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ELEMENTARY DYNAMICS

CAMBRIDGE UNIVERSITY PRESS

C. F. CLAY, Manager

LONDON : FETTER LANE, E.C. 4

:b^EWYORK : THE M ACMILLAN CO.
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CALCUTTA I MACMILLAN AND CO., Ltd.
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ELEMENTARY DYNAMICS

A TEXT-BOOK FOR ENGINEERS

jrwr LANDON, M.A.

FELLOW OF CLARE COLLEGE, AND UNIVERSITY LECTURER
IN MECHANICAL ENGINEERING, CAMBRIDGE

CAMBRIDGE

AT THE UNIVERSITY PRESS

1920

PREFACE

THOSE who have had experience in teaching elementary
dynamics to students of Engineering will agree that the
majority find considerable difficulty in grasping the fundamental
principles on which the subject is based. There are new physical
quantities to be understood, and new principles to be accepted
which can only be expressed in terms of these quantities. Un-
fortunately, in many cases, this subject has to be introduced at a
stage of development in mathematics at which the student expects
some proof of what he is told to believe. He is not so prepared
to accept things without proof as he was when first told that
2x3 = 6. Even at that early stage, probably an attempt was
made to prove to him that 2x3 = 6, yet, in the end, he merely
accepted the fact and memorised it.

The author believes that the difficulty experienced is partly,
though by no means entirely, due to the way the subject is often
presented. In the student's mind it is associated with branches
of pure mathematics such as trigonometry, analytical geometry,
or infinitesimal calculus, and he is inclined to think that salvation
lies in memorising a number of formulae which are to be used in
solving problems, instead of looking upon dynamics as a funda-
mental branch of physical science, in which mathematics is of
secondary importance and the physical ideas of primary import-
ance. Quite commonly in elementary text-books, the second law
of motion is at first summed up in the form, force = mass x ac-
celeration, and the student is then given a number of examples
to work out, most of which consist in substituting numbers in a
formula. This very successfully disguises the true meaning of
momentum, and the extraordinary generality of the second law
of motion. The same applies to problems dealing with motion.
The student is generally presented with certain formulae for
motion involving a constant acceleration. These formulae, are of

VI ELEMENTARY DYNAMICS

very little, if any, use to him later on, and have merely enabled
him to get answers to certain problems without thinking.

At first sight the remedy would appear to lie in teaching
dynamics experimentally, but the author's experience is that this
is not so for the majority of students. The phenomena of every-
day life provide innumerable qualitative experiments, and to
most students quantitative laboratory experiments in dynamics
are neither interesting nor convincing.

In the following pages an attempt has been made to present the
principles of elementary dynamics, and to explain the meaning
of the phj^sical quantities involved, partly by definition and
description, but mainly by worked examples in vyhich formulae
have been avoided as far as possible. By continually having to
think of the principle and the physical quantities involved, the
student gradually acquires the true meaning of them, and they
become real to him.

It will be observed that the first of Newton's Laws of Motion
is expressed in a somewhat different form from that in which it
is usually given, and the laws are called the Laws of Momentum.

In working examples the absolute unit of force has generally
been adopted, and, where applicable, the answers have been
reduced to units of weight. It matters little, in the author's
opinion, whether absolute or gravitation units are used, so long
as mass is not defined as weight divided by the acceleration due
to gravity. To say that the engineer's unit of mass is 32*2 lbs.
is almost to suggest that he is rather lacking in intelligence,
and cannot be expected to understand the difference between
equality and proportionality. If weight is introduced in the early
conception of mass, the student's conception of mass is extremely
vague, and his conception of momentum as a physical quantity is
even more vague or erroneous. A student who cannot understand
the difference in the two units of force, and who has merely to rely
on formulae expressed in one particular set of units, is not likely
to get any knowledge of dynamics which will be of real use to him.

PREFACE VU

A number of graphical examples have been worked out in the
text, and a number are included in the examples to be worked
by the student. These frequently require more time than analytical
examples, but they are more useful and instructive. This is par-
ticularly the case with the engineer, who is so frequently faced
with problems which can only be solved graphically.

Probably the majority of students will be learning differential
and integral calculus at the same time as dynamics, and they
should be encouraged to use the calculus in working examples,
although all the examples given can be worked without its use.

The examples at the ends of the chapters are arranged more
or less to follow the text, and students should work them as they
proceed with the reading, and not wait until they have completed
the chapter. The miscellaneous examples, at the end of the book,
are intended for revision, and for this reason they are not arranged
either in order of difficulty or in the order of the chapters dealing
with the principles involved. The answers have mostly been
obtained by means of a slide rule. It is hoped that the errors in
them are not numerous.

Though primarily written for engineering students the book
may be useful to some others. The course covered is approxi-
mately that required for the Qualifying Examination which
Cambridge students have to pass before their second year, if
they wish to take an honours degree in Engineering.

The author wishes to thank Mr J. B. Peace, Fellow of Emmanuel
College, for valuable suggestions and for having contributed a
large number of examples, also Mr W. de L. Winter of Trinity
College for very kindly reading the proofs and for useful criticism
and suggestions.

J. W. LANDON.

Cambridge,

August 1920.

CONTENTS

PAGE

PREFACE V

CHAP.

I. INTRODUCTORY 1

II. MOTION 20

III. LINEAR MOMENTUM 57

IV. ANGULAR MOMENTUM 88

V. CENTRIFUGAL FORCE— CENTRE OF MASS. 110

VI. WORK— POWER^ENERGY . . . .128

VIL UNITS AND DIMENSIONS .... 162

VIIL SIMPLE HARMONIC MOTION . . . . 166

IX. MISCELLANEOUS 187

MISCELLANEOUS EXAMPLES . . « . . .219

ANSWERS TO EXAMPLES 241

INDEX 245

CHAPTER I

Introductory

The subject commonly called Dynamics is a part of the much
bigger subject called Mechanics. In its broadest aspect, me-
chanics deals with bodies or parts of bodies which are acted upon
by certain forces, and analyses and examines the effect of these
forces in producing motion, or in maintaining a state of rest.
For the present purpose mechanics may conveniently be divided
into three branches as follows :

Kinematics

Mechanics-

Kinetics

.Statics

Kinematics is the branch of the subject which deals with the
motions of bodies. The bodies may be of any size or shape, and
at times it may be convenient to consider them indefinitely small,
i.e. as points.

Kinetics deals with the causes of the motions of bodies, and
attempts to find a definite relationship between these causes
producing or maintaining motion, and the motions themselves.

Statics treats of bodies which are at rest and examines how
this state may be maintained.

The subject of dynamics, as will be seen, does not in itself form
one of the main branches of mechanics, but it may be said gene-
rally to include kinetics and a part of kinematics. It deals with
motions in so far as they are required in the examination of the
forces producing them, and it also deals with the general con-
sideration of the mechanical energy possessed by bodies, either in
virtue of their position or of their motion.

Now there are certain fundamental conceptions, and there are
also certain principles or laws, which form the basis of the whole
subject. Neither the conceptions nor the laws are numerous.
L. B. D. 1

2 ELEMENTAEY. DYNAMICS

Fundamental Conceptions

These consist of the ideas of space, mass and time. It would
be difficult, if not impossible, to define accurately either of these,
or to explain exactly how the human mind understands their
meaning. The conceptions are acquired in childhood or are
born in us. Space, mass, and time are the three fundamental
physical quantities, and all the other physical quantities we shall
deal with may be defined in terms of them. It is easy to realise
that before much practical use can be made of these, we must
decide on some unit of measurement of them. We shall here only
concern ourselves with two systems of units :

(1) Foot, Pound, Second system, (f.p.s.)

(2) Centimetre, Gram, Second system, (c.g.s.)

Space. We may say that this is what possesses length, breadth
and thickness. Each of these is a length and hence space is most
conveniently measured in units of length.

In the F.p.s. system the unit of length is the foot.

In the C.G.S. system the unit of length is the centimetre.
1 foot = 30 '5 centimetres.

Mass. This may be defined as the quantity of matter or the
quantity of stuflT in a body. Although it is difficult to define pre-
cisely what is meant by mass, we have no difficulty in realising
what we understand by the term. If, for example, we ask for
half a pound of tobacco, we are not probably interested in the
amount of space it occupies, or even its weight, so long as we get
the correct quantity of stuflf. It may be compressed in the form
of a cake and occupy little volume, or it may be loose and occupy
a considerable volume. We shall see later how we can compare
masses. Probably our earliest conception of mass in childhood is
when we try to throw things about. We find that some are more
difficult to move or throw than others, and we soon discover
that this does not depend upon the size. We consider that

FUNDAMENTAL CONCEPTIONS 3

those which are more difficult to throw or move have more stuff
in them.

In the P.p.s. system the unit of mass is the pound (1 lb.). The
standard pound is a particular lump of platinum deposited in the
Exchequer Office.

In the C.G.S. system, the unit of mass is the gram. Originally
this was intended to be the mass of a cubic centimetre of pure
water at 4 degrees centigrade, but the standard is now one of
platinum like that of the pound.

1 pound = 453-6 grams.
In dealing with mass we might conveniently define what is
meant by density. The density of a substance is the mass per
unit volume.

This should not be confused with specific gravity. The
specific gravity of a substance is the ratio of the mass of a given
volume of the substance to the mass of an equal volume of water.
For example:

The density of water = 62*3 lbs. per cubic foot.
The specific gravity of steel = 7-78.
.-. The density of steel =- 7-78 x 62*3

= 485 lbs. per cubic foot.
Time. The unit of time which is adopted in both systems of
units is the second.

Vectors

The various physical quantities which we have to deal with
can be divided into two classes :

(1) Scalar quantities.

(2) Vector quantities.

A scalar quantity is one which possesses magnitude only, for
example, an interval of time, as 3 seconds. There is here no idea

1—2

4 ELEMENTARY DYNAMICS

of direction. Again, 2 lbs. of bread. This is merely a definite
quantity of stuff and has no connection with direction.

In dealing with scalar quantities we add and subtract by the
ordinary rules of arithmetic. For example, in making a certain
article in a workshop, work may have to be done on it in three
different machines, and the lengths of time in these may be
15 minutes, 40 minutes, and 10 minutes. The total time for
machining is then, (15 + 40 + 10) = 65 minutes.

A vector quantity is one which possesses both magnitude and
direction^ for example, the weight of a body. We know that this
acts vertically downwards, and that it is generally easier to push
an object along than to raise it up.

Suppose we are dealing with the displacement of a body from
a given position. Here we want to know, not only how far the
body is from its original position, but also in what direction this
distance is. Let us take an actual example.

A ship travels 2 miles due east^ and then travels 1^ miles in a
direction north-east. What is the final displacement of the ship ?

The easiest wa)^ to solve this is to draw a diagram representing
the motion of the ship. Draw
OA due east to represent the
2 miles displacement. Make it
1 inch long. Now draw AB in
a direction north-east, and make
it j inch long to represent IJ
miles displacement.

The final position of the ship is obviously represented by the
point B, and the displacement from the starting-point is given by
OB, both in magnitude and direction. By measuring we find OB
is 1-62 inches, and we find also that the angle BOA is 19 degrees
9 minutes. We can say, therefore, that the ship is at a distance
from the starting-point of 3*24 miles, and that its displacement is
19-15 degrees north of east.

Here we are dealing with vector quantities, since in order to

VECTORS O

define them we have to state both magnitude and direction.
It is obvious that a vector quantity can conveniently be repre-
sented by a straight line, since the line may be drawn of a length
to represent the magnitude of the quantity, and also be drawn in
a definite direction to represent the direction of the quantity. In
fact, this is why such quantities are called vector quantities. A
vector is a straight line of definite length, drawn in a definite
direction. In order to indicate the starting-point of the vector,
or to indicate what is called the sense, it is often convenient to
put an arrow-head on the vector as is shewn in fig. 1.

Addition and Subtraction of Vectors

Suppose we have two vectors as shewn in fig. 2. Let us call
them X and y. If we want to find the vector sum of x and y,
we take a line OA equal and parallel to x, and then a line AB
equal and parallel to y. The sum of the two vectors is given by
OB. We write this :

OB = OA + AB,

the line over the top representing the fact that the addition is
vector addition. The order of the letters gives the sense.

Fig. 2.

If we wish to find the vector difference of x and y we take
the line OA as before, and then draw the line AB of the same

ELEMENTARY DYNAMICS

length as y^ in the same direction, but in the opposite sense as
shewn in fig. 3.

We have OB' = OA + AB' = OA-AB = cc-y.

Now in many cases it is convenient to draw the figure accu-
rately to scale, and to ^ measure the
vector sum or difi'erence. Sometimes,
however, it is more convenient merely
to sketch the figure and to calculate the
true length and direction of OB or OB'.
We shall generally know ^, the angle
between the vector quantities we are
dealing with.

Draw OM perpendicular to AB as
, then
0B2 = 0M2+ MB^

==0M2 + (AB-MA)2
= OM^ + AB^ - AB . 2MA + MA^
= OA^ + AB^ - 2AB . OA cos Q,
z^ = 01? + y^ — 2xy cos 6.
OB'2 = OA^ + AB'2 + 2AB' . OA cos 0,
or z"^ = ar^ + 2/^ + 2xy cos 6.

shewn in fig. 4

Fig. 3.

Similarly

Fig. 4.

The vector OB is called the resultant of OA and AB.

TJie resultant of a number of quantities represents the single

VECTORS

the two
esultant

quantity which is exactly equivalent in every way to the number
of quantities.

For example in fig. 1, OB is the resultant of OA and AB, and
the ship would be in exactly the same position if it had moved
3*24 miles in a direction inclined at an angle 19 '15° N. of E., as
it is by making the two different movements.

Sometimes we shall find it convenient to think of a single
quantity as made up of two quantities. In this case
quantities are called the components of the single
quantity.

For example, consider the case of a body being pulled in a
certain direction by a pull of magni-
tude R (see fig. 5). We might produce
exactly the same effect on the body
by applying two pulls P and Q. say.
These are then called components of
the pull R.

If we fix the directions of P and
Q, we can at once obtain their mag-
nitudes, by remembering that the
vector sum of P and Q. must be
equal to R.

Let OB represent R. From B draw BA parallel to the direction
of Ql to intersect P.

Then OB == OA + AB,

.'. OA represents P, and AB represents Ql.
P _ OA _ sin <^
R ~ ~ ~

Now,

sin (d + <!>)'
sin <f>

and

sin (O + cf))'
AB sin 0

sin (0 + (f>)'
sin^

sin {0 + (f>)'

8 ELEMENTARY DYNAMICS

Generally we employ rectangular components, i.e. components
which are at right angles to one another
as shewn in fig. 6.

In this case we have

Component AB = OB sin 0,
„ OA = OBcos^.

There is, however, more advantage in
choosing rectangular components than

the mere simplification of resolution, if the angle OAB is a right
angle, then the components OA and AB represent the whole of the
quantity OB which is effective in the two directions OA and AB
respectively. On the other hand, if the angle OAB is not a right
angle, as in fig. 5, then OA and AB are not the whole component
of OB in these directions, since component AB has an effect in the
direction OA and component OA has an effect in the direction AB.
The whole component, or resolved part, of OB in direction OA is
given by OM in fig. 5, where BM is at right angles to OA.

OM = OA + ABcosBAM
= 0A + ABcos(^ + <^)
= OA + the rectangular component of AB
in the direction OA.
In the case of rectangular components, neither component has
any effect in the direction of the other. We shall better realise
the importance of this later on when we have dealt with more
concrete cases, but we will take one illustration here. Suppose a
motor-car is travelling due east, and there is a north-east wind
blowing at 20 miles per hour. The north-east wind is equivalent
to two winds, one blowing from the north at 20 x sin 45°, i.e.

20 20

—7^ miles per hour, and one blowing from the east also at —r^

miles per hour. Now the northerly component will have no ap-
preciable effect on the motion of the car, whereas the easterly
component will produce a direct resistance to the car's motion.

VECTORS 9

It may be noted that whereas there is only a single resultant
of two quantities, there are an infinite number of pairs [of com-
ponents which will give the same resultant.

This is shewn in fig. 7.

The components (OAj, A^B), (OA2, AgB), (OA3, A3B) all have the
same resultant OB.

Fig. 7.

Average Values

When we are dealing with two quantities, say y and a?, which
are related to, or depend upon, one another, we shall at times
use the expression, '•Hhe average of one quantity, y say, with
respect to the other x" Let us make quite sure we understand
exactly what we mean.

If we were talking about cricket there would be no need to
explain what was meant by the average of a batsman during the
season. This is simply the total number of runs made divided by
the total number of innings played. Other average values are
obtained in a similar way.

Suppose for example we travel by motor-car between two

places distant 150 miles and we take 5J hours to do the journey,

150
we say the average speed was — — , i.e. 27 3 miles per hour.

ELEMENTARY DYNAMICS

There again we simply take the total value of the one quantity
and divide it by the total value of the other quantity.

We must, however, be careful to state what two quantities we
are considering. For example, we may state the average cost of
an army to the country during a war as so many pounds per
day, or we may state the average cost per soldier. The values
will be entirely different. In each case- we find the total change
in one quantity and divide by the total change in the other
quantity. We shall have many illustrations of this as we proceed.

Rate of Change

Another idea we have to get hold of clearly is the rate of change
of one qy^ntity with respect to another.

Suppose for example we have two quantities, denoted by y and
X, and that y varies with x in some
particular manner. We may ex-
press this by giving a table of
simultaneous values of the two
quantities, as is done in a table of
logarithms. Here the two quanti-
ties are the numbers x and their
logarithms y.

We may also express the vari-
ation by representing simultaneous

Fig. 8.

values of the two quantities on a curve as in fig. 8.

For any point A on the curve, ON represents the value of x and
NA or OM represents the corresponding value of y.

Frequently we want to know how rapidly one quantity, y say,
will change as the other quantity x changes. For example, we
may have a motor the speed of which is changing, and we want
to know what is the actual speed at different instants of time.
This is given us by a speedometer. If a speedometer is not avail-

RATE OF CHANGE

able, we may get the speed approximately by noting the time
taken to travel from one mile-post to the next. This will of
course only give us the average speed for the number of minutes
required to travel the mile, since the speed of the car may be
increasing or decreasing during this time. We should obviously
get more nearly what we wanted if we had posts at short in-
tervals of distance, and an accurate stop-watch to measure the
time taken between two consecutive posts. ^ If we have a cyclo-
meter, we may use this to measure our distances and thereby get
still more accurate values of the speed. Now suppose we take
the times for different distances from our starting-point, and
express the result by a curve as in fig. 9 where y represents the
distance from the starting-point and x represents the time taken.

Time C D

Fig. 9.

To find the speed of the car after a time given by OC we draw
CP perpendicular to Ox, then PC is the distance from the starting-
point.

Now find the distance from the start when the time has
increased by a small amount represented by CD. QD drawn
perpendicular to Ox will give the required distance. Draw PK
perpendicular to QD. Then the extra distance moved in the
small time is given by QD - PC = QK.

Let us represent the small change of distance by hs and the
small change in time by 8^.

12 ELEMENTARY DYNAMICS

Then the average speed for the small interval of time

OK
~ PK

= tanQPK

= tan 0.

Now we shall get nearer to the true value of the speed at P,
the smaller we make PK; we shall only get the true value when
we take the interval of time indefinitely small. If we do
this we can no longer measure QK and PK, but we can still
measure tan ^, because the line joining PQ becomes' the tangent
to the curve at P, when PK becomes indefinitely small. This is

shewn in fig.

10.

Speed at P = tan 0, i.e. the value of ^ when St is made inde-
ed

finitely small
We write
change of distance with respect to time

We write this -j- = tan 6, where -r- stands for the rate of
at at

RATE OF CHANGE

We have taken a special case, but of course the same holds
generally for any quantities x and y and we ma^ write

tan Q.

When Q is greater than one right angle, and less than two

right angles, tan 6 is negative. This means that the rate of change

is negative, i.e. y is decreasing as x increases, see point T (fig. 10).

Now in certain cases, instead of accurately drawing the curve

to scale we can express the relation between y and x analytically.

In such cases we can also determine the value of -

dx'

J. the

rate of change. This is done for us in the differential calculus.

Fig. 11.

We will find the value of the rate of change of y with respect
to x for two simple relationships. This will save us a good deal
of trouble later on.

(1) Suppose the relation is given by

y = a sin hx.

This is shewn plotted in fig. 11.

dii
Now we want to find the value of -^ , i.e. the value of tan 0 at

any point.

ELEMENTARY DYNAMICS

We may obtain it thus: Let the radius of the circle in fig. 12
be made equal to a and let the
angle AOB be equal to hx. Let,
also, the angle BOC be the small
angle representing a small incre-
ment of a?, namely hx, i.e.

AOB = hxj

AOC = h(x+ hx),

.'. BOC = 6. 8a;.

The value of y corresponding to
x = asinhx = BN.

The value of y corresponding to " ^ig- ^2.

(x + hx) = asmh{x-¥hx) = CM.

The increase of y for increase of x equal to hx is OK,
hy _CK
' ' hx hx'

But if COB is very small then the chord CB will be very nearly
at right angles to OB, i.e.

A A

CBK = complement of KBO

= complement of BOA.
Also CB = OB X angle COB

= a X b . hx,

and CK = CB sin CBK

= CB cos BOA

= a.b .hx. cos bx.
, dy ab .hx. cos bx
' ' dx hx

- ab cos bx.

Again, suppose that

y = a cos bx.

RATE OF CHANGE 15

Using fig. 12, we have,
Value of y corresponding to aj -a cos 6aj = ON,

„ „ „ (a3 + Sec) ==acos6 (£c + 8a?) = OM.

Increase of y for increase of x equal to Saj is - MN (since y
really decreases),

, hy _ MN
' ' hx Sx '

But MN=:KB = CBcosKBC

a.b.Sx.sin BOA,

- — ab sin bx,
ax

We have then, if

y = a sin bx, -f = ab cos bx.
^ ' dx '

and if y - a cos bx, —- — — ab sin bx.

^ ' dx

Area Curve

In some cases we are given the rate of change of one quantity
with respect to another, and we want to find the total efiect of
this rate of change. For example, we may be given the speed of
a body at different instants of time, and want to find how far the
body has travelled in different intervals of time. It will be shewn
in the next chapter that the distance is given by the area under
the curve representing the speed and time. We will now examine
a graphical construction which enables the area curve to be
drawn.

Suppose the given curve is OPQ and we want to find another
curve Opq%\xc\v that the ordinate at' any point, say qlA, represents
the area between the curve OQ and the axis of x. Take any
portion of the curve OP and let KL be the mid -ordinate. Draw
KK' parallel to Ox and join K' to any point B on Ox produced.
Draw Op parallel to BK', cutting FN at ;?. Then /)N is propor-
tional to the area OPN.

ELEMENTARY DYNAMICS

l.e.

The triangles OpN and BOK' are similar
pN OK' _ KL
" ON ~ OB ~ OB '

jt)N =^^ X KL. ON

OB
_1_
OB

(area OPN),

i.e. jo is a point on the area curve.

For the area under PQ take the mid-ordinate RS. Draw RR'
parallel to Ox and join BR'. Draw pq parallel to BR', and draw
pv parallel to Ox.

O L N S M

Fig. 13.
The triangles pvq and BOR' are similar,
vq _ OR' _ RS
' ' pv~ OB OB

_ J^
~0B
Hence Mq=pN + vq

X (RS X pv)
(area NPGtM).

(area OPQM),

i.e. 5' is a point on the area curve.

EXAMPLES. CHAPTER I 17

Similarly we may find as many points as we like.
Scales. The scale of the area curve depends upon the distance
OB.

Let OB = A inches.

Take 1 square inch of area under the original curve. This, in

the area curve, is represented by an ordinate of length y inches.

.*. the scale for ordinates of the area curve is,

1 inch = h sq. ins.
Example. A speed-time curve is drawn to the following scales^
1 inch = 5 seconds, for the x axis,
and, 1 inch = 4 feet per second, for the y axis.

It is required to find the value of h in inches, so that the scale
for the ordinates of the area curve is

1 inch — bOfeet.
1 sq. in. under the speed-time curve = 4x5 feet.
.*. for the ordinates of the area curve we have,
A X 4 X 5 = 50,
or, h — 2^ inches.

Examples. Chapter I

X. A sailing boat travels 400 yards in a direction north-east and then
500 yards in a direction 60° west of north. How far has it travelled from the
starting-point and in what direction?

If the boat now returns to its starting-point by moving, first due south and
then due east, how far will it travel in each of these directions?

2. Three vectors are of magnitude 3, 2, and 1, and their directions are
parallel to the sides AB, BC, CA of an equilateral triangle, taken in order.
Shew that the resultant vector is perpendicular to the direction of BC, and
find its magnitude.

Find also the values of two vectors, one acting in direction AB and the
other at right angles to AB, which have the same resultant as the three given
vectors.

L. E. D. 2

ELEMENTARY DYNAMICS

3. Three wires radiate from a telegraph pole, in a horizontal plane. The
first runs east, the second north-east and the third north-west. The tensions
in the wires are, respectively, 200 lbs., 150 lbs, and 300 lbs. Calculate the
magnitude and direction of the resultant tension. The direction of the
resultant is to be specified by the angle ifc makes with the first wire.

Verify your result by a graphical construction.

4. A vector (r, d) is a straight line of length r which makes an angle 6
with some reference line, the angle being measured in a counter-clockwise
direction from the reference line.

If A, B, C represent the vectors (3, 40°), (5, 115°), and (4-5, 260°) respec-
tively, find graphically, (a) A -f B -f C, (b) A - B, (c) B - C - A.

Check the results analytically by resolving the separate vectors in two
directions, along and perpendicular to the reference line. •

5. The depth of a trench, measured from the surface of the ground, at
different distances along the bottom, is given in the table. Plot a longitudinal
section of the trench and estimate the average depth.

Depth in feet . . .

3-0

6-2
5

8-1
24

6-7
40

5-1

6-0
60

5-6
70

6-0

5-2

4-3

100

Distance in feet

6. A body is moved from rest by a pull P which changes with the distance
(s) moved and the time {t) taken, according to the equations

irt

P = 80-5s, and P = 80cos

the value of P being in pounds when s is measured in feet and t is measured
in seconds. Find, graphically, the space-&\era,ge of the pull, and the time-
average of the pull, for the first 8 feet of motion.

7. The' total number of letters collected from a certain district in one
year was 854,200. What was the average number of letters collected per day?

If there are 42 post boxes in the district what was the average number of
letters posted per week in each box?

The population in a certain district at intervals of 5 years was as shewn

below.

EXAMPLES. CHAPTER I

Plot the figures, and from the graph estimate the population in 1908 and
1917, Find also the rate of increase per year for the same years.

Population in ]
thousands j"*

20-8

45-7

71-2

79-3

Year

1900

1906

1910

1915

1920

9. Plot the curve given by ^ = 2 8in-^, between the values a; =0 and

x=S. Measure the rate of change of y with respect to a; for a; = 1 and a;=2'5.
Check your results analytically.

10. Draw the circle x^ + y'^ = 4:, and from it find the value of -p, (1) for
a; = l-5, (2) fora:=-l-5.

11. Draw a semicircle with a diameter of 4 inches. Taking one end of the
diameter as the origin, and the axis of x along the diameter, draw the area
curve.

What is the area under the semi-circular curve from x=0, to a; = 1-5?

12. Plot the curve y = l'5 sin — , between the values x = 0 and a; = 4, and

draw the area curve. When the ordinates of the original curve are negative
the area is to be considered negative.

13. Plot the curve y= -j, between the values x==0 and a; = 4, and draw the

area curve.

2—2

CHAPTER II

MOTION

Speed and velocity

Speed. The meaning of this is quite generally understood. It
is merely the rate of travel qf_a_bgd:y. For example, if we state
that a train is travelling at 25 miles an hour, all we mean is, that
if the train continues to run at the same speed it will pass over
25 miles of line in one hour. Here we are not concerned with the
direction of travel. The direction may be changing continually,
or it may be either in a straight line or along any curved path.

Velocity. The velocity of a body is the rate of change oj
'^osUwnBiidi is always measured ina^^traight line. It is essentially
a vector quantity.

Consider a point on the flywheel of an engine. The flywheel
may be rotating uniformly at 240 revolutions per minute, and we
may say, therefore, that a point on the rim has a constant speed,
but we cannot say that it has a constant velocity, in fact, the
velocity is continually changing, since its direction is changing,
being always tangential to the wheel. This is a very important
distinction, as will be seen later.

Velocity may be either uniform or variable. If a body is
moving with a uniform velocity then it will pass over equal dis-
tances in equal intervals of time, no matter how short the intervals
may be, and the distances will all be in the same direction. In the
case of a varying velocity, the distances passed over in equal in-
tervals of time will be different, or it may be that the directions
will be different.

SPEED AND VELOCITY

- 21

Unit of Velocity. In the p. p. s. system this is 1 foot per second.

In the c. G. s. system the unit is 1 centimetre per second.

If we say that a body is moving with a velocity of 5 feet per
second due east, we imply that if it continues for 1 second to have
this velocity it will have moved 5 feet due east in the course of
the second. If the velocity be varying, then we have two ways of
dealing with it ;

( 1 ) We may state the ac^wa? velocity it has at each instant, or,

(2) We may state the average velocity in respect to time for
the interval under consideration.

Let us look at these things graphically.

a ■■

Fig. 14.

Consider a body K moving in a straight line, and let us start
measuring the time when the body is at A.

Now we may represent the position of the body K at any in-
stant by drawing a distance-time
curve as shewn in fig. 15. After a
time t seconds represented by ON,
the distance from A will be given
by the ordinate PN, say s feet.

If the body had a uniform ve-
locity it is easy to see that the
distance-time curve would be a
straight line, since s has to increase
uniformly with t.

It is also evident, in this case, that the velocity will be repre-

PN
sented by — -, i.e. by tan 6^, (fig. 16).

^^^

P^^^

^^<8

(

O N Time

Fig. 15.

ELEMENTAEY DYNAMICS

The velocity-time curve would be a straight line as shewn in

tan 6

Time

Fig. 16. Fig. 17.

Suppose the velocity is varying as shewn in fig. 15, then for any

interval of time, t say, we can find the average velocity.

_,, . .„ , , distance passed over PN s

This will merely be -. — - — -, = ■ — = - .

time taken ON t

We may want to know the velocity at any particular instant,
and this we can find from the space-time curve as follows :

Suppose we require the velocity after a time t, when the distance
is given by point P on the curve.

Draw TP the tangent to the curve at the point P ; then the
velocity at P will be given by tan PTN, i.e. tan 6.

We have.

velocity

= tan

We may now plot a velocity-time curve, fig. 18, the ordinates
of which represent the values of tan 6 for the space-time curve.

>»

•fj

^^Q

-8t

— -

O N M A Time

Fig. 18. Fig. 19.

Agliin, we might have been given the velocity-time curve and
want to use it to find the distance passed over.

SPEED AND VELOCITY 23

It is easy to see, fig. 19, that for a small interval of time 8t,
the distance passed over is nearly equal to the velocity at P x 8^
= NPx MN

= area of rectangle PM, approximately.
For the true distance we must take St indefinitely small.
In this case we cannot draw the rectangle as it becomes a line,
but we can clearly see that the sum of the rectangles, for all the*
small intervals of time from O to A, becomes the area under the
curve on the base OA.

Having obtained the distance passed over in any time t by means
of the area under the curve, we can now obtain the average velocity
with regard to time by dividing this distance by the time.

^- , .^ area above OA
I he average velocity = —

= the mean height of the curve.

Example (1). A train has a speed of 60 miles jjer hour, what
is the speed in feet per second?

60 miles per hour

= 60 X 1760 X 3 feet per hour

60x1760x3^.
^ 60x60 .f^^^P^^«^<^^^^d

= 88 feet per second.

Example (2). The vanes of a de Laval Steam Turbine are at
a mean distance o/*3J inches from the centre of the rotor which runs
at 30,000 revolutions per minute. Find the speed of the vanes.

Speed = 27r X 3-5 inches in _ minute

Stt X 3-5 X 30,000 ^

— leet per minute

12
27r X 3-5 X 30,000 X 60

12x5280
625 miles per hour.

miles per hour

ELEMENTARY DYNAMICS

Example (3). A shell is fired at a target 2000 yards away,
and explodes at tJie instant of hitting. At a point distant 1800 yards
from the gun and 400 yards from the target, the sounds of firing
and exploding of the shell arrive simultaneously. Taking the velocity
of sound in air as \0^0 feet per second, and assuming the path of
the shell straight, find its average velocity.

In fig. 20, let A be the position of the gun, B the position of the
target and C the point where the sounds are heard.

2000 yds

^800 V^s

Fig. 20.

Time for sound to travel from A to C == time for shell to travel
from A to B 4- time for sound to travel from B to C.

Let V — the average velocity of the shell in feet per second, then,
1800 X 3 2000 X 3 400 X 3

1080
2000

1400
1080

1080

or,

V = 1542 feet per second.

Example (4). At a particular instant a body is moving with a
velocity of b feet per second and 3 seconds later its velocity is \Ofeet
per second. If it is known that the speed
is increasijig uniformly with the time,
find the distance parsed over in the three
seconds.

The velocity-time curve is as shewn
in fig. 21. It is obvious that the time
average of the velocity

= J(10 + 5) = 7*5 feet per vsecond.

Distance passed over = 7*5 x 3

= ?2-5feet.

Time

Fig. 21.

SPEED AND VELOCITY

Example (5). The speed-time curve for a motor omnibus^
obtained hy means of a speed recorder, is given in the following
table :

Speed
m.p.h.

6-1

8-1

9-4

10-7

12-8

14-4

15-9

16-8

17-7

18-3

18'7

19-2

Time
sees.

It is required to draw the distayice-time curve.
The speed- time curve is shewn plotted in fig. 22, the scales
being :

1 small division = 1 mile per hour,

and r small division = 1 second.

jst ^

^^" "^

v> /

.£.

I't |2'

-., 2

20 -,

:§:::: :: :: ::: :::

^" " - =

:t : : "

--;='-*?---

15= ^-'-''

/''

-£- -iS- ^-

^<t^ ^^^

•^ii'^ ^ ^^

~ /

-n T V

10-©- ,^- ^

« ^'^ ^

-.h- ^^ -^^^

vr 7* p'^

- Z ^''

5 -I -A-^q-

'■i ,'^$-

^ '*= +

:^^r^-— +— -"^'r'Vf

3e(por^djS

8I«
II
>

15 20 25

Fig. 22.

The distance passed over is represented by the area under the
.curve to a scale such that

1 small square = ^|^ x 1 feet.
Take the distance passed over in the first 10 seconds.

26 ELEMENTARY DYNAMICS

The number of small squares under the velocity-time curve = 82,

i.e. distance = — wk — feet.
bU

For the distance scale take

1 small division = ^-^- feet.

The distance for first 10 seconds = 4-1 divisions.

This is shewn by point A.

The distance-time curve obtained in this way is shewn in fig. 22.

If squared paper is not available the graphical method given in
chapter I may be used for drawing the area curve.

Referring to fig. 13, p. 16, if we keep the same scales as in
fig. 22, we find the distance OB thus.

Let h = the number of small divisions in OB.

The scale for distances is, 1 small division = hx^^ feet,

• 7, V 88 _ 88

i.e. h= 20 small divisions.

Example (6). The supply pipe to a tap is f inch bore, and the
nozzle of the tap at exit has a bore of J inch. The water leaves the
tap in a direction inclined at 90 degrees to the supply pipe.

If the tap discharges 3*6 gallons per minute, what is the velocity
of the water at exit, and the change of velocity as it passes through

the tap.

1 cubic foot of water contains 6 J gallons.

Let V = the velocity of water at exit in feet per second.
The discharge = ^ ^-[aa ^ ^ ^^^^^ ^®®* P®"^ second

= - —~~ XV X 6-25 X 60 gallons per minute.

16 X 144

Hence rir^,^ x 6-25 x 60 x -y = 3*6,
16 X 144

3-6x16x144

^^ "" ~ 'tt X 6-25 X 60

= 7*05 feet per second.

SPEED AND VELOCITY
li u = the velocity in the supply pipe, then

TT X9

4 X 64 X 144

~xu X 6-25x60 = 3-6,

and

u = 7-05 X ^ X J
= 12*5 feet per second.

In fig. 23 let OA represent u and OB represent v, then AB repre-
sents the change of velocity, i.e. the
velocity which has to be vectorially
added to the velocity u to change it
to velocity v.

Change of velocity = Ju^ + v^

= n/12-52 + 7-052

= V205T ° ^ig- 23.

= 14-3 feet per second.
The direction is given by the angle ^, i.e.

tan~^ - = tan ^ -^ ^
u 12-5

= 291" (nearly).

Angular Velocity

When a body is revolving about an axis it is often more con-
venient to express its speed in terms of the angle turned through
in unit time, instead of the distance moved in unit time. The
former is called the angular velocity of the body and is usually
measured in radians per second or revolutions per minute.

It is easy to establish a connection between the angular
velocity (w) and the speed, or velocity (v) at any instant (t). Let
a point A rotate about centre O, and let its distance from the

28 ELEMENTARY DYNAMICS

centre be r. Suppose it turns through 6 radians in time t and
arrives at the point B (fig. 24).

n
Angular velocity = cd

But speed, v

t
arc AB vQ

t t

Fig. 24.
= (u . r.

Hence the velocity at any instant or the speed = wr.

Example (7). The flywheel of an engine is 7 fleet in diameter
and rotates at 240 revolutions per minute. Find the angular
velocity in radians per second^ and the linear speed ofl a point on
the rim..

Angular velocity = 240 revs, per min. -f

240 '.^

-j^- revs, per sec. ^

240 xW ,.

radians per second

= Stt rads. per sec.
Speed of a point on the rim = cor

= 87rx|-

— 88 ft. per sec.

Example (8). A nut is rotated on a flxed screw at N revolutions
per minute. If the screw has n threads per inch and is ofl effective
diameter d inches^ find an expression flor the speed ofl sliding ofl
the nut and screw.

Suppose we strip off one thread and flatten it out, we should

get an inclined plane as shewn in fig. 25, where CB equals the

circumference of the screw, i.e. ird, and AB equals the pitch of the

. 1
screw, I.e. -.

ANGULAR VELOCITY

Let u = the speed of sliding along AC in feet per minute. We
may resolve this into two compounds (a) vertical v^f (b) hori-

zontal Vg.

Fig. 25.

In one revolution the horizontal travel is ird inches.
.'. the horizontal speed = 7ro?N inches per minute.

In one revolution the vertical travel equals - inches.

^ n

N
.*. the- vertical speed = - inches per minute.

The velocity of sliding is equal to the resultant of v^ and v^,

vl + {-TrdnY inches per minute.

Acceleration

In precisely the same way as velocity is the rate of change
of position, so acceleration is the rate of change of velocity :
i.e. the change of velocity in unit time. The velocity may be in-
creasing uniformly, in which case we get a constant acceleration.
For example, a motor car increased its speed uniformly from

ELEMENTARY DYNAMICS

4 miles per hour to 12 miles per hour in 10 seconds. Its accelera-
tion is (12 — 4), i.e. 8 miles per hour in 10 seconds,

= ^ miles per hour in 1 second

— Tu ^ f & ^^®^ P®^ second in 1 second

= 1 '17 feet per second per second.

