PAGES MISSING WITHIN THE BOOK ONLY DRENCHED BOOK 1 64598 00 P 750-28-4-8 1 - -U),UOO. OSMANIA UNIVERSITY LIBRARY Call No. ^ryJu 4 Accession No. Author Title .*3t marked This book should be returned on or before the <** ELEMENTS OF PROPERTIES OF MATTER WITH TYPICAL NUMERICALS SOLVED I FOR DEGREE CLASSES ] by D. S, MATHUR S . C H AW-D & 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, 9-00 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 time-period 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. 274-341 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 cross-section of the jet Rocket planes Rocket fuel Specific impluse Shape of the Rocket The Multi-stage 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 Yi|S&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 Freezing-point 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 Surface-Film 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 a-ray 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 581-588 Appendix IV Integral Calculus 589-596 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 602-605 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 clear-cut. 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 well-defined 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 Centimetre-Gramme-Second system or the C G.S. system, (ii) the Foot-Pound-Second system or the F. P. S. system and (/w) the Metre-Kilogramme-Second 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 one-hundredth part of a metre, 'which is the distance, at a temperature ofOC, between two lines on a platinum-iridium 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 one-thousandth part of a lump of platinum-indium, 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 platinum-indium 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 platinum-indium 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 non-dimensional (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) Non-dimensional 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) Non-Dimensional 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. =453-6 gms., and 1 ft. = 12 x 2-54 <w.=30-48 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 = (453-6M)(30-48 L)T~*. ~ . , 7 453-6MX 30-48 .r-*^ <-, tct-i Ur, 1 poundal =* ,> -r --- log 453*6 = 2-6567 ' ^ MxLx log 30-48 - 1 4840 Antilog 4- 1407 = l'382xl0 4 s=s 1-382 x 10* units in the C.G.S. system. Thus, 1 poundal = l-382x 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 32-2 ft. poundals I sec. and g = 32-2 /*. /sec. 2 y Again, as shown in Ex. (1), M units in F.P.S. system = 453-6 M units in C.G.S. system, L units = 30-48 L and T units . = T - Since the dimensional formula for power is AfL 2 r~ 8 , we have H.P. = 550x32-2(453-6M)x(30-48L) 2 xr-. F P - 550 x 32-2 x 453 ' 6M X ( 30 ' 48L ) 8 x r - . H.P. 550 x 32 2 x - - - - log 550 = 2-7404 = 550 x 32-2 x 453-6 x (30-48) 2 ergs/see. log 32 2 = T5079 log 453'6 SB 2*6567 550 X 32-2 x 453-6 x (30'48) 2 J^fySL 2 log 30 48 - 2 9680 10 7 9-8730" = 746-4 watts. 7 i g 10 = 7-0000 Antilog 2-8730 746-4 [Because, as we know, 1 joule/sec. =* 1 watt.] Thus, 1 Horse Power = 746-4 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 time-period 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 time-period of a simple pendu- lum. The factors on which the time-period (/) 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 time-period 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 non-dimensional 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 cross-section of their limbs, we have the ratio between their densities equal to one and similarly that between their strengths per unit area of cross-section 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//-f-|a/ 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 non-dimensional 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 non-dimensional 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 wave-length 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 wave-length 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 pound-weight, 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 = 3-5564 5-0643 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, 32-2- 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 3-058 x 10 7 =9 174x 10" cms. log 13-6= 1-1 335 j And, from (c), we have log(9'174x 10 17 ) ; unit of mass, (M) = ML- 3 xL 8 =13'6x(9-174x 10 17 ) 3 . =53 8878 --- Antilog 55-0213 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 time-period 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, y-i . 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 i-f c=0, or c= |. Also |4-A-(~i) = 0, or f-h&~h4 = 0, i.e., l-t-6 = 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 /j-hc^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 a-36+c 0, 1) = 1, and -a-2c = 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 drop-radius 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 foot-pounds 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 time-interval 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 cross-section 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 time-period Tof its oscillation. (Assume the area of cross-section of its neck to be a.) Ans 7=2" Vm/tW *20. A (/-tube of uniform cross-section 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 time-period of oscillation. Ans r [Hint Just put T=*k.d*tfg and show that where T is the time-period 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 actio-i 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 anti-clockwise, 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 centre-seeking 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 mid-point 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 drum-shaped '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 discharge-tube 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 small-sized stones etc,, 20. The Projectile Motion of a Projectile in a non-resisting 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 w-f 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 (180-25), 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(/^ 0-tan 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/usmti-gty+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 0-2 /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 non-resisting 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 = 1-3424 - . ,. 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 one-half 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 - 2-5623 ..(/) log 2 - 0-3010 | 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. 3-84 xlO 10 time taken & 2*x3-84xl0 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 3-84 xlO 10 & (27-3 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 3-84 3 log 10 10 0-9944 =1-7529 =30-0000 33-3494 (') log 6-4 0-8062 8 log 10 =8-0000 log 27-3 = 1-4362 \ log 24 = 1-3802 log 3600 = ^5563 ; 15 : 1789 I 2x15-1789 =30'3578..(/0| Subtracting (//) from (/), we have 2-9916 Antilog 2-9916 = 980-9 * Or, g F L (6-4 xlO 8 ) 2 J' K- 6-4 x 10* J A L(27'3x24x3600) 2 J __ 4ir^x_(3-84xl0 1( V ** (6*4 x "iOx 27 : 3x 24 x 3600) 2 ' 980-9 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 side-pressure on the rails. (Bombay) 20x1760x3 88 f , Here, we have v 20 m.p.h. = and, log 121 log 10800 Antilog And, Or, = 2-0828 4-0344 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 38-33'. 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 . 320-320 = 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, =-1280-1-5120 = 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 164-2560. = 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 pounds-weight 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 mid-point 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 = 0-9031 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 =1-7982 Antilog 1-8250" 66-83 2nxlOOO = 2x3-1416x1000. 6283-2 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=0-9542 log 5=0-6990 log 980=29912 Or, whence, Or, 4-6444 X 4-6444 log w 0-4972 Antilog T8250 =66-83 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. OB-CB OB' (I) l-h 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 > -2r--i3v- 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 counter-balanced 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. 15-43 Ib. wt. 2. A half-pound 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 merry-go-round 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 sea-plane 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 sea-plane, 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 stone-wt. 12. Calculate the angle at which a curve of radius 352 //. should be banked so as to avoid side-slip 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 'loop-the-loop* 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 side-slip, 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 motor-cycle 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 Three-Dimensional Body*. Suppose we have a cubical or a three-dimensional 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 x-axis Fig. 23. and MN, parallel to the j-axis ; also, from P draw PjR, parallel to OM. Then, clearly, the co-ordinates of the point P are x ;= ON = QM ; y = OQ = NM and z = M P = OR. Since the plane xy is perpendicular to the z-axis, any straight line drawn in this pLane is also perpendicular to it, and, therefore, OM and PR are bath perpendicular to the z-axis, (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 x-axis. Similarly, in the right-angled &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 z-axis =*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 z-axis, i.e., I = Zm.OM 2 . Or, /, = Zm(x*+y*). Similarly, the moment of inertia of the body about the y-axis, i.e., L = Zm.PQ 2 . p-' PQ is the J_ dis- Or / Sm (x 2 4-z 2 \ tance between ^e ur ' '* - Zm '\ X + Z > L mass and the axis. And, the moment of inertia of the body about the x-axis, i.e. 9 4 = Zm.PN\ p.- P-V 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 three-dimensional 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 three-dimensional 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 Three-Dimensional 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 right-angled &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 gram-centimetre 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 Pound-feet*, (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 three-dimensional 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 co-ordinates 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 half-truth 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 = M|8 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 mid-point of one side and perpendicular to its plane. (a) Suppose the axis of rotation passes through the mid-point 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 cross-section, 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 co-ordinate 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. 