension. Thus, the string of a violin, apiece ojf
spring, a strap, a band&nd a chain all come under the definition of a
rope.
When one. end of a rope is connected to a body and its
*>ther end pulled, stresses are caused at every crosssection of it,.
FRICTION
401
in exactly the same manner as in a rigid rod. The stresses in neigh
bouring crosssections neutralise each other and the pull is thus
directly exerted on the body at the other end. A rope can thus trans
mit a tension, undlminished in magnitude, from one end to the other.
Due to its flexibility, however, it cannot transmit any compressional
forces along it ; but, at the same time, its flexibility is of great
advantage in many ways and makes it particularly suitable as a
means of transmitting tensional force, unaffected even by a change of
direction*, as in the case of a pulley etc.
On the other hand, when coiled round a cylindrical body, a
rope can exert a very large couple on the body, due to frictiojn
between itself and the body, so
much so indeed, that a man pulling
at one end may even hold a ship,
fastened to its other end. Let us
see how this comes about.
Let A BCD be the cross 
section of the cylindrical surface,
with its centre at O and let a rope
PABCQ, coiled round it, leave its
surface at points A and C, (Fig.
244).
Consider an infinitesimal por
tion EF of the rope at B and let
the mean tension over this portion
be T, with the angle subtended by
it at O equal to eld.
Then, representing the tensions T, at E and F, by the tangents
EG and FH respectively, we have their resultant force represented by
BJ (by the simple application of the parallelogram law of forces)",
where BJ is clearly equal to BG.dO = T.dO, [v BG represents T
and BGJ = d6], in the direction BJ, normal to the section EF.
at B.
If, therefore, \i be the coefficient of friction for the ropo and the
c3 T lindncai surface, we have
frictional force between the rope and the surface = /i.T.rftff*
in the direction of the rope at B.
Due to this frictional force, there comes about a change in the
tension at the two ends of the rope, which, in the absence of any
friction, would have been the same, (see above). In fact, the differ
ence in the tensions at the two ends is just equal to this frictional
force. So that, if dT be the difference in tensions at E and F, wo
have
Fig. 244.
dT =
Or, l T .dT
"This is possible only so long as the body, (e.g., the pulley or the ring
etc.) over which the rope is passed does not interfere with its freedom of motion
i.e., is perfectly smooth and round.
tBecause frictional force J*x normal reaction R* and here, R
402 PROPERTIES OF MATTER
integrating which, we have
<..dd. Or, log, T = nO+K, (0
where g is the angle EOA and K, a constant of integration.
Clearly, if Q = 0, T = T l9 the tension at A so that, we have
log, 2\ = nxQ+K = K.
Substituting this value of K in expression (i) above, we have
log, T = ne+logt T,. Or, log, Tlog, T, = rf.
Or, log, 7) 7\ = ^0. Or, 7)^ = e^,
whence it is clear that the tension T increases as increases.
Obviously, therefore, if <f> be the total angle AOC, and T 2 , the
tension at C, we have
It is thus clear that if ^ bs large, i.e , if the rope be coiled many
times round the cylinder, 7 1 ,/ 7\ is also very large and a small tension
applied at P rcuiy be made to exert a large pull at Q.
(3) The Band Brake. The principle, discussed in 196, (2)
above, is made use of in many forms of brakes, for the measurement
of the power (or the rate of
doing work) of machines, one
of the simplest of which is the
Band brake.
It is a simple device,
consisting of a pulley fixed on
to the rotating shaft of the
machine whose power is to be
determined. A band (or a
cord) is passed round the
pulley, once, twice or thrice
etc., and has its two ends
attached to two spring
A and B suspended from a rigid support, as shpwn in
Fig. 245.
balances
Fig. 245.
Then, if 7\ and T 2 be the readings in the two spring balances
respectively, we have
couple exerted on the pulley due to friction = (T^T^.R,
where R is the radius of the pulley.
Now, work done by or against a couple = couple x angle of rota
tion (in radians).
And, therefore, work done against a couple per second
= couple x angle turned through in one second.
Or, rate of doing work, i.e., power = couple x angular velocity.
Thus, work done per second by the machine against the couple
due to friction ^(T^T^.R x 2nn,
where n is the number of rotations made by the shaft or the axle of
the pulley per second and hence 2;r, the angular velocity of the shaft.
MECHANISM OF FBIOTION 4t)3
This, then, gives the work done per second by the machine, or
the power of the machine. Thus, we have
power of the machine = (T^ r i )./?x27m.
Thus, knowing T t and T z (from the two spring balances), R and n, we
can easily calculate out the power of the machine.
Further, we have the relation T Z IT : = e* ,
where ^ is the coefficieat of friction for the cord or band in question
And the surface of the pulley and 0, equal to TT, STT, or STT, efc. accord
ing to the number of times the cord or the band passes round the
pulley. So that, knowing 7\, T 2 and 0, we can also easily calculate
out the value of M.
197. Mechanism of Friction. Since the laws of solid friction were
enunciated by Amontons and Coulomb, much has come to be known as to the
jfiow and why of friction, thanks to^he work of Hardy, Bowden and others.
According to Dr. Bo