Mathematical SciencesFrom its pre-historic roots in simple counting to the algorithms powering modern desktop computers, from the genius of Archimedes to the genius of Einstein, advances in mathematical understanding and numerical techniques have been directly responsible for creating the modern world as we know it. This series will provide a library of the most influential publications and writers on mathematics in its broadest sense. As such, it will show not only the deep roots from which modern science and technology have grown, but also the astonishing breadth of application of mathematical techniques in the humanities and social sciences, and in everyday life.

DynamicsA.S. Ramsey (1867-1954) was a distinguished Cambridge mathematician and President of Magdalene College. He wrote several textbooks ‘for the use of higher divisions in schools and for first-year students at university’. This book on dynamics, published in 1929, was based upon his lectures to students of the mathematical tripos, and reflects the way in which this branch of mathematics had expanded in the first three decades of the twentieth century. It assumes some knowledge of elementary dynamics, and contains an extensive collection of examples for solution, taken from scholarship and examination papers of the period. The subjects covered include vectors, rectilinear motion, harmonic motion, motion under constraint, impulsive motion, moments of inertia and motion of a rigid body. Ramsey published a companion volume, Statics, in 1934.

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DynamicsA Text-Book for the use of the Higher Divi-

sions in Schools and for First Year Students at the Universities

Arthur Stanley R amsey

CAMbRID gE UnIvERSIt y PRESS

Cambridge new york Melbourne Madrid Cape town Singapore São Paolo Delhi

Published in the United States of America by Cambridge University Press, new york

www.cambridge.orgInformation on this title: www.cambridge.org/9781108003148

© in this compilation Cambridge University Press 2009

This edition first published 1929This digitally printed version 2009

ISbn 978-1-108-00314-8

This book reproduces the text of the original edition. The content and language reflect the beliefs, practices and terminology of their time, and have not been updated.

DYNAMICS

Cambridge University PressFetter Lane, London

New YorkBombay, Calcutta, Madras

TorontoMacmillan

TokyoMaruzen-Kabushiki-Kaisha

All rights reserved

DYNAMICSA Text-Book for the use of the

Higher Divisions in Schoolsand for

First Year Students at the Universities

byA. S. RAMSEY, M.A.President of Magdalene College,

Cambridge; and University Lecturerin Mathematics

CAMBRIDGEAT THE UNIVERSITY PRESS

1929

PRINTED IN GREAT BRITAIN

PREFACE

This book is intended primarily for the use of students in thehigher divisions in schools, particularly for those who intend totake an Honours Course of Mathematics at a University, andalso for University students preparing for a first HonoursExamination. It is based upon courses of lectures given duringmany years to first-year students preparing for the MathematicalTripos, and it is assumed that the majority of readers will alreadyhave acquired some knowledge of elementary dynamics. Althoughthe book contains chapters on Orbits and the dynamics of RigidBodies, none the less it may claim to be a text-book on ElementaryDynamics, for there is probably no branch of elementary Mathe-matics the content of which has expanded so greatly in the lasttwenty years.

One of the changes that accompanied the reform of the Mathe-matical Tripos was the removal of the restriction that ElementaryMechanics meant Mechan ics without the Calculus. This restrictionset well-defined and narrow bounds to the subject and the newregulations which gave teachers and students freedom to use anyanalytical methods in their work have been far reaching in theireffect. Though the schedule in Dynamics for Part I of thenew Tripos has remained unaltered, successive Examiners haveadded considerably to the interpretation of its contents. To giveone instance only—the phrase 'motion under gravity' is nowunderstood to mean 'in a resisting medium'—and it would beeasy to give other examples of the elasticity of interpretation towhich the schedule lends itself. The result of this change is thata first-year course in Dynamics at the University now' includesall the easier problems of two-dimensional dynamics stoppingshort of the use of moving axes and Lagrange's Equations. Thisgrowth in the content of Elementary Dynamics has been agradual process and undoubtedly beneficial to the study of thesubject and stimulating to the average student. It is inevitablethat its effect will extend to the schools, if it has not alreadydone so; and it is not unreasonable to suppose that before many

VI PREFACE

years have passed, candidates for Scholarships in Mathematicswill be expected to possess a wider knowledge of dynamics em-bracing such parts of the subject as 'motion under simple centralforces' and the elements of uniplanar rigid dynamics. The objectof this book is to assist in this development. It is hoped that thepresentation of the subject will prove sufficiently simple. Anattempt has been made to preserve the conciseness of lecturenotes and at the same time to give detailed explanations whereexperience has shewn that students find difficulties. Besidesexamples for solution the book contains a large number of workedexamples; some of these are of purpose very simple illustrationsof the theory, while others are of a more difficult kind for theassistance of readers who wish to learn how to work harder ex-amples. The examples are nearly all taken from Scholarshippapers or Tripos papers and the source is indicated by the lettersS. and M. T. No attempt is made to exhaust the subject andthe later chapters are only intended to be suggestive of the kindsof problems that can be solved, without elaborate analysis, asexamples of the fundamental theorems; some few of these mayprove to be too difficult for weaker students and they are intendedrather to introduce abler students to more advanced work.

In conclusion I desire to express my thanks to the printersand readers of the University Press for their excellent work inthe setting up of the book and the elimination of mistakes, andalso to say that if the book contains errors I shall be grateful toanyone who will point them out.

A. S. EAMSEY30 Nov. 1928CAMBRIDGE

CONTENTSChapter I : INTRODUCTION

ART. PAGE

l ' l . Kinematics and Kinetics 11*2. Frames of Eeference 11-3. Unit of Time 11-4. Rates of Change 11"5. Differential Equations 21-6. Equations of the First Order 21"7. Equations of the Second Order 3

Chapter I I : VECTORS

2*1. Vectors and Scalars 62'2. Composition of Vectors 72-3. Orthogonal Projections 72-4. Analytical Method of Composition 72'5. Multiplication and Division by Scalars . . . . . 82-6. Centroids or Mean Centres 82-7-2-9. Centroid Method of Compounding Vectors . . . . 9

Examples 10

Chapter I I I : RECTILINEAR MOTION. KINEMATICS

3"1. Velocity. Space-time curve 113'2. Acceleration. Velocity-time curve 113-3. Acceleration represented as a space rate of change . . . 1 23-31. Velocity-space curve 123-4. Units 133-41. Change of Units 133-42. Units in Graphical Work 143-5. Uniformly Accelerated Motion 143'51. Acceleration due to Gravity 163-6. Applications 163-7. Graphical Methods 183-8. Worked Examples 19

Examples 21

Chapter IV: RECTILINEAR MOTION. KINETICS

4'1. Newtonian Mechanics 244-11. Force. Newton's First Law 244-12. Mass 25

VU1 CONTENTS

ABT. PAGE

4-13. Material Particle 254'14. Momentum 254-15. Measurement of Force. Newton's Second Law . . . . 2 54-16. Force as a Vector 26417 . Weight 274-18. C.G.s. Uni t s 284-2. Impulse 284'21. Force-time Curve 294-3. Work 294-31. Foot-pound 304-32. Power. Horse-power. Erg 304*4. Energy, Kinetic and Potential 314"41. Formula for Kinetic Energy 314-42. Conservation of Energy 324-43. Force-space-Curve 324-44. Efficiency 334-45. Examples 334'5. Locomotive Engines and Motor Cars 344-54. Effectiveness of Brakes 364-55. Motion on an Inclined Plane . . . . . . . 384-6. Resistance Depending on Velocity . . . . . . 384-62. Fall of a Heavy Body in a Resisting Medium . . . . 4 04-7. Motion of a Chain 414-71. Fall of a Chain on to a Table 414-8. Uni ts and Dimensions 424-81. Change of Uni ts 42

Examples 43

Chapter V : K I N E M A T I C S I N T W O D I M E N S I O N S

5-1. Velocity 525*11. Acceleration 535-2. Relative Velocity 555-3. Angular Velocity 565-31. Examples 565-32. Motion in a Circle 575-33. Relative Angular Velocity 575-4. Centre of Rotation 585'41. Instantaneous Centre of Rotation 595-42. Examples 605-43. Pole Curves 605-5. Angular Velocity of a Body 625-51-5-54. Worked Examples 62

Examples 66

CONTENTS IX

Chapter VI : DYNAMICAL PROBLEMS IN TWO DIMENSIONS

ART. PAGE

6'1. Equivalence o f ' f o r c e ' a n d ' m a s s x acceleration' . . . 696-2. Motion of Projectiles 696-21. Range on an Inclined Plane 716'22. Geometrical Construction 726-3. Resisting Media 746-31. Example 756-32. Resistance x Square of Velocity 766-4. Principle of Work 77

Examples 78

Chapter V I I : HARMONIC MOTION

7 -l . Simple Harmonic Motion 837"11. Periodicity. Amplitude. Phase 847'2. Geometrical Representation 857'3. Elastic Strings. Hooke's Law 867"31. Work done in Stretching an Elastic String . . . . 877'4. Heavy Particle suspended by an Elastic String . . . 887'5. Applications 907'6. Simple Pendulum 937-61. Equivalent Simple Pendulum 937-62. Example of Finite Oscillations 947-7. Disturbed Simple Harmonic Motion 967 7 1 . Forced Oscillation 967-72. Example 977'8. Damped Harmonic Oscillations . . . . . . 987'9. Damped Forced Oscillations 99

Examples 100

Chapter V I I I : MOTION U N D E R CONSTRAINT

8-1. One-sided and two-sided Constraints 1058-2. Motion on a Smooth Curve 1058-3. Motion on a Smooth Circle 1068-31. Heavy Particle tied by a Fine String 1078-4. Cycloidal Motion 108

Examples 110

Chapter I X : T H E LAW OF REACTION. GENERAL P R I N C I P L E S

9-1. Newton's Third Law 1149'2. Motion of a System of Particles. Conservation of Momentum . 1149 -21. Effective Forces 116

X CONTENTS

ART. PAGE

9-3. Motion of the Centre of Gravity. Independence of Translationand Rotation 116

9"4. Conservation of Energy ; . 1189-5. Kinetic Energy in reference to Centre of Gravity . . . 1189-6. External and Internal Forces 1199-7. Rigid Bodies 1199"8. Applications. Car rounding a Curve : 119

Examples 121

Chapter X : GENERAL PROBLEMS

10-1. Pulleys 12310-2. Relative Motion 12310-3. Motion on a Wedge 12410-4. Examples of Conservation of Momentum and Energy . . 12510-5. Conical Pendulum 12610-6. Transmission of Energy by a Belt 12710-7. Further Problems on Strings and Chains . . . . 1 2 810-71. Belt Running at Uniform Speed 12910-8. Problems on Changing Mass 130

Examples 132

Chapter XI : IMPULSIVE MOTION

l r i . Impulse and Impulsive Force 13811 "2. Equations of Motion for Impulsive Forces . . . . 13811-3. Impact of Smooth Spheres 14011-31. Direct Impact 14011-32. Poisson's Hypothesis 14111-33. Oblique Impact 14211-34. Kinetic Energy lost by Impact 14211-35. Generalization of Newton's Rule 14311-4. Examples of Impulsive Motion 14411-5. Kinetic Energy created by Impulses 14611-6. Elasticity and Impulses 147

Examples 148

Chapter XII : POLAR COORDINATES. ORBITS

12-1. Velocity and Acceleration in Polar Coordinates . . . 1 5 512-2. Central Orbits 15612-21. To determine the Law of Force 15612-3. Circular Orbits 15812-4. Elliptic Orbit. Force directed to the Centre . . . . 1 5 912-41. Law of Force fir. Find the Orbit 16012-5. Elliptic Orbit. Force directed to Focus 161

CONTENTS XI

ART. PAGE

12-51. Parabolic Orbit. Force directed to Focus . . . . 1 6 212-52. Hyperbolic Orbit. Force directed to Focus . . . . 1 6 312-53. Law of Force ^/r2. Find the Orbit 16312-54. Velocity Components 16412-55. Velocity from Infinity 16512-56. The Hodograph 16612-6. Kepler's Laws of Planetary Motion 16712-61. Modification of Kepler's Third Law 16912-7. Use of u, 6 Formulae . . . . . . . . . 17112-71. Inverse Cube 17212-72. Apses and Apsidal Distances 17312-74. Einstein's Law of Gravitation 17512-75. Principles of Energy and Momentum applied to Central

Orbits 17712-8. Repulsive Forces 17812-9. Motion of Two Particles 179

Examples 181

Chapter X I I I : MOMENTS OF INERTIA

13-1. Moment of Inertia. Radius of Gyration . . . . 188132. Theorem of Parallel Axes 188133. Plane Lamina 189134. Reference Table. Routh's Rule 18913-5. Momental Ellipse 19213-51. Principal Axes 19313-6. Equimomental Bodies 194

Examples 194

Chapter X I V : MOTION OF A RIGID BODY. ENERGYAND MOMENTUM

14-1. Rigid Body. Fundamental Principles 19714-2. Kinetic Energy of a Rigid Body 19914-21. Examples of Conservation of Energy 20014-3. Momentum of a Rigid Body 20114'31. Examples of Conservation of Momentum . . . . 20214-4. Examples of Conservation of Energy and Momentum . . 204

Examples 206

Chapter X V : EQUATIONS OF MOTION OF A RIGID BODY

151. The Equations of Motion 20815-2. Applications of the Equations of Motion . . . . 20815-3. Equations of Impulsive Motion 213154. Examples of Impulses 214

xii CONTENTS

ABT.

15'5. Motion about a Fixed Axis . . . . . . . 21615'51. Compound Pendulum 2161552. Pressure on the Axis 21715-53. Examples . 2 1 815-54. Axis non-horizontal 21915-55. Centre of Percussion . . 22015-6. Examples of Motion about an Axis 22115-7. Moment of Momentum 22215-8. Moments about a Moving Axis 223

Examples 224

Chapter X V I : MISCELLANEOUS PROBLEMS

16-1. Rolling and Sliding 23116-2. Two Spheres in Contact 23316-3. Initial Motions and Stresses 23516'4. Bending Moments in Bodies in Motion 23616-5. Steady Motion in Three Dimensions 23816'6. Use of the Instantaneous Centre of Rotation . . . . 239

Examples 242

Chapter X V I I : SMALL OSCILLATIONS

17-1. Small Oscillations . 2 4 717'2. Application of the Principle of Energy 24717-3. Examples 24817'4. Use of the Instantaneous Centre of Rotation . . . . 25017'5. Oscillations of a Particle Constrained to move on a Revolving

Curve 25117-6. Stability of Steady Motion 25317'7. Oscillations about Steady Motion 25417'8. Example 25417-9. Steady Motion of a Particle on a Sphere . . . . 256

Examples 257

DYNAMICS

Chapter I

INTRODUCTION

1-1. The subject of Dynamics is generally divided into twobranches: the first, called Kinematics, is concerned with thegeometry of motion apart from all considerations of force, massor energy; the second, called Kinetics, is concerned with theeffects of forces on the motion of bodies.

12. In order to describe the motion of a body or of a pointtwo things are needed, (i) a frame of reference, (ii) a time-keeper.It is not possible to describe absolute motion, but only motionrelative to surrounding objects; and a suitable frame of referencedepends on the kind of motion that it is desired to describe.Thus if the motion is rectilinear the distance from a fixed pointon the line is a sufficient description of the position of themoving point; and in more general cases systems of two or ofthree rectangular axes may be chosen as a frame of reference.For example, in the case of a body projected from the surface ofthe Earth a set of axes with the origin at the point of projectionwould be suitable for the description of motion relative to theEarth. But, for the description of the motion of the planets, itwould be more convenient to take a frame of axes with an originat the Sun's centre.

1"3. It is important to realize that there is no such thing asabsolute time, but the period of rotation of the Earth relativeto the fixed stars provides a unit of time, the sidereal day, which,so far as it can be tested with other time measures, is constantand therefore adequate for the purposes of ordinary dynamics.

14. The functions involved in dynamical problems are for themost part differential coefficients with regard to 'time,' 't,' asthe independent variable. Thus 'motion' is 'change of position'or 'displacement,' 'velocity' is 'rate of displacement' and

2 INTRODUCTION [i

'acceleration' is 'rate of change of velocity.' Hence, if $ denotesa distance, dsc/dt denotes a velocity and d?a>/dt2 denotes anacceleration. The formulation of a dynamical problem thereforein general consists of one or more relations between certainvariables (coordinates of position) and their differential co-efficients with regard to time. Such relations are calleddifferential equations.

NOTE ON DIFFERENTIAL EQUATIONS

1*5. It is assumed that the reader is acquainted with the elementaryprocesses of differentiation and integration.

A differential equation is a relation between an independent variable t,a dependent variable x, and one or more of the differential coefficients of xwith regard to t. The order of a differential equation is that of the highestdifferential coefficient that it contains. A solution of a differential equationis a relation between x and t that satisfies the equation, and the completesolution of a differential equation is a relation between x, t and one ormore arbitrary constants of integration, the number of such constantsbeing equal to the order of the equation.

For example:

(i) g-8,-0

is a differential equation of the first order. It will be found on substitutionthat x=e<u is a solution; and the complete solution is x=Cen, where C isan arbitrary constant.

d2r() + 0

is a differential equation of the second order. It has solutions

x = aint and * = cos t,

and the complete solution is

x = A sin t + B cos t,

where A and B are arbitrary constants.

1*6. The differential equations of dynamics are of either the first orsecond order.

Equations of the First Order.We may have to deal with equations in which the variables can be

separated. Such equations can be put in the form

Mdxjdt=N (1),

1*4—1-7] DIFFERENTIAL EQUATIONS 3

where M is a function of x only (or a constant) and N is a function of tonly (or a constant). The complete solution is

\Mdx=\Ndt+C (2),where C is an arbitrary constant.

For example, the equationdx _ , 2

is solved by writingxdx ,

g-^&=9dt'so that

~2kis the complete solution.

1*61. Another type of equation that sometimes occurs in dynamics isthe linear equation of the first order. A differential equation is said to belinear when it does not contain powers or products of the dependentvariable x and its differential coefficients. Thus the linear equation of thefirst order is

where M, N are functions of t or constants.The solution is effected by first multiplying both sides of the equation

by e l and then integrating; because it can easily be verified that

Hence xJM<u=\JMtUNdt+C (4),where C is an arbitrary constant.

We note that if M is a constant the solution is

xem=\emNdt + C (5).

For example, the equationdx ,Tt+kx=gt

can be integrated if both sides are multiplied by ew, giving on integration

•-!(<-»(6).

1*7. Equations of the Second Order.A common type of differential equation of the second order is

where a and b are constants.

4 INTRODUCTION [i

It is easily seen by substitution that x=emt is a solution of this equation,provided that

m? + 2am + b = 0 (8),so that, if m-i, m2 are the roots of this quadratic, the complete solutionof (7) is

x=C1emit+C2e

nhfi (9),where Ci, (72 are two arbitrary constants.

The roots of (8) are —a + sj{cfi — b), and there are three cases to be con-sidered :

(i) Real roots. ai-b=ni, say; then (9) may be written

A-=e-«(C1e"!+C2e-ll() (10).

(ii) Equal roots. a2-b — 0. The form e~at(C1 + C2) is inadequate for acomplete solution, since C1 + C2 can only be regarded as one arbitrary con-stant, and, since the differential equation is of the second order, the completesolution should contain two. It is easily verified however that, whena? = b, the form

z = e-"t(C1 + C2t) (11)

satisfies equation (7), and since it contains two arbitrary constants, it isthe complete solution.

(iii) Imaginary roots. a2 — b2— -n2, say; (9) may now be written

or x=e-at{(Ci + Cz)cossnt+iid- C2)ainnt};

which again may be written in the more convenient form

x-e-at(Gcosnt+Csinnt) (12).

Special cases of the foregoing. When a = 0.

(a) The complete solution of

(firn 2 x 0 (13>

is x = Aei»*+£e*t \

or X = C cosh nt+Daxahnt> '

where A, B or G, D are the arbitrary constants.

(/3) The complete solution of

is x=Acosnt+Bsmnt\

or x = C cos (nt + a) I (16),

or x = C sin (nt + a!) J

where A, £ or C, a or C", o' are the arbitrary constants

1-7-1-71] DIFFERENTIAL EQUATIONS

1*71. Numerical Examples.

(i) ; p - 4 ^

(ii) J - 4 j + 4*=0; x

(iii) ^ - 4 ^ + 13^=0; x=e2t{A cosZt + BsmZt).

(iv) ^ - 4 a ; = 0; «(Ait

(v) ^2+4^=0; x= A

Chapter IIVECTORS

21. The physical quantities or measurable objects of reason-ing in Applied Mathematics are of two classes. The one class,called Vectors, consists of all measurable objects of reasoningwhich possess directional properties, such as displacement, velocity,acceleration,movientum, force, etc. The other class, called Scalars,comprises measurable objects of reasoning which possess no direc-tional properties, such as mass, work, energy, temperature, etc.

The simplest conception of a vector is associated with thedisplacement of a point. Thus the displacement of a point fromA to B may be represented by the line AB, where the length,direction and sense {AB not BA) are all taken into account.Such a displacement is called a vector (Latin veho, I carry).A vector may be denoted by a single letter, e.g. as when wespeak of 'the force P,' or 'the acceleration /,' or by namingthe line, such as AB, which represents the vector. When it isdesired to indicate that symbols denote vectors it is usual toprint them in Clarendon type, e.g. P, and to write them with a barabove the symbol, e.g. P, AB.

Since the displacement from B to A is the opposite of a dis-placement from A to B, we write

BA=-ABand take vectors in opposite senses to have opposite signs. Sincetwo successive displacements of a point from A to B and fromB to G produce the same re-sult as a single displacementfrom A to C, we say that thevector A G is equal to the sumof the vectors AB, BG andwrite

(1),

and further, if A, B, C... K, L Aare any set of points

~ = AB + B~C+...+KL (2).

2-1-2-4] COMPOSITION OF VECTORS 7

Vectors in general are not localized; thus we may have adisplacement of an assigned length in an assigned direction andsense but its locality not specified. In such a case all equal andparallel lines in the same sense will represent the same vector.On the other hand, vectors may be localized, either at a point,e.g. the velocity of a particle; or in a line, as for example a forcewhose line of action (but not point of application) is specified.

2-2. Composition of Vectors. A single vector which isequivalent to two or more vectors is called their resultant, andthey are called the components of the resultant. Vectors arecompounded by geometrical addition as indicated in formulae(1) and (2) of the last Article.

A vector can be resolved into two components in assigneddirections in the same plane; for if AG be the vector, andthrough A, G two lines are drawn in the assigned directionsmeetings in B, then AB, BC are the components required.

When a vector is resolved into two components in directionsat right angles to one another, eachcomponent is called the resolvedpart of the vector in the directionspecified. Thus if a vector P makesan angle o with a given directionOx, the resolved parts of P in thedirection Ox and in the perpen-dicular direction Oy are

P cos a and P sin a.

23. Since the algebraical sum of the orthogonal projectionson any straight line of the sides of a closed polygon is zero, itfollows that the orthogonal projection of the resultant of anumber of vectors is equal to the algebraical sum of the projec-tions of the component vectors.

24. Analytical Method. To compound n vectors Pi, P2... Pn.Let the vectors make angles oti, a%... an with an axis Ox. Eachvector may be resolved into two components, one in the directionOx and the other in the perpendicular direction Oy. The com-ponents in direction Ox are equivalent to a single vector

X = Pi cos cti + P2 cos «2 + ... + Pn cos an = S (P cos a),

X

8 VECTORS [II

and the components in direction Oy are equivalent to a singlevector

Y = Pi sin <*! + P2 sin a2 + ... + P« sin an = 2 (P sin a).The two vectors X, Y can now be compounded into a single

vector R making an angle 6 with Ox, such thatR cos 6 = X and R sin 0 = Y,

and therefore Ri = X2+ Fa and tan 6 = Y/X (1).

25. Vectors may be multiplied and divided by scalar numbers.Thus if we take n equal vectors AB and compound them togetherwe get a vector AC, such that AC = nAB; and, conversely,

AB = -AC.n

Note that relations of the form AC = nAB, ovpAB + qAC = 0imply that the points A, B, C are in the same straight line.

26. Centroids or Mean Centres. If mx, m2, m3... mn be aset of scalar magnitudes associated with a set of points Aly A%,As... An, the centroid or mean centre of the points for the givenmagnitudes is the point obtained by the following process: Dividethe line A^A^aX £j so that m1A1B1 = m1,B1A2; divide BjA3 ati?2 so that (m1 + m2) B-,B2— m3B2A3; divide B2Ai at B3 so that(mi + mi + ma)B2Bs = miBsAi. Proceed in this way until allthe points have been connected then the last point of divisionBn_i, usually denoted by the letter G, is called the centroid ormean centre.

2-61. In order to shew that this process leads in general to aunique point, i.e. that the pointdetermined by the process isindependent of the order inwhich the points Alt A^... An

are joined, we shall first provethat

Assume that this formula is true for the first r points, i.e. that-i + ... + mr ArBr_1 — 0.

2'4-2"7] CENTROIDS 9

Now the next step in the process is to divide Br-.xAr+x a t Br

so thatm2+ ... + <mr) Br_x Br = mr+1BrAr+1,

therefore by adding the last two lines

mx AxBr + m2 A2Br + ... +mr ArBr + mr+1Ar+1Br = 0.

It follows that if the formula (1) is true for r points it is alsotrue for r + 1; but it is true for two points, since, by hypothesis,mx AXBX + m2 A2Bi = 0. Therefore the formula (1) is true forany number of points.

Now if by taking the points in a different order we arrive ata centroid G' we can shew similarly that

... + mnAnG'= 0 (2);

and by subtracting (1) from (2) we get

m2+ ...+ mn) GG' = 0.

Hence G' must coincide with G unless mx + m2 + ... + mn = 0.In the latter case there is no centroid at a finite distance, be-cause the last step in the process of finding the centroid consistsin dividing a line in the ratio mx + m2 + ... +mn_1:mn, i.e. inthe ratio 1 : — 1.

27. Centroid Method of Compounding Vectors. To shew,with the notation of the last Article, that, if 0 be any other

point, the resultant of n vectors m^OA-i, m2OA2... mn0An is(m1 + m2+ ... + mn) 0G, where G is the centroid of the pointsAlt A2 ... An for the magnitudes m1; m2 ... mn.

This follows at once by substituting

OAX = 0G + GAX, 0A2 = 0G + GA2, etc.,so that

+ ... + mn0An

... +mn) OG

and by 2-61 (1) the sum of the terms in the last bracket is zero,therefore

mx0Ax +m20A2+ ... + mn0An = (mx + m2 + ... + mn) 0G.

10 VECTORS [II

2'8. When reference is made to the centroid of a set ofpoints without mention of any associated magnitudes it isunderstood that the magnitudes are equal; thus the centroidof a triangle ABC is a point 0 such that

A~G + BG + CG = 0.

29. It may be noticed that if P, Q, R are vectors in thelines OA, OB, OC then the resultant vector is

where G is the centroid of the points A, B,G for the magnitudep

P/OA, Q/OB, EIOC; for a vector P is the same as ^-r OA.

EXAMPLES1. Prove that, if O is the middle point of AB and O' is the middle

point of A'B', then AA'+ BE'= 2GG'.

2. Prove that, if O is the centroid of n points At, A2... An, and G' isthe centroid of n points Blt B2...Bn, then

3. Prove that, if H is the orthocentre and 0 is the ciroumcentre of atriangle ABC, then

AMta,n A +~BH tan B + OH tan G=0,

and AOsin f2,A + Wsin^B + COsin 2C=0.

4. Shewthat, if mOA+nOB~+pOC=0 a,ndm + n+p = 0, then thepointsA, B, C are collinear.

Chapter III

RECTILINEAR MOTION. KINEMATICS

31. Consider the motion of a point along a straight lineor axis Ox on which 0is a fixed point. Let 5 P F ~xP, P' denote the posi-tions of the moving point at times t, If; and let OP = x, OP' = x'.The displacement of the point in time t' — t is x' — x, and(x' — #)/(£' — t) is the average rate of displacement or the averagevelocity during the interval t' — t. If this ratio be independentof the interval t' — t; i.e. if it has the same value for all intervalsof time, then the velocity is constant or uniform, and equaldistances will be traversed in equal times.

Whether the ratio (x —x)/(t' — t) be constant or not, itslimiting value as f tends to t is defined to be the measure ofthe velocity of the moving point at time t. But this limitingvalue is the differential coefficient of x with regard to t, so thatif we denote the velocity by v, we have

v = dx/dt.If on squared paper we plot a curve in which abscissae

represent time and ordinates re-present distances traversed, thecurve is called the space-timecurve. The curve gives a graphicalrepresentation of the motion,because it exhibits graphicallythe relation between the timeand the distance traversed inthat time. Also the gradient of the curve, i.e. the tangent ofthe angle that the tangent to the curve makes with the time-axis, gives the value of the velocity dx/dt.

3"2. Acceleration is similarly defined as the rate of changeof velocity. Thus, if v, v denote the velocities of the movingpoint at times t, t', then 1/ —v is the change of velocity in time

x

12 RECTILINEAR MOTION. KINEMATICS [ill

t' — t, and (v' — v)j(t' — t) is the average rate of change of velocityduring the interval t' — t. If this ratio is independent of theinterval t' — t, then the acceleration is constant or uniform, orequal increments of velocity take place in equal intervals.

Whether the ratio (V — v)/(t' — t) be constant or not, itslimiting value as t' tends to t is defined to be the measure ofthe acceleration of the moving point at time t. But thislimiting value is the differential coefficient of v with regardto t, so that if we denote the acceleration by / , we have

Following Newton, it is usual to denote differential co-efficients with regard to time by dots; thus x means dx/dt, andx means d2x/dt\

If on squared paper we plot a curve in which abscissaerepresent time and ordinates represent velocity, the curve iscalled the velocity-time curve. The gradient of the curve givesthe acceleration dv/dt at any instant.

Also the area under the velocity-time curve

= I vdt = I -jr dt = I dx = [x]

taken between proper limits= the distance covered in the corresponding time.

3 3. Acceleration represented as a space rate of change.Since v = dxjdt and / = dv/dt,,, „ . dv dx dvtherefore / = -^—T- = v -y-.

J dx dt dxThis formula for acceleration is very important, as it has to beused in all problems in which the velocity is given in differentpositions rather than at different times.

3"31. If on squared paper a curve be plotted in which abscissaedenote spaces traversed and or-dinates represent velocities, thecurve is called a velocity-spacecurve. The gradient of the curveis dv/dx, and in the figure, inwhich PN is ordinate v and POis normal at P, we have

t&n NPQ = dv/dx,

3-2-3-41] UNITS 13

so that t h e subnormal NG = vdv/dx= the acceleration.

Note that if the acceleration is constant the velocity-spacecurve is a parabola, since this is the only curve for which thesubnormal is constant.

34. Units. Measurements of physical quantities must neces-sarily be expressed as multiples or submultiples of certain units.The choice of the units of length and time is arbitrary; thus theunit of length may be a foot or a centimetre, and the unit oftime may be the mean solar second or it may be the siderealday (1'3), the former being the more convenient unit for purposesoutside an observatory as it is mean solar time that is recordedby ordinary chronometers, watches and clocks.

The units of velocity and acceleration are derived units in thatthey depend on the units of length and time, which may be re-garded as fundamental units. The unit of velocity is a unit oflength described in a unit of time. The unit of acceleration is aunit of velocity added in a unit of time. The unit of velocitytherefore varies directly as the unit of length and inversely asthe unit of time, and the unit of acceleration therefore variesdirectly as the unit of length and inversely as the square of theunit of time. These facts may also be expressed by saying thatthe unit of velocity is of" one dimension in length and minus onedimension in time, and that the unit of acceleration is of onedimension in length and minus two dimensions in time; or, ifL, T denote the units of length and time, the unit of velocity isLT"1 and the unit of acceleration is LT~2.

The measure of any given quantity varies inversely as theunit chosen; e.g. a velocity measured in yards per second is one-third of the same velocity measured in feet per second.

341. Change of Units. If v denote the measure of a velocityand f the measure of an acceleration when L, T are the unitsof length and time, and v', f denote the measures of the samevelocity and the same acceleration when L', T' are the unitsof length and time, then it is clear that

and

14 RECTILINEAR MOTION. KINEMATICS [ill

since the expressions equated are equivalent representations ofthe same velocity and of the same acceleration.

For example, if L = 1 foot, and T = 1 secondand L' = 1 mile and T' = 1 hour,then L' = 5280L and T' = 3600T,and, if v' is a velocity of 60 miles per hour,

v = 60x5280-f- 3600= 88 ft. per sec.

342. Units in Graphical Work. In using the graphicalmethods indicated in 31, 32 and 33 attention must be paid toscales of measurement.

Space-time curve.If one inch in the abscissa t represents a seconds and one inch

in the ordinate * represents b feet, then the measure of the

velocity is - --=- ft. per sec. where dxjdt is the actual gradient in

the diagram.

Velocity-time curve.If one inch in the abscissa t represents a seconds and one inch

in the ordinate v represents a velocity of b ft. per sec, then the

measure of the acceleration is - -rr ft. per sec. per sec. wherea at r r

dvjdt is the actual gradient in the diagram; and the area underthe curve represents distance covered on the scale

one square inch = ab feet.Velocity-space curve.If one inch in the abscissa x represents a feet and one inch in

the ordinate v represents a velocity of b ft. per sec, then the

measure of the acceleration is r~ where vdvldx is the actuala dx '

subnormal in inches; i.e. the subnormal gives the accelerationon the scale

1 inch = b2/a ft. per sec. per sec.35. Uniformly Accelerated Motion. In uniformly accele-

rated motion the acceleration / is constant, so that by integratingthe relation dv/dt=fwe get v =ft + G,

S'4il~S-51] UNIFORM ACCELERATION

where G is constant, and if the velocity at time t = 0 be u (calledthe initial velocity), we must have G — u, and therefore

v = u+ft (1);though it must be remarked that this relation is otherwiseobvious, because, when the acceleration f is constant, thevelocity is increased by f in each unit of time.

Again, since dx/dt — v = u +ft,therefore by integration

x = ut + ^ft2 + C,where C is constant, and if the origin from which x is measuredis taken at the position of the moving point when t = 0, we havex = 0 when t = 0, so that C = 0 and

as = ut + $ft* (2).

If we now eliminate t between (1) and (2) we getV — U + LJX \ii).

This last result can also be obtained by integrating thealternative expression for acceleration, vdvjdx =f, which, whenf is constant, gives 2 _ . „„

where G" is constant; and since v = u when x = 0, thereforeG" = u2 and „ „ OJ,

The student will do well to note that the formulae of thisarticle are only true when the acceleration is constant.

3'51. Formulae (1) and (2) of the last Article follow graphically fromthe velocity-time curve. Since the gradient of the curve represents theacceleration and this is constant, there-fore the curve in this case is a straightline inclined at an angle tan"1/ to thet axis.

Let PQ be this line and Q the point(t, v), P the point (0, u), where 0M=t,MQ=v, 0P=u. Then

(MQ - O/')/Oi/=gradient =/, °therefore v = u+ft.

Again space described in time < = area under curve= 0MQP

M

16 RECTILINEAR MOTION. KINEMATICS [ill,

351. Acceleration due to Gravity. It is an experimentalfact, first observed by Galileo, that all bodies which are allowedto fall freely at a given place possess the same constant accele-ration, usually denoted by g.

The value of g varies slightly, increasing with the latitudeof the place; but the difference between the values at theequator and the poles amounts only to about one-half per cent.The value in the latitude of London is approximately 981 cm.per sec. per sec. or 32-2 ft. per sec. per sec.

3'6. Applications. The law of acceleration in a particular problemmay be given by expressing the acceleration as a function of (i) the time t,or (ii) the distance x, or (iii) the velocity v. The problem of furtherinvestigating the motion can then be solved as follows :

(i) Acceleration a given function of the time, say <j>(t).

We have v — cj>(t)

therefore, by integration, v=$<j>(t)dt + C,

or x = f(t) + C, say,

where i (<) is the integral of <f>(t). Then another integration gives

where the constants C, C" can be determined if the velocity and position ata given time are known.

(ii) Acceleration a given function of the distance, say <f>(x).

In this case, there are two ways of proceeding:

(o) "We use the form vdvjdx for acceleration and write

so that, by integration, v2=2$(j>(x)dx + C.

Then, if we put TJr(x) for %j<f>(x)dx and x for v,

we have x = ± {^(x)+C}*

and therefore

(/3) Alternatively,' we write

x = (f,(x)

and multiply both sides of the equation by 2ir, so that

2xx=2<t>(x)x.

Then integrate with regard to t and we get

x2 = 2$<t>(x)dx + C,.

and from thence we proceed as in (a) above.

3-51-3-6] VARIABLE ACCELERATION 17

This method of procedure, in which, we multiply both sides of the equationby 2x in order to make it integrable, is of frequent application and deservesspecial notice.

As an example let the acceleration be — n2x where, as usual, x denotesdistance from a fixed origin and suppose that at time £ = 0 we have x=aand £ = 0; i.e. a point starts from rest at a distance a from the originwith acceleration n2x towards the origin. We put

x = -n2x,then multiply by 2x and we get

2xx= -which gives on integration

x*=-But x = 0 when x=a, therefore C=n2a?, and

x2=n2(a2-x2),dx ,

therefore cos ~x - = nt + C,a '

and, since x = a when t=0, therefore <7'=0, andx=aoosnt.

Note that the choice of the minus sign on taking the square root isdetermined by the fact that the point starts from rest with an accelerationdirected towards the origin, so that x decreases as t increases, or dx anddt have opposite signs.

In this particular example we might also obtain the same solution byproceeding as in 1*7 (16), where the solution is obtained in the form

If we proceed in this way, we have to determine the constants C and afrom the conditions that When i = 0

x=a or Ccosa = ffl,and x = 0 or - « C s i n a = 0.

These give a = 0 and C=a, making x=a cos nt as before.

(iii) Acceleration a given function of velocity, say <j>(v).

In this case we may either connect velocity with time by writing

and integrating in the form

--t+C;

or we may connect velocity with distance by writing

and integrating in the form'vdv

18 RECTILINEAR MOTION. KINEMATICS [ill

Whether it is then possible to proceed to a further integration afterputting x for v will depend on the form of the functions.

Example. A particle moves in a straight line under a retardation kvm + 1,where v is the velocity at time t. Shew that, if u is the velocity when t=Q,then

TO \Vm Um

and find a corresponding formula for space in terms of v.

We have v=-kvm + 1;

therefore

or — C=kt,mvm

but v = u when t=0, therefore C=l/mum

and fc-

Again v^-= -kvm

axtherefore \~fih^'^—~ {

; — C= kx,

and, if x be measured from the position in which v = u, we have x=0when v = u, therefore C—ll{m-l)um~l and

31. Graphical Methods. If a table of corresponding valuesof any two of the four magnitudes x, t, v, f is given, it is ingeneral possible to determine the other two by plotting suitablegraphs, based upon the relations v = dx/dt, f= dv/dt = vdv/dx.

For example, suppose a table of corresponding values of v and/ t o be given. Since/= dv/dt,

., c ,, dvtherefore 0,1 — -^, or t

[dv

Hence if we plot a curve in which the abscissae denote v andthe ordinates denote 1//, then the area under the curve up toany point will represent the time t.

The corresponding values of the distance x can then be foundfrom vdv/dx = / , which gives

and therefore x = -?dx = —j- and therefore x = -? dv.

3-6-3-81] GRAPHICAL METHODS 19

So that if we plot a curve in which the abscissae denote v andthe ordinates denote vjf, then the area under the curve up toany point will represent the distance x.

Alternatively, so can be found by using the formula v = dxjdt,which gives x=Jvdt, if we first tabulate the values of t foundabove and then plot a velocity-time curve. The area under thecurve up to any point will represent the distance x.

3*8. E x a m p l e s , (i) A train passes a station A at 40 miles per hourand maintains this speed for 7 miles, and is then uniformly retarded,stopping at B which is 8 miles from A. A second train starts from A theinstant the first train passes and being uniformly accelerated for part ofthe journey and uniformly retarded for the rest stops at B at the same timeas the first train. What is its greatest speed on the journey ? [S. 1910]

This example illustrates the utility of the velocity-time curve. Theunits chosen are a mileper hour for v, and an hourfor t. For the first trainthe velocity-time curve con-sists of the straight line PQor v = 40, and the straightline QR along which thevelocity is uniformly re-tarded to zero.

OM, MR are the timesrequired for the first sevenmiles and the eighth mile respectively; and the areas under the curve,i.e. OMQP and MRQ represent 7 miles and 1 mile respectively.

Therefore 4 0 0 ^ = 7 and 201ffl = l.Again, for the second train which is uniformly accelerated and then

uniformly retarded, starting from rest and ending at rest and covering thesame distance in the same time, the velocity-time curve consists of twostraight lines OS, SR with the condition that the area OSR represents8 miles. Hence the greatest speed SN is given by

%8N. 0R = 8 ; but 0R = ^,therefore S1T=7\$ miles per hour.

3*81. In the graphical solution of examples it is sometimes useful toremember that if the acceleration increase or decrease steadily with thetime then the velocity-time curve is a parabola. For if

i>=at + b,where a and b are constants, then

which represents a parabola if v, t are used as coordinates.

20 RECTILINEAR MOTION. KINEMATICS [III

(ii) A train starts from a station A with acceleration "8 ft.-secr2, and thisacceleration decreases uniformly for 2 minutes, at the end of which the trainhas acquired full speed which is maintained for another two minutes. Thenbrakes are applied and produce a constant retardation 3'5 ft.-secr^ andbring the train to rest at B. Draw an acceleration-time curve and a velocity-time curve. Find the maximum velocity and the distance AB.

v\

Acceleration-time curve

60 120 18Q 240 sees.

Velocity-time cufve

The acceleration-time curve consists of portions of three straight linesas shewn in the first diagram.

The velocity at any time during the first 2 minutes is measured by thecorresponding area under the acceleration-time curve :e.g. velocity at one minute = area of trapezium of base 60 and mean height

•6=36 f.s.and velocity at two minutes = area of triangle of base 120 and height

•8 = 48 f.s.In this way any number of points on the velocity-time curve during the

first two minutes may be plotted. At two minutes the velocity-time curvebecomes a straight line parallel to the time axis. At four minutes there isa uniform retardation 3-5 f.s.s which will destroy the velocity of 48 f.s. in48/3-5 sees, or 13"7 sees., so that the last part of the velocity-time curve isanother straight line meeting the time axis at 2 = 253'7 sees.

The total distance travelled is given by the area under the velocity-timecurve. The area under the curved portion can be estimated by countingsquares or by dividing it into parallel strips and using mean ordinates;or, more accurately from the fact that the curve is a parabola with axisvertical and vertex at (120, 48), so that the area is § of the rectangle whosesides are 120 and 48 = 3840. The remaining area is

a rectangle 48 x 120 = 5760,

and a triangle \ x 48 x 13-7 = 329,

making the total distance traversed 9929 ft.

3-81] EXAMPLES 21

This problem may also be solved by direct analysis. Thus, since theacceleration falls steadily from 8 f.s.s to zero in 120 sees, therefore

acceleration after t sees. = '8

or *=-8fl-I

(1 — r-g- ) ,\ 120/

( t2 \« -—- j ft.-sec."1,

and - - 8 g -no constants of integration being required since x = 0 and # = 0 when t — 0.

It follows that when « = 120 we have a; = 48 ft./sec. and # = 3840 and therest of the solution is as above.

EXAMPLES

Uniform Veloc i ty1. Two trains take 3 seconds to pass one another when going in

opposite directions, but only 2'5 seconds if the speed of one is increasedby 50 per cent. How long would one take to pass the other when going inthe same direction at their original speeds ?

2. A steamer takes m minutes to go a mile downstream and n minutesto go a mile upstream. Find the speed of the current and of the steamerrelative to it.

3. A line of men are running along a road at 8 miles an hour behindone another at equal intervals of 20 yards. A line of cyclists are riding inthe same direction at 15 miles an hour at equal intervals of 30 yards. Atwhat speed must an observer travel along the road so that whenever hemeets a runner he also meets a cyclist 1

Uniform Acceleration4. A point moving with uniform acceleration describes distances

*i, «2 fee* i n successive intervals of tt, t2 seconds. Prove that theacceleration is

5. A point moves with uniform acceleration and vly v2, v3 denote theaverage velocities in three successive intervals of time tlt t2, t3. Provethat

6. A heavy particle is projected vertically upwards. Shew that if il5 t2

are the times at which it passes a point at a height h above the point ofprojection in ascending and descending, then t1t2=2h/g.

7. A point moves from rest with uniform acceleration. Shew that inany interval the space-average of the velocity is 4/3 of the time-average.

22 RECTILINEAR MOTION. KINEMATICS [III

8. An express train is sent off in two parts at an interval of 5 minutes.If both parts are uniformly accelerated and attain their maximum speed of60 miles per hour in a mile, prove that the first part has gone 4 milesbefore the second starts and that at full speed they run 6 miles apart.

9. A body moving in a straight line travels distances AB, BC, CD of153 feet, 320 feet, 135 feet respectively in three successive intervals of3 sees., 8 sees., and 5 sees. Shew that these facts are consistent with thehypothesis that the body is subjected to uniform retardation. On thishypothesis find the distance from D to the point where the velocityvanishes, and the time occupied in describing this distance. [M. T. 1915]

10. A train which starts from rest and accelerates in 1 mile to a fullspeed of 40 m.p.h. and stops in \ a mile, has, in the course of a journey,to be slowed down to 20 m.p.h. for a distance of 2 miles. Shew that inconsequence it arrives at its destination 3 ^ mins. late, assuming thatacceleration is always at the same uniform rate and that retardation isalso always at the same uniform rate.

11. Two stopping points of an electric tramcar are 440 yards apart.The maximum speed of the car is 20 miles per hour and it covers thedistance between stops in 75 seconds. If both acceleration and retardationare uniform and the latter is twice as great as the former, find the valueof each of them, and also how far the car runs at its maximum speed.

[S. 1924]Variable Accelerat ion

12. A body moving in a straight line describes the following distancesin the given times :

Time in seconds

Distance in feet

0

5

5

13-75

10

60

15

176-25

20

400

25

768-75

30

1320

Deduce approximately the velocity-time and acceleration-time curvesfor the same period. [S. 1911]

13. The relation between acceleration and time for a car starting fromrest is given by the table

Time in seconds

Accel, in ft./sec.2

0

•5

10

1

20

1-2

30

•5

40

- • 3

50

- • 4 5

60

- •45

70

- • 3

Draw the acceleration-time curve and deduce the velocity-time anddistance-time curves.

EXAMPLES 23

14. The relation between acceleration and distance for a car startingfrom rest is given by the table

Distance in feet

Accel, in ft./sec.2

0

2

100

1-7

200

1-4

400

•85

600

•2

800

- • 2

1000

0

Draw the acceleration-distance curve and deduce the velocity-distancecurve.

15. If the relation between x and t is of the form t=ax2 + Px, find thevelocity v as a function of x, and prove that the retardation of the particleis 2av3.

The following observations were made on a rifle-bullet:

X

t

0

0

150

•0(398

300

•1408

ft.

sec.

Assuming that the relation between x, t is that stated above, find a, (3and shew that the velocity at the instant of the second observation was2131 ft. per sec. approximately. [S. 1912]

16. By proper choice of units the curve on a time base representing theacceleration of an electric train is a quadrant of a circle, whose centre isthe origin. The initial acceleration is 2-5 ft. per sec. per sec, and theacceleration falls to zero in 20 seconds. Calculate the velocity acquiredand the distance described in that time. [S. 1917]

17. A particle moving in a straight line is subject to a resistance whichproduces the retardation kv3, where v is the velocity and h is a constant.Shew that v and t (the time) are given in terms of s (the distance) by theequations v = uj{\+ksu), t = iks2+sju,

where u is the initial velocity.As a result of certain experiments with a rifle, it was estimated that the

bullet left the muzzle with a velocity of 2400 ft. per sec. and that thevelocity was reduced to 2350 ft. per sec. when 100 yards had beentraversed. Assuming that the air-resistance varied as v3, and neglectinggravity, calculate the time of traversing 1000 yds. [M. T. 1913]

A N S W E R S

2. 30 (n—m)/mn, 30 (n + m)/mn miles per hour.1. 15 sees.3. 6 m.p.h. 9. 121 ft., 11 sees.16. 39-27 f.s.; 452 ft. 17. 1-38 sees.

ft 4 4 0 f t -

Chapter IV

RECTILINEAR MOTION. KINETICS

4"1. In the last chapter we discussed the measurement of thevelocity and acceleration of a point moving in a straight line.We must now begin to study the branch of Dynamics calledKinetics, which is concerned with the effects of forces on themotion of bodies.

The definitions and laws of motion enunciated by Newton in theseventeenth century form the foundation of a science of dynamicswhich has been developed by many other mathematicians. Thisdynamics, now commonly called 'Newtonian Mechanics,' isthe basis of all the theoretical work in applied mechanics orengineering and the results of the theory have been and arestill being confirmed every day by numerous appeals to experi-ment; so that nowadays no one questions whether the theory isadequate to furnish reliable results in common matters. It isonly when on the one hand we begin to investigate what is rela-tively very small, e.g. the interior of an atom, or on the otherhand when we leave the Earth and apply the Newtonian theoryon a much wider scale that discrepancies can be detected, andeven here it may be remarked that Newtonian Mechanics hasproved adequate to enable astronomers to predict the time ofeclipses with an accuracy that would hardly have been possiblehad the foundations of dynamics been radically at fault.

From a philosophical standpoint Newton's definitions and lawsof motion offer much scope for criticism, particularly his assump-tions of absolute time and space. But it is no part of our purposeto discuss this aspect of the matter; we intend to leave to thereader his primary conceptions of time and space and adopt aNewtonian basis for the development of the subject.

4'li. Force. It is a fact of everyday experience that bodiesmove more often in curved paths and with varying speed thanin straight paths at uniform speed. We assume that it is theaction of other bodies that cause the speed of a body to vary or

41-415] FORCE 25

its path to bend. We describe this process by saying that thebodies are exerting forces on one another, and we define forceas that which changes or tends to change the state of motion ofa body.

Newton's First Law of Motion that Every body perseveres inits state of rest or of moving uniformly in a straight line, except inso far as it is made to change that state by external forces isimplied in the foregoing assumption and definition of force. Thislaw is also commonly called the Law of Inertia.

4"12. Mass. Matter is one of the primary conceptions of themind, it cannot be defined satisfactorily, but it is clear that anydefinition of matter would have to embrace all things that canbe perceived by the sense of touch and some things which areperceived by other senses. The mode by which the presence ofmatter is most easily perceived is through the effort requiredto produce in i t a sudden change of motion. This propertypossessed by matter is called inertia. The measure of the inertiaof a body is called its mass.

The familiar process of 'weighing' on a common balance is theprocess by which masses are compared. A scale of comparisonis instituted in which we fix upon a certain body A as a unitand determine tha t another body B is equivalent, in the senseof balancing, to some multiple or sub-multiple of the unit A.In this way every body has associated with it a number calledits mass.

4 13. A Material Particle is defined to be a body so smallthat, for the purposes of our investigation, the distances betweenits different parts may be neglected*.

4'14. The Momentum of a material particle is the productof its mass and velocity.

415. Measurement of Force. Newton's Second Law ofMotion states that Change of motion is proportional to theimpressed force and takes place in the direction in which theforce is impressed. In modern phraseology we substitute thewords rate of change of momentum for change of motion,

* Clerk Maxwell, Matter and Motion, Art. vi.

26 RECTILINEAR MOTION. KINETICS [IV

and we see that the law is an assertion of the mode of measure-ment of force. It is proportional to rate of change of momentum.Hence in the rectilinear motion of a particle of mass m, if v beits velocity and / its acceleration it is acted on by a force in

the direction of motion proportional to -=- (mv) or mf. And if weCtt

take the unit of force to be the force which acting upon a particleof unit mass causes it to have unit acceleration we may writethe force P = mf. The acceleration lasts so long as the force actsand ceases when the force ceases to act.

4"16. Force as a Vector. The second law of motion impliesmuch more than is asserted in the last article. It implies thatif a body A is in motion in the presence of several other bodiesB, C, etc. which exert forces P, Q, etc. upon it, either in the sameor in different directions, then each force produces its owncontribution to the acceleration of the body A, and thiscontribution is the same in magnitude and direction as itwould be if the force considered were the only force acting uponA. This implication of the law is usually called the Principleof the Physical Independence of Forces.

Further, if the body A under observation be a materialparticle whose velocity is accelerated, at any instant itsacceleration must possess a definite magnitude and directionand therefore be such as would be produced by a single force ofthe proper magnitude in the assigned direction. It follows that,if such a particle be acted upon by several forces, P, Q, etc.,simultaneously, the combined effect of these forces is the sameas that of a single force. That is to say, the separate forces P, Q,etc. are components of a single resultant force and the accelera-tions they severally produce are components of a single resultantacceleration. And we assert as a fundamental axiom that forcesare compounded by the vector law of addition. The structure ofdynamical theory is built upon this axiom, and the justificationfor the hypothesis is not to be found in any attempts at formalproof but in the general agreement of the theory with practicalapplications as the results of everyday experience and observa-tion.

415-4-17] FORCE 27

4'161. We will now change the order of the steps in the lineof argument and state it concisely as follows:

(i) we assert the fact that forces obey the vector law ofaddition;

(ii) we assert the second law of motion with the implica-tion that when a particle is acted upon by several forcessimultaneously its rate of change of momentum is proportionalto the resultant of the forces compounded vectorially;

(iii) we also ascribe the vector property to acceleration,and assert the exact equivalence of the force vector and thevector ' mass x acceleration/ so that, when resolved in anyassigned directions, the corresponding components of the twovectors are also equal.

417. Weight. The weight of a body is the force with whichthe Earth attracts it. It has already been remarked (3"5i) thatbodies fall to the Earth with a constant acceleration g. Itfollows that, if W is the weight of a body of mass m, then

W= nig.

The unit of force to which reference was made in 4*15 is calledthe absolute unit of force. In the British absolute system ofunits the unit of mass is called a pound. It is the mass of acertain piece of platinum deposited in the office of the Exchequerand defined by Act of Parliament as 'the Imperial StandardPound Avoirdupois.' When pound, foot, second units are em-ployed the unit of force is called the poundal. This is theabsolute unit of force in the pound, foot, second system. It isthe force which when acting on a body of mass one pound givesit an acceleration of one foot per second per second.