1-2 '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 diamet-er. 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 MR-J4. 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, face-area of the annular disc = face-area 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, face-area 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 - (IP-r) 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 74-7- M( * 2 + r *> /* 2/- Aff/P+l * ) ur, /-j-7 = - , 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 4---J the disc x(radius)^. -?*' f ' And, its mo- iftent of inertia about the axis YY' 9 Y /_ . . _ t/g- _a> ?\ -4V-4- / ""\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 semi-vertical 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, face-area 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?2-Z7 a ) ~ (^^H7^J r x 5Jf r^-iJ 2jj/ n^- 14 )" V-OL 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, face-area 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 -"'-T-Tr-P-* 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, 1-2 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 -J-Trr 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 * (R-r) 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 ball-bearings (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 )/4-2 plane. (i v) Through the mid-point of one side (/ or b) and per- peedicula to it* plan* * M(6*/3-|-/ 2 /U^ or M(/ 2 /3-J| 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* mid-point 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 one-third of the sum of the squares of the other two semi-axes, in the case of a rectangular lamina or para- llelopiped ; (//) the products of its mass and one-fourth of the sum of the squares of the other two semi-axes, in the case of a circular or an ellipti- cal lamina ; (til) the product of its mass and one-fifth of the sum of the squares of the other two semi-axes, 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 semi-axes 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 semi-axes 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 semi-apes 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 semi-axes 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 ball-bearings (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 mg-T. 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 time-interval 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 (to4-0)/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 = 4-nnjt'. 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 paper-strip, 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/2-nr = N ; and further (11) that t = 25/v 25/ro, where o> = 4-nn/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.S-lmv*, 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, / = ~ . (*A-Jv) Or, / - ^ (*gh-v*). 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 K-E. =* 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 4-r 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 two-dimen- 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 rotation-axis 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. r----r ----- ---; *{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, o-bout 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 rotation-axis 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 rotation-axis 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 rotation-axis 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 time-period 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.dtfi-dt == 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 -f-ve 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 I--A.\ 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 =z-MK* 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 right-angled &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 right-angled 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 Gyro-Compass. 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 Gyro-Compass. The above arrangement, with a small rnass, suitably suspended below the rota- ting disc or flywheel, constitutes what is called the pendulum Gyro-compass, 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 well-known 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 gm-cm*. (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 = 9-0000 log 1250 = 3-0969 2 log n = 9944- -. ' 13-7903 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 = 1-5051 = * 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 dynes-cm. Now, /.rfco/cfr = Tr. 2xl0 7 a 5xlOOOte-a)x2 - 10000^-100000. Or f 2xlO T a-HOOOOa = 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 = 2-0000 , 10 * io a log ir - 0-4972, - 1000x7r = - k.gm.wt.-cm. ADtilog^ ^1-5028 j =, 31 . 83 ^ m .^ cm . Hence, a couple of moment 31 '83 k.gm.wt-cm. 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 poundal-ft. 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 = 40-npoundal'ft. [ v couple = Ldujdt. total couple required to be applied log 45 1-6532 \ = this couple of 40rc 4 a couple of SK (to overcome fric- log TT = 0-4972 tional couple)* Antilog^2-T504 i Or, total couple required = 40:r-}-5rr = 45?r ^14\'4 poundal-ft. = 141-4 ' 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 - 400nr-407r = 360rr radians. And, if 0, be the angle described by the flywheel in the next 20 sees., we have, as above, 0, - (27tx4)x20-f * 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\ = (5-2) 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 dynes-cm. 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 dynes-cm. 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, applymg-the 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 dynes-cm. log 327-25145 ! ^ n 65400x18x18x77: log 63-17993 ! 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=(Mg-T). Or, T^M(g-a) = 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(32-ra). r poundal-ft. 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 poundal-ft We have couple due to tension T rotational couple. 90(32-0) r = 450 a.r. Or, 90(32-) = 450 a. Or, 2880-90 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 640-2-K062 3=0 4771 1 x 2-3291 Antilog 1-1645 -14-61 [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 = 14-61 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(20M-5 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 -J-125xl0 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". = 10x-4755. 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 => 2-9320 4 log 10 = 40000 2 log 4-755 = "1-3544 Antilog 82864 B 19-34X10 7 log 5 - 6990 3 log 10 = 3-0000 log 25 = 1-3979 2 log 4*755 = T354I Antilog 6-4513 = 28-27 xlO 5 of the wheel and axle is given by i/.w a = }xl71xl0 5 xo)*. = 855xl0 4 x(4-755) 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 23-775 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(Af-f 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= 1-2304 Antilog l : 983f = 96-18 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 96-18 cms. I sec*. v 2 /25. Or, the acceleration of the system of weights is 96-18 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 cross-section r, about (/) the axis of the cylindrical symmetry ; (//) a generating line ; (*//) a diameter of cross-section 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 cross-section, (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 centre-f 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 cross-section = 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 [l-r 5 /K 5 ]_ 2 ri-r*/*'! ~ 5 >[l--r 3 /#<J " 5 U 1 -'' 3 /* 8 J Obviously, r a /K 3 < r 3 //^ 3 - And /. (1 r 5 //? 6 ) > (l-r 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 (R-jR'+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 . ; M-L 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., 13-80 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. 2-413. 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) 980-7 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 2-0 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. 8-9 and 17-8 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 time-period. 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) Gyro-compass, (ii) f'endulum Gyro-compass, (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 time-period, 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 time-period 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 U-tube, 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 = w-v/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. Time-Period 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 lime-period, period of vibration or periodic time, usually denoted by the letter T. Obviously, it is the same as the time-period 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 2-7T r v to = **, the acceleration of M per unit displace- l~ mei ment. whence, Or, time-period 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 time-period 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 one-fourth of the total angle 2-7T, i.e., an angle 2?r/4 or Tr/2 ; or that one-fourth of the time-period, 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, = (wf-e). 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>f-f 2ir-e), 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 time-period 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, 2n-rr, etc., with y assuming the same value over and over again. In other words, the time-period 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 time-periods 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 time-axis and the direction of vibration of the prong as the vertical or the displacement-axis. 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 time-period 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 sine-curve for the first motion, and DffJK for the second, we obtain the sine-curve 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, time-periods (or frequencies) and the phase difference of tee two compo- nent motions. Let us consider the different cases that arise. (a) When the time-periods (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 time-periods 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 arrow-heads. 74. (c) When the time-periods 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 one-eighth 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 arrow-bead. 75. S1MPLB HARMONIC MOTION 125 (d) When the time-periods 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 three-fourths 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 time-periods 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 time-periods 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 time-periods 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 time-periods 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 time-preiod. 