Again, the velocity may be changing at a variable rate, e.g. in
the first second it may increase 1 mile per hour, in the second
4 miles per hour and so on. The acceleration in this case is said
to be variable.

It is easy to see that acceleration is related to velocity in
exactly the same way as velocity is related to position.

Any propeHies or formulae which hold for velocity and position
will hold also for acceleration and velocity.

Velocity-time curve

Suppose we are given a velocity- time curve as shewn in fig. 26.
The acceleration at any time t will be given by the slope of the
curve at P, i.e. by the value of tan 6.

Fig. 26.
Again, if we are given an acceleration-time curve it follows that

ACCELERATION

the change of velocity in time t is given by the area under the
curve from A to P, shewn shaded in fig. 27.

Fig. 27.

Velocity-space curve

In certain practical problems we can obtain the figures for a
velocity-space curve and it is often necessary to estimate the ac-
celeration. If V denote the velocity, and t denote the time, the

ov
acceleration is given by the value of ^ when ht is indefinitely

diminished, i.e. by — . Suppose we wish to find the acceleration

at distance s, i.e. at point P on the curve, fig. 28.

Space

Draw the tangent PT to the curve at P, the ordinate PN and
the normal PG, i e. the perpendicular to the tangent.

ELEMENTARY DYNAMICS

OV 6S

Now the acceleration = ^ • ^ j when U is indefinitely diminished.

But the limit of
and the limit of

8s
It
Bv
Ss

= tan 6.

Hence the acceleration = v^r

= PN . tan e

= PN . tan NPG

= NG.

Or, the acceleration is given by the subnormal NG.

, Space curve

Velocity

Suppose we are given a velocity-space curve and we wish to find
the space-time curve.

We have v = -^ , nearly,
bt

i.e. -Ss = Bt.

Draw the -, s, curve (fig. 29).

Space

Fig. 29.

Then, the time for any distance is
given by the area U7ider the curve.

In many problems, it is not necessary to actually draw the
graphs representing the motion to scale, as we can readily
calculate the quantities we require. It is, however, often very
helpful to sketch the graph roughly.

Example (9). A tramcar starts from rest and accelerates uni-
formly for 8 seconds to a speed o/ 10 miles per hour. It then runs

ACCELERATION

at a constant speedy and finally is brought to rest in iO/eet with a
constant retardation. The total distance passed over is 250 yards.
Find the value of the acceleration^ the retardation^ and the total
time taken.

k-8->K t,

Fig. 30.

The velocity-time curve is shewn in fig. 30, where t^ is the time
during which the speed is constant, and ^2 is the time of retardation.
The maximum speed attained = 10 miles per hour

= -\®- feet per second.
Area BEC = the distance passed over during the retardation, i.e.
||x^,= 40,

12x40

= 5*45 seconds.

Also, area OABC = the total distance passed over, i.e.

88 X 8 + ^8- X ^1 + 40 = 750.

.-. ^, = (750 -40- 58-7) ^3

= 44*3 seconds.

We have then,

88
Acceleration = ^ — 3 = 1 -83 feet per sec. per sec.

Retardation

6 X 5-45

Total time taken = 8 + 44-3 + 545
= 57*8 seconds.

L. E. D.

2*69 feet per sec. per sec.

34 ELEMENTARY DYNAMICS

Example (10). Using the speed-time curve for the motor omnibus
given on p. 25, it is required to construct an acceleration-time curve.

We have seen that the acceleration at any point P is given by
tan PTN, where FT is the tangent to the curve at P, fig. 31.

Take the acceleration at 20 seconds from the start,

A FN

tan FTN = —

= .^^r— (measuring in divisions).

Now each vertical division = 1 mile per hour

= |§ feet per second,
and each horizontal division = 1 second.

Hence to get the true acceleration we have to multiply the

value of tan FTN by |-| or ff .

.*. acceleration at P = .r=-^ x -—
37-5 15

= 0'43 ft. per sec. per sec.

Take for the acceleration scale, 1 small division = 0*1 ft. per sec.
per sec, then the acceleration at 20 seconds is given by the point Q.

Repeating this method for times 2, 5, 10, 15, etc. seconds, we
get the acceleration-time curve shewn in fig. 31.

Example (11). A body moves in such a way that its velocity in-
creases uniformly with the distance passed over, and at a distance
of bOfeet the velocity is 2^ feet per second. What is the acceleration
at the instant when the body has moved 15 feet?

The velocity-space curve is given in fig. 32. PN gives the
velocity at 15 feet. Draw PG perpendicular to OP.

ACCELERATION

Speed in miles per liour

-* ro w

Acceleration In ft per sec. per sec.

3—2

ELEMENTARY DYNAMICS

The acceleration = NG

- PNtan NPG
= PN tan PON

= PN X ^ —
ON

/20 X 15y 1

V 50 J ""15

24 feet per sec. per sec.

Example (12)
tances is given in the table below,
curve, and the speed-time curve.

G 50 Space

Fig. 32.
The speed of a ship in . knots at different dis-

Draw the acceleration-space

Speed (knots)

10-0

12-3

14-1

15-5

16-4

17-0

Distance (nautical
miles)

A nautical mile=6080 feet, 1 knot=l nautical mile per hour.
.'. 1 knot = 100 feet per minute, approximately.

The velocity-space curve is shewn in fig. 33. The acceleration-
space curve is shewn dotted on the same figure. This is obtained
by drawing subnormals. The ordinates are each made equal to
twice the subnormals.

ACCELERATION

Speed in knots,
o

1/veloc

81-

ity

. __ -_-

""I"" ~J

V-

S " "

"~r

> - ■

V - -

♦^•-* - -- t- -

_ :^

/<—

5 '

J<1-

-_lj?.

L_.

1 11

f J

I- i 5

\! 1

: V

: x^L :

W-*

.-*

±1^- -.

il^-

5i^

1 1

L::±±:il

Time in minutes

Acceleration in ft./sec.

38 ELEMENTARY DYNAMICS

Scales : 1 division = 1 knot = \^^ feet per second.
1 division = ^ nautical mile =150 feet.

We saw on p. 31, that the acceleration was given by v . -^ . If

cts

we measure the distances in feet and the velocities in feet per

second, the actual length of the subnormal NG in divisions will

1002
have to be multiplied by :r— r — ^r^ , to obtain the true accelera-

loO X 60^

tion. Hence the scale for acceleration is
1 division

300 X 60-

1
108

^rpr^ feet per second per second.

We have seen that the area under the -, s, curve gives the

time. This curve is shewn in fig. 34, where the scale for - is,
1 division = 2^ J^ (mile-hour units).
Scale for distance, 1 division = ^^ mile.
The dotted line shews the time-distance curve.

This is obtained by plotting as ordinates the area under the

-, s, curve.

The time for distance 8s = - 8s.

Hence, 1 small square ^ -^^ x -^ hour = g^^g-Q hour.
In the curve drawn, 1 division = 40 small squares

= 18 seconds.

ACCELERATION DUE TO GRAVITY S\)

Acceleration due to gravity

The number of cases of motion in which the acceleration re-
mains constant are not numerous in Engineering, as in most
practical problems we either have a constant velocity or a varying
acceleration. One case in which it is usual to assume a constant
acceleration is that of a body falling freely. Experiment shews
that if the distance from the earth's surface is not large, then the
acceleration is nearly constant at any place on the earth. It
varies slightly at diiferent points on the earth's surface, but it
will be sufficiently near for our purpose to take the accelera-
tion as constant, and equal to 32 feet per second per second, or
981 cms. per second per second. This assumes that the resistance
of the air may be neglected, which is only true in the case of quite
small velocities. The numerical value of the acceleration is usually
denoted by the letter g.

We shall discuss this further when we are dealing with the
forces causing motion and the law of gravitation.

Projectiles

We will now work out a few examples on projectiles, making
the assumption mentioned above, but it must be borne in mind
that the results have very little practical value.

Example (13). A body is projected tvith a velocity of ufeet per
second in a direction inclined at angle 0 to the horizontal. It is
required to find the total time of Jiight, the 'maximum horizontal
range, and the maximrUm. height reached.

Here it is convenient to treat the velocity as consisting of
a vertical component and a horizontal component.

Horizontal component = u cos 0.

Vertical component = u sin 6.

If we neglect air resistance the horizontal component of the

ELEMENTAEY DYNAMICS

velocity will remain constant, and the vertical component will
decrease uniformly by g feet per
second per second.

The path of the body will be as

shewn in fig. 35 and it is evident
that the total time of flight will
be twice the time taken to reach
the maximum height. The time
to reach the maximum height is
equal to the time for the vertical
velocity to become zero

u sin 6

Fig. 35.

.'. the time of flight

2u sin 6

The range is given by the distance moved horizontally in this
time,

2usm. ^

(A^

7. G

u cos 6 X

-$-

/_«#sin2^

This is a maximum when sin 26 = I, i.e. when 0= 45°.

To find the maximum height, we note that since the vertical
velocity is decreasing uniformly, the average vertical velocity

u sin 6

.'. the maximum height

u sin 0 u sin 0
u^ sin^ 0

Path of the Projectile. We can easily find the equation re-
presenting the path of the projectile.

PROJECTILES

Let a; = the horizontal displacement at instant of time t from
the start. Then
" x = ucos6.t (1).

For the vertical distance,

Velocity at beginning = u sin 6.
Velocity at end = (u sin 0 — gt).

Average velocity = J (2?^ sin ^ - gt).

.'. 2/ = I (2u sin 0 — gt) . t,

or, y = u^iii6 . t — ^gt"^

Eliminating t from (1) and (2) we get,
u sin 6.x g a?

(2).

2 u^ cos2 6 '
9

2u^ cos2 6

a^.

i.e. y — ^' tan 6

which is the equation of a 'parabola.

Example (14). A rifle hullet is fired at a target, on the same
level as the rifle and distant 900 yards, with a muzzle velocity of
2000 feet per second. Neglecting air resistance, find the angle of
elevation. Find also the limits, between the point of firing and the
target, within which a man 6 feet high can stand without being hit,
assuming the target at the same level as the ground.

V cos 9

Fig. 36.

Let 6 = the angle of elevation, and t = the time of flight.
Horizontally we have

2700 = 2000 cos (9 . < (1).

42 ELEMENTARY DYNAMICS

Time to reach the maximum height = - seconds.

Retardation upwards = g ft. per sec. per sec.
2000 sin e t

9 2

From (1) and (2), by eliminating t we have,
2000 cos e X 4000 sin (9

(2).

2700

/. 2^=1° 14',
or ^ = 37 minutes.

Let a = the safe zone, and t = the time for the bullet to reach C.
For the vertical motion,

/4000 sin e - 32^^

6 =

For the horizontal motion,
2700 -a

/4000sin^-32A

V 2 )'' (^^•

2000cos^.« (4).

From (3) we get, by substituting for sin 6 its value 0-0108,
6 = 21-6^-16«2,

21-6-v/21-62-24 xl6
' = —32

12-55

seconds.

32
From (4) we get

.-. a = 2700 -1569
= 1131 feet.

CHANGE OF DIRECTION 43

Change of Direction

In dealing with acceleration, we have up to the present tacitly
assumed that the direction of motion remained unchanged, and
although dealing with vector quantities we have merely considered
the magnitude. We must now consider the case where the direc-
tion changes.

Suppose, for example, that a body at a particular instant is
moving with velocity 1.1 in a direc-
tion given by OA (fig. 37) and at
t seconds later it is moving with
a velocity v and in a direction
given by OB. Let us make OA
and OB of such lengths that they
represent u and v in magnitude.
Now in order to change from
velocity represented by OA to Fig. 37.

velocity represented by OB, it is

obvious that, we must add vectorially a velocity represented by
AB, call it q. Then the average acceleration during the time
= average rate of change of velocity
_ AB

~ T

Circular Motion

Let us apply this to a very common case, viz. that of a body
moving in a circular path with a constant speed. Suppose that
in time t the body moves from A to B (fig. 38). Draw Oa to re-
present the velocity at A, and Ob to represent the velocity at B.
The added velocity, required to change from velocity v at A to
velocity v at B, is given by AB, i.e. the average acceleration for

time ht is given by -k- .

ELEMENTARY DYNAMICS

But since the triangles Oab and CAB are similar,
ah Oa
AB ~ CA *

Oa. AB

.'. the acceleration

CA. Bt

Now

Fig. 38. -

When St is indefinitely diminished 89 becomes zero also, and we
see that the acceleration is always axiting towards the centre C and

is of magnitude

This is an important result and should be remembered.

Example (15). A motor-car travelling at \0 miles per hour
takes a corner of 10 yards radius. What is the acceleration of the
car in feet per second per second?

The speed = 1^ feet per second.

The acceleration

10^ X 88^
602

X — feet per sec. per sec.

= 7 '2 feet per sec. per sec.

CHANGE OF DIRECTION 45

Example (16). The crank of an engine has a radius of 9 inches

and is rotating at 300 revolutiaiis per minute, what is the linear

speed, and the acceleration of tJie crank pin ?

27r X 300
The angular velocity = — ^ —

= IOtt radians per second.
The linear speed of the crank pin

:=107rx^

= 23-6 feet per second.
The acceleration of the crank pin towards the centre of rotation

9
740 feet per sec. per sec.

Relative Velocity

Up to the present, in dealing with motion, it has been assumed
that we have some fixed starting point for reference, and that
we have measured our displacement from this point. In actual
practice, a little thought will shew that we have no really fixed
point, but so far as we know any body or point may be moving.
It is frequently the case that we imagine a point on the earth's
surface as at rest, but in reality this is moving. Again, take the
case of an engine on a steamer or motor-car; when we talk about
the speed of any moving part of the engine, such as the piston, in
estimating its speed we usually imagine the steamer or car at rest.
In other words, in dealing with the motion of a body we do not
really know anything about the true motion but only the motion
refative to some other body which we imagine at rest. Thus all the
motions we deal with are really only relative motions. This being
so, it will be as well to carefully define what we mean by relative
motion. Suppose we have a body or a point A moving relatively
to another body or point B, then the relative motion of A with

46 ELEMENTARY DYNAMICS

respect to B is the motion which A appears to have when we view
it from B. Take as an example two trains passing one another
on parallel tracks. Suppose the train A is moving at 10 miles per
hour and the train B at 1 5 miles per hour. Then to a person in train
A viewing train B the latter would appear to be moving with a
velocity of (15 — 10) miles per hour, i.e. the relative velocity of B to
A is 5 miles per hour. To a person situated in B and viewing A's
motion the latter will appear to be (10- 15), i.e. — 5 miles per
hour, or the relative velocity is 5 miles per hour in the opposite
direction.

Now let us see how we can always measure relative velocity.
It is obvious that if we give both bodies the same velocity their
relative velocities will be unaltered, e.g. the two trains both have
the velocity of the earth, but this does not affect their relative
velocities.

Suppose, now, a body A is moving as shewn in fig. 39, with
velocity u, a body B is moving with
velocity v, and we wish to find the
velocity of B relative to A. Give to
both bodies a velocity equal and op-
posite to that of A. This brings A to
rest, but does not aflfect the relative
velocity. It is obvious that B's resul-
tant velocity will be the velocity of B , ^

relative to A. This is shewn by bA; in A vel. u

fig. 39, where ^n represents the velocity ^^' '

of B, and nk represents the velocity of A reversed. Looking at

triangle Bn>fc, we see that kn represents the velocity of point A,

and we have

Bn = kn-\- Bk.

Or remembering that we are dealing with vectors :

The velocity of the body B is equal to the velocity of
the body A plus the relative velocity of B to A.

RELATIVE VELOCITY

Similarly we may state :

The velocity of f< is equal to the velocity of B 'plus the relative
velocity of ^ to B.

Two points rigidly connected. Instead of A and B being
two bodies, they may be two points in one rigid body. In this
case the relative motion is given in direction, since it must be
perpendicular to the line joining the two points. If this were not
so, there would be a component of relative velocity along AB, or
the points A and B must be either closing in or separating, which
is contrary to the assumption that they are rigidly connected.
This is shewn in fig. 40.

Relative velocity

Fis. 40.

Another way of stating this is by saying that the relative
motion of B to A is one of rotation of B about A. If to is the
angular velocity of B about A, then the relative velocity of B to
A = « . AB.

Example (17). The maximum speed of an airship in still air is
40 miles per hour. What is the shortest tim,e in which the ship can
travel a distance q/" 10 whiles due norths at a constant altitude^ if
there is a north-west wind blowing a^ 18 miles per hour? In what
direction will a flag attached to the airship fly ?

ELEMENTARY DYNAMICS

The actual velocity of the airship will be equal to the vector
sum of the velocity of the wind and the velocity of the ship rela-
tive to the wind, i.e. the velocity in still air.

Scale
l" = 10m.p.h.

S A

Fig. 41.

Draw OA in a direction south-east to represent to some scale the
velocity of the wind (18 miles per hour). Take a length AB to
represent, to the same scale, the velocity of the ship in still air.
Let this cut the north line through O in the point B.

RELATIVE VELOCITY 49

Then OB represents the actual velocity of the airship.
This by measurement = 25*2 miles per hour.
.'. The shortest time to travel 10 miles due north
10x60 . ,

= 23*8 minutes.

The flag will fly in the direction of the resultant wind on it,
that due to its motion with the airship, and the north-west wind.

This is given by the vector sum of BO and OA, i.e. BA.

.". The flag flies in the direction BA, which is inclined to the
north at an angle of 19°.

Example (18). The crank of a steam-engine is 9 inches and is
rotating a^ 360 revolutions per minute. The connecting rod is 30
inches long. Find the velocity of the piston when the crank is in
the position shewn in the figure below.

OTT

— Q

. Fig. 42.

In 6g. 42, CP represents the crank, PD the connecting rod, and

DC the line of stroke. The crosshead D is fixed to the piston and

will therefore have the same motion as the piston.

The velocity of P - wr

27rx360 9 ^ ,
= — ^.^ — X —r teet per second
60 12 -

= 97r feet per second.

Let us give to D and P a velocity equal and opposite to that of

P. This brings P to rest and D's resultant motion will be the

L. E. D. 4

ELEMENTARY DYNAMICS

relative velocity of D to P, which must be perpendicular to DP.
Draw po to represent a velocity equal and opposite to
that of P. Draw od parallel to DC, and 2>d perpendi-
cular to PD. It is obvious that opd is, a triangle of
velocities, and that od represents the velocity of D,
and pd the velocity of D relative to P.
Velocity of D

OtT ~op~Qb
i.e. the velocity of D = 18'3 feet per second.

— = ^ (by measurement),

Rolling wheel

A wheel rolls along a plane, without sliding, at a constant rate;
it is required to find the velocity of any point A on the rim.

Let V = the velocity of the centre C, w = the angular velocity of
the wheel, and r = the radius of the wheel.

V77777777777777777777777777777777777r

Fig. 44.

For one revolution of the wheel the centre C moves a distance
equal to the circumference of the wheel, i.e. the speed of C is the
same as the speed of a point on the circumference of the wheel.

.*. V = con-

The velocity of A = the vector sum of the velocity of C, and the
relative velocity of A to C.

The latter equals cor and is perpendicular to CA.

In fig. 44 let AE represent the velocity of C, and EF represent
the velocity of A relative to C.

EXAMPLES. CHAPTER II

The actual velocity of A is represented by AF.
.". The velocity of A = 2o>r . cos 6
= (o . PA.
It is seen from the figure that A F is at right angles to PA.
.'. The point A is for the instant turning about the point P with
an angular velocity w.

Examples. Chapter II

1. Express the following in metres per second: (1) The speed of a runner
who does 100 yards in lOi seconds. (2) The speed of a train running at
60 miles per hour. (3) The speed of a point on the rim of a flywheel 7 feet
in diameter rotating at 250 revolutions per minute.

2. A popular method of estimating the distance away in miles of a flash
of lightning is to divide the time in seconds between the flash and the first
sound of the accompanying thunder by 5. What velocity of sound in feet
per second does this assume ?

3. The speed of a ship used to be measured by dropping overboard a log
attached to a line and measuring the speed with which the line ran out, it
being assumed that the log remained stationary. The line was divided into
sections of equal length. Find the length of a section in order that the number
of sections running out in 28 seconds should be equal to the speed of the ship
in knots.

1 knot = 6080 feet per hour.

4. The data for a distance-time curve of a train starting from rest is
tabulated below. Plot the curve and from it deduce the speed-time curve.

Distance x 10-=^
in feet

0-6

2-0

4-4

7-5

11-0

14-8

19-0

Time in minutes

Distance x 10"=^
in feet

23-0

27-8

32-5

37-6

48-4

54-2

Time in minutes

4—2

52 ELEMENTARY DYNAMICS

6. A cam in the form of a circular disc is keyed eccentrically on a hori-
zontal shaft which rotates at 120 revolutions per minute. The diameter of
the disc is 6 inches and its centre is one inch from the axis of the shaft.
A rod presses against the edge of the disc, and is constrained to m.ove vertically
in the plane of the disc along a line passing through the axis of the shaft.

Draw the displacement-time curve of the rod for one revolution of the shaft,
and from this curve deduce approximately the velocity-time curve, stating in
each case the scales adopted. The end of the rod pressing on the disc may
be considered a point.

6. The distance (s) of the piston of an engine from the end of its stroke
is given by the equation, s = 6-5 - 6cos25f -O'Scos 50t, where s is in feet
and t is the time in seconds. Find the velocity at the times given by
t=^V second, t=^ second, and f = |^ second.

7. A point moves uniformly round a circle of 1 metre radius, makihg
100 revolutions per minute. Express its linear velocity in feet per secdlid,
and its angulur velocity in radians per second.

The wheels of a motor-car are 28 inches in diameter. How many revolutions
per minute does each wheel make when the car is running at 35 miles per hour?

8. A pulley wheel of 20 inches diameter is connected by a belt to a flywheel
of 5 feet diameter, which is rotating at 280 revolutions per minute. What is
the speed of the belt if there is no slipping?

Find tjie speed of rotation of the pulley (1) in revolutions per minute,
(2) in radians per second.

0. The spiral grooves in a rifle barrel make one complete turn in 10 inches.
Find the speed of rotation of a bullet when it leaves the muzzle with a velocity
of 2500 feet per second.

10. A fixed screw of ejffective diameter 2 inches has 3 threads per inch,
A nut on the screw is rotated at 300 revolutions per minute. What is the
speed of sliding of the nut in feet per second ?

11. At a particular instant a motor car is travelling at a speed of 10 miles
per hour, when it starts accelerating at a uniform rate of 2 feet per second
per second. Shew that the distance in the 20th second is 53-7 feet.

12. A train starting from rest at one station comes to rest at the next
station 6 miles off in 10 minutes, having first a uniform acceleration, then a
uniform velocity for 8 minutes and then a uniform retardation ; shew that
the greatest velocity attained is 40 miles an hour.

EXAMPLES. CHAPTER H

13. The table below gives the speeds at different times of an electric train
running between two stations. Plot the speed-time curve and from it find the
total distance passed over, the mean speed, the initial acceleration, the ac-
celeration at the end of 40 seconds, and the final retardation.

Speed in miles
per hour

11-2

19 8

29-2

24-8

23-8

21-4

8-5

Time in
seconds

120

130

150

160

166

14. The velocity (v) of the piston of an engine at any time is given by the
equation, v = 10*47 sin 25 « + 0-87 sin 50 f, where v is in feet per second and t
is in seconds. What is the equation giving the acceleration of the piston in
terms of the time ?

15. The velocity of the ram of a slotting machine for different positions
during the cutting and return stroke is given in the table below. Plot the
velocity-space curve, and from it deduce the acceleration-space curve.

The length of the stroke is 9 inches.

Fraction of stroke
in^th

Velocity in ft. /sec.
Cutting stroke

27-5

29-5

Velocity in ft./sec.
Keturn stroke

Fraction of stroke
in iVth

Velocity in ft./sec.
Cutting stroke

29-5

27-5

Velocity in ft./sec.
Return stroke

16. A body moves along a straight path in such a way that its velocity,
in feet per second, is related to its displacement, in feet, from a fixed point
by the equation, v^=lS5-Q0s^. Plot the v, s, curve for the complete
motion, and from it deduce the acceleration-space curve.

Draw also the space-time curve.

54 ELEMENTARY DYNAMICS

17. A stone is dropped from a balloon which is rising with acceleration a,
and ti seconds after this a second stone is dropped. Prove that the distance
between the stones at a time t after the second stone is dropped is

^{a + g)t^{t^ + 2t).

18. A stone falling freely under gravity is observed to pass from top to
bottom of a window 8 feet high in i second. Find the distance from the top
of the window to the point from which the stone fell.

19. A shell is fired so that at the highest point of its flight it just passes
over a mountain half-a-mile high and distant 5 miles from the point of pro-
jection. Shew that the horizontal component of the velocity of projection is
to the vertical component in the ratio of 5 to 1, and find the values of these
components.

20. Water is issuing from a fire-hose nozzle with a speed of 120 feet per
second. The jet is to pass through a window which is distant 35 feet vertically
and 30 feet horizontally from the nozzle. What must be the inclination of
the nozzle if the jet when it reaches the window is (1) to be rising, (2) to be
falling ?

21. A body at a particular instant was moving due east with a velocity
of 5 feet per second, and at a later time it was moving east-north-east with a
velocity of 8 feet per second. Find, graphically, the change of velocity in
magnitude and direction.

If the time which elapsed between the first and second velocities was
7 seconds, what was the magnitude of the average acceleration during this
time?

22. Water is flowing round a small pipe which is bent in an arc of a circle
of 18 inches radius, with a speed of 10 feet per second. What is the ac-
celeration of the water while passing round the bend ?

23. A cutter can be rowed in still water at 2 knots. It is rowed across
a river 400 yards broad, which is flowing at the rate of 1| knots, starting from
one side and reaching the opposite bank 100 yards further up stream. Find
the least time in which the passage can be made, and the direction in which
the boat will point while going across. Find also the least time and the
direction for the return journey.

1 knot =6080 feet per hour.

24. A train A moving with a constant speed of 30 miles per hour is passing
another train B which is at rest on a parallel track. At the instant when the
engines of the two trains are opposite one another, the train B starts with a

EXAMPLES. CHAPTER II 55

constant acceleration of 2 feet per second per second. What length of time
will elapse before the two trains again occupy the same relative position ?

If the train B is 100 yards long, at what distance from the station will it
just have completely overtaken the train A ?

25. Two ships, A and B, start simultaneously from the same point. A
steams north-east at 15 knots, and B steams south at 10 knots. Shew that
the line AB moves parallel to a fixed direction, and find at what rate the
distance AB increases.

After one hour's steaming A changes its course to south ; after what time
will it be due east of B? At this instant it again changes its course and steams
west. At what distance from B will it cross B's course? The distances are
to be reckoned in sea miles.

26. Two straight railway tracks, OA and OB, intersect at O and the angle
AOB is 60°. A train, P, is running in the direction OA at 40 miles per hour,
and another, Q, in the direction OB at 25 miles per hour. Find the magnitude
and the direction of the velocity of Q relative to P.

27. A gun pointed at an inclination d to the horizontal is mounted on a
carriage which can run on a horizontal rail. If the shot leaves the gun with
a relative velocity v, the gun recoiling along the rail with velocity u, shew
how to find the range on a horizontal plane through the point of projection,
treating the shot as a projectile in vacuo. By how much is the range shortened
by the recoil ?

If i; = 1000 and w=10 feet per second, find whether the range is greater
for ^ = 45° or ^ = 46°.

28. The driving wheels of a locomotive are 7 feet in diameter. How many
revolutions does each wheel make in a minute when the locomotive is running
at 50 miles per hour ?

O is the centre of one of the wheels, A its point of contact with the rail,
and P a point on the rim. Find the magnitude and direction of the velocity
of the point P in each of the positions in which the angle AOP is 120°.

29. A ladder 30 feet long rests with one end on a vertical wall and the other
end on the ground at a distance of 12 feet from the wall. The lower end
begins to slide horizontally away from the wall with a velocity of 1-5 feet
per second. Determine graphically the velocity of the end on the wall, and
the relative velocity of the two ends.

30. A cable drum 3 feet in diameter, with side flanges 5 feet in diameter,
rests on the road. The free end of the cable, which comes from the under-
side of the drum, is pulled forward with a velocity of 2 feet per second. How
long will it take to wind up 30 feet of cable, and how far will the drum have
rolled during this time ?

ELEMENTARY DYNAMICS

31. A slider is driven along a straight guide by means of a crank and
connecting rod. The length of the crank is 3 inches and the connecting-rod
12 inches. The line of stroke of the slider is 5 inches below the centre of the
crank shaft. The crank rotates uniformly at 40 revolutions per minute. Draw
a line diagram of the mechanism to a scale ^ full size, and determine the
length of stroke, the times of travel of the slider during the forward and return
strokes, and the velocity of the slider at the middle of each of the two strokes.

32. The figure below shews two sliders B and C with the same line of
stroke BCO. They are driven from the same shaft by means of two cranks
OA and OD. If the crank shaft rotates at 75 revolutions per minute, find
graphically for the position shewn the relative velocity of B and C.

0A = 6", AB^12", 00 = 4", DC = 8".

Fig. 45.

CHAPTER III

Linear Momentum

In the last chapter we dealt with motions, but we did not con-
sider how these motions were produced or changed. We have
now to investigate the causes of motion, and for this purpose, we
shall use two principles upon which the whole subject of mechanics
depends :

(1) The Conservation of Momentum.

(2) The Conservation of Energy.

We will consider the first here, postponing the second for a
later chapter. First of all we must introduce, and make ourselves
quite familiar with, a new physical quantity, viz. Momentum.
It is really the product of two physical quantities and we may
define it thus :

The Momentum of a body is equal to the product of the mass and
the velocity with which it moves.

If a body of mass m lbs. is, at a particular instant, moving with
a velocity of v feet per second, then the body is said to possess
a quantity of momentum equal to 'tnv ft. lb. sec. units.

Sometimes we shall be considering the change of momentum of
a single body, and at others we shall. find it convenient to think
of the increase or decrease of the resultant momentum of a system,
or it may be the component of momentum in any particular direc-
tion. Since velocity is a vector quantity so momentum must be
considered a vector quantity, the direction being that of the
velocity.

The quantity of momentum of a oody or system may be changed
in various ways :

(1) The mass may remain constant and the velocity be changed
in magnitude, e.g. a train moving along a straight track may in-
crease or decrease its velocity.

58 ELEMENTARY DYNAMICS

(2) The mass may remain constant and the velocity be changed
in direction without any change in magnitude, e.g. the speed of
the train may keep constant, but the direction may be changed.

(3) The mass may be changing with the velocity remaining
constant, e.g. a locomotive picking up water from a trough along
the track.

(4) Any combination of the above three, e.g. a rocket when
fired. Here the mass is changing as the products of combustion
are blown out, and also the velocity may be changing both in
magnitude and direction.

We will now state the laws or principles of momentum*.

1st Law. In any body or system the total momentum
remains constant unless the body or system is acted upon
by some external force.

2nd Law. If there is a change of momentum, then the
force producing it is proportional to the rate of change
of momentum and acts in the same direction.

' ThQ first law introduces a new term, viz. Force, which may for
the present be defined thus :

Force is that which produces or tends to produce a change of
momentum.

The law is the result of observation. If the momentum of
a body changes we immediately look for some cause, and this
cause we call force. For example, in cycling we may find that
our velocity has increased in magnitude, or our direction of
motion may be changed, and we immediately look for some cause.
It may be that a wind has sprung up and is exerting a force in
the direction of motion, thus increasing our velocity, or it may
be that we have met an obstacle in the form of a stone with the
result that a side force has been exerted and this has changed
our direction.

The second law is really based upon experimental observation,
* See Preface.

LINEAR MOMENTUM 59

and provides a means of measuring the force which is producing
a change of momentum.

Let us make the matter clear by taking a simple concrete case.
Suppose we have a mass M acted upon by a constant force F, and
that this force, during a time t, produces an increase of velocity
from Vi to fg.

— >^

'-^

///////////y/y

^ / / / / /

Fig. 46.
The momentum at the beginning of the time=: \Av^,
The momentum at the end of the time = M^g-
.'. the change of momentum = M-yg— Mv^.
From the 2nd law,

Foe — ^ ^.

.*. we may write.

/MVg— M-yA

^^ M^.^,-^ („),

where A; is a constant^ the value of which depends upon the units
which we adopt for measuring the force.

Absolute System of Units

The unit force is that force which produces unit change of mo-
mentum in unit time.

Taking M = 1, -yg - '^i = 1) ^^^ t = l, we have,

I =k . —^ , i.e. ^ = 1.

Hence with this unit we may write,

t
or, the force is equal to the change of m.omentum in unit time.

60 ELEMENTARY DYNAMICS

In the F.p.s. system the absolute unit of force is often called
the Poundal, and is that force which acting on a mass of 1 lb.
produces a change of velocity of 1 foot per second in one second.

In the c.G.s. system the absolute unit of force is called the
Dyne, and is that force which acting on a mass of 1 gram pro-
duces a change of velocity of 1 centimetre per second in one
second.

Gravitation Units

Newton discovered that any two masses attract one another
with a force which varies directly as the product of the masses,
and inversely as the square of the distance between them.

Thus, there is a force acting between every body and the earth
and this force is called the force due to gravity. For any particular
body, this force is not really constant except for a definite position,
but near the earth's surface the variation is so small that it may
be neglected. The force due to gravity on a body is called the
weight of the body.

We may take the unit weight, i.e. the force of gravity on unit
mass, to be our unit of force. This, as noted before, will not
strictly be constant unless we specify a definite position or place
on the earth's surface. The position is generally stated to be at
Greenwich, where the value of the acceleration of a freely falling
body is 32-19 feet per second per second, or 981 cms. per second
per second.

Denote the numerical value of this acceleration by g.

If we have M lbs. of mass falling freely the acceleration is
g feet per second per second, i.e. in unit time there is a change of
velocity equal to g, and therefore a change of momentum M^ units.

The force acting is M lbs. wt.

From (a) above, we have :

M=k.M.g.

GRAVITATION UNITS 61

Hence with this unit we write,

^_M{v.,-Vi)

or, Force equals the change of momentum per unit time divided
h9'

In what follows we shall generally use the absolute unit of force
when working problems, and if necessary, we can express the
forces at the end in terms of weight.

It is obvious that :

The absolute unit of force

= - {Units of weight},

or 1 lb. wt. =32 poundals,

and 1 gram wt. =981 dynes.

It is very important to distinguish between mass and weight.

As we have previously stated, m,ass is merely the quantity of
matter, and is a scalar quantity ; weighty on the other hand, is
Si. force J and is a vector quantity (its direction being always towards
the centre of mass of the earth).

We are often, in problems, given the weight of a body, when
what we really want is the mass. This is due to the fact that
the commonest way of comparing masses is by weighing. Since
in any one place the rate of change of velocity due to gravity is
constant, it follows that the force due to gravity will only vary
with the mass. If we use gravitation units for force, the weight
and mass are numerically equal, but it must not be forgotten that
they are two entirely different physical quantities. The mass of
a body is constant no matter where the body may be, but on the
other hand, the weight will vary with the distance of the body
from the centre of the earth.

ELEMENTARY DYNAMICS

Suppose, for example, we use a spring balance for measuring
our weight, and that this spring balance is calibrated at the
equator. Here using foot, second units,

^ = 32-091.

Now suppose we use the spring balance to weigh stuff at one
of the poles. Here

^ = 32-252.

The spring balance requires the same force to produce a definite
reading wherever it may be, hence for the same weight, the quan-
tities of stuff at the equator and poles will be in the ratio of
32-252: 32-091,

, „ , 32-252-32-091 ,„„ ^ ,

I.e. we shall measure q^TaoI x 100 per cent., or ^ per

32*091

cent, more stuff at the equator than we do at the poles.

Example (1). A cage weighing 2 J tons is raised and lowered
in a coal mine shaft by a steel cable. Find the tension in the cable,

(1) When the'hage is raised or lowered with a constant velocity,

(2) When the cage is lowered with the speed increasing uniformly
from 0 to 1000 feet per minute, in the first bOfeet.

Let T be the tension in the cable in poundals, just above the
cage. Then the resultant force acting on the
cage is,

(T- 2-5 X 2240 x^)pdls.

(1) If the velocity is constant, there is
no change of momentum.

.*. T-2-5 X 2240x^ = 0,
or T = 2 J tons wt.

(2) The resultant force downwards

= (2-5 X 2240 x^-T)pdls.
The average velocity = 500 feet per minute. 2-5

The time to move 50 feet

-^ minute.

2240 xg pdts

Fig. 47.

GRAVITATION UNITS

The change of momentum downwards per second

2-5 X 2240 X 1000

From the 2nd Law,

2-5 X 2240 x^-T

177

X 60 X 60 •
2-5 X 2240 X 10000

60 X 60

T = 2-5 X 2240 (32 - 2-8) pdls.

2-5 X 2240x29-2.,

— lbs. wt.

2-28 tons wt.

Example (2). A train weighing 300 tons is being pulled up
an incline of 1 in 150 bi/ an engine
which is exerting a pull in the
coupling of 16 ions wt. If the
tractive resistance is constant and
equal to -^ of the weight, find the
acceleration in feet per second per
second. ^^^' ^^'

Let R be the resistance in tons wt. and P be the pull in the
coupling in tons wt.

The resultant force up the incline = (p — R - 300 sin 6) tons wt.

150'

tan 6 1 , - .

tan 6 =
.'. sinO

+ tan^ 0

If a = the change of velocity per second, i.e. the acceleration,
in feet per second per second

(15-^00-2) x32
• • "^ ~ 300

5-5 X 32

300
0*585 ft. per sec. per sec.

ELEMENTARY DYNAMICS

Example (3). In the arrangement shewn in fig. 49 heloWj the
mass of the pulleys and friction may he neglected, and the string
m,ay he taken as inextensihle.

Find the ratio of — in order that M may ascend with a constant

acceleration of 2 feet per second per second.

The forces, in absolute units, acting on the masses are shewn on
fig. 49.

Consider each of the masses separately.
Take mass M. The resiiltant force acting
upwards in the direction of the accelera-
tion = Tj — M^,

.-. T^-Mg=^ Ma (1).

For mass m, we have

m^-Tg-M/S (2).

Also, since no resultant force is re-
quired to accelerate the pulleys we have,

Ti = 2T2,
and T2 = Tg.

Now imagine M raised 1 foot, then if
the string remains taut, m must be lowered
2 feet, and we get :

/3 = 2a.
Substituting in (1) and (2), we have :
2T2 - Mg == Ma,
and mg — Tg = 2m,a.