We observe that in accordance with the formula W = mg,the weight of a body of mass mlb. is mg poundals, and theweight of a body of mass 1 lb. is g poundals, so that a poundal isequal to a little less than the weight of half an ounce.

There is another system of units, in use among engineers, inwhich the unit of force is the weight of one pound, the unit ofmass being the mass of g lb., and the mass of Wlb. weight beingWjg units of mass. The argument at the foundation of this

28 RECTILINEAR MOTION. KINETICS [IV

system of units is that if different forces act in turn upon thesame body the forces are proportional to the accelerations theyproduce in the body. Thus if a force equal to the weight of Ppounds produces an acceleration f ft.-sec.~2 in a body of weightW pounds then

W g-W

It follows that P = — /

where W is the weight of the body and P the force acting uponit both measured in terms of the same unit, which may clearlybe the weight of one pound, or the weight of one ton or theweight of any other mass that may be found convenient.

4-i8. c.G.s. units. In France the standard of mass is a pieceof platinum called a kilogramme, the thousandth part of which,or gramme, is the unit of mass, the centimetre being the unitof length and the second the unit of time. In this c.G.s. systemthe absolute unit of force is called a dyne. It is the force whichacting on a mass of one gramme gives it an acceleration of onecentimetre per second per second. Since g = 981 cm.-sec.~2,therefore the weight of one gramme is 981 dynes.

4-i9. It is to be noted that the relation

force = mass x acceleration

does not require that the force and acceleration shall be constant.It is a relation that holds good at every instant during motionand may be expressed in the form

X = m'x,

where X is the force tending to increase x.

4"2. Impulse. The impulse of a constant force during a giveninterval of time is denned to be the product of the force andthe time during which it acts; i.e. Pt if P is the constant forceand t the interval.

The impulse of a variable force during a given interval of timeis defined to be the time integral of the force for that interval;i.e. JPdt integrated through the given interval.

4-17-4-3] IMPULSE 29

In every case the

impulse of a force = the change of momentum produced by it.

Thus, using the formulae for constant force and acceleration,

the impulse of the force = Pt

= mft= m(v — u)

= change of momentum.

Again, for a variable force X = mob,

the impulse of the force = Xdt

= m'xdt

= mas2— mxlt

where xi, *2 are the velocities at the beginning and end of theinterval; and, as before, this is the change of momentum pro-duced.

4-21. Force-time curve. If we plot a curve in which abscissaerepresent time and ordinates represent force the area under thecurve will represent the impulse of the force, or the change of'momentum produced in the given interval of time.

4*22. Example . A water jet issues from a nozzle of 2 square inchessection "with velocity 60/.s. and strikes a plane surface placed at right anglesto the jet. Find the force exerted on the plane.

The number of cubic feet that strike the plane per second=Tf¥x 60,and the mass of a cubic foot of water = 62-5 lb.

Therefore the momentum destroyed per second= ih X 60 x 62-5 x 60 = 3125 absolute units.

But if P is the force exerted in poundals its impulse in one second= P absolute units,

therefore P=3125 poundals= 97-65 lb. weight.

4-3. Work. A force is said to do work when its point ofapplication undergoes a displacement in the line of action ofthe force. Thus, if a constant force P has its point of applica-tion advanced through a distance s in its line of action, theforce is said to do work Ps. If, further, the force P remaining

30 KECTILINEAE MOTION. KINETICS [IV

M P

constant in magnitude and direction, the displacement of thepoint of application A be inan arbitrary direction AA',the work done by the force ismeasured by the product ofthe force and the projectionAM of the displacement uponthe line of action of the force,i.e. in either figure work done= P.AA' cos 6, giving a ne-gative result when 6 is anobtuse angle, and no work done when the displacement is per-pendicular to the direction of the force P.

More generally, when the point of application of a variableforce is displaced along any curve,let A A' denote an element ds ofthe curve, P the force at A and6 its inclination to the tangentto the curve. Regarding the forceas constant during the infinitesi-mal displacement AA', the workdone in this displacement isP . AA' cos 6 or P cos Ods; wetherefore represent the work done in any finite displacementby an integral JPCOS0ds,

taken along the curve of displacement from the initial to thefinal position.

4"31. Taking one foot as the unit of length and the weightof one pound as the unit of force, the unit of work is called thefoot-pound. It represents the work that would have to be donein order to lift one pound vertically through one foot.

Similarly, if one poundal be taken as the unit of force, theunit of work is called the foot-poundal.

4'32. Power is defined as the rate of doing work.The practical unit of power is called the horse-power and

represents the doing of 33,000 foot-pounds per minute.In the absolute c.G.s. system of units, the unit of work is

4-3-4-41] WORK AND ENERGY 31

called the erg, and represents the work done when a force ofone dyne has its point of application advanced through onecentimetre in the direction of the force.

The unit of power in the same system is called the watt andrepresents the doing of 107 ergs per second.

4-4. Energy is denned as capacity for doing work. Thus,when a body is in motion, work can be done in overcomingresistance before the body is brought to rest, and this amountof work is called the Kinetic Energy of the body.

A body may also possess energy in virtue of its position,represented by the work that would be done by the forcesacting on the body if it moved from its stated position to somestandard position. This is called Potential Energy.

In measuring potential energy it is necessary to choose astandard position in which a body may be considered to possesszero potential energy, and the choice of this standard positionis to some extent arbitrary. For example, if we consider theweights of bodies as forces capable of doing work as the bodiesdescend to a lower level, we may choose to consider bodies onthe floor of the room as possessing zero potential energy and thenall bodies at a higher level possess a positive amount of potentialenergy, measured in each case by the weight of the body multi-plied by the height above the floor. On the other hand, wemight choose the ceiling of the room as the level of zeropotential energy, and then all bodies below the ceiling wouldpossess negative potential energy. In a particular problem itis not the absolute value of the potential energy that is im-portant but the change of potential energy that takes place ina movement of the body under consideration, and therefore thearbitrariness of the choice of the zero of potential energy sim-plifies rather than complicates a problem.

441. Formula for Kinetic Energy. Suppose that the motionof a body of mass m moving with velocity u is opposed by aconstant force P and brought to rest in a distance x. The workdone against the resistance = Poc

= mfx,when f is the constant retardation produced by P.

32 RECTILINEAR MOTION. KINETICS [iV

But since the body comes to rest in a distance x, therefore

0 = w2 - 2fx.

Hence we have Px — ^mv?,

i.e. |mw2 is the energy possessed by the body in virtue of itsmotion; so the Kinetic Energy of a body of mass m movingwith velocity u without rotation is defined to be \muz.

442. Principle of Work. Conservation of Energy.

In any displacement of a body the change in the KineticEnergy is equal to the work done by the forces.

This is a general proposition of which a proof will be givenin a later chapter. At present we confine ourselves to recti-linear motion, and write the connection between accelerationand force in the form

mvdv/dx = X;

therefore jrnvdv=JXdx + C,

or, if u, v denote the velocities in the positions x = xy, x = x2,we have

^mvi—^mu2=\ Xdx,JXi

which is the required result.Since 'work done' means an equivalent loss of Potential

Energy, the last result implies that in any displacementgain of Kinetic Energy = loss of Potential Energy,

so that the sum of the Kinetic and Potential Energies is con-stant throughout the motion. This is the Principle of Con-servation of Energy. The principle applies to dynamicalsystems of the most general kind, but we observe that so farwe have only proved it for the case of rectilinear motion of asingle body.

443. Force-space Curve. If we plot a curve in whichordinates represent the force acting on a body in the directionof its motion, and abscissae represent distances traversed, thearea under the curve, e.g. JXdx, represents the work done or theincrease in the Kinetic Energy in traversing the distance underconsideration.

4-41-4-45] ENERGY 33

250220

4-44. Efficiency. In most machines some part of the workdone is expended in overcoming friction, or resistances of thatnature; this portion of the work is regarded as 'wasted' and theremaining portion is described as ' useful work.' The ratio ofthe useful work to the whole work done is called the efficiencyof the machine.

4*45. Examples , (i) A weight of 200 Ib. hangs freely from the end ofa rope. The weight is hauled up vertically from rest by winding up the rope.The pull starts at 250 Ib. and diminishes uniformly at the rate of 1 Ib. forevery foot wound up. Find the velocity after 30 ft. have been wound up,neglecting the weight of the rope.

The figure shows the force-space dia- X [gram, the force decreasing from 250 to220 pounds weight in a distance 30 ft.

The work done is therefore=4(250 + 220) x 30 ft. Ib.= 7050 ft. Ib.

But an amount of work= 200x30 = 6000 ft. Ib.

is done in lifting 2001b. through 30 ft.Therefore the additional work 1050 ft. Ib.is converted into kinetic energy; and if vis the velocity acquired

£ X 200 v2= 1050g foot-poundals,therefore v= 18'3 f.s.

This problem may also be solved by the consideration that the upwardpull after ascending x feet is (250 — x) pounds weight; and therefore, sub-tracting the weight of the body, the resultant upward force on the body is(bO — x)g poundals.

Hence 200vdv/dx=(50 - x)g,and, by integrating, 100«2=(50:» — \x2)g ; the constant of integration beingzero because « = 0 when x=0. For x = 30 this gives v = 18-3 f.s. as before.

(ii) A stream of water, 1 square foot in section, flowing at the rate of16 ft. per sec. enters a turbine. At what rate, in horse-power, does the waterdeliver energy ? {Ic.f. of water=62'5lb.)

What fraction of this energy is used when the water power drives a shaft,at a speed of 100 revolutions per minute, with a couple of which the momentis 140 with the pound weight and foot as units ? [M. T. 1909]

16 c.f. of water have a mass of 1000 Ib.; and in each second this massof water with velocity 16 f.s. enters the turbine. The kinetic energy ofthis mass

= £ x 1000 x 162=128000 foot-poundals.RD 3

30Ol x

34 RECTILINEAR MOTION. KINETICS [IV

This is the energy delivered per second, and this rate of working isequivalent to 128000/550^ horse-power

= ' f tH.P.

The work done by a couple in any displacement is the product of themoment of the couple and the angle through which its arm is turned.For since all couples of the same moment in the same plane are equivalentit is immaterial about what point the arm turns. Let the couple, ofmoment O, be represented by equal forces P at the ends of an arm AB.Let the arm turn through a small angle d6about the end A, so that B moves to B'.The work done = P.BB'

= P.ABd8=G.d6.And, by summation, the work done for anyfinite rotation 8 is 06.

Hence in the particular case consideredthe work done on the shaft = 140 X 200n- foot-pounds per minute and thisis equivalent to 140 X 200TT/33000 H.P.

= 28TT/33H.P.

It follows that the fraction of the energy used is 7TT/60 or aboutJJ of the whole, and this fraction represents the efficiency of themachine.

45. Applications. Locomotive Engines and Motor Cars.The propulsive force in the case of a locomotive engine or amotor car is the friction between the driving wheels and theground. The driving wheelsare made to rotate by meansof a crank attached to theaxle. If there were no frictionbetween the wheels and theground the car would remainstationary while the wheelsslipped round, but the slippingof the wheels on the ground is Frictionopposed by friction and this frictional force causes the forwardmotion of the car. If R denote the pressure of the drivingwheels on the ground and /x the coefficient of friction, the pro-pulsive force cannot exceed fiR.

In the case of the other wheels of the car or train, the wheelswhich run freely without compelling cranks, it is the friction

4-45-4-51] FRICTION 35

Friction

between the wheels and the ground which causes the rotation.For if there were no friction thewheels would slide forward with-out revolving; the friction onthem is therefore in the oppositedirection to the motion of thecar, but its amount is small com-pared to the frictional force onthe driving wheels being no morethan sufficient to cause the rota-tion of the wheels.

In order to stop a car or train brakes are applied to the wheels.This is effected by pressing brakeshoes either on to the rims of thewheels or on to the rims of smallerconcentric wheels rigidly attachedto the actual running wheels. Thisresults in frictional forces f beingset up opposing the rotation ofthe wheels, and this necessitates alarger frictional force F between thewheels and the ground if the same speed is maintained. Butthe friction F opposes the motion of the car as a whole and theresult is a reduction of the speed. The maximum value of F isfiw where w is the load carried by the wheel, hence assuming thesame coefficient of friction at all contacts the pressure of the brakeshoes must not exceed w or the wheel will become locked andskidding will ensue. The maximum retarding force that brakescan produce is therefore /A times the weight of the car or train.

4*51. Example . The loeight of a train is 400 tons, the part of theweight of the engine carried by the driving wheels is 30 tons, and thecoefficient of friction between the driving wheels and the rails is -16. Shewthat at the end of a -minute after starting on the flat the velocity will be lessthan 15-8 miles per hour. [M. T. 1909]

The propulsive force is '16 x 30 tons weight, so if v ft. per sec. be thevelocity acquired when all frictional resistance to motion is neglected, byequating the momentum set up to the impulse of the force in 1 minute,we obtain

400#= "16 x 30g x 60 using 1 ton as unit of mass,therefore z> = 23"04 f.s. = 15-7 miles per hour.

36 RECTILINEAR MOTION. KINETICS [iV

452. Problems on the running of trains with constant pro-pulsive force, road resistance and brake resistance may con-veniently be solved by equating the momentum to the impulseof the forces, and equating the kinetic energy to the work done,without introducing the question of acceleration.

Thus if m be the mass, P the propulsive force, and Rx theroad resistance, the equations for getting up a speed v in adistance x and time t from rest are

the impulse (P — Rx) t — mv the momentum generated,and the work done (P — R{) x = %mv2 the kinetic energy created.

Similarly, after steam is shut off and brakes are applied thetotal resistance to motion is _RX + R2 and if this brings the trainto rest in a distance x and time t', then since

momentum destroyed = impulse of the retarding force,therefore mv = (Rx + j?2) t';and sincekinetic energy destroyed = work done by the retarding force,

therefore J mv2 = (P^ + _R2) *'•4*53. Example . A train can be accelerated by a force of 55 Ib. per ton

weight and when steam is shut off can be braked by a force of 440 Ib. per tonweight. Find the least time between stopping stations 3850_/i. apart, thegreatest velocity of the train and the horse-power per ton weight necessaryfor the engine. [M. T. 1912]

In this case road resistance is neglected and with the notation of thelast Article we take TO = 2240, P-hbg, iJ2 = 440# and .»-(-#'= 3850; andthe equations are

55#C = 2240»> = 4405-1!'and bbgx=\ x 2240 v2 = 4 4 0 ^ ,therefore « = 1^-D2 and xl=-gsv

2.Consequently f$v2 = 3850, giving « = 73Jf.s.Again * = l t " aQd t'*=i!xvi

therefore < + i '=ff« = 105 sees.The maximum horse-power per ton weight is P«/550g>=7J.

4-54. Effectiveness of Brakes. The effect of applying brakesto one pair of wheels of a four-wheeled car in motion is to alterthe division of the weight between the wheels. Let the centreof gravity G be at a height h above the road and at horizontaldistances a from the rear axle and a' from the front axle.

Neglecting the inertia of the wheels the friction forces

4-52-4-54] EFFECTIVENESS OF BRAKES 37

between them and the ground are negligible when they arerunning freely.

Let W be the weight of the car and suppose that a retardation/ is caused by applying brakes to the rear wheels. This meansthat friction to an amount Wf/g must act on the back wheelsopposing the motion of the car. Let R and S be the verticalreactions of the ground on the rear and front wheels.

By resolving vertically we haveR + S-W=0.

Neglecting the rotatory inertia of the wheels and takingmoments about G* we get

aR-a'S + hWf/g = O.

Hence a n d SsmW=Wg(a + a)

whereas if the brakes were not in use we should haveR:S = a':a.

The maximum retardation that can be produced by applyingbrakes to the rear wheels without causing skidding is foundfrom the condition

or fie /J.a'g/(a + a'If on the other hand brakes are applied to all four wheels the

friction force Wf/g will be divided between front and backwheels, but the two equations for R and S will remain the sameand the maximum retardation will be given by the condition

or / $ fig.

The justification for this step will appear in Chapter ix.

38 RECTILINEAR MOTION. KINETICS [iV

4*55. Motion on an inclined plane. If a body of mass m isplaced on an inclined planewhich makes an angle a withthe horizontal, its weight mgcan be resolved into com- mo sinceponents mg cos a at rightangles to the plane andmg sin a down the plane. The former component represents thepressure of the body on the plane and the latter componentmg sin a will, if there be no other force along the plane, causean acceleration g sin a down the plane. Consequently a particlefreely projected up or down the plane has an acceleration g sin adown the plane.

If however there is friction between the particle and the planewith coefficient /A, since the normal pressure is mg cos a, there-fore there is a frictional force fxrng cos a opposing the motion.Hence if the particle moves up the plane it has a retardationg (sin a + p cos a), but if it moves down the plane it has anacceleration g (sin a — fi cos a).

4*56. Example . A train of 100 tons is ascending an incline of 1 in224 with an acceleration of 1 f.s.s. Find the resistance to motion in poundsweight per ton of the train if at the speed of 15 miles per hour the horse-power developed is 360.

Since the speed of 15 m.p.h. = 22 f.s., therefore if P is the propulsive forcein poundals,

22P=rate of working=360 x 55O#,therefore P=9000#r. The equation of motion is

where R is the total resistance in poundals.Therefore R = 9000c/ -M£<«"ty-224000

= 8000^-224000= (8000 - 7000) g = 1000 lb. weight= 10 lb. weight per ton of the train.

46. Resistance depending on velocity. When a body ofmass m moves subject to a resistance proportional to somepower of the velocity v, e.g. a resistance kv2, we have an equationof motion ,

atwhere P is the force in the direction of motion.

4-55-4-6] RESISTANCE DEPENDING ON VELOCITY 39

When P is constant, this may be integrated by first writingit in the form

mdvP-kv2

which gives on integration

= dt,

m , dP + v»Jk ,log }L-n -pr = t + const.

Alternatively, we may connect v with distance x, by writing

dv D , ,mv -j- =P — lev2,ax

therefore =:—r-7r = 2dx,P — kv2

and on integration — -j- log (P — kv2) =2x + const.

If on the other hand the force P be not constant, but beworking at a constant rate R, we have Pv = R where R isconstant.

Therefore m T- = ^ 2 .dt v

or mv T- = It — kv6.dt

This may be integrated by substituting - for v, which leads toz

mdz

and is then integrable by partial fractions.A simpler result is obtained connecting v and the distance x,

for in this casedv R , „mv -Y- = ktr,ax v

Tndv3

so that -jg—=-r = 3dx,R — kv3

mleading to — -r log (R — kv3) = fix + const.

40 RECTILINEAR MOTION. KINETICS [IV

4*61. Example . Tlie resistance to the motion of a car varies as thesquare of its speed and the effective horse-power exerted at the road wheelsis constant and equal to 22. If the weight of the car is 1 ton and themaximum speed of the car against its resistance is 45 miles an hour,determine the distance in which the car can accelerate from 15 to 30 milesper hour. [M. T. 1917]

The equation of motion is

mvdvjdx = P— kv2, where m = 2240 lb.,

and the rate of working i*» = 22 horse-power

= 22 x 550 r foot-poundals per second,

therefore P—l2\00g/v=n/v say.

Hence mv2dv/dx=n — kv3, so that by integrating we getTO,

— —f\og(n — kvi) = x+C, where C is a constant (1).

Now at the maximum speed of 45 m.p.h. or 66 f.s. there is no accelera-tion, so that vdv/dx = 0 and n=k. 663, but ?j = 12100(7, therefore

k = 12100 x 32/663=400/297.

Again the initial and final speeds are 15 m.p.h. and 30 m.p.h. or 22 f.s.and 44 f.s., and in (1) we may find C by putting x — 0 when v = 22, andthen find the value of x when •o = 44, namely

_ m . n— k.44?

m 663-443 TO, 2 7 - 8z=z, — — J.OET = LOS

3v& 66^ 22 3 A 27 1

= 2240 x | 5 log, g

= 173-9 ft. by the help of tables.

462. Fall of a heavy body in a resisting medium. Supposethat the resistance is proportional to vn, where v denotes thevelocity. Measuring x vertically downwards, we may write

dv ,v - j - = g — kvn.

djX

The acceleration therefore decreases as the velocity increases,and the acceleration vanishes when the velocity attains the

Ivalue (gjh)n. From this instant the velocity will remain constant.

Since (g/k)n is the greatest velocity attained it is called thelimiting or terminal velocity.

4-61-4-71] MOTION OF A CHAIN 41

47. Motion of a Chain. The principles developed so farcan be applied to the rectilinear motion of a uniform chain.Thus suppose a chain of length I and mass m per unit lengthto be placed on a smooth horizontal table in a line at rightangles to the edge of the table, having at time t a length xhanging over the edge. Let v denote the velocity at time t, andlet T denote the tension at the edge of the table. The equa-tions of motion for the horizontal and vertical portions of thechain considered separately are

m(l — x)vdv\dx = T,

and mxv dv/dx = mgx — T;so that by adding we get

mlvdvjdx = mgx,and on integration lv2 = gx2 + G, where G is a constant dependingon the initial conditions.

This result may also be obtained by the principle of con-servation of energy, for the kinetic energy of the whole chainis ^mlv2, and, if we take the zero level for the potential energyto be the upper surface of the table, the potential energyis — mgx.^x. Then, since the sum of the kinetic and potentialenergies is constant,

J mlv2 — \ mgx2 = const.,

which is equivalent to the last result.

471. Pall of a chain on to a table. A uniform fine chainof length I is suspended with its lower end just touching ahorizontal table and allowed to fall; to find the pressure on thetable when a length x has reached it.

Let m be the mass of unit length. When the velocity is v alength vSt is brought to rest in a short interval of time Bt; i.e.a mass mvSt is brought to rest and therefore the amount ofmomentum destroyed is mv2ht in time St. Hence the pressureon the table due to the rate of destruction of momentum is mv2.But, since the upper end falls freely, the velocity when a length* has reached the table is given by v2 = 2gx; and since the weighton the table is mgw, therefore the total pressure on the table isSmgx. This only holds good so long as x is less than I; as soon

42 EECTILINEAR MOTION. KINETICS [iV

as x becomes equal to I the force due to impacts ceases and thepressure becomes mgl.

4'8. Units and Dimensions. The fundamental units arethose of mass, space and time, denoted by M, L and T. All otherdynamical units are derived from and expressible in terms ofthese in the same way as we expressed the units of velocity andacceleration in 34; the dimensions of any physical quantity interms of mass, space and time being indicated by the indicesattached to M, L and T. Thus momentum being the product ofmass and velocity, the unit of momentum is denoted by MLT-1.Force is rate of change of momentum or the product of massand acceleration, so the unit of force is denoted by MLT~2.

Impulse is force multiplied by time, so that the unit of impulseis denoted by MLT"1, like the unit of momentum, since animpulse is equal to the change of momentum it produces. Workor potential energy is force multiplied by distance. Hence theunit of work is ML2 T~2; and the unit of kinetic energy has thesame dimensions either because kinetic energy is equivalent towork done or because it is measured by \ mv2.

Power is rate of doing work or work divided by time. There-fore the unit of power is denoted by ML2 T~3.

4'81. Change of Units. If a certain physical quantity is offi dimensions in mass, A, dimensions in length and T dimensionsin time, and V is its measure when M, L, T are the units of mass,length and time, then if the units of mass, length and time arechanged to M^L^T! the measure of the quantity becomes Viwhere

since these are equivalent representations of the same thing.

4'82. The consideration of dimensions is a useful check indynamical work, for each side of an equation must represent thesame physical thing and therefore must be of the same dimen-sions in mass, space and time.

Sometimes a consideration of dimensions alone is sufficientto determine the form of the answer to a problem. For example,suppose that a particle starts from rest at a distance a from

4-71-4-82] UNITS AND DIMENSIONS 43

a fixed origin 0 and moves with an acceleration njx* towards 0,where x denotes distance from 0. The time taken to arrive at0 can only depend on the given constants a and p. Supposethat the expressions for the time contain as factors <z" and /jft.Then aafjf is of one dimension in time; but /ijx^ is acceleration,i.e. of dimensions LT~2 and x being of dimensions L, it followsthat fi is of dimensions L3T~2. Hence a*/*3 is of dimensionsLa+S3T-^, but it is also of dimensions T, therefore /3 = — \ anda = f. Consequently the time taken to arrive at 0 is proportionalto a}fi~*.

EXAMPLES

[In numerical work take g to be 32 /.«.«.]

Uniform Acceleration1. Find the minimum horse-power that a fire engine must have if it is

to project 150 lb. of water per second with an initial velocity of 100 feetper second. [S. 1911]

2. If a train travelling at 30 miles an hour picks up 10000 lb. of waterin a quarter of a mile, find the back-pressure thereby produced on thetrain and the extra horse-power required on this account to maintain thespeed. [S. 1910]

3. A column of water 30 feet long is moving behind a plug piston in apipe of uniform diameter with a velocity of 15 feet per second. Provethat the time average of the pressure of the water on the piston, causedby its stoppage in one-tenth of a second, is 610 lb. per square inch.

[S. 1915]

4. Shew that the horse-power required to pump 1000 gallons of waterper minute from a depth of 50 feet, and deliver it through a pipe of cross-section 6 square inches, is about 34J. (Assume that 1 cubic ft. of water is6j gallons, and that 1 gallon of water weighs 10 lb. and neglect the frictionlosses.) [S. 1912]

5. Find the horse-power required to lift 1000 gallons of water per minutefrom a canal 20 feet below and project it from a nozzle of cross-section2 sq. inches. [S. 1926]

6. An engine of 400 horse-power is drawing a train of 200 tons mass upan incline of 1 in 280 at 30 miles per hour ; determine the road resistancein pounds weight per ton mass. [S. 1910]

44 RECTILINEAR MOTION. KINETICS [IV

7. An engine working at 500 horse-power pulls a train of 200 tons alonga level track, the resistances being 16 lb. per ton. When the velocity ofthe train is 30 miles per hour, find its acceleration.

At what steady speed will the engine pull the train up an incline of 1 in100 with the same expenditure of power against the same resistances ?

[S. 1917]

8. An engine of weight W tons can exert a maximum tractive effort ofP tons weight and develop at most H horse-power. The resistances tomotion are constant and equal to R tons weight. Shew that starting fromrest the engine will first develop its full horse-power when its velocity is

. ,. . . , bbWH ,, f.s. after at least _„ .„ .„—=r. seconds.224P

What is the greatest velocity which the engine can attain ? [S. 1923]

9. A car weighing 3 tons will just run down a slope of angle a (= sin ~J £$)under its own weight. Assuming that the forces resisting its motionremain constant, and that the engine exerts a constant tractive force, findto the nearest unit the horse-power of its engine if it can attain a velocityof 30 miles per hour in 4 minutes on the level. [S. 1922]

10. An engine driver of a train at rest observes a truck moving towardshim down an incline of 1 in 60 at a distance of half a mile. He immedi-ately starts his train away from the truck at a constant acceleration of0'5 ft./sec.2. If the truck just catches the train find its velocity when firstobserved. Assume that friction opposing the truck's motion is 14 lb.weight per ton. [S. 1924]

11. The weight of a train is 200 tons, the part of the weight of theengine supported by the driving wheels is 25 tons and the coefficient offriction between the driving wheels and the rails is -18. Prove that at theend of a minute after starting on the flat the velocity will be less than29^T miles per hour. [S. 1911]

12. A particle is projected directly up a plane inclined at an angle a tothe horizon, with initial velocity u given by

u? = 2gh (sin a + p cos a),where /i is the coefficient of friction. Shew that it traverses a distance A,and that it stays at the highest position if tan a < p. If tan a > fi, find itsvelocity when it returns to the starting point, and explain the differencebetween the initial and final kinetic energies. [S. 1923]

13. A train of 200 tons, uniformly accelerated, acquires in two minutesfrom rest a velocity of 30 m.p.h. Shew that, if the coefficient of friction be•18, the part of the load carried by the driving wheels of the engine cannotbe less than 12-7 tons. [S. 1921]

EXAMPLES 45

14. The tractive effort of an electric train is uniform and equal to theweight of 4 tons. The road resistance is 40 1b. weight per ton of the train,and the brake resistance is an additional 200 lb. weight per ton. The trainis taken from one station to the next, distant half a mile, in 1£ minutes,full power being kept on until the speed reaches 30 miles an hour, whenthe train ' coasts' at a uniform speed until power is shut off and thebrakes are put on. Shew that the mass of the train is approximately85 tons. [S. 1917]

15. An engine weighing 96 tons, of which 40 tons are carried by thedriving wheels, exerting a uniform pull gives a train a velocity of 25 milesper hour after travelling for 50 seconds from rest against a resistance of10'5 lb. weight per ton. If the friction between the driving wheels andthe rails is 0-2 times the pressure, find the tension in the coupling betweenthe engine and the first carriage. [S. 1923]

16. A train weighs 200 tons and the engine exerts a constant pull of45 lb. per ton, resistance to motion being 10 lb. per ton. The train startsfrom rest; after a certain time steam is turned off and the brakes put on.The train comes to rest at a distance of 1050 yards from the startingpoint 2 mins. 20 sees, after it started. Find the retarding force per ton ofthe brakes, and also the greatest horse-power developed. [S. 1915]

17. A train travels from rest to rest between two stations 5 miles apart.The total mass is 200 tons ; there is a road resistance of 12 lb. weightper ton, and the engine exerts a uniform pull of 5 tons weight until themaximum speed of 30 miles per hour is reached. This speed is maintaineduntil, steam being shut off, an additional resistance equal to -075 theweight of the train is applied to bring the train to rest. Find the timebetween the stations. [8. 1918]

18. A motor-car has its centre of gravity at a height h ft. midwaybetween the axles, the wheel-base being I ft. Shew that the ratio betweenthe least distances in which the car can be stopped by brakes acting

(a) on the front wheels, (6) on the back wheels, is . " , , where /J. is thet+/Aft-

coefficient of adhesion between the tyres and the ground. The rotaryinertia of the wheels may be neglected. [S. 1917]

19. A mine cage, weighing with its load 5 cwt., is raised by an enginewhich exerts a constant turning moment on the rope drum which is 16 ft.in diameter. The speed rises until the engine is running at 60 revolutionsper minute, when its output is 55 horse-power.

Find the acceleration and the time that elapses before the cage reachesfull speed : also find how far the cage rises in that time. [S. 1915]

46 RECTILINEAR MOTION. KINETICS [IV

20. An engine moves at a steady velocity v along level ground whenworking at a constant horse-power H. When moving up a plane inclinedat a small angle to the horizontal its steady velocity under the samehorse-power is v'. If the engine starts down the same incline with velocityv' and moves for t seconds with a constant acceleration until it reaches itssteady velocity down the plane corresponding to the same horse-power, H,shew that the distance travelled in these t seconds is v'2tl(2v' - v). Assumethat the frictional resistance is constant throughout. [S. 1926]

21. A 20 horse-power motor lorry, weighing 5 tons, including load,moves up a hill with a slope of 1 in 20. The frictional resistance isequivalent to 13 lb. weight per ton, and may be supposed independent ofthe velocity. Find the maximum steady rate at which the lorry can moveup the slope, and the acceleration capable of being developed when it ismoving at 6 miles per hour. [S. 1923]

22. A load W is to be raised by a rope, from rest to rest, through aheight h; the greatest tension which the rope can safely bear is n W.

( 2nh ) iShew that the least time in which the ascent can be done is \- TT— \ .

[S. 1926]

Variable Acceleration23. A tramcar starts from rest and its velocities at intervals of 5 sees,

are given in the following table :

Time in seconds

Velocity in milesper hour

0

0

5

8-1

10

11-8

15

14-6

20

16-3

25

17-7

30

19

Calculate the distance in yards travelled in the above time. Also, if thecar weigh 8 tons, estimate the effective pull exerted on the car at the endof 20 seconds. [S. 1910]

24. A car whose mass is 2000 lb. starts from rest; the resistance to themotion is equal to 50 lb. weight. When it has travelled S feet the forceexerted by the engine is P lb. weight where

s

p

0

644

10

634

20

622

30

607

40

587

50

565

60

537

70

509

80

475

90

440

100

404

Find, approximately, the velocity after the car has travelled 100 feet.[M. T. 1914]

EXAMPLES

25. A body of mass 1 lb. is projected on a rough plane surface with avelocity of 10 feet per second, and its velocity after time t is given forvarious values of t by a smooth curve passing through the points dennedby the following table :

Time in seconds

Velocity in feetper second

0

10-0

5

9-0

10

8-2

15

7-4

20

6-7

25

6 0

30

5-4

Derive the curve connecting the retardation with the distance travelled,and by determining the area of this curve verify that the energy lostduring the 30 seconds is 1"1 ft.-lb. approximately. [S. 1915]

26. The speed of a motor cycle is observed, as it passes five posts placed50 yards apart on a level track, to be 14'0, 26-4, 33-3, 37-l, 38"9 miles anhour respectively. Assuming the resistance in pounds weight due tomechanical and air friction to be 6+0O2i>2, where v is expressed in milesan hour, calculate the horse-power actually developed by the engine whenthe speed is 35 miles an hour, the total weight of machine and riderbeing 400 lb. [S. 1921]

27. A locomotive of mass m tons starts from rest and moves againsta constant resistance of P pounds weight. The driving force decreasesuniformly from 2P pounds weight at such a rate that at the end ofa seconds it is equal to P. Find the velocity and the rate of workingafter t seconds (t < a) and shew that the maximum rate of working is1-54 x 10~6 aPtjm horse-power. [S. 1924]

28. The acceleration of a tramcar starting from rest decreases by anamount proportional to the increase of speed, from 1*5 f.s.s. at starting to05 f.s.s. when the speed is 5 m.p.h. Find the time taken to reach 5 m.p.h.from rest. [S. 1925]

29. A train of weight if lb. moving at v feet per second on the level ispulled with a force of P lb. against a resistance of R lb. Shew that inaccelerating from v0 to vx feet per second, the distance in feet described by

the train is— I „—^. If the resistance R=a + bv2, find an expression9 J vo " — ti

for the distance described when the power P is shut off and the velocitydecreases from v0 to v1. [S. 1925]

30. Shew that a motor-car, for which the retarding force at V milesan hour when the brakes are acting may be expressed as (1000 + 008F2)pounds weight per ton of car, can be stopped in approximately 57 yardsfrom a speed of 50 miles an hour. [log,. 10 = 2-30.] [S. 1927]

48 RECTILINEAK MOTION. KINETICS [IV

•31. The resistance of the air to bullets of given shape varies as thesquare of the velocity and the square of the diameter, and for a particularbullet (diameter 0-3") is 40 times the weight at 2000 f.s. For an exactlysimilar bullet of the same material (diameter 0'5") shew that the velocitywill drop from 2000 f.s. to 1500 f.s. in about 500 yards, assuming thetrajectory horizontal. [loge 10 = 2-30.] [S. 1923]

32. A motor-bicycle which with its rider weighs 3 cwt. is found to runat 30 miles per hour up an incline of 1 in 20 and at 50 miles per hourdown the same incline. Assuming that the resistance is proportional tothe square of the velocity and that the engine is working at the samehorse-power, find the speed that would be attained on the level, and shewthat the horse-power is 2^ nearly. [S. 1924]

33. The resistance to an aeroplane when landing is a+bv2 per unitmass, v being the velocity, a, b constants. For a particular machine,&=10~3 ft.-lb.-sec. units and it is found that if the landing speed is50 miles per hour the length of run of the machine before coming to restis 150 yards. Calculate the value of the constant a. [S. 1927]

34. A particle is projected vertically upwards with velocity V, andthe resistance of the air produces a retardation hv2, where v is the velocity.Shew that the velocity V with which the particle will return to the pointof projection is given by

1L = ± + . [S.1925]

35. OAB is a vertical circle of radius a. 0 is its highest point; 0Asubtends angle a at the centre; AB subtends angle 2/3. (a+/3<^?r.)Shew that the time taken for a particle to slide down the chord AB fromrest at A is 2 x/(a cos ajg), when the angle of friction is also o.

Shew that if the motion is also subject to a resistance proportional tothe velocity, the time of descent is still independent of /3. [S. 1926]

36. In starting a tram of mass 3200 lb. the pull exerted by a horse isinitially 200 lb. and this pull decreases uniformly with the time until atthe end of 10 sees, it has fallen to 40 lb., an amount just sufficient toovercome the frictional resistance of the tram. Shew by means of a curvethe variation in the velocity and find the distance run during this periodof 10 sees.

Shew that the horse-power is a maximun\at the end of 5 sees., and findits maximum value. [M. T. 1913]

37. In starting a train the pull of the engine on the rails is at firstconstant, and equal to P; and after 1;he speed attains a certain value u,the engine works at a constant rate R=*Pu. Prove that when the engine

EXAMPLES 49

has attained a speed v greater than u, the time t and the distance x fromthe start are given by

where M is the mass of the engine and train together.

Calculate the time occupied in attaining a speed of 45 miles an hour,when the total mass is 300 tons, the engine has 420 horse-power and canexert a pull equal to 12 tons weight. [M. T. 1914]

38. The horse-power required to propel a steamer of 10,000 tonsdisplacement at a steady speed of 20 knots is 15,000. If the resistance isproportional to the square of the speed, and the engines exert a constantpropeller thrust at all speeds, find the acceleration when the speed is15 knots.

Shew that the time taken from rest to acquire a speed of 15 knots isabout l£ minutes, given loge7 = r946, one knot = 100 ft. per min.

[M. T. 1916]

39. The resistance to the motion of a train for speeds between 20 and30 miles per hour, may be taken as -j^V2 + 9 in pounds weight per ton,where V is the velocity in miles per hour. Sketch a curve shewing howthe horse-power per ton, necessary to overcome the resistance, increaseswith the speed as the speed rises from 20 to 30 miles per hour, the trainbeing on the level.

Steam is shut off when the speed is 30 miles per hour and the trainslows down under the given resistance. In what time will the speed fallto 20 miles per hour 1 [M. T. 1918]

40. A locomotive drawing a total weight of 264 tons on the level isexerting a tractive force of 20,000 pounds weight at the speed of 15 milesper hour. It works at constant horse-power until its speed is 60 milesper hour, when it is just able to overcome the resistance to motion, whichmay be taken to vary as the square of the velocity. Shew that it reachesthe speed of 45 miles per hour from the speed of 15 miles per hour in adistance of approximately 5080 feet. [M. T. 1919]

41. A train of mass 300 tons is originally at rest on a level track. It isacted on by a horizontal force F which uniformly increases with the time,in such a way that F=0 when t=0, F=b when i = 15 ; F being measuredin tons weight, t in seconds. When in motion the train may be assumedto be acted on by a frictional force of 3 tons, independent of the speed ofthe train. Find the instant of starting, and shew that at i=15 the speedof the train is 0-64 foot per second, whilst the horse-power required at thisinstant is about 13. [M. T. 1920]

50 RECTILINEAR MOTION. KINETICS [IV

42. A train of weight W lb. moving at V feet per second on the levelis pulled with a force of F lb. against a train resistance of R lb. Shewthat in accelerating from Vo to V1 ft. per sec, the distance in feet describedby the train is

W f^ VdV

If TF=300 tons, i i = 2160 + 15F2, shew that the distance described inslowing down on the level from 45 to 30 miles per hour, with the powershut off, is about 537 feet, [log,, 10 = 2-303.] [M. T. 1921]

43. A cyclist works at the constant rate of P horse-power. Whenthere is no wind, he can ride at 22 feet per second on level ground, and at11 feet per second up a hill making an angle s i n " 1 ^ with the horizon.The total mass of man and cycle is 180 lb. The resistance of the air ishi2 lb. weight, when the velocity of the man relative to the air is v feetper second ; the other frictional forces are. negligible.

Find P, and shew that the speed of the cyclist when riding on levelground against a wind of 22 feet per second is between 10 and 10'5 feetper second. [M. T. 1924]

44. The external resistance to the motion of a bicycle and rider may besupposed to consist of two parts, a constant force and a force varying asthe square of the speed. A rider observes that his speed when free-wheeling down a hill of slope 1 in 50 is sensibly constant when it reaches10 feet per second, and that on a slope of 1 in 25 it becomes constant at20 feet per second. The mass of the bicycle and rider is 200 lb. Find thepower expended by the rider in maintaining a steady speed of 15 feetper second on the level, assuming that when the rider is propelling thebicycle 10 per cent, of the work he does is lost in internal friction in thepedalling gear. [M. T. 1926]

45. A uniform chain 30 centimetres long, having a mass of 1 grammeper centimetre, lies partly in a straight line along a rough horizontal tableperpendicular to the edge. The portion hanging over the edge is justsufficient to cause the chain to begin to slip. The coefficient of frictionwith the table being \, find the velocity of the chain and its tension atthe edge of the table when x centimetres have slipped off. [M. T. 1919]

ANSWERS TO EXAMPLES 51

ANSWERS. (flr = 32.)

1. 42fJ. 2. 458J lb. wt.; 36§ H.P. 5. 180f§. 6. 17 lb. wt.per ton. 7. $$ f.s.s., 24T

6^ m.p.h. 8. 55Jfir/224iJ f.s. 9. 30.

10. ^880 f.s. 11. 2-25f.s.s. 12. {Zgh (sin a- / i cos a)}*.15. 11984 lb. wt. 16. 53 lb. wt. per ton ; 736-ft H.P. 17. 10 mins.43f sees. 19. 239 f.s.s.; 21 sees.; 531 ft. 21. 12 m.p.h.;l\l f.s.s. 23. 191 yds.; 2546 lb. wt. 24. 40 f.s. 26. 3-4.27. P (t-P/2a)/70m f.s.; PH (2 - t/ay/UOm ft. lb. per sec.

28. ^ log, 3 = 8-06 sees. 29. ^ - l o g ^ j ^ . 32. ^ v"5To=58-6 f.s.

33. l o g Y l + ^ W g ; a = 3-68. 36. 53Jf t . ; l£J . 37. 3 mins.

21-35 sees. 38. ^ j f.s.s. 39. 97"13 sees. 41. « = 9. 43. P=-2. . . .44. 93'5 ft. lb. per sec. 45. 1x cm. per sec.; 49 (20 — x) (10+.J?) gms. wt.

4-1

Chapter V

KINEMATICS IN TWO DIMENSIONS

5*1. We now proceed to consider the motion in one plane ofa particle and of a rigid body, confining our considerations inthis chapter to Kinematics or the geometry of the motion apartfrom the forces that cause it.

We require general definitions of velocity and accelerationapplicable to curved paths.

The velocity of a point is defined to be a vector drawn throughthe point and such that its resolved part in any direction is therate of displacement of the point in that direction.

Let the point be moving along a curve APQ, and let P, Q beits positions at times t, t + Bt, and let (x, y), (x + Bx,y + By) be the

y

o X

coordinate of P, Q referred to rectangular axes. Then the resolvedparts of the velocity parallel to the axes are

.. Bx dxlim -=- or T7 or x

ot at

and

Again 8a?, By are the resolved parts of the vector PQ, so thatthe resultant velocity as defined above is

.. By dyhm -£ or -£ or y.

l i m ^ .

51-5-11] ACCELERATION 53

But if s denotes the arc AP measured from a fixed point A onthe curve and s+ Bs denotes the arc A Q, we know that

.. chd PQhm —K—^ = 1,

OS

so that the resultant velocity is

.. chd PQ Bs dshm —j—-5 ,T = — or s,st^.o os bt at

and its direction is along the tangent at P.Further, if V denotes the resultant velocity and i|r its inclina-

tion to the axis of x, we have, asa verification, that the velocity / Vparallel to &r. T, , dsdx dxux= Vcosvr = -=- -=- = -=• orx,T at ds dt

and the velocity parallel to

x

5"il. The acceleration of a point is a vector drawn throughthe point and such that its resolved part in any direction is therate of change of velocity in that direction. (See 4*161.)

Thus if we use rectangular axes and u, v denote the resolvedparts of the velocity parallel to the axes at time t, and u + Bu,v + Bv denote the resolved parts at time t + Bt, then the resolvedparts of the acceleration are

,. Bu du

u-^.0 ot at

. .. Bv dvand hm ^i =-r =v=y.

st-*o ot at

It does not follow however that the resultant acceleration isdirected along the tangent to the path of the point.

Let V,V+BV denote the velocities at the points P, Q of thepath, where PQ is the small arc described in time Bt, and letBifr denote the angle between the tangents at P, Q.

From any point 0 draw vectors Oa, Ob to represent V andV+BV. Then ab represents the change of velocity in time Bt.

54 KINEMATICS IN TWO DIMENSIONS [V

Draw am at right angles to Oa meeting Ob in m. Then

ab = mb + am.

in.

a

Therefore the acceleration, being lim ^- both as regards8«-*o or

magnitude and direction, has components

lim -=— = -5- or V in the direction of the tangent to the pathj £ o ot at

-=— = - 5 -ot at

and.. amInn -«r- = li - = lim hs V*

where p is the radius of curvature of the path, and the directionof this component is that of am, i.e. along the inward normalto the path.

We note that the tangential component of acceleration dV/dt

may be written -^--j-or V - 1 - .J ds at ds

5-12. Hence we have equivalent representations of velocityas shewn in the diagrams.

y y

X

y

o

v=s

X

5-11-5-2] RELATIVE VELOCITY 55

Also equivalent representations of acceleration as shewn inthe diagrams

V

X X

and we note that we do not represent acceleration by a singlesymbol unless the path is a straight one ; in fact, bending of thepath involves a normal component of acceleration.

52. Relative Velocity. Let two points P, Q move in thesame plane with velocities repre-sented by the vectors PR, QS. R ^Complete the parallelogramPQR'R. Then QR' is a vectorrepresenting a velocity equal inmagnitude and direction to that p Qof P. But QS = QR' + R/S,therefore R'S represents a velocity which when compoundedwith the velocity of P gives the velocity of Q. Hence R'S is thevelocity of Q relative to P. Similarly SB,' is the velocity of Prelative to Q.

We may also represent the same thing analytically. Thus it(x, y), (oc, y) are the coordinates of P, Q referred to any frameof axes in a plane, we may call

£ = x — x and v = y' ~ythe coordinates of Q relative to P, and by differentiation wefind the relative velocity components

£ = x — x and V ~y' ~ V-

In other words the velocity of Q relative to P is to be foundby subtracting the velocity of P from that of Q, either geo-metrically in the form of vectors, or algebraically by subtractingcorresponding components.

56 KINEMATICS IN TWO DIMENSIONS

therefore

or

53. Angular Velocity. If a point P is moving in a plane theangular velocity of P about afixed point 0 in the plane isthe rate of increase of theangle that the line OP makeswith a fixed direction in theplane.

Thus if a point P be movingin any direction with velocityV, and OP makes an angle 0with a fixed direction Ox, theangular velocity of P about0 = ddjdt. But if OP = r and the tangent to the path makesan angle cj> with OP we have

rdO/ds = sin tf>,

dO ds sin <fr _ F s i n <£dt~dtr r

angular velocity = (component velocity atright angles to OP)/OP.

We may also write this result rO = Fsin <f>,or multiplying by r we get r26 = Vp,where p is the perpendicular distance of the tangent from theorigin. That is r2 6 = moment of the velocity about the origin.

Again, a sectorial element of area of a plane curve is repre-sented by %r2&0, so that \r2Q is the rate of increase of a sectorialarea as P moves along the curve, and therefore Vp is twice therate at which the radius vector OP sweeps out area.

5*31. Examples , (i) A point is moving in a circle. Shew that at anyinstant its angular velocity about a point on the circumference is half itsangular velocity about the centre.

This follows from the fact that in the y ^S.Pdiagram the angle

POC=\PCx

so that -rXPOC) = \

(ii) A point moves in a parabola withconstant velocity. Shew that the angularvelocity about the focus varies as cos3i#,where 6 is the angular distance from the vertex.

5-3-5S3] ANGULAR VELOCITY 57

~We have 6 = ~, but in the parabola with the focus as origin

o2=ar, so that 6 oc - j ;

but

therefore 8 x cos3^0.

532. Motion in a Circle. At time t let Vbe the velocity ofa point P moving in a circle ofradius r, centre 0. Let 0 be the v >angle that OP makes with afixed direction Ox.

Let s be the arc APmeasuredfrom a point A on O». Thenthe velocity

Also the acceleration com-ponents are s or rd along thetangent and V2jr or r62 towards the centre.

If the motion be uniform, i.e. V constant, then there is ofcourse no acceleration along the tangent, but there is in everycase the acceleration V2/r or rB2 towards the centre.

5'33. Relative Angular Velocity. If two points P, Q aremoving in a plane with velocities u, v, then either point in generalpossesses an angular velocity relative to the other, which may berepresented thus :

Let the directions of thevelocities u, v make angles6, <£ with the line PQ.

The component velocity ofQ relative to P at right angles to PQ is v sin <£ - u sin 6, andtherefore the angular velocity of Q relative to P is

(v sin<f> — u sin 6)/PQ.

Whenever v sin 0 = u sin 6, then the line PQ is moving parallelto itself.

58 KINEMATICS IN TWO DIMENSIONS

5*34. Example . Two points describe concentric circles with velocitiesvarying inversely as the square roots of the radii of the circles; to findpositions in which the relative angularvelocity vanishes.

Let 0 be the centre of the circles,a, 6 the radii, and u, v the velocities.If P, Q are positions of the points whenthe relative angular velocity vanishes,the velocities resolved at right anglesto PQ must be equal. Let P0Q = 6.

Therefore u cos OPQ + v cos OQP=O,

or a~ * cos OPQ-b~ I coa (6+OPQ)=0,

or (a~i-b'i cos 0) cos OPQ

But a sin OPQ - b sin OQP=O,

therefore a sin OPQ - b sin {6 + OPQ)=0,

or (a - 6 cos 6) sin OPQ - b sin 6 cos OPQ=0.

Eliminating the angle OPQ we get

(a~* - b ~i cos 6){a-b cos 6) + 5* sin2tf = 0,

therefore cos 6=aibi'/(a- a*5* + 6).This problem has a practical application. If we regard the points P, Q

as representing the earth and an inferior planet describing circular orbitsround the sun 0 in the same plane, we have found the relative positionsof P, Q in their orbits at which the planet Q would appear to be 'stationary'as seen from the earth ; since in this position there is no relative angularvelocity. Since v>u, therefore when the planet Q is on the line OP, thejoin PQ must be turning in a clockwise sense; the motion of the planetamong the stars as seen from the earth is then described as 'retrograde.'The position found above is that in which the clockwise rotation of PQceases and a counter-clockwise rotation is just about to begin. The visiblemotion of the planet then becomes 'direct,' and continues to be directuntil the relative position is such a position as Q'P in the diagram, whereQ'OP=d, this being another 'stationary' position, after passing which themotion again becomes retrograde.

54. Displacement of a Plane Rigid Body in its Plane.By a rigid body we understand a body in which the distancebetween any two particles remains invariable. The position inits plane of a plane rigid body is therefore determined when thepositions of any two points A, B of the body are known. Forif one point A of the body is fixed the only possible motion is a

O-34-5-41] INSTANTANEOUS CENTRE 59

rotation about A, and if a second point B is fixed no motionis possible.

We shall now shew that any displacement of such a body inits plane consists in a rotationabout a point in the plane.

Let A, B be any two pointsof the body which becomedisplaced to A', B'. Draw theperpendicular bisectors ofAA\ BB' meeting in / .

Then I A = I A', IB = IFand AB = A'B'. Thereforethe triangles IAB, IA'B' are A

identically equal and the second is obtained from the first by arotation about / through an angle AIA' or BIB'. I may becalled the centre of rotation for this displacement.

If the displacement be a pure translation A A' and BB' areparallel and the point I is at an infinite distance.

541. Instantaneous Centre of Rotation. Let a plane bodybe moving in any manner in its plane, .and consider thedisplacement that takes place in a short interval of time St.Any two points A, B of the body have definite paths of motionand the displacements A A', BB' that take place in time St arechords of small elements ofthese paths, and the pointI, obtained as the inter- / " ^ ^ J&section of the perpendicularbisectors of AA', BB', will,as 8t—*- 0,tend to coincidencewith the intersection of thenormals to the paths at A and B. Hence it follows that at anyinstant the body is turning about an instantaneous centre ofrotation I, which is found by taking any two points A, B of thebody and drawing normals to the paths of A and B throughtheir instantaneous positions.

It follows that if P be the position of any other point of thebody at the same instant, the direction of motion of P must be

60 KINEMATICS IN TWO DIMENSIONS

/O X

at right angles to IP; and, if S0 be the angle turned through bythe body in time St, the displacement of P in time Bt is IP. SO.

5'42. Examples, (i) The ends of arod AB are constrained to move along twogiven lines Ox, Oy in the same plane.

The instantaneous centre of rotationis the point / found by drawing AI, BIperpendicular to Ox, Oy respectively.

(ii) A circular lamina rolls along astraight line. Prove that the instantaneouscentre of rotation is the point of contact.

Let 0 be the centre, and A the pointof contact. If by rolling through a smallangle hd the point P on the circle isbrought into coincidence with P' on thestraight line, then PP' is of order (8<9)2.Now since the path of 0 is parallel tothe given line, therefore the instan-taneous centre is on OA, and it is alsoon the limiting position as 88-*-O of theperpendicular bisector of PP'. But^4P' = arc AP, and therefore as dd-*-0 the ratio AP'jchAAP-*-l, thereforethis perpendicular bisector ultimately passes through A which is thereforethe instantaneous centre of rotation. Another proof of this theorem willbe given in 5'51.

543. Pole Curves. When a lamina is in motion in its planethere is a locus of the instantaneous centre of rotation in thelamina and also a locus in the fixed plane. These loci are calledthe 'pole curves. Thus in the last example, of a circular laminarolling along a straight line, the locus of the instantaneouscentre in the lamina is the circular boundary, and the locus inspace is the fixed straight line. And this example illustrates thegeneral theorem that the motion of the lamina can be producedby the rolling of one pole curve on the other.

Without attempting a rigorous demonstration of this theorem,•we can give an explanation that may satisfy the reader as to thetruth of the theorem.