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, one-fourth part ahead. Then, proceeding as before, we obtain, in this case, a parabol-a 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 time-periods 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 time-poriods 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 time-periods 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 (wf-f <). = 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 time-period, 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 (wf-f ^) -|-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 time-period 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. -i-a* 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 ^i-f-co5 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>i-f*\ 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 co-ordinates '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, and-c05^=== 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 time-period 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 x-axis at an angle tan^ bja ; so that, the resultant motion is again a S.H.M. , with the same t'me-period 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 semi-axes are equal to b and a respectively. The resultant path is, therefore , an ellipse, (Fig, 84), which it describes once in the time-period 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 time-periods or frequencies are nearly equal: In the cases, dealt with above, where the time-periods of the two component vibrations are identical, the elliptical paths of the resultant vibrations remain fixed. But if the two time-periods 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 $+cos 2wt sin <f>. Or, y\b s= 2 sin wt cos <*>t cos ^+(12 sin 2 a*t) sin <f>. [v sin 2a>t = 2 sin a>t.cos wt and cos 2wt = (12 sin* wt)]. .'. substituting the values of sin wt and cos wt from above, we have JC* \i- / X^ \ I t jcos<f>+( 1 2-~2-Jsw ^. I i .pj* cos f-^.$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 time-periods 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 x-axis at an angle, ta/r-i-A/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 semi-axes are equal to b and a (Fig. 84). The resultant motion is thus an ellipse here,' describ- ed once in the time-period 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 time-period 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 time-period (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 = fyna-to 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 time-period 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 time-period 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 time-period ? Let original volume of the gas be = V and its pressure = P. Let a be the area of cross-section 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 * <*tl1-sin 2o>//4 ~- ^ dot. " T the pa$c with wV *9f * w^pjc time-pcri0d 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(V-x.a) - PV-Px.a+pV-p.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 cross-section 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 P-a-> 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 time-period, 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, time-period 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, . - 1-432 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 - 1-432 - -716//. So that, v - ~ V(f : 432) 2 ^716)^ = \/(T432 1 716)(i'432-'71Q. - -V/2T481T71T-. 12-99 /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 12-99 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 time-period of its vibration. [Radius of the eartl* 4009 miles, aqd value of f op its surface - 32 ft. per sec. per $ec.J SIMPLE HABMONIC MOTION 139 We know that the force with which a body is attracted by the earth to- wards its centre is equal to the weight of the body, (m^), and also equal to G.m.MIR*, where m is the mass of the body ; M, the mass of the earth ; JR, the radius of the earth ; g, the acceleration due to gravity on the surface of the earth and (7, the gravitational constant. nig G.m. MIR 9 . Or, g - G.MJR*. Since the earth is supposed to be a sphere, its volume 4* 8 /3, and, there- fore, if A be its density, we have its mass, M * 4* R*.&/3. So that, * -~ . ^ ;~. G ** 4.7t/?A.C7/3. . .(0 If the value of acceleration due to gravity at a distance r below the surface of the earth, (Fig. 88), be g' t we have, as above, Dividing (//) by (i), we have 8'lg- l.*.(R-r)&.Gl.*.R.&.G. Or, g'Ig=(R~r)lR. Or, # ' = g(R-r)IR - (R-r).glR. Thus, the acceleration of a body at a distance (# r) from the centre of the earth is equal to (R-r).glR ; and since g/R is constant, /A/5 acceleration is proportional to (/ r), //ie displacement of the body from the centre O of the earth. The body, therefore, executes a S.H.M., and its time-period is given by_ _ " and Now, IT//? R 4000 m//e5 - 4000 x 1 760 x 3 //. ^ =* 32 ft. I sec 1 . substituting the values of /? and ^ in relation (111) above, we have 2 A/ V 32 5105 sec*. . 2ff A / 12^176x107 V 32 85-07 m/roife* Thus, the time-period of vibration of the body would be 85'07 minutes. 4. If a body executes a simple harmonic vibration in time TI, under one constraining force, and in time T 2 , under another, what will be its time-period under both forces together ? Let rrass of the body be m, and let its acceleration under the first force F Y be a l$ ,* 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 time-period 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 time-period 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 - -f-MT - ur ' or r = \ I ^ ^ 8 - r, r. A / ~ ' u ' V TV+r,'" l ' V v+r,' ' Thus, the time-period 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 cross-section 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.H-M. Hence, its time-period 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 U-tube of uniform cross-section 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 U-tube, 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 U-tube will thus oscillate with a 09S, time-period equal to 8. Show that the time-period 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 time-period 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 time-period 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 centre-point 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 V-9). ..(//) /. dividing (/) by (//), we have **- - *. Or, 25(a a -9)~16(a 2 -4). Or, 25a 3 -225 - 16a'-64. Or, 25a a ~16a* ~ 225-64. Or. 9a'=161, Or, ' - ^ And/. - <yi|i-4-23/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 23-2 \/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 - 4-23 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 time-pwiod 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 cross-section 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>. i|3.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. 1-974 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 time-period. 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. 14-41 //. 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. (/)3-142, (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 time-period 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 time-period 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 1-66 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, twenty-eight times its weight on earth, on the sun and about one-sixth 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 the-horizonUl axis provided by a knife-edge 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 side-windows 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 left-hand 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 divisio-s from the central to the end knife-edge. With the rider at the 100th division, the effect is equivalent to placing a centigram weight in the right-hand 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 knife-edge, 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 knife-edge 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 knife-edges in one plane. Let Fig. 94 represent a vertical section of the balance through the centre of the beam, passing through the three knife-edges 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 knife-edge C and Jf, its mass. Further, let S and (8-i-m) 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 knife-edge. 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 air-resistance 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 \ \J--r- < 6 S 6 7 8 - 10- ~s- (8- Fig, 95. -S ^ 5~-2 Or, (^-2) 2 = (10-5)(8-5) O r> 5 2 45+4 = 5 2 -185'+80. Or, 145 = 78, whence, S = 76/14. = 5-43. Thus, the pointer will ultimately come to rest at the 5*43r</ division on the scale. (//) Case of a Balance with the end knife-edges in a different plane from the central one The balance, diseased above, with all the three knife-edges 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 knife-edg^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 knife-edges 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.6-2S h.$ = 0(Md-2Sh) t n. . whence, sensitiveness, = -^ ~ ... (iv) Now, three possible cases arise : (/) When h = o, i.e., when the three knife-edges are co-planar. 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 knife-edges are higher than the central knife-edge. 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 knife-edges 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 knife-edges 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 knife-edge, 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 knife-edges 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 knife-edge 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 knife-edge 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 knife-edge, 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 -f-u\),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 Knife-Edges. 