Eliminating T^ we get,

(2m - M)^ = (M -f- 4w) a.
g = S2 ft. per sec. per sec, and a = 2 ft. per sec. per sec,
.'. (2m- M) 16r=M-h4m,
or 17M=:28w,

M 28
i.e. — =Y=.

m 17

EXAMPLES

Example (4). Clay is raised from a pit in a truck which is
wound up an incline of 1 in 10. The winding engine is such that
the pull in the rope is kept constant and equal to 550 lbs. wt. The
weight of the loaded truck is 2 tons and the total resistance is equal
to 40 lbs. per ton.

Find the velocity of the truck after it has moved from rest up
a distance of bO feet measured along the incline.

Fig. 50.

tan6' = y\y,

sin 6

= ^ nearly.

VioT 10

The resultant force up the track

= 550^ - 80^ - 4480 sm 6 . g
= (550-80-448)5^
= 22 X ^ pdls.

Let V — the velocity gained in time t.

Then by the 2nd law of momentum,

22^ = 2 X 2240 X-,

or.

22x32
2x2240'

(1).

L. E. D.

ELEMENTARY DYNAMICS

The gain of velocity will be uniform since the force applied is

constant, and therefore the average velocity = - .

Hence
and from (1)

50,
550

To"

/no

= 3*95 feet per second.

Varying Force

In order to derive our units we have, up to the present, assumed
a constant force and a constant
mass, but in many cases the
force will not be constant and
the momentum may be chang-
ing in various ways. Suppose
the force is varying with the
time in the manner shewn in
fig. 51. Here we can either Time

measure the rate of change of -^^S- ^-'^•

momentum at definite instants of time, or, ,we can find the total
change of momentum for a definite interval of time.

Let 8 (m^•) be a small change of momentum produced in time
St starting at time t. Thus for this interval of time we have the

8(mv)
It

time average of the force equal to

, and the true force at

time t is the value of — ^ — - when ht is indefinitely diminished,

I.e.

F =

d(mv)

~dr '

dt '

or if m is constant,

where a — the acceleration at the particular instant of time t.

VARYING FORCE

Again we have

B(niv)

nearly,

Bt
or F,8t = 8 (mv).

For any time T we may find the sum of each of these terms.
We get ^Fht = ^8(mv).

Now 2f8^, when St is made indefinitely small, is represented
by the area under the force-time curve from t — 0 to t = T, and
this area, therefore, represents the total change of momentum
produced by the varying force.

Suppose we find the value of

^FSt

. This obviously gives us"

the time average of the force. Let us call it F.

Then, i^. T - the total change of momentum.

We must now see how these laws are applied to practical pro-
blems, and chiefly we shall deal here with the second law, which
we will re- write in this form :

Force in Absolute units

= Change of momentum in unit time.

Accelerating Force on a Piston

Take the case of a petrol engine. Suppose we wish to find the
thrust in the connecting rod for a given position of the crank.
Let P =: the total force in lbs. wt. due to the gases on the piston
for the position considered.

y / / y / y y

Fig. 52.

5—2

68 ELEMENTARY DYNAMICS

In example (18), p. 49, we have seen how the velocity of the

piston may be found, and we can therefore draw a velocity-space

curve. From this curve we can construct an acceleration-space

curve. Let a be the acceleration, measured towards the crank shaft

of the piston, in the given position. Let Gl=the reaction, in lbs. wt.,

of the connecting rod on the gudgeon pin. Consider the motion

of the piston. The resultant force in the direction of the motion

= (P — Q cos <^) g pdls.

This must equal the rate of change of momentum.

.'. (P-Qcos<^)^= Ma,

where M is the mass of the piston and gudgeon pin in lbs.,

r Ma") 1 ,,
i.e. Q = j P - -\ T lbs. wt.

To get some idea of the possible magnitude of the quantities
involved, take the following example.

Example (5). In a jmrticular engine the mass of the piston and
gudgeon pin is 5 lbs. When the engine is running at 1500 revolu-
tions per minute the acceleration at the beginning of the working
stroke is 7500 feet per second per second. The diameter of the
cylinder is 4 inches, and the pressure at the beginning of the stroke
is 300 lbs. per square inch.

To find the value of Q at the beginning of the stroke we have
P = TT X 4 X 300 lbs. wt.
cos ^ = 1.

.-.Q^ {1200.-^^^} lbs. wt.

= 3768-1172
= 2596 lbs. wt.

Example (6). In starting a train, the engine driver opens the
throttle so that the tractive force increases uniformly from zero to
7 tons wt. during the first 20 seconds. The total weight of the train

EXAMPLES

is 400 to7is, and (he resistance at starting increases up to 17 Ihs. a
ton, and then r&inains constant for the remainder of the 20 seconds.

Find the instant the train starts to move, and draw a velocity-
time curve for the tims considered.

Let P = the tractive force, and R = the maximum tractive resist-
ance. The train will start when P = R.

In fig. 53, OAB represents the tractive force-time curve and OAC
represents the tractive resistance-time curve.

The maximum value of R = AN

17x400,
= -2240-*""^"^-
= 3 '04 tons wt.

Ayff

1
7 tons

/I ' IG

/ 1
/ |N

20 sees —
Fig. 53.

^ Time

The time at which the train begins to move is given by ON
= — =— X 20 seconds

= 8*7 seconds. -

The resultant force acting to produce momentum is given at
every instant by the difference between the ordinates of OAB and
OAC. Take T seconds from N.

Area A KG represents the total momentum generated.

ELEMENTARY DYNAMICS

li v = the velocity in feet per second, and if we measure our
mass in tons, we have

400^? = area KAG
= J KG. AG,
KG _ 7
AG ~20'

2" 2^(J ^ "^

0-014 Tl

32.

.-. 400v-i
.". V

The velocity-time curve for the 20 seconds considered is shewn
in fig. 54, where the zero for the time in the equation above is
8-7 seconds.

1 1

0 5 10

Time in Seconds
Fig. 54.

Example (7). The curve below shews the total tractive resistance
per ton at different speeds for a motor lorry on a road. The motor
of the lorry is producing a constant total tractive force of 217 lbs.
wt. If the weight of the lorry is 7 tons find the maximu7n speed
attained. Find also the time taken to increase the speed from 9 to
1 3 miles per hour.

The tractive force per ton is equal to ^y^=31 lbs. This is
shewn by the dotted line in the figure.

At maximum speed the change of velocity will be zero and
hence the tractive force = the tractive resistance.

EXAMPLES

From the figure the maximum speed = 15-1 miles per hour.
To find the time we have, if Sv is the gain of velocity in a small
time 8tj

Acceleration (a) = ^ (nearly),

a
Draw a curve representing - and v, then the area under this curve

gives the time.

|-24

4-1

i
•" 22

9.0

9 10 11 12 13 14

Speed, miles'per hour
Fig. 55.

If R = the tractive resistance in lbs. wt. per ton,
Accelerating force per ton - (31 — R) lbs. wt.

.*. Acceleration = ^— — ^^ /^ ft

2240
2240

per sec. per sec.,
70

a (31-R)x32 31

72 ELEMENTARY DYNAMICS

Using fig. 55 we find

Speed in miles

Tractive Force

Accelerating Force

per hour

per ton in lbs.

in lbs.

Accel, in ft./sec.^

7-78

22-8

8-2

8-54

23-75

7-25

9-66

11-7

26-5

4-5

15-5

13-5

27-5

3-5

28-5

2-5

The curve is shewn in fig. 56.

4l ±

on - 1 -

O J

* 1^

in- - I.-****

J __-_

10 n 12 13 14

Speed in miles per hour

Fig. 56.

Scales 1 division = imile per hour = f § foot per second.

1 division =

1 foot per sec. per sec.

EXAMPLES 73

The area under the curve between 10 miles per hour and 13 miles
per hour

= 167 squares
= 167 X ^1 X 1 seconds
= 49 seconds.
.*. The time required = 49 seconds.

Example (8). A machine-gun is mounted on an aeroplane, and
when the latter is travelling at 50 miles per hour the gun is fired in
the direction of travel for 1 5 seconds. Find the reduction in speed
of the aeroplane due to this, and the force te7iding to move the gun
relative to the aeroplane.

Total weight of aeroplane 1800 lbs. Rate of firing 600 bullets
per minute. Weight of bullet \oz. Muzzle velocity of bullets
2000 feet per second.

The reaction of the force acting on the bullets is at every instant
acting on the gun, and therefore on the aeroplane.

The time average of the force on the bullets = the change of
momentum per second.

The mass discharged per second = ^^ lb.

The change of velocity = 2000 feet per second.
/. Force = -{^ x 2000 pdls.
= 19-5 lbs. wt.

This is the force tending to move the gun relative to the aeroplane.

Now let us examine the effect of the reaction of the force on
the aeroplane. We may neglect the small change of mass due to
the discharge of the bullets.

The force of y^^^ x 2000 pdls. acts for 15 seconds. If v = the
velocity of the aeroplane, in feet per second, at the end of the 15
seconds,

1800(55ji^-.)=A.2000><15,

or, v=: 73-3 -5-2

= 68*] feet per second
= 46-5 miles per hour.

74 ELEMENTARY DYNAMICS

Sxample (9). A balloon of total mass 620 lbs. is d^'ifting hori-
zontally token 40 lbs. of sand are suddenly released. Find the
acceleration imtnediately after the sand is released.

Since initially there is no vertical acceleration, the upthrust
equals 620 lbs. wt. This upthrust is equal to the weight of the
volume of air displaced by the balloon. The latter will be practi-
cally the same before and after the release of the sand.

After the release of the sand we have the net resultant force
upwards

= {620 - 580} lbs. wt.

= 40 lbs. wt.

Let a = the vertical acceleration upwards immediately after
release of the sand, then

40^ - 580a,

40 X 32
l.e. a

580
= 2-2 feet per sec. per sec.

As the balloon begins to ascend there will be an increasing
resistance due to the motion, and the acceleration will decrease. .

Example (10). Water issuing from a nozzle^ of 2 inches dia-
meter, ivith a velocity of 50 feet per
second, impinges on a vertical wall, the

jet being at right angles to the wall. Ij / /

there is no splash find the pressure 0"""-^"-— >___ ^ -^ "^/ /

exerted on the wall. ( TD^¥^--^

\ \

If there is no splash the water will \ \

flow along the surface of the wall after ^

impact. Let F be the pressure pro- I

duced on the wall. This must equal Fig. 57.

EXAMPLES

the change of momentum per second in a direction perpendicular
to the wall

= the mass impinging per second x the change of velocity in
the direction of the force

= pav . V, where p is the density of water,
= pav^.
.'. Pressure = 62*5 x x 50 x 50 pdls.

62-5

2500

32 X 144
107 lbs. wt.

lbs. wt.

Example (11). ^ifly cubic feet of water are flowing per minute
along the fixed vane AB shewn in fig. 58 below. The speed along
the vane is constant and equal to 20 feet per second. Find the
magnitude and direction of the resultant force produced on the vane.

.^ '^

Fig. 58.

The mass flowing per second

50 X 62-5

Fig. 59.

60
= 52 lbs.
The magnitude of the momentum per second at A or B
- 52 X 20
= 1040 units.
Let OP represent 1040 units at A, and OQ represent 1040 units
at B (fig. 59).

76 ELEMENTARY DYNAMICS

The change of momentum per second is represented by PQ,

-2x1040 sin 221°

= 2080 X 0-382

= 795 units

= Force of the vane on the water.
The resultant force on the vane

= 795 pdls. -

= 24-8 lbs. wt.
The direction of the force is given by the angle a,
' a = 90-22i

= 671°.

Impulse

When two bodies impinge on one another, or a moving body
impinges on a fixed object, there is a more or less sudden change
of momentum. In many cases, it is impossible to measure the time
of the impact or the rate at which the momentum of either body
is changed, and in such cases, we can only measure the final effect
of the stress which exists between the two bodies while they are
in contact.

We have already seen that, F .t= Mv, Where F is the average
force acting, t the time, and Mv is the total change of momentum
produced.

Since we cannot measure the force itself, in such cases we have
to be content with estimating the value of the product, which
can be obtained by measuring the value of the total change of
momentum. This product is called the Impulse. There is no
special name for the units in which it is measured. In the absolute
system the unit will be the same as that of momentum, in the
gravitation system the unit will be that of momentum multiplied
b5^ g (the numerical value of the acceleration due to gravity).

IMPULSE 77

Example (12). A mass of clay weighing 5 lbs. is thrown
against a fixed wall with a velocity of 10 feet per second and sticks
on the wall. What is the impulse of the blow?

The impulse = the total change of momentum
= 5 X 10 F.p.s. units.

Here we cannot find what is the average pressure on the wall, or
the pressure at any instant, since we know nothing about the time
taken for the clay to come to rest. It may be asked, what happens
to the wall, since it receives the same impulse, but in the opposite
direction, as the clay. Theoretically there is the same change of
momentum produced, but the mass of the wall and the earth, to
which it is fixed, is so great that the velocity is negligibly small.

Magnitude of impulsive force. It may be noted that,
generally, the time of the impact will be very short, and hence
we may expect that the time average of the force will be large.

To give some idea of this, we may take two cases which have
been investigated.

(1) Two ivory billiard balls impinge with equal velocities of
8 feet per second. It has been shewn that the time of the impact
is 4 oVo ^^ ^ second, and that the maximum total pressure between
the balls, which occurs at the instant when they are at rest, is
equal to 1300 lbs.

(2) A leaden bullet weighing 0*03 lb. hits a steel target with
a velocity of 1800 feet per second. The time required to stop
the bullet is about ysJxjT^ second.

Hence the time average of the force on the target

0-03 X 1800 ^,
= 1 pdls.

18000

0-03 X 1800 X 18000

= 22i0T^2 '^^^"'-

= 13'5 tons wt.
In this case the pressure is probably nearly uniform during the
stopping of the bullet.

In cases where the time during which the force is acting is very

78 ELEMENTARY DYNAMICS

small we speak of the impulse as a blow. As we have already seen,
the time may be very short indeed and the average force will
therefore be very large, since the product of the two is equal to
the definite and finite change of momentum which occurs. In
such cases, we can generally omit the effect of steady forces which
may also be acting during the time of the blow, since the changes
of momentum produced by them in the short time will be negli-
gibly small.

Take the example of the bullet given on p. 77, and suppose it

is fired vertically downwards at the target. By neglecting the

steady force, namely the weight, in considering the impact, we are

only neglecting 0*03 lb. wt. in comparison with 13 "5 tons wt.,

^ 0-03 X 100 1

I.e. we are making an error or ^,,^ — tttt or tt^t^ttp^ per cent.

2240 X 13-5 10000

Case of two bodies impinging. When two bodies impinge
we may if we like consider the two as a single system. Hence in
all cases of bodies impinging there is no total change of momen-
tum due to the impact since there is no external force acting, or
as it is often stated, the total momentum before impact equals the
total momentum after impact. Another way of looking at the pro-
blem is to consider each body separately. Since the total change of
momentum is zero, one body must lose exactly the same quantity
as the other body gains. But the forces acting on each body are
proportional to the change of momentum of each body, and since
these are equal in magnitude and opposite in sign, the forces on the
two bodies are equal and opposite. This is true whatever be the
length of time during which the bodies are in contact. The two
equal and opposite forces which are called into play together form
what is called a stress.

Equilibrium. The second law of momentum is quite general
in its application, and we may use it to investigate the forces acting
on a body or system of bodies which is in equilibrium, i.e. at rest.
In this case, the rate of change of momentum is zero, and there-
fore the resultant force, on the system or any part of it, must also
be zero. By applying the law, firstly to the whole and afterwards

IMPACT

to the different parts, we can shew that, in every case, the inter-
action of any two bodies, or parts of a body which are connected
or in contact, consists of a stress in which the two components
are always equal in magnitude and opposite in sign. This fact is
often stated in the form, " to every action there is an equal and
opposite reaction," and is called Newton's third law of motion. As
we have seen, it really follows directly from the second law of
momentum. It forms the basis of investigation of the internal
and external forces in statical problems.

Example (13). A railway truck of mass 12 tons, moving with
a velocity of 6 whiles per hour, impinges on another truck weighing
10 tons, and moving in the same direction with a velocity of 2 miles
per hour. When impact occurs the two trucks are automatically
coupled together. Find the velocity of the trucks after impact.

12 tons

10 tons

Fig. 60.

Let F . i5 be the impulse between the two trucks, and let v — the
final velocity in the original direction of motion.

For the 12 ton truck,

F.^ = 12(6-'y).
For the 10 ton truck,

F.«5 = 10(v-2).

.'. 72-12^=.10v-20,
or v = ^^ = 4*17 miles per hour.

The impulse of the blow =F .t

.^,^ . 2240x88
= 12(6-.) X— ^^-

= 7220 F.p.s. units.
Or, Momentum before impact = Momentum after impact,
i.e. 12 X 6 + 10 X 2 = (12 + 10) V,

.". V = 4:'l7 miles per hour.

ELEMENTARY DYNAMICS

Sxample (14). A shell of mass m, lbs. impinges obliquely on
a fixed target and ricochets. If the striking velocity is u feet per
second inclined at an angle 6 to the normal at the point of contact,
and the velocity of 7'ebound is v feet per second inclined at an
angle <{> to the 7iorm,al, find the I'esultant blow on the target.

Resolve the blow into two components, P and Q say, tangential
and normal to the plane of the target.

normal

/////////////////////A

^////////^////y//////

• \

a \

Tangential^
Normal.

Fig. 61.

p = m,u sin 6 — mv sin ^
= m{u sin 6 ^v sin ^).
Q = mu cos 6 + Tnv cos <^

= m (u cos 6 + V cos <t>).

Resultant (R) = x/p^ + Q^

= m {u^ sin^ 9 + v^ sin^ <j5> — 2uv sin 0 sin <f>
+ u^ cos^ 6 + v^ cos^ <^ + 2uv cos 6 cos <^p

= m {^2 + vV 2uv . cos (0 + <f>)}^.

This makes an angle a with the normal such that

P
^ tan a = -

u sin 6 — v sin (f>

u cos d + v cos ^ '

Or using vectors : The change of momentum equals the vector

difference of mu and mv.

IMPACT 81

This is given by OB in fig. 62, where OA represents the initial
TO omentum mu of the projectile, and BA represents the final mo-
mentum mv.

The impulse of the blow = OB

= m {u^ + v^ ^ 2uv cos (0 + <j>)]^y
and the direction may be calculated from the figure.

It may be noted that generally we shall not be able to deter-
mine the values of v and <^, if we are merely given u and 0.

Fig. 62.

Example (15). A hammer head weighing 1*2 lbs., arid moving
with a velocity of 16 feet per second, strikes a nail of Q'l lb. weight
and drives it | inch into a piece of wood. Assuming no rebound
of the hammer and the resistance to penetration of the nail constant
find its magnitude.

Let V = the velocity in feet per second with which the nail and
hammer move immediately after the impact. Then by the con-
servation of momentum,

l-2xl6-(l-2 + 0-l)v;
16x1-2

•'• ^ = -T3-

= 14'8 feet per second.
Since the resistance to penetration is constant the time rate of
change of velocity will be uniform.

L. E.D. 6

82 ELEMENTARY DYNAMICS

.'. the average velocity of penetration = — ^r—

= 7'4 feet per second,

and the time of penetration = —^ — -— second.

24 X 7-4

The resistance = rate of change of momentum

= l-3x 14-8x24x7-4pdls.

1-3 X 14-8 X 24 X 7-4 „
= 32 l^^-^^-

= 1071bs. wt.

The magnitude of the blow between the hammer and the nail
= the change of momentum of the hammer

= 1-2 (16 -14-8)

= 1-2 X 1-2

= 1-44 F.p.s. units.

Example (16). A bar of inetal 3 inches wide, 1 inch thick, and
1 5 feet long is to he passed through a rolling mill, the area of section
being thereby reduced by \. The diameter of the rolls is 12 inches
and they rotate at 200 revolutions per minute. Assuming that the
speed at which the bar leaves the rolls is the same as that of the
surface of the rolls, shew that the tangential impulse on the rolls
when the bar is first gripped is 1180 absolute f.p.s. units. Shew
also that, apart from the foirce required to overcom,e friction and
to do the work of deforming the metal, a tangential force of
6 '4 lbs. wt. is required to maintain the motion.
The density of the metal = 4:80 lbs. per cubic foot.
The speed of the surface of the rolls

_ 27r X 200

"60x2

= 10-47 feet per second

= the speed at which the bar leaves the rolls.

EXAMPLES. CHAPTER III 83

Since the same volume of metal passes in and out of the rolls
in a unit time, the velocity at which the bar is fed into the rolls

= f X 10-47

= 7*85 feet per second.
At starting, the whole bar is suddenly given this velocity,

I.-, the impulse = 480 x ^f ^ x 15 x 7*85
= 1170 abs. F.p.s. units.
The steady tangential force required to maintain the motion
= the change of momentum per second
^ = the mass per second x the change of velocity
i 480x3x7-85 ,,^,^ ^„^,
; = 144 X (10-47 -7-85)

; = 204 pdls.

= 6-4 lbs. wt.

Examples. Chapter III

1. What do you understand by momentum ? Define force in terms of this
quantity.

In 5 seconds a car weighing 30 cwt. changes its speed from 20 feet per sec.
to .27 feet per sec. ; what uniform force must have been acting if it requires
50 lbs. per ton to just move the car steadily on the track ? If the power is
cut off at the end of the 5 seconds, what constant deceleration will result, and
how far will the car run before coming to rest ?

2. Express a force equal to the weight of 1 ton in (1) absolute f.p.s. units
[(pouudals), (2) absolute c.g.s. units (dynes).

1 inch = 2 -54 centimetres. 1 pound = 453-6 grams.

Acceleration due to gravity = 981 cms. per sec. per sec.

3. A car weighing 12 tons is ascending a slope of 3 in 100 against a
frictional resistance equal to 1 per cent, of its weight. What pull is required
in order that the car may increase its velocity by 1-5 miles per hour in one
second ?

4. Explain fully how you infer that at a given place on the earth a body's
weight is proportional to its mass.

6—2

84 ELEMENTARY DYNAMICS

Two weights W, W are connected by a light string passing over a light
pulley. If the pulley moves upwards with an acceleration equal to that of
gravity shew that the tension of the string is

4WW'
W+W'

5. A cage weighing 2000 lbs. can be raised or lowered in the shaft of a
mine by a cable. Find the tension in the cable (1) when the cage is rising
or falling with a uniform velocity, (2) when it is rising with an acceleration
of 2 feet per second per second, and (3) when it is falling with an acceleration
of 5 feet per second per second.

«. A tram-car weighing 17,000 lbs. is ascending a gradient of 1 in 20 with
an acceleration of 1-2 feet per second per second. If the resistance is equal
to 0*011 of the weight find the tractive force on the rails. If the tractive force
and resistance remain the same, what would be the acceleration when
travelling down a gradient of 1 in 30 ?

7. A truck is running on the level with a velocity of 20 miles per hour
when the wheels are locked by the application of the brakes. If the coefficient
of sliding friction between the wheels and the rails is 0*08, for how long and
for what distance does the truck move before coming to rest ?

Find the corresponding time and distance if the truck is moving down a
slope of 1 in 100.

8. Find the magnitude and direction of the force which, acting on a mass
of 4 pounds which is moving with a velocity of 8 feet per second, will in
4 seconds cause its velocity to be 8 feet per second in a direction at right
angles to the original direction of motion.

9. A fire engine is delivering 312 gallons of water per minute through a
nozzle of 1|- inch diameter, fixed to the engine and inclined upwards at 30°
to the horizontal. Find the vertical and horizontal reactions produced on
the engine.

1 gallon of water weighs 10 lbs. 1 cubic foot of water weighs 62*5 lbs.

10. A motor-car has a machine-gun mounted on it, and when the car is
at rest the gun is fired horizontally and straight to the front for 15 seconds.
Find the velocity of the car at the end of this time.

Bate of firing, 600 bullets per minute ; muzzle velocity of bullets, 2400 feet
per second; weight of bullet, J oz. ; weight of loaded car, 18 cwt.; resistance
to motion of the car, 16 lbs.

11. A mass of 10 lbs. is acted upon by a force P which varies with the

time t according to the law, P=2 sin —t lbs. weight, where t is in seconds,

Plot the force-time curve from i=0 to t = 30, and from it deduce the velocity-
time curve, being given that the velocity is zero when the time is zero.

EXAMPLES. CHAPTER III

12. A boat is sailing at a constant velocity of 6 knots before a following
wind of twice the velocity of the boat. The sail area is 450 square feet, and
may be assumed a plane area perpendicular to the direction of motion of the
boat. Find the total resistance to motion of the boat.

One cubic foot of air weighs 0*08 lb.
One knot equals 6080 feet per hour.

13. The relation between the total tractive resistance and the speed for a
locomotive and train weighing 246 tons is given below. The total load on
the driving wheels of the locomotive is 33 tons, and the coefficient of sliding
friction between the wheels and the rails is 1/5, The train is accelerated from
Test as rapidly as possible on a level track. Plot a curve shewing the value of

r : — for different speeds, and from it estimate the time for the train

acceleration

to attain a speed of 45 miles an hour.

Speed in "feet
per sec.

7-5

Eesistance in
lbs. per ton

10-9

7-3

6-6

6-5

7-a

8-4

9-9

11-8

13-9

16-3

19-3

14. A torpedo boat fitted with hydraulic propulsion took in 1 ton of water
per second in a direction perpendicular to the direction of motion of the boat,
and discharged it horizontally astern with a velocity of 37*25 feet per second,
relative to the boat. With this discharge the steady speed of the boat was
12-6 knots. Find the resistance to motion of the boat.

15. A locomotive while travelling at 40 miles per hour scoops up 760
gallons of water from a trough in a length of 500 yards. The water enters
the scoop horizontally and is discharged into the tank of the locomotive
vertically downwards. The delivery pipe has a diameter of 4 inches. Due
to the change of momentum of the water, find (1) the horizontal resistance
to the train's motion, (2) the reduction of pressure on the rails.

16. If the resistance to the motion of a train running at V miles per hour
be 0-3(V + 10) lbs. per ton, find, graphically or otherwise, how long it will
take to reduce the speed from 50 to 30 miles per hour in the case of a train
running freely on the level. How far will the train run in the time ?

17. A balloon weighing 800 lbs. is descending with a constant acceleration
•of 1 foot per second per second, when 50 lbs. of ballast is suddenly released.
Find the magnitude and direction of the acceleration immediately after the

86 ELEMENTARY DYNAMICS

18. It is found that a bullet weighing 0*4 ounce and travelling at 2400 feet
per second will just penetrate 36 inches of a certain wood. If a similar bullet
with the same initial velocity is fired through 18 inches of the same wood,
what will be the velocity when it emerges, and what is the force of resistance
to penetration ? Neglect the spin of the bullet.

19. Find the minimum plan area a parachute may have to enable a man
weighing 10 stone to descend vertically at a final speed of not more than
25 feet per second. The weight of the parachute may be taken as 10 lbs.,
and the weight of one cubic foot of air 0*08 lb.

20. The velocity of flow in a water main of 6 inches diameter is 5 feet,
per second. At one place the main is bent through an angle of 30". Find
the resultant force on the bend.

21. The inlet valve of a petrol engine is held on its seat by a spring which
exerts a force of 1^ lbs. If the valve opens downwards, weighs 3 ounces, and
has a lift of 0*2 inch, find the time occupied in closing, and the velocity at
the instant of closing.

22. A plumb-line, 7 feet long, is suspended from the roof of a railway
carriage which is travelling along a straight level track. The plumb-bob is
observed to be displaced 3 inches from the vertical through the point of sus-
pension in the opposite direction to the direction of motion of the train.
What is the acceleration of the train?

23. The acceleration of the reciprocating parts of a steam engine is given
in feet per second per second by 120 cos ^ + 20 cos 2^, where d is the angle
which the crank makes with the inner dead centre. If the weight of the
reciprocating parts is 8 tons, find the accelerating force required in the
direction of the line of stroke for the positions given by values of d, 0°, 45°,
90°, 135°, 180°.

24. A small mass is suspended by two equal strings each inclined at 30°
to the vertical. Shew that if one of the strings be suddenly cut the tension
in the other is immediately increased by 50 °/o-

25. Using a pile-driver with a hammer weighing 1 ton, it is found that a
pile weighing 4 cwt. is driven 5 inches into the ground when the hammer
falls a distance of 10 feet before striking. Assuming that there is no rebound
of the hammer, and that the mean resistance and the drop remain the same,
find the distance which the pile would be driven if the weight of the hammer
were increased by | ton.

EXAMPLES. CHAPTER III

26. The muzzle velocity of an eigh teen-pounder shell is 1600 feet per
second, and the gun recoils in its cradle a distance of 40 inches. The force
exerted on the gun during recoil is 1^ tons weight. Find the mass of the gun
and its momentum at the beginning of recoil.

27. A sledge hammer of mass 14 lbs. falls freely from a height of 4 feet
on an anvil and rises 6 inches after the blow. If the time of contact be ^^jt
of a second, find the mean force between the hammer and the anvil.

28. Two masses A and B, weighing 3 and 4 kilograms respectively, are
connected by a light inextensible string which passes over a light frictionless
pulley. The strings are vertical and just taut, the mass A resting on the
ground and the mass B being supported 1 metre from the ground. If the
support of B is suddenly removed, find (1) the time before B first reaches the
ground, (2) the impulse in the string when B is first jerked off the ground,
(3) the greatest height to which B subsequently rises.

29. A freely suspended steel bar, 1^ inches diameter, was subjected to a
pressure on the end by means of a charge of explosive. The pressure varied
with the time in the manner shewn by the table below.

Find the magnitude of the impulse produced. If the bar weighs 12 lbs.
find the velocity acquired.

Pressure, tons per
sq. inch

£0

37
5

8
20

3
30

0
40

Time, millionths ' ^
of a second

2-5

30. Two equal masses A and B, moving in the same straight line and in
the same direction, collide. Before the collision the distance of A behind B
is diminishing at the rate of 6 feet per second and after the collision the
distance of B ahead of A is increasing at the rate of 2 feet per second ; also
at one instant during collision both are moving at 10 feet per second. Find
their velocities before and after collision.

CHAPTER lY

Angular Momentum

In the chapter dealing with motion we saw that when a body was

rotating about an axis it was convenient to use the term angular

velocity to express the motion.

We must now investigate how

forces ma}' produce a change in

angular velocity. In doing this,

we do not have to introduce any

new principles, but merely to

apply those we have already

stated. We will first of all deal

with cases where the rotation

is about a fixed axis, and where

the forces applied are in a plane

perpendicular to the axis' of

rotation. ^ig- ^3.

Suppose we have a body, as shewn in fig. 63, which is free to
rotate about a fixed axis through O perpendicular to the plane
of the paper. Let this body be acted upon by a set of forces Pj,
P2, etc.

Consider any small particle A of mass m situated at a distance
r from O. If at a particular instant the angular velocity of the
body is w, then the linear velocity of A will be tor in a direction
perpendicular to OA.

The particle at A will also have an acceleration of magnitude
cD^r towards O as found in Chapter II.

Now there will be certain forces acting on the particle at A, and
we may conveniently consider these as having a total component,
/ say, perpendicular to OA and ^, say, along AO.

We will deal first of all with the force /^ acting perpendicular

ANGULAR MOMENTUM 89

to OA, leaving the force p for future consideration. It is obvious
that the force p will have no effect on the rotation.

Let the angular velocity at the instant considered be increasing

by an amount Sw in a time U, or at the rate -7- . The linear ac-
celeration of A perpendicular to OA is r — .

Applying the second law of momentum to the particle at A and
measuring forces in absolute units we have

f= the rate of change of momentum,

i.e. f— m.r .-J- •> where m is the mass of the particle.

Multiplying by r we have

•^•^='"'•'•5 (^^-

Now we may imagine the whole body to consist of an infinite
number of particles such as the one at A, and for each particle we
shall get an equation similar to equation (1).

Adding the right and left-hand sides of all these equations we
may express the sum thus

But -J- is the same for each of the particles, and hence we may

write diM

-^ dt

Ifr = the algebraic sum of the moments* of all the forces acting
on the particles about O.

These include both external and internal forces. But the in
ternal forces balance, and hence

2/r = the algebraic sum of the moments of all the external forces
about the axis O.

This is often called the torque or turning moment on the body
and is denoted by the symbol T.

* The moment of a force about a point is equal to the product of the magni-
tude of the force and the perpendicular distance from the point to the force.

2/r — ^[ tnr^ . -^ ) .

90 ELEMENTARY DYNAMICS

%mr^ = the sum of the products of the mass of each particle
multiplied by the square of its distance from O.

This is called the moment of inertia* of the body about the
axis O, and is usually denoted by the symbol I.

The above equation may be written

dt
Now just as the mass multiplied by the linear velocity is called
linear momentum or more commonly momentum, so the product
Id) is called the angular momentum, or sometimes the moment
of momentum.

I -^ is the same thing as \r^ , i.e. the rate of change of angular
momentum.

Using absolute units for the torque we may write the second law
of momentum as applied to rotation thus :

Torque = Rate of change of angular momentum.
If the torque is measured in gravitation units we must write
™ _ Rate of change of angular momentum

Varying Torque

If the torque is varying, we can, as we did in dealing with linear
momentum, either measure the rate of change of angular momen-
tum at definite instants of time, or, we can find the total change
of angular momentum for a definite interval of time.

Suppose the torque at different instants of time is given by the
curve in fig. 64.

We have. Torque, T = —j- .

* Inertia may be defined as that property of a body in virtue of which it
will not change its state of motion or rest unless acted upon by some external
force. For translational problems it is measured by the mass.

VARYING TORQUE

If we take a narrow strip as shewn shaded in the figure, the
area = T 8^

= 8{l<o),
i.e. the area represents the change of angular momentum during
time St.

.*. For any time 6 from the starts the area under the torque-time
curve from < = 0 to t — 9, measures the total change of angular
momentum.

Time

Fig. 64.

If the moment of inertia is constant, then the area under the
curve represents the total change of angular velocity.

As in dealing with linear momentum, we may find the time-
average of the torque. This is given by the area under the torque-
time curve divided by the length of the base. Thus, .

The time-average of the torque x the time

= the total change of angular momentum.

Impulsive Torque

In many problems we cannot measure the torque at each instant
of time, either due to the total time being so short, or for other
reasons. In such cases, all we can do is to measure the total effect
produced by the torque during the time concerned. This is given
us by the total change of angular momentum, and gives us the

ELEMENTARY DYNAMICS

product of the torque and time, or the area under the torque-time
curve. This we may call the impulsive torque but we must
remember that it is not a true torque but a product of force^ dis-
tance and time, i.e. a product of an impulse and a distance.

Example (1). Two masses M and m are supported by an in-
extensible string which passes over a pulley as
shewn i7i Jig. 65. Find the acceleration.

Let I be the moment of inertia and r be the
radius of the pulley.

The forces acting are shewn on the figure.

Let a = the acceleration of M downwards.

We have for M,

Mg-Tj=Ma (1).

Form, T^-mg = ma (2).

For the pulley the resultant torque

= (Ti-T,)r.

The angular acceleration

.-. (T,-T,)
Adding (1) and (2) we get,

,(3).

Fig. 65.

From this and (3),

(T2 - Tj) =3 (M + m) a - (M - 7n) g.

la /

-72=(M+»^)a-(M-m)5r,

I.e.

The acceleration = ~ JL .

M ->-m +

We can also find Tj and T^ if desired.

It will be noted that here T^ and T2 are different, since the
pulley requires a turning moment to change its angular velocity.

EXAMPLES 93-

In example (3) on p. 64, we assumed that the moments of
inertia of the pulleys were negligibly small, and in that case we
had the tensions in the string on the two sides equal.

Example (2). A gas engine works against a constant torque of
525 Ihs. ft. If the gas supply is suddenly cut off and the resisting
torque remains the same, find in how many revolutions the speed
will falljrorn 250 to 240 revolutions per minute.

The fiy wheel of the engine has a mass of 3 tons and it may he
considered concentrated at a radius of 3 feet.

Since the resisting torque is constant the speed will decrease
uniformly.

rPk ^' 4J.U A 250 + 240

Ine time average of the speed ==

= 245 revolutions per minute.
If t is the time in seconds for the fall in speed, then the angular

retardation = — revolutions per minute per second

207r ,.
= -^TT- radians per sec. per sec.

The torque = the change of angular momentum per second.

525 X 32 = 3 X 2240 x 3= x

t =

3t'
9 X 2240 X TT

525 X 32
= 3-77 seconds.
.'. The number of revolutions required

= 15-4 revolutions.

Example (3). A flywheel, of moment of inertia 900 Ih. ft. units,
has a fan attached to its spindle. It is rotating at 60 revolutions
per minute ivhen the fan is suddenly immersed in water. If the

ELEMENTARY DYNAMICS

resistance of the water he proportional to the square of the speed, and
if the speed he halved in 3 minutes, find the initial retarding couple.

The retarding torque = Aw^, where \ is a constant to be deter-
mined, and 0) is the angular velocity in radians per second.

By the second law of momentum we have,

, dia

-Xc

dt '

where I is the moment of inertia.

_dt^_I 2

dlO X. CO'

This gives,

Plot-^,i.e.— -=— -
doi retardation

and speed curve, using any suitable

scale for - ^- . This is done in fiff. 66.
do) ^

■"

"■

■""

■o

£.

*■

■~"

3ir

speed, radians per second
Fig. 66.

2Tr

EXAMPLES 95

The scale for speed is, 1 division = ^ radians per second.

The scale for ^ — ; — is such that its value for o) equal to

retardation

27r radians per second is represented by 5 divisions,

i.e. 6 divisions = T- x — — ,

A 4o7r^

.*. 1 division =

207r2X •
Now if we take a narrow vertical strip between the base and

the curve the area = ^r- x Sw

OO)

= the time to change speed by amount Sa>.
. . The area under the curve between w =77 and <a = 2Tr represents
180 seconds.

The area= 198 small squares

= 0-157 Y seconds.

.-. 0157 X 5^ = 180,
A

i.e. A - 0-785.

The initial retarding couple = 0*785 x 47r^ pdls. ft.

= 0 967 lb. ft.

Example (4). A tope is being wound on to a drum from a coil
on the ground, as shewn in Jig. 67. The rope weighs 1 Ih. a foot
and the radius of the drum, which may also he taken as its radius
of gyration"^, is 2 feet. The mass of the drum is 80 lbs. When the
drum starts there is one complete turn of rope on it, and it
accelerates uniformly at the rate of 20 revolutions per m,inute per
second. Find the couple at the end of 4 seconds. Neglect the
diamieter of the rope.

* For definition of radius of gyration, see p. 105.

ELEMENTARY DYNAMICS

Let Ti and Tg be the tensions in the rope at the points shewn
in the figure.

The weight of the hanging part of the
rope will be 20g pdls.

The average speed during the first
4 seconds

= 40 revolutions per minute.

The mass of the rope wound on in
4 seconds

40x4

X 47r lbs.

The mass of the rope on the drum at
the end of 4 seconds
= (f +l)47rlbs.

= -^r- lbs.

Fig. 67.