Let P, Q, R, S, T be a succession of points on the locus of theinstantaneous centre in space. Consider the results of makinginfinitesimal displacements of the body about the points Q, R,

5-41-5-44] POLE CURVES 61

S, Tin succession. The first of these displaces QP to QPlt thesecond displaces i^QPi to RQxP^, the third displaces 8RQXP2 to

SR^Ps, and the fourth displaces TSR^Ps to 2 ^ J?2Q3P4.Now Slt R2, Q3, Pi are a series of positions on the lamina ofthe instantaneous centre of rotation, i.e. they are points on thelocus on the lamina of the instantaneous centre and they can inturn be brought back into coincidence with the point S, R, Q, Pby the rolling of one polygon on the other. Hence by regardingthe pole curves as the limits of polygons the motions of thebody can be produced by rolling the pole curve fixed on thelamina on the pole curve fixed in space.

5'44. Example . A triangular lamina ABC moves in its plane so thatthe sides AB, AC always pass through two fixed points P, Q. Find the polecurves.

The point P on the lamina can only move along AB, for if it had anycomponent of displacement at rightangles to AB the side AB would nolonger pass through the point P fixedin the plane.

Hence the instantaneous centre / ison the perpendicular to AB through P,and in like manner on the perpen-dicular to AC through Q.

Therefore the angle PIQ is thesupplement of A and a constant angle;and P, Q are fixed points, therefore B Cthe pole curve fixed in the piano is a circle PIQ.

Again AT is a diameter of this circle and is a fixed length (in factPQ cosec A). Hence the pole curve on the lamina is a circle of centre Aand radius PQ cosec A, i.e. twice the size of the former circle ; and themotion of the triangle could be produced by rolling the larger circle onthe smaller.

62 KINEMATICS IN TWO DIMENSIONS [V

55. Angular Velocity of a Body. When a plane body ismoving in its plane we may speak of the 'angular velocity of thebody' without specifying any particular line or point of refer-ence, meaning thereby the rate of increase of the angle betweenany line AB fixed in the body and any line Ox fixed in theplane of motion : i.e. if at time t the angle between AB and Oxbe 8, then 6 is the angular velocity of the body. Also if anyother line CD fixed in the body makes an angle 6 + a with Oxat the same instant, then a is the constant angle between ABand CD and unaltered by the motion, so that 0 is also the rateof increase of the angle between CD and Ox. Consequently inmeasuring the angular velocity of the body, it is immaterialupon what line fixed in the body we concentrate our attention.

Further, the angular velocity of the body is independent ofany motion of translation of the body as a whole, for such amotion would not alter the direction of any line fixed in thebody. For example, when we say that a wheel has an angularvelocity a> in its plane this statement is independent of whetherthe wheel is turning about a fixed point, such as its centre, orrolling along in contact with a fixed line, or possessing any othertranslational motion, a> being the rate of increase of the anglebetween any line fixed in the plane of the wheel and a line fixedin the plane of motion.

5"51. It is important that the reader should appreciate thedifference between the angular velocity of one point aboutanother as defined in 53 and the angular velocity of a body asdefined in the last article, we therefore propose to illustrate thelatter by considering a simple problem from more than onestandpoint.

A circle rolls along in contactwith a straight line; to find thevelocity and acceleration of anypoint on the circumference.

Take the given line as axis of x.Let C be the centre of the circle,a the radius, P the point of con-tact with Ox at time t, A thepoint on the circumference whosevelocity and acceleration are re-

y

5-5-5-52] ANGULAR VELOCITY OF A BODY 63

quired. Let 6 be the angle ACP, then 8 is the angular velocity of thecircle. For convenience take the origin 0 so that OP=axc AP, i.e. 0 isa position of the point A in the rolling motion. Taking an axis Oy atright angles to Ox, let x, y be the coordinates of A, then

y = CP - AC cos 8 = a { \ - )Hence the components of velocity are

x = a8 (I —cos 8),and y = a8 sin 8;and the components of acceleration are

x = a'6{\ - cos 6) + a82 sin 6,and y — ad sin 8 + a82 cos 8.

As a special case we may put 6=0, so that A is the point of contact ofthe circle and the line; then we find that

shewing that the point of contact of the rolling circle has no velocity, i.e.it is the instantaneous centre of rotation as proved previously in 5*42. Inthe same case the acceleration components are x = 0, y = a82, shewing thatthe point of contact of the rolling circle has an acceleration a82 towardsthe centre, as proved previously in 5*32.

5'52. We may also obtain the foregoing expressions for velocity andacceleration by compounding the velocity and acceleration relative to thecentre of the circle with the velocityand acceleration of the centre. Thus,let the circle be rolling with angularvelocity a> and let V be the velocity ofits centre G. Every point on the cir-cumference has a velocity aa> relativeto the centre, so that the total velocityof the point P on the circle is V— aa>.But assuming 'rolling' to mean thatP is the instantaneous centre of ro-tation it follows that P has no velocity and therefore V=aa.

The velocity of a point A on the circumference is therefore compoundedof aa> along the tangent relative to the centre and V or aa> parallel to Ox,giving as before components

aa>(l —cos 8) parallel to Ox,and am sin 8 parallel to Oy,where a=8.

Again for accelerations we have that the accelerations of A relative toCare

ai> along the tangent and aa>2 along AC (5*32),and the acceleration of C is V parallel to Ox.

But V=aa>, therefore V=ai>.

KINEMATICS IN TWO DIMENSIONS

Hence the accelerations of the point A are compounded of ai> along thetangent, aa>2 along 4Cand ai> parallel to Ox, and these are equivalent tocomponents

a i (1 — cos 8) + aa>2 sin 8 parallel to Ox,and ad> sin 8 + aw2 cos 8t parallel to Oy, as before.

5*53. We may also obtain the velocity components of the point A byconsidering the motion relative to theinstantaneous centre of rotation P.

The point A is moving at right anglesto AP with velocity a>AP, and sinceAP=%asin\6, this gives components

2am sin2 \8or aa> (1 — cos 8) parallel to Ox,and 2a<»sin^#cos£#or aa> sin 8 parallel to Oy,as before.

But the finding of the accelerations of the point A by reference to theinstantaneous centre would be more cumbersome than the process of 5*52,because it would involve compounding accelerations of A relative to Pwith accelerations of P, and the latter accelerations would have to befound by compounding accelerations of P relative to C with the accelerationof C, as in 5*52.

5*54. Examples , (i) Prove, by considering a point on the circumferenceof a circle rolling uniformly along a straight line, that the radius of curvatureof a cycloid at any point is twice the length of the line joining that point tothe point of contact of the generating circle with the base. [M. T. 1908]

A cycloid is the curve traced out by any point on the circumference ofa circle which rolls along a straight line.

Let C be the centre of the circle, G thepoint of contact with the line, and P the pointthat is tracing the cycloid. Then since G isthe instantaneous centre of rotation the pointP is moving at right angles to PG with velocityv=aP0, where <a is the angular velocity ofthe circle. Now if p is the radius of curvatureof the cycloid the normal component of the acceleration of P is v^jp. Butthe acceleration of P may also be represented by its acceleration relativeto C, i.e. a?PC (5*32) compounded with the acceleration of C. But G hasno acceleration because the motion is uniform. Therefore the resultantacceleration of P is a?PC along PC.

Hence v*/p = co2 PC cos CPG = |o>2 PG.

But v = aPG, therefore p = 2l'G.

5'52-5'54] ANGULAR VELOCITY OF A BODY 65

(ii) A circle A of radius a turns round its centre with uniform angularvelocity o). A circle B of radius b rolls on the circle A and its uniformangular velocity is a'. Find the time taken

(1) for the point of contact to make a complete circuit of A,(2) for the centre of B to return to a former position.

Determine the accelerations of the common point of the two circles and thegreatest acceleration of a point on the circle B.

Suppose that at time t = 0 the points M, N on the circles are in contactand on the fixed line Ax throughthe centre A of the circle A. LetP be the point of contact at time t.Since AM, BN are lines fixed inrelation to the circles, in time tthey turn through angles at, at.Thereforeif BNmeetsAx in if, theangle BKx^a't while MA x=at.Again the arcs PN, PM are equal,so that if PAM=6 then theangle PBN=aQ\b.

Therefore a>'t =

b (a' — a)t

SC

. b6= 5 - ^

a+6

b(a' — a)a+b

And the time taken for P to make a circuit of the circle A is2n/6 = 2ir (a + b)/b (a - co).

Again the time taken for the point B to describe a circle round Adepends on the angular velocity of the line AB, which=$ + &>.

Hence the time required =2n/(6 + a)

For the accelerations we have that the acceleration of the point P onthe circle A

I' d \ 2

= a ijPAx) along PA by 5*32

and the acceleration of the point P on the circle B

= accel. relative to centre .B+accel. of B

= -ba'* + (a+b)(6+a)2 along BP

= - 6<»'2 + (aa + ba'fl(a + 6).

Also the acceleration of any point on the circle B is compounded of itsacceleration relative to the centre B, i.e. W 2 along the radius towards B,and the acceleration of B, so that the greatest acceleration will be whenthese components coincide in direction, i.e.

66 KINEMATICS IN TWO DIMENSIONS [V

EXAMPLES

1. A ship sailing N.W. by compass through a tide running 5 knots findsthat after 2 hours it has made 4 nautical miles S.W. Determine thedirection of the current and the speed of the ship.

2. An aeroplane which travels at the rate of 80 miles per hour in stillair starts from A to go to B which is 200 miles distant N.E. of A. Ifthere is a wind blowing from the North at 20 miles per hour, determinethe direction in which the aeroplane must move, and the time required.

If at the end of an hour the wind drops to 5 miles per hour, determinethe position relatively to B of the aeroplane at the time when it shouldhave arrived at B.

Prove that provided the velocity of the wind remains fixed in directionand magnitude, all points attainable by an aeroplane in a given time lieon a circle whose radius is independent of the wind. [S. 1924]

3. The velocity of a stream between parallel banks at distance 2a apartis zero at the edges and increases uniformly to the middle where it is u.A boat is rowed with constant velocity v{>u) relative to the water, andgoes in a line straight across. How are the bows pointed at any point ofthe path and how long will it take to get across 1 [S. 1921]

4. An aeroplane has a speed of v miles per hour, and a range of action(out and home) of R miles in calm weather. Prove that in a north windof w miles per hour its range of action is

R (v2 - w2)/v (v2 - w2 sin2 0)*

in a direction whose true bearing is <$>. If .ft = 200 miles, u = 80 milesper hour and w=30 miles per hour, find the direction in which its rangeis a maximum, and the value of the maximum range. [M. T. 1921]

5. If a point moves so that its angular velocity about two fixed pointsis the same prove that it describes a circle. [S. 1903]

6. If two particles describe the same circle of radius a, in the samedirection with the same speed u, shew that at any instant their relativeangular velocity is u/a. [S. 1910]

7. A particle P is moving in a circle of radius a centre O with uniformspeed u. AB is a diameter of the circle and AP=r. Find the angularvelocity of P about A, B and C. What is the angular acceleration of Pabout the same points? [S. 1909]

EXAMPLES 67

8. A particle Pmoves in an ellipse whose foci are /Sand H, and centre C.The velocity at any point of the path varies as the square of the diameterconjugate to CP. Prove that the angular velocity of P about 8 variesinversely as its angular velocity about H. [S. 1911]

9. The line joining two points A, B is of constant length a and thevelocities of A, B are in directions which make angles o and /3 respectively

with AB. Prove that the angular velocity of AB is •——- , where u* acos/3

is the velocity of A. [S. 1918]

10. Two points are describing concentric circles of radii a and a' withangular velocities a and a> respectively. Prove that the angular velocityof the line joining them when its length is r is

11. The end P of a straight rod PQ describes with uniform angularvelocity a circle whose centre is 0, while the other end Q moves on a fixedline through 0 in the plane of the circle. The end Q' of an equal straightrod PQ1 moves on the same fixed line through 0. Prove that the velocitiesof Q and Q' are in the ratio QO : OQ1. [S. 1925]

12. A circular ring of radius 6 turns round a fixed point 0 in its cir-cumference with uniform angular velocity Q. A smaller ring of radius arolls on the inside of the larger ring with uniform angular velocity a>, theangular velocities being in the same sense. Find the velocity of any pointof the smaller ring in any position. Also shew that, if aa> = b&, then inevery position of the smaller ring one point on it is at rest. Indicate theposition of this point for a general position of the rings. [S. 1921]

13. C is the centre of two concentric circles A, B, and a line CPQ meetsthe circles in P, Q. Tangents XPX, YQY are drawn to the circles atP, Q. The circle A rolls along the line XX carrying the circle B with it,so that C, P, Q are always collinear, until the point P is again on the lineXX and Q is consequently again on the line YY. The distance betweenthe two positions of P is equal to the circumference of the circle A.Investigate the fallacy in the assertion that the distance between the twocorresponding positions of Q is equal to the circumference of the circle B.

[S. 1924]

14. If P is any point on the circumference of a circle, centre C, whichrolls with angular velocity <» on a fixed circle of centre 0 and radius a,prove that the angular velocity of P about 0 is

a~m>ooscop- tCo11- Exam- 1 9 1 2 J5-2

68 KINEMATICS IN TWO DIMENSIONS [v

15. A particle moves in the curve y=a log sec - in such a way that theQj

tangent to the curve rotates uniformly ; prove that the resultant ac-celeration of the particle varies as the square of the radius of curvature.

[S. 1925]

16. A circle and a tangent to it are given. A rod moves so that ittouches the circle and one end is upon the tangent. Shew that the loci ofthe instantaneous centre in space and relative to the rod are both parabolas.

[S. 1925]

17. Shew that if two given points of a lamina describe coplanar straightlines, any point on a certain circle fixed in the lamina will also describe astraight line. [S. 1917]

18. Prove that the motion of a rigid lamina moving in its own plane isat any instant (in general) equivalent to a rotation about a certain point / .What is the exceptional case 1

Prove that, if the vectors Oa, Ob represent the velocities of two pointsA, B, the triangles Oab, IAB are directly similar and that their corre-sponding sides are perpendicular.

Given Oa, Ob find a geometrical construction for the vector Oc whichrepresents the velocity of a third point C. Shew in particular thatAC : CB = ac : cb, and that, if ABC is a straight line, so is abc.

Four rods are freely jointed together so as to form a quadrilateral PQRS.Shew that if PQ is fixed the angular velocities of QR, PS are in the ratioPT : QT, where T is the point of intersection of PQ, RS. [S. 1923]

19. If A and B are points on a rod which is moving in any way in aplane, and if Oa and Ob represent the velocities of A and B at any instant,prove that ab is perpendicular to AB. If C is any other point on the rodand if c divides ab in the same ratio as that in which C divides AB, provethat Oc represents the velocity of C at the same instant.

PQ, QR, RS are three rods in a plane jointed together at Q and R, andwith the ends P and S jointed to fixed supports. If a triangle Oqr is drawnwith Oq, qr, rO perpendicular to PQ, QR, RS respectively for any positionof the rods, prove that as the rods move through this position Oq and Orrepresent on the same scale the velocities of Q and R. [S. 1915]

ANSWERS

1. 22-2 hours. 2. 34° 49' E. of N.; 3h. 5f m.; 31-44m. N. of B.3. At inclination cos"1 (uxjav) to the bank, where x is distance from the

nearer bank. —sin" 1 - . 4. E. or W. 185'4m. 7. u/2a, ul2a,u/a;0.

12. If B, A are the centres of the larger and smaller rings and P any pointon the latter, the vel. of P rel. to A is aa>, of A rel. to B is bQ — aa>, andof B is 6Q. The point required is the end nearer to 0 of the diameter ofthe smaller circle parallel to OB.

Chapter VI

DYNAMICAL PROBLEMS IN TWO DIMENSIONS

6'1. In the early part of Chapter IV we interpreted Newton'slaw that rate of change of momentum is proportional to the im-pressed force and takes place in the direction in which the forceis impressed as implying the equivalence of two vectors, the' force' acting on a particle, and the product' mass x accelera-tion '; and the rest of Chapter iv consists for the most part ofexamples of this equivalence in the case of rectilinear motion.

In the present Chapter we shall consider examples of thisequivalence when a particle is free to move under the action offorces in one plane; the equivalence implying that the resolvedparts of the two vectors in any assigned direction are equal.Consequently, if m be the mass of the particle, x, y its coordi-nates, and X, Y the sums of the resolved parts parallel to rect-angular axes of all the forces acting upon the particle, we havethe equations

m'x = X and my = F.In a large class of problems in dynamics of a particle the force

components X and T are given, and the solution of the problemconsists in integrating these equations in order to determinethe path of the particle.

6'2. Motion of Projectiles. Consider the case of a particleof mass m freely projected under theaction of gravity in a non-resistingmedium. Take for axes Ox, Oy thehorizontal and upward vertical linesthrough the point of projection 0, ~7Q ' ' 1 ~xand let the particle be projectedwith velocity V in a direction making an angle a with thehorizontal.

We have X = 0 and Y= — mg.

Therefore x = 0 and y = —g.

70 DYNAMICAL PROBLEMS IN TWO DIMENSIONS [VI

Integrate and introduce the initial values of x, y, namely Fcos a,Fsin o, and we get

x=V cos a and y = V sin a — gt

therefore x=Vtc,osa and y = Vt sin a-%gt2 (1).

Eliminate t and we obtain the equation of the path of the particle1 ox2

y = x tan a — „ ^ sec2 a (2).

This represents a parabola and by writing the equation in theform

F2 . \2 2F2cos2a/ F2sin2a\/ F2 . \2 2K2cos2a/[x sin a cos a = y •V g ) g V

it is seen that the latus rectum is 2 F a cos2 a/g, the vertex/ F2 F 2 sin2 a\

the point I—sin a cos a, ^ J and the axis vertically

downwards.The directrix is therefore horizontal, and its height above the

point of projection is equal to the height of the vertex plus one-fourth of the latus rectum, i.e.

F2sin2a F2cos2a F2

2g + 2g °* 2g'

Again we can shew that the velocity at any point of the pathis, in magnitude, the velocity that would be acquired in fallingfreely from the directrix. For, if v is the velocity at time t,

y {2g= 2g x depth below the directrix.

The time of flight before the projectile again reaches thehorizontal plane through the point of projection is got byputting y = 0 in (1), which gives t = 2 V sin a/g.

The range of the projectile on the horizontal plane throughthe point of projection is obtained as the value of x in (1) when

6"2-6*21] MOTION OF PROJECTILES 71

for t we substitute the time of flight, i.e. F2sin2a/<7; and thegreatest range for a given F is therefore V*/g got by taking

Since sin 2 (J TT — a) = sin 2a, therefore in general there aretwo directions of projection with a given velocity which givethe same horizontal range, viz. those of inclination a and \TT - a,the two coinciding when a = \ir.

6"2i. The range on an inclined plane through the point ofprojection may be found from theequation of the path 6'2 (2) bywriting y = x tan /3, where /3 isthe inclination of the plane tothe horizontal, i.e.

1 ox2 „

therefore x = (tan a — tan

and the range required= x sec /3

2Fa

cos a sin (a — /3) sec2 /3

F2

= — sec2 /3 jsin (2a - /3) - sin ft).

For given values of V and /3 this expression is greatest when2a —/3= ^7r, and this makes the maximum range

We may write the last resultF2

— = 1 + sin /3,grand we know that the polar equation of a parabola with thefocus as pole is

- = 1 + cos 6,r

where 6 is measured from the vertex and I is the semi-latusrectum.

A comparison of the last two equations shews that, if weconstruct a parabola with focus at 0, latus rectum 2V2/g, axis

72 DYNAMICAL PROBLEMS IN TWO DIMENSIONS [VI

vertical and vertex A upwards, then any radius OP drawn from0 will give the maximum range in direction OP for a particle

M

projected from 0 with velocity V. This parabola therefore is anouter boundary to the region that can be reached by projectilesstarting from 0 with velocity V.

622. Geometrical Construction. A particle is to be pro-jected from a point P so as to pass through a point Q. Itmay be shewn that with a given velocity of projection theremay be two possible paths, one possible path or no possiblepath.

Thus if V be the given velocity, from P draw PM verticallyupwards and of length V2/2g, thenM is a point on the directrix. Thedirectrix of the path is therefore ahorizontal line MN. Draw QN atright angles to the directrix. Thenfrom the focus-directrix property ofthe parabola, the focus must lie on acircle of centre P and radius PM, and also on a circle of centreQ and radius QN. If the circles intersect in two points S, S'either of them is the focus of a possible path and the bisectorof the angle MPS, or MPS' gives the required direction ofprojection. Alternatively the circles may touch at a point S onPQ, which is then the focus of the one path possible; or thecircles may not meet one another, in which case the velocity ofprojection is inadequate. The least velocity that will carry theparticle from P to Q will correspond to the lowest possibleposition of the directrix, i.e. the position which makes

PM + QN = PQ.

G^l-G^] MOTION OF PROJECTILES 73

Thus if v be the least velocity and PQ = h, and k be theheight of Q above P, we have

PM + QN=h,

and PM-QN=k,

so that 2PM=h+k,

and v* = 2gPM = g(h + k).

6*23. Examples , (i) Shew that the product of the two times of flightfrom P to Q with a given velocity of projection is 2PQ/g. [M. T. 1916]

Let V be the velocity of projection, o the inclination to the horizontalof the direction of projection, and a, b the horizontal and vertical distanceof Q from P.

Then a= Vtco&a,and 6 = Vt sin a — ^gfi;therefore a2 + (6 + \gfif = F21\or %gH* + (gb- F2) t2 + a2 + 52=0.

Hence, if tt, t% are the two times of flight,

so that t1t2 = 2PQIg.

(ii) A gun fires a shell with a muzzle velocity 1040 feet per second.Neglecting the resistance of the air, what is the farthest horizontal distanceat which an aeroplane at a height of 2500 feet can be hit and what gunelevation is required? Shew that the shell would then take approximately44'2 seconds to reach the aeroplane. [S. 1926]

This problem can be solved by regarding the equation of the path 6*2 (2)as giving the range x as a function of o, when for y the height of theaeroplane is substituted. We may then obtain another relation betweenx and o by differentiating with regard to a and using the fact that for amaximum value of x the derivative dx/da must vanish. We can thensolve the two equations for x and a, and also find the time taken from therelation Vt cos a=x .

A simpler solution is obtained from the consideration that the quadraticequation for fi

igW+igb- F2)*2+a2 + &2=0 (1),obtained in the last example must have real roots. In this case 6 denotesthe height of the aeroplane and a denotes its horizontal distance. Thecondition for real roots is

The greatest value of a is therefore - (F 2 - 2^6)*, and, taking F= 1040

and 5 = 2500, this gives a = 31200 ft.

74 DYNAMICAL PROBLEMS IN TWO DIMENSIONS [VI

Also for this value of a the quadratic (1) has equal roots in t2, namely

and for the given values of V and b this makes t = 44"2 sees.Reverting now to the equation for the horizontal range

a= Vtco&a,we find that sec a = 1 '473,so that the elevation required is 47° 17'.

63. Resisting Media. So far we have assumed that gravityis the only force acting on a projectile. Now suppose that themotion is opposed by a force proportional to the velocity. Thusif m denote the mass and F t h e velocity, let mkV denote themagnitude of the resistance. Therefore the components of theresistance parallel to horizontal and vertical axes Ox, Oy are

— mkx, — mky.

Let u, v denote the initial horizontal and vertical componentsof velocity. The equations of motion give

x = — kx or x + kx = 0,therefore x + kx = u, since initially x = 0 and x= u.

Now multiply by the integrating factor ekt and integrate bothsides as in 161,

but x = 0 when t = 0, therefore G = — u/k; and, dividing by ekt,

x^l-er*) (1).

Again y = -g-ty,therefore y + ky = v — gt, since initially y — 0 and y = v.

Now multiply by the integrating factor ekt and integrate bothsides

but y — 0 when t — 0, therefore C'= — T — ~ ; and, dividing by e**,fc /c

Results (1) and (2) express x and y in terms of t, and if t iseliminated the resulting equation gives the path.

623-631] RESISTING MEDIA 75

From (1) it follows that, as t becomes large, x tends to a

constant value u/k, so that the path approaches a vertical

asymptote.

6*31. Example. A particle subject to gravity is projected at an anglea with the horizontal in a medium which produces a retardation equal tok times the velocity. It strikes the horizontal plane through the point ofprojection at an angle o>, and the time of flight is T. Prove that

t a n o _ ekT-\-kTtana ~ e~kT- 1+kT'

and deduce that a, > a. [M. T. 1924]

Taking u— Fcos a and v— Fsin o as the initial components of velocityin the equations of the last article, the time of flight T is obtained byputting y=0 in (2); i.e.

(3).

Again tan a> is the value of — dyjdx, when y = 0 and t—T, so that wehave

y_ Vsma-gT_ Vain a-gTx~ Fcos a — kx Fcos a— Fcos a (1— e~kT)

Vsaaa-gTV coaae-kT '

Eliminate V by means of (3), first multiplying numerator and denomi-nator by (1 — e~kT), and we get

tan co = •- l (!-«-**•)} cot a '

, , tantherefore tana e-™ (kr-l+e-kT)

~e-*T-l+kT'

If we expand ekT and e~kT, the numerator becomes

2! + 3 T + "" '

a series of positive terms, and the denominator becomes

2 ! ~3T+'"'the same series, save that the terms are alternately positive and negative.Therefore the numerator is greater than the denominator and co > a.

76 DYNAMICAL PROBLEMS IN TWO DIMENSIONS [VI

6*32. Resistance proportional to the Square of theVelocity. We shall illustrate another law of resistance bytaking an example in which the resistance varies as the squareof the velocity.

A particle is projected vertically upwards under gravity. The resistanceof the air produces an acceleration opposite to the velocity and numericallyequal to kv2, where v is the velocity and k a constant. If the initial velocityis V, and the square of k V2/g can be neglected, shew that the particle reaches

V kV3

its highest point in time 5-= , and that the greatest altitude reached is

V2 kV*~%> 4gr2 •

If the initial velocity, in addition to the vertical component V, has a smallhorizontal component U, and the resistance follows the same law, shew thatwhen the particle returns to the original level its horizontal velocity isapproximately P i r ^ V [M. T. 1925]

For the upward vertical motion, measuring y vertically upwards, wehave

dv 7 , dv2 , „v -= -= —q — KVA. o r —r- = — 2(7— zkv*.

dy y dy y

Therefore

and, by integration,log(2fo2 + 25- )=-2%+C (1),

where C is a constant. Putting v= V when y = 0 gives

and substituting this value for C in (]),

log^2ty.5 kV2+g d

The highest point is reached when v = 0, and then

P o w e r s ;yi fey *

therefore the greatest altitude is ^ --~j •To find the time, we write

dv ,

dv-—

g + kv2

1 IkTherefore - -= tan"1 W - v= C -1, where C is a constant.

slkg v 9

6-.32-64] RESISTING MEDIA 77

But when (=Owe have v= V, therefore C'=~-== tan"1 K I - V

*Jkg V g

Hence -= tan"1 J - v= —= tan"1 U - V-1.s/kg V 9 *Jkg V g

The time of reaching the highest point is found by putting v=0, andthen ,

*]kg V 9

The function on the right-hand side can be expanded by Gregory's series..., provided - 1 <#^1] , so that

'g *\g.V kV3

= 7r-£, neglecting higher powers.9 °9

When there is also a horizontal velocity, we have for the horizontalacceleration

where yfr is the inclination of the path to the horizontal; but v=i and

cos ty = dx fds,

therefore x= —k?dxjds=— kax,

x ..or r = — ks.

xHence, by integration, log x=C"- ks, and x— U when «=0, so that

C"=log U.

Therefore x=Ue~k».Now the horizontal velocity is small and decreasing, therefore the

horizontal distance travelled is small, and as an approximation to thevalue of s when the particle again reaches the plane we may take twicethe greatest altitude and write

9 WBut by hypothesis the last term is negligible, so that ks — kV2/g, and therequired value of x is Ue~kr'lf.

6-4. Principle of Work. Reverting to the equations of motionof 6-1, viz.

mic = X, my = Y,if we multiply these equations respectively by x and y and add,we get

m {xx + yy) = Xx + Yy.

• ° DYNAMICAL PROBLEMS IN TWO DIMENSIONS [VI

Now integrate this equation with regard to t for any interval oftime from t0 to ti, and it follows that

\m (x2 + y2)t — \m (a? + y2)0 = f(Xdx + Ydy),

or \ m«i2 — \ mv<? = J(X dx + Ydy),

where the suffixes denote the values at the beginning and endof the interval. This result shews that the increase in the kineticenergy of the particle in any interval is equal to the work doneby the forces acting on the particle in that interval.

Alternative proof. Let F be the resultant force acting on theparticle Avhen it is at any point P ofits path, and e the angle that the forcemakes with the tangent to the path.

Resolving along the tangent, wehave

dv „ Amv ~T~ = F cos e.asIntegrating along the path between any two points A, B,

we get

\mv2 — \mv<? = \Fcos eds,

where v0, v^ are the velocities at A and B, and by 4'3 the integralrepresents the work done by the resultant force.

EXAMPLES1. Shew that if a gun be situated on an inclined plane, the maximum

range in a direction at right angles to the line of greatest slope is aharmonic mean between the maximum ranges up and down the planerespectively. [S. 1910]

2. P, Q are two points distant a apart, and at heights h, h above agiven horizontal plane ; prove that the minimum velocity with which aparticle must be projected from the plane so as to pass through P and Qis V{g(a + h+k)}. [S. 1909]

3. A particle is projected with given velocity from a point P so asto pass through a point Q. If S is the focus of either of the possibletrajectories, shew that the times of flight in the two trajectories are

{(SP+ SQ+PQ)i ± (SP+ SQ - PQ)l}/gl.[S. 1917]

6'4] EXAMPLES 79

4. Prove that the velocity required to project a particle from a heighth to fall at a horizontal distance a from the point of projection, is at leastequa l t o J[g {J{a? + A2) - h}\

The hammer was thrown at Cambridge a distance of 122 ft. in 1911; ifthe head of the hammer was 8 ft. from the ground at the instant of pro-jection, calculate the least velocity required and the corresponding angleof projection. [S. 1912]

5. A particle is projected under gravity from A so as to pass through B.Shew that if B has horizontal and vertical coordinates x, y referred to A,and the velocity of projection is /\/%gh, the angle between the two paths atB is a right angle if B lies on the ellipse x2 + 2y2 = 2hy. [S. 1917]

6. Three particles are projected simultaneously and in the same verticalplane from a point with velocities »i, v^, v3 in directions making angles«i> "2> <*3 with the horizontal. Shew that the area of the triangle formedby the particles at any time is proportional to the square of the timeelapsed from the instant of projection, and that the three particles willalways lie on a straight line, if 2v2vs sin (o2-a3) = 0. [S. 1912]

7. A particle P is describing a parabola freely under gravity. Shewthat the angular velocity of the line joining P to the focus is Zgujv2, wherev is the velocity of the particle at P and u is the horizontal componentof v. [S. 1923]

8. A particle is projected from a given point with a velocity whosevertical component is given. Prove that the initial angular velocity aboutthe focus of the path is greatest when the angle of projection is 45°.

[S. 1910]

9. Particles are projected simultaneously from a point under gravity invarious directions with velocity V. Prove that at any subsequent time tthey will all lie on a sphere of radius Vt, and determine the motion of thecentre of this sphere. [S. 1927]

10. A projectile is fired from a point 0 with a velocity due to a fall of100 feet from rest, and hits a mark at a depth of 50 feet below 0 and at adistance of 100 feet from the vertical line through 0. Shew that the twopossible directions of projection are at right angles, and find to the nearestminute their inclinations to the horizontal. [M. T. 1915]

11. A particle is projected at time t=0 in a fixed vertical plane from agiven point S with given velocity ij(2ga), of which the upward verticalcomponent is v. Shew that at time « = 2a/« the particle is on a fixedparabola (independent of v), that its path touches the parabola, and thatits direction of motion is then perpendicular to its direction of projection.

[M. T. 1922]

80 DYNAMICAL PROBLEMS IN TWO DIMENSIONS [VI

12. A particle is projected from a point A so as to pass through a givenpoint B and to return to the same level as A at a point C. Shew that thevelocity with which the particle must be projected is

J{g(AiV2. NC2+AC2. NB*)\ZAN.NB. JVC},where N lies in A C directly below B.

13. Shew that all points in a vertical plane, which can be reached byshots fired with velocity v from a fixed point at a distance c from the

2v2

plane, lie within a parabola of latus rectum — , whose focus is at aC2<7 ^

distance =- vertically below the foot of the perpendicular on the planefrom the point of projection. [S. 1925]

14. A fort is on the edge of a cliff of height h. Shew that there is anannular region in which the fort is out of range of the ship, but the shipis not out of range of the fort, of area 8TTM, where tj2gk is the velocity ofthe shells used by both. [S. 1926]

15. A fort and a ship are both armed with guns which give theirprojectiles a muzzle velocity sltyk, and the guns in the fort are at aheight h above the guns in the ship. If d1 and d2 are the greatest(horizontal) ranges at which the fort and ship, respectively, can engage,prove that

16. From a gun placed on a horizontal plane, which can fire a shellwith velocity \/2gH, it is required to throw a shell over a wall of height h,and the elevation of the gun cannot exceed a, where a < 45°. Shew thatthis will be possible only when A<i7~sin2a, and that, if this condition besatisfied, the gun must be fired from within a strip of the plane whosebreadth is 4cos a Vtf ( # sin2« ^X). [S. 1911]

17. A shell fired with velocity V at elevation 6 hits an airship atheight H which is moving horizontally away from the gun with velocity v.Shew that if

(2 V cos e - v) (F 2 sin2 6 - 2gH)i = v V sin 6,

the shell might also have hit the airship if the latter had remainedstationary in the position it occupied when the gun was actually fired.

[M. T. 1917]

18. A battleship is steaming ahead with velocity V. A gun is mountedon the battleship so as to point straight backwards, and is set at an angleof elevation a. If v is the velocity of projection (relative to the gun), shew

that the range is — sino (vcos a—V); also shew that the angle of elevation

for maximum range is cos"1 {(V+ \ / F 2 + 8W2)/4«}. [S. 1925]

EXAMPLES 81

19. An aeroplane is flying with constant velocity v and at constantheight h. Shew that, if a gun is fired point blank at the aeroplane after ithas passed directly over the gun and when its angle of elevation as seenfrom the gun is a, the shell will hit the aeroplane provided

2 (V cos a — v)v tan2 a=gh,

where V is the initial velocity of the shot, the path being assumed to beparabolic. [S. 1915]

20. The range of a rifle bullet is 1200 yards when a is the elevation ofprojection. Shew that if the rifle is fired with the same elevation from acar travelling at 10 miles per hour towards the target the range will beincreased by 220 tani a feet. [S. 1917]

21. A shot is fired with initial velocity V at a mark in the samehorizontal plane ; shew that if a small error e° is made in the angle ofelevation, and an error 2e° in azimuth, the shot will strike the ground at

a distance from the mark ——.

Shew also that if the angle of elevation is less than about 31J° an errorin elevation will cause the shot to miss the mark by a greater amountthan an equal error in azimuth. [S. 1923]

22. A shell bursts on contact with the ground and pieces from it fly inall directions with all velocities up to 80 feet per second. Shew that aman 100 feet away is in danger for § J2 seconds. [S. 1921]

23. A particle moving under gravity describes a parabola of vertex Aand focus S, the velocity at A being u. Prove that, when the particle isat P, where SP=r, the components of its velocity along and perpendicularto SP are respectively equal to the vertical and horizontal components ofits velocity.

Shew that the component in the direction SP of the acceleration of theparticle is g - u'/r. [M. T. 1923]

24. An aeroplane travelling at a speed V relative to the air experiencesa resistance R = aV2 + b/V2, where a and b are constants within certainlimits of V. Shew that, within these limits of V, the power absorbed inair resistance has a minimum value Ha, at a speed Fo, where

Assuming that the effective thrust power of the propeller Z is inde-pendent of V, find the greatest rate of gain of height, and shew that the

7 T-f

aeroplane is then climbing at an angle sin~x wl/

a to the horizontal,

where W is the weight of the aeroplane. [M. T. 1925]

RD 6

82 DYNAMICAL PROBLEMS IN TWO DIMENSIONS [VI

25. A particle is projected horizontally with a velocity v0 in a mediumin which the resistance to its motion is kv2 per unit mass, when its velocityis v. If \ff is the downward inclination of its path to the horizontal whenit has traversed an arc s, shew, by resolving along the horizontal, that

and by resolving along the normal to the path, that

eiu dz_ = vjt s e cs,/ , . [M. T. 1926]

26. A particle P moving along a horizontal straight line has retardationhu, where u is the velocity at time t. When t = 0, the particle is at 0 andhas velocity «0. Shew that ua — u is proportional to OP, and that the finalvalue of OP is tco/k.

A particle subject to gravity describes a curved path in a resistingmedium which causes retardation h x velocity. Shew that the resultantacceleration / has a constant direction, and that /=/oe~*S where f0 is theacceleration when < = 0. [M. T. 1923]

A N S W E R S

3. 60-4 f.s.; 43° 7' with horizontal. 10. 76° 43'; - 13° 17'.

Chapter VII

HARMONIC MOTION

71 . When a particle moves in a straight line so that itsacceleration is always directed towards a fixed point in the lineand is proportional to the distance from that point its motionis called Simple Harmonic Motion.

Let 0 be the fixed point and P the position of the particle attime t. Let OP = x, then theacceleration may be denoted •

. O Pby fix in magnitude, where fiis constant. But the acceleration is directed towards 0, thereforethe acceleration in the positive direction of the axis of x is — fix,and this will be so whether the particle is on the right or leftof 0. Hence in all positions of the particle we have

x = — fix (1).The solution of this equation may be written down as in

17 (16) or, as in 36 (ii) (/3), we may multiply both sides by 2xand integrate, thus

2xic = — Ifixx,

therefore x2 = G — fix2 where G is a constant, which may bedetermined if the velocity is known in any one position.

This relation is equivalent to

dx"777=; i \ ~~ ± **>V(C - fix2)

where the positive or negative sign must be taken according asx is increasing or decreasing with t.

On integration we get

where a is a constant,

or x = A/-sin(±V fi

6-2

84 HARMONIC MOTION [VII

The solution contains two arbitrary constants C and a and maybe written in the simpler form

x = A sin \/fx t + B cos 'J fit (2).

This is the general form of solution of equation (1) and there-fore includes all special cases.

For example : (i) Let the particle start from rest at a distancea from 0; i.e. when t = 0, x = a and x = 0.

Then, since x= A V/xcos \/fit — jBvVsin V/*£ (3),

on putting t — 0 in (2) and (3) we get

a — B, and 0 = ^.,

therefore x = a cos *J fit (4)

gives the position at time t in this case.(ii) Let the particle be at a distance a from 0 and moving

towards 0 with velocity v when t = 0.Here we have x — a and x = — v when t = 0. Therefore, from

(2) and (3), a — B, and — v = A \'/t, so that in this case theposition at time t is given by

x=—r=sin Vfit + a cosVfit (5).VfM

7*11. Periodicity. The solutions obtained in the last articleall represent periodic motions, for, if t is increased by 27r/V/i orby any multiple of 27r/V/i, the values obtained for x and x in(2), (3), (4), (5) remain unaltered, shewing that after a time27r/V/x the particle is again in the same position and movingwith the same velocity as before.

The interval 27r/V/i is called the period of the harmonicmotion or oscillation; and the frequency of the oscillationsmeans the number of complete oscillations per second, i.e.

The distance through which the particle moves away from 0on either side is called the amplitude of the oscillation. Wenotice that the period only depends on the given constant ft,,so that the period is independent of the amplitude.

Further when a coordinate of position is represented by aformula a cos (nt + e) e is called the phase at the instant from

7-1-7-2] PERIODICITY 85

which t is measured. If two different harmonic motions withthe same period 27r/re are represented by

a cos (nt + e), b cos (nt + e'),they are said to differ in phase. The difference in phase e — e'may be spoken of as a certain number of radians, or it may bemeasured in terms of the time in which either of the anglesconcerned would increase by this amount, i.e. (e — e')/n seconds,or it may be measured as the fraction (e — e')/27r of a period.Thus if two harmonic motions are represented by

a cos nt, b sin nt,

it follows that, since sin nt — cos (nt — -= J, these motions differ

in phase by one-quarter of a period, i.e. one particle is at theorigin when the other is at an extreme position and vice versa.

72. Geometrical Representation. Let a point Q describea circle of centre 0 with uniformangular velocity a. Let P be L^_ • „the projection of Q on a fixeddiameter AA'. As Q movesround the circle P will oscillateto and fro in the line AA', andwe can shew that the motion ofP is simple harmonic motion.

Since Q moves uniformly in acircle its only acceleration is a>2QO along QO. The accelerationof P is the resolved part along A A' of the acceleration of Q,i.e. a)2P0. Hence the acceleration of P is always directedtowards 0 and proportional to its distance from 0 and thereforethe motion of P is simple harmonic.

The period of oscillation of P is the time taken for Q todescribe the circle, i.e. 2"jr/co. This agrees with the 2TT/V/J, of7-11 because, by comparing the formulae for acceleration, we seethat a>2 = /i.

The velocity of Q in the circle is coQO along the tangent TQ.The velocity of P is the resolved part along AA' of the velocityof Q, i.e. coQO cos QTP = wQO cos OQP = coPQ in magnitude.The velocity of P is therefore proportional to the ordinate PQand has its greatest value when P is at 0.

86 HARMONIC MOTION [vil

7"21. A particle starts at a distance a from 0 with velocity vtowards 0 and acceleration = fi x distance directed towards 0;to find the uniform circular motion of which this motion is aprojection.

Take OM = a, and erect a perpendicular MN so that

Join ON, and this will be the radiusof the required circle. For if a pointQ moves round the circle wil h an-gular velocity V/x then by 72 its AMprojection P on OM has accelerationfiPO towards 0, and its velocitywhen P is at M is '•J/iMN, which wehave made equal to the given velocity v.

We can deduce formula (5) of 7'1 thus: Measuring t fromthe instant when the particle is at M, in time t OQ turnsthrough the angle NOQ= Vfit, and, if MON= e, we have

x = OP = OQ cos {^1 fit + e)

= ON cos e cos Vytt t — ON si n e sin V/i t

= OM cos \l fit — MN sin V'fit,

therefore x = acos^fit^Vf

7"22. The geometrical representation of simple harmonicmotion is very useful in the solution of examples. The re-volving radius may be regarded as a time-keeper of the motion,since the angle through which it turns between any two posi-tions is proportional to the time taken.

7*23. It is to be noted that an equation of the form* + fix = 0,

where fi is a positive number, always represents a periodicmotion, of period 2TT/\/fi independent of the amplitude.

73. Elastic Strings and Spiral Springs. Hooke's Law.The ' extension' of a stretched elastic string means the ratio

7-21-7-31] ELASTIC STRINGS 87

of the increment in length to the unstretched length. Thus ifIt, is the natural or unstretched length and the stretched lengthis I then the extension is (I — Z0)/70*.

Hooke's Law is that the tension of the string is proportionalto the extension. If T denote the tension and we state the lawin the form

then X is called the modulus of elasticity of the string.The extension or compression of a spiral spring follows the

same law, but in this' case the length is measured along theaxis of the helix and not along the wire that forms the spring;and when the spring is extended or compressed the force exertedby the spring is a tension or a thrust in the direction of theaxis. The formula above may be used for compression as wellas extension provided we regard a negative tension as a thrust.For when the spring is compressed the length I is less than thenatural length l0, so that the formula would give a negativetension, i.e. a thrust of magnitude X(lo — l)llo.

7-31. Work done in Stretching an Elastic String. Let OArepresent the natural length of a string. Suppose the end 0 tobe fixed and the string stretched by moving the end A.

To find the work done as the length is increased from OBto 00.

c

Let P be a point between B and G and at every such pointerect an ordinate Pp to represent the tension. Since the tensionis proportional to the extension, therefore the ratio Pp: APis constant. Hence the upper ends of the ordinates lie on astraight line Abpc; and this line is a 'force-space curve,' 4-43.

* The above is the proper mathematical meaning of the word extension, it ishowever frequently used to denote increment of lenyth.

HARMONIC MOTION [VII

In order to stretch the string the experimenter has to apply tothe end A a force equal to the tension; the line BPG is the'space' through which the point of application of the graduallyincreasing force is moved, and the area under the 'force-spacecurve' represents the work done. Therefore the work done inextending the string from a length OB to a length OG ismeasured by the area

BCcb = i(Bb + Gc) BG— increase in length x mean of initial and final tensions.

7-32. The same result may be obtained by using Hooke'sformula for tension. Thus if T be the tension when the lengthis x, we take T= \ (x — lo)llo.

Also the work done when a force T moves its point of applica-tion through a distance dx in its line of action is Tdx; thereforein increasing the length from I to I' the work done is

v x 'V

To. (x — la) dx

= increase in length x mean of initial and final tensions.

74. The motion of a heavy particle suspended from a fixedpoint by an elastic string or a spiral spring is simple harmonicmotion, because the tension of the string causes an accelerationproportional to the distance from a fixedpoint and directed towards it.

Let A be the fixed point and AB theunstretched length of the string. Let m bethe mass of the particle and let the weightof the particle be such as to extend thestring to the length AO when the particlehangs at rest.

Let AB = l0, AO=l and measure x verti-cally downwards from 0. In the equilibriumposition the weight is equal to t'he tension, so that

(1).

7-.31-7 4] ELASTIC STRINGS 89

Let the particle be in another position P where OP = x; itthen has a downward acceleration given by

mx = downward force

= mg-T (2);

where T is the tension when the length is AP=l+x, sothat

lo)/lo (3).Using (1) and (3), equation (2) becomes

mx = — Xx/l0 (4).

This equation represents a simple harmonic motion of period

Up to this point it has not been necessary to specify in whatway the motion began. Equation (4) is independent of initialconditions.

As a particular case let us suppose that the particle is pulleddownwards to a point G, where OC = a and then set free. IfOC*;, OB the equation (4) will apply to the whole motion, whichwill be a purely oscillatory motion about 0 as centre througha distance a below and above 0. But if OG > OB, the stringwill become slack when the ascending particle passes B. Drawa circle of centre 0 and radius OG. Equation (4) can be writtenx = — (o2x where a = *J(\/ml0); and the simple harmonic motionis the projection on the vertical of the motion of a point de-scribing the circle with angular velocity m. This motion lastsuntil the particle reaches B, and the time from C to B

= the angle COD/a

= — I 7T — COS"1 — .

o) \ a /At B the velocity is v = a>BD = a> \J{a? — (I — lof}, and the

particle then moves freely upwards under gravity, returningto B in time 2vjg. Hence the whole time of ascent anddescent

= — \/{az — {I —10)2} H— \ir— cos"1

where m = >J(\/mlo).

90 HARMONIC MOTION [vn

75. Further Applications. In the working of examples onelastic strings or springs it is not necessary that the data shouldinclude the natural length and the modulus of elasticity pro-vided that the ratio of an additional force to the additionallength it produces is given.

Example , (i) A body of mass 5 lb. is hung on a light spring and isfound to stretch it 6 ins. The mass is then pulled down a further 2 ins. andreleased. Find the period of the oscillations and the kinetic energy of themass as it passes through its equilibrium position.

Since a stretch of "5 ft. is due to a force of 5 lb. weight therefore astretch of x ft. would correspond to a tension of ICte lb. weight.

In the equilibrium position the weight is balanced by the tension, andwhen the body is at a distance x ft. below the equilibrium position thereis therefore an extra tension of \Qxg poundals acting upwards on the body;and it is this extra unbalanced tension which causes acceleration, so that

5i?= — lOxg,

or x=- 64%.

Hence the period = 2n-/x/64 = |n- =-785 sec.Again the velocity is greatest as the body passes through the equilibrium

position and is then a> x amplitude, where a>2=64 and amplitude = 2 ins.=Jft.

Therefore the velocity i> = 4/3 f.s., and the kinetic energy

\mvi=|x 5 x Jge- absolute units = ^ x ^ g = ^- foot pound.

(ii) A cage of mass Mlb. is being pulled up with uniform velocity u bya long steel cable when the upper end of the cable is suddenly fixed. Havinggiven that a weight of m lb. would extend the cable 1 ft., shew that theamplitude of the oscillation of the cage is u*J(M/mg).

If the cage were hanging from the cable at rest, the cable, thoughextended by the weight of the cage, would have a definite length whichwe may call the equilibrium length, and such a position of the cage maybe called an equilibrium position. When the upward motion is uniform,since there is no acceleration, the length of the cable will remain theequilibrium length and the position of the cage relative to the upper endof the cable will still be the equilibrium position.

We assume that any vertical displacement of the cage which extends orcompresses the cable results in an extra tension or thrust proportional tothe extension or compression. Hence, after the fixing of the upper end ofthe cable, the motion of the cage becomes a simple harmonic motion aboutits equilibrium position, starting from this position with velocity u.

Since a weight of m lb. would extend the cable 1 ft., therefore adisplacement of the cage through x ft. from the equilibrium position

7-5] ELASTIC STRINGS 91

would be opposed by an unbalanced force of mx pounds weight, sothat

Mx = — mgx,

or x= —a>2x,

where o^ = mgjM.But if a is the amplitude of the oscillation the velocity at the centre of

the harmonic motion is aa>.Therefore aa>=u ; but o) = »

so that a=Un

(iii) A heavy particle is supported in equilibrium by two equal elasticstrings with their other ends attached to two points in a horizontal planeand each inclined at an angle of 60° to the vertical. The modulus ofelasticity is such that when the particle is suspended from any portion of thestring its extension is equal to its natural length. The particle is displacedvertically a small distance and then released. Prove that the period of itssmall oscillations is 27r\/2l/5g, where I is the stretched length of either stringin equilibrium. [S. 1923]

Let m be the mass of the particle and X the modulus of elasticity. Thenby supposing the particle to besuspended from any portion of thestring, since the extended length isdouble the natural length we findthat \=mg.

If l0 be the natural length ofeither string, we have, in the equi-librium position,

but \ = mg, therefore lo = ^l.

Let x denote the vertical displacement and y the length of either stringat time t. To find the period of small oscillations we want to obtain anequation of the form

X=-flX,

where fi is a constant. It will therefore be sufficient for our purpose towrite down the equation of motion at time t and neglect all powers of xhigher than the first.

We ha ve mx—mg — 2X -j—°- cos OP A,'o

where P is the particle at time t, 0 is its equilibrium position andPA = PB—y are the strings.

92 HARMONIC MOTION [VII

Now f=>

therefore

correct to the first power of x, and

cos

to the first power of x.„ .. 2\(l+$x-M) 1 / , SxHence mv = mg ^ S {l

therefore x=g-g

5 qx

which represents a simple harmonic motion of period 2n- \jzljbg.

(iv) A warship is firing at a target 3000 yards away dead on the beam,and is rolling {simple harmonic motion) through an angle of 3° on eitherside of the vertical in a complete period of 16 sees. A gun is fired during aroll 2 sees, after the ship passes the vertical. The gun was correctly aimedat the moment of firing, but the shell does not leave the barrel till 0'03 sec.later. Shew that the shell loill miss the centre of the target by about 4 feet.

[S. 1923]

Let 8 denote the angle turned through by the ship in t seconds afterpassing the vertical. Then the change in 8 is simple harmonic, so that itis connected with t by an equation

8=-n28 (1),where 27r/ra=the complete period = 16 sees.

The complete solution of (1) is8 = A sin nt + B cos nt,

but 8 vanishes for i = 0, therefore B=0, and8 = A sin nt.

Also A is the amplitude of the oscillation, i.e. an angle of 3° or»r/60 radians.

The angular velocity is therefore given by8 = nA cos nt;

and 2 sees, after passing the vertical the value of this is

or taking TT2 as equal to 10, the angular velocity of the ship at the instantof firing the gun is 1/48 ^/2 radians per sec.

7-0-7-61] SIMPLE PENDULUM 93

It follows that during the 003 sec. before the shell leaves the barrelthe gun receives an additional angular elevation 5 = '03/48^/2 radians.

Now if F be the velocity and a the angular elevation of projection,(F2sin2o)/0=9OOO feet;

and, if A denote the additional range when the elevation is o + 8,

F 2 sin 2 (a + t)lc/ = 9000 + A ;

or F 2 (sin 2a + 28 cos 2a)Iff = 9000 + A,

so that A = 180008 cot 2o.

Also the shell will pass over the centre of the target at a heightA tan (a+ 8) approximately; and if we neglect the square of the angularelevation, this is equal to 90008, and substituting the value found for 8this gives 4 feet as the approximate result.

7-6. Simple Pendulum. The simple pendulum consists of aheavy particle suspended from a fixed point by a fine threadmoving in a vertical plane.

Let m be the mass of the particle and I the length of thethread. If at time t the thread makes an 0

angle 6 with the vertical the accelerationof the particle along the tangent to itspath is Id. The forces acting on the par-ticle are its weight mg and the tension ofthe thread and the former alone has acomponent along the tangent. Hence byresolving along the tangent we have

mid = — mg sin 6. 'If the oscillations are so small that we may put 0 instead of

sin 6, the equation becomesW = -gd.

This equation represents a harmonic oscillation of period

A " seconds pendulum" is one of which a single swing orhalf-period occupies one second, so that the length of theseconds pendulum is given by ir >J{ljg) = l, or l=g/ir2.

7-61. Equivalent Simple Pendulum. Any simple harmonicmotion may be compared with the motion of a simple pendulumand such motions may be regarded as equivalent if their periodsare the same.

94 HARMONIC MOTION [VII

Thus, if we compare the equations

ic = — fix and 18 = — g6,

they represent equivalent motions if /A = g/l; and we maytherefore say with regard to the motion x = — fiai that the' length of the equivalent simple pendulum' is g//i.

E x a m p l e . A particle of mass m is attached to the middle point of astring of length 26 which is tightly stretched between two fixed points. To

fmd the length of the equivalent simple pendulum, when the particle is dis-placed at right angles to the string, neglecting the force of gravity in com-parison with the tension of the string.

If x denotes the displacement of the particle, the increase in length ofthe string is 2 {v/(62 + x2) — 6} and, if we neglect all powers of x above thefirst, this is zero. Therefore the tension T remains constant and byresolving at right angles to the string we get

or, neglecting x2, mx=-2Txjb.