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 counter-balanced 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 -, .8-g \ 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 scale-pan 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 left-hand and right hand arms be a and b respectively. Then, since equilibrium is attained with the body in the left-hand pan and a weight W in tho right-hand pan, the moments on either side of the central knife-edge 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 right-hand pan and a weight W % in the left-hand 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 -r-2H' 1 W'-h2H' a W-r W l = 4w l w^2wJV n i-2w l lV l f Or, ^ 2 -h2^(>Vt 4- w 2 ) * Adding Ovj-Hv;) 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 knife-edge is a distance x vertically above the middle of the line joining the knife-edges 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 knife-edge, 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 right-hand pan and the lighter weight w,, in thr left-hand 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 mid-point of the line joining the knife-edges of the two scale pans). Let the central knife-edge 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 knife-edge 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 (x-f y) sin 9, and DF^OF-OD^OB' cos 9 OC sin = (a cos 0-x sin 0). So that, CPH-w> 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 0-f 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. extra-load 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 co-planar. And, since, in the question, every other factor is given except the depth of the e.g. of the beam below the central knife-edge 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 knife-edge 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 = -AfJ-lOOA .......... (//) dividing relation (/) by (//), we have 3 MJ-100/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 knife-edges are below the central knife-edge. 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 ~ Md-2i)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 1--00123 x - 300 300(1- '00008902) - 300--026706 - 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 knife-edges be co-planar. 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 knife-edges are in a plane and the centre of gravity ot the beam is 0*02 cm. below the centre knife-edge, 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 knif-edges carrying the pans 15 cms from the central knife-edge. What is the depth of the centre of gravity of the beam bslow this knife-edge, 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) kiufe-edge 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 point-mass), 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 time-psriod 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 time-period 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[(SO-8C)-(SO-SD)]. 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[(l-l 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 6-cos 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 time-ppriocUs 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 time-period 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 time-period (/), 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 suspension-thread 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 time-period //+ 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 2-75 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 time-period 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 time-period 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 time-period 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 time-period 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 time-period 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 time-period 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 time-period 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 time-period 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 time-period 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 (a-f fe) and (a b) from the e.g., for which the time- period is the same. ACCELERATION DUE TO GRAVITY 173 even less than one-thousandth 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 knife-edg^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 in-between the two knife-edges f The pendulum is first suspended from the knife- edge F,, and its time-period determined. It is then sus- T pended from the knife-edge F 2 , and its time-period deter- K X'-/ B mined again. If there be a divergence in the two values of the time-period, the heavier weigtit W l is moved up or down and a proper position found for it so that the time-period 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 knife-edge at the centre of oscillation of the other. The distance between the two knife-edges 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 time-periods about the two knife-edges, 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 knife-edges lying alongside, and on a level with, a standard steel scale, and the positions of the knife-edges 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 knife-edges 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 cross-wires then gives the distance between the two knife-edges. 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 time-periods of the pendulum Is obtained about either knife-edge, by adjusting W l and W % , as ex- plained above, the time-period 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 knife-edges, 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 time-periods about the two knife-edges takes place. The time-periods 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 time-periods 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 time-period 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 knife-edge 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 eye-piece 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 time-periods, 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 time-periods 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 time-periods, 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 time-period 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 time-periods 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&2l|L/ ~~~ 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 re-arranging (/) and (//), we have / 1 2 /g=4-7r 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 knife-edges, 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 knife-edge, 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 time-period 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 time-poiiod The observed time period may be corrected for this error by multiplying it with (1 ^.^g/IH), where a, and 2 are half-swings (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 0-cos 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 time-period of the pendulum, we have - O - o (cos Q-cos*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 time-period 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 <.- (//) Air-effects. There are three distinct ways in which the pre- sence of air affects the time-period 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 (ml-m'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(ml-m'h). This is obviously the equation of a simple harmonic motion, of time- period t t given by / : (ml-m'hg) which is clearly greater than/ 2w \/(* +/*)//*, the expression for the time- period in vacuo. The time-period 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 air-efFectf 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 time-period 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 time-period 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 P-M 2 tn^h f Neglecting second Ur> 4** - / + ml f 7 * ml ' L order terms ' assuming the time-period 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 time-periods 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 right-hand 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 time-period 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 viscosity-drag 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 time-period ' _ Now,*2rr/v ^ r , the time-psnod 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. (///) Non-rigidity of the Support. Due to the yielding of the support, the time-period 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 time-periods 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 time-period 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 co-penodicaily 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 time-period 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, ). 6-f /^. /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/P-F/ 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 timf-r}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 time-periods 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 l-rigths 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 correc-ion 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 time-periods to very near equality. The correction >y is thus bcsi eliminated as explained in 72, (page 187). Fig. 116. ('V) The knife-edges 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 (/ 1-r). Since, 'however, the instantaneous axis of rotation of the pendulum stiH passes through the bottom of the knife-edge (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 time-period is given by t whence. . = /4-r Or. If we find two lengths of the pndulum, say / t and /, on the two sides of its c.g , such that the time-periods ( 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 knife-edges 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, i-W [ (/i-M.) - (Ijzjl)^!-^) ] Or, have Again, putting J ~] ~ f - = T 1 , where T is the computed time-period, 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 knife-edges have the same radius of curvature. Since it is difficult to make the two knife-edges of exactly the same radius of curvature, the suggestion at once comes to the mind that the same knife-edge 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 air-corrections. This difficulty may be tided over by ananging two knife-edges 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 knife-edge with the other exactly in its true or original position. This difficulty too may be got over, however, by performing the experiment four-times, first taking the two observations for / x and / 8 with one position of the knife-edges and! then two similar observations with the knife-edges inter changed . Thus, if T! and TJ be the respective computed time-periods 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 knife-edges 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 knife-edge. 