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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. BOMBAY '\

CALCUTTA I MACMILLAN AND CO., Ltd. MADRAS j

TORONTO : THE MACMILLAN CO. OF CANADA, Ltd.

TOKYO : MARUZEN-KABUSHIKI-KAISHA

ALL RIGHTS RESERVED

' Jee>» .

ELEMENTARY DYNAMICS

A TEXT-BOOK FOR ENGINEERS

BY

jrwr LANDON, M.A.

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

CAMBRIDGE

AT THE UNIVERSITY PRESS

1920

i

I

I

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

/

I

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

I

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

I

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.

or

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,

OB

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

and

sin (O + cf))' AB sin 0

OB

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.

10

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

11

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

13

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

dx

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

dx

any point.

14

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.

A

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'

A

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

A

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

= a X b . hx,

A

and CK = CB sin CBK

A

= 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.

16

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.

or

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

1

X (RS X pv) (area NPGtM).

OB

(area OPQM),

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

i

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

18

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 54	6-0 60	5-6 70	6-0 80	5-2 92	4-3 100
Distance in feet	0

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

30

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

19

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	29	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

o

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.

x^

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.

K

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).

«
o	^^
c	^^^
i	P^^^
(0 B	/^	1
^^<8	(	!

O N Time

Fig. 15.

22

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.

ds

We have.

velocity

dt

= 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.

>»		p
•fj	^^Q
>	-^	i	-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

24

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

V

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

25

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.	2	4	6	8	12	16	2Q	24	28	32	36	40

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'
3	-., 2
O'	7
20 -,	7
:§:::: :: :: ::: :::	^" " - =
: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 >

10

30

35

40

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

27

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

60

= 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

29

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^,

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

30

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

31

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

I

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

U

dv

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

dt

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.

32

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

as

= PN . tan e

= PN . tan NPG

= NG.

Or, the acceleration is given by the subnormal NG.

1

, Space curve

Velocity

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

Ss

We have v = -^ , nearly, bt

i.e. -Ss = Bt.

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

01

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

33

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

88

= 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 = —

NT

= .^^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

A

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

35

Speed in miles per liour

-* ro w

Acceleration In ft per sec. per sec.

3—2

36

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.

15

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	17-0
Distance (nautical miles)	0	i	i	I	1	H	H

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

37

Speed in knots, o

1/veloc 81-	ity
. __ -_-	""I"" ~J
V-	t_
i	1
	t
^
^	7
	/
S " "	/
	T
	i~
	"~r
	/
	7
> - ■	1
	l/
V - -	J
	k
	r
^
♦^•-* - -- t- -	1
_ :^	/<—
5 '	J<1-
	-_lj?.
L_.
\	I
	1
	I
	J
	1 11
	J
	f J
	jy
	Ix
0)	it
	1
	I- i 5
	L
X	"^
	\! 1
	\
	: V
	\
	: x^L :
.^	1
W-*	.-*
	±1^- -.
	\-
	il^-
	^^
	5i^
	n
	^
1 1	u
Li	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

1

-, s, curve.

V

The time for distance 8s = - 8s.

V

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

40

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

M,

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

PROJECTILES

41

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).

y

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

32

/. 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).

2

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- .

44

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

1^

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

60

r

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

47

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 ?

48

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.

N

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

>d

— 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

50

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

51

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	0-6	2-0	4-4	7-5	11-0	14-8	19-0
Time in minutes	0	1	2	3	4	5	6	7
Distance x 10"=^ in feet	23-0	27-8	32-5	37-6	43	48-4	54-2
Time in minutes	8	9	10	11	12	13	14

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

53

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	0	11-2	19 8	28	30	29-2	27	24-8	23-8	21-4	8-5	0
Time in seconds	0	10	20	40	50	60	90	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	0	1	2	3	4	5	6	7	8
Velocity in ft. /sec. Cutting stroke	0	14	21	24	26	27-5	29	29-5	30
Velocity in ft./sec. Keturn stroke	0	19	32	42	49	55	59	61	62
Fraction of stroke in iVth	9	10	11	12	13	14	15	16
Velocity in ft./sec. Cutting stroke	29-5	29	27-5	26	24	21	14	0
Velocity in ft./sec. Return stroke	61	59	55	49	42	32	19	0

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 ?