Comparing this equation with l6=—gd, we see that the length of theequivalent simple pendulum is given by

I mbg = M"

or l=mgbl2T.

7-62. As an illustration of the finite oscillations of a simplependulum we will take the following:

Example . A particle moving along the axis of x has an accelerationXx towards the origin, where X is a positive function of x which is unchangedwhen —x is put for x. The periodic time, when the particle vibrates betweenx=-a and x = a, is T. Shew that

where Xx, X2 are the greatest and least values of X within the range x= — ato x=a.

Shew that, when a simple pendulum of length I vibrates through 30° oneither side of the vertical, T lies between

tin \fljg and 2JT tjljg x yV/3-

7-61-7-62] FINITE OSCILLATIONS 95

We have x — — Xx, throughout the motion, therefore

2xx = — 2Xxx.

Integrate this, remembering that the velocity x vanishes when x=a, andwe get

x2=-2 [xxdx = 2 ("Xxdx.J a J x

Therefore dt = —j-—j 7 > ^n e negative sign since we have supposed

the motion to begin from x = a, so that x is decreasing as t increases. Theperiod is four times the time from x = a to x=0, therefore

dx

'£-7Now X is a positive function, and

Xl>X>Xi,

/

"a fa fa

X1xdx>2 I Xxdx > 2 I X2xdx;X J X J X

and Xi, Xi are constants, so that the last line givesXj (a2 - x2) > 2 ["Xxdx > X2 (a2 - X*).

J xHence

dx <i(a dx

but 4 / — ^ - = 4 sin->- =2ir,7o v/(a2-^2) L aJo

therefore 2n-/\/Zi < T7 < 2TT/\/T2.

Again the equation of motion of the simple pendulum

16=

may be written 6 = — j . . 6,

and sin 6\6 is positive and unchanged when — 6 is put for 6, so that thelast result may be applied. Also as 6 increases from 0 to n/6, sin#/0decreases from 1 to 3/TT, and therefore the greatest and least values of

"- —j— are 2 and f - . Hence the period lies between

2TT \lljg and 2?r V% x \lnjZ.

96 HARMONIC MOTION [VII

7-7. Disturbed Simple Harmonic Motion. Suppose thatin addition to the force varying as the distance there is a forcein the line of motion producing a constant acceleration f inthe positive direction of the axis of x. Then

or x = — fi (% —

If we put x — x' +f/fi, which is equivalent to moving theorigin to the point of equilibrium x =fjfi, the equation becomes

x = — fix'.

This shews that the effect of the constant force is merely tomove the centre of the simple harmonic motion a definitedistance f/fi in the direction of the constant force; i.e. thecentre of the simple harmonic motion is the point of equi-librium, or the point where the variable force is balanced by theconstant force. We have had applications of this in problemsof masses suspended by elastic strings or springs (7*4, 7"5)where the constant force of gravity causes the extension of thestring or spring and the oscillations take place about anequilibrium position.

7-71. Periodic Disturbing Force. Forced Oscillations.The equation id = — fix +fcos pt (1)

represents harmonic motion disturbed by a periodic force pro-portional to cos pt.

To find a particular integral of this equation, substitutex = G cos pt

and we get — Cp^cospt = — /iG cos pt +f cos pt;

so that x = G cos pt is a solution provided that

is a particular integral. The complete solution is found byadding to the particular integral the complete solution of

x + fix = 0,which by 1-7 (16) or by 7-1 (2) is

A sin Vfit + B cos "Jfit,

so that * = A sin Vui + -Bcos"/ui+- —• (2).fip2

7-7-7-72] FORCED OSCILLATIONS 97

The two parts of the solution represent what are called the'free' and the 'forced' oscillations, the latter being of the sameperiod 2w/p as the disturbing force. Either free or forced oscil-lations can exist independently and the actual motion may be acombination of both, depending on the initial circumstances.

When _pa is nearly equal to p, i.e. when the forced and thefree oscillations have nearly the same period, the amplitude//(/A—jo2) of the forced oscillations is large. When p2 = i^,equation (1) no longer has a solution of the form x= C cos pt;but if we substitute % = Ct sin pt in (1) we get

C(—p2t sin pt + 2p cos pt) = — fiCt sin pt + / cos pt,which gives C=//2p, so that the forced oscillation is now

represented by x = J- t sin pt; shewing that the amplitude of

the forced oscillation increases continuously with t.The phenomenon of the large amplitude of vibration of a

body forced to vibrate with a period nearly equal to that of itsfree vibrations is a familiar one in the theory of sound and ofelectromagnetism and is known as resonance. The reason whytroops are ordered to fall out of step in crossing a bridge is lestby forcing upon the structure vibrations of period nearly thesame as that of a free vibration, displacements of large ampli-tude should be set up and cause the bridge to break.

7*72. Example . Consider again the problem of 7*61, supposing thatone end of the string is fixed while the other end has a periodic motion ofsmall amplitude at right angles to the string such that the displacement attime t is given by %=asinpt.

As before, if x denotes the displacement of the particle, and we neglectx" and a2, we can shew thatthe tension T remains con-stant. Resolving in the di-rection of x, we have, for themotion of the particle,

or, putting %T\mb = %

The solution is, as

X

R D

i2, we getx= -n2x+]

in 7-71,

= Asmnt + Bcoa

tn2 a sin pt.

nt , i n2a8vnpt

7

98 HARMONIC MOTION [VII

Suppose that when t = 0, x = 0 and x = 0; the first of these conditions

requires that 5 = 0 , and the second makes 0 — nA+i " „• In this case

therefore the solution (3) takes the form

x — \ ~jj 2

and represents a combination of forced and free oscillations.If however we suppose that when t = 0, x=0 and that the velocity x is

then i 4 - ^ 5 . w e find t h a t b o t h A a n d 5 a r e ze r0 ' s o t h a t w i t h t h i s s e t of

initial conditions (3) becomes, n2a sia.pt

shewing that now there are no 'free' oscillations but the 'forced' oscillationsalone.

78. Damped Harmonic Oscillations. When a particlemoves in a straight line under the action of a force towards afixed point in the line, varying as the distance, and the motionis resisted by a force proportional to the velocity, we may writethe equation of motion

x = — ixx — 2ktb,

or x + 2kx + fix = 0 (1).

This equation is solved in 1*7. There are three cases.If /A < £2, the solution is of the form

x = Ae-^"1^-^ * + .Be-*'-*"*2-*)« 17 (10),

and the motion is non-oscillatory.If fi = k2, the solution is of the form

<c = e-*t(A+Bt) 1-7(11),which again does not represent oscillations; but if //, > k2, thesolution may be written

x = e~M(A cosnt + Bsinnt) 17(12),

where n2 = fi — k2,

or x = Ge~u cos (nt + e),

where G and e are arbitrary constants.This last result represents harmonic oscillations with ampli-

tude Ce~kt decreasing steadily with the time. They are calleddamped oscillations and e~kt is the damping coefficient. Theoscillations die away as t increases.

7-72—7*9] DAMPED FORCED OSCILLATIONS 99

7*9. Damped Forced Oscillations. The equation

x + 2kx + fix =f cos pt

represents damped forced oscillations. The complete solution ofthis equation is got by adding a particular integral to thesolution of equation (1) of 7'8.

To find a particular integral we substitute

x = E cos pt + F sin pt

as a trial solution, and we find that it satisfies the differentialequation if

E(—p2 cos pt — 2kp sin pt + /J, cos pt)+ F(—p2 sin pt + 2kp cos pt + /J, sin pt) = / cos pt.

Hence we must choose E and F so as to make the last equationan identity for all values of t, i.e. equate to zero the coefficientsof cos pt and sin pt. Therefore

(lx-p2)E + 2kpF-f=0,

and -2kpE+(p-p*)F=0;

E _ F _ fwhence 2 ^ (^ _ ^ 2 +

Therefore the required integral of the given differential equa-tion is

_/.( /* — P2) cos pt + 2kp sin pt

j . cos(»i + a)or x = f v^ '

u i. 2kPwhere tan a = —~—

Hence, in the case in which p > k2, the complete solution is,from 7-8,

n w / _, \ f COS (pt + 0)x = Ge~u cos (nt + e) H — — ; ,

{(2Y+4k22}t{((

where n2 = /A — k2.

The two terms represent the free and the forced oscillations,and owing to the damping the free oscillations die away andthe forced oscillations alone persist.

7-2

100 HARMONIC MOTION [ v n

EXAMPLES

1. A particle of mass m is moving in the axis of x under a central forcefimx to the origin. When t—2 seconds, it passes through the origin, andwhen t = 4 seconds, its velocity is 4 feet per second.

Determine the motion and shew that, if the complete period is 16 seconds,

the semi-amplitude of the path is feet. [S. 1926]7T

2. A mass of 1 lb. is hung on to a light spiral spring and produces astatic deflection of 1£ inches. A mass of 1 lb. is suddenly added to theoriginal mass, (i) Find the maximum elongation produced; (ii) shew thatthe time of an oscillation of the whole mass is approximately f- sec.

[S. 1911]

3. A body is suspended from a fixed point by a light elastic string ofnatural length I whose modulus of elasticity is equal to the weight of thebody and makes vertical oscillations of amplitude a. Shew that, if as thebody rises through its equilibrium position it picks up another body ofequal weight, the amplitude of the oscillation becomes (P + £a2)*.

[S. 1921]

4. A mass m hangs from a fixed point by means of a light spring, whichobeys Hooke's law, the mass being given a small vertical displacement. Ifn is the number of oscillations per second in the ensuing simple harmonicmotion, and if I is the length of the spring when the system is in equi-librium, find the natural length of the spring, and shew that, when thespring is extended to double its natural length, the tension is m (Air^rfil—g).

[S. 1925]

5. A heavy particle of mass m is attached to one end of an elastic stringof natural length a, whose other end is fixed at 0. The particle is let fallfrom rest at 0. Shew that part of the motion is simple harmonic, andthat, if the greatest depth of the particle below 0 is a cot2 8, the modulusof elasticity of the string is \mg tan2 8, and that the particle attains thisdepth in time

— {\ + (n-8)oot6},

where 8 is a positive acute angle. [M. T. 1915]

6. A horizontal board is made to perform simple harmonic oscillationshorizontally, moving to and fro through a distance 30 inches and making15 complete oscillations per minute. Find the least value of the coefficientof friction in order that a heavy body placed on the board may not slip.

[S. 1918]

EXAMPLES 101

7. An elastic string is stretched between two fixed points A and B inthe same vertical line, B being below A. Prove that, if a particle is fixedto a point P of the string and released from rest in that position, it willoscillate with simple harmonic motion of period t^/p and of amplitude pa,where t is the period and a the amplitude when P coincides with the mid-point of AB, and n = AP. PB/AB2. The string may be assumed tautthroughout. [S. 1924]

8. A string AB consists of two portions AC, CB of unequal lengths andelasticities. The composite string is stretched and held in a verticalposition with the ends A and B secured. A particle is attached to C, andthe steady displacement of C is found to be 8. Shew that a further smallvertical displacement of C will cause the particle to execute a SimpleHarmonic Motion, and that the length of the equivalent simple pendulumis 8. Both portions of the' string are assumed to be in tension throughout,and the weight of the string may be neglected. [M. T. 1920]

9. A light elastic string is stretched between two points in the samevertical line, distant I apart. The tension in the string is F. A body,whose weight is small compared with F, is attached to the mid-point ofthe string, causing it to sink a distance d. Shew that the periodic time, Tlt

of small vertical oscillations of the body is the same as that of a simplependulum of length d.

If the periodic time of small horizontal oscillations of the body is T2,shew that the mass of the body is approximately

Fd/T,

10. On a given day the depth at high water over a harbour bar is 32 ft.,and at low water 6J hours earlier it is 21 ft. If high water is due at3.20 p.m., what is the earliest time at which a ship drawing 28 ft. 6 ins.can cross the bar, assuming the rise and fall of the tide to be simpleharmonic ? [S. 1917]

11. A particle is constrained to move along a straight line, and isattracted towards a fixed point 0 in that line by a force proportional toits distance from 0. It is subjected in addition to a constant force X,acting in the same straight line away from 0, and of magnitude sufficientto hold the particle in equilibrium at another point A. Shew that themost general motion possible to the particle is a simple harmonic oscillation,of arbitrary magnitude and phase, about the point A as centre.

The particle being initially at rest at 0, the force X is applied andmaintained constant for an interval t, after which it ceases. For whatvalue of t (expressed as a fraction of the natural period T) will the particlearrive at A with zero velocity ? [M. T..1927]

102 HARMONIC MOTION [VII

12. A ship is rolling with a period of 10 sees. A man at the masthead100 feet above the deck is swung to and fro 25 feet on either side of thevertical with a motion which is approximately horizontal and simpleharmonic. The man weighs 200 lb. and his horizontal hold failing at501b. he is thrown off the mast. The width of the deck being 80 feet, provethat he falls clear of the ship.

[Assume TT2 = 1 0 and # = 32 f.s. units.] [M. T. 1913]

13. A ship is making n complete rolls a minute and the motion of themasthead h feet above sea level may be taken as a horizontal simpleharmonic motion of total extent 2a. When at a distance x from the meanposition a weight falls from the masthead. Find where it will hit thewater, and prove that the distance of this point from the ship will be amaximum when

( ) ~ * a p p r o x - [ S - 1 9 2 V ]

14. A particle when hanging in equilibrium at the end of a light elasticstring stretches it a distance a. Prove that the period of vibration of theparticle in a vertical line through its equilibrium position is the same asthat of a simple pendulum of length a.

A light endless elastic string of unstretched length 26 passes over twosmall smooth pegs on the same level distant b apart. A particle is attachedto a point on the string and when the particle is in equilibrium the stringforms the three sides of an equilateral triangle. Prove that the period ofvibration of the particle in a vertical line is the same as that of a pendulum

of length ^ 6 . [S. 1915]

15. A mass of 12 lb. hangs from a long elastic string which extends0-25 inch for every pound of load. The string and the given mass aremoving upwards in relative equilibrium with uniform velocity 2 feet persecond, when the upper end of the string is suddenly brought to rest.Find the distance through which the mass will oscillate. [S. 1918]

16. According to Hesiod the anvil of Vulcan would take 9 days and9 nights to fall from the Earth to the realms of Hades. Placing Hades atthe centre of the Earth and assuming that the acceleration downwardsvaries directly as the distance from the centre (and is 32 ft./sec.2 at theEarth's surface), shew that Hesiod's figures would give a value of about15 x 108 miles for the Earth's radius. [S. 1924]

17. Prove that if a point move in an arc of a parabola having the vertexas middle point so that the motion of the projection of the point on theaxis of the parabola is simple harmonic, then the motion of the projectionof the point on the directrix is also simple harmonic and of double theperiod. [S. 1924]

EXAMPLES 103

18. A heavy particle hangs at one end of a light elastic string which issuch that the period of a small vertical oscillation of the particle is 2irT.The string is moving vertically upwards with uniform velocity gT0 and theparticle is in relative equilibrium. Shew that, if the upper end of thestring is suddenly fixed, the string will become slack if TQ is greater thanT, and that in this case the new motion has a period

2 (ff - cos "1 T\ To) T+ 2 (T<? - T2)*. [S. 1926]

19. A ring of mass m can slide on a smooth circular wire of radius a ina horizontal plane. The ring is fastened by an elastic string to a point inthe plane of the circle at a distance c (> a) from its centre. Shew that ifthe ring makes small oscillations about its position of equilibrium the

period is 2n- \ ~\ , where X is the modulus of elasticity of the

string and I < (c - a) is its natural length. [S. 1924]

20. A particle of mass m lies upon a smooth horizontal table and isattached to three points upon the table, at the vertices of an equilateraltriangle of side 2a, by means of three strings of natural lengths I, I' and I'and of moduli X, X' and X' respectively. Shew that if the particle can restin equilibrium at the centre of the triangle, then

2a (\/l - \'IV) = (X - X') J3.

Find also the period of a small oscillation of the particle in the line ofthe string of natural length I. [S. 1926]

21. A particle of unit mass is tied by four equal elastic strings ofnatural length I and modulus of elasticity X to the corners of a square.If the particle is displaced a small distance towards one of the corners and

then set free, prove that the time of a small oscillation is n*/ .T,• A (cc-~~C)

where a is the length of the diagonal of the square and a is so muchgreater than I that the strings remain stretched. [S. 1909]

22. It is required to bring to rest a weight W which has fallen freelyfrom a height h by means of the direct pull of a rope of modulus X, oneend of which is attached to it and the other to a point at a variable heightvertically above. Find the minimum length of rope if the tension is notto exceed a given value T. Shew that with this length of rope the distancein which the weight is stopped is

23. A particle is hung at the end D of a light string CD knotted at Cto two equal light strings AC, CB fastened at points A, B at the samelevel. Find the equations of motion for small oscillations of the particle

104 HARMONIC MOTION [VII

in the vertical plane through AB, and in the vertical plane through Cperpendicular to AB, and integrate them.

If a is the depth of G below AB in equilibrium, and CD = b, shew that,when a=36, the particle may be made to describe an arc of a parabola.

[M. T. 1922]

ANSWERS

2. 4-5 ins. 4. l-gjA^nK 6. -098. 10. Oh. 56m. 57s.

11. IT. 15. 4-24 ins. 20. 2, U f) + f -22. 2\Wh/(T22TW)

Chapter VIII

MOTION UNDER CONSTRAINT

8*1. A particle may be constrained to move along a givencurve or surface, and the constraint may be one-sided, as forexample when a heavy particle slides on the inside of a sphericalsurface and is free to break contact with the surface on theinside of the sphere but cannot get outside. There will then bea normal pressure inwards exerted by the sphere on the particleso long as contact persists, and the pressure will vanish at thepoint where the particle leaves the surface. On the other handif the constraint is two-sided as when a particle moves in a finetube, or a bead moves along a wire, then the normal reactionmay vanish and change sign but the particle persists in theprescribed path.

82. Motion of a Heavy Particle on a Smooth Curve ina Vertical Plane. The motion is determined by the tangentialand normal components of acceleration. The beginner may findit useful in such problems as this to make two diagrams, oneshowing the components of acceleration multiplied by the massand the other showing the forces. It is then only necessary torealize that the two diagrams are equivalent representationsof the same vector, so that the resolved parts in any assigneddirection in the two diagrams are equal.

ymvdv

d$ V

o X

If m is the mass of the particle the forces acting on it arethe weight mg and the reaction R along the normal. The com-ponents of acceleration are vdvjds along the tangent and v2/p

106 MOTION UNDER CONSTRAINT [VIII

along the inward normal (5'11). Hence, by resolving along thetangent, we get

mvdv/ds = — mg sin i|r = — mgdy/ds,therefore, by integration,

or, if v0 is the velocity when the ordinate is y0, we have%m(v2-v0

2) = mg(y0-y) (1).This is the equation of energy and might have been written

down at once; for since the curve is smooth no work is doneby the reaction R in any displacement, so the increase in kineticenergy is equal to the work done by the weight.

Again, resolving along the normal, we getmv2/p — R — mg cos yjr (2).

Substituting for v from (1), we haveR = mg cos -*jr+ m{v0

2+ 2g(yo-y)}/p (3).Assuming that the form of the curve is given, the values of p

and -v/r at any point can be determined, and thus R is known;and if we equate to zero the value of R we shall have an equationto determine the point, if any, at which the particle leaves thecurve.

83. Motion of a Heavy Particle, placed on the outside ofa Smooth Circle in a Vertical Plane and allowed to slidedown. If the particle starts from Q at an angular distance afrom the highest point A, and Aa is the radius of the circle andv the velocity at P where theangular distance from A is 6,then, from 82 (1),

v2 = 2ga (cos a — cos 0).Also by resolving along the inward normal

mv2/a = mg cos 6 — R,where R is the outward reaction of the curve.

Therefore R = mg (3 cos 0 — 2 cos a),shewing that the pressure vanishes, and that the particle fliesoff the curve, when cos 8 = f cos a.

8-2-8-31] MOTION IN A VERTICAL CIRCLE 107

8-31. Motion in a Vertical Plane of a Heavy Particleattached by a Pine String to a Fixed Point. Suppose thatthe particle starts with velocity u from its lowest positionB. If v is the velocity at P and 0is the angle that the string makeswith the vertical, the equation ofenergy is\m(y2 — u2) — — mga(l — cos#)...(l),and by resolving along the inwardnormal

mv2/a = T— mg cos 8,where T is the tension of the string.

Therefore

T = m (3g cos 6 - 2g + u2/a) (2).

To find the height of ascent we put v = 0 in (1), and get

2gacos0 = 2ga-u2 (3),

and by putting T = 0 in (2), we find that the tension vanisheswhen

cos 0 = — u2 (4).We have the following cases:

(i) If u2 < 2ga, the string does not reach the horizontalposition and the tension does not vanish.

(ii) If v? = 2ga, the string just reaches the horizontal position,the tension vanishes for 0 = ^ IT, and the particle swings througha quadrant on each side of the vertical.

(iii) If 2ga <u2< 5ga, we find that there is a value of 0, anobtuse angle, given by (4) smaller than that given by (3), sothat the string becomes slack before the velocity vanishes andthe particle will fall away from the circular path and move ina parabola till the string again becomes taut.

(iv) If u2 = bga, the tension just vanishes in the highestposition, but v does not vanish, so that circular motion persists.

(v) If w2 > 5ga, neither v nor T vanish.This is an example of a one-sided constraint; if instead of the

problem of a particle attached to a string, we consider that of

108 MOTION UNDER CONSTRAINT [VIII

a bead sliding on a wire, we find that if u2 = 4<ga the bead willreach the highest point of the wire and for any greater valueof u it will describe the complete circle.

8'4. Cyeloidal Motion. A cycloid is a curve traced out bya point on the circumference ofa circle as the circle rolls alonga straight line. Let P be thepoint on the circumference, Gthe centre of the circle, AGDthe line on which it rolls, 0 thepoint of contact, i.e. the instan- ^taneous centre of rotation. PO ° T xis therefore normal to the path of P and the tangent PT passesthrough the other end of the diameter through G. Let A bethe position of P when it is on the given line, then since thecircle rolls we have A G = arc GP, and if PGP' is a diameterand GD = arc GP', then P' will coincide with D when the circlehas turned through an angle ir, and P will then be at 0, whichis called the vertex of the curve, the line AD being called thebase. As the rolling proceeds the curve is repeated with acusp at A and wherever P reaches the fixed line.

Take axes Ox parallel to DA and Oy along 0D, and let x, ybe the coordinates of P and a the radius of the circle. Let theangle PTx — yjr. Then the angle PGT in the alternate segmentis also T|T and PGT is 2ifr, so that

x = 0T+GP sin 2f = a (2f + sin

and y = GT-GP cos 2f = a(l- cos 2i|r).

These are the ' parametric' equations of the cycloid. Forthe intrinsic equation, if s denote the arc OP, we have

= 4>a2 {(1 + cos 2i/r f + sin2

= 8a2 (1 + cos 2f) difr2 = 16a2 cos2fdf2,

so that ds = 4a cos

and by integration s = 4a sin yjr (]);

8-31-8-4] CYCLOIDAL MOTION 109

and this is the intrinsic equation, no constant of integrationbeing required since s and yfr vanish together.

We notice that the radius of curvature

p = dsjd^r = 4a cos ty = 2PG,

so that if PG be produced to Q sothat GQ = PG, then Q is the centreof curvature.

Also if we draw an equal circleto pass through Q and touch AGat G, it is easily seen that the locusof centres of curvature or evoluteof the cycloid could be constructedby rolling this second circle along aparallel line EB. A cusp of onecycloid corresponding to a vertex of -the other and vice versa.

Now let us consider the motion of a particle under gravityon a smooth cycloid in a vertical plane with its base horizontaland vertex downwards.

Resolving along the tangent we have ms = — mg sin yfr; but

s = 4<z sin y}r,

therefore s = — gsj^a. R '

This equation represents a peri-odic motion, the time of oscilla- ^y? irmgtion being 2v »J(4<a/g). It follows ^that the particle oscillates on °either side of the vertex with a period that is independent ofthe amplitude. This property is called the ' isochronism of thecycloid.'

We recall that the formula 2ir *J(l/g) obtained for the periodof vibration of a simple pendulum depended on the amplitudebeing so small that 6 can be used for sin 6, and we now seethat if the ' bob' of the pendulum could be made to move ina cycloidal path the formula for the period of vibration wouldbe independent of the amplitude. This can be attained bymaking the string that supports the bob wrap and unwrapitself on cycloidal arcs.

110 MOTION UNDER CONSTRAINT [VIXI

Thus in the second figure, from (1)

the arc AQ = 4a sin QGD

= QP,while arc AB = 4a.

Hence if a string of length 4a had one end fastened at Band were wrapt round the curves BA and BA' alternately, the

other end P would trace out the cycloid AOA', and a particleattached to P moving under gravity and the tension of thestring would oscillate in the time 2TT V(4a/(?) whatever be theamplitude of the oscillation within the limits AOA'.

E X A M P L E S

1. A heavy particle of weight W, attached to a fixed point by a lightinextensible string, describes a circle in a vertical plane. The tension inthe string has the values m W and n W, respectively, when the particle isat the highest and lowest points in its path.

Shew that n = m + 6. [M. T. 1927]

2. A heavy particle slides under gravity down the inside of a smoothcircular tube held in a vertical plane. The particle starts at the highestpoint with the velocity it would acquire if it fell down the radius, provethat when in the subsequent motion the vertical component of its accele-ration is a maximum the pressure on the curve is equal to twice the weightof the particle. [S. 1919]

8-4] EXAMPLES 111

3. A heavy particle, hanging from a fixed point by a light inextensiblestring of length a, is projected horizontally with velocity V. Shew thatduring the circular motion the tension of the string at any time isproportional to the depth of the particle at that moment below a certainhorizontal line ; and find the values between which V must lie that thestring may become slack. [M. T. 1916]

4. A particle is projected along the inner side of a smooth circle ofradius a, the velocity at the lowest point being u. Shew that if u2 < bgathe particle will leave the circle before arriving at the highest point andwill describe a parabola whose latus rectum is 2 (u2 — 2ga)sj27gsa2.

[S. 1916]

5. A particle slides down the surface of a smooth fixed sphere of radiusa, being slightly displaced from rest at the highest point. Find where itwill leave the sphere, and shew that it will afterwards describe a parabolaof latus rectum ^f a, and that it will strike the horizontal plane through

the lowest point of the sphere at a distance —— from the27

vertical diameter. [S. 1918]6. Two beads connected by a string are held at rest on a vertical

circular wire with the string horizontal, and above the centre. Theirmasses are m, m', and the string subtends an angle 2a at the centre. Ifthe beads are released, shew that the tension of the string when it makesan angle 8 with the horizontal is

t a n a c o s ^ r g ? ,L • Jm+m' '

7. Two equal particles are tied together by a light string of length 7ra/2and rest in equilibrium on the surface of a smooth circular cylinder in aplane perpendicular to the axis of the cylinder, which is horizontal, theradius of the cylinder being a. The particles are then slightly displacedin a plane perpendicular to the axis of the cylinder. Prove that when thelower of the two particles leaves the surface, the pressure on the other israther more than three-fourths of its weight. [S. 1912]

8. A particle starts from rest at any point P in the arc of a smoothcycloid whose axis is vertical and vertex A downwards; prove that the

time of descent to the vertex is n- » / - , where a is the radius of the

generating circle.

Shew also that if the particle is projected from P downwards along thecurve with velocity equal to that with which it reaches A when startingfrom rest at P, it will now reach A in half the time taken in the precedingcase. [S. 1915]

112 MOTION UNDER CONSTRAINT [VIII

9. A particle is constrained to move under gravity on a parabola withaxis vertical and vertex upwards. Shew that the pressure on the curve isnumerically m{u^—gl)jp, where u0 is the velocity at the vertex, %l thelatus rectum and p the radius of curvature at any point.

What is the meaning of the special case u^=gl ? [M. T. 1927]

10. A heavy particle P slides on a smooth curve of any form in avertical plane. The centre of curvature at P is Q, and R is on the samevertical as Q and at the level of zero velocity. Shew that the accelerationmakes with the normal an angle tan"1 (J tan PRQ). [S. 1912]

11. A smooth wire is bent into the form y=s in# and placed in avertical plane with the axis of x horizontal. A bead of mass m slides downthe wire starting from rest at x = j n. Shew that the pressure on the wireas the bead passes through the origin is mg/^2, and find the pressure asit passes through x= —\TT. [S. 1921]

12. A bead moves on a smooth circular wire under the action of forcestending to the corners of a regular polygon concentric with the circle.The forces vary as the distance and are equal at equal distances. Provethat the pressure on the wire is constant. [S. 1901]

13. A particle of mass m is attached by a string to a point on thecircumference of a fixed circular cylinder of radius a whose axis is vertical,the string being initially horizontal and tangential to the cylinder. Theparticle is projected with velocity v at right angles to the string along asmooth horizontal plane so that the string winds itself round the cylinder.

Shew (i) that the velocity of the particle is constant,(ii) that the tension in the string is inversely proportional to the

length which remains straight at any moment,

(iii) that if the initial length of the string is I and the greatest tensionthe string can bear is T, the string will break when it has turned throughan angle

l/a-mv^/aT. fS. 1927]

14. A switchback railway consists of straight stretches smoothly joinedby circular arcs, the whole lying in a vertical plane. Shew that a carstarted on a level stretch and running freely will leave the track if thedownward gradient exceed cos"1! at any point; but that if braking isavailable up to half the weight of the car, gradients of about 77° areadmissible.

A level and a straight descending stretch are smoothly joined by an arcof radius a. Two equal cars without brakes are joined by a cable oflength 2a. Shew that the greatest admissible gradient such that thesecond car does not leave the track is cos"1 ^f.

[Neglect the size of the cars, and resistance to motion other thanbraking.] [S. 1924]

EXAMPLES 113

15. A small ring fits loosely on a rough spoke (length a) of a wheelwhich can turn about a horizontal axle and the ring is originally at restin contact with the lowest point of the rim : if the wheel is now made torevolve with uniform angular velocity a, prove that the angle 6 throughwhich the wheel will turn before the ring slides is given by the equation

cos(#—A)/cosX + (B2a/<7 = 0,

where X is the angle of friction. [S. 1910]

ANSWER11.

Chapter IX

THE LAW OF REACTION. GENERALPRINCIPLES

91. So far we have been concerned with the motion of asingle particle. When two or more particles are moving in sucha manner that the motion of any one is affected by the presenceof the others we have to make use of another law enunciatedby Newton, viz. Action and Reaction are equal and opposite,or, the actions of two bodies on one another are always equal andopposite. In explicit terms this means that if a body A exertsa force F on a body B, then B exerts an equal force F on A butin the opposite direction. Consequently the momentum com-municated to A by the action of B is equal and opposite to themomentum communicated to B by the action of A.

Consider the case of a system of bodies, attracting or repellingeach other or acting on one another by contact, or through con-nections by means of strings or rods, either for a finite time orby instantaneous impulses. In this case any momentum whichis produced or destroyed in any assigned direction is accom-panied by the production or destruction of an equal momentumin the opposite direction.

Hence it follows that, if no external forces act on a system ofbodies, the total •momentum, of the system, in any assigned directionremains constant.

This is the principle of conservation of linear momentum.

92. Motion of a system of particles. Let (xlt y{), (00%, y2),etc. be the coordinates of a system of particles of masses m1( m-ir

etc. Let the particles be subject to given external forces whosecomponents parallel to the axes are

X1; F,, Xt, Y2, etc.and also to internal actions and reactions due to the mutualactions of the particles upon one another of which the com-ponents on m1, w2, etc. are

Zi', Y{, X,', F2', etc.

9'l-9'2] CONSERVATION OF LINEAR MOMENTUM

Writing down the equations of motion for the separateparticles, we have

(1).m2y2=Y2+Y2'etc.

" l

etc.Whence by addition

2and

Now by the law of reaction the internal actions and reactionsare equal and opposite in pairs, so that the sums of their re-solved parts in any direction must vanish, i.e. 2X' = 0 and2 F ' = 0.

Hence equations (2) reduce toXmS = XX, 2my = 2 F (3),

or, in words, the rate of change of the linear momentum of thewhole system in any prescribed direction is equal to the sum ofthe resolved parts of the external forces in that direction.

It follows that if there is a direction, say the axis of a, inwhich the sum of the resolved parts of the external forces iszero, i.e. 2X = 0; then, by integrating Xmx = 0 we get

2 mx = const.,i.e. the linear momentum in that direction is constant. This isagain the principle of conservation of linear momentum.

Again, from equations (1), by multiplying each y equationby the corresponding x and each x equation by the correspondingy and subtracting, we deduce that

mx (xxyx - ?/i

«l2 (#2^2 — Vi

whence by additionXm (xy - yx)

= %(xY-yX)-Now the moment

«i) = *iFt-yiZ14x.i) = x2Y2-y2Xi +

etc., etc.,

ht(xY'-yX').about the origin

of a vector whose components are X, Ylocated at the pointhence %{xY'-yX')

(x,y) is xY-yX;is the sum of the

•xxYx -yxXj >

x2Y2' - y2X2',

y *

X

i

i

o X

116 THE LAW OF REACTION. GENERAL PRINCIPLES [iX

moments about the origin of the internal actions and reactions,which are equal and opposite in pairs. Therefore

2 O F ' - 2 / X ' ) = 0,and we have

tm(xy-yx)=^(xY-yX) (4).

This may also be written

j t tm (xy ~yx)=t(xY- yX),

or, in words, the rate of change of moment of momentum of thesystem about any fixed origin (or axis) is equal to the sum of themoments of the external forces about that origin (or axis).

If the sum of the moments of the external forces about anyfixed axis is zero, it follows that the moment of momentum of thesystem about that axis is constant. This is the principle ofconservation of moment of momentum.

Moment of momentum is frequently called angular mo-mentum.

9'21. Effective Forces. The product of the mass and theacceleration of a particle is called the effective force of theparticle, and the principles embodied in equations (3) and (4)of the last article are that for any system of particles theeffective forces are the exact equivalent of the external forcesacting on the system.

93. Motion of the Centre of Gravity. Independence ofTranslation and Rotation. If M be the whole mass of thesystem of particles and x, y the coordinates of the centre ofgravity G, the usual formulae for the position of the centre ofgravity are

Mx = "tmx, My = %my (1).

By differentiating these equations we get

Mx = "Zmx, My = %my,shewing that the linear momentum of the system is the sameas that of a particle whose mass is the whole mass, movingwith the velocity of the centre of gravity.

9'2-9'3] INDEPENDENCE OF TRANSLATION AND ROTATION 117

A second differentiation givesMx = limx, My = %my,

therefore from (3) of 9-2M = 2X, My =2,7, (2),

and these equations shew that the motion of the centre of gravity0 is the same as if all the mass were collected into a particle atG and all the external forces were moved parallel to themselvesto act at G.

It follows that if the system is not acted upon by externalforces its centre of gravity is either at rest or moving withconstant velocity, for in this case the integration of equation (2)gives Mx = const., My — const.

Again in equation (4) of 9'2 we putX=x+x, y = y+y',

so that x, y denote coordinates of the particle of mass mrelative to axes through the centre of gravity G; then, from (1),

2ma/ = 2,my = 0,so that by differentiation 2m»' = "%my = 0, and (4) of 92becomes

2m {(x + x') (y + y) - (y,+ y') (^ + x))

Multiplying out and observing that such terms as

2/rrixy' = *2my' = 0,there remains

M {xy - yx) + 2m (xtf - y'x') = *2 Y - ylX + 2 (x' Y - y'X).

In virtue of (2) this equation reduces to

2,mW-y'x') = 2,(af7-i/X) (3),shewing that, in the motion of the system relative to the centre ofgravity, the rate of change of moment of momentum about thecentre of gravity is equal to the sum of the moments of theexternal forces about the centre of gravity.

The results of this article shew that the motion of the centreof gravity and of the system relative to the centre of gravity areindependent of one another and this constitutes the principleof the independence of translation and rotation.

118 THE LAW OF REACTION. GENERAL PRINCIPLES [iX

94. Conservation of Energy. Reverting to equations (1)of 9'2, by multiplying each equation by the correspondingvelocity component and adding, we get

Xm (xx + yy) = t {(X + X') A + (Y+ Y')y],

or | {

The left-hand side is the rate of increase of the kinetic energyof the system, and the right-hand side is the rate at which allthe forces external and internal are doing work. From this weconclude that the increase of kinetic energy in any time isequal to the whole work done. In all cases in which the potentialenergy depends on the configuration of the system and in whichthe change of potential energy due to a change of configurationis independent of the manner in which that change is made,the work done will be equal to the loss of potential energy.Whence it follows that

kinetic energy + potential energy = constant.

Kinetic energy may be created by explosions, in which casethe visible kinetic energy together with the heat developed arethe equivalent of the chemical energy stored in the explosivesubstance. Similarly kinetic energy may be dissipated byfriction when it is converted into heat.

95. Kinetic Energy in reference to Centre of Gravity.The kinetic energy of a system of particles is equal to the kineticenergy of the whole mass moving with the velocity of the centre ofgravity together with the kinetic energy of the particles in theirmotion relative to the centre of gravity.

Using the notation of 9-3, we have that the kinetic energy

= \M (x2 + f) + xXmx + ytmy + \ Xm (x2 + y'2).

But %mx' = Xmy' = 0, so that Xmx' = %my = 0, and thekinetic energy = £ M (x2 + y2) + $ £m (a/2 + y'2), which proves theproposition.

94-9-8] RIGID BODIES 119

9 6. In the last few articles we have distinguished betweenexternal and internal forces. Whether any particular force isto be classed as external or internal depends on the way inwhich we regard the particles composing the system. Thus ifwe are considering the motion of a single particle A, then allthe forces acting upon it are classed as ' external forces,' but ifwe are considering a system composed of two particles A andB then the forces exerted by B on A and A on B are internalforces. The force of gravity is an external force in consideringthe motion of a body relative to the earth, but in consideringthe motion of the earth and moon regarded as one systemmoving about the sun the gravitational pull of the earth onthe moon or the moon on the earth is an internal force.

97. Rigid Bodies. A rigid body is considered to be anaggregation of particles bound together by forces of cohesionand internal mutual attractions which are in all cases equal andopposite. The results obtained in this chapter for a system ofparticles are therefore true for a rigid body. Consequently themotion of the centre of gravity of such a body does not dependupon its size or shape but only upon its mass and the resultantof the external forces acting upon it. It is only when we areconcerned with rotational motion relative to the centre ofgravity that the size and shape come into consideration, forupon them depend the moment of momentum and the momentsof the forces.

9"8. In the chapter on rectilinear motion when we appliedthe second law of motion to rigid bodies or cars and trainswe considered every particle of the body to be moving withthe same acceleration, and any assumptions there made whichmight seem to need further justification find it in the principlesestablished in 93 of this chapter. Moreover many problemsabout rigid bodies not involving rotational properties may besolved by elementary methods in virtue of the principle thatthe motion of the centre of gravity 0 is the same as if all themass were collected into a particle at Q and all the externalforces were moved parallel to themselves to act at G.

120 THE LAW OF REACTION. GENERAL PRINCIPLES [IX

As an example let us find the condition that a car may upset whenrounding a curve.

Let v be the velocity and m the mass.We are concerned with the forces inplanes at right angles to the directionof motion.

If r is the radius of curvature of thecurve described by the centre of gravityO, then G has an acceleration v2/r inwardsalong the normal to its path. Thereforethe resultant of all the forces perpendicular to the direction of motionmust be mv2jr. But the only force in this direction is friction on thewheels. So if F is the total friction across the track we have F=mv2/r.Let R, S be the upward pressure of the ground on outer and inner wheels,h the height of O and 2a the distance between the wheels.

Then by resolving vertically, R + S=mg, and by taking moments aboutG (since there is no moment of momentum about G) we get

{R-S)a=Fh,

therefore R - S=mv2h/ra.

Hence 2R = mg+mv2h/ra,

and 2S=mg — mv2h/ra.

We conclude that if v2 > grajh the car will overturn outwards, since thiscondition implies a negative pressure of the inner wheels on the groundwhich is an impossibility.

9'81. The necessity for sideways friction F can be obviated by bankingup the track. If the track bebanked up to make an angle 6with the horizontal and thereis no sideways friction (i.e. notendency to side-slip), we get,by resolving horizontally andvertically,

and

) sin (9,

= (R+S) cos 6.

Therefore v2 = rg tan 6 gives the velocity at which tendency to side-slipis eliminated for a track of slope 6.

EXAMPLES 121

E X A M P L E S

1. The wheel axles of a motor car are 4 feet long and the height of thec. a. is 2 feet. Find the speed of the car if in going round a level track of400 feet radius the inner wheels just leave the ground. [S. 1916]

2. A skater describes a circle of 40 feet radius with a velocity of 15 feetper second. At what angle must he lean inwards ? [S. 1914]

3. Prove that, in order to allow properly for a curve on a railway lineof radius 1320 feet for a train moving at 45 miles an hour, the outer railmust be raised above the inner rail by 5'8 inches. (The rails are 4 feet8 | inches apart.) [M. T. 1921]

4. A railway of gauge 5 feet is taken round a curve of \ mile radius.What ' superelevation' must be given to the outer rail in order that atrain travelling round this curve at 30 and at 60 m.p.h. may impose thesame side pressure on the inner and outer rail respectively 1 [M. T. 1928]

5. A car takes a banked corner of a racing track at a speed V, thelateral gradient a being designed to reduce the tendency to side-slip tozero for a lower speed U. Shew that the coefficient of friction necessaryto prevent side-slip for the greater speed V must be at least

( F « - V) sin a cos o/(F2 sin2 a + U2 cos2 a). [S. 1925]

6. A pile-driver weighing 200 lb. falls through 5 feet and drives a pilewhich weighs 600 lb. through a distance of 3 inches. Find the averageresistance offered to the motion of the pile, assuming that the two remainin contact after the blow.

How many foot-pounds of energy are dissipated during the blow ?[S. 1926]

7. A shell of mass (TOi-t-m2) is fired with a velocity whose horizontaland vertical components are U, V, and at the highest point in its paththe shell explodes into two fragments OTJ, OT2- The explosion produces anadditional kinetic energy E, and the fragments separate in a horizontaldirection: shew that they strike the ground at a distance apart which isequal to

I k ( ) \ [S. 1924]

8. A gun of mass M fires a shell of mass m horizontally, and the energyof the explosion is such as would be sufficient to project the shell verticallyto a height h. Shew that the velocity of recoil of the gun is

i. [S. 1927]

122 THE LAW OF REACTION. GENERAL PRINCIPLES [IX

9. A set of n trucks with s feet clear between them are inelastic and areset in motion by starting the end one with velocity V towards the next.Find how long it takes for the last truck to start and the value of thefinal velocity. [S. 1927]

10. A battleship of symmetrical form and mass 30,000 tons is movingat 10 miles per hour and fires a salvo of all its eight guns in a directionperpendicular to its motion. If the shells weigh 15 cwt. each, have amuzzle velocity of 2000 feet per second, and are fired at an elevation of30°, shew that the direction of motion immediately after firing makesan angle of about 1° 21' with that before. [S. 1922]

11. A ball is dropped from the top of a tower 100 feet high. At thesame moment a ball of equal mass is thrown from a point on the ground50 feet from the foot of the tower so as to strike the first ball when justhalf-way down. Find the initial velocity of projection of the second balland the direction of projection. If the two balls coalesce how long willthey take to reach the ground 1 [S. 1926]

12. A shell of mass 1120 lb., and velocity 1350 feet per second, is firedinto a railway truck (containing sand) of mass 20 tons, the direction ofmotion being parallel to the rails. If the shell fails to penetrate the sand,find the velocity given to the truck and account for the conservation ofenergy in the phenomenon, specifying how much remains kinetic. Howfar will the truck run against a constant retarding force of 30 lb. weightper ton ? [M. T. 1921]

13. A pile-driver weighing 2 cwt. falls through 5 feet and drives a pileweighing 6 cwt. through a distance of 4 inches. Find the average resistanceto the pile in cwt., assuming the two to remain in contact. Find in foot-pounds the energy dissipated in one stroke. [S. 1917]

A N S W E R S

1. 113-13 f.s. 2. 9° 58' to vertical. 4. 3-45 ins. 6. 1000 lb. wt.;750. 9. %n(n-l)s/V; Vjn. 11. 20^/10 f.s.; tan"12 above horizontal;5 (^5- l ) /4V2 . 12. 32ff f.s.; 777,896 ft.-lb.; 1265 ft. 13. 7'5cwt.;280 ft.-lb.

Chapter X

GENERAL PROBLEMS

101. We shall now apply the principles already establishedto a few simple problems.

The figure shews an arrangement of pulleys, two fixed and two mov-able. The latter are of masses mi, m2 and •£ .the strings passing round them can slipwithout friction, so that the tension of eachstring is constant throughout its length.Particles of mass mz, «i4, ro5 are attachedto the free ends of the strings. Let thetensions be denoted by T, T'. It is notnecessary to use five unknown quantitiesto denote the accelerations of the fivemasses, for if / , / ' , / " denote the accelera-tions with which the strings are slippingover the pulleys as indicated in the figure,then the upward accelerations of »ii, ro2,ms, m4, mb are

m1f=T' -{2.T-m1g,Therefore

nii{f-f)=T-rrng,

m6f"=m5g-T.

These equations are sufficient to determine the five unknown quantities

/, / , /", T, r.10'2. A fine smooth wire of mass M forms an equilateral triangle ABC.

The triangle can move horizontally in a vertical plane, the uppermost sideBC passing through smooth fixed rings in a horizontal line. Beads of massesm and in' are free to slide on the wires BA, CA. The system begins to movewith the beads at B and G respectively. Prove that the velocity of the wire atany instant, while both the beads are moving on it, is equal to the differenceof the speeds of the beads relative to the wire, and that the acceleration ofthe wire is

J3(m'~ m) gl(4:M+ 3m + 3m'). [S. 1927]

124 GENERAL PROBLEMS

At time t from the start let V be the velocity of the wire and v, v' thevelocities of m, m! relative to the wire.

Since there is no external horizontalforce acting on the system as a whole thehorizontal momentum remains zero.

ThereforeMV+m(V+v cos60")

+ro'(F-v'cos60°) = 0 .(1).Resolving along BA for m and along CA

for m', we have

m(v+V cos 60°) = mg cos 30°]and m'(v'- V OOB 60")=m'g cos 30" I'

Therefore by subtraction v' — i= V.Integrating this, and noting that the velocities are all zero initially so

that there is no constant of integration required, we get V=v' — v.Again differentiating (1) gives

(M+ m + m!) V= \m'v' - \and substituting for v and v' from (2) we get

so that

...(2).

V=s/z(m'-m)g.

10'3. A particle of mass m slides down, the face of a smooth wedge ofinclination a to the horizontal. The wedge is of mass M and it rests on arough horizontal table. Shew that the pressure on the table is (M+ m cos2 a) gor (M+m) Mgj{M-\-mB\na (sin a - / J cos a)}, according as the coefficient offriction ji is greater or less than

m sin a cos a/(M+m cos2 a).Let R be the reaction between the particle and the wedge, £ and F the

vertical reaction and the friction actingbetween the table and the wedge.

(i) If the wedge does not move wemust have F< pS. </ §\c.

But, resolving vertically and hori- s^ m9 |zontally for the wedge, we get t MS=Mg+R cos «, F=Rsma (1),

and by resolving at right angles to the wedge for the particle, we get

R

Therefore S'= (M+m cos2 a) g and F= mg sin a cos a.Hence this value of S is the value for the pressure provided that

y. > m sin a cos aj{M+m cos2 a).

10-2-10-4] CONSERVATION OF MOMENTUM AND ENERGY 125

(ii) If jti is less than this value there is not enough friction to preventmotion, so motion takes place and the friction F=pS. Let / be theacceleration of the wedge (from left to right in the figure) and / ' theacceleration of the particle relative to the wedge.

Resolving horizontally for the wedge, we get

Again, the acceleration of the particle m at right angles to the face ofthe wedge is the same as the acceleration of the wedge in this direction,viz. /sin a; therefore

m/sin o = mg cos a — R (3).Eliminate / from (2) and (3), and we get

R (M+ TO sin2 a) = pSm sin a + Mmg co$ a;but from (1) Rcoaa = S — Mg,therefore S {M+ m sin a (sin a — p cos a)} = (M + TO) Mg,which is the required result.

We note that ifp.=m sin a cos a/(M-\- m cos2 a),

the two values of S are the same.

104. Examples of conservation of momentum and energy.(i) A bead of mass M can slide on a smooth straight horizontal wire and aparticle of mass m is attached to the bead by a light string of length I. Theparticle is held in contact with the wire with the string taut and is then letfall. Prove that when the string is inclined to the wire at an angle 8 thebead will have slipped a distance ml (1 — cos 8)j{M+m) along the wire, andthat the angular velocity to of the string will be given by the equation

(Jf+TOCos2 8) lto2=2(M+m)g sin 8. [M. T. 1915]Suppose that the bead moves a distance x while the string turns through

an angle 8. Then the velocity ofthe bead is x and the particle mhas a velocity 18 relative to thebead in addition to the velocity ofthe bead.

Since there is no external hori-zontal force, the total horizontal ••- \TT •momentum remains zero, therefore

Mx + m(x-ld sind) = 0 (1).Integrating and observing that initially x and 8 are both zero, we get

Mx+m (x + lcos 6) = ml,therefore x = ml (1 - cos 8)/(M+ TO).

This result could also be obtained from the consideration that the centreof gravity of the masses M and m undergoes no horizontal displacement.

126 GENERAL PROBLEMS [x

Again, by equating the kinetic energy to the work done by gravity we get

$Mx2+im(x2 + l2d2-2xW8md) = mglsin6 (2),

and eliminating x from (1) and (2) gives

(M+m cos20)l62=2(M+m)gsin 6.

(ii) A hemisphere of mass M is free to slide with its base on a smoothhorizontal table. A particle of mass m is placed on the hemisphere at anangular distance a from the vertex; to determine the motion, the surface ofthe hemisphere being smooth.

Suppose that .at time t the particle is at an angular distance 6 from thevertex, and that the hemisphere has acquireda velocity v, the velocity of the particle beingad relative to the hemisphere, where a is theradius.

Since there is no external horizontal force acting on the system as awhole the total horizontal momentum remains zero throughout the motion,therefore

Mv+m (v-a6 cos 6)=0 (1).

Also the kinetic energy created is equal to the work done, therefore

\ Mv2+\m (v2 + a2 61 - 2va6 cos 6) = mga (cos a -cos 6) (2).

The result of eliminating v between (1) and (2) is

a82=2g (cos a-cos 6) (M+m)l(M+m sin2 6) (3),which gives the velocity of the particle relative to the hemisphere, andthen (1) gives the velocity of the hemisphere.

To find at what point the particle leaves the hemisphere, let R denotethe mutual reaction between them. Then by resolving horizontally forthe hemisphere, we get Mv=R sin 6.

But from (1)

(M+ m)v=ma(6 cos 6 - 62 sin 0)

and R vanishes when v vanishes, i.e. when

6 cos 6 = 62 sin 6.By differentiating (3) we find 8\ then by equating 6oos0 to 8'2sind we

find the equation for 6, viz.

m cos3 6 - (M+ m) (3 cos 6 — 2 cos a) = 0,

which determines the point at which the particle leaves the surface.

105. Examples of Circular Motion. Conical Pendulum.A particle suspended by a thread from a fixed point is projectedso as to describe a horizontal circle with uniform velocity; toprove that the time of revolution is lir \l(h\g), where h is thedepth of the circle below the fixed point.

10-4-10-6] CONICAL PENDULUM 127

If co is the angular velocity, I the length of the string and aits inclination to the vertical, the accele-ration of the particle is aP-l sin a towardsthe centre of the circle. Therefore if mis the mass of the particle the effectiveforce is m<oH sin a towards the centre ofthe circle and this is the resultant of theweight mg and the tension T. Hence byresolving horizontally and vertically

mcoH sin a = T sin a,and mg = T cos a.Therefore a>2l = g sec a, or &>2 = gjh;and the time of revolution = 2ir/a> = 2ir \l{hjg).

1051. The problem of a particle describing a horizontal circle on the in-side of a smooth surface of revolution witha vertical axis is solved in the same way.

Thus if P be the particle, PN perpen-dicular to the axis is the radius of thecircle described by P. And if the normalto the surface at P meets the axis at G,the forces on the particle in this caseare its weight rag and the reaction R ofthe surface along PO, and as in the lastarticle <»2. NG=g.

10-6. Transmission of Energy. Power is sometimes trans-mitted from one rotating shaftor pulley to another by meansof a belt passing round both.Thus if a shaft A be made torotate by an engine, a belt roundit and another shaft B will causethe latter to rotate. Let b be the radius and w the angularvelocity of the shaft B, and let Tt, T2 pounds weight be thetensions in the tight and slack sides of the belt. The momentabout the centre of the forces producing the rotation of theshaft B is (2\ — T2) b, and the work done per second by thiscouple is (Tt — T2) bco, or (JPJ — T2) v if v is the linear velocity ofthe belt. Hence the rate of working is (Ti - T2) v foot-poundsper second, or the horse-power transmitted is (Tx — T2) w/550.

mg

128 GENERAL PROBLEMS [X

107. Further problems on strings and chains. Let sdenote the length of a curved string or chain measured from afixed point A on the string up to a point P, and let Ss denote ashort length PQ. Let the tangents at P, Q make angles yfr,yfr + Stfrwith a fixed direction. Suppose that the tension varies along

the string being T at P and T+ST at Q. Then, no matterwhat are the forces that make the string assume a curved formand whether it be at rest or in motion, the effect of the tensionsof the rest of the string on the element Ss is represented by aforce T along the tangent at P and a force T+ST along thetangent at Q. Resolve the latter into components along thetangent and the inward normal at P, and we get along thetangent

-T + (T+ST)cosSf

and along the inward normal

(T +ST) sin Sf.Neglecting squares and products of 8T and Si|r, these com-ponents are ST along the tangent in thesense in which s increases and TSty alongthe inward normal.