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 time-period 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 knife-edges 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 knife-edge, 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 knife-edges, the true time-period / 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 Kater-type pendulum but is svmmetrical in geometrical form^about its mid-point, (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 knife-edges E l and 2 inside the rings and carrying two bobs B l and B 2) one solid and the other hollow*. The time-period of the pendulum can be made nearly equal about either knife-edge 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 knife-edges) and TJ and T 2 , their computed times, then \/(T 1 2 ~T 2 2 ) gives the correct time-period 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 knife-edge, (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 time-periods of the two pendulums respectively. Thus, subtracting the second expression from the first, we have, straightaway, ** ' \ V-T 2 2 )=L i -L z . This removes at one stroke the errors due to yielding of tke support and curvature of the knife-edges, 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 time-period 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 time-period 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 time-period 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 distance-time 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 distance-axis. So that, sub- stitu ing tha co-ordinates 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(SQ-SP). Now, SQ = (R-r).cos and SP = (Rr).cos 0'. And .-. SQ-SP = (Rr).(cos 6-cos 0'). So that, work done by the force of gravity in moving the ball from A to B is equal to mg.(R-r) (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 (R-r).(cos B -cos 0') = J/o/ 2 . If co be the angular velocity of the ball about a horizontal axis (jy . _H_ Va,. .-. mg.(R-r).(cos 0-cos 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.(R-r)jin 6 = 7 . . ~ . ^ wS'r*(R-~ r ) sin 9 mg.r 2 .s-m w#.r 2 . p.- * I(R-rT ~~ /.(.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. Re-arranging the expression for t 2 , obtained above, we have 5f 2 = 28w 2 jR-287T 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 ribbon-type of machine, [Fig. 122 (a)], a strip or ribbon R is passed round the flat rim of a light and frictionless pulley, (running on ball-bearings), 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 time-per.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., time-periods 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, \Lv-iR- 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 Re-arranging 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' = (m-rn f ). Or, g(Aw/a - m//') == (m-m'), ^ , , ~ ,'/ m m' \* whence, g = (/H-TH ) ^ _- 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 reta-ded). 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(Af-Af' )/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 time-period 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 in-between 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 cross-section. 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~ . j-g = .. - 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 re-arranging 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 time-period 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 time-periods 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 re-arranging, 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 Non-Parallel Threads. Let the rod or cylinder AB, [Fig. 128 (#)], be suspended symmetrically by two equal but non-parallel 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 ro-arran.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 the-e 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 k-HOPKBTIBS fflATTKB (b) relation (v/ff) reduced to T = 27r.~r-A/ ,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 re-arranged, gives / = - -. n 2/ and relation (ix) 9 when re-arranged, 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 3-39 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 nsa-er to its centre than those near th^ equator ; (//) to the non-homogeneity of its comyosilio'ii tho density of its different layers b3in-z 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 one-tenth 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(l-vA/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)(9SO-61~-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_ '-l-j } 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 over-estimated 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' = (9SO-6l--0:55 cos 2x)(l-5/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 p-ge 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 knife-edges 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 half-second 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 knife-edges gsts automatically eliminated. What is done, in actual practice is tint three half-second 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 invar-steel^ ', (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 quartz-crystal 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 time-period is one second] ~"~ fit is an alloy of nickel and steel, whose co-efficient 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 place-to-place 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 invar-steel, 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 invar-steel, (with its negligible coefficient of ex- pansion), are further corrected for by a direct determination of the change in time- period with temperature. The time-period 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 time-periods 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 time-signals broadcast at frequent intervals. In the ultimate analysis, however, the results obtained will be restricted to the same order of accuracy to which the time-period of the pendulum and the other constants involved have been determined at the base station. The use of the time-signals 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 time-periods of the pendulums can then be compared with the equal time-intervals 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 direct-reading 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 gravity-meter, 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 time-period 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 b-idy 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 ribbon-spiral 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 eye-piece. 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 platinum-indium, 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 x-axis, (lying along the north-south 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 mid-point O of the beam, we have, from relation (1) above, Si-So = 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 time-period !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 two-fold, viz., (i) the main one being to determine the value of the force of gravity and its direction at various points of the sea-level 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 north-south 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 north-south 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 dome-like 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 well-trained 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 time-period 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 (Af-f- 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 time-period 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 1-412 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=*U-0)=*. 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 (l-o?s 2 0/288). Now, at the equator, ^ = 0, so that, cos <f> and .'. cos 2 $ 1. Hence ^ = #(1-1/288), -where gp is the value, of '#' at the equator. And, at the poles, ^ = 90, so that, cos $ and .*. cos 2 $ = 0. Hence gp = #(1-0) =-- 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^(1-1/288) = ^(1-1/288), (/) 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) = g-g #/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'1248-O 1116 - 0-0132//./WC*. .*. decrease in the weight of the train = mass x decrease in acceleration due to gravity. = 100x2240x0-0132 poundah - 100x2240x0-0132/32 Ibs. wt. - 100x2240x0-0132/32x2240 = 1 00 xO'Ol 32/32 ton wt. - 0'0412 ton wt. .'. apparent weight of the train = 100-0-0412 = 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 - - 40UOX-17605T Hence, "the resultant velocity of the train, say, v" =* 1536-88 = 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 - 0-1116-0-0993 = 0-0 123 //./sec 2 . And .*. increase in apparent weight of the train 100x2240x0-0123^^^/5 = 1 00 x 2240 xO'Ol 23/32 Ws. wt^ 100x2240x0-0123/32x2240 = lOOx '0123/32 ton wt. C '03 844 ton wt. Thus, the apparent weight of the train - 100+0-03844 - 100-03884 100'04 tons weight. EXERCISE VI 1. What is a simple pendulum ? Is it obtainable in actual practice 7 Deduce an expression for its time-period 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 parallel-knife-edges. Indicate the sources of error in an experimental determination of ^. (Bombay, 1940-41 ; 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 knife-edjes 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 time-period 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 time-period 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 half-way 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 time-periods are four.d to be 1 42 sees in each case. If the distance of the two axes be 10-8 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 knife-edges 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/l-414 "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 knife-edges are t and (f-f!T), where T is a smaJl quantity. The knife-edges 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 time-period 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 time-period 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 well-equipped 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 time-period, 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 = 27-3 days = 27-3x24x60x60 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 = 32-2/60* = -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 point-mass, 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 Q-e 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 lime-period (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 semi-major and semi-minor 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 6-669 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 the-n, 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 mountain-masses. 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 mountain-mass, (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 north-south 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 object-glass 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 mountain-mass The valo-e 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 5-0 gms.lc c. after a careful re-survey 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 time-periods 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 wndulum-bob 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 - , (Af-m'> -, t>that, *--Br.0 and g' - ="' G> hence, g' M-m' f_R_ V T~ ST' \R-itS ' 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' = tolP-GR-Wp. ...