The moment of inertia of the drum and rope on it

447r
= 80 X 22 + ^ X 22

= 504 lbs. ft.2
The speed of the rope at the end of 4 seconds

80x47r. ^ ,

= — TTTT — leet per second.
oO

Let C = the couple in pdls. feet units on the drum. For the
drum we have,

C-2T, = 504x2-?A?5 (1)

= 1055 pdls. ft.
For the portion of the rope between the drum and the ground,

T,-20^-T, = 20x^x2 (2)

= 84 pdls.

EXAMPLES

Considering the coil, the pull Tj has to supply momentum to
the new portion of the rope which is being jerked from the coil.

Ihe mass per second = — ^ — lbs.

The change of velocity = — ^tt" ^^®^ P^^ second.

T, =

From (2) and (3),
From (1),

_ /80 X 47ry

^"1^60"-; ■

= 280 pdls.

T2 = 84 + 280 + 640
= 1004 pdls.

C - 1055 + 2008
- 3063 pdls. ft.
= 96 lbs. ft.

.(3)

Example (5). In a press for stamping medals a flywheel is
keyed to a vertical screw which rotates in a nut fixed to the frame.
If M is the mass, k the radius of gyration of the screw and wheel,
n the number of threads per unit length, r the m^ean radius of the
screw, and fx the coefficient of friction, find an expression for the
vertical acceleration of the screw and wheel when left to themselves.

Imagine one complete turn of the thread stripped off from the
nut and opened out as shewn in fig. 68.

•^l^

1
n

^^^^^^v'

< 2i,

-->

Fig. 68.

L.E.D.

98 ELEMENTARY DYNAMICS

Let R = the normal pressure between the nut and screw. The
friction fxR will act at right angles to this and so as to oppose the
motion.

The resultant force vertically downwards

= M^ - R cos 0 — fiR sin 0.
The horizontal force acting at a radius r

= R sin 6 — fxR cos 0.
Let a — the linear axial acceleration, and A ::= the angular ac-
celeration. Then

an 1

A 27r 27m*
We may apply the second law of momentum to the two com-
ponents of the motion and we have

M^ — R (cos 0 + fjL sin 0) = Ma (1),

and R(sin^-/xcos^)r= MF. A (2).

Eliminating R from (1) and (2) we get

Mk^ . A /cos 0 + iJi sin ^\

Mo . ( -T—^ — ^ I = Ma.

r \sin d — fx cos 0/

.'. The vertical acceleration

ik^ 27rn /27rnr + /aX ]
\ r \2'7rnr — fxJ J

If /x, = 0 this reduces to - — p

— — + 1

Example (6). In an inward-flow water turbine the water
enters the wheel at a radius of \\ inches^ with a velocity of ^Q feet
per second, and is inclined at 10° to the tangent at the point of
entry. The water leaves the wheel with a velocity of 10 feet per
second at a radius of 5 inches, and in a direction inclined hack-

EXAMPLES

wards at 50° to the tangent at the point of exit. If 400 cubic feet
of water per minute pass through the wheel, and there is no shock
at entry, find the turning moment produced.

Fig. 69.

The mass of water per second

400x62-5

-417 lbs.
The moment of momentum initially ~ 450 x 36 x OC.

„ „ finally =-417xlOxOD,

where OC and OD are the perpendiculars from O on to the actual
directions of motion of the water at entry and exit.
The turning moment

= Change of moment of momentum per second
= 417 X 36 X OC + 450 x 10 x OD
= 417 {36 X ii cos 10° + 10 X 3% cos 50°} pdls. ft.
= ^y. {36 X 1-1 X 0-985 + 10 x ^^ x 0-643} lbs. ft.
= 458 lbs. ft.

7—2

100

ELEMENTARY DYNAMICS

500 r.p.m.

Fig. 70.

Example (7). Two toothed wheels, of ■moiments of inertia 4
and 2 lbs, ft. units respectively, are
arranged on -parallel shafts. The
first wheel is rotating uith a speed
of 500 r.p.m. when the second wheel,
which is initially at rest, is suddenly
made to mesh with it, by sliding the
second tvheel along tlie shaft. If the
number of teeth on the wheels are 28
and 20 respectively, find the speed of each wheel immediately after
they are in mesh.

When the wheels are in mesh, since the circumferences have
the same speed, we have

r^~ 20~n^'
where n^ and n^ are the speeds, in revolutions per minute, after
meshing.

Let P = the tangential component of the impulse between the
teeth of the wheels during the period of meshing.

|(500-«.).

For (1),
For (2),

P . ri = 4

P.i-

0 2t

i.e.

ri 2 (500 - ni)

1000 - 2n,

i.e.
and

or
and

28
20

^2= l-4ni,
1000-2??.! = 1-42x^1,
1000

3-96
1400
3-96

253 r.p.m.
354 r p.m.

MOMENT OF INERTIA

101

Moment of Inertia

The values of the moment of inertia of various bodies can be
obtained by employing the integral calculus to effect the sum-
mation required, or we may employ certain graphical methods.
Here we will state the values for a few bodies which are frequently
met with in rotation problems.

^ BoiUh's Rule for Moment of Inertia

The following simple rule for rapidly obtaining the moment of
inertia of many bodies dealt with in dynamics will be found very
useful.

The moment of inertia about a symmetrical axis through the
centre of gravity

Sum of the squares of the perpendicular semi-axes
3, 4, or 5

3, is to be used for a rectangular or square body.

4, is to be used for an elliptical or circular body.

5, is to be used for an ellipsoidal or spherical body.
Let us see how this is applied :

Rectangular plate

Suppose we want the moment of inertia about ox. Here the
only axis perpendicular to the axis con-
sidered is the 2/-axis, the thickness being
supposed negligibly small.

/l2

Similarly,
and

rVMA^

where oz is the axis perpendicular to the
plane of the figure.

K-- b--

Fig. 71.

102

ELEMENTARY DYNAMICS

Circular disc or cylinder
Let r = the radius.

_ M (r^ + 7-2) _ Mr2

For a thin disc

Sphere

I =!^'-i

L.= M

7^ + r

Moment of Inertia about any A ocis

For determining the moment of inertia about an axis which
does not pass through the centre of gravity we may employ the
following :

The moment of inertia of a body about any axis is equal to the
mom,ent of inertia about a parallel axis through the centre of
gravity plus the product of the mass and the square of the distance
between the two axes.

Fig. 73.

Referring to fig. 73, let CD be an axis through the centre of
gravity parallel to the axis AB. Then

MOMENT OF INERTIA

103

= 2m (z + hy

= 2mz^ + ^mh^ + Sm . 2hz
= ^m^ +Mh^+2h '%mz,
where M = the total mass.

Now %mz^ = IcD >
and "Xmz = the moment of the mass of all the particles about CD
= zero, since the centre of gravity lies on CD. See p. 117.

Plane Lamina
Another proposition which is often useful in determining
moments of inertia is as follows :

In the case of a plane lamina or flat body in which the thickness
is negligibly small the moment of inertia about an axis perpendicular
to the plane is equal to the sum, of the moments of inertia about any
two axes in the plane mutually at right angles to each other and
intersecting on the first axis.

Fig. 74.

104 ELEMENTARY DYNAMICS

Take the axis of oz perpendicular to the plane, and the axes ox
and 07/ in the plane.

If m is the mass of a particle A, fig, 74, then

= ^m 0A2

= 2r/i (ON^ + NA2)

= "^mx^ + ^my^

Hoop about a diameter

The moment of inertia I^^ for a thin-rimined hoop of radius r
and mass m equals mr^.

But lox = loy,

Hollow cylinder about its axis

Let R and r be the external and internal radii respectively.
Let h be the height, and p be the density.

Treat the hollow cylinder as the difference between two solid
cylinders of radii R and r respectively.

R2 ^ r^

I = p-rrR^h ^ — pirr^h -

p.(R^-^).(?!±l)

R^ + r'

where M is the mass.

Thin rod

This may be treated as a thin rectangular plate in which two
of the axes are negligibly small.

Let I be the length of the rod and m be the mass.

RADIUS OF GYRATION 105

The moment of inertia about an axis through the middle

= 7n.' (Routh'
o

s rule)

-12-

The moment of inertia about one

end

Radius of Gyration

In the case of a thin-rimmed hoop all the mass may be con-
sidered to act at the same radius, viz. the radius of the hoop r, and
hence the moment of inertia about an axis through the centre
perpendicular to the plane = mr^, where m is the total mass.

In the case of a body where the mass is at different distances
from the axis of rotation, it is often convenient to imagine it re-
placed by a simpler body having all the mass concentrated at the
same radius. This body must have the same dynamical effect so
far as rotation is concerned as the original body.

If M = the mass of the original body,

I = the moment of inertia of the original body,

and k = the radius at which we may imagine the whole mass to
be acting, then

U¥ = I,

or k^ = — .

k is called the radius of gyration.

For example, the moment of inertia of a flywheel = 6750 f t.^ lbs.

106

ELEMENTARY DYNAMICS

The mass of the wheel = 1100 lbs.

.'. The radius of gyration = sj^\

50
TOTT

= 248 feet.

Example (8). The figure shews a sheave of an eccentric, the
sheave consisting of a steel disc of radius 6 inches and thickness
1 irtch, with a hole of 4 inches diameter, the centre of which is
3 inches from the centre of the disc. Find the moment of inertia of
the sheave about the axis of the hole.

the

Fig. 75.

The moment of inertia about A = the moment of inertia of the
complete disc about B + the mass of the complete disc x AB^
moment of inertia of the part removed for the hole about A

= ^^'^^^12 1 2 H4)r^"^36^12^^>
where p is the density, i.e. 480 lbs. per cubic foot,

_307r_107r

= 5-8 lbs. ft.2

EXAMPLES. CHAPTER IV 107

Examples. Chapter IV

1. Define the term moment of inertia as applied to a revolving mass.

Awheel of mass 50 lbs., rotating at 400 revolutions per minute, is brought
to rest in 10 seconds by the application of a brake. Find the average
frictional torque during the 10 seconds, if the whole mass of the wheel may
be considered as situated at 18 inches from the axis.

2. A wheel of 200 lbs. weight and radius of gyration 2 feet is mounted on
smooth bearings. The axle is 3 inches in diameter, and a weight of 40 lbs.
hangs from a string wrapped round the axle. With what acceleration will
the weight fall ?

If after the weight has fallen through 6 feet from rest the string is cut and
a retarding tangential force of 20 lbs, applied on the axle, how many revolu-
tions will the wheel make before coming to rest ?

3. A uniform circular disc of weight W is mounted with its axis horizontal
and a light string passing over it carries weights M and M + m, at its ends.
If the string does not slip on the disc, shew that when the system is in motion
the acceleration of the weights is equal to

2m
^4M+2m + W

Find also the values of the tensions in the two portions of the string, and
the resultant thrust on the bearings of the disc.

4. The axle of a flywheel is 6 inches in diameter. A weight of 20 lbs.,
hung at the end of a fine flexible wire which is wrapped round the axle, is
just sufficient to overcome friction, and when put in motion, falls with a
constant velocity. When an additional 50 lbs. is hung on the wire it descends

■mth a constant acceleration equal to :^. Find the value at the end of

10 seconds, of (1) the velocity of the weight, (2) the angular velocity of the
flywheel. Find also the moment of inertia of the flywheel.

5. In the previous question if instead of the additional 50 lbs. being hung
on the wire, the weight of 20 lbs. is pulled downwards with a steady force of
50 lbs. wt., what will be its acceleration ?

6. A propeller and shaft, the moment of inertia of which is 500 lbs. ft.2
units, is observed to slow up in the manner given in the table below. Find

108

ELEMENTARY DYNAMICS

the retarding torque at speeds of 800 and 300 revolutions per minute, and
the number of revolutions made during this fall of speed.

Speed,
revs, per min.

1000

875

765

645

550

460

375

305

240

Time,
seconds

105

120

7. The drum of a winch has an effective diameter of 12 inches. The axis
is horizontal and 40 feet above the ground. One end of a chain, weighing
2 lbs. per foot, is fixed to the drum, and the chain hangs vertically to a loose
heap ou the ground where there are 40 feet of chain. The drum is rotated at a
constant speed so that 60 feet of chain are wound up per minute. Neglecting
friction, find the couple required to rotate the drum. Draw a curve shewing
the couple during the last 40 seconds before the chain is completely wound up.

8. An oscillating rotary valve has a motion given by the equation,

^ = 3sin4<,
where 6 is the angle turned through in time t. Shew that the torque varies
directly as the angle turned through.

©. A disc, the moment of inertia of which is 300 lbs. ft.^ units, is retarded
from a speed of 50 revolutions per minute by a resisting couple which varies
directly with the speed. If the time for the speed to be reduced to 20 revolu-
tions per minute is 4 minutes, find the initial retarding couple.

10. In a rifle barrel the grooves make one complete turn in 10 inches. A
bullet of mass 0-4 ounce and radius of gyration 0*11 inch has a muzzle velocity
of 2000 feet per second. What are the values of the effective impulse and
impulsive couple exerted on the bullet during its travel along the barrel ?

11. In order to determine the moment of inertia of a flywheel and its shaft
it was speeded up by a belt which was then thrown out of gear. When coming
to rest under its own friction it was found to take 42 seconds to change the speed
from 200 to 180 revolutions per minute. With a brake giving a constant torque
of 18 lbs. feet it took only 18 seconds to change from 200 to 180 revolutions
per minute. Find the moment of inertia of the flywheel and shaft.

12. A cylindrical nut, whose internal and external radii are 1 inch and
4 inches respectively, works along a vertical screwed shaft. The pitch of the
screw is | inch and the coefficient of friction J . If the nut be set turning so
as to travel down the shaft, shew that its linear retardation will be approxi-
mately 0-05 foot per second per second.

EXAMPLES. CHAPTER IV 109

13. A cylinder rolls down a plane inclined to the horizontal at an angle 6.
Shew that the acceleration of the centre of the cylinder is | gr sin a.

14. As a flywheel rotates it winds up on its axle a light inextensible string
which is attached to a weight of 50 lbs. resting on the ground in such a
position that the string is vertical when it becomes tight. The moment of
inertia of the flywheel is 200 lbs. ins.^, and the diameter of the axle is 2 inches.
Shew that at the instant when the weight leaves the ground the angular
velocity of the flywheel is reduced in the ratio 4 to 5.

15. A uniform circular trap-door, 2 feet in diameter, is to be provided with
a stop in such a way that when the door is thrown open against the stop there
shall be no jar on the hinge. Find where the stop must be placed.

16. A water turbine is taking 500 cubic feet of water per minute. The
water enters the wheel at a radius of 10 inches and with a velocity of 25 feet
per second inclined at 20° to the tangent at the point of entry. If there is no
shock at entry, and the discharge is radial, find the turning moment on the
wheel.

17. A uniform circular disc of diameter 1 foot is rotating about an axis
through its centre perpendicular to its plane at a speed of 100 revolutions per
minute ; this is brought up to a stationary disc of the same mass but of
diameter 2 feet, which is free to rotate about the same axis. After rubbing the
two rotate together ; find the common angular velocity.

18. A door is 6 feet 6 inches high and 2 feet 6 inches wide and weighs
40 lbs. Find the moment of inertia about the hinges.

19. A hollow sphere has an external diameter of 9 inches and an internal
diameter of 8 inches. It is made of metal the density of which is 480 lbs. per
cubic foot. Find the moment of inertia about a diameter.

20. The rim of a cast-iron flywheel is rectangular in cross-section, the
thickness being 6 inches. The outside and inside diameters of the rim are
4 feet and 3 feet 3 inches respectively. If the density of cast-iron is 460 lbs.
per cubic foot, find the moment of inertia of the flywheel. The hub and
spokes may be omitted.

21. A round rod, \ inch in diameter, is screwed into a solid sphere of
6 inches diameter, the axis of the rod being along a radius of the sphere. The
length of the rod from the end to the centre of the sphere is 3 feet, and both
the rod and sphere are of steel the density of which is 490 lbs. per cubic foot.
Find the moment of inertia about an axis (1) through the centre of gravity,
(2) through the end of the rod.

CHAPTEE V
CENTRIFUGAL FORCE AND CENTRE OF MASS

Centrifugal Force

When we were applying the second law of momentum to rotation
we dealt only with the tangential components of the forces acting
on the particles ; we must now deal with the normal forces.

Suppose we have a mass m rotating in a circle of radius r with
an angular velocity w. We have already

shewn that such a mass has, continually, an y- ^ \

acceleration towards the centre of magnitude / w \

(oV. This means that there is a rate of change | '\ x

of momentum towards the centre equal to * ^/ r

?no>V. In order to produce this rate of change \ /

of momentum we must have a force F acting •- — --^

always towards the centre. This force is some- - ^ ^ig- 76.
times called the centripetal force.

Thus, F = m.(i>V...> (1).

The force may be provided, in the case just considered, by
attaching the mass m to the centre of rotation by means of a string.
The tension in this string will be equal to mwV.

In the case where m is a small particle of a body, the force
is provided by an internal stress being set up in the material of
the body. That this is so, is well known, since it is possible by
rotating bodies at sufficiently large speeds to cause the internal
stresses set up to be greater than the material can withstand, with
the result that the body flies to pieces. Many cases have occurred,
frequently with disastrous results, where engine flywheels, for

CENTRIFUGAL FORCE 111

example, have suddenly flown to pieces due to the governor sticking,
and the speed increasing greatly beyond the normal.

In dealing with cases of rotation where the angular velocity is
constant it has become customary to use the principles of statics
to solve problems. There is a good deal to be said for this method,
but unfortunately it is often a stumbling block to beginners.
However, the method has become so usual that it is not desirable
to try to change it, and we must adopt it. In statical problems the
velocity is zero and the rate of change of momentum is zero. The
resultant force is therefore zero. Now, taking the case of the
rotating mass just considered, let us imagine that there is a force
equal to mwV acting away from the centre instead of towards the
centre, and that instead of the body having a rate of change of
momentum it is in equilibrium. The resultant force on the mass
must be zero,

.*. F - y?icoV = 0, '

where F is the pull in the string towards the centre.

This imaginary force equal in magnitude to mm'^r and acting
away from the centre of rotation, which we apply to make the
problem a statical one, is called the centrifugal force.

It is exactly equal in magnitude to the true force which must act
to produce the acceleration but is in the opposite direction. It is
obvious that all dynamical problems might be treated as statical
problems if we introduce imaginary forces equal and opposite to
the rates of change of momentum. It is often convenient to do
this.

Example (1). A mass of m lbs. is suspended fro7n a string of
length I and is rotating in a horizontal circle of radius r. Find
the time of one revolution and also the tension in the string.

Such an arrangement is called a conical pendulum.

We will treat this problem firstly as dynamical and secondly
as statical.

112

ELEMENTARY DYNAMICS

(1) Dynamical.

Let T be the tension in the string, and w the angular velocity.

The resultant force vertically = T cos 6 — mg.

The resultant force horizontally = T sin 6.

The rate of change of momentum vertically = 0.

The rate of change of momentum
horizontally (towards C) — moi^r.

From the second law of momentum

T cos 6 — mg — 0 {a),

and T sin 6 = mwV ...{b).

From {a) and (6),

^ a ^^'^
tan ff - — ,

I.e.

Fig. 77.

rcot Q

= ^ , where h - OC,

J\-

For one revolution the angle turned through = 27r.

The time for one revolution

-v/l

From (a),

cos Q

CENTRIFUGAL FORCE

113

(2) Statical.

Apply a force nua^r as shewn in fig. 78. We may now treat the
mass m as in equilibrium under
the action of the three forces,
T, mwV, and mg.

Resolving vertically and
horizontally, we have

T cos ^ — mg = 0,
and T sin 9 — rrn^r = 0.

From these, as before, we get

I
■T = mgy

ma)V

and

-A-

Fig. 78.

Example (2). A simple governor for operating the throttle valve
of a steam engine is shewn in fig. 79. The vertical spindle^ to which
is fixed the piece AB, derives its rotation from, the crank shaft of
the engine. The sleeve DC can slide up and down the spindle, and
moves a lever, the forked end of which fits in a groove on the sleeve.
This lever opens or closes the throttle valve.

It is required to find the total mass W of the loaded sleeve so that the
angle CAH may be 30°, when the engine is moving at its normal speed
and rotating the governor spindle at \20 revolutions per minute.

Take AB = CD = 3 inches.

CH - HA = BG = GD = 10 inches.

The weight of each hall = 3 Ihs.

The radius of the ball path = (1 -5 + 10 sin 30°)

= 6*5 inches.

The centrifugal force per ball —

3xl67r"x6-5
12

pdls.

= 267r2.
Let P equal the pull in each of rods CH and DG,
and Q „ „ „ HA and GB.

L. E. D.

114

ELEMENTARY DYNAMICS

Consider one ball and resolve the forces vertically and hori-
zontally.

Fig. 79.
We have P cos 30' - Q cos 30° + 3g = 0^

(P + Q)sin30°-267r2:=0/'

or p - Q = -^

P + Q = 267r2 X 2

rri 9 96 X 2
.-. 2P = 527r2-

= 512-111
= 401 pdls.
200-5

P==

6-26 lbs. wt.

CENTRE OF MASS

116

Revolving vertically for the sleeve we have
2Pcos30° = W^.

401 X n/3

2x32
= 10-85 lbs. wt.

Centre of Mass, or Centre of Inertia

In dynamical problems when we are dealing with a body of
finite size and* not merely a small particle, we may consider the
body as consisting of a very large (infinite) number of small par-
ticles. These particles may be moving with different velocities,
e.g. a body rotating about an axis, and in estimating the total
change of momentum we shall require to find the vector sum of
the changes of momentum of all the particles.

Take the case of a body the mass of which may be considered
concentrated in one plane, e.g. a thin sheet of material. Let the
body be rotating about a fixed axis with an angular velocity w.

Fig. 80.

Suppose we take two axes Ox and Oy in the body, for convenience
at right angles, and passing through the axis of rotation. Let x
and y be the coordinates of a particle of mass m (fig. 80).

8—2

116 ELEMENTARY DYNAMICS

The rate of change of momentum is rrnn^r^ where r = OA, and is
in direction AO.

The rate of change of momentum in the direction Ox

= moi^r cos 6

= moi^x.

The rate of change of momentum in the direction Oy

= moy^r sin 6

= TtHiP-y.

For the whole body we have,

Total force in direction xO = ^nm^x = P.

„ „ yO = %mi^^y = Q.

/. P = o>^'^mx,

and Ql = oy^%my.

Now we can obviously find a point, coordinates S, y, say, in the
body such that

M5 = ^mx

and My = ^my,

where M = the total mass of the body,

%mx

and 2/=-^.

The point whose coordinates are x and y is called the centre
of mass or centre of inertia of the body.

The resultant force in the direction xO and yO is given by

and Q = M2/ . w^

and instead of thinking of the individual particles we may imagine

the whole mass of the body concentrated at the centre of mass.

CENTRE OF MASS 117

The single resultant force required to produce the change of
momentum

= \/p2 + a2

= Ma)2 OG^

where G is the centre of mass of the body.

The centre of gravity of a body is the point through which
the resultant force on the body due to gravity always acts, no
matter what the position of the body.

It is easy to shew that the centre of gravity coincides with the
centre of mass.

In fig. 80 imagine the weight of each particle, such as m at A,
to act at right angles to the plane of the figure.

Let X and y be the coordinates of the centre of gravity. Taking
moments about the axes Oy and Ox, we have

M^ xx' = ^mgx,

i.e.

and

i.e.

From the previous article we see that, x=x, and y' = y, and
hence the centre of gravity coincides with the centre of mass.

Example (3). A motor om,nibus, when fully loaded^ weighs
6^ tons and the height of the centre of gravity is 4 feet 10 inches
above the ground, and may be assumed to be in the vertical plane
midway between the wheels. The effective breadth of the wheel base
is 6 feet 8 inches. Assuming no side slip, what is the maximum,
speed at which the omnibus can take a corner of 5 yards mean
radius without beginning to overturn ?

%m,x
~ M '

xy'

= ^fngy,

^my
M

118

ELEMENTARY DYNAMICS

Th^, omnibus keeps the middle of the road so that the wheel base
is horizontal.

What minimum, coefficient of friction is required to prevent side
slip?

4' 10'

x«

1^ 6'8" ^

Fig. 81.

Let V — the maximum gpeed in feet per second, and m = the
mass of the loaded omnibus.

The acceleration of the centre of gravdty = —

feet per sec. per sec.

The centrifugal force acting at the centre of gravity = ^k" •

When overturning is about to begin the pressure between the
inner wheels and the road will be zero.

Let R = the normal pressure on the outer wheels, and P — the
tangential pressure preventing skidding.

CENTRE OF MASS 119

Taking moments about A we have
'^'x58-m^x40 = 0,

, 40 X 15 X 32
i.e. ^ = gg

= 333,
.•. V = 18-2 ^eet per second
= 12*4 miles per hour.
For P we must have

pdls.

6-5 X 2240 X 18-2^
" 15

= 4*5 tons wt.
Similarly R = mg pdls.

= 6*5 tons wt.

The coefficient of friction = -

_4;5

"6-5

= 0-69.

Cant on Railway Curves

When a train travels along a curved track a force will be re-
quired towards the centre of curvature in order to provide the
necessary change of momentum. If the track is level, as is usual
in the case of tramways, the flanges on the wheels bear against
the outer rail. In railway curves the outer rail is raised above
the inner rail, and by this means, for a definite speed, all side
thrust on the flanges may be avoided. The amount the outer rail
is raised above the inner is called the cant. If the speed of the
train exceeds that for which the cant was calculated the outer
flange has to transmit some thrust ; if the speed is less than that

120

ELEMENTARY DYNAMICS

for which the cant was calculated the inner flange will bear
against the rail.

We will investigate this.

Fig. 82.
In fig. 82 let G be the centre of gravity, P and Q tlie normal
thrusts on the wheels, a the mean distance between them, S the
flange thrust, and v the speed in feet per second.

The acceleration towards the centre of curvature is - , where R

is the radius of curvature. Put on a centrifugal force , and

we may then treat the problem as a statical one.

Resolving all the forces parallel and perpendicular to the track,
we have :

Perpendicular to the track

P + Q - Mg cos

0^0

(1).

CANT ON RAILWAY CURVES 121

Parallel to the track

S + M^sin^-— -cos^ = 0 (2).

If S is zero, i.e. no side thrust on the flanges, we get from (2)
tan ^ = — .

Let h = the cant, then tan 6 = , = - , since h is small

compared with a,
i.e. the cant = — .

Also S == M ( — cos^ — ^ sin^

^ . - j , if ^ is small.

Example (4). Let R = 660 feet,

V =: 40 miles per hour

= — 5 — feet per second,
o

a = i feet 8 J inches

56-5 _ ^
^-^feet.

The cant of the outer rail {h) for no side flange thrust

56-5x88^x4
~ 12x660x9x32
= 0-765 foot
= 9|^ inches. ,
For a speed of 60 miles per hour, the side thrust of the flanges
(-882 ^^ 0-765 X 121

= M {11 -7 -5-2 1 absolute units.

122 ELEMENTARY DYNAMICS

Or taking M = 2240 lbs. we get,

The thrust per ton

_ 2240x6-5

~ 32

= 455 lbs. per ton.

Stress in the Rim of a Flywheel

Let us consider a flywheel in which the thickness of the rim is
small compared with the mean radius, the flywheel consisting of
a heavy rim connected to the hub by spokes.
Let p = the density of the metal of the wheel,
Q) = the angular velocity of the wheel,

r^the mean radius, i.e. the radius of the centre of mass
of a cross- section of the rim.
Take a small part of the rim subtending an angle SO at the
centre of the wheel (fig. 83).

Let a = the area of cross-section of the
rim.

P = the total pull on the area at A or B
in absolute units.

The mass of the element AB = par W.

The acceleration towards the centre C

= wV.

The centrifugal force on the element

= p.a.rKoi\8e,

and acts at the centre of mass of the ele-
ment.

This centrifugal force has to be balanced
by the components of the forces P at A
and B,

on • ^^

2P. sm —

par'^ (o^ 80.

STRESS IN FLYWHEEL RIM 123

But since BO is very small, we may write
. SO SO

and we get P . 80 = par^m^SOy

or P = par^tt)^.

Let/= the internal force per unit area at A or B, then

P
a'

= pv^ absolute units,

where v is the velocity of the centre of mass of a cross-section
of the rim.

EiXaonple (5). A flywheel is made of cast-iron which breaks
when subjected to a pull of 10 tons per sq. inch. The external
diameter is % feet and the thickness of the rim, is 6 inches. What is
the speedy in revolutions per minute, which will cause the flywheel
to fly to pieces ?

The density of ca^st-iron = 470 lbs. per cubic foot.

Let N = the required speed.

Then c. = -g^,

and V = -^jr- X -^ feet per second.

The breaking stress = 10 x 144 x 2240 x g pdls. per sq. foot.
Using the formula, y^p-w^, we have

470 X 47r2x n2x7-52

10 X 144 X 2240 X 32 =

1202

470 x4x 3-142 X 7-52
= 1195 revolutions per minute.

124

ELEMENTAEY DYNAMICS

Examples. Chapter V

1. A bucket of water is swung round in a vertical circle of 26 inches radius.
What is the minimum speed of rotation if none of the water is spilled ?

2. A particle is attached to the end of a string of length I, the other end
of which is fixed, and moves as a conical pendulum making n revolutions
per minute. Find the radius of the circular path which it describes.

A mass of 10 lbs. rotates as a conical pendulum at the end of an elastic
string, of unstretched length 3 feet, and makes 40 revolutions per minute.
If the string is stretched ^^^ of its length by the weight of 1 lb. find the length
of the string during the motion.

3. The spindle EB shewn in fig. 84 receives vertical support only at E,
and is supported horizontally by collars at A and B, DC is a stiff arm rigidly
attached to the spindle and carrying a weight at C. If A B is 8 inches, DC
10 inches, D midway between A and B, and the weight at C 10 lbs., find the
horizontal forces at A and B.

Also find these forces when the spindle is rotating freely at 100 revolutions
per minute, and determine the speed of rotation for which the reaction at A
is zero.

Ej,

mm//////

Fig. 84.

4. A particle of mud, sticking to the rim of a motor-car wheel travelling
at 20 miles per hour, leaves it when situated at a point 30° behind the vertical
line between the centre of the wheel and the road. Determine its velocity
relative to the road, the mudguard, and the top of the wheel respectively.

To what forces and to what accelerations is the particle subjected before it
leaves the wheel ?

EXAMPLES. CHAPTER V

125

6. The governor of a steam engine, when making 210 revolutions per
minute, takes up the position shewn on the sketch.
If each ball weighs 1 lb., determine the weight W.

Supposing the speed of the engine to increase 2
per cent, before the throttle valve moved, what pull
would the governor exert on the throttle valve
lever?

6. A steam roundabout revolves four times a
minute. A wooden horse on the roundabout is
suspended from the roof by an iron rod, and the
centre of gravity of the horse and the man on it is
at a distance of 20 feet from the axis about which
the roundabout turns. Find the angle which the
suspending rod makes with the vertical. Find also
the horizontal displacement of the centre of gravity
caused by centrifugal force, supposing this centre of gravity to be at a dis-
tance of 5 feet from the point of suspension.

Fig. 85.

7. A car weighing 30 cwt. is running at 15 miles per hour round a curve
of 60 feet radius on a level road. What horizontal force perpendicular to the
direction of motion must be exerted by the ground on the wheels of the car?

Assuming that the grip of the outer wheel is sufficient to prevent skidding,
find at what speed the inner wheel will begin to lift off the ground. The width
of the wheel base is 4 feet, the centre of gravity is 3 feet above the ground,
and the given radius, 60 feet, is the radius of the circle midway between the
tracks of the inner and outer wheels.

8. A motor-car track is designed to allow of cars running round a curve
of 500 feet radius at 60 miles an hour without any frictional side pressure
between the tyres and the road surface. Find the slope to which the curved
part of the track must be banked up, and the side pressure produced per ton
of car when the speed is 80 miles an hour.

9. A uniform disc is mounted on an axle which passes through its centre
O. A mass of 25 lbs. is clamped to the disc, its centre of gravity being at a
point A distant 2 feet from the centre O, and another mass of 40 lbs. is
clamped with its centre of gravity at B distant 2*5 feet from O. The angle
AOB is 120°. Find the resultant force on the axle due to the rotation of
these masses, when the disc is making 200 revolutions per minute.

126

ELEMENTARY DYNAMICS

lO. In the shaft governor, shewn diagrammatically in fig. 86, the
ball levers are pivoted to a piece D
which is fixed to the shaft F. For the
position shewn, the pull in the springs
connecting the balls is 150 lbs., and there
is a force on the sleeve along the axis of
the shaft equal to 115 lbs. In addition to
this there is a maximum axial force on
the sleeve due to friction of 10 lbs. , which
may act in either direction. Find the two
extreme speeds at which the shaft may run
without the sleeve of the governor moving ^*

along it. The length of AB is 6 inches; the length of BC is 3| inches;
the balls are 12 inches apart and each weighs 15 lbs.

11. A cast-iron flywheel has a mean diameter of 4 feet. If the mean tensile
stress in the rim is to be limited to 1000 lbs. per sq. inch find the maximum
allowable speed of the wheel. A cubic foot of cast-iron weighs 470 lbs.

12. What is the side pressure between a train weighing 300 tons and the
rails, when the train is going round a curve of 120 yards radius at 30 miles
per hour, the rails being on the same level? What should be the cant for no
side pressure with a speed of 30 miles per hour if the gauge is 4 feet 8^ inches?
With this cant, what will be the side pressure if the speed is 45 miles per hour?

13. A belt the mass of which is p lbs. per foot length passes round half the
circumference of a pulley wheel, the speed of the belt being v feet per second.
If Tj and Tg be the tensions in lbs. in the two sides of the belt, shew that,
neglecting the weight of the belt, the total pull on the pulley is equal to

( 2pv^ '

Ti-fT,

lbs.

14. The wheel base of a steam tractor weighing 16 tons is 6 feet 10 inches
across and the distance between the axles when they are parallel is 10 feet
10 inches. When going round a curve the distances between the points of
contact of the wheels and the ground are 11 feet 4 inches for the outside
pair, and 9 feet 4 inches for the inner pair. If the speed is 6 miles per hour,
find the force tending to shift the tractor sideways.

15. A railway line over a bridge is curved, the radius of the curve being
400 yards. Find the side thrust on the bridge when a locomotive weighing
90 tons passes over at a speed of 30 miles per hour.

EXAMPLES. CHAPTER V 127

If the gauge is 4 ft. 8| inches, find how much the outer rail must be raised
above the inner rail in order that there shall be no thrust on the flange.

16. A casting weighing 6 lbs. is bolted to the face-plate of a lathe, for the
purpose of machining, in such a position that its centre of gravity is at a
distance of 2 inches from the axis of the mandrel. The centrifugal force is to
be balanced by two 3 lb. masses placed on two radii on opposite sides, and
inclined at 45° to the diameter passing through the centre of gravity of the
casting. Find how far from the centre the masses should be placed.

CHAPTER VI
WORK, POWER AND ENERGY

Work

We have now to introduce some new physical quantities, and
also another principle or law, mentioned in Chapter III, which is
very far reaching in its applications and of particular importance
to engineers.

Fig. 87.

If a force acting on a body causes a displacement of it then the
force is said to do work. Also, if a body is moved in the opposite
direction to a force acting on it, work is said to be done against
the force.

We see that there are two things necessary before work can be
done, viz. force and motion.

The quantity of work done is measured hy the product of the
component of the force in the direction of the displacement^ and the
displacement.

For example, in the case shewn in fig. 87, if the force P moves
the mass a distance s then.

The work done = P cos 6 . s
= F.s,
where F is the component of the force in the direction of motion.

WORK

129

The absolute unit of work in the f.p.s. system is the/oot-poundal.

One foot-poundal is the work done by 1 poundal of force acting
through a distance of 1 foot.

In the c.G.s. system the absolute unit is the erg.

One erg is the work done by 1 dyne of force acting through a
distance of 1 centimetre.

1 joule =r 10'' ergs.

Frequently the gravitation unit is used, and we speak of the
number of foot-lbs. of work done.

One ft. -lb. is the work done by a force equal to the weight of 1 lb.
acting through a distance of 1 foot.

In the simple case shewn in tig. 87 we considered the force
constant, but in many cases the
force varies as the displacement
increases. In such cases we may
conveniently draw a Force-Space
curve such as is shewn in fig. 88.

The work done for a small dis-
placement 8s will be represented
by AB . Ss, i.e. the cross shaded
area, and the total work done is
represented by the area under the
force-space curve, shewn shaded
vertically.

Or, we may find the space-average of the force (F) from the
graph, or otherwise, and the work done will then be equal to F . s,
where s is the total displacement.

Note. This space-average of the force must not be confused
with the time-average of the force which we used in applying the
principle of momentum. In very few cases will the two averages
have the same value.

Example (1). The weight of the centre span of the Quebec bridge
is 5400 tons. This span was raised into position through a vertical
li. E. D. 9

Space

Fig. 88.

130

ELEMENTARY DYNAMICS

height o/* 150 feet hy means of hydraulic jacks. Find the useful
work done in foot-tons.

During lifting the vertical force downwards was constant and
= 5400 X 2240 lbs. wt.

The distance through which this force was overcome

- 150 feet.

.-. The work done = 5400 x 2240 x 150 ft. -lbs.

- 8-1 X 10^ ft.-tons.

Power

Power is a name given to the time-rate of doing work, or is the
amount of work done -per unit time.
Absolute System of Units.

F.p.s. system. The unit of power is 1 foot-poundal per second.
c.G.s. system. The unit of power is 1 erg per second.
1 watt is 1 joule per second, i.e. 10'' ergs per second.

Practical Unit. The British unit, most frequently employed
for measuring power, is the horse-power originally introduced
by Watt.

1 horse-poiver (h.p.) is a rate oj
working equal to 550 ft.-lhs. of
work per second, or 33,000 ft.-lbs.
of work per minute.

It may be noted that power
multiplied by time gives us work
done, or

W - H . ^,

where W = the work done, H - the
power, and t = the time.

In dealing with heat engines we

I Ti

Fig. 89.
frequently use this as a basis for a large unit of work called the

POWER 131

horse-power-hour. This is the work done by an agent, working
at 1 horse-power, in one hour.

i.e. 1 horse-power- hour = 33,000 x 60 ft. -lbs.

If the power is varying we may find the work done by drawing
a power-time curve, fig. 89.

The work done in time t is obviously represented by the shaded
area under the curve.

Example (2). An electric crane raises a load o/S tons through
a vertical distance of 20 feet, in 12 seconds, at a uniform speed. If
23 per cent, of the power supplied is wasted in friction, what is
the horse-power which has to be supplied to the motor ?

The work done in 12 seconds = 3 x 2240 x 20 ft.-lbs.

rw., , n , 3 X 2240 x 20 ^, „
The work done per second = yY) rt.-lbs.

.*. The horse-power used in raising the load
3 X 2240 X 20
12 X 550

= 20-4 H.p.
Let H =: the horse-power supplied to the motor.
H = 20-4 + ,?^H,

20-4
" = 077
= 26-5 H.p.

Energy

The energy of a body is its capacity to do work.