Statical problems on strings or chains canbe solved by resolving along tangent andnormal for an element 8s and including inthe equations along with the external forces acting on theelement the forces ST and TSyjr. Dynamical problems can besolved in like manner by equating the components along tan-gent and normal of the effective forces of the element Ss to thecomponents of the external forces together with ST and

10-7-10-71] STRINGS AND CHAINS 129

For example, an endless chain under the action of noexternal force can run with uniform speed along a curve of anygiven form. For, the velocity v being constant, there is noacceleration along the tangent, and no external force; thereforeBT = 0, or T is constant.

And if m is the mass per unit length, the mass of the elementBs is mBs; and, if p is the radius of curvature, by consideringthe motion of this element and resolving along the inwardnormal we get

but p = Lim Bs/By]r,

therefore T = mv2.

If the speed of an endless chain be high the tension may beso great that external forces such as the weight of the chainare negligible in comparison, and then the chain will continueto run in the same curve for a long time, whatever form thecurve may have.

1071. Belt running on a Pulley at Uniform Speed. Supposethat the friction is limiting, i.e.just sufficient to prevent slippingof the belt on the pulley. / / \ .

Let v be the velocity of the belt,mBs the mass of a length Bs andlet a be the radius of the pulley.The normal reaction of the pulley T*on the element Bs of the belt may be denoted by RBs, and thenthe friction is fiR Bs if /J, is the coefficient of friction. The tensionson the element Bs are equivalent to BT along the tangent andTBifr along the inward normal. Resolving along the tangent andnormal we have

O = BT-fiRBs,

anda

But Bs = aByfr, therefore5>/T7

(T-mv^Bf^—

130 GENERAL PROBLEMS [X

and, by integrating,log (T— mi)2) = /ui/r + const.,

or T - fflji2 = Get1*.

Hence if A, B are the points where the belt leaves the pulley andTA, TB the tensions there, and we measure i|r from the tangentat-A, we have T—TA when T|T = 0, so that G=TA — mvz, and

T-mvli=(TA-mv2)e'l+.If the belt is in contact with the pulley round a semicircle

we have

If the inertia of the belt is neglected the result becomesTB — Tj^ev, just as if the belt were at rest.

10*8. Problems on Changing Mass, (i) If a man on a truckrunning on smooth level rails is throwing out m Ib. of sand per second in adirection parallel to the rails, and thus doing H ft.-lb. of work per second,prove that the velocity of the sand relative to the truck is */(%E~g/m).

[M. T. 1903]Let M be the mass of the truck and its contents at time t; V, V+ 8 V

the velocities of the same at times t, t + 8t and -1) the velocity of the sandrelative to the truck.

In time 8t a mass m8t lb. of sand is thrown out with velocity V—v.The linear momentum is constant and, by equating the momentum attime t + 8t to its value at time t, we get

or, neglecting the product 8t 8 V,

M8V-mv8t=O (1).

Again the work done in time 8t is Hg8t foot-poundals and this is equalto the increase in kinetic energy.

Therefore Hg8t=£ (M- m 8t) ( V+ 8 Vf + $ m 8t.(V- vf -1MV*

= MV8V-m8t.

and by (1) this = \m8t. v2,

so that v

(ii) A mass m of water issues per unit time from a pipe with uniformvelocity u, and strikes a pail which retains it, there being no elasticity.Initially the pail is at rest, and at a subsequent instant is moving in thedirection of the steam with velocity V. Prove that

dV _m(u-Vfdt

1O71-10-8] PROBLEMS ON CHANGING MASS 131

and that the loss of energy up to this instant is

\MuV,

where Mis the mass of the pail, and gravity is omitted from consideration.[S. 1910]

Let V denote the velocity of the pail and M' the mass of water in it attime t.

If a denotes the cross-section and p the density of the stream of water,since the velocity is u the length of the stream that emerges in unit timeis u, the'volume is ua and the mass pua. Therefore pua—m. But when thepail has velocity V the velocity of the stream relative to it is u — V andtherefore the length of stream entering the pail in time bt is (u — V) bt and

its mass is p (u— V) abt==— bt, and its momentum is m{u— V) 8t.

But the momentum of the pail and its contents at time t is {M + M') Vand in time bt this increases to (M+M' + bM')(V+bV).

The increase must be equal to the momentum of the water that entersthe pail, by conservation of linear momentum.

Therefore {M+M')bV+ VW = m(ii- V)bt (1).

But bM' is the mass of water added in time bt, so that

8JT = — y?~— 'dt... (2).

Substituting in (1) we get

(M+M1) 8 F = ^ ^ bt (3).

Eliminate bt from (2) and (3) and we get

bM' _ bVM+M'~u-V'

therefore (M+M')(u— V) = const.

= Mu (4),because initially M' = 0 and V—0.

Dividing (3) by (4) gives the result

dV_m(u- V)3

dt JMit

Again, the loss of energy up to this point is

and on substituting the value of M' from (4) this gives

\MuY.

9-2

132 GENERAL PROBLEMS [X

EXAMPLES

1. Two weights W, W balance on any system of pulleys with verticalstrings. If a weight w be attached to W, shew that it will descend withacceleration

W(W+WT\'w ] '

neglecting the inertia of the pulleys. [S. 1918]

2. A light string passes over two smooth pulleys in the same horizontalline and carries masses m1, «i2 at its extremities and a smooth ring ofmass m3 free to slide on the string between the pulleys. If all parts of thestring hang vertically, prove that the ring will remain at rest provided

+

) = <±m1m2,

and that in that case the acceleration of the centre of gravity of the threemasses is

(mt-mg)2 TS 19001

3. A mass m lying on a smooth horizontal table is attached to a stringwhich after passing over the edge of the table hangs in a loop on which aheavy smooth ring of mass M is threaded and then passes over a smoothfixed pulley and supports a mass m'. If the free portions of the string arevertical and the whole system lies in a vertical plane, determine the tensionof the string, and shew that the mass M will remain at rest provided that

2/J/"=l/m + l/TO'. [8. 1924]

4. Two particles of masses m and 3m are connected by a fine stringpassing over a fixed smooth pulley. The system starts from rest and theheavier particle, after falling 8 feet, impinges on a fixed inelastic support.Find the velocity with which it is next jerked off the support; and shewthat the system finally comes to rest 3 seconds from the beginning of themotion. [S. 1917]

5. A light string passes over a fixed smooth pulley and carries at oneend a mass 6m, and at the other a smooth pulley of mass 3m over whichpasses a second light string carrying masses 2m and m at its ends. As-suming that the system moves from rest, obtain expressions for the velocitiesand accelerations of the movable pulley and the masses. Use this systemto verify the principle of the conservation of energy, and explain why itdoes not illustrate the principle of the conservation of linear momentum.What distribution of masses between the various parts of the system wouldcause it to do so ? [S. 1924]

EXAMPLES 133

6. Two equal particles A and B are connected by a light inextensiblestring of length a which is stretched at full length perpendicular to theedge of the table. The particle A is drawn just over the edge of the tableand is then released from rest in this position. Describe the nature of thesubsequent motion and shew that after B leaves the table the centre ofinertia of the two particles describes a parabola of latus rectum a/2.

[S. 1917]

7. From a gun of mass M which can recoil freely on a horizontal plat-form a shell of mass m is fired with velocity v, the elevation of the gunbeing a. Determine the direction in which the shell is moving when itleaves the gun, and shew that, if the shell strikes at right angles the planethat passes through the point of projection and is inclined to the horizontalat an angle /3, then

). [S. 1924]

8. A particle of mass 2 lb. is placed on the smooth face of an inclinedplane of mass 7 lb. and slope 30°, which is free to slide on a smoothhorizontal plane in a direction perpendicular to its edge. Shew that if thesystem starts from rest the particle will slide down a distance of 15 feetalong the face of the plane in 1-25 seconds. [S. 1923]

9. A smooth wedge of mass M and angle a is free to move on a smoothhorizontal plane in a direction perpendicular to its edge. A particle ofmass m is projected directly up the face of the wedge with velocity V.Prove that it returns to the point on the wedge from which it was projectedafter a time

2 V (M+ m sin2 o)/{(w+M) g sin a}.Also find the pressure between the particle and the wedge at any time.

[S. 1917]

10. A smooth wedge of inclination o is placed on a horizontal table ; astring, to the extremities of which masses m, m! are attached, passes round asmooth peg which projects from the upper face of the wedge, the particlesbeing in contact with the same face of the wedge. Prove that, if M be themass of the wedge, its acceleration is

(»i - m')2g sin a cos aM{m + m') + (m — m')2 sin2 a + imm!'

assuming that the motion is in planes of greatest slope. [S. 1899]

11. A particle of mass m is placed on the smooth slant surface of awedge of mass M and angle o. The wedge moves (edge foremost) along asmooth horizontal plane under the action of a constant horizontal force Pperpendicular to its edge, all the motion being parallel to the same vertical

plane. Prove that the acceleration of the wedge is —=j-^ ~—- , andcc

find the pressure between the particle and the wedge. [S. 1907]

134 GENERAL PROBLEMS [X

12. A particle of mass m slides down the rough inclined face of a wedgeof mass M and inclination o, which is free to move on a smooth horizontalplane. Shew that the time of describing any distance from rest is less thanthe time taken when the wedge is fixed in the ratio

f\

( 1M+m J

13. A smooth wedge of mass M resting on a horizontal plane is subjectto smooth constraints so that it can only move along the plane in adirection at right angles to the intersections of its slant faces with theplane. A particle of mass m is moving along a face of the wedge which isinclined to the horizontal at an angle a so that the component velocity ofthe particle perpendicular to the line of greatest slope is «. Shew that thewedge moves with constant acceleration and that the path of the particleon the surface of the wedge is a parabola of latus rectum

2M2 J/"+msin2ag {M+m)sm.a'

Verify that the principle of conservation of energy holds good in thiscase.

Solve the same problem when the wedge, instead of being free to move,is made to move in the same direction as before with constant velocity V.

[S. 1925]

14. Two particles of mass M and m (M>m) are placed on the twosmooth faces of a light wedge which rests on a smooth horizontal plane.The faces of the wedge are inclined to the horizontal at angles a and /3,respectively. If the system starts from rest, shew that the smaller particlewill move up the face on which it is placed if

, . i /s in a cos a r o , . . . ,tan /3 < v/"---9 . [S. 1915]^ J/sin2a+m L J

15. A particle of mass m is placed on the inclined face of a wedge ofmass M which rests on a rough horizontal table. Prove that, if the particleslides down, the wedge will begin to move provided that

m cos X sin X'-17 >"

where a is the inclination of the face of the wedge to the horizontal, X isthe angle of friction for the particle and the wedge, and X' is the angle offriction for the wedge and the table. [S. 1924]

16. A tube ABC of mass rn is bent at right angles at B. The part ABis horizontal and slides freely through two fixed rings; the part BC isvertical. Particles P, Q, each of mass rn, move without friction in AB, BC,and are connected by a string passing over a smooth pulley of negligiblemass at B. The system is released from rest Apply the principles of

EXAMPLES 135

momentum and energy to shew that, when Q has fallen a distance y fromits initial position, its vertical velocity is ij(6gyl§).

Shew that the vertical and horizontal components of the accelerationof Q are 3#/5 and g/5. [M. T. 1922]

17. A smooth straight tube BAC is bent at A and is fixed in a verticalplane so that AB, AC make angles a, /3 on opposite sides of the vertical.A heavy uniform string PAQ in the tube is slightly displaced from theposition of equilibrium ; shew that when a length x has passed over A inthe direction of P both the velocity and the acceleration vary as x, andthat the tension of the string at A varies as (p +x) (q- x), where p, q arethe lengths AP, AQ in the position of equilibrium.

Shew also that the resultant vertical pressure on the tube is

/ x2 \W I 1 cos o cos B),

\ PI Iwhere W is the weight of the string. [S. 1917]

18. A light string ABCDE, whose middle point is C, passes throughsmooth rings B, D which are fixed in a horizontal plane at a distance2a apart. To each of the points A, C, E is attached a mass m. Initially0 is held at rest at 0, the middle point of BD, and is then set free. Shewthat 0 will come instantaneously to rest when 0C=4a/3. [The total lengthof the string is greater than 10a/3.]

Shew that when C has fallen through 3a/4 from 0, its velocity is[M. T. 1922]

19. Two particles, masses M and m {M> m), are attached to the ends ofa string, length 21, which passes over a smooth peg at a height I above asmooth plane inclined at an angle a to the vertical. The particles areinitially held at rest on the plane at the point vertically below the peg,M being below m. Prove that, if the particles are released, m will oscillatethrough a vertical distance 2M (M-m) ll(m2 sec2 a — Jf2), provided thattan2 o is greater than (3M+m) (M- ro)/m2. [S. 1923]

20. A ring of mass m slides on a smooth vertical rod; attached to thering is a light string passing over a smooth peg distant a from the rod,and at the other end of the string is a mass M(>m). The ring is held ona level with the peg and released: shew that it first comes to rest afterfalling a distance

W^rf- [S. 1915]

21. A mass m is suspended at the lower end of a vertical elastic wire ofmass m and length L, suspended at its upper end. The system is causedto execute small vertical oscillations. Assuming that the wire can betreated as uniformly stretched throughout the motion, shew that the

136 GENERAL PROBLEMS [X

kinetic energy of the system is \xi(M+\m), where x is the displacementof the mass M from its equilibrium position. Hence shew that the time of acomplete oscillation is 2irJ{kL(M+^m)}, where k is such that unit forceproduces an extension of k units in unit length of the wire. [M. T. 1921]

22. Two equal particles A, B attached to the ends of a light string oflength a are placed on a smooth horizontal table with the string ABperpendicular to the edge of the table and B hanging just over the edge.The system is released from rest in this position. Prove that when firstthe string is horizontal the distance of B from the vertical through theedge of the table is Ja (n - 2), and find the tension in the string. [S. 1923]

23. A horizontal bar AB of length a is made to rotate with a constantangular velocity a about a vertical axis through the end B. If a particleis attached to A by a string of length I, the string makes an angle 6 withthe vertical when the motion is steady. Prove that

Icosd+acot6=gl<o2. [M.T. 1914]

24. A particle of mass m' is attached by a light inextensible string oflength I to a, ring of mass m free to slide on a smooth horizontal rod.Initially the two masses are held with the string taut along the rod andthey are then set free. Prove that the greatest angular velocity of thestring is

Also shew that the time of a small oscillation about the vertical is

2TT \lm/g (m + ro')}*- [Coll. Exam. 1914]

25. Two particles of masses 7m and 3m are fastened to the ends A, Brespectively of a weightless rigid rod, 15 ft. long, which is freely hinged ata point 0, 5 ft. from A,: if the rod is just disturbed from its position ofunstable equilibrium, prove that the velocity with which A will passthrough its position of stable equilibrium is ^ \f'Z80g. [S. 1910]

26. Two particles, each of mass m, are joined by a rod of negligible massand of length a. One particle rests on a smooth horizontal plane and theother is vertically above it. The upper particle is given a small displace-ment so that the rod begins to fall. Shew that when its inclination to thevertical is 60° the velocity of the lower particle is */(ga/l4:). [M. T. 1926]

27. A heavy ring of mass m slides on a fixed smooth vertical rod and isattached to a fine string which passes over a smooth peg distant a fromthe rod and then after passing through a smooth ring of mass M is tied tothe peg. Prove that, if m is dropped from the point in the rod in the samehorizontal line as the peg, then, provided M> 2m, it will oscillate througha distance ArnMaj^M2 — 4wi2) and find the greatest velocity of m. [S. 1924]

EXAMPLES 137

28. The driving wheel of a motor cycle has a diameter of 28 ins.; thebelt pulling on the driving wheel has a diameter of 20 ins. If 3 H.P. is therate of working when the cycle is going 20 miles an hour, and in order thatthe belt may not slip the tension on one side may not exceed 2£ times thaton the other, shew that the tensions are 131J lb. wt. and 52^ lb. wt.

[S. 1914]

29. Power is delivered by an engine through ropes passing round theflywheel of the engine and over pulleys in the mill. If the flywheel is30 ft. in diameter and turns at 90 revolutions per minute, determine howmany ropes of l j inch diameter will be required to transmit 2000 horse-power. Assume the tension on the tight side to be twice the tension onthe slack side, and allow not more than 300 pounds per square inch pullin the ropes. [M. T. 1918*]

30. A belt-driven pulley of diameter 2 feet transmits 10 horse-powerwhen running at 240 revolutions per minute. Shew that if the belt is juston the point of slipping, and subtends an arc of 180° of the pulley, thebiggest tension in the belt is 306 lb. nearly. The coefficient of frictionbetween belt and pulley can be taken at 0'4. [M. T. 1920]

31. A machine gun of mass i f contains a mass M' of bullets which itdischarges at the rate m units of mass per unit time, V being the velocityof the bullets relative to the ground. Shew that, if p be the coefficient offriction between the gun and the ground, the whole time of recoil of thegun will be

(2m V- ngM') M'/Z/igmM. [S. 1921]32. A horizontal rod of mass M is movable along its length, and its

motion is controlled by a light spring which exerts a restoring force Exwhen the rod is displaced through a distance x. A spider of mass m standson the rod, and everything is initially at rest. The spider then runs adistance a along the rod, and then stops, his velocity relative to the rodbeing constant and equal to u. Shew that the total energy of the systemafter the run is

. 2 / a /S m

1,2M V J/

M+m S m 1,2M V JU+m) 'and find the amplitude of the final motion. [S. 1925]

33. A machine gun of mass M stands on a horizontal plane and containsshot of mass M'. The shot is fired at the rate of mass m per unit of timewith velocity u relative to the ground. If the coefficient of sliding frictionbetween the gun and the plane is p, shew that the velocity of the gunbackward by the time the mass M' is fired is

ANSWERS22. Half the weight of either particle. 29. 30.

Chapter XI

IMPULSIVE MOTION

11 "1. The state of rest or motion of a body sometimes under-goes an apparently instantaneous change owing to the suddenapplication of a force which acts for a very short time only.For example—a ball struck by a bat, or the collision of twobilliard balls.

In such cases it is not possible to measure the rate of changeof momentum because a finite change of momentum takesplace in an infinitesimal interval of time.

In 4-2 we saw that the change of momentum produced by a

variable force X acting from time t = tx to t = t2 is X dt.

Now it is possible for the force to increase and at the sametime the interval t2 — tx to decrease in such a way that theintegral tends to a finite limit, although we have no meansof measuring the exact value of X at any instant during theinterval. It is usual to call such a force an Impulsive Forceor Impulse and measure it by the change of momentum itproduces, i.e. we measure it by its ' impulse' as defined in 42.

It is to be noted that an impulsive force or impulse isnot of the same physical dimensions as ' force.' The latter is ofdimensions MLT~2 and the former of dimensions MLT"1 (4r8).

112. Equations of Motion for Impulsive Forces. Theequations of motion for a system of particles acted upon byfinite forces were found in 92 to be

~Zmx = %X, Xmy = 2 Y,

and T- 2TO (xy — yx) = %{xY — yX).

If we integrate these equations with respect to t through aninterval from t0 to t, we get

rt= £ I Xdt,

Jt.rt

y — %my0 = 2 Ydt,

1 1 1 - 1 1 2 ] EQUATIONS OF MOTION 139

and( rt rt •)

Sm (xy — yx) — %m (xy0 — yxa) = 2 •<x\ Tdt — y\ Xdt>,

where x0, y0 denote the values of x, y at time to.Now if we suppose that we are concerned with impulsive

rt rtforces as defined in l l ' l , then Xdt and I Ydt are the

Jt0 Jt,measures of the components of the impulse and may be denotedby P, Q respectively. Hence the last three equations may bewritten

"tmx — 2wi«0 = SP"iSmy — 1my0 = 2Q;

and 2m (xy — yx) — 2m (xy0 — yx0) = 2 (xQ - yP) .. .(2).

These equations express the facts that

(1) the instantaneous increase in the linear momentum in anydirection is equal to the sum of the externally applied impulsiveforces in that direction; and

(2) the instantaneous increase in the moment of momentumabout any axis is equal to the sum of the moments about thataxis of the externally applied impulsive forces.

Further we observe that with the notation of 9'3 equations(1) may be written

and

Also the equations confirm the principles of conservation oflinear and angular momentum, in tha t if there be a directionin which the external impulsive forces have zero componentthere is no change of momentum in that direction; and if therebe an axis about which the external impulsive forces have zeromoment there is no change of moment of momentum abouttha t axis.

I t is to be observed that, in writing down equations for theinstantaneous change of motion produced by impulsive forces,all finite forces such as weight are to be neglected, because if F

is a finite force then I Fdt vanishes when the interval t —10J t0

tends to zero.

140 IMPULSIVE MOTION [XI

11 "3. Impact of Smooth Spheres. The problem of themotion of smooth spheres after impact is determined partlyby the principle of conservation of linear momentum andpartly by an experimental law due to Newton, viz. ' The relativevelocity of the spheres along the line of centres immediately afterimpact is — e times the relative velocity before impact,' where eis a constant depending on the substances of which the spheresare composed, called the coefficient of restitution. For hardsubstances like steel or ivory e is nearly unity, but for a softsubstance it is small. When a substance is described as perfectlyelastic it is to be understood that e — 1, and, when inelastic, e = 0.

The value of e may be determined experimentally by sus-pending two spheres by equal finethreads, so that when at rest thespheres are in contact and the threadsvertical. Then one of the spheres isdrawn back and released so as tostrike the other. From 8'2 the square -of the velocity of either sphere at thelowest point of its path, i.e. at striking,is proportional to the vertical height through which it hasdescended or to which it ascends. Hence by measuring thevertical heights all the velocities immediately before and afterimpact can be found, and so e is determined.

11*31. Direct Impact. Let m, m be the masses of the twospheres, u, u' their velocities before impact and v, v' theirvelocities after impact and let the motion be along the line ofcentres.

The momentum in the line of motion is unaltered by theimpact, so that

mv + mV = mu + m'u' (1);and, by Newton's rule,

v — v' = — e (u — u') (2).These equations determine the velocities after impact, namely

mu + m'u' — em' (u — u')m + m

and, mu + m'u'+ em(u — u)

m + m

H'3-11-32] IMPACT OF SPHERES 141

The impulse between the spheres which reduces the velocityof the first from u to v is in (u — v), which is equal to

(1 + e) mm (u — u')m + m'

and this is (1 + e) times what the impulse would be if thecoefficient of restitution were zero.

li"32. Poisson's Hypothesis. This hypothesis is that whenthe bodies come into contact there is a short interval in whichthey undergo compression followed by another short interval inwhich the original shape is restored. At the instant of greatestcompression the bodies have a common velocity along the lineof centres, and the impulsive pressure between the bodiesduring restitution is less than the impulsive pressure duringcompression in the ratio e : 1. Now if the bodies were inelastic(i.e. if e = 0) they would acquire a common velocity

(mu + m'u')/(m + m')

and there would be no restitution, so the hypothesis is thatwhen the bodies are elastic the whole impulsive pressure duringimpact is (1 + e) times the impulse during compression, andfrom 11'31 this is in accordance with Newton's rule.

In fact we may deduce Newton's rule from this hypothesis.If with the notation of the last article we denote by I theimpulse during compression and by U the common velocity atthe instant of greatest compression, then

m(u-U) = I, and m'{U-u) = I.

Eliminate U, then

u-u' = ~ +— (1).m m x '

But, by hypothesis, the whole impulse = (1 + e) / , therefore

m (u - v) = (1 + e) I, and rri (V - u') = (1 + e) / .

Whence we get

and from (1) = (1 + e) (u — «');

therefore v — v' — — e{u — u),

which is Newton's rule.

142 IMPULSIVE MOTION [XI

11 33. Oblique Impact. When the directions of motion ofthe spheres are not along the lineof centres their velocities can beresolved along and perpendicularto the line of centres. We mayuse the symbols of 11 "31 to denotethe velocities along the line ofcentres, and they will satisfy theequations of that article. When we consider the motion ofeither sphere at right angles to the line of centres, we see that,since the spheres are smooth, there is no impulsive action thatcan affect the velocity in this direction. Hence if U, U' denotethe component velocities at right angles to the line of centresbefore impact, they also denote the components in the samedirections after impact.

1134. Kinetic Energy lost by Impact. In general there isa loss of kinetic energy in the cases we are considering. Weneed only consider the case of direct impact, because, in obliqueimpact, the square of the velocity being the sum of the squaresof its perpendicular components and components at right anglesto the line of centres being unaltered by impact, it follows thatit is only the velocity components in the line of centres thatcan effect a change in kinetic energy. With the notation of1131 and using the theorem of 95 we write the kinetic energybefore impact in the form

\mu2 + \mu2 = | (m + m) ^——~ r

mu + m'u'\2 , , I . mu + m'u'\2

) +\m [ u )m+m J * \ m+mJ

where {mu + mu')j{m + m') is the velocity of the centre ofgravity and the terms in the last two brackets represent thevelocities of the spheres relative to the centre of gravity. Thisreduces to

{mu + m'u')2 . mm' (u — u')2

* m + m' * m + m'Similarly the kinetic energy after impact

_ ^ (mv + m'v')2 . mm' (v — v')2

2 m + m' m + m'

11-33-11-36] KINETIC ENERGY LOST BY IMPACT 143

But mv + m'v = mu + m'u,

and v — v' = — e (u — u');

therefore there is a loss of kinetic energy equal to

\mm' (1 - e2) (u - w')2/(m + m'),

or (1 — e2) times the kinetic energy of the spheres relative tothe centre of gravity before impact.

11 35. Generalization of Newton's Rule. For impacts ofbodies other than spheres we may generalize Newton's ruleand say that the velocities, before and after impact, of the pointsof two bodies that come into contact resolved along the commonnormal at the point of contact are in the ratio 1: — e.

Thus suppose a sphere moving with velocity u strikes a smoothplane in a direction making anangle 6 with the normal to theplane, and that it rebounds with w

velocity v making an angle <f> withthe normal.

The above rule makes

v cos <{> = eu cos 0;

and since the velocity parallel to the plane is unaltered

v sin <\> = u sin 6,

therefore cot (f> = e cot 6.So that the angles of incidence and reflection of such a sphereimpinging on the plane are governed by this law of reflection.

1136. Example . Two equal spheres of mass m! are suspended byvertical strings so that they are in contact with their centres at the same level.A third equal sphere of mass m falls vertically and strikes the other twosimultaneously so that their centres at the instant of impact form an equi-lateral triangle in a vertical plane. If u is the velocity of m just beforeimpact, find the velocities just after impactand the impulsive tension of the strings.

After impact the spheres of mass m! beginto move horizontally because of the constraintsof the strings; let v be the velocity of either;and let u' be the velocity of m, which bysymmetry must be vertical. Let 1 denotethe impulse between the upper and either of

144 IMPULSIVE MOTION [XI

the lower spheres, T the impulsive tension of the strings and e thecoefficient of restitution.

By Newton's ruleu' cos 30° — v cos 60° = —eu cos 30",

or \J?Lu'-v=-\IZeu (1).Resolving vertically for m,

m(M-M') = 2/cos30° (2),and horizontally for m', m'v = IcoB 60° (3).

Eliminating / from (2) and (3), we get

m (w — u') = 2'\/3m,'v;

and from (1) =6mf (u' + eu),

therefore u' = u (in, — 6em')l(m + 6m');

and from (1) v=\f3um(l+e)l(m + 6m').

Again, resolving vertically for m', we get

T = / cos 30°=«/3 m'tf;

and from (3) =3mm'ii(l+e)l(m+fym,').

11'4. As further examples of impulsive action we take the following:

E x a m p l e s , (i) A body of mass m1 + m2 is split into two parts of -massesm1 and m2 by an internal explosion which generates kinetic energy E.Shew that if after explosion the parts move in the same line as before, theirrelative speed is

sl{^E(m1^-m2)lm-im^. [M. T. 1902]Let u be the speed of the body before explosion, and »i, v2 the speeds of

m1, m2 after explosion.There is no change in linear momentum, therefore

The kinetic energy is increased by E, therefore

\ mx V + \ m2 v%2=-| (Mij + m2) «2 + E.

Eliminate u, therefore

(m1 vx2 + m<L v2

2) (m-i + m2) = (TOtvx + m21»2)2 4- 2 (m-i + m2) E,

or

and

(ii) A light rigid rod ABC has three particles each of mass m attached toit at A, B, C. The rod is struck by a blow P at right angles to it at a pointdistant from A equal to BC. Prove that the kinetic, energy set up is

ipz2 ~m

where AB = a, and BC=b.

11-36-11-4] IMPULSIVE MOTION 145

The motion of the rod is completely defined by its angular velocity a>and the linear velocity u of any onepoint, say C. The velocity of B is ^ ^ Uj.then u + bco and the velocity of A is / I . J

(a + b)a>.

To find the two unknown quantities u, a we resolve at right angles tothe rod and equate the total momentum to the impulse, i.e.

or 3u + a>(a + 2b) = P/m (1).We also take moments, and the point 0, at which the blow is applied, is

the most convenient point about which to take moments, because P has nomoment about 0, so that the total moment of momentum about 0 is zero.Therefore

m{u + (a+b)a>} b — m (u-\-ba>) (a— b) — mua — 0,or « ( a - 6 ) = 62<B (2).

From (1) and (2)P 1

P 1m ' P

And the kinetic energy created by the blow

2 m

(iii) Three equal particles A, B, C of mass m are placed on a smoothhorizontal plane. A is joined to B and C by light threads AB, AC and theangle BAC is 60°. An impulse Us applied to A in the direction BA. Findthe initial velocities of the particles and shew that A begins to move in adirection making an angle tan~l sJ'Ajl with BA. [S. 1924]

Let u, v denote the initial velocities of A along and at right angles toBA, and let T, T' denote the impulsivetensions in AB, AC.

Since B and A are joined by an inex-tensible string their velocities along BAmust be the same, therefore the velocityof B is u along BA. For a similar reasonthe velocity of C is u cos 60° — v cos 30°or ^(u — sfZv) along CA.

Now, writing down the equations of impulsive motion of the particlesin turn, we have for A .

2for B mu=T,

and for C im(w-«/3») = Z".

146 IMPULSIVE MOTION [XI

Eliminating T' gives

-rsv, or

Then by eliminating T and T' from the first equation we get

so that M = 7//15TO,

and v = */3I/l5m.Also the direction of motion of A makes with BA an angle

li'5. Kinetic Energy created by Impulses. Consider asystem of particles of masses mi, m2, m3, ... moving withvelocities whose components parallel to the axes Ox, Oy are(«i> Vi), (ui} v2), (MS, V3), Let them be acted upon byimpulses whose components in the same directions are (Xi, Fi),(X2, F2), (X3, Y3), ..., including the ' internal ' as well as the' external' impulses on every particle ; and let the velocities bechanged by the impulses to («/ , vx'), (w2'> va')» (W, Vz)> •••• Thenthe equations of impulsive motion of the particles are

u1) = X1, mx (vi - vt) = Yx,

l — u2) = X2 , m2 (v2 — v2) = F2,

3 (u3r - M3) = X3, m3 (v3

r - v3) = F3,

Multiply each of these equations by half the sum of thevelocity components that it involves and add them all together,and we get

This equation expresses the fact that the change of kineticenergy created by a set of simultaneous impulses is the sum ofthe products of each impulse into the mean of the velocities of itspoint of application resolved in its direction.

11'51. Let us consider how the last theorem is illustrated by examples(i) and (ii) of 11-4.

(i) Here the impulse I is an internalone which increases the velocity of (say) ^ _ __^mi from u to »!, and decreases the ' | | 'velocity of m2 from u to v2, acting inopposite directions on the two masses. ^ W

Hence 1

H-4-11 '61] KINKTIC ENERGY CREATED BY IMPULSES 147

And, by the last theorem, the energy created by the simultaneousimpulses is

E

as is otherwise obvious.(ii) The velocity of the point 0 is zero initially, and after the blow it is

v say, wherepm

Therefore the kinetic energy created by the blow

11 6. Elasticity and Impulses. It is to be observed that anelastic or deformable body as distinct from a rigid body yieldsto an impulsive action and does not begin to offer resistanceuntil a finite deformation has taken place. Thus if an impulseis applied to an elastic string there is no impulsive tension setup in the string, and it is not until a finite elongation has takenplace that a finite tension is produced. For example in 11*36, ifthe strings were elastic, the initial velocities of the spheres m'would not be horizontal but along the lines joining the centreof m to their centres.

This property of an elastic string is also illustrated in thefollowing example.

11*81. Example . Two particles of masses m^ and m2 are connectedby a fine elastic string of natural length I and modulus of elasticity X. Theyare placed on a smooth horizontal table at a distance I apart, and equalimpulses I in opposite directions act simultaneously on them in the line ofthe string so as to extend it. Prove that the greatest extension of the string inthe ensuing motion is

I {{mi + m2) Ijmi ?»2A}4 >and that this value is attained in time

£»r {mjwia J/(m, + m2) \}k [M. T. 1916]Let x, y denote displacements of mx, m^ in opposite directions in time t,

and let x+y — z. The tension at time t is therefore \z/l.The equations of motion of the particles are therefore

—\z/l, and

By adding we get ^ Q ^

As(m1 + TO2)

148 IMPULSIVE MOTION [XI

This represents a simple harmonic motion so long as the string remainsstretched and the time during which the velocity z changes from itsmaximum to zero is one quarter of the period, i.e.

Again the initial velocities of mi and m2 are 1/m.i and I/m2, since thestring being elastic has no impulsive tension. Therefore the kineticenergy created

2 mi m2

Now when the string has its greatest extension the particles aremomentarily at rest and the kinetic energy has been converted into thepotential energy of the stretched string. Hence if z is the greatestextension and T the greatest tension, the work done in stretching

and by equating this to the kinetic energy we get

z

We might also obtain this result by multiplying equation (1) by 2£ andintegrating, and so finding the velocity z in terms of z; then find the valueof z for which z vanishes.

E X A M P L E S

1. A smooth sphere impinges on another one at rest; after the collisiontheir directions of motion are at right angles. Shew that if they areassumed perfectly elastic, their masses must be equal. [S. 1926]

2. Two smooth elastic spheres (coefficient of restitution e) impingeobliquely in any manner; one of them being initially at rest, it is foundthat the angle between their subsequent directions of motion is constant.Find the ratio of the masses of the spheres, and the angle. [S. 1921]

3. Two equal spheres of mass 9ra are at rest and another sphere ofmass m is moving along their line of centres between them. How manycollisions will there be if the spheres are perfectly elastic? [S. 1923]

4. n equal perfectly elastic spheres move with given velocities under noforces in the same straight line. Shew that the final velocities of thespheres depend only on their order and not on their initial distances apart.

[S. 1926]

11-61] EXAMPLES 149

5. Shew that if a smooth sphere of mass mx collides with another smoothsphere of mass m^ at rest, and is deflected through an angle 6 from itsformer path, the sphere of mass mt being set in motion in a direction <f>

with the former path of ?»,. then tan 6= ^ - r , both spheresTOi-m2Cos20

being perfectly elastic. [S. 1925]6. A body of mass m rests on a smooth table. Another of mass M

moving with velocity V collides with it. Both are perfectly elastic andsmooth and no rotations are set up by the collision. The body m is drivenin a direction at angle 6 to the previous line of the body M'a motion.

Shew that its velocity is -. Fcos 6. [S. 1921]

7. A sphere collides obliquely with another sphere of equal mass whichis initially at rest, both spheres being smooth and perfectly elastic. Shewthat their paths after collision are at right angles.

The centres of two such spheres, B, C, each of 3 cms. radius, areat E, F, where EF= 16 cms. An equal sphere A is projected withvelocity u at right angles to EF, and strikes first B and then C. Its finalpath is at right angles to EF. Find the point of contact between A and B.Shew that

vA = 9uJ25, vB—20ul25, vc=12u/25,where vA, vB, v0 are the final velocities. [M. T. 1922]

8. The masses of three spheres A, B, C are 7m, 1m, m; their coefficientof restitution is unity. Their centres are in a straight line and C liesbetween A and B. Initially A and B are at rest and C is given a velocityalong the line of centres in the direction of A. Shew that it strikes Atwice and B once, and that the final velocities of A, B, C are proportionalto 21, 12, 1. [M. T. 1915]

9. Two equal marbles, A and B, lie on a smooth horizontal circulargroove at opposite ends of a diameter. A is projected along the groove andat the end of time t impinges on B; shew that the second impact willoccur after a further time 2t/e, where e is the coefficient of restitution.

[M. T. 1921]

10. Two smooth and perfectly elastic spheres of masses 1 and 4respectively are initially at rest under no forces. The more massive sphereis then projected in such a direction as to strike the other sphere andrebound. Prove that the direction of motion of the more massive spherecannot be deflected by the collision through an angle greater than 14° 29'.

[M. T. 1925]

11. Two equal pendulums OP, CQ, of length I, are suspended from twopoints 0, C in a horizontal line, such that when the bobs are hanging atrest they are just in contact. The bob P is projected horizontally withvelocity \/gl from the point at height I vertically above 0, and strikes the

150 IMPULSIVE MOTION [XI

bob Q which was previously hanging at rest. Shew that the string of Qwill become slack before Q reaches its highest point, if the coefficient ofrestitution lies between Vl-6 —1 and unity. [M. T. 1924]

12. Two small spheres A and B of equal mass m, the coefficient ofrestitution between which is e, are suspended in contact by two equalvertical strings so that the line of centres is horizontal. The sphere A isdrawn aside through a small distance and allowed to fall back and collidewith the other, its velocity on impact being u. Shew that all subsequentimpacts occur when the spheres are in the same position as for the firstimpact, and that the velocity of the sphere A immediately after the thirdimpact is \u (1 — e3). Shew that the kinetic energy of the system tends tothe value \mv?. [M. T. 1926]

13. Two imperfectly elastic particles of equal mass, whose coefficient ofrestitution is e, are suspended from the same point by light strings of equallength. One particle is drawn aside a small distance x0 and then released.Shew that, between the mth and (« + l)th impacts, the particle originallydrawn aside swings through a distance J {l+( — e)n}x0 on one side of thevertical through the point of suspension. [S. 1923]

14. A spherical ball of mass m suspended by a string from a fixed pointis at rest, and another spherical ball of mass m' which is falling verticallywith velocity u impinges on it so that the line joining the centres of theballs makes an angle a with the vertical. Prove that the loss of energy

= J (1 - e2) mm'u? cos2 a/(m + m' sin2 a),where e is the coefficient of restitution. [S. 1924]

15. Three equal similar spheres of mass m! are suspended by equalvertical threads so that their centres are at the corners of an equilateraltriangle in a horizontal plane. A fourth smooth sphere, of mass m, fallsvertically so as to strike the other spheres simultaneously. Determine thevelocities immediately after impact, having given the velocity u of thefourth sphere and the angle 6 which the lines of centres make with thevertical at the instant of striking. [S. 1925]

16. A steel ball is released from rest and falls upon a fixed steel anviland rebounds, the coefficient of restitution being 0'9. The lowest point ofthe ball is initially at a distance of one foot above the anvil. Find theposition and velocity of the ball half a second after its release.

Shew that the ball finally comes to rest on the anvil 4'75 seconds afterits release and that the total distance travelled is 9Jg feet. [M. T. 1914]

17. A spherical particle is let fall vertically under gravity and afterdescribing a distance h impinges at a point A on a smooth plane inclinedat an angle a to the horizontal. Shew that the particle ceases to reboundfrom the plane when it reaches a point B such that

where e is the coefficient of restitution. [M. T. 1920]

EXAMPLES 151

18. Two smooth spheres of equal mass, whose centres are moving withequal speeds in the same plane, collide in such a way that at the momentof collision the line of centres makes an angle %n — e with the directionbisecting the angle o between the velocities before impact. Shew that afterimpact the velocities are inclined at an angle tan"1 (cos 2c tan o), thecollision being perfectly elastic. [M. T. 1923]

19. Two particles of masses i f and m are connected by a fine inextensiblestring passing over a fixed smooth pulley, and the motion of the heavierparticle, M, is limited by a fixed horizontal inelastic plane, on which it canimpinge. The system starts from rest with M at a given height above theplane; shew that the successive heights of M at which it comes toinstantaneous rest form a geometrical progression of ratio {m/(M+ TO)}2,and that the whole time of motion is three times the interval from thebeginning of the motion to the first impact on the plane. [M. T. 1916]

20. A bucket of mass ml is joined to a counterpoise of mass m2 by alight string hanging over a smooth pulley. A ball of mass m is droppedinto the bucket. Shew that the ball will come to rest in the bucket at a

time -.—•—\ after the first impact, where v is the velocity of the ball

relative to the bucket immediately before the first impact, and e is thecoefficient of restitution.

Shew that the sum of the upward momentum of the system on one sideof the pulley, and the downward momentum of that on the other side,increases at a uniform rate, and determine this rate. Hence or otherwiseshew that the velocity of the system so soon as the ball has come to restin the bucket is

mm<i + e (mmi + m-f — m22)

( 1 — e ) m ( m + m + ™) 'where u is the downward velocity of the bucket immediately before thefirst impact. [S. 1925]

21. A railway truck is at rest at the foot of an incline of 1 in 70. Asecond railway truck of equal weight starts from rest at a point 1000 feetup the incline, and runs down under gravity. The trucks collide at thefoot of the incline, the coefficient of restitution being J-. Find how far eachtruck travels along the level, the frictional resistances for each truck being16 lb. wt. per ton, both on the incline and on the level. Where the inclinemeets the level, the rails are slightly curved, each in a vertical plane, sothat there is no vertical impact, and at the instant of collision both trucksare on the level. [S. 1925]

22. A heavy elastic particle is projected from a point 0 at the foot of aninclined plane of inclination a to the horizon. The plane through thedirection of projection normal to the inclined plane meets the inclinedplane in a line OA which makes an angle <f> with the line of greatest slope

152 IMPULSIVE MOTION [XI

and the direction of projection makes an angle 6 with OA. Find equationsto determine the position of the particle after any number of rebounds andshew that the particle will just have ceased to rebound when it againreaches the foot of the plane if

tan 8 tan a = (1 - e) cos <£,

where e is the coefficient of restitution. [S. 1926]

23. Shew that after an elastic collision (e = 1) between two equal smoothspheres, one of which is initially at rest but free to move in any direction,the directions of motion of the two spheres are at right angles.

Shew further that if the mass of the resting sphere is greater in theratio 1 + ( : 1 (e small), then the angle between the directions of motion willexceed a right angle by J c tan <j> approximately, where (f> is the deflectionof the moving sphere. [M. T. 1927]

24. A heavy perfectly elastic particle is dropped from a point P on theinside surface of a smooth sphere. Prove that the second point of impacton the surface of the sphere will be in the same horizontal plane as thefirst if the augular distance of P from the highest point of the sphere is

cos-»{(2* + l)*/2}. [S. 1919]

25. A sphere of mass m impinges directly on a sphere of mass m' at reston a smooth table. The second sphere then strikes a vertical cushion atright angles to its path. Shew that there will be no further impact of thespheres if m (1 + e' + ee') < em!; where e, e' are the coefficients of restitutionbetween the spheres and between the sphere and the cushion. [S. 1921]

26. Two equal flat scale pans are suspended by an inextensible stringpassing over a smooth pulley so that each remains horizontal. An elasticsphere falls vertically and when its velocity is u it strikes one of the scalepans and rebounds vertically. Shew that the sphere takes the same timeto come to rest on the scale pan as it would if the scale pan were fixed.

[S. 1924]

27. A ball is projected on a pocketless billiard table. Shew that, if theeffect of friction and rotation be neglected, it will travel always parallel toone of two fixed directions so long as it strikes the four cushions in order:and that the velocity is decreased in the ratio e2:1 after each completecircuit, e being the coefficient of restitution. [S. 1922]

28. A bullet of (H lb. weight is fired with a speed of 2200 feet persecond into the middle of a block of wood of 30 lb. weight, which is at restbut free to move. Find the speed of the block and bullet afterwards, andthe loss of kinetic energy in foot-pounds. What becomes of this energy 1

[M. T. 1917]

EXAMPLES 153

29. A bullet of mass m is fired into a block of wood of mass M, which isfree to move on a smooth horizontal table, and penetrates it to a depth a.Shew that, at the instant when the bullet comes to rest relative to theblock, the block has moved through a distance ma/(M+m), the stressbetween the bullet and the block being assumed constant, so long as thereis any relative motion. [S. 1911]

30. A, B, C are three equal particles attached to a light inextensiblestring at equal intervals a. The system is placed on a smooth horizontalplane with the three particles in a straight line. A blow P is applied tothe middle one B in a direction perpendicular to the string. Describe thenature of the subsequent motion, and shew that the angular velocity of ,42?or BC is P/rna (2 +cos 0)i where 6 is the angle ABC and m the mass of aparticle. [S. 1913]

31. Four particles, each of mass m, are connected by equal inextensiblestrings of length a and lie on a table at the corners of a rhombus the sidesof which are formed by the strings. One of the particles receives a blow Palong the diagonal outwards. Prove that the angular velocities of the

strings after the blow are equal to P sin a/2ma, where 2a ( o < -j) is the

angle of the rhombus at the particle which is struck. [S. 1926]

32. Three particles A, B, C each of the same mass rest on a smoothtable at the corners of an equilateral triangle; AB and BC being tightinextensible strings. A is given a velocity v in the direction CB. Shewthat when the string AB again tightens C starts off with velocity ^gv.

[S. 1921]

33. Three equal particles A, B, C connected by inelastic strings AB,BC of length a lie at rest with the strings in a straight line on a smoothhorizontal table. B is projected with velocity V at right angles to AB.Shew that the particles A and C afterwards collide with relative velocity

2F/V3.

If the coefficient of restitution is e, find the velocities of the threeparticles when the string is again straight. ^ [S. 1927]

34. Three masses mu ra2 and m3 lie at the points A, B and C upon asmooth horizontal table; A and B, B and C are connected by light in-extensible strings, and the angle ABC is obtuse. An impulse / is appliedto the mass m3 in the direction BC: find the initial velocities of themasses and shew that the mass m2 begins to move in a direction makingan angle 6 with AB where

m2 tan 6 + (mx + m2) tan B = 0. [S. 1926]

154 IMPULSIVE MOTION [XI

35. Four equal masses are attached at equal distances A, B, C, D atpoints on a light string, and so placed that L ABC=LBCD = 12O°, and thevarious parts of the string are straight; an impulse / is given to the massat A in the direction BA, shew that the impulsive tension in AB is Jf/.

[S. 1910]

36. A light string passing through a smooth ring at 0 on a smoothhorizontal table has particles each of mass m attached to its ends A and B.Initially the particles lie on the table with the portions of string OA, OBstraight and 0A = 0B. An impulse P is applied to the particle A in adirection making 60° with OA. Prove that when B reaches 0 its velocityisP\/22/8?». [S. 1923]

37. Two equal particles connected by an elastic string which is at itsnatural length and straight, lie on a smooth table, the string being suchthat the weight of either particle would produce in it an extension a.Prove that if one particle is projected with velocity u directly away from

the other, each will have travelled a distance nu I jr- when the string

first returns to its natural length. [S. 1925]

38. A man of mass m is standing in a lift of mass M, which is descendingwith velocity V; the counterpoise being of mass M+m. Suddenly the manjumps with an impulse which would raise him to a height h if he werejumping from the ground. Calculate the velocities of the man and the lift,immediately after the impulse; and find also their subsequent accelerations.Deduce that the height in the lift to which he jumps is h (M+m)/(M+^m).

39. A particle of mass M is at rest at a point A on a smooth horizontalplane. It is attracted towards another point B in the plane by a forceproportional to its distance from B. At the instant at which it is releasedit is given an impulse / in the direction AB. If AB=Xo, and if the initialvalue of the attractive force is Po , shew that the particle reaches B aftera time

P<sFind an expression for the rate at which the force attracting the particle

towards B is doing work, and shew that it has a maximum value

2. « : 1 ; in-. 3. 3.

i) [M-T.1926]

A N S W E R S

u;

(m sin2 6 - 3m'e cos2 6) mu (1 + e) sin 6 cos 6i 2 ' 2 0 ' 2 l

16. '8 ft. above anvil, downward velocity -8 ft.21. 160 ft.; 360 ft. 28. 7-3 f.s.; 7537'3 ft.-lb.; converted into heat.33.

X

Chapter XII

POLAR COORDINATES. ORBITS

121. Velocity and Acceleration in Polar Coordinates.Let the position of a point P bedefined by its distance r from a fixedorigin 0 and the angle 8 that OPmakes with a fixed axis Ox. Thecartesian coordinates (x, y) of P areconnected with the polar coordinates(r, 8) by the relations x = r cos 8,y — r sin 0.

Let u, v denote the components of velocity of P in the directionOP and at right angles to OP in the sense in which 0 increases.The resultant of the components u, v is also the resultant of thecomponents x, y. Therefore by resolving parallel to Ox and Oywe get

u cos8 — vsin8 = x = -.-(rcosd) = rcos8 —rdsin8,

and u sin 8 + v cos 8 = y = -j- (r sin d) = r sin 0 + rd cos 8.at

Solving these equations for u and v clearly gives

u = r and v — rd,and these are the polar components of velocity.

In like manner i f / i , / 2 denote the components of accelerationalong and at right angles to OP, since these have the sameresultant as x and y, we get

d2

/ i cos 8 - / z sin 8 = x = -^ (r cos 8)

- (r - rd2) cos 0 - (rd + 2rd) sin 8,

and " ' / . . - - . / . •• d 2

= (r - r62) sin 8 + (rd + 2r0) cos 0;

giving on solution fi = r — 2rd.

156 POLAR COORDINATES. ORBITS [XII

These components constitute a third representation of thevelocity and acceleration of a point moving in a plane (see 5"12);they are sometimes called radial and transverse components,and we note that the transverse component of acceleration may

also be written - -7- (r20).rat

12-2. Central Orbits. If a particle is describing an orbitunder the action of a force directed to a fixed point

(i) the orbit must be a plane curve; because at any instantthe particle is moving in the plane through the tangent to itspath and the fixed point and the only force acting on theparticle lies in this plane, therefore the particle continues tomove in this plane.

(ii) the rate of description of area by the radius vector drawnfrom the fixed point to the particle is constant; for there is noforce at right angles to the radius vector so that the transversecomponent of acceleration is zero throughout the motion,

i.e. —JT (r2d) = 0, therefore r28 = constant. Since a sectorialr dtelement of area is ^r2B6, it follows that r2d is twice the rate ofdescription of area by the radius vector. It is usual to denotethis constant by h; and we note that r2d is also the moment ofthe velocity about the fixed point and that the following formsare equivalent:

h = r^d = pv = xy — yx.

y y

P

12-21. Given the Orbit and the Centre of Force todetermine the Law of Force. There are two common formulaefor the law of force, one for use when the orbit is given by its(r, p) equation, where r denotes the radius vector and p theperpendicular from the origin to the tangent, and the other for

121-12-21] CENTRAL ORBITS 157

use when the orbit is given by its polar equation, i.e. a relationbetween r and 9.

(i) Orbit given by (r, p) equation. Let / denote the requiredforce per unit mass, i.e. the accelerationtowards the centre 0.

Then, if v is the velocity, the accele-ration along the inward normal istherefore by resolving along the normal

r Pm

But p = rdr/dp and vp — h, therefore

J p3p p3 dr *- '"

(ii) Orbit given by (r, 6) equation. If we write u for - , and

make use of the formula

P \de)'to be found in books on Calculus, we have

_ ± ( _ + (/p3 dr dr VpV d6 \ \deJ )/d6'

and since -^ = 5 -j^, therefore the last relation reduces todd M2 d6'

1 dp 2 ( d*up 3 dr {

and (1) may be written

Alternatively, we may start from the polar components ofacceleration and write

and the latter giving r2<? = h, it follows that

d h d _ , 2 dJt^^d6= dd'

where w = - .r

158 POLAR COORDINATES. ORBITS [XII

Substituting

r — r82, we get

Substituting this operator for ~ and - for r in the expressionCbv It

/ = —to2-— Ihu2 JTJ (-)\+h2u3J dd { d6 W j

h*

or /

as before.

12 22. The converse problem—given the law of force to findthe orbit—is solved by substituting the given expression for fin formula (1) or (2) and then integrating the resulting equation,and provided that the orbit can be identified by its (r, p)equation it is clear that (1) is the simpler formula to use asthe orbit will be found by a single integration.

12-3. Circular Orbits. Since a particle describing a circlewith uniform velocity v has a constant acceleration v2/a towardsthe centre, where a is the radius, it follows that a particlecan describe a circle under any constant force per unit mass ftending to the centre provided that it is projected at rightangles to the radius with velocity *J(af).

Circle with any internal point 0 as centre of force. Let a bethe radius and c the distance of 0from the centre C. Then if P isany point on the circle and p theperpendicular from 0 to the tan-gent at P, and OP = r, it is easyto see from the triangle OOP that

so that dp/dr = r/a.Therefore the force towards 0 under which the particle

would describe the circle is. h2 dp _ h2r

^ ps dr paa'

the velocity at P being given by vp = h, or v = - .

12-21-12-4] CIRCULAR ORBITS 159

In the special case in which 0 is on the circumference, c — aand r2 = 2ap, so that

2 , 2ah

'and U = T T ;

or and » =

Conversely a particle projected from a point P at a distance rfrom 0 with velocity V^ytt/r2 under the action of a force /i/r5

per unit mass towards 0 will describe a circle passing through0, and the position of the centre of the circle and therefore alsoits radius depends only on the direction of projection; for weconstruct the position of G by drawing PG at right angles tothe direction of projection and making the angle POG equalto OPG.

124. Elliptic Orbit. Force directed to the Centre. LetP be a point on the ellipse whose centreis G and semi-axes a, b.

If p is the central perpendicular onthe tangent at P and CD is the radiusconjugate to CP, from the properties ofthe ellipse, we have

and p.CD = ab.

Hence, if GP = r, we have

a2b2

.(1).

This is the (r, p) equation of the ellipse when the centre isthe origin. By differentiating (1) we get

a2b2 dpp3 dr

Hence the force to the centre necessary for the description ofthe ellipse is

, h*dp A2

where ft is a constant h?/a2b2, or where h — a6 V'/*•

160 POLAR COORDINATES. ORBITS [XII

Again the velocity at P is given by

p ab'and is parallel to CD.

Also since h is twice the rate of description of areas, thereforethe time taken to describe the ellipse or the periodic time inthe orbit = twice the area divided by h

= 2-n-ab/h = 2-Tr/s/fi.