(//) 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 copper-mine, 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 mid-point of the cross-bar was a torsion head M which could also be worked from outside, from which was suspended a deal-rod 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 deal-rod 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& deal-rod 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 air-draughts, 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 time-period 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 quartz-fibre, 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 quartz-fibre 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 2-5 cms. long which acts as the torsion bar, is suspend- ed by means of a fine quartz-fibre, 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 quartz-fibres, two small gold* balls A and B, about 0'5 cm. in diameter and 2-65 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 mid-point of the mirror strip, and let / be its half-length (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 right-angled 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 suspension-fibre, 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 suspension-fibre*, 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 time-period. 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 original-positions and can- be easily read on a circular scale attached to it. Since a quartz-fibre 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 6-6576 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 a-h/ 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 gun-metal beam, provided with steel knife-edges 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 /"" turn-table, 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 turn-table 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 turn-table 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 turn-table 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 is-also 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 half-milli- 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 6-6984 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 2-44 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 0-25 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 constant-temperature 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 time-periods 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 time-period 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 $, d 1 d* Fig. 153. where = LOA'D Now, since And, therefore, LO'B'J - (90-a). LA'OP = LB'O'P = y, we have (0+y). F coi a = M.m.G . ,. . ^ , r -.sin (0-fy). Or, F cos (I) r$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 time-period 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 actiiT-Dfbh) 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 (5-5150'004)^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 5-52 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 to-day ! 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 in-between 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 other-workers, 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 time-periods 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 Boys-Cavendish 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 l-6xlO-* 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 spaoe-timo co-ordinatea 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 one-hundredth 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 well-equipped 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 f-dx. 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 ](d-a) d d ](d-a) </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 (a-f 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 foot-note, 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 (FO-CO). X(MO-x). [v FO - MO, and CO x, Hence . - -* . 2ir r I _ Or, o * 2n(l- cos a), where a is the semi-vertical 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 side-view 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 re-adjustmont 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<*) (1-2<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 wave-length, 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 world-wide 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 present-day 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 frame-work, 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 moving-coil 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 horsa-shoe 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 semi-cylindrical lens and allowed to fall on a sensitized (i.e., photographic) paper, wrapped round a rotating drum, worked by a clock-work arrange- ment and moving uniformly along its axis, with a peripheral speed of 3 cms. per minute. Time-signals 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, or-nearly 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., dead-beat), 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 horse-shoe 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 time-intervals S P. Thus, from the seismogram of an earthquake, obtained at an Observing Station, we can determine the time-interval 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 North-South 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 in-between 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^T-TO)*. 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 quake-proof 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 2-7 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 ground-region to be surveyed for the purpose, by detonating an explosive, like gun-cotton 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 'quake-proof 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 one-tenth 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.(6-4xl0 8 )x5-5x6*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(ll-5>*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 5-33 -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 2TTX1-49X10 1 ' distance covered by it in 1 sec., or, v Hence, centripetal force on the Earth = mv*/r. /2nxl-49xlO 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 6-66xlQ- " (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 (a-x) 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 - 0-009189 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 x|760_x 3 V 1 / 1 V 32 * V2406o6xl760xV " V 60 / ' Or, taking logarithms, we have 4 4582 - n(2-2218). 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 = 6-66 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 3x6-37xl0 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 = 6-66 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 6-37 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. 1-044 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 6-37 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. l-5xlO- e cmi. 16. Describe one of the most accurate methods of measuring the constant of gravita-tion. 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 cross-section of a steel bar which could just with- stand the gravitational pull between Sirius and the earth. (G = 6'67xlO~ 8 dyne-cm*. 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 non-homogeneous) 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 extension-axi 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 cross-section 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 clastic-after 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 moving-coil galvanometers etc. etc. As a natural consequence of the elastic after-effect 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 ninety-two 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 cross-section 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 cross-sectional 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 one-fiftieth 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 mono-atomic 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 co-efficient of rigidity of the material of the body, denoted by n. Thus, tangential stress = Fla, and shear strain = 6 = //L. And, therefore, Co-efficient 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 right-angled * F j g i 72 ~ triangles, (i.e., right-angled 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'-OA-AA'-^-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 F-F ^ 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 cross-section, 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 4-v. 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+arva-2vp-r..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(a-2p)-l == 3T(a-2(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(a-2p)' "* (/) 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, a-2p = 1/3AT. ...(/) And, from relation (77), a-f 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 cross-section (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 3ai-2a ~" 3l-*o L <C a whence, Y = 3^(l-2j),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"(l-l>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 wire-drawing 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 = Q-5 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 spirit-level 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 spirit-level 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 cross-section (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 F-shaped 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 cross-section 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 screw-point 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 metre-long 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 cross-section, 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 = (A-dA)(L+dL). [ v "S - x '\^ h = AL+A.dL-dA.L-dA.dL Or, V+dV = V+A dL-dA.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 cross-section (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 cross-section 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 co-axial, 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 = 2-nn. ~.-.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 co-axial, 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 time-period 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 side-pull 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 side-pull 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 time-perioi 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, /, - M-a^-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 4-7T 3 (tf-V) = - c .(mi-m^a*. Further, substituting the value of C, we have 4-7T 2 , % . 