'It is measured by the amount of work which can be done, and
therefore has the same units as work. So long as we deal with
the energy of a body, or system of bodies, and not with energy

9—2

132 ELEMENTARY DYNAMICS

in the abstract, for mechanical problems we may divide energy
into two classes :

(1) Potential Energy.

(2) Kinetic Energy.

The Potential Energy of a body is the energy a body possesses
in virtue of its position.

The commonest case is the energy due to a body's position
relative to the earth.

Suppose we have a mass of M lbs. raised a distance h feet above
the ground, then if we allow the body to fall the force of gravity,
M^ absolute units, will act through the distance A, and the work
it does will be equal to Mgh absolute units.

Mgh is a measure of the potential energy, in absolute units, of a
mass M raised a height h.

Another example of potential energy is afforded by a mass of
iron which is in the neighbourhood of a fixed magnet. If there
is no resisting force such as friction, the mass of iron, if freed,
will be moved to the magnet. Therefore, when separated from the
magnet, it possesses Potential Energy due to its position. Again,
a helical spring or elastic body which is stretched, has potential
energy stored up in it, and this can be got out when the spring or
body is allowed to return to its normal length.

The Kinetic Energy of a body is the energy it possesses in
virtue of its m,otion.

That there is energy possessed by a body in motion is easily
seen when we realise that, in order to stop a body which is moving,
we shall have to exert a force, and work will be done against this
force while the body is being brought to rest.

We must find an expression to represent the quantity of kinetic
energy possessed by a mass M say, moving with a speed v.

Suppose we apply a constant force F, in a direction opposite to

ENERGY 133

that of the motion, and that this force stops the body in a
distance s.

The kinetic energy of the body

= the work done against the force

= F . s.

Now this constant force will produce a uniform rate of change

1)
of speed given by - , where t is the time taken to come to rest.
t

.'. We have F = — - .

Also, the time-average of the speed will be ^ ,

The kinetic energy = F . s

Mv V

= 1 Mv\
This is the expression required for the kinetic energy.
The kinetic energy in absolute units is equal to one-half the pro-
duct of the mass and the square of the speed.

Suppose now the force is not constant. Let it be F when the
velocity is u, and suppose that it produces a small decrease of
velocity 8w, and that the distance moved is 85. Then

F . 8s = the change of kinetic energy for the distance 8.9

^ 8(lMt)2)

or F = -~ .

This is only true when 8s is made indefinitely small, as F may
vary over the distance 8s. When 8s is made indefinitely small
we get

_d{^Uv^)

134 ELEMENTARY DYNAMICS

or, the force for any position of the body is equal to the rate of
change of the kinetic energy with respect to the distance.

If F is the space-average of the force, and s is the total distance,
F.s = the total change of kinetic energy.

Gravitation Units.

Mv^
The kinetic energy = J — , where g is the numerical value of

the acceleration due to gravity.

If we use this unit, it is clear that in equating the work done
to the kinetic energy the force must be measured in Gravitation
Units.

Elxample (3). A motor-car weighing 18 cwt. and travelling at
10 rniles per hour is brought to rest in 20 feet by the application of
its brake. Find the space-average of its retarding force.
The kinetic energy of the car

= ^mv^ abs. units
= 1 X 18 X 112 X (If X lOy ft.-pdls.
-2-17 X 10^ ft.-pdls.
Let F be the space-average of the retarding force in abs. units.
The work done against this force F by the car is equal to F.s,
i.e. F X 20 ft.-pdls.

.-. Fx 20 = 2-17 X 10^

2-17 X 10« J,
F=-^^— pdls.

~ 20x32 ^^'•
= 3^0 lbs.

Comparison between Momentum and Kinetic Energy

If a constant force F acts on a body of mass M and produces a
change of velocity from u to v, in a distance s, and a time t,
then,

F X ^ = the change of momentum = (^Mv— Mu),

F X s = the change of kinetic energy = (| Mv^ — ^ Mu^).

ENERGY AND MOMENTUM 135

If the force is varying, for estimating the change of momentum
we must use the time-average of the force. *

For estimating the change of kinetic energy we must use the
space-average of the force.

Example (4). In the problem on p. 69 we had the accelerating
force-time curve for the first 11*1 seconds after the starting of a
train.

Find the time-average and the space-average of the force.

The force increased uniformly with the time and at the end of
11-1 seconds was 3 -96 tons wt.

3'96
.*. The time-average of the force = —^

= 1-98 tons wt.
To find the space-average of the force we want to know the
distance moved. This we can obtain from the velocity-time curve
shewn on p. 70. The area under the curve gives the distance. By
counting squares we find there are 128.

.-. The distance = 128 x O'l x 0-5 = 6-4 feet.

The velocity at the end of the time = 1-72 feet per second.

Equating the work done to the kinetic energy generated, we
have

F . s = i Mt;^, where F is the space-average of the force.
F X 6-4 = 1x400 X 1-722.
400 X 1-722

2 x6-4
400 X 1-72-

tons wt.

2x6-4x32

= 2-89 tons wt.
We thus have,

Time-average of the force = 1-98 tons wt.

Space-average of the force = 2-89 tons wt.

136

ELEMENTAEY DYNAMICS

It should be noted that energy is a scalar quantity and not
a vector quantity. The force in the case of potential energy
and the velocity in the case of kinetic energy will have definite
directions, but this does not affect the energies. We have already
defined the energy of a body merely as its capacity for doing work.
By suitable means we can arrange for the direction of the work
done by a given quantity of energy to be what we wish. For
example in the case shewn in fig. 90, we may use the potential
energy of the mass m in doing work against a resistance F acting
on M.

hzs

y/////////////?////////

Fig. 90.

Or again, take the case of a motor-car going round a corner at
a constant speed. In order to change the momentum we know we
must have a force acting towards the centre of rotation for a given
time, but this force does no work since it always acts at right
angles to the motion, and the kinetic energy remains constant
although the direction is changing.

In dealing then with energy, in order to ohtaiii the total we add
the quantities algebraically and not vectoHally.

Conservation of Energy

So far we have only been dealing with two forms of energy,
potential and kinetic, but, apart from these mechanical forms, there
are, in nature, several forms of energy, heat, light, electrical,
chemical and so on.

Now the various forms of energy can be converted one into
another. When a piece of iron is hammered on an anvil the

CONSERVATION OF ENERGY 137

potential energy of the hammer is first changed into kinetic energy.
This is changed into heat when th^ hammer strikes the iron. In
the case of a steam engine driving an electrical generator we have
a case of heat energy of the steam being used to produce electrical
energy. The heat of the steam is produced from chemical energy
in the coal.

In all these conversions of energy from one form to another
which are either arranged by man, or occur in nature, we assume
that there is no net loss or gain of energy. When a definite
quantity of one form of energy, A say, is used up in producing
another form of energy, B say, the quantity of B produced is
exactly equal to the quantity of A which has disappeared. To put
it in another way, man finds he cannot create or destroy energy
and he believes that the Deity does not do so.

The whole of physical science rests on this principle of the con-
servation of energy. Whenever physical phenomena have been
discovered, which appeared at first to violate this principle, further
experiments have always shewn, either an error of observation,
or that a form of energy has been discovered which, up to that
time, was unknown to man. It seems most probable that there
are still forms of energy which exist unknown to man, or which
man cannot make use of by converting them to other forms.

Now in mechanical problems we shall always be dealing with
a definite body, or system of bodies, and generally we need only
concern ourselves with the mechanical forms of energy.

In such cases the body or system of bodies may have the total
energy altered by work being done on or by the body or system.

We may express this as follows :

Whenever a body or system of bodies has its energy changed the
increase of energy is exactly equal to the net work done on the
system by the external forces. Or,

Work put in - work got out = Gain of energy.

If the left-hand side is negative the gain of energy will be
negative, i.e. there will be a loss of energy in the body or system.

138 ELEMENTARY DYNAMICS

Example (5). A shell weighing 2,60 lbs. , which did not explode,
was found to have penetrated the ground in a certain place a dis-
tance of ^ feet in the direction of impact. The striking velocity was
820 feet per second. What was the space-average resistance to
penetration of the ground? Neglect the energy due to rotation.

If the velocity of the shell had been 1000 feet per second at striking,
how far would it have penetrated ?

Let R — the average resistance to penetration in lbs. wt.

We may neglect the change of potential energy of the shell and
may say,

The work done against the resistance of the ground = the loss
of kinetic energy of the shell, i.e.

Rx^x8 = ix260x 820^
. „ 260x820%,

2 X 32 X 8

= 342,000 lbs. wt.

= 153 tons wt.

Let 37 = the distance of penetration for a striking velocity of
1000 feet per second.

Then, since R is constant,

X 10002

8 8202 '
.-. ^=11-9 feet.

Example (6). A spring is such that it extends J inch for a pull
of 1 lb. wt. A mass of 4 lbs. is suspended by the spring, being
fixed to the end, and is pulled down until the spring is extended a
total distance of 5 inches. If the mass is suddenly let go, how
high will it rise and what will be its maximum velocity ?

The mass of the spring may be neglected.

The energy stored in the spring, when stretched, is equal to
the work done in stretching it.

CONSERVATION OF ENERGY

139

//////////////

The pull-extension curve will be a straight line as shewn in
fig. 91.

For 5 inches extension the pull of
the spring =10 lbs. wt.

The energy in the spring initially
= area OAB

Let h = the maximum height in
inches which the mass rises.

When the mass is at its maximum
height, the spring is extended (5 - A)
inches.

The pull of the spring in this posi-
tion

= 2 (5 -A) lbs. wt.

The energy stored in the spring in
this position

Extension B

Fig. 91.

2 (5 -Ay

2 X 12

ft.-lbs.

The gain of potential energy of the mass = ^ ft.-lbs.

1 z

The gain of potential energy of the mass =*the loss of strain
energy of the spring.

4A_10x5 2(5-A)2
*• 12 ~2Vl2~ 2712 '
or h^-eh=0,

i.e. h = 0, or A = 6 inches.

It is obvious that the spring will be in compression.

To find the maximum velocity of the mass. The velocity will
continue to increase so long as the pull of the spring is greater
than the weight of the mass, i.e. until the spring's extension is
reduced to 2 inches. We have then.

140 ELEMENTARY DYNAMICS

The gain of potential- and kinetic energy of the mass = the loss
of strain energy in the spring,
or, using absolute units,

4x3x32 1/10x5 ^x2=x

2'*" -V2~>n2~2TT2J"'-''

12 2

i.e. 32 + 2!)2 = If X 32,

18o< 32
Ux'2 '
or -y =^ 3'46 feet per second.

Example (7). A railway siding is level for the first 50 yards,
and then rises at a slope of ~~. A wagon weighing 10 tons is
shunted on to the siding with a velocity o/* 1 8 Tniles per hour, and
is observed to reach the foot of the incline in 6 seconds. If the
resisting force due to friction is constant, what is its magnitude ?

How far up the incline will the wagon travel before coming to rest ?

Let V — the velocity at the foot of the incline in feet per second.
The average velocity for the first 6 seconds

. 18 X 88

+ v

. 26-4 + 2;
. . ^ — X b = 150,

i.e. v = 50 -26-4

= 23"6 feet per second.
Let F — the force due to friction in abs. units.
By the 2nd law of momentum we have,

m (v, - V,)

10 X (26-4 -23-6)

10x2-8^

tons wt.

abs. units

6 X 32
= 0*146 tons wt.

CONSERVATION OF ENERGY 141

The total retarding force up the incline

= 0-146 + ^V X 10 (approx.)

= 0-396 tons wt.
Again, by the 2nd law of momentum for the incline,

0-396 X 32 = 10 X *^ — ,
t

where t is the time taken to come to rest, i.e.

10 X 23-6

seconds.

0-396 X 32

The average velocity = -^— feet per second.

03. g 236

.*. The distance up the incline = ^^-^7— x — - — ^-^77,-7^
* 2 32 X 0-39o

= 220 feet.

Or, we may use the energy principle.

The work done against friction = the loss of energy.

For the first part of motion,

Fx 150 = 1 X 10(26-42-23-62),

5(26-42-23-62) ^

•*• F = ^ n^Tv abs. units

150

5 X 50 X 2-8

"-^32^150 ^^"^"^-

= 0-146 tons wt.

For the motion up the incline, let s = the distance in feet.

10 X s

The gain of potential energ}^ = —rfr- (nearly)

= — foot-tons.
4

.-. 0-146 x^xs = |- lOx 23-62- ^x^,

5 X 23-62

0-396 X 32
= 220 feet.

142 ELEMENTARY DYNAMICS

Example (8). 200 gallo^is of water are to he pumped per
miriute through a 3-inch pipe up a height of 40 feet. The useful
horse-power of the pump is 42 per cent, of the horse-power supplied^
and the estimated loss due to friction in the pipe is equivalent to
an additional height o/ 16 feet. What horse-power will have to be
supplied to the pump ?

1 cubic foot contains 6^ gallons.

1 gallon of water weighs 10 lbs.

Let V = the velocity of delivery in feet per second, and d = the
diameter of the pipe in feet.

^, Trd' 200 1

Then, ^x^ = — x^,

_ 200 X 4 X 16
^•®' "" ~ 60 X 6-25 X TT

= 10 '85 feet per second.
The mass of water delivered per second == -J^- lbs.

-33-3 lbs.
The potential energy supplied per second = 33*3 x 40 ft.-lbs.
The work done against friction per second = 33*3 x 16 ft.-lbs.
The kinetic energy supplied per second
1 X 33-3 X 10-852

^2 ^32 ^'-^^'-

The total energy supplied per second

r 10-8521

- 33-3 J40 + 16 +^^^^1 ft.-lbs.

= 33-3 {40 + 16 + 1-84}

- 1930 ft.-lbs.

The horse-power given to th& water = -V\V"-
.'. The horse-power to be supplied to the pump = -^fjj- x -W~

= 8-35.

CONSERVATION OF ENERGY 143

In applying the principle of conservation of energy to mecha-
nical problems, we must be careful to see that we are not getting
energy developed which we are neglecting. For example, take
the case of an impact between two bodies where we are getting
a sudden change of momentum. Most frequently in such cases
we shall get energy dissipated in the form of heat. Work may
actually be done in damaging the bodies which impinge, or
vibrations may be set up by the impact, the energy of which is
ultimately changed to heat. In such cases we cannot apply the
principle stated above, but we can still apply the principle of con-
servation of momentum.

Consider the example of two trucks on p. 79. '

Before impact the total kinetic energy

= (i X 12 X 36 + i X 10 X 4) g()|^ ft.-tons

= 15-85 ft.-tons.

After impact the total kinetic energy

= ^x22x4-172x

60^ X 32
= 12-9 ft.-tons.

.'. The energy lost during impact =^(15-85 — 12*9)

= 2-95 ft.-tons.

Example (9). In a ballistic pendulum, for estimating the
velocity of rifle bullets, a block of wood weighing 10 lbs. was sus-
pended by two parallel strings as shewn in fig. 92. When a bullet
weighi7ig 0*4 oz. was fired into the block in a horizontal direction,
passing through the centre of gravity of the block, it was observed
that the block rose 6*95 inches. Find the velocity of the bullet.

In this case we shall get a considerable quantity of heat
developed. We may assume that the block does not appreciably
move until the bullet has penetrated its greatest distance.

Let V = the initial velocity of the bullet in feet per second, and

144

ELEMENTARY DYNAMICS

V = the velocity with which the block begins to move in feet per
second.

The total momentum of the bullet and the block in a horizontal
direction must remain constant, since there is no resultant force
in that direction.

i.e.
or

0-4 / _ 0-4\ ,

v=: 10-025 X 40v,
V-401V

.(1).

/////////////////////////

Fig. 92.

The block and the bullet now swing forward, and all their
kinetic energy is changed to potential energy when they reach
their highest point.

.'. 1 (10-025) V2= 10-025 X 32 x ^^^ (2),

, 64 X 6-95
i.e. V^ = ___^-_,

or V = 6*06 feet per second.

.*. from (1), '

The velocity of the bullet = 401 x 6-06

= 2430 feet per second.

CONSERVATION OF ENERGY

145

The fraction of the initial energy dissipated in heat
^ (9^\ X 2430^ - 1 (10-025) X 6-062

= 1

10025 X 40 X 6-06»

2430^

= 0-9975.

Example (10). Loads M, m, and P are suspended by a string
which passes over a light frictionless
pulley as shewn in fig. 93. A ring fixed
at A is large enough for in to pass through
but stops P. The string connected to m
passes through a hole in P. M is greater
than m and less than P + m. Shew that^
if the masses start from rest in the
position shewn in the figure, after the
Tnass m has passed twice through the
ring the system will, for the instant,
come to rest again with P at a height

TETX

/ M+m V ^ , .

I above the ring.

\P + m+Mj ^

Fig. 93.

Let h' be the required height above
the ring when the system is again at rest.

The total energy lost is all potential
and is equal to

(P + m-M)g{h-h').
This energy is lost due to the two impacts at the ring.
On the first passage through the ring the velocity v is given by

<,(P + m-M)A = (^±^)«^ (1).

When P strikes the ring the energy lost

^Pv\

L. E. D.

.(2).
10

146 ELEMENTARY DYNAMICS

On the return journey, just before m strikes P, the velocity of
m will again be v, since there is no loss of energy.

Let T = the impulse in the string when P is jerked into motion,
and let u = the velocity after the jerk.

Then for P and m,

T = (P + m)u — 7nv,
and for M, T = M^; — Mu.

Eliminating T, we get,

/ M+m \
u= V (3).

The energy lost

= i(M +7)i)v^-l(M + m+P}u" (4).

The total energy lost from (2) and (4)

--i(P + m+ M)(v^-u'')

= (P +m-M)g{h-h').
Since the masses came to rest at a height h',

(P + ?n - M) gli = (P + m + M) — .

From equations (3) and (1),

(M + mf v^
gh -

2(P + m + M)(P + m- M)'
M+m

\P + M + mj

Energy and Momentum Principles

It should be noted that since in defining work and mechanical
energy we have introduced no new physical facts, but merely
definitions in terms of physical quantities already measurable, we
cannot really obtain any new results by the application of the
principle of the conservation of energy in the restricted form given
above. The same results can always be obtained by the principle
of momentum, but frequently the labour may be very considerably
reduced by using the principle of energy.

ENERGY DUE TO ROTATION 147

This of course does not apply to the general principle of con-
servation of energy where all forms of energy are taken into account.
We will emphasize the two principles as applied to translation
by restating them.

(1) "/?! any system, the total momentum remains constant unless
the system is acted upon by external forces "

If external forces are acting, then the resultant change of
momentum is equal to the resultant external force multiplied by
the time during which it acts.

(2) "/rz. any sy stein the total' energy remains constant unless the
system is acted upon hy external forces "

If external forces are acting, then the total change of energy
is equal to the resultant external force multiplied by the distance
moved by the point of application in the direction of the force.

Kinetic Energy due to rotation about a fixed axis
Bearing in mind that kinetic energy is a scalar quantity, it is
easy to estimate the kinetic energy
of a body due to rotation.

Suppose a body is rotating about
a fixed axis O with an angular velo-
city o> radians per second. Consider
a small particle of mass m at a dis-
tance r from the axis. This has a
speed inr, and its kinetic energy

Fig. 94.

= -2-- t---

.'. For the whole body the kinetic
energy = ^S^^mw^r^, the summation
being effected for all the particles of the body,
i.e. The kinetic energy = ^ co^ !Smr^

where I equals the moment of inertia of the body about the axis
through- O.

10—2

148

ELEMENTARY DYNAMICS

Fig. 95.

Work done by a couple or torque

Let a body, pivoted about a fixed axis, be subjected to a turning
moment or torque T. Consider a particle
at A, distant r from the axis, and let f
be the force acting on the particle in a
direction perpendicular to AO. The com-
ponent of force acting on A in the
direction AO will do no work, since there
is no motion towards or away from O.
If the body rotate through a small angle
8^, the work done on the particle ~f. rSO.

For the whole body we shall have,

The work done = 2/r8^

= 8^ X y^r,

since S6 is the same for all the particles.

Now 2/r = the algebraic sum of the moments of all the forces,
external and internal, about the axis through O. The internal
forces mutually balance.

.-. 1fr = the algebraic sum of the moments of all the external
forces about the axis through O, i.e.

2/r = the torque.
.*. The work done=^ tSO.
Constant Torque.

If the torque is constant then the work done for an angle of
rotation 6 = TO,

Varying Torque.

If the torque be a varying one, the work done may be obtained
from the curve connecting the torque and angle turned through.

The work done = 2x8^

= The area under the torque-angle curve.

WORK DONE BY A COUPLE OR TORQUE

149

Energy principle applied to rotation

The energy principle applied to rotation may be stated thus,

Ths work done by the resultant torque = the gain of rotational
kinetic energy.

Example (11)- A clock spring is such that the turning moment
required to wind it up varies as the a^igle of wind. The turning
moment exerted at the end of the first revolution equals 3 inch-lbs.
Find the work done in making the third revolution.

The torque-angle curve will be a straight line as shewn in fig. 96.

Let OC = 47r radians, and OD = 67r
radians.

The work done during the third

revolution

area ABDC

/AC+BD\

xCD.

AC = 6 inch-lbs. BD = 9 inch-lbs.
.-. The work done

0 c

Fig. 96.

6 + 9 _

" 2 ""

= IStt inch-lbs.

= 3-93 foot-lbs.

angle

Example (12). In a De Laval Steam Turbine giving 5 horse-
power the rotor runs at 30,000 revolutions per minute. The vanes,
on which the steam, acts, are at a mean distance from the axis of
31 inches. Neglecting frictional losses, find the tangential force
exerted by the steam on the rotor.

150 ELEMENTARY DYNAMICS

Let P = the force in lbs. weight acting tangentially on the rotor.

P X 3*5
The torque on the rotor = — — — lbs. ft.

The angle turned through per minute

= 30,000 X 27r radians.
The work done per minute by the steam

Px 3-5 X 30,000 x27r

12
Horse-power x 33,000.

ft.-lbs.

.-. P X 3-5 X 5000 X TT = 5 X 33,000,

r, 33

or P =

3'5 X TT

= 3 lbs. wt.

Example (13). A uniform trap-door, 2 feet hy 2 feet, and
weighing "20 Ihs., falls from the vertical position and strikes the
floor only along the edge opposite the hinges. Find the angular
velocity of the door just before it strikes the Jioor, and the magnitude
of the blow on the edge.

(1) To find the angular velocity we may equate the loss of
potential energy and the gain of kinetic energy.

Since the centre of gravity falls 1 foot, the loss of potential
energy

= 20 X ^ ft.-pdls.

The moment of inertia about an axis through the hinges

= 20{^+l}

= -^lbs. ft."

EXAMPLES

151

The gain of kinetic energy == J x »/ x w^ where o> is the angular
velocity just before striking the floor.

.-. */-o)2z= 20x32,
or, CO = \/48

= 69 radians per second.
(2) To find the impulse on the edge, we have, the impulsive
torque equals the change of angular momentum.
Let P = the impulse in absolute units.
Then P X 2 = the impulsive torque.

or.

P X 2 = 8/ ^ 6-9^

P =1 92 abs. units.

Example (14). The turning moment on the crank shaft of an
engine for different positions of the crank is given in the figui'e
below. The scales are such that 1 inch = 150 Ihs. ft. of turning
moment, and 1 inch = 135 degrees of angle turned through by the
crank. The mean speed of the engine is 150 revolutions per
minute, and the moment of inertia of the flywheel is 944 Ibs.ft.^
units.

If the resisting moment is constant, find apj^roximately the greatest
fluctuation of speed.

200 -

100

180"
Crank angle
Fig. 97.

152 ELEMENTARY DYNAMICS

The area under the curve gives the work done per revolution.

This area =1*16 square inches.

. . 150 X 135 .^ „
1 sq. in. = — -—- — TT ft.-lbs.
loU

= 1127r ft.-lbs.

The work done per revolution = 112 x l*167r= ISOtt ft.-lbs.
If T is the resisting torque, we have
Tx27r=1307r,
.-. T = 65 lbs. ft.
This torque is represented in fig. 97 by the line abed.

Now from a to h, since the turning moment is greater than the
resisting moment, there will be a resultant moment increasing the
speed of the engine. The gain of kinetic energy will be represented
by the shaded area above ab.

Similarly from 6 to c the speed will be decreasing, and the loss
of kinetic energy will be given by the shaded area below be.

From c to d the speed will again be increasing, the gain of
kinetic energy being given by the shaded area above cd.

From d to a the speed will be decreasing, the loss of kinetic
energy being given by the shaded area below da.

The greatest change of energy will be given by the greatest of
the four shaded areas. By trial it will be found that the area above
ab is the greatest.

This area = 0*227 square inch
- 25-57r ft. -lbs.

Let o>i be the angular velocity at a, and w.^ be the angular
velocity at b. If I is the moment of inertia of the flywheel, we
have

^ Io>2^ - i Iwj^ = 25'57r X g,

!•«' ("'^ "'^ ) ^ — 944 (^)-

EXAMPLES 153

Now we do not exactly know the mean speed in terms of Wj
and (Uj since the speed is not changing uniformly. We may, how-
ever, as a close approximation, near enough for practical purposes,

assume that ^-^r — - is equal to the mean speed, i.e.

CO2 + a>i
2

'Zir X it>U

60 •

equation (1)

we get

(a>2 - coi) (0)2 +

1 °^

XTT X 32

944

(0,2-

. 51
<">)=9|

X TT X 32 X 60

4x47rx 150

i.e. (wo — tuj) = ^— ; — -— - radians per second.

^ ' 944 X 47r X 150 ^

Or, if Ni and Ng are the extreme limits of speed in revolutions
per minute,

(a>2 — 0)1) X 60

(N2-N1)

27r

51 X 16 X 60 X 60

944 X 47r X 150
= 1 '65 revolutions per minute.
The percentage fluctuation in speed

= ^'^^00

= 1*1 per cent.

Example (15). Taking the figures given in the example on
p. 100, find the energy lost in heat due to the impact of the two
wheels.

The kinetic energy before impact

'27r X 500\2

/27r X 0UU\2

^^ V 60 7 ^^-'P^^^'

154

ELEMENTARY DYNAMICS

The kinetic energy after impact
1 47r2

2 3600
1 47r2

2 3600
.'. The energy lost in heat
1 47r2

(4 X 253^ + 2 X .3542)

X 5-07 X 10^ ft.-pdls.

2 3600
^^'^x 10^x4-93
9T64

=-. 84-3 ft.-lbs.

{10«- 0-507 X lO*'} X ^ft.-lbs.

Examples. Chapter VI

1. A cage weighing 3 tons is supported by a steel rope which weighs 20 lbs.
per yard, and is being lifted at a uniform speed. How much work is done
when the free length of the rope is shortened from 1500 feet to 300 feet ?

If the uniform speed is 2 feet per second, at what rate is work being done
in lifting (1) when the depth is 1500 feet, (2) when it is 300 feet ?

2. How many foot-pounds of work are done in pumping 1 ton of water
into a boiler against a pressure of 150 lbs. per square inch? If the pump has
a solid ram of 2 inches diameter, with a 10-inch stroke, how many strokes
are required and what is the thrust on the ram ?

1 cubic foot of water weighs 62*5 lbs.

3. The values of the net pressure on the piston of an engine during one
stroke are given in the following table. The length of the stroke is 16 inches,
and the distances are measured in inches from one end of the stroke. The
pressures are given in lbs. per square inch.

Distance

8
32

10
26

12
22

16
10

Pressure

EXAMPLES. CHAPTER VI

155

Draw a graph connecting the pressure and the distance and find the space-
average of the pressure.

If the piston diameter is 14 inches, how much work is done during the
stroke ?

4. What do you understand by the expressions " the work done by a force,"
and "the power supplied by an agent"? In what units are these quantities
usually measured ?

A truck, weighing 8 tons, is being drawn by a horse along a siding on a down
grade of 1 in 400, the horse exerting a constant horizontal pull of 120 lbs. on
a chain inclined at 30° to the rails. The frictional resistances to motion of
the truck amount to 12 lbs. per ton. Find how far the truck will be moved
from rest in one minute, and the power which the horse is exerting at the
end of the minute.

5. The table below gives the horse-power trariamitted by the propeller
shafts for different speeds in the case of the " Mauretania " Atlantic Liner.

Draw a graph shewing how the resistance overcome varies with the speed.

Shaft
horse-power

12,500

20,000

27,300

37,500

51,000

75,000

Speed in
knots

!
16 18

24-

6. A planing machine is working with its stroke set at 4 feet, and is planing
a piece of work 3 feet long, the tool clearing the work 6 inches at each end.
There is a constant frictional resistance equivalent to 45 pounds at the tool,
and the cutting resistance is 560 lbs. more. The machine makes 20 complete
strokes per minute. Find its average rate of working, in horse-power.

If the cutting stroke take twice as long as the return stroke, find the average
speed of each, and if the highest cutting speed is 1| times the mean, find the
greatest rate of working.

7. A wagon weighing 10 tons is running at 6 miles per hour when the
brakes are applied and it is brought to rest in 15 yards. How much work is
done in bringing the wagon to rest, and what is the space-average of the total
braking force ?

156

ELEMENTARY DYNAMICS

8. A car, for which the frictional resistance is 30 lbs. per ton, is travelling
on the level at the rate of 30 feet per second, when it comes to a hill of 1 in
20, and after the speed has become steady on the hill the engine is working
at the same power as on the level. Find this steady speed.

If while the car is slowing down the tractive force of the engine increases
uniformly with the distance from the bottom of the hill until it reaches its
steady value on the hill, find how far the car travels before reaching its steady
speed.

9. The road and wind resistance to the motion of a motor-car weighing
30 cwt. is given by (40 + 0*03 v^) lbs., where v is the speed in miles per hour.
What horse-power will be required to drive the car up a hill of 1 in 30 at a
steady speed of 20 miles per hour ?

If the maximum tractive force the motor can supply is 220 lbs., what is the
maximum speed attainable up the hill ?

10. A body is given kinetic energy by a force acting on it. How is the
force related to the kinetic energy generated ?

A railway wagon weighing 12 tons runs into a stop when travelling at
6 miles per hour. The buffer springs of the wagon are such that each exerts
a force of 4 tons per inch of compression. Assuming the stop does not yield,
find how much the springs are compressed when the wagon is for the instant
brought to rest.

11. A motor-car, weighing 1 ton, ascends a hill of slope 1 in 12 at a
uniform speed of 6 miles per hour. When it reaches the top it begins to
travel down a uniform slope of 1 in 40. If the tractive force produced by the
engine and the road-and-wind resistance are constant, find what the velocity
and the kinetic energy will be when the car has travelled 25 yards down the
slope.

12. A 5 ton truck is pulled from rest up a slope of 1 in 50 ; the pull to just
move it on the level is 126 pounds ; the tension in the rope varies with the
distance (along the slope) as given below. Find the total work done, the
kinetic energy at the end, and the horse-power exerted at the distance of
15 feet.

Distance
in feet

Pull in lbs.

450

570

620

640

620

550

470

410

350

EXAMPLES. CHAPTER VI

167

13. A body is acted upon by a force which causes motion. How are the
momentum and the kinetic energy generated, each related to the force acting?

It is estimated that the actual work done in accelerating a steamer of
1000 tons displacement from rest, for the first 210 seconds, is 4500 foot-tons,
and the distance travelled is 2000 feet. Find,

(1) the speed attained,

(2) the space -average of the accelerating force,

(3) the t^me-average of the accelerating force.

14. The total pressure on the base of a shell during its passage along the
bore is given in the table below. The total travel is 4^ feet and it may be
assumed that 6 per cent, of the pressure is used in overcoming friction and
imparting rotation to the shell. Plot the pressure- distance curve and estimate
the muzzle velocity of the shell, given that it weighs 12 lbs.

Distance
in feet

Total pres-
sure, tons

123

130

124

106

15. A train, running on the level, with steam shut off, has its speed
reduced from 52 miles per hour to 48 miles per hour in 800 yards. Assuming
that the resistance to motion is constant, find its value per ton of the train.

If the train weighs 200 tons, at what rate, in horse-power, must work be
done to keep it moving at 50 miles per hour against the same constant re-
sistance?

16. A motor lorry, of total weight 8 tons, starts from rest and is observed
to have the accelerations at different points given below :

Distance, feet

Acceleration,
ft. per sec. per sec.

21 1-57

1-2

0-87

0-54

0-30

Find the accelerating force for each distance, and plot a force-distance
curve. From the curve, find the total work done in accelerating, and the
velocity at each distance.

158

ELEMENTARY DYNAMICS

17. State the principle of conservation of energy as it applies to mechani-
cal problems.

A mass is suspended by a string ABC, 3 feet long, from the point A, and
is swinging through an angle of 30° on either side of the vertical. Find the
velocity of the mass when it is vertically below A.

If at this instant the point B in the string is suddenly fixed, find the new
angle of swing, AB being equal to ^ BC.

18. A light spring, whose inertia may be neglected, is such that a | lb.
weight will compress it 1 inch. It is compressed 2 inches and placed on a
table so that it will expand in a vertical direction. If a ^ lb. weight is put
upon it and the spring is released, what is the velocity of the | lb. at the
instant that it leaves the spring?

19. In fig. 98 A is a hard steel shaft which can slide horizontally in the
bearings BB. It is provided with stops which are just in contact with the end
of the spring bujffer. A lead bullet moving with a velocity of 1000 feet per
second strikes the end of the shaft in the centre. The mass of the shaft is
5 lbs., that of the bullet 0*04 lb., and the spring compresses 1 inch under a
load of 20 lbs. The force required to overcome the friction of the shaft is
1 lb. Assuming that the mass of the spring may be neglected, find to the
nearest hundredth of an inch by how much the spring is compressed by the
blow.

^^:::i

f b"

■^

Fig. 98.

20. A bullet, weighing 0*025 lb. and moving with a velocity of 2000 feet
per second, strikes a ballistic pendulum weighing 20 lbs., and remains in it
after the impact. Determine the percentage of kinetic energy lost, and the
average pressure exerted between the bullet and pendulum, supposing the
former to come to rest after travelling six inches.

21. A railway wagon of total load 20 tons is shunted on to a siding and
reaches a hydraulic buffer stop at a speed of 6 miles per hour. The buffer
stop is such that it exerts a substantially steady force of 40 tons while the
buffers are being pushed in, but only exerts a negligible force while returning.
The wagon buffer springs each require a force of 4 tons to compress them
1 inch.

EXAMPLES. CHAPTER VI 159

Find (1) the distance the wagon moves before coming to rest after striking
the buffer stop, (2) the velocity of the wagon at the instant it leaves the
buffer stop.

22. A steamer of 10,000 tons, when its engines are not working, slows
down from 7 to 5 feet per second in 90 feet. Find the average resistance to
motion.

If the engines are started and the speed is uniformly increased from 5 to
7 feet per second in 120 feet at what average rate in horse-power are the
engines working during this motion, the resistance to motion being taken as
uniform, and equal to the average resistance in the first part of the question?

23. A train weighing 200 tons has its speed reduced by a constant braking
force from 60 miles per hour to 40 miles per hour in 4 seconds. Determine
the change in the momentum of the train, the change in its kinetic energy,
and the distance travelled. Also find the value of the retarding force, taking
it as (1) the change of momentum per second, and (2) the change of kinetic
energy per foot of distance.

24. The propulsion horse-power required to drive a steamer of 12,000 tons
displacement at a steady speed of 20 knots is 15,000. Assuming that the re-
sistance is proportional to the square of the speed, and that the engine exerts
a constant propeller thrust at all speeds, find the initial acceleration, when
the steamer starts from rest, and the acceleration when the speed is 10 knots.

(Take 1 knot = 100 feet per minute.)

25. Shew that the twisting moment in a shaft which is transmitting

power is given by , where H is the horse-power transmitted and N

27rlN

the number of revolutions per minute.

A torsion dynamometer, which registers the angle of twist per foot length
of the shaft when running, was used to determine the horse-power given out
by a steam turbine. The mean speed of the shaft was found to be 2000 r.p.m.
and the dynamometer registered 8i^'25 minutes. Find the horse-power given
out by the turbine, having given that a couple of 72 inch-lbs. applied to the
shaft produces an angle of twist per foot length of one minute.

26. The drum of a capstan has an effective diameter of 10 inches. Eight
men on the capstan bars, walking round three times a minute, at a mean
radius of 5^ feet, can produce a pull of 2i tons in the rope being wound up.
Find the horse-power got out, and the horse-power wasted in friction etc.,
assuming each man exerts a force of 70 lbs. on the bars. The tension in the
rope coming off the drum may be neglected.

160

ELEMENTARY DYNAMICS

27. What is understood by the moment of inertia of a body about an axis?

Shew that the kinetic energy of a body rotating with angular velocity w
about an axis is 1 1 w^, where I is the moment of inertia of the body about the
axis.

A rectangular door of sides 10 and 6 feet and weighing 500 lbs., swings
about a horizontal axis fixed along its greater edge ; find its angular velocity
as it reaches the vertical from rest in the horizontal position. What is the
pull on the axis as it reaches the vertical? What would be the angular
velocity if the hinges exert a constant frictional couple of 100 inch-lbs.?

28. An engine works against a constant load which absorbs 25 horse-
power when the speed is 250 revolutions per minute. The flywheel of the
engine weighs 3 tons, and its radius of gyration is 3 feet.

Find (1) the work done against the load in one revolution, (2) the energy
stored in the flywheel when the speed is 250 revolutions per minute.

If the source of power be cut off while the load remains constant, in how
many revolutions will the speed drop from 250 to 240 revolutions per minute?

20. The figure below shews the torque exerted on the crank of a two-
cylinder tandem gas engine for different cranks angles during one revolution.
The torque scale is such that 1 inch = 40,000 lbs. feet.

The moment of inertia of the flywheel is 90 tons ft.^ units, and the speed
is 200 revolutions per minute. The engine is working against a constant re-
sisting moment, and giving the same torque-angle diagram every revolution.

Find graphically the resisting torque and estimate the maximum percentage
fluctuation of flywheel speed.

360'

Fig. 99.

EXAMPLES. CHAPTER VI 161

30. A 6-inch shaft is being tuwied in a lathe and it is found that the
tangential pressure on the tool at the cutting edge is 785 lbs. The traverse
of the saddle is ^V inch per revolution of the mandrel and the speed of the
latter is 20 revolutions per minute. Find the energy usefully expended in
cutting per foot of traverse. Find also the efficiency of the machine if 1*3
horse-power is being supplied by the belt.

81. Two uniform discs of steel are each 3 feet in diameter. They are
mounted side by side on the same shaft, and are free to move indepen-
dently and without friction. One disc A is rotating at 500 revolutions per
minute, the other, B, is at rest. A projection on the disc A impinges on
a similar projection on the disc B and the blow dissipates one-fifth of the
kinetic energy. Find the angular velocities of the two discs after the blow,
and determine how long it will be until the projection on B overtakes the
projection on A.

Also find the magnitude of the blow, if each disc weigh 250 lbs. and if the
projections on the disc are each 15 inches from the centre of the shaft.