Let A be an end of the major axis and Q the point on theauxiliary circle corresponding to the point P on the ellipse;then if t denotes the time taken for the particle to move fromA to P we have t = (2 sector ACP)/h.

But the ellipse is the projection of the circle, so that

area A GP: area A GQ = 6 :a,

and area A GQ = | a 2 . A GQ;

therefore area A GP = \ab.A GQ,

cUid it ^— ~z—, / L o u = = -ti.\j\j)I\f LL.

or the angle AGQ = V'fit. It follows that as P moves round theellipse the corresponding point Q moves round the auxiliarycircle with uniform angular velocity *//"••

Further, the coordinates of the point P in terms of theeccentric angle ^1 fit are given by x = a cos 'Jfit, y — b sin Vfit.This shews that the elliptic motion can be compounded of twosimple harmonic motions along lines at right angles having thesame period 2n/n/fj, and differing in phase by a quarter of aperiod 7*11.

1241. Law of Force ftr. Find the Orbit. The law of forceh? dp ,

gives - s ~ = fir, and on integration

But the (r, p) equation of an ellipsewith the centre as origin is

so that the orbit is in general an ellipse.

124-12 - 5] ELLIPTIC ORBIT. FORCE DIRECTED TO FOCTTS 161

The constants involved are determined by the initial circum-stances. Thus, if the particle be projected with velocity ufrom a point P at a distance c from the centre of force C in adirection making an angle a with CP, we have

h = moment of velocity about G = we sin a;and since h/p always measures the velocity, therefore substi-tuting in (1) gives w2 = A — /MC* or4=M2 + fie2.

Now by comparing (1) and (2) we get„ ,„ A i

andh? u2c2 sin2 a.

These equations determine the lengths of the semi-axes ofthe ellipse in terms of the data /n, u, c, a. The inclination 6 ofthe major axis to GP is then to be found from the polarequation of the ellipse, viz.

cos2 6 sin2 6 _ 1_a2 b* r*

125. Elliptic Orbit. Force directed to Foeus. Let S, Hbe the foci of an ellipse and SY,HZ the perpendiculars to thetangent at P. To find the (r, p)equation with 8 as origin weassume three properties of theellipse

(ii) 8Y.HZ=b\(iii) the tangent is equally inclined to the focal distances

so that SPY, HPZ are similar triangles. Hence, if SP = r,HP = r',SY=p and HZ = p\ we have

b

Hence -

£=^7, therefore also = . / 4^- = 7 r — ^ .r r V rr *J{r(2a — r)\

2a-r\1, or, squaring,

62 = 2a

jp2 r

p V I.(1).

162 POLAR COORDINATES. ORBITS [XII

Therefore -5 —- = -5, and the force to the focus 85pi necessaryJ

for the description of the ellipse is given by. _ h2 dp _ h2a 1 _ n

•(2),

where p = h2a/b2 or h2 = fil, if I denotes the semi-latus rectum b2ja.Again the velocity v = h/p, therefore

„ A2 h2 (2a

(bu t h2 — ub2la, therefore i>2 = — ( 1 ) ,' a \r J'

orr a

Again, as in 124, the periodic time = 2irabjh

.(3).

(4).

12-51. Parabolic Orbit. Force directed to Focus. The(r, p) equation of a parabola isobtained from the facts that if thetangent at P meets the tangent atthe vertex A in Y and S is thefocus, then SY is at right angles toPY and the triangles ASY, YSPare similar. This gives

P2 = ar (1),1 1

or -g = — ,

so that by differentiation2 dp _ 1

p3 dr ar2'Hence the force to the focus necessary for the description ofthe parabola is given by

h? dp _h2 1 _{ h

where /J, = h2/2a, or h2 = /J, x semi-latus rectum.Also, if v is the velocity at P,

a _ _ ^

p2 ar.(3).

1 2 5 - 1 2 5 3 ] LAW OF FORCE fj,/r2. FIND THE ORBIT 163

1252. Hyperbolic Orbit. Force directed to Focus. Tofind the (r, p) equation of ahyperbola we assume the cor-responding properties as forthe ellipse, viz.:

(i) SY.HZ=b2,(ii) r'-r = 2a,(iii) the tangent bisects the

angle between the focal dis-tances.

Hence as in 12'5 we have

so that

and by differentiation

p=p= /PP_ = or r' V rr' Vr(2a +

-s = h ipi r

•(1),

Therefore •(2),

b2 dp _ ap3 dr r2'

_h2 dp _h2a 1 _ /x.p3 dr b2 ' r2 r2' ^

where /M = h2a/b2, or A2 = yu.?, if I denotes the semi-latus rectum62/a.

In this case2_h2 _h2 (2a \_fi(2a \

7)2 ~~ b2 \ r 1 a \ r /

or «2 = -£-4-C (3)r a v 7

gives the velocity.

12-53. Law of Force /t/r2. Find the Orbit. We now haveh2 dp _ (ip3 dr r2'

and, by integration, thereforeTO o

—a = — + C, where C is a constant (1).p* r v '

Comparing this with the (r, p) equations of the ellipse,parabola and hyperbola obtained in the last three articles, viz.:

b2 _ 2a _ 2 _ b2 _ 2ap2 r ' " ' p2 r '

164 POLAR COORDINATES. ORBITS [XII

we see that the orbit required is an ellipse, parabola orhyperbola according as the constant of integration G in (1) isnegative, zero, or positive.

Now hjp always denotes the velocity, so that if the particleis projected with a velocity V from a point at a distance c fromthe centre of force, by substituting in (1) we have

(2),

and therefore C is negative, zero or positive according as

F 2 < , = or >2/*/c.

Hence the required orbit is an ellipse, parabola or hyperbolaaccording as

F 2 < , = or >2fi/c.

Also by comparing (2) of this article with (3) of the last threearticles, we see that

in the ellipse G = — fija,

in the parabola C = 0,

in the hyperbola G = p/a.

Therefore the semi-major axis of the ellipse or hyperbolais given in terms of the initial velocity and distance by theequations

c a'

Also when the initial circumstances are given h, the momentof the velocity about the centre of force is known, and then ineach case the semi-latus rectum = h?/fi, so that the dimensionsof the orbit are completely determined. The determination ofthe position of the major axis is left as an exercise for thestudent.

12-54. Velocity Components. When a particle describes anellipse about a centre of force in the focus the velocity can beresolved into two components of constant magnitude, one perpen-dicular to the radius vector and the other perpendicular to themajor axis.

12-53-12-55] VELOCITY COMPONENTS 165

Let SP meet HZ in H', then since the tangent is equallyinclined to SP, PH it is easy toprove that HZ=ZH', but HC=CS,therefore CZ is parallel to SP.

Again, the velocity at P in theellipse is

A hand its direction is perpendicularto HZ.

But a vector HZ is equivalent to vectors HC, CZ; thereforethe velocity at P may be represented by components

™ HC perpendicular to HC,

and f^ CZ perpendicular to CZ.

Since HC = ae where e is the eccentricity, and CZ = a and is

parallel to SP therefore the velocity components are -^- or -|-

perpendicular to the major axis, and y- or r perpendicular

to SP.It is easy to see with the help of a

figure in what sense these componentsmust lie in order always to give aresultant in the direction of themotion of the particle, and we noticethat the component e/xjh always actsin the same sense and the componentfi/h is always directed in front of theradius vector SP as it revolves.

12 55. Velocity from Infinity. In connection with centralorbits the phrase 'velocity from infinity' at any "point of anorbit means the velocity that a particle would acquire if itmoved from infinity to that particular point under the action ofan attractive force in accordance with the law associated withthe orbit.

166 POLAR COORDINATES. ORBITS [XII

Thus if f stands for acceleration towards the origin, byresolving along the tangent to the path, we get

dv ArVds = ~-'ds'

and therefore the velocity from infinity is given by

\v*=-\r fdr.J 00

For example, if / = p/r2,v2 = 2fi/r,

so that the orbit is an ellipse, parabola or hyperbola accordingas the velocity at any point is less than, equal to or greaterthan the velocity from infinity (12'53).

1256. The Hodograph. It is convenient here to mentionthe hodograph, for though it has no special connection withcentral orbits they afford some of the simplest illustrations of it.If from a fixed origin 0 a line Oa is drawn to represent thevelocity of a moving point P the locus of a is called the hodographof the path of P.

Referring to the figure of 5'11, when the points P, Q aresufficiently near to one another, ab may be regarded as anelementary arc of the hodograph. It is there shewn that the

acceleration of P is lim -=-, but this is the velocity of the pointM-»O ot

a in the hodograph, so that the velocity with which the point adescribes the hodograph is a measure in magnitude and directionof the acceleration with which the point P describes its path.

As examples of the hodograph consider the two cases ofelliptic motion:

(i) when the centre of the ellipse is the centre of force, thevelocity is V/xCD in magnitude and direction (12'4), therefore thehodograph is a similar ellipse;

(ii) when the focus is the centre of force, we saw (12-54) that

the velocity is -^ HZ perpendicular to HZ. But the locus of Z

is the auxiliary circle, therefore in this case the hodograph is acircle.

12-55-126] KEPLER'S LAWS 167

126. Kepler's Laws of Planetary Motion. The followinglaws were announced by Kepler (1571-1630) as the result ofobservations of the planets:

I. The planets describe ellipses with the sun in a focus.

II. The areas described by radii drawn from the sun to aplanet are proportional to the times of describing them.

III. The squares of the periodic times are proportional to thecubes of the mean distances of the planets from the sun.

The discovery of these laws was followed some sixty yearslater by Newton's enunciation of the Law of Gravitation:

Every particle in the universe attracts every other particlewith a force which is directly proportional to the product of theirmasses and inversely proportional to the square of the distancebetween them. Thus, if m, m' denote the masses of two particlesand r their distance apart, the force of attraction between them

mmis 7 —g-; where 7 is a constant known as the gravitation

constant, representing the attraction between two particles ofunit mass at unit distance apart.

In his Principia Newton proved a series of propositions inMechanics which enable us to follow a process of reasoningfrom Kepler's Laws to the Law of Gravitation.

Thus Newton proved that Every body, which moves in anycurved line described in a plane, and describes areas proportionalto the times of describing them about a point either fixed ormoving uniformly in a straight line, by radii drawn to thatpoint, is acted on by a centripetal force tending to the samepoint {Principia, Sect. II, Prop. 11).

Hence by combining this proposition with Kepler's first andsecond laws, it follows that the planets, which describe ellipsesabout the sun in a focus, and describe areas proportional to thetimes of describing them by radii drawn to the sun, must beacted upon by forces directed to the sun.

Newton also proved that If a body describes an ellipse underthe action of a force tending to a focus of the ellipse, the forcemust vary inversely as the square of the distance {Principia,Sect, in, Prop. xi).

168 POLAR COORDINATES. ORBITS [XII

Hence by Kepler's first law the sun must attract a planetwith a force varying inversely as the square of the distance.

Then in the same section of the Principia, Sect. Ill, Props, xivand xv, Newton shewed that If any number of bodies revolve abouta common centre, and the centripetal force vary inversely as thesquare of the distance, the latera recta of the orbits described willbe in the duplicate ratio of the areas, which the bodies willdescribe in the same time by radii drawn to the centre of force.And On the same supposition, the squares of the periodic times inellipses are proportional to the cubes of the major axes. Thesetwo propositions are demonstrated on the hypothesis that thereis a common law of force for all the different bodies, i.e. thatif the force be denoted by /i/r2 then fi is the same for all thebodies revolving about the common centre, and only on thishypothesis would these theorems be a logical consequence ofwhat precedes. But Kepler's third law states as a matter ofobservation that for the planets the squares of the periodictimes are proportional to the cubes of the mean distances (ormajor axes). The inference is that for the attraction of the sunon the planets there is a common law of force /n/r2 per unitmass, where p is the same constant for all the planets. Andsince this expression /i/r2 is the force per unit mass or theacceleration imparted to the planet by the sun, therefore thewhole force exerted by the sun on the planet varies as (massof planet)/?-2. It is only a short step further to infer that theforce is also proportional to the mass of the sun, and the law ofuniversal gravitation is a generalization from these inferences.

Calculations based upon the law of gravitation were sufficientto enable the astronomers, Adams and Le Verrier, independentlyto determine the existence and position of the planet Neptunebefore it had been actually observed, and predictions of thereturn of comets have been fulfilled. In fact the law has provedadequate as a basis for dynamical astronomy generally, thoughthere are a few known phenomena where calculations shewminute differences from the results of observation, and thesehave been explained and accounted for by the substitutionof Einstein's theory of Relativity for the absolute Newtoniandynamics.

12-6-12-61] KEPLER'S LAWS 169

1261. Necessary Modification of Kepler's Third Law.In the preceding argument the acceleration produced by theaction of the sun on a planet is considered and the accelerationof the sun produced by the action of the planet is ignored.Both should be taken into account, for the observations madeare based on the motion of the planet relative to the sun andnot on an absolute motion of the planet.

Assuming that the sun and planet attract one another liketwo particles, if S and P denote their masses and r theirdistance apart, the force on each towards the other is ySP/r2,where 7 is the gravitation constant.

• P

jSPjr2

Therefore their accelerations are as shewn in the nextdiagram,

s • *• -*—• p

yP/r2 yS/r*

and to find the acceleration of the planet relative to the sun wemust add to the acceleration of both that of the sun reversed indirection,

S • « • P

so that the acceleration of the planet relative to the sun is7 (S + P)/r2, and this is the /a/r2 of our theory. Hence theperiodic time for the relative motion is

and Kepler's third law should read ' The square of the periodictime is proportional to the cube of the mean distance andinversely proportional to the sum of the masses of the sun andthe planet.'

Since the mass of the largest planet, Jupiter, is less thanone-thousandth of the sun's mass it is clear that this correctionto the law is a small one.

170 POLAR COORDINATES. ORBITS [XII

It must be remembered that the units are such that ymm'/r2

is a force, so that in terms of fundamental units

therefore <y = M~\and this makes 2irefi/\/{y(S + P)} of one dimension in time.

12*62. Examples , (i) The eccentricity of the earth's orbit round the sunis 1/60; prove that the earth's distance from the sun exceeds the semi-axismajor of the orbit during about 2 days more than half the year.

[M. T. 1908]If S be the sun in the focus of the ellipse ABA'B', then, since £5=semi-

axis major, therefore the distance from S ofa point P on the curve exceeds the semi-axismajor so long as P is on the arc B'A'B, andsince time of describing an arc is proportionalto sectorial area, therefore the time required

area, SB1 A'BS , ,= T of a year, where a, 6 are the

semi-axes,. , SC.GB , e , , .

=$H ,— = § + — of a year, where e is eccentricity,irao 7r

a year + _ x : days,

= \ a year + about 2 days.(ii) The greatest and least velocities of a certain planet in its orbit round

the sun are 30 and 29'2 kilometres per second. Find the eccentricity of theorbit. [M. T. 1919]

Since the moment of the velocity about the focus S is constant, thegreatest and least velocities are where thedistances from S of the tangent to the pathare least and greatest, i.e. at the ends A, A'of the major axis.

Hence

or

therefore 59i2e = '8 or e = l/74.

(iii) The acceleration of a particle in the gravitational field of a staris fi/r2 towards the centre of the star, where p. is a constant and r is thedistance from the centre of the star. A particle starts at a great distancewith velocity V, the length of the perpendicular from the centre of the staron the tangent to the initial path of the particle being p. Shew that theleast distance of the particle from the centre of the star is X, where

12-61-12'7] EXAMPLES. LAW OF INVERSE SQUAEE 171

If H = l, F=2, in the system of units chosen, and if the radius of the staris 0 005, prove that the particle will strike the star if p is less than 0'05.

[M. T. 1925]

Since the particle starts from a great distance, we may assume thatF2 > 2fi/c, where c denotes this great dis-tance, so that by 12*53 the orbit is ahyperbola, with the centre of the star atthe focus S.

Initially the particle is moving alongthe asymptote at a great distance, and p,the perpendicular SF from the focus tothe asymptote, is known to be equal tothe semi minor axis b. This also followsfrom the properties of the hyperbolic orbit;for V2p2 = h2 = jxtfja (12'52), and the ve-locity v at distance r from 8 is given by v2= — + - , but at a great distance

T OJ

we may put r = oo and v — V, so that F2 = pja; therefore from above p2 = b%.Now the least distance of the particle from S is AS ; therefore, if e

denotes the eccentricity,

X = SA= CS - CA = a (e -1) ,

but p2= 62 = a2 (e2 -1 ) , and V2a=p,

therefore Vipi = H2(e2-1) or fi

2e2=/i2+ Vlp2.

Hence F2X

or F2X = ( F2 + F V ) * - F -

Substituting the numerical values fi = l and F=2 we have

and the particle will strike the star if it would pass so that its leastdistance from the centre is less than the radius, i.e. if X < '005, or if4X < "02. Therefore the necessary condition is

(1 + 1 6 ; B 2 ) * - 1 < - 0 2 ,

or 1 + 16^2 <(l-02)2,

or p < '05.

127. Use of u, 6 Formulae. So far we have only made useof the formula which expresses the law of force in terms ofr, p, and this method is adequate for most simple problems.We shall now make use of the alternative formula

172 POLAK COORDINATES. ORBITS [X I 1

Law of force /AM2. Find the orbit.Since f= /nu2, therefore

d?u fi

Putting u =u — /A/h?, the equation becomesd V , „

the integral of which is u' = A cos (6 — •&) (see 17 (16)), or

u=~ + A cos(O-vx),

where A, vr are constants of integration. This equation may bewritten

AW a

1 | ( 0 )(1),| ( )

r fju

which we recognize as the polar equation of a conic with thefocus as origin and semi-latus rectum A2//*, since the typicalform of the equation of a conic is

- = 1 + e cos 6.rThe constants of integration can be found when the circum-

stances of projection are known and the form of the conic canbe determined, but the details are not so simple as in 1253,where the same problem is solved by the (r, p) equation.

1271. Inverse Cube. Let the force to the centre be fiu3,then we have

d2u _ fid&+U = h?U-

The solution of this equation is different in form according as/A/A,2 > = or < 1.

First let /j,/h? —l = n2, then the equation is

the solution of which is u = AenB + Be~n0 (17 (14)), where theconstants A, B depend on the circumstances of projection; ifthese are such that either A or B is zero the path is an equi-angular spiral.

12-7-12-7 2] APSES 173

Secondly, when /j,/h2=l, the equation is -^ = 0, the solution

of which is u = Ad + B, the curve known as the reciprocalspiral.

In the third case, putting 1 — fijh? = n2, the equation is

and the solution isu — A COB nd + Bsxund 1-7(16),

a curve with infinite branches.

1272. Apses and Apsidal Distances. An apse is a point ina central orbit at which the normal to the curve passes throughthe centre of force, and the length of the radius vector at sucha point is called the apsidal distance.

Whenever the central force is a function of the distance,the velocity and the inclination of thepath to the radius vector are alsofunctions of the distance; for by re-solving along the tangent to the pathwe get

dv _ j.drVte~~~fds''

therefore \v* — G — j/dr,

which is a function of r alone.Again, if <f> is the angle between the tangent and radius

vectorsin <f> = pjr, but pv = h,

therefore sin <f> = h/rv,

which is also a function of r.It follows that at all points in the orbit which are at the

same distance from 0 the velocity v and the inclination <f> havethe same values. Hence an apse line, Le. a line from the centreto an apse must divide the orbit symmetrically, as is alsoobvious from the fact that if particles were projected from anapse in opposite directions at right angles to the apse linethey would describe the same curve in opposite senses.

174 POLAR COORDINATES. ORBITS [xil

There can therefore be only two different apsidal distancesthough there may be any number of apses; for if A, B, C, D,etc. are consecutive apses, by symmetry about OB we haveOA — OC, by symmetry about OG we have OB = OD and so on.

The analytical condition for an apse, i.e. that the tangent tothe curve must be at right angles to the radius vector, is thatdujdO vanishes and changes sign as 6 increases through thevalue that indicates the position of an apse.

In the motion of a planet about the sun there is a singleapse line, namely the major axis of the orbit. The apse nearerto the sun is called perihelion and the further apse is calledaphelion. In like manner if the relative motion of the earthand sun be ascribed to the sun the apse nearer to the earth iscalled perigee and the further apse is called apogee.

12-73. Example. When it is required to find an orbit for a morecomplicated law of force it is generally necessary to make use of the (u, 8)equation, and we shall illustrate the general method of procedure by otherexamples.

If the law of force is 5fiv?-\-8fic2u5, and the particle is projected from anapse at the distance c with velocity 3 V ulc; to prove that the orbit is r=ccos 2fl/3.

,„ (dlu \ „ ,We have h21 -j^ + u I = 5/iw + 8nc* u3.

\at> /Multiply by 2du/d6 and integrate and we get

To determine the constant C we substitute the initial values and we may

do this either by observing that since ( ~ \ + «2=—5, therefore the left-\cLu / p

hand side of the equation (1) is h2/p2, i.e. the square of the velocity; or byobserving that at the apse dujd8 = 0 and w = - and that h, the moment of

cthe velocity, is 3 \Jp. Either process gives C—0, and then, since A2=9^,the equation reduces to

This s i

— dror, putting u = l/r, j ^ - - ^ = %d6

Tso that cos ~' - = \6 + a,

c

12-72-12-74] EINSTEIN'S LAW 175

i.e. r= ccos(|# + a), and if we measure 8 from the apse line then r = cwhen 6 = 0, therefore a=0 and r=c cos 25/3.

It should be remarked that on taking the square root in (2) above, thereis a choice of signs, and that whichever sign be chosen we get the sameequation for the path.

1274. Einstein's Law of Gravitation. One of the smalldiscrepancies unexplained by the law of the inverse square is aslight continuous change in the position of the apse line of theplanet Mercury spoken of as 'the advance of perihelion.'

We have seen in 127 that the Newtonian differential equationof the orbit is

d2u _ ft . . .

w + u~w {h

the solution of which as in (1) of 12-7 is of the form

£ - w ) } (2);

and since this makes du/dd = 0 when 6 = ts, therefore OT is theangular coordinate of perihelion.

According to Einstein's Law of Gravitation the differentialequation of the orbit is

The term 3/xw2 is small, for its ratio to /j,/h? is 3h?u2, i.e. threetimes the square of the transverse velocity measured in termsof unit velocity, which in Einstein's theory is taken to be thevelocity of light. For applications in the solar system thisratio is of order 10~8. Hence Einstein's theory adds a smallterm 3fiv? to the right hand side of equation (1). Equation (2)is therefore a first approximation to the solution of equation (3),when the small term on the right is neglected. In order toobtain a second approximation to the solution we next substitutethe value of u given by (2) in the small term on the right of (3)getting

The only one of the additional terms that gives appreciable

176 POLAR COORDINATES. ORBITS [XII

effects is the one that contains cos(0- •or), and neglecting theothers we have

Now it is easily verified that a particular integral of theequation

T™ + u = A cos (0 — tss)

is u = %A0sin(6 — •sr).Hence the solution of (4) is obtained from the solution of (1)

by adding to it a term

~ ed am (6 - *T),

whence we get from (2)

M = ^{l + e cos(0--5r)}+^e0sin(6>- O T ) ...(5).

And this is of the form

g w - 8 t x ) } (6),

*3 2

provided that SOT =-~~ 8 and (Stn-)2 is neglected.ft

Now OT is the angular coordinate of perihelion and 6 is thecorresponding angular coordinate of the planet; and (6) shewsthateat any instant the planet may be considered to be movingin an ellipse but that the ellipse is not fixed, for the coordinate•sr which defines the position of the apse line is undergoing acontinuous change proportional to the angular motion of theplanet in the ratio

SOT _ V s _ S/J,

~W ~ h? ~ a(l-e2)'since h? = fil= /j,a (1 — e2).

This calculated motion of the apse line in the case of theplanet Mercury accords closely with the results of observation.

12*741. Example . Shew that the orbit is a conic when the law of forceis fin?, and find the angular advance of the apse line in one period when thevelocity in the orbit is less than the velocity from infinity and the centralacceleration contains a small extra term, aw3. [M. T. 1927]

12'74-12'75] ROTATING OKBIT 177

We have already seen that but for the small extra term the orbit wouldbe an ellipse

£ - r a - ) } (1).

In this case the differential equation is

From this point we might proceed as in the last article to regard (1) asa first approximation to the solution of (2) and substitute the value of ugiven by (1) in the small term on the right of (2) and follow closely theprocedure of the last article. But inasmuch as (2) can be solved in explicitterms we may proceed directly to the solution. Thus (2) may be written

and, regarding a— ~^— as the dependent variable, the solution of thisft — <x

differential equation is by (16) of 1'7

where A, •m are constants of integration.Neglecting a2 this may be written

= p f i + cos (8 -zt- (5),

provided that — = ~-^.

The form of (5) shews that at any instant the orbit may be regarded asan ellipse whose apse line is advancing at the rate of air jh2 in a revolution.

1275. Principles of Energy and Momentum applied toCentral Orbits. Problems on centralorbits can also be solved by applyingthe prinoiples of energy and mo-mentum. Thus, if f denote the forceper unit mass towards the origin, thework done in a small displacementds is

,dr ,-f-r-ds, or -fdr;

178 POLAR COORDINATES. ORBITS [XII

and since f and rd are the components of velocity we have

(1),where C is a constant.

Also the moment about 0 of the momentum is constant,therefore

r2d = h (2).

Equations (1) and (2) are sufficient for the determination ofthe orbit, if/ is a given function of r.

12*751. Example . A particle moves in an orbit under a central ac-celeration fijr% along the radius vector r; obtain the equations of energy andangular momentum.

If the particle is projected with velocity u at right angles to the radius, atdistance c from the origin, prove that

The equation of energy is

or i(r* + rW) = £+C (1),

and the equation of angular momentum is

r20 = h (2).

Initially r=c, r=0, and rd = u.

Substituting in (1) we find %u? = - + C,c

and in (2) cu — h\

therefore ^ ^

and rz8=cu.

Eliminating 8 we find that

12-8. Repulsive Forces. The motion of a particle under theaction of a central repulsive force can be discussed in likemanner.

12-75-12-9] REPULSIVE FOKCES 179

For example, when the repulsive force is inversely as thesquare of the distance, we maywrite

p^dr = ~ ^ '

so that -^=0—- ...(1).j)2 rNow in 1252 we see that

for a hyperbola

f=r'so that for points on the branch remote from the focus S we have

bz r — 2a „ 2a•(2),

p* r rand a comparison of equations (1), (2) shews that

G = fi/a, and h? = idPja.Hence the particle would describe the farther branch of the

hyperbola under a repulsive force /x/r-2, and from (1) the velocitywould be given by

2 fi 2fi

a r12 9. Motion of Two Particles under their Mutual At-

traction. If two particles are moving under the action of noforces but their mutual attraction, there is no external forceacting upon the system as a whole and therefore the centre ofgravity G is either at rest or moves uniformly in a straightline, 9-3.

If the velocities of the particles at any instant are knownthe velocity of the centre of gravity can be calculated. If thevelocity of the centre of gravity reversed in direction is com-pounded with the velocity of each particle, the centre of gravityis reduced to rest and the velocities of the particles becometheir velocities relative to the centre of gravity. The accelerationof each particle can then be expressed in terms of its distancefrom the centre of gravity and thus everything necessary forthe determination of the orbits relative to the centre of gravityis known.

180 POLAR COORDINATES. ORBITS [XII

1291. Law of Inverse Square. Thus if m, m are the massesof the particles P and Q and G is their centre ofgravity, and PG = r, QG = r' and ymm j(PQ)2 isthe force between the particles, we have

PQ .rm'

rm m

and the acceleration of P is

towards Cr.or

7TO'

(m + TO )2 r2

Hence the path of P relative to G is a conic of which G is afocus. In like manner the path of Q relative to G is a similarconic, as also are the paths of the particles relative to oneanother. The form of the conic depends in each case on theinitial circumstances. For example, if V is the velocity of Prelative to G, its path is an ellipse, parabola or hyperbola,according as

7 2 <, = or > 2m'3/(m + TO'2) r.If on the other hand we wish to find directly the path of P

relative to Q, we can do so by first giving to both particles thereversed velocity and acceleration of Q, so that Q is therebyreduced to rest. Then the path determined for P is its pathrelative to Q.

12*92. Example . Two gravitating particles of masses M, m encounterone another after approach from a great distance. If at a great distance thevelocity of one relative to the other is V along a line distant p from this otherparticle, and <j> is the angle through which the relative velocity is turned bythe complete encounter, prove that

tan I </>=y (M+ m)l V2p.Shew further that if the two

particles have equal masses andone is initially at rest the finalvelocities are

Veos^cj) and Vsin^<p.The force between the par-

ticles is yMmjr2, so that theaccelerations of M and m areym\rh and yM/r2 towards oneanother; and the acceleration ofM relative to m is y(M+m)jr2.And since at a great distance

12-91-12-92] TWO GRAVITATING PARTICLES 181

if r is taken large enough, therefore the orbit of M relative to m isa hyperbola, and M is initially moving in the direction of an asymptotewith velocity V and the asymptote is at a distance p from the focus.Hence p is the semi-conjugate axis of the hyperbola.

Again if v is the relative velocity at distance r

(1),

and by substituting the values r=oo, v= Vwe get

The angle <f> is the angle between the asymptotes indicated in the figure,because after the encounter M will be at a great distance in the directionof the second asymptote.

Hence tan k<h = - = i^—- '-,*T yi

Again let m be initially at rest and let the final velocities of m and Mparallel to the axes of the hyperbola be u, v and u, v' as indicated in thefigure. The magnitude of the final relative velocity is given by (1) whenr -—<*> and is therefore V. Hence u' — u= Vsin ^(j>, v' — v= Fcos^<£. Butthe final momentum must be the same as the initial momentum, and,since the masses are equal, this gives u' + u — — Fsin \ <f> and v'+v= Fcos J <f>.From these equations we deduce M' = 0, M= — F s i n ^ , «/= Fcos ^<j>, v=0,so that the final velocities of M and m are Fcos \<$> and Fsin \<$>,

E X A M P L E S

1. If the angular velocity of a particle about a point in its plane ofmotion be constant, prove that the transverse component of its accelerationis proportional to the radial component of its velocity.

2. The velocities of a particle along and perpendicular to a radius vectorfrom a fixed origin are Xr* and pO2: find the polar equation of the path ofthe particle and also the component accelerations in terms of r and 8.

[S. 1926]

3. A steamer moving with constant speed, v, relative to the water passesround a lightship anchored in a tideway, keeping the lightship always deadabeam. Shew that the path of the steamer is an ellipse whose minor axisis in the direction of the tidal current and whose eccentricity is iijv. (u isthe speed of the tide and we assume u < v.) [S. 1924]

4. A smooth straight wire rotates in a plane with constant angularvelocity about one end. Shew that a particle which is free to slip alongthe wire may describe an equiangular spiral. [S. 1921]

182 POLAK COORDINATES. ORBITS [XII

5. A particle on a smooth table is attached to a string passing througha small hole in the table and carries an equal particle hanging vertically.The former particle is projected along the table at right angles to thestring with velocity *j2c/h when at a distance a from the hole. If r is thedistance from the hole at time t, prove the results

(ii) the lower particle will be pulled up to the hole if the total lengthof string is less than a + \h+V'ah+$k?,

(iii) the tension of the string is \mg ( ] +—5- J, m being the mass of

each particle. [S. 1926]

6. A smooth straight tube rotates in a horizontal plane about a pointin itself with uniform angular velocity o>. At time ( = Oa particle is insidethe tube, at rest relatively to the tube, and a distance a from the point ofrotation. Shew that at time t the distance of the particle from the pointof rotation is

a cosh at.

Find the force the tube is then exerting on the particle. [S. 1923]

7. If two particles P, Q describe the same ellipse under the samecentral force to the centre C, prove that the area of the triangle CPQ isinvariable. [Coll. Exam. 1902]

8. A particle is projected from a given point P under the action offorces whose accelerating effects are uPS, pPS' directed towards fixedpoints S, 8'. Find the magnitude and direction of the velocity of projectionin order that the orbit may be an ellipse with S, S' for foci. [M. T. 1910]

9. Two particles are describing the same ellipse about a centre of forcein the centre in opposite directions, the mass of one being double that ofthe other. If the particles meet and coalesce at an end of the minor axis,shew that the new orbit trisects the major axis of the old. [M. T. 1911]

10. A particle of unit mass describes an ellipse under the action of acentral force fir. Shew that the normal component of the acceleration atany instant is abffijv, where v is the velocity at that instant and a, b arethe semi-axes of the ellipse. [M. T. 1921]

11. An elastic string has one end fixed at A, passes through a smallfixed ring at B and has a heavy particle attached at the other end. Theunstretched length of the string is equal to AS. The particle is projectedfrom any point in any manner. Assuming that it will describe a planecurve, shew that the curve is in general an ellipse. [S. 1926]

EXAMPLES 183

12. A particle is repelled from a centre of force 0 with a force yx per unitmass, where r is the distance of the particle from 0. Shew that, if theparticle is projected from a point P in any direction with velocity OP \f/i,its path is a rectangular hyperbola with 0 as centre. [S. 1925]

13. Three particles P, Q, R, each of mass m, attract one another with aforce fim2 (distance). They move on a smooth horizontal plane and, whent = 0, are at A, B, C and are then moving with velocities equal in magnitudeand direction to \BC, \CA, \AB. Shew that their centre of gravity, Q,remains at rest, and that each particle describes an ellipse about O in theperiodic time T= 2jr/v/(3/im).

Shew that the area of each ellipse is %\ST, where S is the area of ABC.[M. T. 1922]

14. If a particle be describing an ellipse about a centre of force in thecentre, shew that the sum of the reciprocals of its angular velocities aboutthe foci is constant.

15. Prove that the earth's velocity of approach to the sun, when theearth in its orbit is at one end of the latus rectum through the sun, isapproximately 18£ miles per minute, taking the eccentricity of the earth'sorbit as 1/60, and 93,000,000 miles as the semi-axis major of the earth'sorbit. [M. T. 1909]

16. A body moves under a central force varying inversely as the squareof the distance, the accelerating effect of the force at a distance of one footbeing 32 f.s.s. If the body is projected at right angles to the radius at adistance of -75 foot with velocity 8 f.s., determine the major axis andeccentricity of the orbit. What must be the velocity of projection to makethe orbit a parabola 1 [Coll. Exam. 1914]

17. A particle is projected from a point A at right angles to SA, and isacted on by a force varying inversely as the square of the distance towards 8.If the intensity of the force is unity at unit distance, SA, and the velocityof projection is ^, prove that the eccentricity of the orbit is f, and findthe periodic time. [Coll. Exam. 1910]

18. A particle describes an ellipse under the action of a force to a focus;if the particle on arriving at any point in the orbit is diverted by a blowin the direction of the normal so that the tangent to the new orbit at thatpoint passes through the second focus, prove that the latus rectum of thenew path is four times that of the old. [Coll. Exam. 1911]

19. If a particle is describing an ellipse of eccentricity -5 under theaction of a force to a focus and when it arrives at an apse the velocity isdoubled, shew that the new orbit will be a parabola or hyperbola accordingas the apse is the farther or nearer one. [Coll. Exam. 1912]

184 POLAR COORDINATES. ORBITS [xil

20. Two particles describe in equal times the arc of a parabola boundedby the latus rectum, one under an attraction to the focus, and the otherwith constant acceleration g parallel to the axis. Shew that the accelerationof the first particle at the vertex of the parabola is ~^-g. [M. T. 1915]

21. In an elliptic orbit described under the action of a force to a focus S,if the tangent at P meet any line through S in T, shew that the componentof the velocity at P in a direction perpendicular to 8T varies inverselyas ST. Also, find the two points of the orbit at which the component ofvelocity in any direction LM and the component in the opposite directionML have maximum values; and shew that the sum of the two maximumvalues is the same for all directions of LM. [M. T. 1915]

22. A particle is projected from infinity with velocity V so as to pass afixed point at a distance o if undisturbed. If it is attracted to the fixedpoint with acceleration /x«2, where u is the reciprocal of the radius vector,shew that the equation of the orbit is

u cos 8

6 being measured from the apse. [M. T. 1924]

23. Shew that, in elliptic motion about a focus under attraction /*r~2,the radial velocity is given by the equation

5 ) 2 = {a(l+e)-r}{r-a(l-«)}. [M. T. 1916]

24. Shew that, if a particle describes an ellipse under a force to a focus,the velocity at the mean distance from the centre of force is a meanproportional between the velocities at the ends of any diameter.

25. A planet of small mass is describing an elliptical orbit round a sunof large mass. When the planet is at perihelion, the mass of the sun issuddenly doubled. Shew that the planet will continue to describe anelliptical orbit, but that its velocity at perihelion is to its former velocityat aphelion as twice the major axis of the former orbit is to the major axisof the new one. [M. T. 1926]

26. A particle acted on by a central attractive force /rw3 is projected1

with a velocity ~ VM a t a n angle of jn- with its initial distance a from the

centre of force : shew that its orbit is the equiangular spiral r=ae~e.[Coll. Exam. 1909]

27. If a particle is projected from an apse at distance a with the velocityfrom infinity under the action of a central force nr~2n~3, prove that thepath is rn = ancosn6. [Coll. Exam. 1910]

EXAMPLES 185

28. Find the velocity necessary for the description of a circular orbit ofradius a under a central force 2 ia2«6 - jiu3. Shew that the orbit is unstableand that if a slight disturbance takes place inwards the path may berepresented by r=a tanh 6. [Coll. Exam. 1914]

29. A particle of mass m moves under a central force

a being > 6, and is projected from an apse at distance a + b, with velocity

\fp/(a + b); shew that its orbit isr—a + bcoad.

30. Prove that, if the law of force is 3/i«3 + 2/i«V and a particle is pro-jected in a direction making an angle cot"12 with the initial distance a,and with a velocity equal to that in a circle at the same distance, theorbit is

au = tan {6 + ^n).

31. The law of force is fxus and a particle is projected from an apse atdistance a. Find the orbit (i) when the velocity of projection is vWa2 v'2,(ii) when it is \/fi/a2.

32. If the law of force be 2fi (u3 - s ¥ ) and the particle be projectedfrom an apse at distance a with velocity sj\s.]a, shew that it will be at adistance r from the centre after a time

1 ' " " -a 2 ) + a2cosh-ir/a}.

33. From the fact that the velocity in the hodograph of the motion ofa particle represents the acceleration in the path, deduce that, if r—f{8)be the hodograph of a projectile with the initial line horizontal and 6measured positively downwards, the retardation due to the resistance of the

air is g I sin 6 -y- J, and the horizontal range is - jr^dd.

[M. T. 1919]

34. A point describes a circle of radius a so that its hodograph is asecond circle of radius b. If the pole of the hodograph be at distance cfrom its centre, where cjb is small, shew that the time of a completerevolution is approximately

[S. 1915]

35. Prove that the hodograph of the motion of a small ring, movingalong a circular wire under the action of no force, other than the reactionof the wire, is a circle or an equiangular spiral, according as the wire issmooth or not. [Coll. Exam. 1910]

186 POLAR COORDINATES. ORBITS [XII

36. A particle of mass m is connected to a slightly extensible string ofmodulus X, the other end of which is fixed to a point in a smooth horizontaltable. The particle lies on the table and is projected with velocity v atright angles to the string which is initially just taut. Shew that themaximum extension of the string is approximately 2mv2/\. [S. 1923]

37. Two particles of masses m, m' are moving in the plane of xy, underan attraction R along the radius r joining the particles. Shew that thecentre of gravity moves with constant velocity in a straight line; and thatif x, y are the rectangular coordinates of either particle with respect to theother, then

mm' d2x T>.v mm' d2y _ ym + m' dt2 r' m + ni' dt2 r'

If the relative orbit is a circle of radius r, described in a period T, provethat

mm' 4TTV' T2 '

Assuming Newton's law of attraction, and that the moon describes acircle of radius r, relative to the earth, establish the equation

M_ 4r3

+ E~~

where a is the earth's radius, I the length of the seconds pendulum, M andE are the masses of the moon and earth respectively, and N is the numberof seconds in the moon's period. [M. T. 1914]

38. Two gravitating particles of masses m and M, starting from rest atan infinite distance apart, are allowed to fall freely towards one another.Shew that when their distance apart is a, their relative velocity of ap-proach is \/{2y (M + m)/a}, y being the constant of gravitation.

Impulses of magnitude / are applied to the particles in opposite directionsperpendicular to their line of motion at the instant when their distanceapart is a. Shew that p, their distance of closest approach during the sub-sequent motion, is the positive root of a certain quadratic equation, andthat if 1 is small, p is given approximately by

[M.T.1923]L J

39. A weight can slide along the spoke of a horizontal wheel but isconnected to the centre of the wheel by means of a spring. When thewheel is fixed the period of a small oscillation of the weight is 27r/ra: shewthat when the wheel is made to rotate with constant angular velocity <athe period of oscillation is

27r/\/(m2-0)2).

EXAMPLES 187

If the wheel is a light frame whose mass may be neglected, and isstarted to rotate freely with angular velocity Q, shew that if £l = 6n./5Jllthe greatest stretch of the spring is 20°/o of its original length.

[M. T. 1917]

ANSWERS

2. -^-5 = -^ + C; a X V 3 - / * 2 ^ ; ^rB^+^e^jr. 6.f8. J(fySP.S'F). 16. 3ft.; 0-5; 16/\/3f.8. 17. 16TT/7 \/7 sees.21. The points are the ends of the chord through S perpendicular to LM.

28. \ffi/a. 31. (i) r—a cos 6; (ii) tanh<9/V2 = ?-/a or air bothhaving a common asymptotic circle r=a.

Chapter XIII

MOMENTS OF INERTIA

13-1. If the mass of every element of a body or particle of asystem be multiplied by the square of its distance from an axis,the sum of the products is called the Moment of Inertia of thebody or system about that axis. Thus if m denotes the mass ofan element or particle and r denotes its distance from the axis,the moment of inertia is 2mr2.

In like manner we may define the moment of inertia of asystem with respect to a point or plane as Xmr2, where r denotesdistance from the point or plane.

If M denotes the whole mass and k be a line of such lengththat Mk2 is the moment of inertia about an axis, then k is calledthe radius of gyration of the system about that axis.

When rectangular coordinate axes are used, the moments ofinertia of a body about the axes are denoted by

A = 2m(yi + z2), B = tm(z2 + x2), G = %m{a? + y2);

and the sums represented by

D = ~%myz, E = "£mzx, F= ~%mxy

are called the products of inertia of the body with regard tothe axes yz, zx, xy.

132. Theorem of Parallel Axes. The moment of inertia ofa body about any axis is equal to its moment of inertia about aparallel axis through its centre of gravity, together with the productof the whole mass and the square of the distance between the axes.

Let the given axis be taken as axis of z. Let (x, y, z) be thecoordinates of a particle of mass m, (%, y, z) the coordinates ofthe centre of gravity G, and let x = x + x, y — y + y', z = z + z'.

The moment of inertia about Oz

= Zm(x* + if) = 1m {(x + xj + (y+ y'f\

= 2m (x'2 + y'2) + (Sa + y2) 1m + 2x2,mx + tyimy1.

13'1-13'4] REFERENCE TABLE 189

But Xmx = 0 and %my = 0, therefore

The moment of inertia about Oz— 1m (a/2 + y'2) + (a? + ?/2) 1m;

and the first sum is the moment of inertia about an axisthrough 0 parallel to Oz, and the remaining terms are theproduct of the whole mass and the square of the distance be-tween the axes.

133. Plane Lamina. The moment of inertia of a plane laminaabout an axis perpendicular to its plane is equal to the sum of themoments of inertia about any two perpendicular axes in the planethat intersect on the first axis.

Take the plane of the lamina as the plane of xy and theperpendicular axis as Oz. Then, since z = 0 at all points on thelamina, we have

so that G = A + B.

134. Reference Table. The calculation of a moment ofinertia is generally a simple matter of integration and thefollowing results are tabulated for convenience. In all cases Mkz

is taken to be the moment of inertia, where M is the wholemass, and G denotes the centre of gravity. In some of thecases given the result follows from the preceding result by anapplication of 132 or 133.

Value of fc2

Rod of length 2a,about a perpendicular axis through O ^a2

about a perpendicular axis through an end -Ja2

Rectangular lamina of sides 2a, 26,about an axis bisecting the sides 2as ... Ja2

about a perpendicular to its plane through O ... ... J(a2 + 62)about a perpendicular to its plane through a corner ... j («2

Rectangular parallelopiped of edges 2a, 26, 2c,about an axis through its centre perpendicular to the

plane containing the edges 26, 2c ... ... ...Elliptic lamina of axes 2a, 26,

about the axis 2a ... ... ... Jabout the axis 26 Jabout a perpendicular to its plane through O i («2

Ellipsoid of axes 2a, 26, 2c,about the axis 2a

190 MOMENTS OF INERTIA [XIII

The last three results may be included in a single formulaknown as Routh's rule:

A body has three axes of symmetry: about an axis of

symmetry &2 has the value

Sum of squares of perpendicular semi-axes

3, 4 or 5 '

where the denominator is to be 3, 4 or 5, according as the body

is rectangular, elliptical or ellipsoidal.

Thus a circle is a special case of an ellipse, and if we want theradius of gyration of a circular lamina of radius a about adiameter, the perpendicular semi-axis in the plane is of lengtha and that perpendicular to the plane is zero, therefore k2 = Ja2.But for an axis through the centre perpendicular to the plane

k2 = l(a?+a?) = ±a?.

We notice that a sphere is a particular case of an ellipsoidand that its moment of inertia about a diameter is thereforef Ma2, where a is the radius.

13'41. Some of the results of the last article can be obtained by simpledirect methods.

Circular disc of radius a and mass M. First find the moment ofinertia about an axis through the centre perpendicularto the plane. For this purpose divide the disc intonarrow concentric rings. Let r denote the radius anddr the breadth of one of these rings. Its area is iirrdr

and therefore its mass is —xZnrdr and every element

of the ring is at a distance r2 from the axis. Therefore

the ring makes a contribution —^-rsdr to the moment of inertia, and the

moment of inertia of the whole disc

a J o

Secondly, for the moment of inertia about a diameter; since the momentof inertia about every diameter is the same, therefore by 13"3 the momentof inertia about a diameter—\Ma?.

Sphere of radius a and mass M. The moments of inertia about alldiameters are the same, so that

A =B= C=

13-4-1342] ELLIPSES AND ELLIPSOIDS 191

Divide the sphere into concentric shells. Let r denote the radius and drthe thickness of such a shell. Its volume is Anr^dr and therefore its mass

= j 5 4irr*dr = —, r2dr.|wa3 a3

Hence, for the whole sphere,

a3 Joand therefore the moment of inertia about a diameter = §J/a2.

1342. We append two examples of integration:Elliptic Lamina. To find the moment of inertia about the minor axis.

Let M be the mass and 2a, 26 the axes.The area may be divided into narrow stripssuch as PP', in the figure, of length 2y andbreadth dx. The mass of such a strip is

—=• Iiidx.•nab

Hence the moment of inertia

about Oy — —? I x2ydx, wheres irabj_a 3 p'

ThereforeT

Put x = a sin 8, and we get2 /"*"•

A» ^= a I s i^ J - h™

1 / •*" •

2TT JO

ain2 6 cos2 6d6

It follows that the square of the radius of gyration about Ox is J 62 andabout a perpendicular to the plane through 0 it is j (a2 + 62).

Ellipsoid. Let M be the mass of the ellipsoid and 2a, 26, 2c thelengths of its axes. Taking co-ordinate axes Ox, Oy, Oz alongthe axes of the ellipsoid let us findthe moment of inertia about Ox.Let the solid be divided by planesparallel to yz and let QRQ/R' bethe section at a distance 0N=%from 0. The semi-axes of thissection are given by

R1

and RN = i

192 MOMENTS OF INERTIA [XIII

so that the area of the section = -•=- (a2 - *'2), and the volume of a slice ofct

thickness dx is —j- (a2 — x2) dx. But the whole volume is f wabc, thereforeQ •!#•

the mass of the slice at distance x from 0 is -—^ (a2 — «2) dx, and the square

of its radius of gyration about Ox is

I ( § # 2 + iJ#2) = ^ (62 -t-c2) (a2 - ^2).

Hence the moment of inertia of the whole ellipsoid about Ox

= l S (62+°2) [«**-

135. Plane Lamina. Momental Ellipse. Reverting to thecase of a plane lamina we can shewthat the moment of inertia about a yline through the origin making anangle 6 with the axis of x is givenby the formulaI = A cos2 8 + B sin2 6 - 2Fsin 6 cos 9,where A, B, F have the meaningsassigned to them in 131.

For if P (x, y) be the position of an element of mass m andPN be perpendicular to the line, PN — ycosO — x sin 6, and

/ = 2,-mPN2 = 2m (y cos d - x sin Of= cos2 02m2/2 + sin2 6%mx* - 2 sin 6 cos 02mOT/= A cos2 0 + 5 sin2 6 - 2JPsin 0 cos 6.

If along any line through 0 we measure off a length OQ = r,such that the moment of inertia of the lamina about OQ isinversely proportional to OQ2, the locus of the point Q is anellipse called a momental ellipse for the lamina at the point 0.

To prove this theorem we have only to put / = Jie4/r2, andtherefore

A cos2 0 + B sin2 6 - 2F sin 0 cos 0 = Me4/r2,or, if x, y are the rectangular coordinates of Q,

Aa? + By2 - 2Fxy

13-42-13-52] PRINCIPAL AXES 193

which represents a central conic; and, since every radius isreal by construction, therefore the conic is an ellipse. M isintended to denote the mass and e is a length in order to makeboth sides of the equation of the same dimensions.

The constant e is arbitrary, so that we have any number ofsimilar, similarly situated, momental ellipses at the same point.

13*51. Principal Axes. When rectangular axes at a pointare so chosen that the products of inertia %myz, Imzx andtmxy are all zero, these axes are called the principal axesat the point. In this case A, B, C are called the principalmoments of inertia.

It can be shewn that at any point there always is a set ofprincipal axes, but we shall confine ourselves to proving thistheorem for the case of a lamina.

Reverting to the figure of 13*5, the product of inertia withregard to the axis ON and a perpendicular axis is2m (PN. NO) = 2m (y cos 0 — x sin 0) (x cos0 + y sin 0)

— ~S,mxy (cos2 6 — sin2 0) + 2m (y* — x2) sin 0 cos 0= Fcos 20 + KA-B) sin 20,

and this will vanish if 0 = \ tan"1 -=—-j; thus determining the

principal axes at 0 in the plane of the lamina.The existence of the principal axes at 0 also follows from the

fact that when the equation of a momental ellipse is referredto its axes the coefficient of xy vanishes, but this coefficient isthe product of inertia with regard to those axes. Therefore theaxes of a momental ellipse at 0 are the principal axes at 0.

13 52. Geometrical Representation. Suppose that we haveconstructed a momental ellipse ata point 0, and that a, b are itssemi-axes. By hypothesis the mo-ment of inertia about any radiusOP is M^/OP2. But if p is thelength of the perpendicular from 0to the tangent parallel to OP it isa property of the ellipse that p. OP = ab, therefore the moment

RD J3

194 MOMENTS OF INERTIA [XIII

of inertia about OP is directly proportional to^p2; or, on somesuitable scale of measurement, p represents the radius of gyra-tion of the lamina about OP.

13*53. If the principal moments of inertia of a lamina at itscentre of gravity are known, the moment of inertia about anyother axis in the plane can be calculated. For if A, B are theprincipal moments and Ox, Oy the principal axes, the momentabout an axis through 0 making an angle 0 with Ox is by 13'5

A cos2 6 + B sin2 6,and the moment about any parallel axis can be found by thetheorem of parallel axes in 132.

13*54. If the principal moments of inertia of a lamina areequal at any point, then a momental ellipse at that point is acircle and any pair of axes at right angles through the point areprincipal axes through the point. For example, at the centre ofa square the moments of inertia about lines parallel to the sidesof the square are clearly equal, therefore a momental ellipse isa circle and the moment of inertia about every line in the planepassing through the centre is the same.

136. Equimomental Bodies. Two bodies are said to be equi-momental when their moments of inertia about all straight linesare equal each to each.

This will be so if the bodies have the same centre of gravity, thesame mass, and the same principal axes and principal momentsof inertia at the centre of gravity. Thus a straight rod of massM is equimomental with particles of mass \M at each end anda particle of mass \M at the centre.

EXAMPLES

1. A parabolic area is cut ofl' by a double ordinate at a distance h fromthe vertex, shew that for the moment of inertia about the tangent at thevertex &2=f A2.

2. Shew that the square of the radius of gyration for the area of thecurve r2=a?cos28 about an axis through the origin perpendicular to theplane is \na2.

13-52-13-6] EXAMPLES 195

3. A right cone of height h stands on a circular base of radius a. Shewthat for the moment of inertia about the axis of the cone k2=$ja?t andabout a line through the vertex perpendicular to the axis £2 = g (A2 +Jo.2).

4. Shew that, for a thin hemispherical shell of radius a and mass M,the principal moments of inertia at the centre of gravity are

%Ma2; and that for a solid hemisphere the values are

5. Particles of equal mass are placed at the corners of a regular polygon.Prove that the squares of the principal radii of gyration at any point 0 inthe plane of the polygon are a2/2 and (os2 + 2A2)/2, where a is the radius ofthe circumscribing circle of the polygon and h is the distance of 0 fromthe centre of this circle. [Coll. Exam. 1913]

6. Prove that a uniform triangular lamina of mass M, and a systemconsisting of three particles each of mass ^M, situated at the middlepoints of the sides and rigidly connected by light rods, have the samemoment of inertia about any axis in the plane. [M. T. 1925]

7. Shew that, if p, y are the distances of the corners B, C of a triangleABC from any straight line through A in the plane of the triangle, thenthe moment of inertia of the triangle about this line is ^M(fi2 + f3y + y2),where M denotes its mass.

8. Shew that, if o, /3, y are the distances of the corners of a trianglefrom any straight line in its plane, then the moment of inertia of thetriangle about this line is

where M denotes its mass.