47 ^ , , ox ?r..a , % , 7r... 2 -, Or, U 2 f -'i 2 ) = n r * - faa-flM, 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 time-period (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 time-periods 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 cross-section, 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 cross-section. 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 "124-J 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 cross-section of the beam. For a rectangular cross-section, a = bxd, and fc 2 = d 2 /12. Hence, bending moment for a rectangular cross-section = Y.b.d*ll2R. For a circular section, a = 7rr a and A 2 = r 2 /4. I 9 = 0A-2 = 7rr 4 /4, /.., the same as the moment of inertia of a disc about a diameter. .*. bending moment for a circular cross-section = 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.(L-x) = 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 (L-X)W.(L-X).dX Then, dy = (Lx).dO = ~ ---- jTak*~~ * fFrom ( ' 7) above> JV(L-x)*.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 mid-point 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 = (L-x).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 cross-section 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 cross-section of a rectangular beam, bent so as to be concave downwards along its length is convex downwards across its length. Similarly, a circular cross-section 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 cross-section 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 clock-spring, 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(a-x).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(a-x).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 time-period of these vibrations. Sift We have seen above, ( 130, pages 311), how WL* , TT7 y = -s-rrv- , 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 time-period 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 cross-section, we have ak z = 7rr 4 /4, where r is the radius of the cross-section ; 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 knife-edge 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 mid-point W(l-X) W*urL pi 200 C of the beam and taking half-length 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 knife-edge, midway bet- ween them, as shown. The depi ession y of the mid-point, 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 = r-r-ji- 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 cross-section 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 cross-section. 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 mid-point 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 time-period 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 co-efficient 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*(L-x)*.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 mid-point 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 mid-point 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 cross-section 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 mid-point 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 cross-section 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 mid-point 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 end-points. It passes, instead, between the initial un- bent position AB of the column, and its mid-point 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 mid-point 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 one-fourth 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 one-fourth 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)(l-2(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 in-bet- 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 ) = <?K-w 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^ i-m 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 -r-E 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 re-distribution 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 two-fifth 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 slow-moving 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-/) = i-i + 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(V-dv] = PV-P.dvdp.V-dp.dv. Or, PV = PY-P.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, JT-yr. 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 non-conducting 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 288-8^. Therefore, K - 7/3(1-2(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. zaii-t-p/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.(L-x) = 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 cross-section, 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 cross-section, (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 cross-section 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 mid-point of the bar is given by y = jF/ 3 /48 Yf ff . [See page 3 1 6. Now, for a bar of rectangular cross-section, j g = b.cl s l\2. [See foot-note 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 cross-section 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 x-01-x Olx-01 Therefore, work done = ^ - ^ ./ = --. ---- 10 18 x-01 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(a-hl), 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 dyne-cm, 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(crH-l), see solved example 3, above, (page 332), where it is shown that N = Y/2(cr-H), whence, Y/N = 2(cr-h 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 dynes-cm., and angle of twist = 1 radian ^ we have, couple per unit twist = 145/1 *= 145 dynes-cm. 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(a-fl), we have 2(<j + l)=2'858. Or, (a-fl) = 2-858/2 - 1'429. And, .-. a = r429-l = '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, time-period of the bar, i.e., t - 335'7/50 = 6'714 sees. Now, time-period 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 . .'. 6-714 = 2*W ^ 826. ^ 1440 ^ 6301 ) . 826 x ^J- 3 - 826x120-5. 99540 ELASTICITY 33 1 squaring which, we have ^TUM A 2 9954 ^ ^ 4Tr*x99540 (6-714)' -4*'x-c-. Or, C- Now, C is also = .wr 4 /2/, where the symbols have their usual meanings. 77Xnx(-0501) 4 = 4Tt*x 99540 2x37-85 (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 300-301), 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 time-period 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 (M-f 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/ cross-section and y, the Young's modulus for its material. M+m 25 . M-hm-M 25-16 . 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 cross-section 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 cross-section 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 ==2-598x10-*. Since D = 1 mm. -= -1 cm-, we have dl'l - 2-598 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(l-P p) cms. Now, increase in length of the tube = -06 cm. lOO.P.oc = -06, whence, a = 06/100 P. And clearly, initial cross-section of the tube = *r* sq. cms., and, final cross-section 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 (P-jS) 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 100-I--04 cms. Since volume of the water column remains the same, we have Or, 100-4-100-04x2P=100. Or, 100 04 x2Pj8=100'04- 100-00 ='04, whence, ft = 2Px TdOW 04 100P -04x100 == - X a == 2Px 100-04 *06 2xlOO-04x-06 4 ______ 1 ~ 2 x 1 00 : 04 x '06 ~~ 50-02 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 cross-section* =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 X7rx-ll 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 cross-section 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 1-00 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. 4-5Tcxl0 4 er,s, 8. A bar of iron, 0'4 sq. /.- in cross-section 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 0-0000121C. (Institute of Civil and Electrical Engineering) Ans. 5-464 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. 7-96 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 cross-section 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 9-78 x IQ 11 dynes cm.-*). Ans. 4-01 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 cross-section, the cross-section 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. 19-99 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 2-K^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 7-25 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 -i-ive, 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 mid-point to the end of the wire. Calculate (1) Young's modulus, (2) Shear modulus, and (3) Bulk modulus of the wire. Ans Y = 20-81 x 10 11 dynes cmr* ; n = 8-268 X 10 11 dynes cm~ 2 and K = 14 35 = 10 11 dynes cm.- [Hint. (/) Twisting couple = mtQr*!2l = 5 x 981 x 20 dyne-cm., and 9 = 70 x TT/ 1 80 radians. (ii) K nYI(9n-3Y). [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 knife-edges .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 1-26 cms. and the distance between the knife-edges 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. 20-42 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 one-fourth 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 clear-cut distinction between a solid and a fluid, it is not quite so easy to distinguish between the two in many a border-line 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^t|i||, 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 inter-relation 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 liquid-prism 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 non-uniform 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 two-thirds 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.XJ-Xh = 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(XJ-h) = 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~-r-J- 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 gold-content 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 other-f 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 meta-centre (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 non-collinear 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 water-surface. 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 300-301 , 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 wedge-shaped portion of it to rise out of the water, on the left hand side. Let these wedge-shaped 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 wedeje-shaped 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 surface-plane 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=BM--BG' = 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- _r-jr-~ -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 cistern-type 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, p-log, P = K.g.H-K.g.h = K.g.(H-h). Or, log,(^-) ^K.g.(H-h). ...(v) Thus, (//-/( ) = ~ - ...