32. In example (14), p. 109, shew that the kinetic energy of the system is
reduced in the ratio 4 to 5.

L. E. D. 11

CHAPTER VII

Units and Dimensions

In the first chapter we noted that there were three fundamental
conceptions, space, mass, and time, and we saw how these were
measured in certain units. The units were those of length, mass,
and time.

We also called attention to the fact that all physical quantities
could be defined in terms of these fundamental units. In the
previous chapters we have introduced all the physical quantities
we require jn elementary dynamics, and we have based the
measurement of each of these quantities directly on the funda-
mental units, or on phj^sical quantities the measurement of which
had already been based on these units.

It will be useful, as a revision of the previous chapters and also
for other purposes, to examine again, in some detail, how each of
the physical quantities we have introduced and used is related to
the units of length, mass, and time. We shall then be able to see
immediately how the measure of each quantity would be affected
by a change in any or all of the fundamental units. We will de-
note these by L, M, and T respectively.

Let us consider density. This we defined as the mass per unit
volume, i.e. the mass of a body divided by the volume of the body.

Now the volume is equal to the product of length, breadth and
thickness, and each of these is a length. We say, therefore, that
volume has three elements of length or, has the dimensions
(length )^ i.e. L=^.

M
. *. Density has the dimensions -^

= L-MVI . T",
T^ denoting that it has no time dimension.

UNITS AND DIMENSIONS 163

The significance of this will perhaps be realised better if we take
a concrete example.

Example (1). The density of steelin c.G.s. units is 7 '78. Find
the density in f.p.s. units, given that, 1 centimetre = 0'0^2% footf
and 1 ^ram=r 0-00221 Ih.

The dimensions are L~^ . M . T°.

.*. The density in lbs. per cubic foot

= 7-78 x(0-0328)-3x 0-00221

= 484.

We will now find the dimensions of certain quantities in length,
mass, and time.

Dimensions

Absolute System of Units

Velocity (linear) = ^^^ = L . J-^ - L . M^ . T~\

Acceleration (linear) = ^^ ^^^ 7 ^ L . T-^ . J-^ = L . M<* . T'^.

^ ' time

Angle '(radians) = ^^ - L . L'^ = L^ . M^ . T^.

Angular velocity = — ; = T~^ ^ L^ . M^ . T~^.

time

Angular acceleration = ^"^^ ^! ^^ ^^^ ^ = T'^ . T-^ = L° . M^ . T~\

time

Momentum ■= mass x velocity =M.L.T~^=L.M. T~^.

,- momentum , , , .. _^ „

Force = -. = M . L . T-i . T-i = L . M . T-2.

time

Weight = force = L . M . T-2.

Impulse = force x time = L . M . T"- . T - L . M . T~^.
Angular momentum = momentum x distance = L . M . T"^ . L
- L2 . M . T-i.

11 2

164 ELEMENTARY DYNAMICS

Torque = force x distance = L . M . T"^ . L = L^ . M . T'^.

Impulsive torque = torque x time = L^ . M . T~^ . T = L^ . M . T~^.

Moment of inertia = mass x (distance)^ = M . L^ = L^ . M . T^,

Work = force x distance ^L.M.T-^.L^L^.M. T"2.

Energy = capacity for work = L^ . M . T"^.

worlc
Powers -r^ = LMVI.T-2.T-i=L2.M.T-3.
time

Use of dimensions in checking formulae

If we have an equation involving physical quantities, the
dimensions in each of the units, length, mass, and time, must be
the same on the two sides of the equation. If this were not so,
by changing the units of length, mass, and time we should alter
the equation.

Suppose, for example, in an equation, the dimensions of the left-
hand side in mass were M^, and on the right-hand side M, and we
wished to change from lbs. to grams. We should multiply the left-
hand side by (453-6)2 and the right-hand side by 453-6, which
would completely alter the equation.

We will examine a few of the formulae we have derived in the
previous chapters. The student should examine all the remainder.

(1) Net work done = kinetic energy gained,
or p .s = \ Mv\

The left-hand side has dimensions, L^. M . T"^.
The right-hand side has dimensions, M . (L . T"^)^

= L2. M.T-2.

(2) Acceleration = v . -^ ,

The left-hand side has dimensions, L . T-^

L . T~^ X L . T~'
The right-hand side has dimensions, —'

L
L.T-2.

UNITS AND DIMENSIONS 165

3) Torque - moment of inertia x angular acceleration,

T-:I.A.

The left-hand side has dimensions, L^ . M . T"^.
The rightrhand side has dimensions, L^. M . T~^.

Example (2). Find the number of watts in 1 horse-power,
having given: 1 lh. = ^b^'^ grams, ly^.^^ 30*48 cms., and g = ^2'2 feet
per sec. per sec.

\ horse-power = 550 ft.-lbs. per sec.

= 550 X 32*2 abs. f.p.s. units of power.

The dimensions of power are L^ . M . T~^.

/. 1 horse-power = 550 x 32*2 x 30*482 x 453*6

= 746 X 10^ ergs per sec.

= 746 watts.

CHAPTER VIII

Simple Harmonic Motion

In many cases of motion we have a body which moves back-
wards and forwards about a mean position, as, for example, the
piston of a steam engine or the pendulum of a clock.

We will now investigate a particularly simple form of such a
motion called Simple Harmonic Motion.

A body is said to move with a simple harmonic motion in
a straight line when it has an acceleration directed towards some
fixed point in its path, and proportional to the distance of the body
from the fixed point.

Suppose, for example, that a body P moves in the straight line
A'A with simple harmonic motion about a fixed point O.

Fig. 100.

The acceleration will be in direction PO always, and equal to
/A. OP, where /x is a constant*.

If we consider OA to be the positive direction of the displace-
ment, the acceleration along PO will be negative, and we must

write V —

Acceleration - - fi , s,
where s = OP.

* It should be noted that /a is a constant for a definite system of units,
but its value will change if the unit of time be changed. Writing accelera-
tion =/x X distance, we see immediately, that in order that the dimensions of
the two sides of the equation may be the same, /x must have dimensions

LMVI« . T-2.

SIMPLE HARMONIC MOTION

167

Now suppose the body is started with a given velocity, u say,
when it is at O, then, since the acceleration is always towards O,
it follows that the velocity will gradually be diminished to zero,
and the body will for an instant be at rest at the point A, say.
The body P will then start moving back to O with a velocity which
will increase to n, but in the direction AO. Exactly the same type
of motion will now occur to the left of O. We see then, that the
body will continue to move backwards and forwards along the
path AA'.

The velocity-time curve must be something like that shewn in
fig. 101,

Fig. 101.

where OB represents the time to reach the first rest position A,
and 00 represents the time to make one complete reciprocation, i.e.
the time for the body to travel from O to A, A to A' and A' to O.
Also, we can see that the displacement-time curve must be some-
thing like the dotted curve in tig. 101, where BD represents OA.

Now in order to analyse the motion we want to know the exact
shape of these curves.

For the space-time curve let us try the equation

s = rt sin ht.

168 ELEMENTARY DYNAMICS

From the example on p. 14 we have

The velocity, v = -j-^ab cos bt.

dv
The acceleration, a = -y^ = - a6^ sin ht
at

= -b\s.

If, therefore, we make b^ equal to /x, we see immediately that
this displacement is the one w^e want.

We have then for simple harmonic motion

The acceleration — — fx . s.

The velocity = a . Jfx . cos \/fxt.

The displacement = a sin J fit.

Also, we may obtain the acceleration in terms of the time, and
the velocity in terms of the displacement.

The acceleration = — /xa sin ^J/xt

The velocity = ajfx . v 1 - sin^ si fit

av^.yi-(iy

= J jx . sja^ - s\

The periodic time is the time for one complete vibration.

This is the time for the body to move from O to A, A to A' and
back to O.

From the curve we see that this is represented by the length OC,
for which J fit = 27r,

.*. the periodic time = — p .

The greatest displacement (a) on either side of the centre is called
the amplitude of the motion.

SIMPLE HARMONIC MOTION

169

It should be noted that the periodic time is independent of the
amplitude and depends only on the value of /x.

Acceleration
' Displacement *

Hence we may say, for any simple harmonic motion,

The periodic time = 27r

/Di

Displacement •

/////////////

Acceleration

Example ( 1 ). A mass of i lbs. is suspended from the end of a
light helical spring, and when the system is at rest the spring is
found to be extended 3 inches. The mass is now slowly depressed a
further distance of 2 inches and then let go.

Find {a) the periodic time of vibration., (b) the
maxim,um, velocity.

Experiment shews that the pull produced by
a helical spring is proportional to the extension.

If P = the pull, and x — the extension, then
we may write

P = \x, where A, is a constant for the spring.
In this case, when P = 4^ pdls.,
x = ^ feet.

. A ^

Fig. 102.

or, A. == 1 6^.

Let s = the displacement at any instant downwards from the
equilibrium position.

From the 2nd law of momentum we have

(P — ig) = 4a, where a = the acceleration upwards.
P-4^

= ig.s,

170 ELEMENTARY DYNAMICS

or, a varies as the displacement, and therefore the motion is simple
harmonic.

rp, • 1- ,• n /Displacement

ihe periodic time = Stt x . / — ^-

V Acceleration

"•/I

iff

= — 7=r seconds.

4v/2

The number of complete vibrations per minute

= X bO

= 108.

The maximum velocity occurs when the mass passes the equi-
librium position, i.e. when s = 0.

.'. The maximum velocity = Jfx . a

4v^2

feet per second.

Oscillation

Now suppose instead of a mass moving in a straight line we
have a mass oscillating in a circle about a fixed axis. It is easily
seen that all the reasoning we have just employed will hold, so
long as we substitute angular displacement for linear displacement,
angular velocity for linear velocity and angular acceleration for
linear acceleration.

Simple Pendulum

A simple pendulum is a mass, which may be considered small,
attached to a fixed point by a weightless rod or string and oscil-
lating in a vertical plane.

SIMPLE PENDULUM

171

Let m = the mass, I = the length of the string, and let the
string make an angle B with the vertical at
some instant of time.

Treat the problem as one of rotation about
O, and apply the 2nd law of momentum to
the mass m.

The only forces acting on m are the ten-
sion T in the string and the weight mg.

The turning moment about O
— mgl sin 6.

The moment of inertia of m about O - ml^,
since the mass m is small.

.-. mgl sin 6 = — rrd'^ . A,
where A is the angular acceleration in a
counter-clockwise direction.

mgl sin Q

.'. The angular acceleration =

mr-

^ . sin 0,

i.e. the angular acceleration varies as sin 0.

This is not a case of simple harmonic motion, but if we keep
the angle 6 always small, i.e. the amplitude small, then sin 6 = 0,
very nearly. In this case we may write.

The angular acceleration = j.O,

i.e. the angular acceleration varies as the angle of displacement,
and the motion is simple harmonic.

The periodic time = 2i

-2i

W Ac

Displacement
Acceleration

The fact, that for small swings the pendulum has a definite
periodic time independent of the amplitude, is the reason why it is
frequently employed to control clocks.

172

ELEMENTARY DYNAMICS

Example (2). Find the length of a simple pendulum to heat
seconds, i,e. a pendulum which makes one complete oscillation in
2 seconds.

The periodic time = 27r . / - .

or,

V 32'

= 3-25 feet.
It may be noted that if we keep the length of the pendulum
exactly constant, then for different positions on the earth's surface
the periodic time will vary inversely as the square root of the
acceleration due to gravity. This gives us a convenient means of
comparing the values of g at different places.

Compound Pendulum

A body which is arranged to swing like a pendulum, but in
which the whole mass is not concen-
trated at a point, is called a compound
pendulum.

Let O be the centre of suspension, G
be the centre of gravity, and let the body
be at some instant in a position such
that OG is inclined at an angle 6 to the
vertical. See fig. 104.

Let OG = h.

The weight may be considered to act
through G.

The restoring moment about the axis
O =mg xOO sin 6

= mgh . Of
if e is always small. Fig. 104.

COMPOUND PENDULUM 173

The equation of motion becomes

I . A = - mgh . 0,
where I is the moment of inertia of the body about O, or,

i.e. the body has a simple harmonic motion.
The periodic time = 27r ^ — j-

where h is the radius of gyration of the body about O.
For the simple pendulum, the periodic time

Hence the length of a simple pendulum which has the same
periodic time as the compound pendulum

~ h'

This length is called the length of the simple equivalent pendu-
lum (s.E.p.).

Example (3). The centre of gravity of a connecting rod is 4*5
feet from the centre of the small end. When suspended on a knife
edge at the centre of the small end it has the same periodic time as
a plumb-line of Q'2 feet. If the rod weighs 500 lbs,, find the mo-
ment of inertia about an axis through the centre of gravity parallel
to the axis of the end.

We have the simple equivalent pendulum

~ h' ,

174

ELEMENTARY DYNAMICS

where k is the radius of gyration about the axis of the small end,
and h is the distance 4 -5 feet.

6-2 =

4-5'

or.

A;=x/6-2x4-5

= V27^

The moment of inertia about the axis through the centre of
gravity

= Mk^ - MA2

= 500 {27-9 -4-52}
= 500 X (27-9 -20-3)
= 3800 lbs. ft.2

Example (4). A solid cylinder of radius r is free to oscillate
in a vertical plane about a fixed axis A as shewn in the figure below.
21ie cylinder is connected to the link IKE by a pin joint at the axis,
which is horizontal.

Show that if the inertia and weight of the link AB is negligible

Fig. 105.

EXAMPLES

175

and there is no friction, the periodic time for a small oscillation
ivill be given by 2ir .1 — .

Further shew that if the cylinder is fixed to the link the periodic

time will be increased in the ratio

2R-^ + r2

2R^

\st Ca^e. Since the pin joint at B is frictionless, the onlj ex-
ternal forces acting on the cylinder are
the pull P along link BA and the weight
of the cylinder mg acting vertically down-
wards.

Now since there is no couple on the
cylinder, it will have no rotation about
B, and a diameter which is horizontal,
say, will remain horizontal, i.e. there will
be no relative motion between any point
of the cylinder and B.

Therefore the acceleration of every point
of the cylinder will equal the acceleration
of B. The latter may be considered as
consisting of two components,

(1) an acceleration w^BA in the direc-
tion BA, where w is the angular velocity
of AB, and

(2) an acceleration a, say, perpendicular to BA.

Also, a = R . O, where O is the angular acceleration of BA.
Resolving the forces on the cylinder perpendicular to BA
we get

mg sin 6 = — mR . fi.

Fig. 106.

I.e.

- ^ . ^, if ^ is small.

This is a simple harmonic motion.
The periodic time

i.y

176 ELEMENTARY DYNAMICS

2nd Case. If the cylinder is fixed to the link it will rotate with
the link and we get a compound pendulum as treated on p. 172.

YYtV

The moment of inertia about A = mR^ + —^ .

.'. The (radius of gyration)^ about A = R^ + — .

The centre of gravity of the pendulum is at a distance R from A.

2~'

.'. The periodic time _ ^ ,

i.e. the periodic time is increased in the ratio ^ /

^ V 2R2

It will be seen that in the first case the arrangement behaves
as a simple pendulum of length R, and it will be remembered that
in the simple pendulum the mass was to be considered small, i.e.
we were neglecting the rotation of the mass itself about its centre
of mass. This is equivalent, in the above example, to neglecting

the term — ^ .

Example (5). The balance wheel of a ivatch weighs ^i^ ounce^
and its radius of gyration is J inch. It is controlled by aflat spiral
spring and has a periodic time of \ second. What torque per
degree of twist must the spring exert ?

Let X - the torque in lbs. inch exerted by the spring per degree
of twist.

If the wheel is rotated through 6 radians from its mean position,
then

The restoring couple due to the spring

= Tn—^ Pdls. ft.

TT X 12

Let A = the angular acceleration of the wheel in the direction
of twist.

EXAMPLES

177

The equation of motion of the wheel is,

300

1\2 X.^xlSOx^r

X A = — -

A =

TTX 12

A. X 15 X 4800 X 16 X 144 x 32

= -(1-89 X lO^x A).^.

27r 1

The periodic time = , - ^ .

Vl-89xl0»xA 2

.-. X

1-89x10''

8-96 X 10-8 lbs. ins.

Harmonic Motion and Circular Motion

We will now see how we can obtain mechanically a simple
harmonic motion from a circular
motion or vice-versa.

In fig. 107, Q represents a pin
fixed to a circular disc which is
rotating about O. This pin en-
gages with a slotted link K con-
strained by guides to move in a
line passing through O.

As the disc is rotated the pin
Q will slide in the slot, and the
link K will move backwards
and forwards in a straight line.
Draw QP perpendicular to the
line of motion of K, then the
motion of P will be the same as
the motion of the link K.

Suppose the disc is rotated
with a constant angular ve-
locity OJ.

L. E. D. 12

Fig. 107,

178

ELEMENTARY DYNAMICS

The acceleration of Q = w^ . QO, and is always towards O.
The acceleration of P = the component of the acceleration of Q
in the line PO, ^

= (iP . QO . cos POQ

= 0)2 . PO,

i.e. the acceleration of P is always towards O, and varies as its
displacement from O.

.'. P has a simple harmonic motion.

It is easy to see that the periodic time is the time for Q to

make one complete rotation, i.e. — . Here w^ corresponds to /a in
the previous pages.

The other results follow very simply.

The velocity of P = the component of
Q's velocity in the direction OP,

= <o . OQ . sin POQ

= u) . OQ . cos QOC,
where 00 is perpendicular to OP (fig. 107).
If OQ = cr, and if we start measuring
the time when P is at O,
the angle QOC = (at.

.', The velocity of P = wci cos int.
For the displacement, in the same way,
we have

s = OP

= a sin oit.

Crank and con^iecting rod

In the crank and connecting rod
mechanism, used in steam engines and
other machines, the slotted link is replaced by a connecting rod
QA, as shewn in fig. 108. The" crosshead has not a true simple
harmonic motion.

Fig. 108.

HARMONIC MOTION AND CIRCULAR MOTION

179

We can see, however, that if the rod QA is very long compared
with the crank OQ, then the angle <fi will always be small and the
motion of A will be very nearly the same as that of P, where QP
is perpendicular to the line of stroke,

i.e. A has approximately a simple harmonic motion.

Sxample (6). The slide-valve of a steam engine has a total
travel of 4 inches. It is driven from the crank shaft hy an eccentric^
the rod of which is so long com^pared with the travel that the motion
of the valve may he considered simple harmonic. If the crank slwft
rotates at 240 revolutions per mifiute, find (1 ) the maximum, velocity
of the valve, (2) the velocity of the valve at one-third of the stroke,
(3) the time to travel the middle third of its path.

Fig. 109.

Let AA' represent the travel of the valve, i.e. 4 inches. Draw
a circle with AA' as diameter. The motion of the valve will be
given by the component of motion, parallel to AA', of a point Q
rotating round the circle with an angular velocity (w) equal to that
of the crank shaft.

27r X 240
60

= Stt radians per second.

12—2

180 ELEMENTARY DYNAMICS

(1) The maximum velocity will occur when the valve is at O,
i.e. Q is at Qj.

.*. The maximum velocity = Stt x ^^

= 4-19 feet per second.

(2) Let 0P2 = |^AA'.

The velocity at one-third stroke
= 4-19 sin (9

=-- 4-19 X — --

OQ2

_ 4-19 X n/2^-(|)^

2
= 3*94 feet per second.
(3) LetOPg^l^AA'.

The time to travel the middle third of the stroke, i.e. P2P3,
= the time for Q to rotate through an angle (tt — 26)

IT -26 ' ^
= —5 — seconds.

OTT

Now cos 6 = \^
i.e. 6 =- 70-5° = 0-391 tt radians.

. rru .. • A 1-0-782
. . Ihe time required =

= 0-0272 second.

Example (7). The drum of a steam engine indicator is driven
by a cord wrapped round part of the drum and is controlled hy a
helical spring which always exerts a couple opposed to that produced
hy the pull of the cord. The end of the cord has a motion propor-
tional to the piston's motion, and the latter Tnay be assumed to be
simple harmonic. Shew that, neglecting friction and the effect of
stretch of the cord, the spri7ig must he wound up so that the mean

EXAMPLES 181

torque it exerts on the drum is not less than the absolute magnitude

— ^ - 1 L where N= the speed of the engine in revolutions per

minute, n = the number of oscillations the drum would make per
minute if oscillating freely under the control of the spring^ \ = the
torque required to twist the spring through 1 radian, and a = the
amplitude of the motion of the drum.

Fig. 110.

Let 0 = the angle of rotation of the drum measured in a clock-
wise direction from the mean position.

Let C = the couple produced by the spring for this position. '

„ T = the tension in the cord.

,, r = the radius of the drum.

,, I =: the moment of inertia of the drum.

The resultant torque on the drum = Tr-C.

.'. From the 2nd law of momentum we have

Tr-C = I.A (1),

where A is the angular acceleration measured in a clockwise
direction.

182 ELEMENTARY DYNAMICS

Let (f> — the angle of twist of the spring when the drum is in its
mean position.

For angle of rotation 0, the angle of twist in the spring = {<f> + 0).

.'. c = \(<f> + 0).

Substituting in (1) we get

T.r-X(<^ + ^)-I.A (2).

The motion of the end of the string is given by

27rN
x = ra sin wt, where ot = -^^ .

But x = re,

.'. 0 = a sin oit

Also A == — ao)^ sin (lit.

.'. From (2) we get

Tr = \ (<j> + a sin wt) — lacu^ sin w^,

i.e. Tr = \(f> — (law^ — Xa) sin lot.

Now if the drum's motion is to be the same as that of the string
T must always be positive, and if (law^ — Xa) is positive, A^ must
be not less than (laay^ - \a) sin mt.

The maximum value of sin mt^l.

.* . \<f> <^ law" — Xa.
But X<fi = the mean torque due to the spring.

,'. The mean torque <(; Xa \— — IV (3).

Suppose the drum were oscillating with its own spring, and there
were no cord, we should have

-X(9 = I.A,

X ^
I.e. A = - = .d,

which is a simple harmonic motion.

EXAMPLES. CHAPTER VIII 183

The periodic time

-v/;-

But the periodic time

_60
n
I 602

Also

" =60'^ •

.-. From (3),

the mean

torque <j: Xa j — ^ -

-■)

If (laoo^ —\a) is negative, i.e. ( — ^ - 1 j is negative, then we must

take sin co^ = — 1 , and we get

the mean torque <j; Xa -^ 1 ^ r »

i.e. the mean torque must not be less than the absolute magnitude

of Xa -^ —

Examples. Chapter VIII

X. A mass is suspended at the end of a light vertical spring, and when at
rest stretches the spring I feet. Shew that if it is slightly displaced from this
position of rest and then let go it will vibrate with a periodic time equal to

27r . / - , where g is the acceleration due to gravity in feet per second per
second.

2. A body of mass 2 lbs. moves in such a way that the curve connecting
its displacement (in feet) from a given point in its line of action and the time
(in seconds) is represented by the equation s = 2sin3«. Find the time of a
complete oscillation, and the force acting on it when at a distance of 1 foot
from the fixed point. Plot the velocity-time curve for a complete oscillation.

184 ELEMENTARY DYNAMICS

3. A pendulum of a clock beats seconds. If the clock keeps correct time
in one place, how many seconds per day will it lose in a place where the ac-
celeration due to gravity is decreased by y^ per cent.?

4. A part of a machine weighing 60 lbs. describes Simple Harmonic Motion,
making 180 complete vibrations per minute, and the length of the stroke is
2 feet. Find the greatest velocity, and the force acting on the moving part
at one end of its stroke and at ^ of its stroke.

5. A uniform rod, weighted at one end, floats in water with its axis vertical
and part of the rod projecting above the free surface. The total weight of the
rod is 3 lbs. and its cross-section is circular and of diameter 2 inches. If it
is depressed vertically and then let go find the periodic time of vibration.

When a body is in water the upthrust on it is equal to the weight of the
volume of water displaced.

6. A particle of mass m moves in a straight line under the action of a force,
towards a point O in the line, the value of which, at a distance x from the
point, is nifjix. Determine the time of an oscillation.

Shew that the potential energy when the particle is at a distance x from O
is ^nifjuc^.

Shew also that the motion of a simple pendulum making small oscillations
approximates to the same type.

7. A simple pendulum, of length 3 feet, oscillates through an angle of
12 degrees on either side of the vertical. Using the principle of conservation
of energy, find the maximum velocity. Compare this with the maximum
velocity obtained by considering the motion to be simple harmonic.

8. A thin rod, 3 feet long, is pivoted at one end and oscillates through a
small angle in a vertical plane. Find the periodic time of oscillation, and
shew that it will be the same if the rod is pivoted at a point one-third the
length from the end as it is when pivoted about the end.

9. A mass of 5 lbs. has a simple harmonic motion, the period of which is
^ second and the amplitude 6 inches. Draw diagrams to stated scales shewing
(i) the force acting on the mass as a function of the time, (ii) the velocity as
a function of the displacement.

10. A flywheel is hung up with its axis vertical by two long ropes parallel
to and equidistant from its axis so that it can execute torsional oscillations.
It is found that a static couple of 50 lbs. ft. will turn it through an angle of
one-tenth of a radian, and that if it be turned through a small angle and then
let go the period of a complete oscillation backwards and forwards is 5 seconds.
Calculate how much energy will be stored in this flywheel when running
200 revolutions per minute.

EXAMPLES. CHAPTER VIII

186

11. A mass of 5 lbs. is suspended by a spring, and when at rest, just
touches the platform of a spring balance without the latter taking any of the
weight. The spring is such that it extends 1 inch for 10 lbs. and the platform
of the spring balance is depressed 1 inch for 20 lbs. If the mass of 5 lbs. is
slowly depressed | inch and then let go, find the periodic time of vibration.
Find, also, the maximum height the mass rises and the maximum velocity it
attains.

12. A Tee square, of the dimensions shewn in the figure, is suspended
at A, and oscillates through a small angle in its own

plane. The horizontal cross-piece is twice as thick as
the vertical piece. Find the periodic time of oscillation.

13. A rectangular bar magnet, weighing 0*2 lb., is
6 inches long and of cross-section ^ inch by J inch.
It is pivoted in a horizontal plane so that it is free to
swing about a vertical axis through its middle. Find
the periodic time of oscillation in a place where the
earth's magnetic field has a horizontal intensity of
0"19 dyne, if the magnetic moment of the magnet is
400 dyne-cm. units.

14. A rotary valve for an engine makes 500 com-
plete oscillations per minute about a horizontal axis,
the total angle of rotation being 90 degrees. The motion
is obtained from a rack which moves with a simple
harmonic motion, and drives a pinion of effective

diameter 1| inches keyed to the valve spindle. If the K--- 9" >\

moment of inertia of the valve, spindle, and pinion is ' -i?- 1 1 1 '

30 lb. -inch units, find, neglecting friction, the maxi- ^'

mum total pressure between the teeth of the rack and the pinion.

15. A machine is carried by, and in the middle of, two girders, built in
at the ends. The dead weight of the machine causes a deflection of | inch.
Shew that if the girders weigh jijjth of that of the engine, the frequency
of a small vertical oscillation will be between 252 and 265 vibrations per
minute.

Assume firstly, that all the weight of the girders is at the centre, and secondly,
that the weight of the girders may be neglected in comparison with the weight
of the machine. *

16. The stroke of an engine is 6 inches, the connecting rod 10 inches, and
the speed of the crank 720 revolutions per minute. The line of stroke passes
through the axis of the crank shaft.

A
o-

-%-

2'!.

cq
1

1
1

186

ELEMENTARY DYNAMICS

Draw the velocity-space curve for the motion of the piston, and from this

construct the acceleration-space curve-
On the same diagram shew the acceleration- space curve, assuming the

connecting rod infinitely long.

17. A uniform rectangular block of length 12 inches and of height 6 inches
is suspended in a horizontal position by
two vertical strings AB and C D, as shewn ///y
in the figure. The centre of gravity of ^^^^^
the block is 2 feet below thfi points of
suspension. Find the time of a small
oscillation in the plane of the strings.

If the vertical strings are replaced by
two equal strings OB and OD, the centre
of gravity remaining at the same dis-
tance below the point of suspen8ion,what
will be the* periodic time for small oscil-
lations?

18. Shew that the motion of the
crosshead of an engine is approximately
simple harmonic.

Find the force that is necessary to
start the reciprocating parts of an engine
into motion at the beginning of each
stroke, when the engine is running at
120 revolutions per minute. The mass ^^^' ^^-'•

of the reciprocating parts is 1 ton, the stroke is 5 feet and the obliquity of
the connecting rod may be neglected.

CHAPTER IX

MISCELLANEOUS

Transmission of power by belts

Power is frequently transmitted from an engine shaft or motor
shaft to a machine by means of a belt connecting pulleys on the
two shafts as shewn in fig. 113. When the machine is doing work
the tension in the two sides of the belt will be different.

Fig. 113.
Let Tj = the tension in lbs. wt. in the tight side of the belt,
Ts = the tension in the slack side of the belt,
N = the speed in revolutions per minute of the machine

shaft,
R = the radius of the pulley on the machine shaft in feet.
The resultant turning moment on the machine shaft
= TiR - T2R
= (Ti - T2) R lbs. ft.
The angle turned through in 1 minute = 27rN radians.
The work done per minute = (Tj - To) R x 27rN ft. -lbs.

'. The horse-power transmitted to the machine = - ^ ^ .

188

ELEMENTARY DYNAMICS

Example (1). In order to measure the horse-power which can
he given out hy an engine, a brake is arranged, consisting of a belt
on the Jlyivheel of b feet diaineter, as shewn in fig. 114. A iveight
of 50 lbs. is attached to one side of the belt and a weight of 20 lbs.
to the other side. The 20 lbs. weight is partly supported on a spring
balance. When the flywheel is running at 1 50 revolutions per minute
the spring balance reads 9*2 lbs. Find the horse-power absorbed by
the brake.

50 lbs

Fig. 114.

We have Tj = 50 lbs. wt.,

and Ta = (20 - 9-2) = 10-8 lbs. wt.

The resisting moment on the wheel

= (50- 10-8) X 2-5 lbs. ft.
(50-10-8)x2-5x27rx 150

.*. The horse-power

33,000
39-2 X 2-5 X 6-28 X 150

33,000

2-8.

MACHINES 189

Machines

A machine is an instrument for converting energy into useful
work. ]n order to take in the energy supplied the receiving end
of the machine will require a definite motion, depending on the
form in which the energy is supplied. Also, the motion of the
working end will have to be of a definite kind, depending on the
nature of the work which is required to be done.

Take the case of a steam engine. Here the energy is supplied
in the form of the pressure of the steam. This acts on a piston,
and does work by moving it along a straight path. The receiving
end has therefore a reciprocating motion. The working end, on
the other hand, is required to rotate, and the force of the steam
on the piston has to produce a couple causing rotation of the
crank shaft against the resisting torque.

Fig. 115.

If we like, we can consider the torque on the crank shaft
produced by a force R acting on the crank pin at right angles to
the crank. In this case the crank pin becomes the working end
and the piston becomes the receiving end.

Let V = the velocity of the piston or the crosshead
and V = „ „ crank pin.

The velocity ratio of the machine

_ The velocity of the receiving end
The velocity of the working end
_V
V '

190 ELEMENTARY DYNAMICS

If we may neglect friction and also the change in kinetic and
potential energy of the moving points, then we must have for a
definite interval of time^

The work put in at the receiving end

= the work got out at the working end.
Taking a unit interval of time we have,
The power supplied at the receiving end

= the power given out at the working end.
If E = the force exerted at the receiving end
and R = „ overcome at the working end,

E.V^R.-y

R V

or — = - ,

E v'

i.e. t\\Q force ratio is inversely proportional to the velocity ratio.

Example (2). On p. 50 we found, for the position of the crank

V 42

shewn, the value of - = ^ . Suppose the steam pressure = 80 lbs,

per sq. inch, and the diameter of the pisto7i = 10 inches. Find the
crank effort, i.e. the tangential force on the crank pin.

The total pressure (E) = 257r x 80

= 6280 lbs. wt.

V
The crank effort (R) = - x E

= 11 X 6280
= 4050 lbs. wt.

Friction. If we consider friction then there will be some energy

R V

wasted in heat, and the ratio of — will no longer be equal to - .

E ^ ^ V

The ratio - = — =7^; — -, — is called the mechanical advantage of
E Effort

a machine.

EXAMPLES . 191

^, T . 7 /r. • f. 1 . Power given out

Ihe mechanical efficiency of a machine = ^pi -. -. —

*^ ^ Fower given in

_ Rv

~EV

Mechanical advantage
Velocity ratio

Example (3). Fig. \\^{a)onp. 192 shews diagrammatically the
meclianism of a drawing press. ABC is a hell-crank lever pivoted
at B. Tfie end C is connected hy the link CD to the piece D, which
is constrained to move in direction BD. By dratving velocity tri-
angles, find the ratio of the velocity of f< to the velocity of D.

Neglecting friction, wJiat vertical force P applied at A is required
to produce a vertical push of 1 ton on D?

The bell-crank lever ABC is pivoted at B, and therefore the
velocity of A : the velocity of C as AB : BC.

In fig. 116 (6), draw oa perpendicular to AB, and equal to AB,
to represent the velocity of A. Draw oc perpendicular to BC, and
equal to BC, to represent the velocity of C.

The relative velocity of D to C must be perpendicular to CD,
and the velocity of D is along BD.

Draw cd perpendicular to CD, and draw od parallel to BD.

The velocity of D equals the vector sum of the velocity of C and
the relative velocity of D to C.

ocd is a triangle of velocities.
.*. od represents the velocity of D, to the same scale as oa
represents the velocity of A.

Velocity of A oa
Velocity of D od

= ^TKc. (by measurement)

= 1-2.
Now P acts vertically. Draw op vertical, and draw ap perpen-
dicular to op.

192

ELEMENTARY DYNAMICS

Then op represents the component of A's velocity in the
direction of the force P.

Fixed centre

Fig. 116.
Employing the principle, work put in = work got out, we have
p X op = F X oc?,

where F is the force required at D.

op
od
0-98
0^
= 1-18 tons.

= 1 X

EXAMPLES 193

Example (4). The mechanical efficiency of a certain type of

tackle may be taken equal to I = j , where n = the total number

of sheaves round which the rope is wrapped.

A runner tackle^ in which the m,oving block has 3 sheaves and the
fixed block 2 slieaves, is to be used to overcoine a resistance of 5 tons.

What will be the pull in the running end of the rope, and the
mechanical advantage of the tackle ?

.^ . >P

5 tons

.. . ^ Fixed

Moving

Fig. 117.
The sheaves are really all the same size and not as shewn in
fig. 117, and the portions of the rope may be assumed parallel.

From the figure, we see that if the moving block shifts 1 foot
there will be a motion of the running end equal to 6 feet.
.'. The velocity ratio = 6.
Let P = the pull of the running end, in tons wt.

The mechanical efficiency = .^^ — -. — ^

Work put in

~Px6'

or P = fo

Px6 7 + 5'

6.0
42

= ^ tons wt.

The mechanical advantage

5
" P

— 35
-TT7

-3-5.

L. E. D. 13

194

ELEMENTARY DYNAMICS

Example (5). In a steam tractor the motion is transmitted from
the crank shaft A through two parallel shafts B and C. A wheel of
22 teeth on A gears with a wheel of 21 teeth on B. Compound with
this wheel is a wheel of 17 teeth gearing with a wheel of 4:4: teeth
keyed on shaft C. Keyed also on C is a wheel of \2 teeth gearing
with a wheel of 5Q teeth fixed to the road wheel.

Find the velocity o'atio of the gear, and, neglecting friction, find
the coujyle required on the crank shaft to produce a total tractive
force of 1000 lbs., the diameter of the road wheels being 7 feet.

44
C

Road wheel

Fig. 118.

The arrangement of the gear is shewn diagrammatically in
fig. 118,

EXAMPLES 195

For one revolution of the crank shaft A we have

Shaft B makes |f revolutions,

„ C „ If ^ ii revolutions,

» D „ If xjjx If revolutions.

r^, , . ' n .^ Speed of A

/. The velocity ratio oi the gear = ,5^^ — -^— ^ —

21 X 44x56
~22xl7x 12

= 11-53.

Let T = the couple on the crank shaft in lbs. ft.

P = the tractive force in lbs., at the points of contact of the
two road wheels.

The couple on the road wheels shaft D
= P X I lbs. ft.

Neglecting friction, we must have the work put into the crank
shaft in any time = the work given out by the road wheels in the
same time.

.-. p x|--Tx 11-53.

Putting P- 1000 lbs.,

100x3
11^53"

we get T = 15^^' ^303 lbs. ft.

Relative Momentum and Relative Kinetic Energy

In dealing with motion we pointed out that we really do not
know anything about absolute motion, and that all the motions
we deal with are relative motions. In the same way, when we
talk or write about momentum, and kinetic energy, since these
involve velocity and speed, we are really implying relative mo-
mentum, and relative kinetic energy.

13—2

196 ELEMENTAKY DYNAMICS

MomentuTn. Suppose we are determining the force acting on
a body by considering the time-rate of change of momentum. If
the momentum is changing uniformly with the time, we have

m{y — u)
^^ t '

where v and u are the final and initial velocities of the mass m,
measured relatively to some body of reference.

Suppose the body of reference has itself a constant velocity V
the s{
mentum

in the same direction as u and v, then the real change of mo-

= m (v + y) - m (u + W)
= m{y- u).

Now we may assume that the time is the same whatever is our

body of reference,

m (v — u) - „

.*. F = — ^^ , as before.

Kinetic Energy. Suppose we are determining the force acting
on a body by considering the space-rate of change of kinetic
energy. Again taking the force constant, relative to our body of
reference, we have

where s is the distance moved relative to the body of reference.

If the body of reference has a constant velocity V, then the
real change of kinetic energy

2 2

+ mV {v — u).

RELATIVE MOMENTUM AND RELATIVE KINETIC ENERGY 197
Let s' = the total distance moved during this change.
Then Fs' = —^ ' ^m{y- u) V.

Now

I.e.

s =s + yt, and F = -^^ ^ .

.'. Fs+F .Wt = — ^^ ^ + Fi5 . V,

m (v^ — u^) , „
F = — ^-^r- ' , as beiore.

Hence we may use relative kinetic energy if at the same time
we use the relative displacement.

As an example, consider the energy required to accelerate or
retard the piston of a locomotive which is travelling at any given
speed relative to the earth. In estimating this, we are not con-
cerned with the speed of the locomotive, but merely with the
speed of the piston relative to the cylinder, and the distance
moved by the piston relative to the cylinder while this change of
speed has occurred.

In the same way, in order to find the total work done by the
steam, we use the force-space curve, the force being the total
pressure of the steam on the piston and the space being the rela-
tive displacement of the piston in its cylinder.

Indicated Horse-Power

The indicated liorse-power (i.h.p.) of an engine is the horse-
power supplied to the engine by the working substance, e.g. in a
steam engine, the horse-power supplied by the steam as it acts
on the piston. The term indicated is used since the i.h.p. is
usually obtained by means of an Indicator. This is a small in-
strument which autographically records the force-space curve at
the receiving end of the engine. An indicator diagram for a

198

ELEMENTARY DYNAMICS

steam engine in which the steam acts only on one side of the
piston is shewn in fig. 119.

ahc represents the pressure of the steam as the piston is moving
in towards the crank shaft, i.e. during the forward stroke, and cda
represents the pressure during the return stroke.