9. Shew that the principal axes at a corner of a rectangular lamina ofsides a, b make with a side angles 8, 5 n + 8, where

10. Shew that, in the last example, if half the rectangle be removed soas to leave a right-angled triangle of sides <z, b, the value of 8 for theprincipal axes at the right angle is given by

11. A uniform rectangular plate whose sides are of lengths 2a, 26 has aportion cut out in the form of a square whose centre is the centre of therectangle and whose mass is half the mass of the plate. Shew that theaxes of greatest and least moment of inertia at a corner of the rectanglemake angles 6, \n + 8, with a side, where

tan 28 = f ab/(a2 - ¥•). [M. T. 1928]

13-2

196 MOMENTS OF INERTIA [XIII

12. Find the moment of inertia of a uniform rod of mass m and length 2cabout any axis for which the line of shortest distance to the given rod cutsthe rod at its middle point. [Take 6 for the angle between the axis andthe rod.]

Shew by direct integration of the previous result that the moment ofinertia of a plane uniform elliptic lamina of mass M and semi-axes a andh about an axis through its centre in the plane normal to the laminacontaining the axis a and making an angle <j> with that axis is

\M (V + a? sin2 <£). [M. T. 1927]

ANSWER

12. m (dP + \c2 sin2 6), where d is the shortest distance between the rodand the axis.

Chapter XIV

MOTION OF A RIGID BODY. ENERGYAND MOMENTUM

14-1. A rigid body means a body in which the distancebetween each pair of particles remains invariable. The bodieswith which we are familiar are all more or less elastic andcapable of compression, extension or distortion under the actionof external forces, and the problem of the motion of such abody is in general rendered more complicated by its deform-ability. The problem of the motion of bodies is greatlysimplified by the hypothesis that they are rigid in the sensedenned above, and in elementary dynamics we limit our con-siderations for the most part to the problem of the motion ofone or more such bodies.

Further, we regard a rigid body as an agglomeration ofparticles held together by cohesive forces such that the actionand reaction between any pair of particles are equal and opposite(97), and we use the principles established in Chapter IX forthe motion of a system of particles. In the case of the rigidbody however there is an additional fact to be observed, namely,that since the distance between each pair of particles isunaltered in any displacement of the body, therefore the totalwork done by the action and reaction between the particlesis zero.

For if R denotes the mutual reaction between two particlesA, B which undergo a smalldisplacement to A', B', so that A'A'B' = AB, where A'B' makes "~a small angle 6 with AB; then, A R M

if MN is the projection of A'B' on AB, the work done is

R. AM - R. BN = R (AB - MN)

= R.AB(l-cos0);

and this is zero to the first order of small quantities, and by

198 MOTION OF A RIGID BODY. ENERGY AND MOMENTUM [XIV

summation the same result holds good for a finite displace-ment *.

It will be convenient to summarize here the results ofChapter IX in reference to a rigid body, thus :

(i) The rate of change of linear momentum of a rigid body inany direction is equal to the sum of the components of the externalforces resolved in that direction (9'2).

(ii) The rate of change of moment of momentum, of a rigidbody about any fixed axis is equal to the sum of the moments ofthe external forces about that axis (9"2).

(iii) The increase in the kinetic energy of a rigid body in anytime is equal to the work done by the external forces in thattime (9-4).

In 9-4 work, if any, done by the internal forces is included,but in the case of a single rigid body the internal forces do nowork in any displacement and therefore in (iii) we only refer tothe external forces.

The same principles also apply to the motion of a system ofrigid bodies, provided that in (iii) we include the work done, ifany, by the internal actions and reactions between the bodies.

The principles of the independence of translation and rotationestablished in 9 "3 also hold good for a rigid body or system ofbodies.

* A more formal proof of this theorem may be set out as follows :Let (x, y), (x', y') be the coordinates of the particles A, B in any position of

the body, B the mutual reaction and r the invariable distance of the particles,so that

Let X, Y denote the components of B at A and consequently - X, - Y denotethe components of B at B; then

X=B(x'-x)lr, Y=B(y'-y)jr.

The work done by B at A in any displacement of the body is \(Xdx + Ydy)integrated along the path of A, and the work done by B at B is - j(Xdx' + Ydy')integrated along the path of B. Hence the total work done is

' B{(x' - x) (dx' - dx) + (y'-y) (dy' -dy)};

and this is zero, since by differentiating (1) we get

(x - x') (dx - dx') + {y- y') (dy - dy') = 0.

141-14 - 2 ] KINETIC ENERGY OF A RIGID BODY 199

142. In order to apply the principles of the last article, weneed to find the most convenient expressions for the kineticenergy, momentum and moment of momentum of a rigid body,and we shall confine our attention for the most part to two-dimensional motions, noting however that the principles can alsobe applied to motion in three dimensions.

Kinetic Energy of a Rigid Body. The velocity of every pointof the body is determinate provided that the angular velocity ofthe body is known and the linear velocity of some one point ofthe body.

Let us suppose that the centre of gravity G has a velocitywhose components parallel to coordinate axes Ox, Oy are u, v,and let <o be the angular velocity of the body.

If P be the position of an element of mass m of the bodywhose coordinates referred to parallel axes through G are x', y'or r, 8 in polar coordinates, the velocity of P relative to G is ra>perpendicular to GP. Therefore the components of velocity of Pparallel to the axes are

u — rco sin 8 or u — y'a>,

and v+rcocos8 or v+x'a>.

Hence the kinetic energy of the body is

A 2m {(u - y'a>)2 + (v + x'wf)= £ (u2 + v2) 2m + -|o)22m (a/2 + y'2) — uw'tmy' + va> 2,mx'.

But 2m is the whole mass M of the body; 2,mx and ~2my' arezero because G is the centre of gravity, and 2m (x'2 + y'2) isthe moment of inertia of the body about an axis through G

2 0 0 MOTION OF A RIGID BODY. ENERGY AND MOMENTUM [XIV

perpendicular to the plane and may be denoted by Mk2, wherefor a plane body we may speak of k as the radius of gyrationabout G. Hence the kinetic energy may be written

%M(u2 + v2 + k2a>2) (1).The first two terms represent the kinetic energy of the whole

mass moving with the velocity of the centre of gravity and thethird term is the kinetic energy of the motion relative to thecentre of gravity, in accordance with the theorem previouslyproved in 9'5.

As a special case, when the body is turning about a fixed axisthrough 0 perpendicular to the plane xy, let h be the distanceof G from this axis, then ha> is the velocity of G and the formula(1) becomes \ M (h2 + k2) or*, or by the theorem of parallel axesin 132

\I«? (2),where / is the moment of inertia of the body about the fixedaxis. This result is capable of simple independent proof, for if rdenotes the distance of an element of mass m from the fixedaxis, its velocity is ra>, and therefore the kinetic energy of thebody

14*21. Examples of Conservation of Energy. Numerousproblems can be solved by the principle of energy alone; weappend a few solutions.

(i) A uniform disc is free to turn about a horizontal axis through itscentre perpendicular to its plane. A particle of mass m is attached to apoint in the edge of the disc. If motion starts from the position in which theradius to the particle m makes an angle a 'with the upward vertical, find theangular velocity when m is in its lowest position.

Let a be the radius of the disc and Mk2 its moment of inertia about itscentre. Then, if a> is the angular velocity, the kinetic ^—~T-~^energy of the disc is \M1C2B?, and of the particle \ma2a?',since its velocity is aa>; also, since the weight of theparticle is the only force that does work, the workdone in reaching the lowest position is mga (1 + cosa),therefore

| (Mk2 + ma2) »2=mga (1 + cos a).(ii) An electric motor which gives a uniform driving torque drives a pump

for which the torque required varies with the angle during each revolutionaccording to the law To. sin26: the mean speed of the pump is 600 rev.

14-2-143] MOMENTUM OF A RIGID BODY 201

per win. and the mean horse-power required is 8. To limit the fluctuation ofspeed during each revolution, a flywheel is provided between the motor andthe pump which successively stores and gives out energy. Shew that the energythus successively stored and given out by the flywheel is approximately70 foot-pounds. [S. 1925]

' Torque' means couple. Let the torque required for driving the pumpbe given by

r = £ sin2 <9 ft.-lb.,where k is a constant.

The mean speed is 10 revolutions or 20JT radians per second. The meanrate of working is 8 horse-power or 8 x 550 ft.-lb. per second. Therefore themean couple required

=8X550/20TT

= 70ft.-lb. approximately.Now the mean value of sin2 6 taken over a revolution

Hence the mean value of the torque required for the pump is also \k,\therefore \k= 70 ft.-lb., or £ = 140 ft.-lb.

The torque required for driving the pump is therefore given by

r = 140sin20 ft.-lb.,

and varies from 0 to 140 ft.-lb. during a revolution. But the electricmotor gives a uniform driving torque equal to the mean torque requiredby the pump, i.e. 70 ft.-lb., and when the pump requires a smaller torquethan this the surplus energy is stored by the flywheel, to be given outagain when the pump requires a torque larger than the mean. Since thetorque required by the pump varies from 0 to 140 ft.-lb. and the steadytorque supplied by the motor is 70 ft.-lb., it follows that 70 ft.-lb. of energyis successively stored and given out by the flywheel.

143. Momentum of a Rigid Body. If x, y are the coordinatesof an element of mass m of the body and x, y are the coordinates ofthe centre of gravity and M is the whole mass, we have '%mx=Mxand Liny = My. Therefore, by differentiation, 1mx = Mx andSmy = My, so that if u, v denote the components of velocity ofO as in 14-2, the components of linear momentum of the wholebody are Mu, Mv; i.e. the same as if the whole mass were collectedinto a particle at G and moved with the velocity of G.

Now consider the moment of momentum of the body about anarbitrarily chosen origin 0. Reverting to the notation and figureof 142, the velocities of the element m are u — y'co and v+x'm,so that its components of momentum are m (u — y'a>) and

202 MOTION OF A RIGID BODY. ENERGY AND MOMENTUM [XIV

m (v + x'co). Taking moments about 0, the moment of mo-mentum of the whole body

= Xm [x (v + x'a>) — y(u — y'co)},

or, since x = x + x and y = y + y', the moment of momentum

= 2 m {(% + x)(v-+ x'w) — (y + y')(u — y'oo)}.

Multiplying out and remembering that Xmx' = 0 and~Zmy' = 0, there remains

2m (xv — yu) + a %m (x'2 + y'2)= M(xv-yu) + Mk2co (1),

where k is the radius of gyration about G.We observe with regard to this formula that the first term

M (xv — yu) is the moment about 0of a vector localized at G and having ycomponents Mu, Mv; i.e. it is the

G ->.

ymoment about 0 of the componentsof linear momentum of the body. Wealso remark that the second term informula (1), viz. Mk2a>, is independentof the choice of origin 0 and independent of the velocity of G.It is in fact the moment about G of the momentum in themotion of the body relative to G.

Hence the whole momentum of the body may be regardedas composed of a vector with components Mu, Mv localized atthe centre of gravity, together with a couple Mk^a which maybe called the spin couple; for we get the moment of momentumabout any point 0 by taking the algebraical sum of the momentsof Mu, Mv and adding on the spin couple Mk%a>.

1431. Examples of Conservation of Momentum. Numer-ous problems are simple illustrations of the principles of con-servation of momentum. We append the solution of a fewexamples.

(i) A thin straight tube of small bore is movable about its centre on asmooth horizontal table, and it contains a uniform thin rod of the samelength and mass whose centre of gravity is nearly at the middle point of thetube. Prove that, if the system be set in motion with angular velocity a>, theangular velocity of the tube as the rod leaves it is o>.

1 4 3 - 1 4 ' 3 1 ] EXAMPLES OF CONSERVATION OF MOMENTUM 203

Let m be the mass and 2a the length of either body. The initial momentof momentum about 0 is § m a VIf a is the angular velocity re-quired, when the rod is leavingthe tube its centre of gravity Ghas velocity 2aa>', so that the mo-mentum of the rod consists of alinear component 2maa> at rightangles to the rod and a spin couple\ma2a>, hence the moment about 0 of the momentum of the rod is4ma2<o' + J m a V , and the moment of momentum of the tube is now JuwV.And since there is no external force the total moment of momentum about0 remains unaltered, therefore

4moV + § ma2a> = § ma2a,

or a>' = a>.

(ii) A horizontal wheel with buckets on its circumference revolves about africtionless axis. Water falls into the buckets at a uniform rate of mass mper unit of time. Treating the buckets as small compared with the wheel,find the angular velocity of the wheel after time t, if Q. be its initial value;and shew that if I be the moment of inertia of the wheel and buckets aboutthe vertical axis and r the radius of the circumference on which the bucketsare placed, the angle turned through by the wheel in time t is

0 I+mrH

In time t a mass mt of water falls into the buckets and the total momentof inertia about the axis is increased from 1 to / + mr2t. If 8 be the angleturned through in time t the angular velocity is 6, and the moment ofmomentum about the axis at time t is 6 (I+mr2t). But the system is notacted upon by any force which has a moment about the axis, so that themoment of momentum is constant; therefore

H e n O e

(iii) A man of mass m stands at A on a horizontal lamina which canrotate freely about a fixed vertical axis 0. Originally both man and laminaare at rest. The man proceeds to walk on the lamina, ultimately describes{relative to the lamina) a closed circle having OA (=a) as diameter, andreturns to the point of starting on the lamina. Shew that the lamina hasmoved through an angle relative to the ground given by n {1 — V1/(1+ ma2)},where I is the moment of inertia of the lamina about the axis. [M. T. 1921]

204 MOTION OF A RIGID BODY. ENERGY AND MOMENTUM [XIV

Let Ox be a fixed line, and at time t let AOx — 6 and AOP=<f>, whereP denotes the position of the man.The angle xOP increases at the rate<p — 6, therefore the velocity of the man 0 /at right angles to OP is OP (</> - 6) andhis moment of momentum about 0 isWJOP2 (<£ — #). The lamina has a momentof momentum Id in the opposite sense ;and the total moment of momentum in the same sense remains zerothroughout the motion, therefore

But OP=a cos <]>, therefore

( / + ma2 cos2 <f>)6=ma2 cos2 <£<£.

TT m ma2 cos2 d>Hence d6=-f 5—?-7 dd>

1 + ma' cos1* q> r

\ 1+ma2 cos2 <f>JNow as the man walks round the circle from A to 0 the angle <fi increases

from 0 to Jw, and as he walks on from 0 to A the angle <f> increases fromf 7T to 2n. Therefore the whole angle turned through by the lamina

f o I+ma2 cos 2 <j>'

To evaluate the integral put tan <$>=t, and we get

W _2/T—J*__J uI+ma2 + It2

2 /tan"1

= n {1 - V'1/(1+ ma2)}.

14*4. Examples of Conservation of Energy and Momentum.(i) A uniform straight rod is held at an inclination a to the vertical withits lower end in contact with a smooth horizontal plane and let go. Findthe angular velocity when the rod becomes horizontal.

Since there is no horizontal force acting on the rod its centre of gravity Odoes not acquire any horizontal velocityand therefore remains in the same verticalline Oy. If m is the mass, 2a the length,y the height of O above the plane and 6the inclination to the vertical at time t,the kinetic energy is \m(y2 + k282), whereP = ^ a a and y = acos$; therefore the kineticenergy is \nia2 (sin2 6 + \) 62; and the work

y

14-31-14-4] CONSERVATION OF ENERGY AND MOMENTUM 205

done as O descends from a height a cos a to a cos 6 is mga (cos a — cos 6);therefore

\ma2 (sin2 8 + \) S2 = mga (cos a — cos 6)

gives the angular velocity in any position, and for 6 = \n

(ii) A sphere of radius a, and of radius of gyration k about any axisthrough its centre, rolls with linear velocity v on a horizontal plane, thedirection of motion being perpendicular to a vertical face of a fixed rect-angular block of height h, where h<Ca. The sphere strikes the block; thesphere and the block are perfectly rough and perfectly inelastic. Shew thatthe sphere will surmount the block if

(a2-ah + k2)2v2>2gha*(a2+k2). [M. T. 1924]

Since the sphere rolls with linear velocity v, therefore its angular velocityis vja just before striking the block.Let a> be the angular velocity im-mediately after striking the block. If0 is the point of contact of the spherewith the edge of the block the spherebegins to turn about 0 and when itbegins to rise the velocity of its centre Ois aa at right angles to 00, and themoment of momentum about 0 is therefore m(a2+k2)a>, immediately afterthe impact, where in is the mass. Now there is an impulsive reaction onthe sphere at 0, but it has no moment about 0, therefore the moment ofmomentum about 0 is unchanged by the impact. But before the impactthere was a linear momentum mv and a spin couple mk2vja, therefore

TO (a2 + k2)a> = mv (a — h) + mkhi/a.

This equation determines the angular velocity a> with which the spherebegins to rise. Now in order to surmount the block the kinetic energy ofthe sphere when beginning to rise must be more than the work that wouldhave to be done in lifting the sphere through a height h; otherwise thekinetic energy would be destroyed by the time the sphere had risen to therequired height. But since the velocity of O is aa when starting to rise,therefore the kinetic energy of the sphere is \m (a2+k2) a2, and the requiredcondition is \m(a2 + k2)ai2>mgh, and, on substituting for a,, we get

(a2 - ah+k2)2 v2 > 2gha? {a2+k2).

206 MOTION OF A RIGID BODY. ENERGY AND MOMENTUM [XIV

E X A M P L E S

1. A wheel whose moment of inertia about its axis is 200 lb.-ft.2 makes10 revolutions per second. What is its kinetic energy (in ft.-lb.), and whatconstant couple would reduce it to rest in one revolution ?

2. A flywheel of a pressing machine has 150,000 ft.-lb. of kineticenergy stored in it when its speed is 250 revolutions per minute. Whatenergy does it part with during a reduction in speed to 200 revolutionsper minute ? If 82"/o of this energy given out is imparted to the pressingrod during a stroke of 2 inches, what is the average force exerted by therod? [M. T. 1917]

3. AB and CD are two rods of lengths 2a and b respectively, the massof each rod being m per unit length. The rods are rigidly joined togetherat right angles at C, the middle point of AB, and the system is free torotate in a vertical plane about D. If the system is held with AB verticaland then let go, calculate the angular velocity when AB is horizontal.

[M. T. 1912]

4. A wheel and axle has moment of inertia / about its axis. A mass mis suspended from the axle, which is of radius r, by a fine thread wrappedround and fastened to the axle. Rotation is opposed by a constant frictionalcouple O. Find the angular velocity acquired as the mass TO descends adistance h.

5. A uniform sphere rotating about a diameter contracts by cooling.

Shew that when its radius is reduced to —th of its former value the kineticn

energy is multiplied by n2.

6. A wheel 30 inches in diameter, which can rotate in a vertical planeabout a horizontal axis through its centre 0, carries a mass of j lb. con-centrated at a point P on its rim. The wheel is held with OP inclined at30° above the horizontal and then released. Owing to a friction couple ofconstant magnitude L at the bearing, the first swing carries OP to aposition only 45° beyond the vertical. Determine the value of L; andprove that in the next swing OP will come to rest before reaching thevertical. [M. T. 1913]

7. A flywheel of moment of inertia / is rotating with angular velocity Qabout a vertical axis. The flywheel contains a pocket at a distance a fromthe axis into which is dropped a sphere of mass M, moment of inertia iand spin a about a vertical axis, without horizontal motion. Find theangular velocity of the system after the sphere has come to relative restin the pocket. [S. 1927]

EXAMPLES 207

8. A uniform rod of length a and mass m is describing circles on asmooth horizontal table about one end which is fixed. If the rod strikes aparticle of mass m' at distance 6 from the fixed end, and this particleadheres to the rod, find the ratio in which the angular velocity is changed.

[Coll. Exam. 1908]

9. A cube of side 2a slides down a smooth plane inclined at an angle2 tan"1 J to the horizontal, and meets a fixed horizontal bar placed per-pendicular to the plane of motion and at a perpendicular distance \a fromthe plane. Shew that, if the cube is to have sufficient velocity to surmountthe obstacle when it reaches it, it must be allowed first to slide down theplane through a distance J^f a. The obstacle may be taken to be inelasticand so rough that the cube does not slip on it. [M. T. 1918]

10. A uniform cube of edge 2a is placed in unstable equilibrium withone edge in contact with a horizontal plane and allowed to fall. Shewthat, if o> is the angular velocity when a face of the cube comes into contactwith the plane, then

or

according as the plane is smooth or sufficiently rough to prevent sliding.

ANSAVBES

1. 12,500 ft.-lb.; 6250/jr ft.-lb. 2. 54,000 ft.-lb. ; 118£f tons.3. {3gb(4a + b)/(2a3 + 6ab* + V)}K 4. ^A (mgr - 0)/r(I+mr1)}i.

6. 1 5 ( 1 + V2)/88B- ft.-lb. 7. (l£t + ia)l(I+i+Ma2). &. mcfi + Zm'W-.ma?.

Chapter XV

EQUATIONS OF MOTION OF A RIGID BODY

151. Using the expressions found in the last chapter for themomentum of a body, we obtain the formal equations of motionof the body by equating the rates of change of momentum tothe external forces and equating the moment of the rate ofchange of momentum about any axis to the moment of theexternal forces; thus

Mu = sum of the external forces resolved parallel to Ox,Mii= „ „ „ „ „ Oy;

and an equation of moments, viz. algebraical sum of momentsabout 0 of Mu and Mi) plus the couple MJc*(b = algebraical sumof moments about 0 of the external forces, where 0 is any pointin the plane.

If we take 0 to be the point through which the centre ofgravity G is passing at the instant considered, then since Mu andMi) are localized at G they have no moment about 0 and thelast equation reduces to Mk2a> = sum of moments about G of theexternal forces.

152. Applications of the Equations of Motion, (i) Pulleywithinertia. Let two particles of masses m, m' be connected < r

by a fine string passing over a pulley of mass M, radiusa and moment of inertia about its axis MB. Supposethat the pulley is free to turn about its axis withoutfriction and that the string does not slip on the pulley.Let the tensions in the two straight portions of thestring be T, T.

If the angular acceleration of the pulley is d> in thesense indicated in the figure, the linear acceleration ofthe string and therefore also of the particles is a&>.Hence by resolving vertically for the particles we get

ma<Z> = mg- T and m'aa> = T' — rn'g.

And by taking moments about its axis for the pulleyW C g e t MWi,=aT-aT',and if we eliminate the tensions we get

(mai + m'a2+Mk2)a> = (m - m!) ag,shewing that the acceleration is constant.

mig mg

151-15-2] ROLLING SPHERE 209

Alternatively this result might be obtained from the principle of energy.For if we suppose that in descending a distance s the particle m acquiresa velocity v, then the angular velocity acquired by the pulley is y/a, andthe whole kinetic energy is

•but the work done as m descends and m! ascends a distance sis, (m- m')gs;therefore

J (m + n! + Mk2\a2) v2 = (m-m') gs,

and differentiating with regard to s gives

(m + m' + MA;2/a?) v dv/ds — {m — m')g;so that the acceleration vdv/ds has the same value as was obtainedpreviously for aa.

(ii) A circular hoop, a disc or a sphere rolls down an inclined plane.Let a be the radius of the body, m themass, mk2 the moment of inertia, F thefriction, R the normal reaction and athe inclination of the plane to the hori-zontal.

Let u be the velocity of the centre 0and to the angular velocity.

The first of the diagrams shews theexternal forces F, R and the weight mg,the second shews the rates of change ofmomentum, and these are two equivalentsystems.

Resolving along and at right angles tothe plane we get

mil — mg sin a — F (1),and 0=mg cos a- It

Taking moments about O gives(3).

One other equation is required for the determination of the four un-known quantities F. B, u, a>, and it is got from the kinematical conditionthat the body rolls, i.e. that the velocity of the point P in contact withthe plane is zero. The velocity of P relative to & is aa> up the plane andthe velocity of Q is u down the plane, therefore the condition that thebody rolls is

u — aa> = 0 (4).From (3) and (4) we get

F=mk2uja2 (5),and substituting for F in (1) gives

so that the body rolls down the plane with constant acceleration.

ED 14

210 EQUATIONS OF MOTION OF A RIGID BODY [XV

For a hoop ^2 = a2 and u = $g sin a.

„ disc k =\a „ u

„ sphere £2=§a2 „ u

We have assumed that the body rolls and found the acceleration on thishypothesis, but whether it rolls or slides depends on the relation of thecoefficient of friction /i to the slope of the plane. The friction F cannotexceed limiting friction fiR. From (5) and (6) we see that the frictionnecessary to cause rolling is

„ mi?

but

therefore

l = mg cos a,

; = -rr7,2 t a n a>

and the body will slide instead of roll if this fraction exceeds /*; and theacceleration when sliding is g (sin a - jx cos a). (See 4'55.)

(iii) A sphere rolls down on the surface of a fixed sphere. Let a, b be theradii of the moving and fixed spheres. Let G be the centre, m the mass,m$? the moment of inertia of the moving sphere, F the friction and R thenormal reaction. Let the common radius OG through the point ofcontact P make an angle 6 with the vertical at time t. Since O describes

a circle of radius a + b, therefore it has accelerations (a + b) & at right anglesto OG and (a + b) 62 along GO. And if a> is the angular velocity of thesphere the rate of change of the spin couple is mkla>. The first of the twodiagrams shews the external forces acting on the moving sphere, and thesecond shews the rates of change of momentum and these two systems areequivalent.

Eesolving at right angles to and along GO we get

m(a + b)6 =mg sin 6 - F (1),

and m(a + b)62=mgcosd-R (2).

Taking moments about G givesmk2i> = Fa (3)

15-2] KOLLING SPHERE 211

The kinematical condition for rolling, which expresses that the point Pof the moving sphere has no velocity, is

(a + b)8-aa> = 0 (4),

because aa> is the velocity of P relative to O and (a+b) 8 is the velocityof O in the opposite direction.

From (3) and (4), putting fa2 for Jc\ we get

F=\ma&, = %m(a + b) 6 (5)

and by substituting this value for F in (1)

I (a+b) 8'=g sin 8 (6).

Multiply by 28 and integrate,

Let the motion begin when 6 = a, i.e. # = 0 when 8 = a, so that C=2g cos a,a n d | (a + b) 82=2g (cos a -cos 6) (7)

gives the velocity in any position.The sphere will begin to slide when the friction F becomes equal to pR.

But (5) and (6) giveF=%mg sin 8,

and (2) and (7) give

R=^mg (17 cos 8 — 10 cos a);

so that sliding would begin when

2 sin 8 = n (17 cos 8- lOcosa),provided that the spheres are still in contact.

-And the moving sphere will leave the fixed sphere in the position forwhich R vanishes, i.e. when cos 6=\ty cos a.

Beginners in this branch of the subject will find it a help at first todraw separate diagrams to shew the rates of change of momentum and theequivalent system of external forces, as we have done in these last twoexamples.

Beginners should also be careful to remember that the angular velocityof a body moving in two dimensions is the rate of increase of the anglebetween a line fixed in the body and a line fixed in the plane. Thus theangular velocity of the moving sphere in this example is not 8 because OPis not a fixed line in the moving sphere.

We leave it as an exercise to the student to obtain equation (7) by theprinciple of energy, and we also remark that if we are not concerned withthe question of the possibility of sliding we can avoid the introduction ofF into the equations by taking moments about P, and thus getting theequation

ma (a + b) 8 + m£2<i = mg a sin 8,

which with the help of (4) is the equivalent of (6).

14-2

212 EQUATIONS OF MOTION OF A RIGID BODY [XV

(iv) A four-wheeled railway truck has a total mass M, the mass andradius of gyration of each pair of wheels and axle are m and k respectively,and the radius of each wheel is r. Prove that, if the truck is propelled alonga level track by a force P, the acceleration is P/(A1 + 2m£2/ra) / and find thehorizontal force exerted on each axle by the truck. [M. T. 1913]

\_Axle friction and wind resistance to be neglected.]

Let u be the velocity of the truck and <o the angular velocity of thewheels. The condition that thewheels roll is, as in the lastexample,

u-ra> = 0 (1).There must be the same frictionforce F on each pair of wheels,since they have the same angu-lar velocity; and by momentsabout the axis of a pair of wheels

«!.&& = Fr (2).By resolving in the direction of

motion for the truck and wheels-f-

= P-2FBut from (1) and (2)

Fand by substituting this value of F in (3) we get

(3).

(4),

Again considering a pair of wheels alone, if Q be the horizontal forceexerted by the truck, mu=Q — F.

Therefore from (4)

(v) The end of a thread wound round a reel is held fixed and the reel isallowed to fall so that the thread is unwound. Find the acceleration of thereel, assuming its axis to remain horizontal.

Let m be the mass and a the radiusof the reel and mk% its moment of inertiaabout its axis. Let v be the velocityof its centre and a> its angular velocityat time t.

The point A of the reel in contact withthe straight thread is the instantaneouscentre of rotation and has no velocity, butits velocity relative to the centre is aa>,therefore

v — aa=O (1).

mk%( .

two

15-2-15-3] IMPULSIVE MOTION 213

The rates of change of momentum are mi and the spin couple mk2 a>;and the external forces on the reel are its weight mg and the tension T ofthe string.

Therefore by resolving vertically we get

mv = mg-T (2),

and by moments about the axis

mWi> = aT (3).Hence from (1) and (3)

and therefore from (2)v=ga2l(

Instead of equations (2) and (3) we might take moments about thepoint A and thus avoid the introduction of the tension T. The equationof moments about A being

mav + mi:2ai = mga,

and this with (1) gives the same value for v as before.

153. Equations of Impulsive Motion. Reverting to theequations of motion for a rigid body under the action of finiteforces in 15"i, which may be written

Mi = Y,

and Mk2a = L,

where X, Y denote the sums of the components of the externalforces and L the sum of their moments about the centre ofgravity, if we integrate these equations with respect to tthrough an interval from t0 to t, we get

rtM(u-uo)= I Xdt,

J U

M(v-vo)=(t Ydt,

and MB (<o - w0) = I* Ldt,J ta

where u0, v0, a>0 denote the values of the velocities at thebeginning of the interval. If we now suppose that we areconcerned with impulsive forces as defined in 11 "1, the integralsare the measures of the components of the externally applied

214 EQUATIONS OF MOTION OF A RIGID BODY [XV

' impulses' and their moments and may be denoted by P, Qand H, so that we have

M(u-uo) = P,

and Mk2 (to — e»0) = H.These three equations merely repeat what we have already

demonstrated in li'2, that the instantaneous change of mo-mentum both as regards linear momentum and angularmomentum is the exact equivalent of the externally appliedimpulses.

15'4. Examples of Impulses. Two uniform rods AB, BC of lengths2a, 26 are smoothly jointed at B and placed on a smooth table so that ABCis a straight line. An impulse P is applied at A at right angles to AB,determine the initial velocities.

Let m, m' be the masses of the rods AB, BC and O, H their centres. Itis clear that if the angular velocities of the rods are known and the linearvelocity of any one point, then the velocities of every point of either rod

mu m'(u-aw-biv')

can be stated. Suppose then that u is the velocity of the point O and thata, a are the angular velocities of the rods in the senses indicated in thefigure.

By considering the velocities of A and B relative to O we see that thevelocity of A is u + aa> and of B is u — am, and we also see that H has avelocity — ba> relative to B in the same sense, so that the velocity ofH is u — aa> — ba>.

Since the unknown impulse on the rod BC acts at the joint B it is con-venient to take moments about B for BC, giving

m'b (u — aa — ba)-^ m'b2a>'=0,

or u-aa> — ba>' = 0 (1).

We can again avoid introducing the unknown impulse at B by takingmoments about A for the two rods together, getting

mau — JmcAo + ro' (2a + 6) (u — aa>- ba') — \m'b2m'=0,or

u {ma+m' (2a + 6)} - am {J ma + m' (2a + b)} — 2ba>'m' (a+1 b)=0... (2).

15-3-15-41] EXAMPLES OF IMPULSES 215

Then by considering the linear momentum at right angles to the rodsW e g e t mu + m' (u-aa-ba,') = P (3).

Equations (1), (2) and (3) are sufficient for the determination of u, a>, u>.

15-41. The effects of sudden prescribed changes also serve toillustrate the application of the equations of impulsive motion :

(i) A circular disc is revolving in its plane about its centre with angularvelocity <o. What is the new angular velocity if a point on the rim issuddenly fixed ?

Let a be the radius, m the mass and a the new angular velocity.Before the fixture the momen-tum of the disc is simply thespin couple \maru> (since for adisc F = i«2). After the fixture ° | ) I O -the centre of gravity Q has avelocity aa> at right angles to00, where 0 is the point fixed;therefore the momentum consists of a linear component maa> and a spincouple \ma<ias. The impulse that causes the change must act through thepoint 0 that becomes fixed, therefore there can be no change of momentof momentum about 0; hence we have

(ii) A square lamina ABCD rests on a smooth horizontal plane. If thecorner A is made to move with velocity u along the line BA produced, deter-mine the initial angular velocity of the lamina.

Let m be the mass and %a a side of the square. Then, if a is theangular velocity, the centre of gravity O hasvelocity AO.a> relative to the point A.Therefore the velocity components of O are

u — AOa cos \TT,

ora-as parallel to BA, andAOcosin JTT,

or aa> parallel to BA.Therefore the momentum of the square

consists of linear components m(u — aa>)parallel to BA, maa parallel to DA anda spin couple mk2a>, where £2 = §a2. But theonly external impulsive action is appliedat A, therefore the moment of momen-tum about A is zero. Hence we have

ma2a> — ma {u — aa>)+§ ma?a> = 0,

giving a> = 3uj8a.

216 EQUATIONS OF MOTION OF A RIGID BODY [XV

155. Motion about a Fixed Axis. When a body turnsabout a fixed axis the motion is completely determined by theprinciple that the rate of change of moment of momentum isequal to the sum of the moments of the external forces. Theequation of motion takes a very simple form, for if <o be theangular velocity and r denotes the distance from the axis ofan element of mass m, its velocity is r&> and its moment ofmomentum about the axis is mr^co. Therefore the momentof momentum of the whole body is %mr2a> or Ia>, where I isthe moment of inertia of the body about the axis. Hence theequation of motion is

I<b = L,

where L is the sum of the moments about the axis of theexternal forces.

15*51. Compound Pendulum. Any body oscillating abouta fixed horizontal axis, about which it canturn freely under the action of its weight,is called a compound pendulum.

Let m be the mass of the body, G itscentre of gravity at a distance h from thefixed axis. Let GO be drawn at rightangles to the axis to meet it in 0, andat time t let GO make an angle 6 withthe vertical. Also let k be the radius ofgyration of the body about an axis through G parallel to theaxis of rotation, then the moment of inertia about the axis ofrotation is, by 13*2, m(k2 + h2). Hence by the last article

TO (F + h2) 6 = - mgh sin 6,

or h(1).

Now in 7 6 we saw that the equation of motion of a simplependulum of length I is

Id = — g sin 6;

therefore the length of the equivalent simple pendulum is

I = (k* + h2)jh.

15-5-15-52] COMPOUND PENDULUM 217

The period of small oscillations of the compound pendulum,i.e. oscillations of such small amplitude that we may write 9 forsin# in (1), is

27rv/{(P + n W (2).If on OG produced we take a point C such that

then C is called the centre of oscillation, while 0 is called thecentre of suspension of the pendulum.

Since OC = h+k2/h, where h=OG, thereforeOG.GC = P (3),

whence it follows that the centres of suspension and oscillationare interchangeable, i.e. if the body swings about a parallelaxis through G the point 0 will become the centre of oscillationand the period of small oscillations will remain unaltered.

1552. Pressure on the Axis. We can also determine thereaction of the axis on the body. Thus if X, T are the com-ponents of this reaction at 0 at right angles to and along GO(the body is assumed to be symmetrical about the plane of

the diagram), one of the accompanying figures represents theexternal forces and the other the equivalent system of ratesof change of momentum; and by resolving at right angles toand along GO, we get

mhd = X — mg sin 0 (1),mh62= Y-mgcosO (2),

and by taking moments about the axis, as in 15-51,sin <9 (3).

218 EQUATIONS OF MOTION OF A RIGID BODY [XV

By multiplying the last equation by 20 and integrating we get

(Jc* + h?)62=C+2ghcos0,

where the constant of integration can be determined if thevelocity is known in any one position; for example, if a is theamplitude of the oscillation, so that 0 = 0 when 0 = a, thenC= — 2gh cos a, and

{t? + h2) 02 = 2gh (cos 0 - cos a) (4).

Substituting from (3) and (4) in (1) and (2) we get

-„ mgl? . „

i T7- n 2mqh?, n .and I = mg cos v + J , 2 (cos 0 — cos a).

1553. Examples on the Compound Pendulum, (i) An arc ofa circle is formed of thin wire (whose density may or may not be uniform.)and hangs from a point P of the arc. Shew that, if in the position ofequilibrium Q is the point of the circle vertically below P, then PQ is thelength of the equivalent simple pendulum when the wire oscillates about P inits own plane. [M. T. 1916]

Let APB be the arc, G its centre of gravity, 0 the centre of the circleand a its radius. In equilibrium PGQ is a verticalchord of the circle. Let k be the radius of gyrationof the arc about O and m the mass. The momentof inertia about 0 is «i«2, therefore, by the theoremof parallel axes, 13"2,

ma2 = mk2+m. 0 G2.

Therefore k2=a2 -OG2

= PG. GQhy the property of the circle.Hence, by (3) of 15'51, since P is the centre of suspension, Q is the

centre of oscillation and PQ is the length of the equivalent simplependulum.

(ii) A pendulum consists of a thin uniform rod of mass M pivoted at itsmid-point and of a regulating nut of mass m, which can be screwed to anydesired position of the rod. The nut may be treated as a particle. Provethat if M>3m the period is always lengthened when the nut is raisedslightly, but if M < 3m the period is lengthened or shortened according tothe position of the nut. [M. T. 1923]

Let 2a be the length of the rod, x the distance of the nut and h thedistance of the centre of gravity of the whole from the mid-point. Then

15'52-15-54] COMPOUND PENDULUM 219

(M+m) h = mx, and the moment of inertia of the whole about the point ofsuspension is J Ma?+nix2. Therefore by formula (2) of 15'51 the period

= 2n- »/{(4 Ma? + mz2)/(M+ m) gh}

Hence the function of x whose variations are to be considered is_ Ma? .

The derivative is

dxNow since a~£.x it follows that, if M> Zm, dyjdx is negative; so that

if the nut is raised slightly making duo negative, then dy is positive, or theperiod is increased. But if M < 3m, then dy/dx is negative or positiveaccording as Ma? > or < 3mx2, implying that the lengthening or shorten-ing of the period for a slight displacement of the nut depends on itsposition.

1554. Compound Pendulum with Axis Non-horizontal.Suppose that a body is free to rotate about an axis that makesan angle a with the vertical, thecentre of gravity G will then oscil-late in a plane making an angle awith the horizontal.

The weight mg can be resolvedinto mg sin a parallel to the linesof greatest slope in this plane, andmg cos a parallel to the axis. Thelatter component has no momentabout the axis, and the former com-ponent alone is effective in causingoscillation.

Hence if GO is perpendicular to the axis and makes an angle8 with the line of greatest slope, and GO = h and k is the radiusof gyration about an axis through G parallel to the axis ofrotation, the equation of moments about the fixed axis is

m (k2 + h2) 6 = — mg h sin a sin 8,so that the length of the equivalent simple pendulum is

and the period is

By making a. small the period can be increased indefinitely.

220 EQUATIONS OF MOTION OF A RIGID BODY [XV

1555. Centre of Percussion. If a single impulse can beapplied to a rigid body which is free to turn about a fixed axisso as to produce no impulsive stress on the axis, the line of theimpulse is called the line of percussion and the point in whichthis line meets the plane through the centre of gravity G and theaxis of rotation is called the Centre of Percussion.

We shall limit our investigation to the two-dimensional case,i.e. we suppose that the body is symmetrical about a planethrough G perpendicular to the axis and that the impulse actsin this plane.

Let GO perpendicular to the axis be of length h, and let mbe the mass and mk% the moment of inertia about a parallel axisthrough G. Suppose that when an impulse P is applied in a

direction making an angle d with GO there is no reaction at 0and that the body begins to turn about 0 with angular velocity a>.The velocity of G is hw, so that the momentum is representedby mhco perpendicular to GO and the spin couple mk2a.

Since there is no reaction at 0 there can be no momentumset up in a direction at right angles to P, therefore

mhco cos 6 = 0,which requires that 0 = \ TT, or the direction of P must be atright angles to 0G, as in the second figure. Let P meet 0G ata distance p from 0. Since there is no reaction at 0, the equa-tion of linear momentum is

mha> = P;and, by moments about 0, we get

Therefore p = (h2 + W)jh, or the distance from 0 of the centreof percussion is equal to the length of the equivalent simplependulum if the body were set to swing about the given axisplaced horizontally.

15-55-15-6] MOTION ABOUT AN AXIS 221

15-6. Further examples of Motion about an Axis, (i) Theloch of a railway-carriage door will only engage if the angular velocity of theclosing door exceeds a>. The door swings about vertical hinges and has aradius of gyration k about a vertical axis through the hinges, while thecentre of gravity of the door is at a distance a from the line of the hinges.Shew that if the door be initially at rest and at right angles to the side ofthe train which then begins to move with acceleration f the door will notclose unassisted unless f> %a>W/a. [M. T. 1920]

Let 8 be the angle through which the door turns in time t. Let 0

fmaO - ~ "

mad

represent the line of the hinges andlet O be the centre of gravity. Then 0has an acceleration f and O has ac-celerations ad and aS2 relative to 0,therefore the rates of change of mo-mentum are, as in the figure, mf,mad2, mad and mWd, where h' is theradius of gyration about a vertical line through G, so that h2 = cThe only forces on the door in addition to its weight act through the line ofthe hinges, and therefore by taking moments about this line

m (a2+h"z) 8-maf cos 8 = 0 (1),

or mkfy — maf cos 8 = 0.Multiply by 28 and integrate and we get

rnhW - 2ma/sin 8=0.But 8 = 0 when 8=0, since the door is initially at rest; therefore C= 0 and

h282 = 2a/sin 8.When the door is about to close 8=^n and #2=2a//£2. In order that

the door may close we require that6 > a>, or

rmad

(ii) Motion of two heavy rods AB, BC smoothly jointed at B and swingingin a vertical plane about the end Awhich can turn about a fixed point.

Let m, m' be the masses, and 2a,26 the lengths of the rods, and attime t let them make angles 8, cj> withthe vertical. Let O, H be theirmiddle points. The accelerations ofO are a62 and a8 along and perpendi-cular to OA. The accelerations of Hrelative to B are b<f>2 and b(f> alongand perpendicular to HB, and B hasaccelerations 2a82 and 2a8 along andperpendicular to BA which must beadded to the two former components inorder to get the total acceleration of H.

222 EQUATIONS OF MOTION OF A RIGID BODY [XV

The rates of change of momentum for the rod AB are therefore mad2,ma£, and the spin couple \ma?6, and for the rod BG they are m'b^>2, m'b(j>,2m'a62,2m'aff, and the spin couple fyn'Wcj). All these are shewn in the figure.The external forces, not shewn in the figure, consist of the reactionsat A, B and the weights of the rods rag acting at G and m'g actingat H.

In order to avoid introducing the unknown reactions at A and B weform equations by taking moments about B for the rod BC, and about Afor the whole system.

By taking moments about B for the rod BC, we get

| m'b2cf> + m'b2(f> + Zm'abS cos (0 - 6)

+ 2m'ab62 sin (cf> - 6) = -m'gb sin (j> (1).

And by taking moments about A for the two rods we get

%m,'tf<j> + m'b$ {b + 2a cos (# - 6)} - m'bj>2 2a sin (<f> - 6)

+ 2m'aS {2a + b cos {$-$)} + 2m'a02b sin (<£ - 0) + \ma26 + meflS

= -TO'£r(6sin$+2asin 6) — mga&md (2).

These two equations serve for the determination of the angular velocities;and one first integral might be found directly by writing down the equationof energy.

157. The last two examples serve to illustrate an importantpoint. In establishing the theory of moment of momentum, see92, 14*1 and 14r3, we always referred to moments about anaxis fixed in space and the phrase employed was 'rate of changeof moment of momentum.' It is important however to note that,if the axis about which moments are taken is not fixed in space,then the phrases 'rate of change of moment of momentum' and' moment of rate of change of momentum' are not equivalent.The latter phrase was used in 15*1. Thus in the last examplethe components of velocity of H are bcj> relative to B and thevelocity 2ad of B, so that the components of momentum of BGare m'bcj>, 2m'a6, and the spin couple ^m'b2cj>. Hence the momentof momentum about .8 is

£m'&a<j> + m'b2cj> + 2m'a6b cos (<j> - 0),

and the rate of change of this expression is not the same thingas the left-hand side of (1) in the last example, which represents'the moment about B of the rate of change of momentum.'The reason why the expressions differ is because ' rate of changeof moment about a moving point B' must take account of B's

15-6-15-8] MOMENT OF MOMENTUM 223

displacement during the short interval St during which thechange is observed, and is not the same as if the point B werefixed.

15"8. The student who desires to pursue this point further may revertto equation (4) of 9'2, viz.

SOT (xy -yx) = S(xY- yX).

This represents that the sum of the moments about the origin—anyorigin in the plane—of the rates of change of momentum (mx, my) of theparticles is equal to the sum of the moments about that origin of theexternal forces, and when in 15"6 (ii) we took the moments about B of therates of change of momentum we were merely applying this principle andtaking as origin the fixed point in the plane through which the junction Bof the rods is passing at the instant considered.

But in 9'2 we go on to rewrite equation (4) in the form

^ S,m (xy -yx) = 2 (x T-yX),

and read the left-hand side as 'rate of change of moment of momentum1

instead of 'moment of rate of change of momentum,' and the importantpoint is that we may not rewrite (4) in this way unless the point aboutwhich we take moments does not change its position while the operationdjdt is performed.

This will become clear if we suppose that at the instant considered theorigin about which we take moments is moving with velocities u, v parallelto the axes. At time t the momenta of a particle m at (x, y) are nix, my.At time t + bt its coordinates have become x+xbt, y+ybt; but the originhas moved to tcbt, vbt, so that the relative coordinates are

x + xbt-ubt, y+ybt-vbt.

At the same time the momenta have become

m (x + xbt), m(y+ybt),

so that the new moment of momentum about the moving origin is2m{(x + xbt-ubt)(y+ybt)-(y+ybt-vb

and if we subtract the original moment of momentum

and retain only the first power of bt, we get as the increment in themoment of momentum

2wi (xy —yx — uy + vx) bt,so that the rate of change of moment of momentum about the movingorigin is

Sm(xy-yx}- u2my + v2mx,and this is not the same as

224 EQUATIONS OF MOTION OF A RIGID BODY [XV

In conclusion therefore we repeat that it is only permissible to makeuse of 'rate of change of moment of momentum1 when the axis aboutwhich moments are taken is fixed in space; but in every case it islegitimate to take moments about any axis and equate the moment of therate of change of momentum and the moment of the external forces.

E X A M P L E S

1. A reel of radius r with rims of radius R rests on a plane inclined atan angle a to the horizontal. A thread fixed to the reel passes round andunder it and then upwards, parallel to the plane, and over a smooth pulley,a mass m hanging freely from this end. The thread lies in the verticalplane of symmetry of the reel, which is also perpendicular to the inclinedplane. Calculate the tension on the thread and the acceleration of m(i) when the inclined plane is smooth, (ii) when the plane is rough andthere is no slipping of the reel on the plane. [M. T. 1917]

2. A reel of mass M, consisting of a cylinder of radius a connecting twodiscs of radius b(b> a), rolls without slipping on a horizontal plane. Alight thread, wound on the cylinder, passes in a plane perpendicular to theaxis of the reel, horizontally from its under side, over a smooth pulley, andthence vertically downwards; to its free end is attached a mass m.

Shew that the reel will move with acceleration

f=n mb(b-a)J y M(b*+k2)+m(b-a)2'

where h is the radius of gyration of the wheel about its axis. [M. T. 1927]3. A uniform solid cylinder, mass M and radius a, rolls on a rough

inclined plane with its axis perpendicular to the line of greatest slope. Asit rolls the cylinder winds up a light string which passes over a fixed lightpulley and supports a freely hanging mass m, the part of the stringbetween the pulley and the cylinder being parallel to the lines of greatestslope. Discuss the motion of the cylinder, and prove that the tension ofthe string is

(3 + 4 sin a) Mmg3M+8m '

where a is the inclination of the plane to the horizontal.[Coll. Exam. 1913]

4. A solid spherical ball rests in equilibrium at the bottom of a fixedspherical globe whose inner surface is perfectly rough. The ball is strucka horizontal blow of such magnitude that the initial speed of its centre is v.Prove that, if v lies between »JlOdg/7 and */27dg/7, the ball will leave theglobe, d being the difference between the radii of the ball and globe.

[Coll. Exam. 1911]

15-8] EXAMPLES 225

5. A flywheel has a horizontal shaft of radius r; the moment of inertiaof the system about the axis of revolution is K. A string of negligiblethickness is wound round the shaft and supports a mass M hangingvertically. Find the angular acceleration of the wheel when its motion isopposed by a constant frictional couple G.

If the string is released from the shaft after the wheel has turnedthrough an angle 6 from rest, and if the wheel then turns through afurther angle 0 before it is brought to rest by the frictional couple, shewthat

KM9r6 rM T 1915-|

6. A flywheel, turning with average angular velocity p, is acted on by adriving couple A sin2 pt, and has a constant couple \A opposing its motion.Find the least moment of inertia required to make the difference betweenthe greatest and least angular velocities less than p/100. [M. T. 1922]

1. A circular plate of mass M and radius a has a mass m fixed in it ata distance b from the centre. An axis through the centre of the plate andperpendicular to it can slide without friction horizontally, while the platerevolves. If the plate is just disturbed from rest when m is in its highestposition, find the angular velocities when the disc has made one quarterand one half a turn.

Determine the pressure on the axis in each case. [M. T. 1918]

8. Shew that, if a uniform heavy right circular cylinder of radius a berotated about its axis, and laid gently on two rough horizontal rails at thesame level and distant 2a sin a apart so that the axis of the cylinder isparallel to the rails, the cylinder will remain in contact with both rails ifthe coefficient of friction fi < tan o, but will initially rise on one rail if/u > tan a. [M. T. 1919]

9. A uniform trap-door swinging about a horizontal hinge is closed bya spring coiled about the hinge. The spring is coiled so that it is justable to hold the door shut in the horizontal position. The horizontalopening which the door closes is in a body which is mounting with uniform

( 1 *2S\0"57 H J g, the door starting from the

vertical position will just reach the horizontal position, a being the anglethrough which the spring is coiled when the door is in the horizontalposition. [M. T. 1919]

10. An ellipse of axes a, b and a circle of radius 6 are cut from thesame sheet of thin uniform metal and are superposed and fixed togetherwith their centres coincident. The figure is free to move in its own verticalplane about one end of the major axis: shew that the length of the simpleequivalent pendulum is (5ct2-a6 + 262)/4a. [Coll. Exam. 1909]

RD 15

226 EQUATIONS OF MOTION OF A RIGID BODY [XV

11. A rigid pendulum 00 swings about a horizontal axis through 0, itscentre of gravity being at O. The pendulum is released from rest when00 is horizontal. "When 00 is vertical, the pendulum is brought to restby an inelastic buffer B which is such that the line of the reaction betweenB and the pendulum is horizontal and at a distance I below 0. The massof the pendulum is m, its moment of inertia about a horizontal axisthrough G is mh2 and 00 = h.

Shew that, if the impulse of the force exerted by B upon the pendulumduring the impact is P,

lP=m \/Deduce the impulse Q of the horizontal force exerted on the pendulum

during the impact by the axis 0 and shew that it vanishes when I is equalto the length of the equivalent simple pendulum. [M. T. 1914]

12. A straight rod of mass m and length 11 swinging about one end asa compound pendulum starts from rest in a horizontal position and whenvertical is struck a blow at its middle point which reverses and halves itsangular velocity. Prove that the impulse of the blow=»i \/<ogl.

[Coll. Exam. 1912]

13. A flywheel weighing 5 tons is suspended from a pair of centresentering conical holes in the rim so that it can swing in a vertical plane.The line joining the centres is parallel to and distant 4 feet from the axisof the wheel, and the period of a complete swing is 3'2 seconds. Find theradius of gyration of the wheel about its axis, and determine how muchenergy the wheel would lose in falling from a speed of 120 to 90 revolutionsper minute when revolving round its axis. [M. T. 1918]

14. A thin uniform rod of mass m and length 2a can turn freely aboutone end which is fixed, and a circular disc of mass 12™ and radius a/3 canbe clamped to the rod so that its centre is on the rod. Shew that, foroscillations in which the plane of the disc remains vertical, the length ofthe equivalent simple pendulum lies between 2a and 2a/3. [M. T. 1915]

15. A pendulum is supported at 0, and P is the centre of oscillation.Shew that, if an additional weight is rigidly attached at P, the period ofoscillation is unaltered.

16. A rope hangs over a pulley, which is of moment of inertia / ,and perfectly smooth on its bearings, but perfectly rough to the rope.Two monkeys of equal mass m hang one on each end of the rope. Themonkeys can climb with constant speeds % and «2 relative to the rope(«! > «2)- Shew that in a race through a height h the monkey of speed ux

can give the other monkey any start up to

where a is the radius of the pulley. (The system is at rest before themonkeys start climbing.) [S. 1925]

EXAMPLES 227

17. A fine string has masses M, M' (M> M') attached to its ends andpasses over a rough pulley with a fixed centre; shew that if the string doesnot slip, the downward acceleration of M is

g (M- M')I(M+ M'+mW\aF)

where m, mk2, a are respectively the mass, moment of inertia about axisand radius of pulley.

Shew also that to prevent slip the coefficient of friction must be greaterthan

[M.T.1918]

18. A uniform rod of mass m and length 2a is lying on a smooth hori-zontal table and is struck a blow B perpendicular to its length at oneextremity. Find the velocities with which the two ends of the rod beginto move. [M. T. 1922]

19. Two uniform bars of the same material and cross-section, but oflengths a and b, are joined by a smooth hinge and laid out in a straightline. An impulse J at right angles to them is applied at the hinge. Shewthat the hinge takes up a velocity 4Jjm, where m, is the mass of the barstogether. [Coll. Exam. 1927]

20. Two thin uniform rods AO, OB of lengths 2a and 26 and of massesm and M, are smoothly jointed at 0, and lie in a straight line on a smoothtable. An impulse P is applied at A perpendicular to AOB. Find theimpulsive reaction at 0 and the initial velocity of 0.