(v) HYDROSTATICS 361 Or, substituting the value of K 9 i.e., p//>, in (v/), we have (H-h) .P p.g P- Thus, knowing p 9 P, p and g, the altitude (H-h) 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.fa-hJ, and log, (pM = K.g.fa-hJ. 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 = 1-5 ft. -*. pressure at centre of area = h.p.g. = I'5x62'5x32poundals. = 1-5X62-5 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 62-5 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. = 1-443x62-5 Ibs. wt. And /. thrust on the door = 1 '443x62 5 Ibs. X area of the door. = 1 -443 x 62-5 x i base x altitude = 1 "443 x 62 5 x i x 5 x 4-329 Ibs. wt. = 1-443 x 62-5 x 2-5 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. = 1-443x62 5 x2 5 X 4-329x2- J 64 /6s wt. And, if Fbe the magnitude of the force on the hinge, its moment about Ibs. wt. Clearly, therefore, Fx4 329=1-443x62 5 x2'5x 4 329x2 164. Or, F-=l 443x625x25x2-164-487-5 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 0-f/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 0-f-/z) respectively. And, therefore, the moment about the new liquid surface E'F' is also equal to ka*b>cos 6>p-g(% a cos -f h) + ab.h p-g. ($ 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 6-h/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 5r|4, 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. = 196-3 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 = 62-5 XTTX 5/4 = 245-5 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 5-196-3 = 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 ton-feet, 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- Jj-JE 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 = = 5-748 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 c-cs. 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 (0-j-7i)/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 = 1-571 gms. wt. cm. Now, with the whole of the cylinder immersed in the water, the weight of water displaced = 4n gm. wt. = 12-57 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., 12-57 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 surface-plane of the cylinder about its diameter. Sc that, /c 2 =/ 2 /4, k being the radius of gyration of the plane about the surface-line 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 /(1-p) _ r 2 ~2/p/(l-p) _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 V-shaped 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 H-S. Certificate) Ans. 0-0303 //. 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 surface-line 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 one-tenth 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 flying-machines, mz., (/) light er-than-air machines, or Air ships, (//) heavier-than*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 cigar-shaped 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 non-inflamma- 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 mid-line 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 heavier-than-air 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 air-flow 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 air-proof. 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 1-2 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 Air-Screw. 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 air-screw. 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 straight-line 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 straight-line 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 right-hand 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 mid-point, (Big. 236), their up and down movement being con trolled by the side-ways pull on the stick-f, 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 shock-absorbing 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 wide-track 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 spar-like 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 air-fbw, 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 three-wheeled 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 take-offs 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 wing-like structure the 'slat 9 arranged over its leading edge*. Without the slat, if the airplane were to stall, the air-flow 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 'lighter-than-air 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 'heavier-than-air' 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 air-ship 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, jet-propulsion 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 jet-propelled 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 jet-force. 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 cross-section 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 ducted-fan 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 power-unit 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 cross-section 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 mass-flow 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 (u-V)* = \F(u-V) [-.- 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 power-unit) = 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(u-V)} 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 cross-section of the jet. As indicated earlier, the cross-section of the jet in a jet-propelled 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 cross-section 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 cross-section 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 jet-propulsion a better mode of propulsion ? The answer is manifold : (i) Initially, when jet-propulsion 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 fuel-consumption, the turbo-jet 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 over-all 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 fuel-con sumption rate, necessarily implies that if the flight speed bo high, the turbo-jet unit will also be about as economi- cal as the ordinary propeller-engine 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 turbo-jet 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 power-output upto this speed. And this is the point where the jet-unit 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 air-resistance or 'drag* "That is why, in the case of a turbo-jet 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 jet-unit 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 non-rotating 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 over-all 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, oil-cookers 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 fire-work rockets, rising pretty high up in the air, to the amusement of on-lookers, are a common enough sight everywhere and they are obviously jet-propelled 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 turbo-jet plane. In fact, it is a turbo-jet 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 cross-section and its velocity higher. Is this difference in degree, then, the only factor that distinguishes it from a F . 2 , 8 jet-plane? 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 jet-propulsion _____ ^ ^ ^ ~~ 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 rocket-propulsion 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 self-contained 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 rocket-plane, 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 self-contained, 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 mass-flow 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 power-unit, 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 0-2 Ib. at an altitude of 50,000 ft. a turbo-jet unit weighs about 0*3 Ib., a rocket-unit, weighing only 0-1 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 rocket-unit, 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 short-duration flights or remotely controlled missiles, and the latter Is the one commonly used for rocket-propelled air-crafts. 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 fuel-consumption, 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 mass-flow through the nozzle and u, the exhaust velocity of the gases. So that, / = mujmg = ujg. And, therefore, the higher the jet-velocity, the higher the specific im- pulse and the smaller the fuel-consumption for a given thrust. Besides fuel-consumption 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 one-tenth 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 fuel-supply 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 heat-resisting 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 over-all shape being more or less like that of a cigar. 175. The Multi-stage 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 fuel-consumption 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 fuel-stock 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 pay-load 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 over-all 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 multi-stage 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 multi-stage 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 multi-stage rocket, with its fortifications and fasten- ings etc. (iV) Rocket-length. This obviously means the height of the rocket or that of the individual rockets of the multi-stage 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 \\.% fuel-consumption, 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 man-made 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 jet-propulsion (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 inter-planetary 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 83-6 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 pre-determinod and well-calculated orbit by a multi-stage 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 al-o 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 breath-taking ones in thoir wake, with the Russians putting the first cosmonaut of the immortalised name, Major Yuri Gagarin, into space in a much larger space-vehicle or space-ship 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 one-eighth 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 11-2 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 multi-stage 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 air-resistance 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 air-craft, 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 present-day long-range 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 in-between 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 auto-pilot (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 nose-part 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 air-resistance 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 radio-waves. 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 o