Stroke
Fig. 119.

Let Py = the mean pressure in lbs. per sq. inch during the
forward stroke.

Pj, =: the mean pressure in lbs. per sq. inch during the re-
turn stroke.

A = the area of the piston in sq. inches.

L = the length of the stroke in feet.

N = the speed of the engine in revolutions per minute.

The work done by the steam during the forward stroke

= P^..A. L ft.-lbs.

The work done by the steam during the return stroke

= - P,, . A . L ft.-lbs.

INDICATED HORSE-POWER 199

The negative sign is used since the engine has to do work in
pushing the steam out of the cylinder.

.*. The work done by the steam per revolution
= (P^-P,).AL
= P . A . L ft. -lbs.,
where P is the mean effective pressure given by the diagram abcda.

This is obtained from the mean distance between tl^p curves
abc and cda, or the area abcda divided by the length of the diagram.

The work done per minute = P . A . L . N ft.-lbs.

P . L.A.N
.'. The indicated horse-power = ' ' ' .

The brake horse-power (b.h.p.) or the horse-power which is given
out by the. engine may be obtained by using a brake such as is
shewn in fig. 114, p. 188.

The difference between the indicated horse-power and the brake
horse-power represents the horse-power which is wasted in friction
in the engine itself.

The mechanical efficiency

Brake horse-power
Indicated horse-power '

Example (6). The driving wheels of a two-cylinder locomotive
are ^ feet in diawxter. The mean pressure of the steam, in each
cylinder is 60 lbs. per sq. inch. The diameter of the cylinders is
20 inches^ and the stroke 22 inches. Find the indicated horse-power
if the speed of the locomotive is 20 miles per hour.

Find also the draw-bar pull produced, if \2'5 per cent, of the
power supplied is wasted in friction.

The speed of the driving wheels and crank shaft
_ 20 X 5280

60 X TT X 6

= 93*5 revolutions per minute.

200 ELEMENTARY DYNAMICS

i.H.p. for one end of one cylinder

_ 60 X TT X 10^ X 22 X 93-5
~ 33,000 X 12

= 98.

.-. Total I.H.P. =.4 X 98
-392.
The horse-power available for draw-bar pull
-0-875x392
= 343.
If P — the draw-bar pull in tons
The work done per revolution
— P X TT X 6 ft.-tons.
The useful energy supplied per revolution
4x60x7rxl00x22

= ■ 2240x12 ^ ^'^^^ ^^-*^^^'

Equating these, we get

240 X TT X 100 X 22 X 0-875

2240 X 12x7rx6
= 2-86 tons.

Mechanical Losses in a Steam Tractor

As an illustration of the method of applying the principle of
the conservation of energy to a practical problem, we will briefly
discuss an actual test of a steam tractor which was made in order
to determine the mechanical efficiency, and to discover where the
losses of power occurred.

The tractor is driven by a steam engine, fitted with a flj^wheel,
and the motion is transmitted through gear wheels from the crank
shaft to the driving wheels, as described in example (5), p. 194.
A clutch is fitted between the crank shaft and the gear.

MECHANICAL LOSSES IN A STEAM TRACTOR 201

Tests were made while the tractor was drawing a train of loaded
wagons along a level road. The power put in was measured by
taking indicator diagrams from the cylinders. The power trans-
mitted to the wagons was measured by the pull in the spring
coupling between the tractor and the wagons.

For this purpose an apparatus was arranged to autograph ically
draw a pull-distance curve, and at the same time equal intervals
of time were recorded on the diagram.

The following are the figures for one test :

Indicated horse-power =:: 19*1.

Speed of crank shaft = 236 revolutions per minute.

Space-average of pull on wagons = 582 lbs.

Speed of tractor from diagram = 446 feet per minute.

Velocity ratio of gear = 11 'SS.

Diameter of driving wheels = 7 feet.
Horse-power transmitted to the wagons
_ 582 X 446
~ 33,000

= 7-9.

.'. Overall mechanical efficiency

_ 7-9
~19-1

= 41*3 per cent.
Horse-power lost in the tractor

= 191 -7-9

= 11-2.

To find the slip of the driving wheels on the road, we have,

236
Speed of the driving wheels = Z" revolutions per minute

= 20-4 revolutions per minute.

202 ELEMENTARY DYNAMICS

Speed of the circumference of driving wheels

= 20*4 X 7 X TT feet per minute
= 450 feet per minute.

Speed of the tractor = 446 feet per minute.

.*. Percentage slip of the driving wheels

450-446
= -^50—^1^^
= 0-9 per cent.

The total mechanical losses in the tractor consist of the following
items :

(1} Engine friction loss.

(2) Gear friction loss.

(3) Friction loss of the driving wheels on the road and other

resistances to motion (traction loss).

(1) In order to determine the engine friction the clutch was
disengaged and an absorption brake fitted on the flywheel. A
number of tests of the indicated horse-power and the flywheel
horse-power were taken.

These tests shewed that the power lost in engine friction was
practically constant and equal to 3*3 horse-power.

(2) To determine the gear loss the total horse-power lost be-
tween the cylinders and the driving wheels was estimated. This
was eftected by jacking up the tractor, and fitting a brake-ring
and an absorption brake on one of the driving wheels.

From a series of tests it was found that the brake horse-power
was related to the indicated horse-power as follows :

Brake horse-power ==0-925 (indicated horse-power) - 3*06.
For indicated horse-power = 19-1, we have,

Brake horse-power = 0-925 x 19-1 - 3-06
= 17-66 -3-06
= 14-6.

MECHANICAL LOSSES IN A STEAM TRACTOR 203

The horse-power transmitted to the gear

= 191- 3-3

= 15-8.
The horse-power lost in friction of the gear

= 15-8 -14-6

= 1-2.

14-6
The efficiency of the gearing = r-— ^

= 92*5 per cent.

(3) The horse-power given to the wagons

= 7-9.
.'. Horse-power wasted in tractor road-friction and resistance

= 14-6-7-9

= 6-7.

Summary of Losses

Input.

19-1 H.P.

Output.

Engine Friction

3-3 H.P.

Gear Friction

1-2 H.P.

Traction Loss

6-7 H.P.

Transmitted to Wagons

7-9 H.P.

Total

19-1 H.P.

Pelton Water Wheel

In a Pelton Wheel a number of buckets are arranged round the
rim of a disc, and a jet of water, moving with a high velocity,
impinges tangentially on the buckets. The reaction to the tan-
gential pressure required to change the momentum of the water,
provides the turning moment on the wheel.

204 ELEMENTARY DYNAMICS

A diagrammatic sketch of the wheel is shewn in fig. 120, and
an enlarged sectional plan of one bucket is shewn also in the figure.

Let a = the area of the nozzle in sq. feet.

u = linear speed of the buckets in feet per second.

V = the velocity of the water as it issues from the nozzle in
feet per second.

p = the density of the water in lbs. per cubic foot.

Fig. 120.

The tangential pressure on the wheel
= the change of momentum of the water per second
= the mass of water impinging per second x the change of velocity.

The velocity of the water, relative to the bucket, just before it
impinges is {y - u).

In the absence of friction, this will also be the magnitude of
the relative velocity of water and bucket at any point of the
bucket. The direction will be changed by the normal pressure
but this cannot affect the relative tangential velocity.

The final velocity of the water relative to the bucket in the
direction of the jet = - {v — u) cos 0.

PELTON WATER WHEEL 205

.-. The final velocity of the water relative to the fixed nozzle
= — {v-u) cos 6 + u
= u (1 + cos 0) — V.
.*. The change of velocity in the direction of the jet
= v — {w (1 + cos 0) - v]
= 2v-u(l +cos6).

In practice 6 is made very small, just sufficient to allow the
water leaving to clear the wheel. Take ^ = 0. .

The change of velocity in thfe direction of the jet = 2 (v - u).
The quantity of water impinging on one bucket per second
depends upon the relative velocity, and = pa(v — u) lbs.

With a number of buckets, neglecting splash, we may assume
that all the water impinges on the wheel.
.-. The mass of water impinging per second

= pav lbs.
The tangential pressure on the wheel

= pav x2{v -u) abs. units

Jpav{v-u)^^^^^^^

9
The work done on the wheel per second

= -^ — ^ ^ X u ft.-lbs.

The energy supplied to the wheel per second
1 ^

.^Theeffidency = 2^1i!±.-Iib
pair

~2~

4: (v — u) u

206

ELEMENTARY DYNAMICS

The kinetic energy carried away by the water per second
= ^-^^fc^ft.-lbs.

J.i V = 2u, i.e. the velocity of the bucket equals one-half the
velocity of the jet, the kinetic energy carried away equals zero,
and the efficiency is unity.

Taking this condition we get

The maximum horse-power of the wheel

2g X 550 •

Total Kinetic Energy of a Body moving in a Plane

Let the body shewn in fig. 121 have any motion whatever in
the plane of the paper.

Let G be the centre of gravity, and let it have a velocity v at
the instant considered.

Consider the velocity (u) of any point A.
The velocity of A = the velocity of G + the relative

velocity of A to G (vector sum).

Fig. 121.

TOTAL KINETIC ENERGY OF A BODY MOVING IN A PLANE 207

Let o> = the angular velocity of the body at the instant con-
sidered, and r = the distance AG.

Then the relative velocity of A to G = wr, and is perpendicular
to AG.

Draw AB perpendicular to G A to represent tar, and draw BO to
represent v.

Then AC will represent u, and we have
AC = AB + BC,

or u^ = (oV^ + v^+ 2(or . v . cos <j}.

Draw AN perpendicular to the direction of motion of G. Then
AN is perpendicular to BC, and AB is perpendicular to AG.

.-. GAN = <^
and rcos</) = AN.

If Sm is the mass of a particle of the body at A, the kinetic
energy of it

= I Sm . u^.

For the whole body we get
The kinetic energy = ^ ^8m .u^

= ^i8m (o)V^ + v^+ 2wv . AN)

= ^ :S r^Sr/i + ~'$Sm + 2o>v ^ Sm . AN.

^T^Sm = the moment of inertia (I) of the body about an axis

through G.
S Sm = the total mass M.

^Sm . AH - the moment of each particle about the axis GN

= zero, since G is the centre of gravity.
.'. The kinetic energy of the body

208

ELEMENTARY DYNAMICS

We may express this in words thus :

The total kinetic energy of a body moving in a plane equals
the kinetic energy of the whole mass moving with the velocity of
the centre of gravity plus the kinetic energy due to rotation about
the centre of gravity.

Example (7). In the arrangement shewit in fig. 122, each of
the pulleys has a mass w, a radius r, and a
radius of gyration k. If when left to itself
the mass M descends and raises mass m, find
the velocity of M when it has fallen a dis-
tance hfrom rest.

Using the principle of the conservation
of energy we may write,

the loss of potential energy = the gain
of kinetic energy.

Let 17= the velocity of M after descending
a distance h.

The mass m and the pulley B will have
ascended a distance ^ , and their linear

velocity will be ^ .

The angular velocity of pulley A will be

Fig. 122.

Yr'
We have then,

lAgh -{in + w)g ■^ = \ Mv^ -\-^{m + w) -+^wk^ -^ + 1 wk^—^

I.e.

5wk

m w Dw/c^)

M + T-^i + i;^}

v^i

(2M

(2M-

m — w) gh

m> — w) gh,

m w bwk^''

EXAMPLES 209

Example (8). A wagon of total mass M ha^ four wheels^ each
of mass m, radius r, and radius of gyration k. Shew that the effect
of the rotation of the wheels is the same as that of an increase of

mass equal to — ^ .

Let the speed of the wagon be increased from u to v hy the
application of a constant accelerating force F in the direction of
motion.

We may equate the work done by this force to the gain of
kinetic energy.
Initially.

The kinetic energy of the wagon without the wheels
= J (M - 4m) u\

The kinetic energy of the wheels = 4 1 1 mu^ + J m^ . — j- , si

since

the angular velocity of the wheels is - .

:. The total kinetic energy = ^Mu^-¥ 2m — . u"^

Finally.

The total kinetic energy = J ( M + 4m -^ ) v^.

If s is the distance, and t the time for the change of velocity,

f.s = i(m + ^)(«^-«^) (1).

With the wheels locked we should have
F.s=::iM(2;2-w2),
assuming the same accelerating force to act.

.*. The rotational inertia of the wheel is equivalent to an ad-
ditional mass of — r— .

IT
L. E. D. 14

210 ELEMENTARY DYNAMICS

We may find the acceleration tlius :
rrora(l). f.s^^^f

But

V + u

. t = s,

M H — ) ^ acceleration.

The acceleration

M +

^vfik^ '

J Examples. Chapter IX

1. It is required to transmit 50 horse-power from a pulley, 36 inches
in diameter, running at 250 revolutions per
minute. The belt is I inch thick, and the per-
missible stress in the material is 600 lbs. per
square inch. Find the proper width of the belt,
on the assumption that the tension in the tight
portion of the belt is twice that in the slack
portion.

2. Shew that the horse-power delivered by a
belt to a pulley of diameter D feet, running at
N revolutions per minute, is given by

(Ti-T2)D.N
10500
where Tj and T2 are the tensions in lbs. in the
two sides of the belt.

Power is transmitted from a pulley A to a
pulley B, and in order to measure the power the
belt is arranged as shewn in fig. 123.

If the weight W attached to the light jockey
pulley P is 150 lbs. and the spring balance
attached to the light jockey pulley Q reads
70 lbs. , find the horse-power transmitted when the speed of B is 200 revolu
tions per minute and its diameter is 9 inches.

Spring
balance

EXAMPLES. CHAPTER IX

211

3. A flywheel, of mass 10 tons, is rotating 40 times per minute and the
mean diameter of its rim is 19 feet. Express its energy as the fraction of
a horse-power-hour.

Find the difference of the tensions in belting passing over the flywheel
which would reduce it to rest in 10 minutes, taking the outer diameter of
the flywheel as 20 feet.

4. Fig. 124 shews an absorption brake arranged for measuring the horse-
power given out by a water turbine. When the brake was arranged on a
pulley, which was quite free to rotate without friction, the spring balance
read 4 lbs. If the brake pulley of the turbine is rotating at 500 revolutions
per minute, and the spring balance reads 36 lbs. find the horse-power given
out.

//////////////

Fig. 124.

5. The table of a planing machine has a rack fixed to its underside, and
takes its motion, along the bed of the machine, from a pinion which gears
with the rack. The pinion has 14 teeth of 1| inch circumferential pitch, and
to its axle is keyed a wheel of 68 teeth, which is driven from the pulley shaft
through a wheel of 17 teeth. If the pulley shaft is running at a speed of
90 revolutions per minute, what is the linear speed of the table in feet per
minute?

The table weighs 1000 lbs., and the coefficient of friction between the table
and the bed is 0*12. If no power is lost in the gear, what torque must be
applied to the pulley shaft in order to move the table along the bed?

14—2

212

ELEMENTARY DYNAMICS

6. The moving block of a tackle has two sheaves and the fixed block three

sheaves. For raising a load the relation of the effort P to the load W is given

W / w\
by P = Q ( 1 + o ) » where Q is the velocity ratio, and n is the number of

sheaves round which the rope is wrapped.

Sketch the arrangement, and find the ejB&ciency and the effort, when a load
of 10 tons is raised.

What value of P would you expect to be required to slowly lower 10 tons?

7. The sketch, fig. 125, shews the mechanism of a coining press, in which
pressure is exerted on the piece D by a pull
at the end A of a bell-crank lever ABC
pivoted at B. The direction of the pull is
perpendicular to BD, and D is constrained to
move along BD.

Determine, graphically, the velocity ratio
of D to A. Neglecting friction, what pull is
required at A to move D against a thrust of
2 tons?

[Prick the figure through to your paper. ]

8. A winch has four parallel shafts A, B,
C and D. On shaft A there is a pinion with
11 teeth, which meshes with a wheel with 39
teeth on shaft B. Compound with this wheel,
a pinion of 12 teeth meshes with a wheel with
50 teeth on shaft C. This shaft has a pinion
with 11 teeth meshing with a wheel with 70
teeth on the drum shaft D. The external
diameter of the drum is 10 inches, and
the winding rope has a circumference of
2^ inches.

If the pull in the rope being wound on the
drum is 5 tons, and the efficiency of the ^. ^^k

machine is 75 per cent, , what tangential force

will have to be applied to a handle of 15 inches' effective radius (i) if it is
fixed on shaft A, (ii) if it is fixed on shaft B?

9. A flywheel, 6 feet in diameter and weighing half a ton, is supported
midway between two bearings on a shaft of 4 inches diameter. Power is
delivered to the shaft at the rate of 25 horse-power, and drives the flywheel

EXAMPLES, CHAPTER IX

213

at 150 revolutions per minute against the friction of a strap which passes
over the upper half of the circumference, and carries 300 lbs. at one end and
25 lbs. at the other. Find how much of the power is absorbed by the friction
of the strap, and how much by the friction of the bearings. Find also the
total thrust, and the resultant frictional force, on each bearing.

lO. Define the terms velocity ratio, mechanical advantage, efficiency, and
shew that the efficiency is equal to the mechanical advantage divided by the
velocity ratio.

In a test of a crane with a velocity ratio of 280 the following values of load
and effort were observed:

Load in tons

Efforts in lbs.

67-5

104-5

123

Plot the values given and obtain the relation between load and effort,
also a curve shewing the efficiency for the above loads.

Plot

11. Shew how to find the crank effort in a single cylinder engine for any
given crosshead thrust. The crank of an engine is 8 inches, the connecting
rod 30 inches; the crosshead thrusts at quarter stroke, at half stroke, and at
three-quarter stroke are 4000, 2500, and 600 lbs. wt. respectively. Find the
corresponding turning moments on the crank.

12. An electric motor has keyed to its spindle a wheel A of 14 teeth which
gears with a wheel B of 70 teeth. Compound with B is a wheel C of 13 teeth
which gears with a wheel D of 48 teeth. Compound with D is a wheel E of
23 teeth which gears with a wheel F of 54 teeth. Wheel F is keyed to the
axle of the chain barrel of a crane, the effective diameter of which is 15 inches.
Find the speed of the motor when the chain is being wound on the barrel at
a rate of 20 feet per minute.

If a load of 10 tons is being raised at this speed, what power will be taken
by the motor, assuming the combined efficiency of the motor and crane is
45 per cent. ?

13. A steam engine is driving a dynamo by a belt and is running at 120 re-
volutions per minute. The diameter of the driving pulley is 3 feet, and the
tensions in the tight and slack portions of the belt are 2000 lbs. and 800 lbs.
respectively. What horse-power is the engine delivering ?

214

ELEMENTARY DYNAMICS

14. A sliding door of a railway carriage is 3 feet broad and weighs 250 lbs.
It closes in the direction of motion of the train. The door is fully open and
the train travelling at 20 miles per hour when the latter starts slowing up
with a constant retardation of 2 feet per second per second. If the resistance
to motion between the door and the carriage is constant and equal to 4 lbs.,
find the velocity, relative to the train, with which the door closes. If the
catch on the door automatically couples it when it closes, find the impulse
and the energy dissipated.

15. A locomotive with two cylinders each 19 inches diameter, 22 inches
stroke, and driving wheels 5| feet diameter, is working with a mean steam
pressure in the cylinders of 135 lbs. per sq. inch. Find the tractive force at
the rails, assuming that 10 per cent, of the power developed is wasted in engine
friction. Calculate also the greatest load, including the engine, which can be
taken up an incline of 1 in 150 at 10 miles per hour if the resistance per ton
at this speed is 6-9 lbs.

16. The pressure in one end of a steam engine cylinder, during ^ revolu-
tion, is shewn on the indicator diagram below, fig. 126. The piston area is
80 sq. inches, the stroke is 15 inches and the engine runs at 120 revolutions
per minute. Determine the average effective pressure during a revolution, the
work done per minute, and the horse-power.

Fig. 126.

17. The engines of a steamer are developing 2500 indicated horse-power
when the speed is steady and equal to 16 knots. If the overall efficiency is
60 per cent., find the total resistance to motion of the vessel in tons.

EXAMPLES. CHAPTER IX

215

Assuming the total resistance varies as the square of the speed, find the
new steady speed when the horse-power is increased to 3000, the overall
efficiency remaining the same.

[1 knot = 6080 feet per hour.]

18. A gas engine is speeding up under no load, and at the beginning of
one cycle the speed is 100 revolutions per minute. The indicator diagram for
the cycle is given below. Find the work done during the cycle, and the speed
of the engine at the end of the cycle, neglecting friction.

Diameter of piston 18 inches. Length of stroke 2 feet. Moment of inertia
of flywheel 107 tons ft.2 Scale for pressure, 1 inch = 180 lbs. per sq. in.

Fig. 127.

19. In a single cylinder steam engine, the diameters of the piston and the
piston rod are respectively 11 inches and 1| inches, the length of stroke
1 foot, and the speed 197 revolutions per minute. The mean pressure on
each side of the piston is found from the indicator diagrams to be 16*5 lbs.
per sq. in. Find the indicated horse-power.

If the engine is driving a dynamo, the output of which is 10*8 kilowatts
and the efficiency 95 per cent. , find the horse-power wasted in friction in the
engine.

[0*746 kilowatt = 1 horse-power.]

20. In order to determine the efficiency of some gearing, power was
transmitted through it from a steam engine. The indicated horse-power of
the engine and the horse-power given out by the gearing to a brake wheel

216 ELEMENTARY DYNAMICS

were measured. Previous experiment shewed that the horse-power wasted in
friction in the engine was practically constant and equal to 2'6.

Find the eflBciency of the gearing from the following data :
I.H.P. of the engine, 24-2.
Speed of engine shaft, 240 r.p.m.
Velocity ratio (crank shaft to brake wheel), 11-7.
Torque on brake wheel, 4750 lbs. ft.

21. A motor-car delivers 25 horse-power to its shaft. The engine shaft is
geared down to the back axle in the ratio 4 to 1, and 12 per cent, of the
shaft-power is lost in the gear. The car wheels are 28 inches in diameter, and
the car is running at 32 miles per hour. Find the torque on the engine shaft,
the torque on the back axle, and the propulsive force between the road and
the driving wheels.

The car runs up a slope of 1 in 25, and the gear is changed so that the
reduction ratio is 6 to 1, and the gear loss is now 25 per cent, of the shaft
power. If the torque on the engine shaft remains constant, while the
resistance to motion is proportional to the square of the speed, find the
uniform velocity with which the car ascends the slope. The weight of the
car is 2500 lbs.

22. A gas engine is working at 10 indicated horse-power and at 200 revo-
lutions per minute when the gas is suddenly cut off. Assuming the torque
due to the brake and other resistances remain constant, find the number of
revolutions the engine makes before coming to rest. The moment of inertia
of the flywheels is 3500 lbs. feet^.

23. What is the rate in horse-power at which energy can be delivered by
a stream, 2 square feet in section, flowing at the rate of 8 feet per second?
(1 cu. ft. of water weighs 62^ lbs.)

If the stream is used to drive a water wheel turning at a constant rate of
30 revolutions per minute and exerts on it a constant couple of 120 in feet
and lbs., what is the efficiency of the arrangement?

24. Flat vanes are arranged radially round the circumference of a wheel
to form a water turbine. A jet of water impinges on the vanes, the direction
of the jet being tangential to the wheel and at a distance of 9 inches from
the centre. The velocity of the jet is 50 feet per second, and the discharge
is 20 cubic feet per minute. Find what should be the speed of the wheel for
maximum horse-power, and the theoretical horse-power at that speed.

EXAMPLES. CHAPTER IX 217

25. The effective diameter of a Pelton wheel is 40 inches and the nozzle
area 12-5 square inches. It is suppUed with 672 cubic feet of water per minute
and runs at a speed of 380 revolutions per minute. Find the theoretical
efficiency assuming the direction of motion of the water completely reversed.

If the horse-power given out is 250, what is the actual efficiency?

26. A moving staircase has a speed of 100 feet per minute, and rises
30 feet vertically in 50 feet horizontally. 180 people use the staircase per
minute, and the average weight per person is 9 stones. When they step on
at the bottom, they do so at a speed relative to the ground of 80 feet per
minute in the direction of motion of the staircase. They remain at rest
relatively to the staircase while moving horizontally at the. bottom and top,
but during the rise they climb relatively to the staircase at a rate of 60 feet
vertically per minute. At the top they step off in the direction of motion
with a speed of 100 feet per minute relative to the ground.

Find the horse-power required to maintain the motion of the staircase,
neglecting friction.

27. Work is done at the rate of 17*5 horse-power in driving a car, which
weighs 8 tons, up an incline of 1 in 20 at a speed of 6 miles per hour. If the
frictional resistances remain constant, find the acceleration in feet per second
per second when the car is allowed to run freely down the same incline.

28. In a horse car the energy of rotation of the wheels is a negligible
fraction of the total energy of the car, but in an electric car the energy of the
rotation of armatures, gear wheels, etc., is appreciable. Given two cars, one
in which the rotational energy may be neglected, and one in which it amounts
to one-tenth of the total energy of the car, shew that, neglecting friction,
their accelerations down a given slope are in the ratio of 10 to 9.

Call the two cars A and B and suppose that their total weights are equal,
then find the acceleration down a slope of inclination a (1) of car A running
alone, (2) of car B running alone, (3) of A and B running coupled.

29. A pump is driven by an electric motor by means of a belt. When the
inotor is absorbing 20 horse-power, the pump lifts 400 gallons of water per
minute through a height of 80 feet. What is the efficiency of the plant as a
whole ? Neglect the kinetic energy of the water. 1 gallon of water weighs
10 lbs.

If the motor has an efficiency of 88 per cent., find the efficiency of the pump,
nagl&cting any loss of power in the belt.

218 ELEMENTARY DYNAMICS

In addition to raising the water through 80 feet the pump is required to
deliver it against a pressure of 20 lbs. per sq. inch. Assuming that the
efficiency of the plant is the same as in the previous case, find what power
must be supplied to the motor.

30. A steam engine has a stroke of 10 inches, and is running uniformly
at a speed of 250 revolutions per minute. The piston weighs 80 lbs. Neglecting
the obliquity of the connecting rod, find how much energy is stored in the
piston (1) between dead centre and quarter stroke, and (2) between quarter
stroke and half stroke. «

Also find the amount of energy stored in the piston while the crank moves J
through an angle pf 45° from dead centre.

MISCELLANEOUS EXAMPLES

The following examples are not arranged in order of difficulty nor according
to the principle involved.

1. Four members of a frame, A, B, C and D, all lie in one plane, and
meet at a point. Members B, C and D make angles, all measured in the
same direction, of 90°, 210°, 257^° respectively, with the member A. The
pulls in the members C and D are 4'7 tons and 5 tons respectively. Find the
pulls in members A and B (1) graphically, (2) by calculation.

a. In order to find the height of an airship which is travelling over the
sea, a gun is fired from the ship and the time between the firing of the gun
and the receipt of the sound reflected from the sea is noted. The time in one
case was 15*8 seconds. What was the height of the airship ?

If five minutes later the time was 16 "6 seconds, what was the vertical
velocity assuming it to be uniform?

[Velocity of sound in air = 1080 feet per second.]

3. A section of a railway between two points A and B distant 2000 yards
apart, is given by the following table which shews the elevation above A for
every 200 yards :

Distances
in yds.

200

400

600

800

1000

1200

1400

1600

1800

Heights
in ft.

36i

28i

A train, the mass of which exclusive of the engine is 300 tons, passes A at
a speed of 30 miles per hour. The locomotive exerts a constant draw-bar
pull of 7000 lbs., and the train resistance may be taken as 15 lbs. per ton.
Draw a curve shewing the variation in speed as the train advances from A to B.

4. Shew that if a body is moving in a circular path of radius r feet with
a uniform speed of w radians per second, then it has an acceleration towards
the centre equal to w^r feet per second per second.

The mean diameter of a cast-iron flywheel is 4 ft. 3 in. What is the
maximum speed of rotation, expressed in revolutions per minute, if the mean
stress in the rim is not to exceed ^ ton per sq. inch?

[1 cubic foot of cast-iron weighs 460 lbs.]

220

ELEMENTARY DYNAMICS

5. The relation between the tractive force on a car weighing 1 ton and the
time is as follows :

Time in sees.

Force in lbs. wt.

■

110

108

104

If the car starts from rest and the resistances to motion are equivalent to
a force of 40 lbs., draw the acceleration-time curve for the motion and find
the velocity at the end of the 30 seconds.

6. A train A, moving with a constant speed of 30 miles per hour, is passing
another train B, which is at rest on a parallel track. At the instant when
the engines of the two tr|i,ins are opposite, the train B starts with a constant
acceleration of 2 feet per second per second. What length of time will elapse
before the two trains again occupy the same relative position ? If the train B
is 100 yards long, at what distance from its starting point will it have com-
pletely overtaken train A ?

7. Below is an indicator diagram for one end of one cylinder of a loco-
motive, the spring used being such that 1 inch on the diagram represents
150 lbs. per sq. inch. The diameter of the cylinder is 20 inches, the stroke
22 inches, and the speed 240 revolutions per minute. Find the indicated
horse-power represented by the diagram.

[Prick the diagram through to your paper.]

Fig. 128.

8. A uniform disc, 18 inches in diameter, and weighing 30 lbs., is keyed
to a spindle of 1 inch diameter and is mounted on frictionless bearings. A
light cord is wound round the spindle and carries a weight of 1 lb. at its free
end. Find the angular acceleration of the disc, and the linear acceleration
of the weight when the system is in motion.

MISCELLANEOUS EXAMPLES 221

9. One part of the track of a switch-back railway consists of a complete
vertical circle of 20 feet diameter. Neglecting friction, find the minimum
velocity with which a car must start on its journey round the circle if it is to
keep the rails. How high must the starting point of the track be above the
top of the circular portion in order that this minimum velocity may be at-
tained ? The rotational inertia of the wheels may be neglected.

10. A shell is fired so that at the highest point of its flight it just passes
over a mountain 3000 feet high and distant 5 miles from the point of pro-
jection. Find the ratio of the horizontal component of the velocity of pro-
jection to the vertical component, and the values of these components.

If the shell weighs 350 lbs., what is its kinetic energy (1) as it leaves the
gun, (2) as it passes through the highest point of its flight ?

11. The crank of an engine is 9 inches, the connecting rod 3 feet, and the
line of stroke of the piston passes through the axis of the crank shaft. The
speed of the crank is 210 revolutions per minute. Find, graphically, the
velocity of the piston at one-quarter of the distance from each end of the
stroke.

For the same two positions of the piston, find the velocity of rubbing
between the connecting rod and crank pin, if the latter is 2| inches in diameter.

12. In a double-acting steam engine the stroke is 12 inches long and the
connecting rod may be assumed infinitely long. The effective thrust on the
crosshead due to the steam pressure is constant and equal to 2000 lbs. Shew
that the torque on the crank shaft produced by this thrust varies as the sine
of the angle turned through by the crank from the dead centre position.

Plot a curve shewing the turning moment on the crank and the angle
turned through by the crank.

13. In the previous question, the engine is working against a constant
resisting torque and the mean speed is 240 revolutions per minute. Find the
moment of inertia of the flywheel required to limit the speed fluctuation to
3 per cent, above or below the mean speed.

14. A gunboat with hydraulic jet propulsion takes water in through
vertical openings amidships and discharges it astern. A discharge of 5*2 tons
per second at a velocity of 29 feet per second relative to the boat gave a speed
of 9-3 knots. Find the resistance to motion of the boat.

222 ELEMENTAKY DYNAMICS

15. A casting, the shape of which is an equilateral triangle with a cylin-
drical boss at each corner, is bolted to the face-plate of a lathe, the axis of
one of the bosses coinciding with the axis of the face-plate. The length of
the sides of the casting, measured from centre to centre of the bosses, is
10 inches, and the weight is 6 lbs. Find where on the face-plate a mass of 2 lbs.
should be bolted in order to balance the centrifugal force of the casting.

16. A flexible belt weighing | lb. per foot connects two equal pulleys the
centres of which are 10 feet apart. The diameter of the pulleys is 3 feet, and
the moment of inertia of each is 110 lbs. ft. Find the couple required to
uniformly increase the speed from 120 revolutions per minute to 240 revo-
lutions per minute in 30 seconds.

17. A roundabout makes 3 revolutions per minute, and the horses are
suspended from the roof at a distance of 15 feet frona the centre by rods
8 feet long. What is (1) the angular velocity of the roundabout, (2) the speed
of the points of suspension, (3) the speed of the horses?

18. A train weighing 100 tons starts from rest and the engine exerts a
uniform pull of 3 tons against a road resistance of 15 lbs. per ton. In what
time will the speed reach v feet per second, and how far will the train move
in the time ?

If steam is cut off and the brakes applied, with a retarding force of 20 tons,
when the speed is v feet per second, how far and how long will the train run
before coming to rest?

The train has to run a distance of 2 miles from stop to stop and the speed
is not to exceed 45 miles an hour. How long will the journey take?

19. The balance wheel of a small clock weighs jounce and its radius of
gyration is f inch. The periodic time of oscillation is 1 second and its
amplitude 180 degrees. What is the strength of the controlling spring ex-
pressed in inch-lbs. per radian? Find also the maximum energy stored in
the spring at any time.

20. Explain clearly how it is that the pressure of a brake on the rim of
the wheel of a vehicle produces a retarding force on the vehicle considered
as a whole.

A railway truck, weighing 15. tons, is running freely at 20 miles an hour
down an incline of 1 in 100 when brakes are applied equally to all four wheels

MISCELLANEOUS EXAMPLES

223

with a pressure of 1000 lbs. each. Taking the coefficient of friction between
brake-block and wheel to be |, and neglecting other resistances to motion
and also the inertia of the wheels, find how far down the incline the truck
would travel after the application of the brakes. In what manner would the
inertia of the wheels, if taken into account, affect the calculated value of the
retarding force?

21. A cage, weighing 16 cwt. with its load of coal, is lifted from the bottom
of a mine by a rope weighing 6 lbs. per yard. The rope is wound straight on
to a drum 8 feet in diameter, and a constant torque of 4 tons ft. is applied
to drive the drum. At the start there is a length of 210 feet of rope hanging
vertical, and there is one complete turn of rope on the drum. The inertia of
the drum itself may be neglected. Find how much work has been done when
four more complete turns have been wound on the drum, and the velocity of
the cage at that instant.

22. A marine turbine makes 165 revolutions per minute, and carries a
screw whose blades measure 8 feet 8 inches from the centre of the shaft to
the tip of the blade. Find the circumferential speed of the tip of the blade,
and express it in feet per second and in miles per hour.

The wheel of a Laval turbine is 30 inches in diameter, and it runs at
10,500 revolutions per minute. Find the speed of a point on the rim.

23. A car starts from rest and its velocity has the following values at the
times specified:

Time in sees.

Velocity in ft. per sec.

8-0

9-5

Plot the velocity-time curve and find the time-average of the velocity, and
the total distance covered.

Find also the time-averase of the acceleration, and its maximum value.

24. A body weighing 10 lbs. is moving with a simple harmonic motion,
the length of its path being 12 inches, and the maximum velocity 8 feet per
second. Draw the force-distance curve, and the force-time curve for the
motion from the centre to one end of the path, and determine the space
average and the time-average of the force.

224

ELEMENTARY DYNAMICS

25. The figure below shews a governor in which, at a speed of 120 revolu-
tions, the sleeve remains at rest. For this speed, find the forces in the rods
AB and BC, if the ball at B weighs 1 lb.

Assuming the sleeve at C does not move, at what speed will the stress in
BC be zero?

Fig. 129.

26. A btillet weighing ^ ounce and travelling at 2000 feet per second is
just held up by a fixed block of wood of length 36 inches in the direction of
impact. A ballistic pendulum arranged as shewn in fig, 92, p. 144, is made
of the same wood. It is 30 inches long in the direction of impact and weighs
25 lbs. A bullet similar to the above and travelling with the same speed is
fired into the pendulum. Find (1) the speed with which the bullet emerges,
(2) the height the pendulum rises, (3) the kinetic energy imparted to the
pendulum, (4) the energy wasted in heat, (5) the time taken for the bullet to
pass through the pendulum. The pendulum may be assumed to remain at
rest during the passage of the bullet through it.

27. What is the side pressure between a train weighing 300 tons and the
rails when the train goes round a curve of 120 yards radius at the rate of
30 miles an hour, the rails being on the same level? What will the side
pressure be if the rails are banked at an angle of 5° ?

28. A motor running at 600 revolutions per minute is belted to a machine,
the ordinary fast and loose pulley arrangement being used, and the speed
being reduced in the ratio 3 to 1. If the belt is suddenly shifted from the

MISCELLANEOUS EXAMPLES 225

loose to the fast pulley, find the speed of the machine shaft immediately after
slipping has ceased. The moment of inertia of the armature shaft of the
motor is 800 lbs. ft.^, that of the machine shaft 400 lbs. ft.2, and the inertia
of the loose pulley and belt may be neglected.

29. A uniform plank 20 feet long has one end against a vertical wall, the
other on a horizontal plane. The plank just starts skidding when the angle
of inclination to the vertical is 80°. Assuming the resistance to motion
down the vertical wall is negligibly small, and that along the horizontal plane
it is constant, find the angular velocity of the plank at the instant it reaches
the ground.

30. The motion of a point is the resultant of two simple harmonic motions
in two directions at right angles to one another and of the same periodic
time. Shew that the path of the point is an ellipse.

31. A ship lies at anchor, half a mile from the shore, in a stream running
at 8 miles per hour. A boat starts from the shore, 1 mile higher up stream,
and is rowed at right angles to the shore. At what speed must it be rowed
in order that it may reach the ship ?

Find also the speed required when the boat is rowed in a direction at right
angles to the line joining its starting point to the ship.

If the greatest speed at which the boat can be rowed is 5 miles per hour,
shew that it cannot reach the ship unless its starting point be at least
1100 yards higher up stream than the ship.

32. A bullet of mass m travelling at a certain speed is just stopped in a
distance S by a fixed target, the resistance to penetration of which is constant.
Shew that if the same bullet travelling horizontally with the same speed
strikes a target of the same material as before, and of mass M, which is free to

move horizontally, it will penetrate a distance ., S. Find the time of

M+7n
penetration,

33. In a steam engine the crank is 6 inches long, the connecting rod
3 feet long, and the speed 200 revolutions per minute. Shew that the velocity
in feet per second is approximately given by 10*47 sin ^+ '87 sin 2^, where 6
is the angle which the crank makes with the inner dead centre.

Find a similar expression for the acceleration of the piston.

34. In an engine driven by water pressure the crank AB is acted upon
by three connecting rods BC, BD, BE. For the positions shewn in fig. 130,
the thrusts in the rods are 40 lbs., 180 lbs., and 200 lbs. respectively. Find,
graphically, the resultant force produced by these on the crank pin, and
also its components parallel and perpendicular to the crank.