56

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".

A

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.

	— >^	1	>v	2
'-^	M		M
///////////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 — ^ ^.

t

.*. we may write.

/MVg— M-yA

^^ M^.^,-^ („),

t)

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.

9

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.

62

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

63

From the 2nd Law,

2-5 X 2240 x^-T

1

177

X 60 X 60 • 2-5 X 2240 X 10000

or

60 X 60

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

2-5 X 2240x29-2.,

— lbs. wt.

32

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.

64

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.

M

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

65

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

1

= ^ 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'

n

70

(1).

L. E. D.

6Q

ELEMENTARY DYNAMICS

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

V

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 '

dv

dt '

or if m is constant,

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

VARYING FORCE

67

Again we have

B(niv)

nearly,

I

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

69

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.

		>	(t
K	/ ^		1 1
Ayff			1 7 tons
/I ' IG	C	1 1
/ 1 / |N		1

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.

70

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.

o

/

t

/

/

1

f

/

/

/

/

/

/

/

>

/

y

,^

_

_

1 1

J,

15

20

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

71

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.
32							/
2 a 28 © o E c6 I |-24 4-1 i •" 22 9.0						/
				y	/
			y	/
		y	/
	^	^

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

15

16

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	1
per hour	per ton in lbs.	in lbs.	Accel, in ft./sec.^
9	22	9	7-78
10	22-8	8-2	8-54
11	23-75	7-25	9-66
12	25	6	11-7
13	26-5	4-5	15-5
13-5	27-5	3-5	20
14	28-5	2-5	28

The curve is shewn in fig. 56.

	J
	4l ±
	X
	i
	t
on - 1 -	A
	t
e	J
O J	r
tJ	I
* 1^
	/
«	7'
u	^^
in- - I.-****
	J __-_

10 n 12 13 14

Speed in miles per hour

Fig. 56.

15

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

1

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

75

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.

.^ '^

I

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

79

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.

80

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 = -

Q

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,

15

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

85

I

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.	0	5	7-5	10	20	30	40	50	60	70	80
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

87

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	27	£0	37 5	22 10	8 20	3 30	0 40
Time, millionths ' ^ of a second	1	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

91

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

92

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,

la

r

,(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 ==

2i

= 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

TT

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

94

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'

1

This gives,

Plot-^,i.e.— -=— - doi retardation

dt

and speed curve, using any suitable

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

""

■"

""

^

""

"~

■"

""

""

""

"■

■""

"*

^

""

\

\

\

K

\

V

A

\

c

\

V

d

>

I

■o

s,

s

d)

s

£.

t

^

V

%,

Sl

^s

^

**

*1

*■

■~"

_

J

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^

T

.*. 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.

A

.-. 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.

96

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

60

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

97

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.

60

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^
\		A
\	f^	^^		1 n
^^^^^^v'	Mg			1 y
< 2i,	r		-->

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

99

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

60

-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

M

3

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)
mP
-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^ = — .

M

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	0	15	30	45	60	75	90	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

/

I

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.

9.

The time for one revolution

CO

-v/l

From (a),

cos Q

CENTRIFUGAL FORCE

113

I

(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

mg

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==

32

6-26 lbs. wt.

CENTRE OF MASS

116

Revolving vertically for the sleeve we have 2Pcos30° = W^.

401 X n/3

W

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 ?

	x'	%m,x ~ M '
Ug	xy'	= ^fngy,
	y'	^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?

T

1-

i

4' 10'

T

mg

15

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 = —

15

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.

15

6-5 X 2240 X 18-2^ " 15

= 4*5 tons wt. Similarly R = mg pdls.

= 6*5 tons wt.

P

The coefficient of friction = -

R

_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

R

is the radius of curvature. Put on a centrifugal force , and

R

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).

R

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

or

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//////

G

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

01

I Ti

me

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 :