L. E. D. 15

226

ELEMENTAKY DYNAMICS

Check your results analytically, by resolving all the forces along and
perpendicular to the crank.

[The figure may be pricked through to your paper.]

R = 3'34

Fig. 130.

36. An engineer's pocket-book gives the following formula for the re-
tarding force R required to stop a train travelling at V miles per hour in
D feet, R being the force in percentage of the weight of the train.

Prove this formula.

36. In a small shaping machine the mechanism of which is shewn dia-
grammatically in fig. 131, the tool is fixed to the ram C. When the crank AB
is inclined at 45° to the line of stroke the force at the cutting edge of the tool
along the line of stroke is 500 pounds. Find the couple required on the crank,
neglecting friction and the effect of inertia of the moving parts. Find, also,
the length of stroke of the tool, and the ratio of time of cutting to time of
return.

AB= 3"
BC = 15"

AD=: 3"

Fig, 131.

MISCELLANEOUS EXAMPLES 227

37. In an inclined railway the length of the track is 7017 feet and the
gradient is uniform and equal to 1 in 9-75. The weight of an empty train is

19 tons, and it can carry a maximum load of 21 tons. Two trains are run
simultaneously, one ascending and the other descending, the two being con-
nected by a steel-wire rope which passes round a 10 foot pulley at the top.
The pulley is rotated by a stationary engine. The weight of the rope is
2^ lbs. per foot. Neglecting friction, find the varying torque required to
steadily haul a full train up the incline, the descending train being empty.

If the journey takes 15 minutes, neglecting the power to start and stop,
find the average horse-power.

38. In question 37 above, if the trains are started with a uniform ac-
celeration so that they attain their speed in 6 seconds, what torque on the
pulley will be required for accelerating? The moment of inertia of the pulley
and the wheels of the train may be neglected.

39. A disc of 10 inches diameter is keyed eccentrically to a horizontal
shaft, the distance between the centres being 3 inches. The disc presses
against the underside of a horizontal plate, and is used as a cam to give a
reciprocating vertical motion to the plate. Shew that the reciprocating
motion is simple harmonic, and determine the length of the stroke. Find
also the maximum velocity of the plate when the disc is rotating once in a
second.

40. A car starts from rest and moves with a uniform acceleration of
1^ feet per second per second. How long will it take to acquire a velocity of

20 miles per hour, and how many feet will it move in the time ?

If the car weighs 2000 lbs. and the motion is on the level and against an
external resistance of 30 lbs. wt. what is the value of the horizontal force
between the car and the ground ?

41. A weight moves as a conical pendulum, at the end of a string 8 feet
long and makes 40 revolutions per minute. Find to the nearest degree the
inclination of the string to the vertical.

A string whose length is I passes through a heavy ring and has its ends
attached to two points, distant a apart in the same vertical line. Shew that
when the ring rotates in a horizontal circle the portion of string between the
ring and the lower point of support will be horizontal if the angular velocity

is given by 0,2=2^^^^^.

15—2

228 ELEMENTARY DYNAMICS

42. A traction engine weighing 8 tons is hauling a loaded wagon iSreighing
18 tons along a level road, at 5 miles per hour. There is a spring coupling
between the engine and the wagon, and the extension of the spring shews
that the pull on the wagon is 600 lbs. At what rate in horse-power is work
being done in hauling the wagon?

If the resistance per ton is the same for the engine as for the wagon, what
is the total resistance overcome by the engine and at what rate, in horse-
power, is the engine working?

43. The motion of a body moving in a straight path is given by
s = 10 + 8t + 6t^, s being measured in feet and t in seconds. Plot the distance-
time curve, the velocity-time curve, and the velocity- space curve, for the first
6 seconds. For this period find the time-average and the space-average of the
velocity.

44. A body swings about a fixed horizontal axis through an angle a on
either side of its position of equilibrium. Prove that for different values of

a the maximum angular velocity of the body is proportional to sin - .

A body makes complete revolutions about a fixed horizontal axis, about
which its radius of gyration is A;, and the centre of gravity of the body is at
a distance c from the axis. If the greatest and least angular velocities are

^ per cent, greater and ^ per cent, less than w, prove that w = */ ,f .

45. What do you understand by a curve of crank effort? Shew how to
determine the crank effort corresponding to a given piston-thrust in any
position of the crank of a single-cylinder engine.

Determine the crank effort for a piston-thrust of 100 lbs. when the crank
makes (I) 45°, (2) 90°, (3) 135° with the line of stroke, the lengths of the
crank and connecting rod being 6 inches and 24 inches.

46. A cage weighing 1 ton is being raised from the bottom of a shaft
200 feet deep by means of a drum and chain. The chain weighs 5 lbs. per
foot. Draw a curve of the effort exerted at the lifting drum, throughout the
motion, and find the whole work done during the lift. If the cage is raised
at a uniform speed in 2| minutes, find the maximum and the mean values
of the horse-power exerted at the drum.

47. Two links AB and CD are pivoted at A and C and a weight of 10 lbs.
hangs from point E in the link BD, as shewn in fig. 182. Prick the figure
through to your paper, and taking 2 inches to represent the velocity of the

MISCELLANEOUS EXAMPLES

229

point D, find the angular velocity ratio of AB and CD, and the position of
the point in BD which has a vertical motion.

Find also what couple must be applied to CD in order to just support the
weight.

-> Couple

27r

Fig. 132.

48. A uniform heavy rod is supported in a horizontal position, being
pivoted at one point in its length and attached at another point to the end
of a spring which hangs from a fixed point vertically above the point of
attachment. If a is the distance between the points of support, k the radius
of gyration about the pivot, and d the extension of the spring when it carries
a weight equal to the weight of the rod, shew that the period of small oscilla-
tions in a vertical plane is

If a sliding weight is moved from the pivot towards the spring how is the
period affected ?

49. The engines of a steamship develop 20,500 i.h.p. Of the power de-
veloped 15 per cent, is lost in engine and shaft friction, and 30 per cent, in
slip and friction of the propeller. If the speed of the ship is 20 knots what
is the total resistance to its motion?

If the resistance vary as the square of the speed, what i.h.p. is required for
a speed of 22 knots?

1 knot is 6080 feet per hour.

50. A centrifugal pump lifts water at the rate of 6000 gallons per minute
and discharges it at a level of 10 feet above the level of suction. The dis-
charge pipe is 18 inches in diameter and runs just half full. Find how much

230

ELEMENTARY DYNAMICS

work is spent per minute (1) in lifting the water, (2) in imparting kinetic
energy to it.

The pump runs at 460 revolutions per minute and one-fifth of the power
supplied to it is wasted in friction, slip, etc. What is the torque on its shaft?

51. A train starts from rest and its velocity during the first 90 seconds of
its motion is given in the following table:

Time from start in sees.

75
19-3

90
20

Velocity in ft, per sec.

5-2

14 1

17-3

Plot the velocity-time curve, and determine the distance travelled in the
given time.

From the curve deduce the value of the starting acceleration and plot a
curve shewing approximately how the acceleration diminishes.

62. In a gas engine the maximum amount of energy which the two fly-
wheels have to absorb and give out during running is equal to 3550 ft.-lbs.
The mean speed of the engine is 200 revolutions per minute, and the moment
of inertia of each wheel is equivalent to a mass of 692 lbs. acting at a radius
of 2^ feet. Find the greatest fluctuation of speed of the engine.

53. What is the length of a simple pendulum whose periodic time is
2-3 seconds?

How will the time-keeping of a pendulum clock be affected if it be taken to
a place where "/7" is decreased by O'Ol per cent.?

54. A balloon weighing 5 cwt. is moving horizontally at 100 feet from the
ground when 50 lbs. of ballast is suddenly released. Neglecting the friction
of the air, find how high the balloon will be when the ballast reaches the
ground.

55. A shell is fired at an elevation of 15° so as to hit a mark 900 yards
distant on a horizontal plane. Find its velocity of projection and its time of
flight. Find also its height above the ground when it has travelled 600 yards
horizontally.

56. A lifting gear consists of a train of toothed wheels as shewn in fig.
133, the intermediate wheels being keyed to the same axle. The gear is
driven by a force P applied at the end of the handle AP, which is 18 inches
long and is fixed to the wheel A. The load W is supported by a rope wound

MISCELLANEOUS EXAMPLES

231

round a drum of 4 inches diameter. The numbers of teeth are, on A 21, on
B 60, on C 18, on D 70. Find the velocity ratio of the machine.

If a force P of 3 lbs. raises a load W of 200 lbs. , shew that the efficiency is
66*7 per cent. ,

Fig. 133.

57. An engine running at 120 revolutions per minute has a pulley of
36 inches diameter which is belted to a pulley of 20 inches diameter on
a countershaft. A second pulley on the countershaft is belted to a pulley of
12 inches diameter on a dynamo, and the dynamo is required to run at 720
revolutions per minute. Find the diameter of the second pulley on the
countershait.

The engine is delivering 50 horse-power to its belt, and 2 horse-power is
lost in friction at the countershaft. Neglecting other losses, find the tension
in each belt, assuming that in each case the tension in the tight portion of
a belt is double that in the slack portion.

58. In an inward flow water turbine the external diameter of the rotor is
15 inches and the vanes are radial at the point of entry. The velocity of the
water just before entry is 50 feet per second, and the speed of the rotor is
380 revolutions per minute.

At what angle of inclination with the tangent should the water enter if
there is to be no shock ?

Assuming no shock, the supply 500 cubic feet^per minute, and the discharge
radial, find the turning moment on the rotor.

232 ELEMENTARY DYNAMICS

59. Two trains start from a station along the same track, one 8 minutes
after the other. Each has a constant acceleration until the maximum speed
of 50 miles per hour is reached, 1 mile from the start. How far has the first
train gone when the second train starts, and how far apart are they when
both are running at the same speed?

If the second train, with the same acceleration, reaches a maximum speed
of 60 miles an hour, at what distance from the starting point will it overtake
the first ?

60. A car weighing 22 cwt. is running on a level road at 12 miles an hour
round a curve of 40 feet radius, measured to the centre of gravity of the car.
What horizontal force perpendicular to the direction of motion must be exerted
by tie ground on the wheels of the car?

Assuming that the grip of the outer wheels is sufficient to prevent skidding,
find at what speed the inner wheels will begin to lift off the ground. The
width of the wheel base is 3 feet 9 inches and the centre of gravity is 3 feet
above the ground.

61. What do you understand by the principle of conservation of momen-
tum ? Shew how this principle is based on the laws of momentum.

A railway truck, weighing 10 tons, moving with a velocity of 4 miles per
hour, impinges on another truck, weighing 8 tons, and moving in the opposite
direction with a velocity of 1 mile per hour. When impact occurs the two
trucks are automatically coupled together. Find the velocity of the trucks
after impact, and the number of foot-tons of energy lost.

62. A cage is being pulled up a shaft with a velocity of 10 feet per second.
A block of stone, weighing 50 lbs., is dropped down the shaft and after falling
through 60 feet it meets the cage. The cage continues to ascend without
change of velocity, carrying the stone with it. What is the impulse of the
blow between the stone and the cage ?

If the motion of the stone is completely reversed in half a' second, what is
the average value of the thrust between the cage and the stone ?

63. On a railway line there are two points of observation 1 mile apart.
A train passes the first point at 30 miles an hour, and after an interval of
90 seconds it passes the second point at the same speed. What is its average
speed between the points ? If the train had uniform acceleration during the
first half of the interval and uniform retardation during the second half,
what was the highest speed reached during the interval, and what was the
value of the acceleration ?

MISCELLANEOUS EXAMPLES

233

64. Simultaneous values of speed and time for a train are given jn the
table below.

Find graphically the acceleration in feet per second per second at the end
of the 1st minute and at the end of the 5th minute.

Find also the distance passed over in attaining a speed of 55 miles pei

hour.

Time
sees.

0
0

100

150

200

250

300

350

400

450

500

•

Velocity
m.p. h.

24-6

36-3

43-7

48-3

51-3

53-5

55-1

56-2

57-6

57-9

65. In a steam engine mechanism, AB represents the crank, BC the con-
necting rod, and T is the point of intersection of the connecting rod and the
line through A perpendicular to line of stroke. Shew that if M is the couple
on the crank, the thrust on the crosshead in the direction of motion is given

M M

by ^^ , and the normal thrust on the guides of the crosshead is given by ^ .

Shew, also, that the angular velocity ratio of the connecting rod and crank is
equal to ^^ .

66. An electrically propelled car is fitted with a recording accelerometer,
the drum of which is driven by gearing connected to the wheels. The record
traced in starting the car from rest is shewn in fig. 134. Find the velocity of
the car after 5 seconds and the horse-power per ton which is then being
expended in acceleration.

•

^ °

10 20 30 40 50 60 70 80
Distance travelled by car. feet
Fig. 134.

90 TOO

234

ELEMENTARY DYNAMICS

V-

67. If a train's resistance to motion under certain conditions be 4 + —— lbs.

I JUL)

per ton, when the speed is V miles per hour, shew that the horse-power ex-
pended in hauling a train of 200 tons is 2-13 V 4-0-0026 V^. Plot a curve
shewing the relation between horse-power and speed between 10 and 40 miles
per hour.

68. A truck moving with a given velocity impinges on another truck of
the same mass which is at rest. Shew that, if there is no loss of energy due
to the impact, the first truck will be left at rest, and the second truck will
move off with the velocity originally possessed by the first truck,

60. Shew how the distance passed over in a given time by a body whose
velocity continually varies with the time, may be estimated.

A body starts from rest with a constant acceleration of 1 foot per second
per second ; after a time the body has a constant retardation and finally
comes to rest 10 miles from its starting point. Find the maximum velocity
and the value of the retardation if the total time taken is 16 minutes.

70. A cage weighing 4000 lbs. is wound up a shaft. The relation between
the tension, T lbs. wt. in the rope, and the distance, x feet, which the cage
has risen, is given in the following table :

6000

5900

5660

5200

•

4500

3500

2650

2130

2000

Plot a curve for tension and distance, and find the work done during the
80 feet given, and the kinetic energy of the cage at the end of that distance.
At what point is the kinetic energy greatest, and what is then its value ?

71. A door I feet wide, of mass m lbs., swinging to with an angular velocity
w radians per second is brought to rest in an angle d radians by a buffer stop

which applies a uniform force P at a distance ^ from the axis of the hinges.

Find the magnitude of P, and the hinge reactions normal to the door, when
the buffer is placed in a horizontal plane half way up.

72. A motor-car has 30 inch wheels, and weighs 2000 lbs. When travelling
at 18 miles per hour, a total braking force of 500 lbs. is applied at a radius
of 6 inches from the centre of the wheels.

Determine the time taken to reduce the speed to 10 miles per hour, and
the ratio between the distances travelled while the speed waa being reduced

MISCELLANEOUS EXAMPLES 235

from 18 to 10 miles per hour, and from 10 miles per hour till the car stopped.
In practice, what other factors would have to be considered in making these
calculations?

73. A 4 inch shaft is to be turned in a lathe. If the cutting speed is not
to exceed 40 feet per minute what is the maximum speed at which the shaft
may be rotated. The speed of the tool along the shaft is so small that it may
be neglected.

74. The stroke of a steam engine is 24 inches, the cylinder diameter is
14 inches. Steam is admitted to the cylinder at 100 lbs. per sq. in. pressure,
during a quarter of the stroke, and then expands, the expansion curve being
the hyperbola pt; = constant. On the other side of the piston there is a
constant pressure of 16 lbs. per sq. inch. Plot a curve shewing the effective
steam thi'ust on the piston during the stroke, and find how much work is
done by the steam on the piston in one stroke.

75. A string 15 feet long has its ends fixed to two points at the same level
and 10 feet apart. A smooth ring is threaded on the string and is initially
held vertically below one fij:ed end of the string by a horizontal force. If the
ring is suddenly released shew that it will move in an elliptical path. Find
its velocity when it has travelled a horizontal distance (1) of 2-5 feet, (2) of
5 feet.

76. A flywheel weighing 10,000 lbs. is suspended from a pair of centres
entering conical holes in the rim, so that it can swing in a vertical plane.
The liue joining the centres is parallel to, and distant 3 feet from, the axis
of the wheel. The period of a complete swing is 2-5 seconds. Find the radius
of gyration of the wheel, and the energy stored in it, when running at 250
revolutions per minute.

77. Two vessels start from the same point: one steams due north at
10 miles per hour, the other steams north-east at such a speed as always to
keep due east of the first. Find the velocity of the second vessel, and the
magnitude and direction of the relative velocity of the two.

At one hour from the start both vessels change their course and steam due
west, maintaining the same speeds as before. After what time will the faster
vessel overtake the slower, and how far has each steamed in the time?

78. The crank of a petrol engine is 3 inches and the connecting rod
12 inches long. The engine runs at 720 revolutions per minute. Find
graphically the velocity of the piston when it has performed (a) one-quarter
of its out-stroke, {b) one-quarter of its in-stroke.

Find also the angular velocity of the connecting rod for each of the two
positions.

236

ELEMENTARY DYNAMICS

79. A pendulum bob, weighing 8 ounces, rotates uniformly in a hori-
zontal circle at the end of a light string 1 foot long, the other end of the
string being fixed. If it make 120 revolutions per minute, what is the tension
in the string, and what angle does the string make with the vertical?

At what angle and with what speed does the pendulum rotate if the tension
in the string is equal to four times the weight of the bob ?

80. A lifting tackle consists of two blocks: the upper one is fixed and
contains two sheaves, the lower one contains one sheave and is fast to one
end of the rope. Sketch the tackle and find its velocity ratio.

The following table gives the pull required to lift various loads :

Load in lbs.

112

168

224

280

336

Pull in lbs.

109

130

150

Find the efficiency of the tackle for each of the given loads, and draw
graphs of pull and efficiency on a load base.

81. A hose is directed perpendicularly against a wall and delivers 10 cubic
feet of water per minute through a nozzle of 1 inch diameter, the water
having no rebound from the wall. Determine the pressure on the wall.

82. A practical formula used for estimating the tractive force F, at the

PLD2

rails in lbs. for a two cylinder locomotive is, F :

, where W is the dia-

meter of the driving wheels in inches, P is the mean pressure of steam in the
cylinders in lbs. per sq. inch, D is the diameter of the cylinders in inches,
and L is the length of stroke in inches. Derive this formula.

83. A jet of water impinges with velocity v feet per second on a serie? of
plane vanes set round the circumference of a wheel and moving with velocity
u feet per second. Shew that for each square inch of section of the jet the
horse-power communicated to the wheel is very nearly 2*45 {v - u)v .ii. 10~^,
and find an expression for the efficiency of the jet.

84. The slide-valve of a horizontal steam engine derives its motion from
a point P in a link AB, where AP = ^AB. The horizontal displacements of
A and B for any crank position are given by the equations

a;i = 2-5sin(^-f27°),
a;2 = 2-6sin(6' + 150°).

MISCELLANEOUS EXAMPLES

237

Shew that the resulting motion of the valve may be expressed by the
equation

x = asm (d + e)
and find the values of a and e.

86. A flywheel possesses 50 foot-tons of kinetic energy when rotating at
200 revolutions per minute. What is its moment of inertia, the unit being
1 lb. at 1 foot radius? How many foot-lbs. of work must be done to increase
its speed from 200 to 202 revolutions per minute ?

86. A steam engine has a crank 9 inches long and a connecting rod
36 inches long. At the instant when the crank is at right angles to the con-
necting rod there is a thrust of 600 lbs. wt. in the piston rod. Find the
turning moment on the crank, and the force on the crosshead guides.

On the return stroke there is a pull of 600 lbs. wt. in the piston rod at the
instant when the crank makes 180° with the position given above. Find in
this position the turning moment, and the force on the guides.

87. Three masses A, B, and C are bolted to the spokes of a horizontal
wheel, the centre of which is O. The masses are 3, 5, and 4 lbs. respectively ;
AOB = 60°, BOC = 90° ; OA =: 24 inches, OB = 20 inches, OC = 18 inches. Find
the magnitude and angular position of a mass D which will produce no re-
sultant transverse thrust on the shaft, taking OD = 15 inches,

88. The speed and power trials of a cruiser gave the following results:

Speed in knots

Horse-power

2600

6000

11,800

23,000

Find the value of the resistance overcome at each speed, and plot curves
of horse-power and resistance relative to speed.

If the vessel weigh 5000 tons, what additional horse-power is required to
give the vessel an acceleration of 1 knot per minute when the speed is 22 knots?
Take a knot as 100 feet per minute.

89. A flywheel and shaft, the moment of inertia of which is 10 lbs. ft.'^,
are mounted on ball bearings so that friction may be neglected. The diameter
of the shaft is 2 inches, and a string wrapped round it carries a weight of
10 lbs. The shaft is horizontal, and the string is so connected to it that, when
completely unwound by the rotation of the shaft, it winds on again. The weight
is supported 4 feet from the ground, with the string just taut, and is then let go.
Find how high the weight will be raised after its first contact with the ground.

238

ELEMENTARY DYNAMICS

90. The mechanism of a small machine for punching holes in paper is
shewn in fig. 135, the links being connected by pin joints, and the punch
constrained to move vertically. Find, graphically, the velocity ratio of the
points C and B.

Estimate the vertical force required at C to produce a force on the punch
of 50 lbs. (a) when ABC is in the position shewn, (b) when ABC is perpen-
dicular to OA.

0A = 1", AB^I", AC=i3i", and O is f" from the axis of the punch.

/////////////

Fig. 135.

01. Two equal toothed wheels A and B are connected by a light link C.
The wheels are rotating with an angular velocity w, with the link C fixed,
when suddenly the link C is released and the wheel A is fixed. Shew that,
if the inertia of the link be neglected, it will begin to rotate with an angular

velocity r-yra 5^ » where k is the radius of gyration of either wheel about its

centre, and a is the effective radius of the wheels.

What is the impulsive couple which will be required on wheel A ?

92. A pendulum consists of a thin rod weighing 2 ounces and a uniform
disc of 3 inches diameter weighing ^ lb. The rod which is 3 feet long extends
to the centre of the disc. Find the periodic time of swing of the pendulum
when suspended from the free end.

MISCELLANEOUS EXAMPLES 239

93. It is required to transmit 25 horse-power from a pulley of 30 inches
diameter running at 350 revolutions per minute, by means of a belt. Find
the required width of the belt, having given the ratios of the tensions on the
two sides to be equal to 2, the thickness | inch, and the allowable stress
500 lbs. per sq. inch.

94. A smooth wire, bent into the form of a circle of radius r, is rotating
above its vertical diameter with angular velocity w. If a mass is strung on
the wire, shew that it will be in equilibrium when its angular distance, 0,
from the vertical diameter is such that

ru}'^cosd = g.
The mass is given a small displacement from its position of equilibrium,
shew that it will oscillate about this position and that the period of oscilla-
tion is

2«-
w sin 0 '

95. For the purpose of conveying small packages from one room to
another vertically above it, the package is enclosed in a small cylinder which
forms a piston for a vertical tube passing from the one room to the other.
The cylinder being fitted into the lower end, the upper portion of the tube is
connected to a reservoir in which a vacuum of 5 lbs. per sq. inch by gauge
is maintained. The cylinder and contents weigh 1 lb. and the diameter is
1^ inches. Taking the friction between the tube and the cylinder to be 1 lb.
find the time for the cylinder to travel vertically upwards 40 feet from the
starting point, and also find the velocity of arrival. The vacuum pressure
may be assumed to remain constant, and the effect of the inertia of the air
below the cylinder may be neglected.

96. A mass vi, whose centre of gravity is at a point B, is supported by
two equal links AB, CB, which are hinged to a vertical spindle at A and C.
Find the forces in the links AB, CB

(1) when the system is at rest ;

2fl

(2) when it is rotating about AC with angular velocity <a and w^ = 7r7» ;

4:0

(3) when it is rotating about AC with angular velocity w and ^2= Ap •

97. A steam engine is suddenly brought up by an unyielding obstacle
interposed in the line of motion of the piston rod ; find the impulse for a
given position of the crank, taking into account the momentum of a flywheel
on the crank shaft, and of a piston and connecting rod, but neglecting the
angular motion of the connecting rod, that is to say treating it as of infinite
length.

240 ELEMENTARY DYNAMICS

98. The displacement s, in feet, of a piston is given in terms of the time

t, in seconds by

s = 0-5 cos Airt + 0-25 cos 8irt.

If the piston weighs 20 lbs. find the force in the piston rod required for
accelerating at the following times, 0, ^V, ^, I second.

Find also the kinetic energy given out or taken in by the piston (1) during
the first half-stroke, (2) during the middle half-stroke.

90. A particle is suspended by three equal strings of length a from three
points forming an equilateral triangle of side 2b in a horizontal plane. If one
string be cut, shew that the tension of each of the others is instantaneously
changed in the ratio

3aP-ib^

lOO. A wheel of 24 inches diameter, and having a radius of gyration of
10 inches, rolls along the ground and meets a step 6 inches high. Shew that
if the speed of the wheel is less than 6*18 feet per second it will not be able
to lift itself on to the step. Assume that the wheel is inelastic, and that there
is no slipping at its point of contact with the square edge of the step.

ANSWERS TO EXAMPLES
CHAPTER I. (Pages 17—19)

1. 554 yards; 15° 45' west of north; 533 yards; 150 yards.

2. 1-732; 1-5; 0-866. 3. 331 lbs.; 73° 34' north of east.

4. (a) 2-1, 106^°; (b) 5-06, 329-4°; (c) 7-97, 118°. 5. 6-1 feet.

6. 60; 66. 7. 2340; 390. 8. 37,600; 76,800; 3600; 1300.

0. -; --^. lO. t1-13; ±1-13. 11. 2-15 square inches.
3 Js

CHAPTER 11. (Pages 51—56)

1. (1) 8-97 metres per second. (2) 26-84 metres per second.
(3) 2796 metres per second.

2. 1056 feet per second. 3. 47-3 feet.

6. 131 feet per second ; 150 feet per second ; zero.

7. 34-4 feet per second; 10-47 radians per second; 420 revolutions per

minute.

8. 4400 feet per minute ; 840 revolutions per minute ; 88 radians per second.
O. 3000 revolutions per second. lO. 2-62 feet per second.

13. 1-05 miles; 22-8 miles per hour; 1*65 feet per second per second;

0-42 foot per second per second ; 2*08 feet per second per second.

14. a = 261-8 cos 25^ + 43-5 cos 50«. 18. 21-2 feet.

19. 2055 feet per second ; 411 feet per second. 20. 51° 20'; 88°.

21. 3-88 feet per second; 52° 12' north of east; 0-554 foot per second per

second.

22. 66-7 feet per second per second.

23. 12-2 minutes ; 151° to the direction of current ;

7-1 „ 123°

24. 44 seconds; 2499 feet. 25. 23-2 knots; 4-12 hours; 7*07 sea miles.

26. 35 miles per hour ; inclined at 82° to OB.

27. Greater for ^ = 45°.

28. 200 revolutions per minute ; 127 feet per second inclined at 150° and

30° with the direction of motion of the train.

29. 0-655 foot per second ; 1*635 feet per second. 30. 10 seconds; 50 feet.

31. 6-72 inches ; 0-69 second ; 0-81 second ; 80 feet per minute ; 63-2 feet

per minute.

32. 324 feet per minute.

L. E. D. 16

242 ELEMENTARY DYNAMICS

CHAPTER III. (Pages 83—87)

1. 222 lbs. wt.; f foot per second per second ; 510 feet.

2. 72,100 poundals ; lO^" dynes. 3. 1-305 tons wt.

5. 2000 lbs. wt.; 2125 lbs. wt.; 1688 lbs. wt.

6. 1674 lbs. wt, ; 3'86 feet per second per second.

7. 11-4 seconds ; 167 feet ; 13-1 seconds ; 192 feet.

8. 0-354 lb. wt. inclined at 45° to the original direction of motion.

9. 98 lbs. wt.; 170 lbs. wt. lO. 1-77 feet per second.
12. 116 lbs. wt. 13. 92 seconds. 14. ^ ton wt.
15. 550 lbs. wt. ; 505 lbs. wt. 16. 138-6 seconds; 10,000 feet.
17. 1-06 feet per second per second upwards.

l'8. 1700 feet per second ; 750 lbs. wt. 19. 96 square feet.

20. 4-97 lbs. wt. 21. 0*0122 second ; 2-73 feet per second.

22. 1-14 feet per second per second.

23. 35,21-2, -5, -21-2, -25 tons wt. 25. 6-6 inches.
26. 0-517 tons. 12-85 tons-feet-seconds units. 27. 950 lbs. wt.

28. 1'19 seconds; 286,000 absolute c.g.s. units; 18-1 centimetres.

29. 46 absolute f.p.s. units ; 3-83 feet per second.

30. (A) 13 feet per second ; 9 feet per second. (B) 7 feet per second ; 11 feet

per second.

CHAPTER IV. (Pages 107—109)

1. 14-7 lbs. feet. 2. 0-025 feet per second per second ; 14-45 revolutions.

{4(M-fm)-hW}M (4M-f-W)(M4-m) , 8M2-f8Mr/i-f W (2Mj-7n)
^ • 4 M -f- 2m -f- W ^ ' ~^lvr+2?M + W ^ ' 7" 4 M -f 2m + W ^'

4. 32 feet per second ; 128 radians per second ; 26-9 lbs. feet-.

6. 0-155^ foot per second per second.

6. 12 lbs. feet; 7-4 lbs. feet; 720 revolutions.

7. 40-03 lbs. feet. 9. 0-186 lbs. feet.

10. 50 absolute f.p.s. units ; 0*032 absolute f.p.s. units.

11. 8660 lbs. feet^. 15. 1 foot 3 inches from the hinge.
16. 318 lbs. feet. 17. 20 revolutions per minute.

18. 83-3 lbs. feet2. 19. 2*65 lbs. feet^. 20. 3360 lbs. feet^.

21. (1) 6-53 lbs. feet^. (2) 298 lbs. feet^.

ANSWERS TO EXAMPLES

243

9.
lO.
11.
14.

CHAPTER V. (Pages 124—127)

86-8 revolutions per minute. 3. \l^ '--i ; 3'58 feet.

Pull of 12-5 lbs. wt. at A and push of 12-5 lbs. wt. at B.

„ 26-8 „ „ pull of 1-7

92-3 revolutions per minute.
10-36 miles per hour inclined at 75° to the road.
20 „ „ „ 150°

38-6 „ „ „ 165°

7-7 lbs. ; 0-4 lbs. wt. 6. 6-2° ; 6^ inches.

847 lbs. wt.; 24-4 miles per hour. 8. 25-8°; 767 lbs. wt. per ton.

1182 lbs. wt.

270 revolutions per minute; 266 revolutions per minute.
473 revolutions per minute. 12. 50*4 tons wt.; 9^ inches ; 63 tons wt.
5-23 tons wt. 15. 454 tons wt. 2-8 inches. 16. 2-83 inches.

8.
lO.
12.
13.
14.
16.

CHAPTER VI. (Pages 154—161)

6800 ft. -tons ; 60-8 horse-power ; 31-7 horse-power.
346ft.-tons; 1970 strokes ; 471 lbs. wt.

40 lbs. per square inch ; 8200 foot-lbs. 4, 169 feet ; 1*07 horse-power.
0-62 horse-power ; 60 feet per minute ; 120 feet per minute ; 1*65 horse-
power.
12-1 ft.-tons ; 0-269 ton wt.

6-3 feet per second ; 538 feet. 9. 8-75 horse-power ; 47*5 miles per hour.
6-6 inches. 11. 16-7 miles per hour ; 9-32 ft.-tons.

9-6 ft.-tons; 3-3 ft.-tons; 5-2 horse-power.
17 feet per second ; 2-25 tons wt. ; 2-52 tons wt.
1880 feet per second. 15. 12-5 lbs. per ton ; 333 horse-power.

17.
19.
21.
22.
23.

Distance
feet

Speed

feet per second

9-2

9-9

10-3

13-4 ft.-tons.

5-06 feet per second; 33-6°. 18, 3-26 feet per second,

2-45 indies. 20. 99-75 per cent., 3125 lbs. wt.

9*8 inches ; 5*15 feet per second.

41-7 tons wt.; 1785 horse-power.

5867 abs. ft.-ton-sec. units; 13,400 ft.-tons; 293 feet; 45-8 tons wt.

244 ELEMENTARY DYNAMICS

24. 0-294 foot per second per second ; 0'22 foot per second per second.

25. 18 '9 horse-power. 26. 1-33 horse-ppwer ; 0-43 horse-power.

27. 4 radians per second; 1250 lbs. wt.; 3-98 radians per second.

28. 3300 ft. -lbs.; 289 f t.-tons ; 3-76 seconds.

29. 8160 lbs. ft.; 1^ per cent. 30. 105-5 ft.-tons ; 57-5 per cent.
31. A, 56-5 revolutions per minute; B, 443-5 revolutions per minute;

0-155 second ; 326 lbs. wt. sec. units.

CHAPTER VIII. (Pages 183—186)

2. 2-09 seconds ; 1% lb. wt. 3. 4-32 seconds.

4. Gtt feet per second; 665 lbs. wt.; 337-5 lbs. wt.; 1-65 seconds.

7. 2-048 feet per second; 2-052 feet per second. 8. 1-57 seconds.

10. 31 ft.-tons. 11. 0-179 second ; 1-3 inches; 3 feet per second.

12. 1-4 seconds. 13. 30-4 seconds. 14. 224 lbs. wt.

17. 1-57 seconds ; 1-59 seconds. 18. 12-3 tons wt.

CHAPTER IX. (Pages 210—218)

1. 9| inches. 2. 0-57 horse-power.

3. 0-28 horse-power-hour ; 44-2 lbs. wt. 4. 5-07 horse-power.

5. 39 '4 feet per minute ; 8-38 lbs. feet. 6. 61*5 per cent.; 3| tons ; f ton.

7. 0-215 ; 0-45 ton wt. 8. (1) 57 lbs. wt. (2) 203 lbs. wt.

9. 23-6 horse-power ; 1-4 horse-power ; 722-5 lbs. wt. ; 147 lbs. wt.

10. P = 18-4 W -I- 12-6, where P-the effort in lbs., and W = the load in tons.

11. 2533 lbs. feet ; 1733 lbs. feet ; 328 lbs. feet.

12. 221 revolutions per minute; 30-2 horse-power. 13. 41-1 horse-power.

14. 0-547 foot per second; 136-5 absolute f.p.s. units; 33-6 ft.-lbs.

15. 6-55 tons wt.; 673 tons.

16. 35-5 lbs. per square inch; 190 ft.-tons; 12-9 horse-power.

17. 13-7 tons wt.; 17 knots. 18. 21 -8 ft.-tons; 106 revolutions per minute.
19. 18*5 horse-power; 3-3 horse-power. 20. 86 per cent.

21. 85-5 lbs. feet; 301 lbs. feet; 258 lbs. wt. ; 30-2 miles per hour.

22. 14-5 revolutions. 23. 1-82 horse-power ; 37-8 per cent.

24. 318 revolutions per minute ; 0*74 horse-power.

25. 100 per cent, (nearly) ; 75-6 per cent. * 26. 9-6 horse-power.

27. 1-25 feet per second per second.

28. (1) .9 sin a. (2) ^^g sin a. (3) f l^f sin a.

29. 48-5 per cent. ; 55 per cent.; 31-6 horse-power.

30. Ill ft.-lbs.; 37 ft.-lbs.; 74 ft.-lbs.

INDEX

{The numbers refer to the pages)

Absolute unit of force 59

Erg 129

Absorption brake 188

Examples Chapter I 17

Accelerating force on piston 67

„ n 51

Acceleration 29

„ III 83

Acceleration due to gravity 39

„ IV 107

,, ,, to change of direc-

„ V 124

tion 43

„ VI 154

Action and reaction 79

„ VIII 183

Amplitude 168

„ IX 210

Angular momentum Chapter IV 88

Miscellaneous 219

,, velocity 27

Area curve 15

Force 58

Average value 9

,, , Unit of 59, 60

,, velocity 21

Forces — external and internal 89

F.P.S. Units 2

Ballistic Pendulum 143

Belts 187

Gearing 194

Brake, absorption 188

Gravitation unit of force 60

Gravity — Acceleration due to 39

Cant on railway curves 119

Centre of gravity 117

Harmonic motion 106

,, ,, mass or inertia 115

,, „ and circular i

Centrifugal force Chapter V 110

tion 177

Centripetal force 110

Horse-power 130

C.G.S. Units 2

Horse-power-hour 131

Circular motion 43
Components 7
Conical pendulum 111
Conservation of momentum 55, 147
„ „ energy 55, 136, 147

Density 3
Dimensions 162
Dyne 60

Efficiency 191

Energy Chapter VI 131

„ , potential and kinetic 132
,, , conservation of 136

Equilibrium 78

Impact 78

Impulse 76

Impulsive force^magnitude of 77

Impulsive torque 92

Indicated horse-power 197

Inertia 90

Joule 130

Kinematics 1
Kinetics 1
Kinetic energy 132

„ ,, due to rotation 147

,, .total 206

246

ELEMENTARY DYNAMICS

Laws of momentum 58

,, ,, motion Preface vi
Length 2

Linear momentum Chapter III 57
,, velocity 20
,, acceleration 29

Mass 2

Machines Chapter IX 189
Mechanical Advantage 19
Mechanics 1

Momentum, Linear Chapter III 57
„ , Angular ,, IV 88

,, ■ , Laws of 58
Moment of a force 89

,, ,, inertia, definition 90,
values of 101
Motion Chapter II 20
,, , Third Law of 79

Newton — Laws of Motion Preface vi

Oscillation 170

Pelton Wheel 203
Pendulum, Simple 170

,, , Compound 172
,, , Simple Equivalent, 173
Periodic time 108
Piston-velocity 49

,, accelerating force 67
Potential Energy 132
Poundal 60
Power Chapter VI 130

,, , Unit of 130
Projectiles 39

Radius of gyration 105
Railway curves 119
Rate of change 10
Relative momentum 195

„ kinetic energy 195

,, velocity 45
Resultant 6

Retardation 33
Rolling Wheel 50
Routh's Rule 101

Scalar quantity 3

Simple equivalent pendulum 173

Simple harmonic motion 166

Space 2

Space-average of force 129

Specific gravity 3

Speed 20

Springs 139, 149, 169

Statics 1

Steam tractor — Mechanical losses

200
Stress 78

,, in flywheel 122
Subnormal 32

Third law of motion 79

Time 3

Time-average of force 129

Torque 89

Transmission of power by belts 187

Units, C.G.S. and F.P.S. 2
Units and dimensions Chapter VII
162

Varying Force 66

Torque 91
Vectors 3
Velocity, linear 20

„ , angular 27

,, curves 22

„ , average — instantaneous 21

,, , relative 45
Velocity-ratio 189
Vibration 167

Watt 130
Weight 60

Work Chapter VI 128
„ , Unit of 129

CAMBRIDGE : PRINTED BY J. B. PEACE, M.A., AT THE UNIVERSITY PRESS

University of Toronto
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