If equal and opposite impulses P are applied simultaneously at A andB, perpendicular to AOB, shew that the impulse at 0 is numerically equalto J P and that the initial velocity of 0 is zero. [M. T. 1924]

21. Two flywheels, whose radii of gyration are in the ratio of theirradii, are free to revolve in the same plane, a belt passing round both.Initially one, of mass mx and radius alt is rotating with angular velocity Q,and the other, of mass m2 and radius <?2> is at r8S*- Suddenly the belt istightened, so that there is no more slipping at either wheel. Shew thatthe second wheel begins to revolve with angular velocity

22. A circular disc of radius a lies on a smooth horizontal table, whena point on the circumference is compelled to move in the direction of thetangent at that point with velocity u. Prove that the disc begins to turnwith angular velocity 2u/3a. [Coll. Exam. 1910]

228 EQUATIONS OF MOTION OF A RIGID BODY [XV

23. A rectangular lamina A BCD of mass M and sides AB, CD of length2a, and AD, BC of length 26, rests upon a smooth horizontal plane. At acertain instant it receives a blow of impulse P applied at A and directedalong AB. Shew that after a time t from this instant the distance of Cfrom the original position of the centre of gravity is given by

\ M2 M

where tan e = a/b. [M. T. 1926]

24. A uniform rod AB of mass M and length 2a is struck by aninstantaneous impulse £, acting at B in direction BP which makes anangle 8 with AB produced. Shew that AB will become parallel to BPafter a time t given by

t=Ma8IZt-f&a8.If the end B, instead of being free, is compelled by a frictionless

constraint to move in the line BP, shew that the impulse £ will beaccompanied by a simultaneous impulsive reaction on the constraint, ofmagnitude

3£ sin 8 cos (9/(1+3 cos28). [M. T. 1927]

25. Two rods are smoothly jointed at A and B to a circular lamina,AB being a diameter of which the rods are continuations. The wholesystem lies at rest on a smooth horizontal table. Shew that a horizontalblow P applied normally to the rim of the lamina will produce in thelatter no initial angular velocity, provided that

where m1; m2 are the masses of the rods, aly a2 are the distances of theircentres of gravity from A and B respectively, and hi, k2 are their momentsof inertia about their centres of gravity. [Coll. Exam. 1911]

26. A free lamina of any form is turning in its own plane about aninstantaneous centre of rotation S. A point P in the line joining 8 to thecentre of gravity O is brought to rest by an impulsive force passingthrough the point. Find the position of P in terms of SG and the radiusof gyration about 6, assuming that the velocity of O is the same as beforebut reversed in direction. [M. T. 1918]

27. If a body can only turn about a smooth horizontal axle, and whenthe body is at rest, the axle is given an instantaneous horizontal velocity vin a direction perpendicular to its length, prove that the centre of masswill start off with a velocity v(k2 — h2)/k2, and that the initial angularvelocity will be vhjk%, where h is the distance of the centre of mass fromthe axle and k the radius of gyration about the axle. [Coll. Exam. 1913]

EXAMPLES 229

28. A gate of length I can swing about a vertical hinge at one end.When it is in equilibrium a particle of mass m and velocity v strikes thegate normally at its middle point. The moment of inertia of the gateabout the hinge is / = j - r f , and the coefficient of restitution is J. Findthe velocity of the particle and the angular velocity of the gate just afterimpact. [M. T. 1928]

29. The speed of a railway truck, weighing 5 tons, is reduced uniformlyfrom 25 to 20 miles per hour on the level in a distance of 695| ft. by thebrakes. Shew that, if no slipping takes place between the wheels and therails, the normal pressure between the rails and each of the front wheelsis 50 lb. weight greater than the corresponding pressures on the backwheels, given that the distance between the axles is 12 ft. and that thecentre of gravity of the truck is 4J ft. above the ground and equidistantfrom the axles, while the diameter of each wheel is 3 ft. and the momentof inertia of each pair of wheels and axle about its axis is 3600 lb. ft.2 units.

[M. T. 1916]

30. A thin uniform rod OA, of length a and mass m, turns in a hori-zontal plane about a vertical axis through 0; an equal rod AB is jointedat A to OA and a smooth guide compels the end B to move along a hori-zontal straight line Ox. The angle A0x = 6. No external forces act uponthe rods except those due to the axis and the guide, and the motion takesplace without friction. When 6=0, d6jdt=a>. Shew that

Prove that, if H be the angular momentum of the system about thevertical axis through 0,

H=ma?ddldt.

Shew that the force which the smooth guide Ox exerts upon the rod AB is

Zma«>2 sin 0 /2(4-3 cos2 6f. [M. T. 1923]31. A rectangular gate is free to swing about one edge which is inclined

at an angle of 5° to the vertical. When the gate is shut the top andbottom lines are horizontal. The gate is then opened through 90° andslightly disturbed so that it shuts. Find its angular velocity in anyposition, and the time occupied in the last 45° of its swing. The gate isuniform and 4 feet wide. [M. T. 1924]

A N S W E R S

/( + l

(ii) T

230 EQUATIONS OF MOTION OF A EIGID BODY [XV

5. (Mgr-G)I{K+Mr>). 6.

(M+m) g - 2m? b2gjMa2 ;

(if+m) g {l+8mWg/[M(M+m) a

18. 45/TO, - 2 5 / T O . 20. MPfi(M+m); 2P/(M+m).

26. B{GP-SG)=2SO.GP\ 28. £»; 2»/J.

31. 8 {2sin 5°(1 -cos 6)}ijk; ,-.—^ loge-—1 o , seconds where 6 iso Y sin o t3>n XX Xo

the angle turned through and k is the radius of gyration about the hinges.

Chapter XVI

MISCELLANEOUS PEOBLEMS

161. Rolling and Sliding. When sliding takes place thefriction F bears a constant ratio /JL to the normal reaction Rbut when rolling takes place the friction has generally a muchsmaller value.

When two bodies in contact at a point A have a relativemotion the process of determining whether this motion involvesrolling or sliding at A is as follows:

Write down the equations of motion which involve the frictionF and the normal reaction R, and assuming (i) that rollingtakes place also write down the kinematical condition whichexpresses the fact that there is no relative tangential velocityat A. If from the solution of these equations we find that F/Ris less than the coefficient of friction /i, the assumption thatrolling takes place is justified and rolling will continue until F/Rbecomes greater than /JL.

(ii) Assuming that sliding takes place, write /JLR instead of Fin the equations of motion and solve the equations without thekinematical condition above referred to. If the solution shewsthat there is a relative tangential velocity at A and the directionof motion is opposed to what has been assumed as the directionof the friction then we have found the true motion and it willcontinue until relative velocity at A vanishes.

1612. A wheel spinning about a horizontal axis is projected along arough horizontal plane, to determine the subsequent motion.

Let m be the mass and a the radius of the wheel, ml? its moment ofinertia about its centre O, and A the point ofcontact with the plane.

Suppose that initially Q has velocity v0 andthat there is an angular velocity o>0 in theopposite sense to what it would be if themotion were one of pure rolling. Let v, <o bewhat these velocities become at time t. Sincethe point A of the wheel has a velocity v + aa>,

232 MISCELLANEOUS PROBLEMS [XVI

sliding is taking place and the friction is /J.R where R is the verticalreaction.

The equations of motion aremv= - JXR, O = R-mg, mkii>=—fiRa,

Thereforev=—fig and k2a>=—figa.

Hence, by integrating and putting in the initial values,v=vo — figt and k2a = k'ia>a — figat (1),

These equations shew that v and a decrease steadily and the subsequentmotion depends on which vanishes first.

v vanishes after a time v^ipg a n d "> vanishes after a time k2a0/figa. Letus suppose (i) that v vanishes first, i.e. that vo<k2a>i)ja. At the instantwhen v vanishes the angular velocity has the value <»i given, from (1), by

or o>i = CD0 — avajk2 (2),

which is positive in the same sense as a>0, so that the point A still hasa velocity aa>i in the same sense, i.e. there is still slipping at A though &has come to rest and there is still friction jiR in the same sense.

This friction will now give Q an accelera-tion u in the reversed direction, and theequations of motion while sliding lasts are

or u = fxg and k2w—-/j.ga

so that u=figt and k2a> — k2a>i — pgat ...(3),

where t is the time since O was at rest.These equations shew that u increases and a decreases, and the velocity

of the point A is1+Tjj) — a&i (4).

This velocity vanishes after a time Waa^/fig (k2+a2) when sliding ceases,and since at that time u is k2aa\j{k2+di)therefore the wheel rolls, and the equationsbecome

mu=-F, mk2i> = Faand the condition for rolling u — «<B=0.

These equations shew that u=Q oru = constant = k2aa1/(k*+a2) where », is A Fgiven by (2), and the wheel continues to roll witk this constant velocity.

Reverting now to the first stage of the motion and suppose alternatively(ii) that av0 > k2a0 so that a> vanishes first. This happens, from (1), whent-=k2a^\figa, and at that instant the point of contact A has the samevelocity as Q, namely, v1 = v0—k2a>0/a from (1), so that sliding continuesand the friction y.R begins to create an angular velocity Q in the

mk2w

1612-16-2] ROLLING SPHERES 233

opposite sense to the former angular velocity,instant at which a> vanishes, the equationsof motion are

mi= —jiR, Q = R — ing, mh2Q = iiRa

Therefore, by integrating and using the initialvalues, we get

v=Vi — jigt and k2Q = ngat.The velocity of A is H1**

Measuring t from the

mv

which vanishes whenv - aQ, = »! - figt (1 + a2jk2)

«*/*»).

Sliding then ceases, and since v is not then zero rolling begins and asbefore it may be shewn that the final velocity is uniform.

16'2. Two Spheres in Contact. A sphere of mass m and radius arolls on a sphere of mass m! and radius b which rolls on a horizontal plane;the friction is sufficient to prevent sliding and the motion is all parallel tothe same vertical plane.

Let A, B be the centres of the spheres and a, a> their angular velocities.Let v denote the velocity of the sphere m', and let BA make an angle 6with the vertical at time t.

Relative to B, A is describing a circle of radius a + b, therefore itsaccelerations relative to B are (a + b) 6 at right angles to AB and (a+b) 62

along AB, and with these must be compounded the acceleration v of B.Hence the rates of change of momentum of the sphere m are mv, m (a + b) 6,in (a + b) 62 and the spin couple mk2a> or §WMZ2O>. Similarly the rates ofchange of momentum of mf are rn'i and the spin couple m'V2a> or |m'62t»'.

There are two kinematical conditions for rolling, viz. the point ofcontact Q of the lower sphere with the plane has no velocity, therefore

y-6<o' = 0 (1).Also the point of contact P of the two spheres must have no relativetangential velocity, or the tangential velocities of the point P on the twospheres must be the same.

234 MISCELLANEOUS PROBLEMS [XVI

On the sphere m, P has velocity aa> relative to A, but A's velocity is(a + b) 8 relative to B and B's velocity is v; therefore on the sphere m, P'stangential velocity is

(a + b) 8 + v cos 8 — am.But on the sphere m', P has velocity bat relative to B so that its tangentialvelocity is ba>'+v cos 6; therefore

(a + b)8-aa = bu' (2).We have now to write down the equations of motion and we may avoid

the unknown reactions by taking moments about P for the sphere m, andabout Q for the two spheres together. Thus we have

ma (a + b) 8 + mav cos 8+:g ma!la=mgasin 8 (3),and m(a + b)8 (a + b + bcos8) — rn(a+b) 62b sin 8 + mv {(a + b) cos 8 + b}

+ f ma2w + m'bv + % m'b2a,' = mg (a+ 6) sin 8 (4).And equations (1), (2), (3) and (4) are sufficient to determine the motion.

We might alternatively have written down the equation of energy forthe whole system. Observing that the velocity of A is compounded of vand (a + b) 8, this equation is

I m {v* + (a + bf 82 + 2 (a + b) 8v cos 8 + f « V }+ \m'(v2 + b2a'2)+mg (a + b) cos# = const (5).

This equation is of course a first integral of the equations of motion.If however we wish to determine the unknown reactions, denoting them

as in the second figure, we can write down the formal equations of motionfor each sphere ; resolving, and taking moments about the centre for eachsphere we have

m (a + b) 6 +mi cos 8=tngsin8 — F,m (a + b) 82 — mi) sin 8 = mg cos 8 — R,lma2i> = Fa,m'i= — F' — R sin 8+F cos 8,0 = 8 - rn'g - R cos 8 - Fsin 8,

These six equations together with (1) and (2) are equivalent to (1), (2),(3) and (4) and also determine the unknown reactions.

16-2-163] INITIAL MOTIONS AND STRESSES 235

16'21. The same problem as the last but no friction between the lowersphere and the plane.

In this ease equation (1) no longer holds, but instead we have the con-dition that there is no external horizontal force acting on the system, sothat if it starts from rest its horizontal momentum remains zero through-out the motion. This fact is expressed by the equation

mv + m (a + b) 8 cos 6+rn'v = 0,and the other equations are as in the last article.

16-3. Initial Motions and Stresses. When one or more ofthe constraints which maintain a system in equilibrium areremoved the initial motion and the initial values of the remain-ing stresses in the system may be determined by writing downthe kinematical conditions that specify the remaining constraintstogether with the equations of motion of the system in itsinitial position, with the simplification that initial velocities areall zero so that radial and transverse accelerations are simplyr and rd and all normal components of acceleration vanish.

Examples, (i) A heavy uniform straight rod is suspended from a pointby two strings of the same length as the rod attached to its ends. If one stringis cut, prove that the initial angular acceleration of the rod is nine timesthat of the remaining string.

Let m be the mass and 2a the length of the rod; and let a>, &>' denotethe initial accelerations of the string and rod.

The middle point O of the rod has accelera-tion ai)' relative to the end A ; but the end Aof the string has acceleration 2aa> perpendi-cular to the string OA, therefore G has also Tthis acceleration, and the rates of change of /momentum are maa, 2maa> and the spin / / Gcouple JfflaV. A | \ , •

If T be the initial tension in the string, by \ ^ 2 ma w

resolving horizontally for the rod, we get mad)'

2mai> cos 30° =Tcos 60° (1),and by taking moments about O

lma^'=TaamG0° (2),whence we get

a>'=9o) (3).To find T in terms of the weight of the rod we resolve vertically and get

mai> + 2maa> cos 60° = mg — Tain 60°,

and with the help of (3) and (1) we find that T=?-^ mg.13

236 MISCELLANEOUS PROBLEMS [XVI

(ii) A solid hemisphere is held with its base against a smooth vertical walland its lowest point on a smooth floor. The hemisphere is released. Find theinitial pressures on the wall and the floor.

Let m be the mass, a the radius, 0 the centre and G the centre ofgravity of the hemisphere. CG = %a. Let 0 bethe point of contact with the ground and let00 = h.

Since the hemisphere begins to turn about 0,if a> is the initial angular acceleration, the ac-celeration of G is ha> at right angles to 00.Hence the rate of change of momentum is re-presented by mhu> and the spin couple mk2d>where k is the radius of gyration about a hori-zontal axis through O parallel to the wall.

By taking moments about the line of intersection of the wall and floorwe get

m (A2 -f- k2) o>=f mga.Now m(h2+k2) is the moment of inertia of the hemisphere about the

line of intersection of the wall and floor. But its moment of inertia abouta diameter of its face is § ma2, and therefore by the theorem of parallelaxes (13'2) the moment of inertia about the required axis is § ma2.

Hence %ma2a> = § mga, or a<b = ^g.

Let X, Y be the pressures of the wall and floor on the hemisphere, thenby resolving horizontally and vertically we get

mha cos COG =X, or mad> = X,

and mhi>sm COG — mg — Y or \ma<o = mg — Y.

Therefore

164. Bending Moments in Bodies in Motion. When a rodAB is in equilibrium under the action of given coplanar forcesthe stresses at any point P of the rod may be determined byimagining the rod to be divided by a cross-section at P. Then

J

by considering the equilibrium of either portion, say PB, wesee that the forces exerted by AP on PB must balance all theother given forces that act on PB. But a system of coplanarforces can be reduced to a single force acting at any assigned

16-3-16-41] BENDING MOMENTS 237

point in the plane together with a couple. Hence the givenforces acting on PB can be represented by a force at P with atangential component T and a normal component S togetherwith a couple 0, and the reaction of AP on PB balances these.T is called the tension, 8 the shearing force and G the bendingmoment, and the latter is taken as the measure of the tendencyof the rod to break at P.

When the rod is in a given state of motion instead of inequilibrium, if we write down the equations of motion for eitherportion these equations will suffice for the determination of T, 8and G, if the accelerations have first been determined byconsidering the motion of the whole rod.

16"41. Example . A heavy rod swings freely in a vertical plane aboutone end; to find the stresses at any point of the rod.

Let I be the length of the rod OA, m the mass of unit length, and 6 itsinclination to the vertical at time t.

By taking moments about the fixed end0 for the whole rod, we get

\mls'6=-\mgP sin 6

or # '= - §gr sin 0 (1),and, by multiplying by 28 and integrating,

162 = $g(cos6 -cos a) (2),when a denotes the amplitude of theoscillation.

Now let the point P at which thestresses are required be at a distance afrom 0, and let T, S, G denote the tension,shearing force and bending moment in thesense in which they act on PA. We have to write down the equations ofmotion of the part PA of the rod. Consider a small element of length dxand mass mdx at Q where OQ=x. The accelerations of this element arex8 and x92 at right angles to and along QO. Hence by resolving at rightangles to and along A 0 and taking moments about P, for the portion PAof the rod, we get

mdoc.xQ

and

mx'8 dx=S-mg (l-a)smd (3),•,

flI mxd2dx=T-mg(l-a)oos 8 (4),

J a

I mx8(x-a)dx= -Q-\mg (l-a)2sin8 (5).J &

238 MISCELLANEOUS PROBLEMS [XVI

These are equivalent to2- a2) 8 = S-mg (I-a) sm 8 (3)',

-a2) 62=T-mg (l-a)cos8 (4)',and \m (l-a)2 (21 +a) 8= - O-\mg (l-a)2 sin 8 (5)'.

Substituting for 8 from (1) and for 62 from (2) we get

„ , (I-a) (I-3a) . .S- i mgv '-^ ; sin 8,

(I2 — a2)T= mg (I - a) cos 8 + f mgy-—-—-(cos 8 - cos a),

„ . (l-a)2 . .and G = f mg -—j-?- a sin 8.

165. Steady Motion in Three Dimensions. A steadymotion is one in which all velocity components are constant inmagnitude. Problems of steady motion are often solved bysimple applications of the principles of energy and momentum.

Examples , (i) A uniform straight rod of length I is free to turn aboutone end. Find the angular velocity with which it can describe a cone ofsemi-vertical angle a.

If <o is the angular velocity then each element of the rod describes ahorizontal circle with angular velocity <». OLet m be the mass of unit length andconsider an element mdx at a distance xfrom the fixed end 0. Its acceleration isa>2x sin a, therefore its rate of change ofmomentum is mdx. a?x sin a towards thecentre of the circle that it describes.

Take moments for the whole rod abouta horizontal axis through 0 perpendicularto the rod, and we get

mdx.iu2xsina

Joix cos a . ma2x sin adx = mlg. il sin a,

I o>2 £3cosasina = J mgl2 sin a,

therefore la2 cos a = § </.(ii) The framework in the figure revolves about the vertical axis and the

balls move outwards for an increase of speed, the weight E sliding up theaxis. Shew that the angular velocity <o with which the frame rotates in theposition in lohich the arm ABD makes an angle a with the vertical isgiven by

(a + l sin a) <a2 = (1 + m'b/ml) g tan a,

where AB=BC=b, AD^l, AA' = CC' = 2a.

The balls are each of mass m and are fixed to the arms: the sliding weight Eis of mass m' and the weights of the arms may be neglected. [M. T. 1917]

16-41-16-6] INSTANTANEOUS CENTRE 239

The point D describes a horizontal circle of radius a+ 1 sin a withangular velocity a>. Therefore the acceleration of D is (a +1 sin a) a2

towards the axis of rotation and the rate of change of momentum of theball is m (a + 1 sin a) a>2. We have now to consider whether to resolve ortake moments and we notice that we cannot resolve without introducing

A'

the stresses in the rod AD to which the ball is fixed, and we cannot takemoments about any point of the rod except the point A without intro-ducing the bending moment of the rod. We therefore take moments forthe ball and rod AD about an axis through A perpendicular to the planeof the diagram. This gives

TO (a+l sin a) a2.1 cosa=mgl aina+Tb sin 2a,where T is the tension in the rod BO.

Next resolving vertically for the weight E, which is at rest in the steadymotion, we have „_ ,

2Tcoaa = m'g.Eliminating T, we get

(a + l sin o) a>2 = (l +m'b/ml)g tan a.166. Use of the Instantaneous Centre of Rotation. In the

motion of a body in one plane, let u, v denote the velocities ofits centre of gravity G parallel to rectangular axes and let co bethe angular velocity of the body. If(r, 6) are the polar coordinates of a pointP of the body referred to G as origin,the velocity of P relative to G is ra> atright angles to GP and this has com-ponents — rco sin 6, rco cos 0 parallel tothe axes of x and y or — ym, xa> respectively, if {x, y) are co-ordinates of P relative to the axes through G. Therefore thewhole velocities of P are

u — yto and v + xa>

240 MISCELLANEOUS PROBLEMS

and if these components vanish the point P has no velocity.Therefore the point whose coordinates are — v/co, uja> is at restat the instant considered. This point / is called the instantaneouscentre of rotation (5'41).

Let us now write down the equa-tion of motion for the body obtainedby taking moments about i". Therates of change of momentum areTOW, mi) and the spin couple mk2a>, andif L denotes the sum of the momentsabout / of the external forces we have

L = moment about / of rate of change of momentum. u= mu . ( v\

mvV «/

= — (uu + vv + k2cow)

a> at

If now r denotes the distance GI, since the body is, at theinstant considered, turning about I, the velocity of G is ra>, sothat M2 + v2 = r2a>2, and therefore

i~im(r2 + k2)^} = L (1).

By the theorem of parallel axes (13'2) if K denotes the radiusof gyration about /, we have

K* = r2 + k\and therefore

^ - 1 (2).

Since the instantaneous centre is generally changing itsposition relative to the body, K is not generally constant butthere are the following special cases:

(i) If the body is turning about a fixed axis K is constantand (2) becomes

mK2i> = L (3),as in 15-5, and the same is true if the instantaneous centre isat a constant distance from G.

166—1661] USE OF THE INSTANTANEOUS CENTRE 241

(ii) If the axis be not fixed but the body starts from rest, since(2) is equivalent to

r77T2

(4),atand initially a> = 0, therefore the initial value of the accelerationis given by

mK2o) = L (3).(iii) In a small oscillation about a position of equilibrium, if

we take moments about the instantaneous centre in a positionslightly displaced from the equilibrium position, then in equation(4) co is small and dK^/dt is of the order of a velocity, and there-fore the second term of the equation is of the order of the squareof a small velocity and can be neglected, so that again

mK2d> = L (3).

To summarize the results, it appears that equation (3) is a validequation if the instantaneous axis of rotation is a fixed axis orat a fixed distance from the centre of gravity, or if we are dealingwith an initial motion or a small oscillation; but in every othercase in which we take moments about the instantaneous centreor axis of rotation we must use equation (2).

1661. Examples , (i) The ends of a heavy rod are constrained to moveon two smooth intersecting wires, one of tvkich is vertical and the otherhorizontal.

Let AB be the rod, 2a its length, m its mass and 6 its inclination to thevertical. The instantaneous centre ofrotation is the corner / of the rectangleOAIB (5'42) where Ox, Oy are the wires.

Also, since 01—a, the moment ofinertia of the rod about /

(13-2)

= %ma. -. _ _

This is a case therefore in which wecan employ equation (3) of 166, taking moments about I and therebyavoiding the unknown reactions at A and B which pass through / . Theangular velocity of the rod is 6 in the counter-clockwise sense, and the onlyforce that has a moment about / is the weight mg in the same sense,therefore

§ma?S=mga sin 6.RD

l 6

242 MISCELLANEOUS PROBLEMS [XVI

On multiplying by 6 and integrating, this gives

Za62=g (cos a —cos 6),

where a is the initial inclination of the rod to the vertical.We leave it to the student to obtain the same result by the principle of

energy.

(ii) The figure represents the cross-section through the centre of a uniformelliptic cylinder which rests on a smooth horizontal plane against a smoothvertical plane. The minor axis of thecross-section passes through the inter-section 0 of the planes. To find theinitial angular acceleration. Q

The normals at the points of con-tact P, Q meet in the instantaneouscentre /, and in the symmetricalposition OGI is a straight line,where G is the centre of gravity,and it can be shewn that

y

where a, b are the semi-axes. P xSince the reactions at P, Q pass through / the weight mg is the only

force that has a moment about / .Therefore if w be the initial angular acceleration

m (P + GI2) a> = mgGIIs/2.

But £2 = (a2 + 62)/4,therefore

(5a4 - 6«262 + 56*) & = 2 \/2g (a2 - ¥) V(a2 + 62).

1662. In the next chapter we shall have occasion to usethe instantaneous centre of rotation in examples of smalloscillations.

EXAMPLES1. A circular hoop of radius a, while spinning in a vertical plane with

angular velocity u> about its centre, is gently placed on a rough inclinedplane which slopes at the angle of friction a for the surfaces in contact.Shew that, if the sense of the rotation be that which causes the slipping atthe point of contact to be down the line of greatest slope, the hoop willremain stationary for a time aa>/g sin a.

2. A uniform sphere, of radius a, is projected with velocity V down arough slope of inclination o having also an angular velocity Q about ahorizontal axis in such a sense as to tend to cause rolling up a line of

16-61-16-62] EXAMPLES 243

greatest slope. Prove that it will never stop slipping unless /* > f tan a

A 5 V \l - s — I.

^ as//[Coll. Exam. 1928]

3. A circular hoop in a vertical plane is projected down an inclinedplane with velocity Vo and at the same time is given an angular velocity<»o tending to make it roll up the plane. Find the relation between Vo, a>0,the slope of the plane and the coefficient of friction, if the hoop comes toa position of instantaneous rest. [Coll. Exam. 1914J

4. A circular cylinder is fixed with its axis horizontal, and a sphere isplaced on the highest generator and slightly disturbed. If the surfaces arerough, find the position of the sphere when sliding begins, and shew thatthe sphere must slide before leaving the cylinder. [Coll. Exam. 1910]

5. A homogeneous sphere of radius r is placed on a smooth horizontaltable and a perfectly rough equal homogeneous sphere is placed on thetop and then slightly displaced. Prove that the same points alwaysremain in contact and that the angular velocity of either sphere is

at the instant of impact with the table. [Coll. Exam. 1909]

6. Investigate the motion of a circular hoop, hanging over a rough pegwhose cross-section is circular, and find the components of the reactionin any position, assuming that the peg is sufficiently rough to preventslipping. [Coll. Exam. 1908]

7. A sphere of mass m rolls down the rough face of an inclined plane ofmass M and angle a, which is free to slide on a smooth horizontal planein a direction perpendicular to its edge. Investigate the motion and shewthat the pressure between the sphere and the inclined plane is

m(2m + 7M) g cos a/{(2 + 5 sin2 a) m + 7M}.

[Coll. Exam. 1912]

8. A sphere of radius a and mass M rests on a smooth horizontal plane,and a second sphere of radius b and mass m is placed upon it, the line ofcentres being inclined at an angle 60 with the vertical; the surfacesbetween the two spheres are rough. The system is allowed to movefrom rest in this position. Shew that as long as the spheres remainin contact

(1M+ 5m sin2 6) (a + b) 62 = 10 (M+ m) g (cos 80 - cos 6),

where 8 is the inclination of the line of centres to the vertical.

[Coll. Exam. 1927]

16-2

244 MISCELLANEOUS PROBLEMS [XVI

9. A perfectly rough solid sphere, of mass m and radius r, restssymmetrically upon a hollow cylinder, of mass M and radius R, free toturn about its axis which is horizontal. If the sphere roll down, shew thatat any time during the contact the angle <£ between the line of centres andthe vertical is given by

(7M + 2m) (R + r)(j> = (5i/"+ 2m) g sin (j>;and find the value of tf> when the bodies separate. [Coll. Exam. 1926]

10. The mass of a sphere is -J of that of another sphere of the samematerial, which is free to move about its centre as a fixed point, and thefirst sphere rolls down the second from rest at the highest point, thecoefficient of friction being /x. Prove that sliding will begin when theangle 8 which the line of centres makes with the vertical is given by

sin0=2/i(5cosd-3). [M. T. 1905]

11. A rough perfectly elastic sphere is dropped without rotation on toa horizontal cylinder which is free to turn about its axis. There is noslipping at the point of contact during the impact, and the sphere startsmoving horizontally after the impact. If 8 is the angle which the radiusof the cylinder through the point of contact makes with the vertical,prove that i

tan2 8=1+—?. *,Did' I B S '

where a and a' are the radii of the cylinder and sphere, / and / ' are theirmoments of inertia about their centres and m is the mass of the sphere.

Shew also that the coefficient of friction between the cylinder andsphere must be greater than J (tan 8 — cot 6) in order that there may beno slipping during the impact. [M. T. 1922]

12. A uniform straight heavy rod AB of mass M is freely jointed abouta smooth horizontal axis at A, and is supported at an inclination 8 to thevertical by a light string which is perpendicular to the rod and attachedto it at B. The string is suddenly cut. Find the pressures on the axis at Abefore and immediately afterwards. [M. T. 1919]

13. A square lamina is suspended by vertical strings tied to twoadjacent corners; two edges of the square being vertical. If one stringis cut, shew that the tension of the remaining string is instantaneouslydiminished to £ of its former value. [M. T. 1911]

14. A uniform rod is held at inclination a to the vertical, with its lowerend resting on an imperfectly rough table. It is suddenly released. Shewthat the lower end will instantly begin to slide if the coefficient of friction

is less than ——^—r~i~ > ar>d hence that if the coefficient of friction4 — 3 sin2 a

exceeds j the rod will not slide initially whatever a may be.[Coll. Exam. 1927]

EXAMPLES 245

15. A uniform heavy rectangular plate is placed with an edge along theintersection of a smooth vertical wall and a smooth horizontal plane andallowed to turn round the edge from a position of rest in a nearly verticalposition: shew that the edge will leave the wall during the motion.

[Coll. Exam. 1915]

16. A uniform circular disc of radius a is rolling without slipping alonga smooth horizontal plane with velocity V when the highest point becomessuddenly fixed. Prove that the disc will make a complete revolution roundthe point if F 2 > Mag. [M. T. 1928]

17. A heavy uniform rod of mass m is supported against a smooth fixedsphere by a horizontal string fastened to its upper end and also to thehighest point of the sphere. Shew that, if the string is cut, the initialpressure of the rod on the sphere is m^cosa/(l+3sin4a), where a is theangular distance of the point of contact from the highest point of thesphere.

18. A smooth ring of mass m slides on a wire bent into the form of acircle of radius r which is made to rotate about a vertical diameter withuniform angular velocity a. Find the position of relative rest of the ringon the wire and shew that the pressure between the ring and the wire isthen m«?r. [S. 1917]

19. Two unequal masses are connected by a string of length I whichpasses through a fixed smooth ring. The smaller mass moves as a conicalpendulum while the other mass hangs vertically. Find the semi-angle ofthe cone, and the number of revolutions per second when a length a ofthe string is hanging vertically. [S. 1927]

20. The ends of a uniform rod can slide without friction on a circularwire which is made to rotate about a vertical diameter with angularvelocity o>. Shew that, if there is a state of steady motion in which therod is not horizontal, its inclination to the horizontal is cos"1 (g/aa>2),where a is the distance of its centre from the centre of the circle.

21. A rigid ring hanging over a smooth peg is set spinning aboutits centre (which remains stationary) in its own vertical plane. It iscompletely fractured at one point A. Prove that the maximum bendingmoment experienced at the diametrically opposite point A' is

and find the corresponding direction of A A'. The ring has unit massper unit length, its radius is r and its angular velocity is a>. [M. T. 1918]

246 MISCELLANEOUS PROBLEMS [XVI

22. The ends of a uniform rod of length I can slide freely along twosmooth wires at right angles to one another, one of which is horizontal andthe other vertical and below the first. Shew that, if the frame rotatesround the vertical wire with uniform angular velocity <o, there is a positionof relative equilibrium for the rod, in which its lower end is at a depth3<7/2a>2 below the horizontal wire.

23. A rectangular lamina of diagonal 26 can move in a vertical planewith adjacent sides in contact with two smooth pegs at a distance a apartin the same horizontal line. Prove that the angular velocity 8 is given by

{a2 +$62 - 2a6 cos (a - 8)} 62+g {26 sin (a + 6) - a sin 28} = C,where a is the inclination of the diagonal to a side and 8 is the inclinationof that side to the horizontal.

A N S W E R S

3. ix. (ao>0 — Fo) = aa>0 tan a, where a is the radius and a the slope.4. 2 sin 6—ix (17 cos 6 -10). 6. The equivalent simple pendulum isof length equal to the difference of the diameters of the hoop and peg.The reactions when the line of centres makes an angle 8 with the verticalare mg (2 cos 6 — cos a), \ing sin 8; where m is the mass and a the extremevalue of 6. 12. The reaction along the rod remains Mg cos 6, thereaction at right angles to the rod changes from \Mg sin 8 to \Mg sin 6.18. The radius to the ring makes an angle cos"1 (a%r\g) with the vertical.

19. cos"1 (m/m1); {m'gl4n2m (I — a)}i, where m, m' are the masses.21. A A' makes an angle tan"1 \TT with the vertical.

Chapter XVII

SMALL OSCILLATIONS

17*1. In Chapter vn we considered the case of harmonicoscillations of finite amplitude of a particle moving in a straightline, the characteristic equation for which is

x + rfioo = 0,

and we remarked (7-23) that an equation of this form alwaysrepresents a periodic motion with a period independent ofthe amplitude. There is a large class of problems wherein theequation of motion can be reduced to this form, though generallywith the limitation that x remains small throughout the motion,this class includes all problems of a system having one degreeof freedom which is slightly disturbed from a position of stableequilibrium and proceeds to oscillate about that position. Theusual method of solving such a problem is to write down theequation of energy, i.e. express the fact that the sum of thekinetic and potential energies is constant, then differentiatethis equation. The method will be explained more fully in thenext article.

172. Application of the Principle of Energy. If a systempossesses only one degree of freedom its position can be definedby the values of a single variable x, which may denote a linearor an angular displacement. Let us suppose that the systemhas a position of stable equilibrium and that the origin fromwhich x is measured is so chosen that * vanishes in the positionof equilibrium. In any other position the kinetic energy Tmust be proportional to x2, say

T=\A& (1),

where A may contain x but does not vanish when x vanishes;and the potential energy V will be a function of x, which weassume can be expanded in ascending powers of x in the form

V= V0 + ax + bx* + cx3 + (2).

248 SMALL OSCILLATIONS [XVII

Now in a position of stable equilibrium the potential energyV is a minimum*, therefore dV/dx = O when x = 0, and thisrequires that a — 0. Further the oscillations that we are aboutto consider are assumed to be so small that powers of x and xabove the second power can be neglected. This means that in(1), if A is a function of x, since A is multiplied by x2, it willbe sufficient for our purpose to retain the part of A that isindependent of x, or put x = 0 in A; and in (2), since a = 0, wemay now write

V=V0 + bx2 (3),neglecting higher powers of x than the second.

The equation of energy T + V = const, now becomes$Ax2 + bx2 = const (4),

where A and b are constants.Differentiate with regard to t and divide by x and we get

Ax + 2bx=0 (5),which represents a periodic motion of period lit »J(A/2b).

17'3. Examples , (i) A solid uniform circular cylinder of mass m andof radius r rolls {under the action of gravity) inside a fixed hollow cylinderof radius R, the axes of the cylinders being parallel to each other and alsohorizontal. At any time t during the motion the plane containing the axesof the cylinders makes an angle 8 with the vertical. Shew that the potentialenergy of the moving cylinder is mg {R — r) (1 — cos ff), and that its kineticenergy is ^m(R-r)i8i. Hence, or otherwise, shew that the time, T, of asmall oscillation is

T=2n- ^{3 (R-r)j2g}. [M. T. 1914]Let C and O be the axes of the fixed and rolling cylinders, and let 0 be

the equilibrium position of the point G. ThenC0 = CO = R — r; and the potential energy isthe weight multiplied by its height above 0,i.e. {CO - CO cos 6), therefore

V=mg {R -r){l- cos 6),and, as far as the second power of 6, this is

Again, let a denote the angular velocity ofthe rolling cylinder, then the point of contactP has a velocity ra> relative to the axis O, but6 has a velocity {R — r)6 in the opposite direction, so that the conditionthat P has no velocity at the instant considered gives

r<o = (R-r)d (1).* See Note on the Energy Test of Stability, p. 257.

172—173] SMALL OSCILLATIONS 249

Since the velocity of the centre of gravity O is (R - r) 8, and the momentof inertia of the cylinder about its axis is £mr2, therefore the kineticenergy is

\m {(R - rf 82 + \r2a2} = f m (R - rf 82.

The equation of energy is therefore| m (R - rf 82 + § mg (R - r) 6* = const.,

when higher powers of 8 are neglected.Differentiating and dividing by m (R - r) 8, we get

%{R-r)6+g6 = Q (2).This represents oscillatory motion with period

Otherwise, the accelerations of O are (R — r)6 and (R-r)82 at rightangles to and along OC; therefore the rates of change of momentum arem(R-r)i), m(R-r)82, and the spin couple \mriu,. Therefore by takingmoments about the line of contact of the cylinders we get

m(R-r) rd + mr2d>= —mgrsm 8,but from (1) r& = (R — r) 8\ therefore the equation reduces to

%{R-r)6= -gaind,which, if we write 8 for sin 8, thus neglecting 83, is the same asequation (2).

(ii) The ends of a uniform rod of length 2a can slide on a smooth circularwire of radius b in a vertical plane. If the rod makes small oscillationsabout its equilibrium position, find the length of the equivalent simplependulum.

Let AB be the rod, m its mass and G its centre; and let 0 be thecentre of the wire. At time t let 00 makea small angle 6 with the vertical. ThenO(?=N/(62 —a2), and the potential energy is

mg00(1 -cos8), or \mgJ(^-a?)8\to the second power of 8.

Again, the velocity of O is 008 or

and the angular velocity of the rod is 8,therefore the kinetic energy is\m {(W- a2) e2+iai82} = m (3b2 - 2a2) 82.The equation of energy is therefore

\m (362 - 2a2) 82+\mg V(62 - a2) 82 = const.,when higher powers of 8 are neglected.

Differentiating and simplifying, we get

Therefore the length of the equivalent simple pendulum is(362-2a2)/3v/(62-a2).

250 SMALL OSCILLATIONS [XVII

174. Use of the Instantaneous Centre of Rotation. Wesaw in 166 that in problems on small oscillations we can takemoments about the instantaneous centre of rotation as thoughit were a fixed point, writing

mK2d> = L,

where mK2 is the moment of inertia and L the moment of theexternal forces about the axis of rotation through the in-stantaneous centre.

As an application consider the last example of 17'3.In this case 0 is the instantaneous centre and the moment of inertia of

the rod AB about 0 is by the theorem of parallel axes (13'2)

=m(|2 + 62-aA = \m(362-2a2).

Also the reactions at A and B pass through 0 and therefore have nomoment about 0, and the moment of the weight is

mg 0G sin 8, or mg s/(b2 — a2) sin 8.

Hence, neglecting 6s, we have

\m (362 - 2a2) 6=-mg v/(62 - a2) 6,

as in 17'3.

17'41. Example! Two uniform rods each of length 2es and of the samemass m are smoothly jointed and placed over two smooth pegs in the samehorizontal line, so that in equilibrium the rods make equal angles a with thevertical. Find the period of the small oscillations in which the joint movesvertically.

Let P, Q be the pegs at a distance 2c apart, and let O be the centre ofgravity of a rod AB, A being thejoint. Since when A moves verti-cally the rod AB slides over thepeg P, therefore the instantaneouscentre / for the rod AB is wherethe perpendicular to the rodthrough P meets the horizontalthrough A. The reaction a t P andthe reaction of the rod AC on ABare along PI and I A respectively. Therefore in equilibrium the verticalthrough G goes through /, and

or c = asin3a (1).

In a slightly displaced position let 8 be the inclination of the rods to

17'4-17-5] USE OF THE INSTANTANEOUS CENTRE 251

the vertical. The vertical through O no longer passes through 7, and by-taking moments about / for the rod A B we have

m(ia2 + GI2) 6'=-mg(aam6-ccosec26).

But GI2 = GP2 + PI2={a-c cosec 6)2 + c2 cot2 8 cosec2 8= a? — 2ac cosec 8 + c2 cosec4 6.

Therefore(§ a2-2ae cosec 0 + c2 cosec4 #)#' = - # (a sin 5 - c cosec2 8) ...(2).

Here 5 is not small, but differs from the constant a by a small quantity,so that we may write 8 — a + x where x is small and the first power only ofX is to be retained in the equation. Since now 8 = x, therefore, in thecoefficient of x on the left-hand side of the equation, it will be a sufficientlyclose approximation to take 8 = a. The right-hand side is a function of 8which by (1) vanishes when 8 is put equal to a. To obtain the value ofthis expression correct to the first power of x, the simplest method is toremember that, if x2 can be neglected,

but by (l)/(o) = 0, therefore f{8)=xf («)•In this case f(6)=—ff(a s i n 6 — ccosec2 8),

therefore / ' (8) = — g (a cos 8 +2c cosec3 6 cos 8),

and / ' (a) = —g (a cos a + 2c cosec3 a cos a)

= - 3ga cos a, from (1).

Hence the right-hand side of (2) becomes - Zga cos a. x a n d we have

(|a2 - 2ac cosec a + c2 cosec4 o) x = — tya cos a. x,or, by substituting from (1),

(f-sin2a) ax + 3gcoa a. x = 0,or (1+3 cos2a)a^+9<7cosa.x=0,making the period

2n </{« (1 + 3 cos2 a)/9g cos a}.

175. Oscillations of a Particle constrained to move on aRevolving Curve. Let a particle P of mass in be constrainedto move on a plane curvewhich revolves with uniformangular velocity m about anaxis in its plane; the field Mof force being the same inall planes through this axis.Take the axis of rotationfor axis of y and a perpen- ° x

dicular for axis of x. Let the coordinates of the particle be x, y

252 SMALL OSCILLATIONS [XVII

at time t. Let X, Y be the components parallel to the axes ofthe forces acting on the particle, including the reaction of thecurve. The effect of the rotation of the curve is to make the xcomponent of acceleration of the particle * — xa>2 instead of x;because, if PM be perpendicular to the axis of y, the radiusvector MP is turning about M in a plane perpendicular to Oywith angular velocity a>, so that the radial component of accele-ration (12'1) is ob — xco2. Hence the equations of motion are

m (x — xa>2) = X, my = Y,

or mx = X + muPx, my = Y.

Therefore the effect of the uniform rotation can be providedfor in the solution of the problem by adding to the forces actingon the particle a force a>2x per unit mass directed from the axisof rotation and otherwise regarding the given curve as at rest.

17'51. Example . A particle is constrained to move on a smooth circularwire which rotates uniformly about a diameter which is vertical. If in theposition of relative rest the radius drawn to the particle makes an angle awith the vertical, find the period of small oscillations about this position.

Let a be the radius of the circle, m the mass of the particle and let 6 bethe angle that the radius drawn tothe particle in any position makeswith the vertical. We may regardthe wire as at rest if we add to theforces acting on the particle a forcemafia sin 8 away from the verticaldiameter. The other forces are theweight mg and the normal reaction.Hence in relative rest

tan 6=ma>2a sin 8/mg, mg

or, since 6 = a in relative rest, therefore aa>2=g sec a. And in any generalposition by resolving along the tangent

ma0=m<o2a sin 6 cos 6 - mg sin 6.

For small oscillations about the position 6 = a, put 8 — a + x where x issmall, and we get

aX=i a>2a s m (2a + 2x) ~ 9 s™ (a + x)= | a?a (sin 2o + 2^ cos 2a) — g (sin a+x cos a).

Then, substituting g sec a for aw2, we haveaX + ffXs*n2 «/cos a = 0 ,

which gives a period2n- *J(a cos a/g sin2 a).

17"5-17"6] STABILITY OF STEADY MOTION 253

176. Stability of Steady Motion. In previous chapters andin the last two articles we have had examples of steady motion,some such were described in 16'5. Such a motion is said to bestable if when owing to any cause the motion is slightly dis-turbed it continues without a wide departure from the originalpath.

Suppose, for example, that a particle is describing a circle ofradius 1/M0 under the action of a force to the centre which wouldproduce an acceleration /(«•) at distance 1/u.

Let h be the moment of the velocity about the centre in thecircular orbit, then hu0 is the velocity, and the acceleration f(uQ)is (vel.)2/(radius), i.e. h2u0

3, or

Now suppose that without altering h, so that l/w0 remainsan apsidal distance, the particle is slightly displaced from thecircular orbit. The equation for the path with the law of force/ ( « ) is, by 12-21,

u - h%u2'

d2u uo3/(u)

ad2 vPf^Uo)

Putting u = uo + x, where * is small, and retaining only thefirst power of x, we get

^ + u , x_ "o3/(«o + a)

Therefore g + ^ ^

If the coefficient of x is positive and we denote it by n2,we have

d2x

the solution of which is * = A cos (nd + a).This shews that x is periodic in 8 and oscillates in its values

so that the motion is stable. Also, if ce vanishes for a given value

254 SMALL OSCILLATIONS [XVII

of 6, it will next vanish when 0 is increased by -jr/n, so that theapsidal angle or angle between consecutive apses is irjn, i.e.

The condition for stability is that the expression under thesquare root must be positive. If the force varies as the kth

power of the distance, i.e. f(u)cc—^, uof'(uo)lf(l<o)=:~k, so

that 3 + k must be positive and k must not be less than — 3.When the force varies as the distance, k = l, and the apsidal

angle = \rn. The disturbed circular orbit becomes in fact anellipse with the centre of force in the centre and there are fourapses at the ends of the axes (12-41).

When the force varies inversely as the square of the distance,k = — 2, and the apsidal angle = ir. In this case the disturbedorbit is an ellipse with the centre of force in a focus and thereare two apses at the ends of the major axis (12'5).

177. The usual method of procedure in solving a problem ondisturbed steady motion is first to write down the equations ofmotion in their general forms for the given system under thegiven field of force together with the kinematical conditions, ifany. Next deduce from these equations the condition for steadymotion by making all the second differential coefficients zero.This means that we are making all velocities constant andfinding the particular value of the variable, 6 = a say, whichdefines the position in steady motion. Next suppose the motionto be slightly disturbed from the steady state and substitute inthe original equations 6 = a + ^ where % is small, then by re-taining only the first power of % and its derivatives we obtainthe equation between % and %, which gives the period of theoscillations.

We shall illustrate this process by some examples.

17'8. The point of suspension of a pendulum is A, and A is caused tomove along a horizontal straight line OX. The centre of gravity of thependulum is O, and AO=l. The radius of gyration about any axis throughGperpendicular to AO is k. The pendulum can move in the vertical plane

17-6-17-8] ABOUT STEADY MOTION 255

containing OX. At time t, OA = x, and the angle between AO and thevertical is 6, supposed positive when GAX is obtuse. Prove that

(P+k2) 6 + Ig sin 6=1 cosd . x.

Ifx has the constant value f, shew that the pendulum can maintain a con-stant inclination a to the vertical, where tan a=f/g, and that the periodictime of small oscillations about this position is

. [M.T. 1924]

The point A has acceleration x, and O has accelerations IB and Iff1

relative to A at right angles to and along OA. Therefore, if m be the massof the pendulum, the rates of change of momentum are mx, mid, ml62, andthe spin couple mk'2i).

Hence by taking moments about A we get

mP6 — mxl cos 6 + mk2d= - mgl sin 6,

(1).

This is the general equation of motion, and if x has the constant value/ , it becomes

(P+i?)6 + lgsm6 = lfcosd (2).In this case we find the possible steady motion by putting $=0, which

gives tan 8 =f/g; and we denote this value of 6 by a. Substituting g tan afor / i n (2) gives

(72 + £2)#+Zg<sin(0-a)/cosa = O (3).To find the period of the small oscillations about this position we now

put 6 = a + x, where x ' s small, and neglect all powers of x above the first.Hence we get

giving a period 2JT cos a

It should be observed that the latter part of this question can be solvedby elementary considerations, for the relative effect on the pendulumof giving its point of suspension a constant horizontal acceleration isequivalent to keeping the point of suspension fixed and applying a forcem/to the pendulum at O in the opposite sense to the acceleration. The

256 SMALL OSCILLATIONS [XVII

equilibrium position of AO is then the direction of the resultant of / andg, and its time of oscillation is got from the ordinary formula for theperiod of a compound pendulum by writing \/(g2+f2) or gseca instead

17'9. A particle moving on the inside of a smooth fixed sphere describesa horizontal circle. Shew that if a> be the angular velocity the depth of thecircle below the centre of the sphere is g/a>2, and find the period of smalloscillations about this steady motion.

Let m be the mass of the particle and a the radius of the sphere. Letthe radius to the particle make an angle 6with the downward vertical, and let themeridian circle make an angle <f> with a fixedvertical plane.

In the general motion of the particle on thesphere its velocity along the vertical circle isa9 and along the horizontal circle a sin 8. <j>.Therefore the equation of energy is

\ma2 {82 + sin2 8$) + mga (1 - cos 6)= const....(l).

Again the particle is acted on by its weight mg, and the reaction of thesurface which passes through 0, neither of which has any moment aboutthe vertical diameter. Therefore the moment of momentum about thevertical diameter is constant, i.e.

ma2 sin2 8. <£=const.

Hence if there be a steady motion in which 8 — a and <j> = a>,

sin2 <?.$ = o> sin2 a (2).

Eliminating <f> between (1) and (2), we get

\a82+\aa>2 sin4 a cosec2 6-g cos 0 = const.

By differentiating this equation it follows thata6-aa>2 sin4 a cosec3 0cos0+#sin 0 = 0 (3).

The condition for steady motion with 8 = a is got by putting £ = 0, whence

aa>2 cos a~g,

or the depth of the horizontal circle below the centre of the sphere isa cosa=g/a>i (4).

Substitute g sec a for am2 in (3) and we getad—g sin4 a sec a cosec3 9 cos d+g sin 8 = 0 (5).

Then for small oscillations about the horizontal circle we put 8 = awhere ^ is small, and retain the first power of ^ only.

Now /W=/(«+x)=/(«) + x/'(«)>and to ge t / ' (a) we first differentiate f{6) and then put 0 = a.

17-8-179] ABOUT STEADY MOTION 257

In this case f(8)=—g sin4 a sec a cosec3 8 cos 8+g sin 8,

and / ' (6) = 3g sin4 a sec a cosec4 8 cos2 8 +g sin4 a sec a cosec2 6 +g cos 8.

Therefore/(a) = 0, and

/ ' (a) = 4g cos a +g sin2 a sec a=g(l+3 cos2 a) sec a.

Hence (5) becomesax+g (1 + 3 cos2 o) seco.^ = 0,

and the period of the small oscillations is2?r J{a cos a/g (1+3 cos2 a)}.

Note on the Energy Test of Stability. The proposition thatpositions of minimum and maximum potential energy are positions ofstable and unstable equilibrium respectively is generally found in bookson Statics; it is however an application of the principle of conservationof energy.

In the first place if the potential energy of a system is stationary invalue for small permissible displacements from an assigned position, thenno work is done by the external forces in any such displacement so that,by the principle of virtual work, such a position is one of equilibrium.

Now consider a body in a position of minimum potential energy. If thebody undergoes a small but finite displacement in any direction compatiblewith its geometrical connections and is then set free so that motion ensues,it begins to acquire kinetic energy and therefore to lose potential energy.It moves therefore so that its^potential energy decreases, i.e. back towardsthe position of minimum potential energy, and that position is thereforeone of stable equilibrium. In like manner it can be shewn that a positionof maximum potential energy is unstable.

EXAMPLES

1. A circular disc of mass M lies on a smooth horizontal table; if aparticle of mass m resting on the disc is attached to the centre by a springwhich exerts a force px when extended a length x, prove that the periodof oscillations when the spring is extended and then set free is

2w{Mml(M+m)ii}i. [S. 1914]

2. A weightless rod of length 3 feet, with equal heavy rings at the ends,one of which can slide on a smooth horizontal wire, is describing smalloscillations in the vertical plane containing the wire. Shew that the periodof oscillation is about T36 seconds. [S. 1916]

3. A running watch is placed on a smooth horizontal surface so that itmay be regarded as free to rotate about its own centre of gravity. It iscomposed essentially of a balance-wheel of moment of inertia i, and a body

258 SMALL OSCILLATIONS [XVII

of moment of inertia / connected by a hair spring. The inertia of the hairspring and other parts of the mechanism may be ignored. Determine theangular motion, and shew that in general, in addition to its oscillations,the body of the watch rotates steadily with a small uniform spin.

If / = 110 gm. (cm.)3, ?'=0'05 gm. (cm.)2, shew that the watch gainsnearly 20 seconds a day in this position, if its rate is correct when thebody is rigidly held. [M. T. 1920]

4. A uniform bar AB of mass M is suspended from a point 0 by twoequal elastic strings OA, OB. In the position of equilibrium the length ofeach string is I, and its inclination to the vertical is a. The increase in thetension of either string when its length is increased by x is Ex, where Eis constant. The bar vibrates vertically, remaining horizontal. Shew thatthe period of a small oscillation is 2?r/p, where

2iEeoa« + g r in« tgna j [M.T.1924]

5. A light rod AB, carrying a heavy particle at B, is connected by asmooth hinge at A to a second light rod AO. OA, AB are initially in thesame straight line, rotating with uniform angular velocity about a fixedcentre 0.

AB is now displaced through a small angle, relatively to OA, in theplane of rotation. Shew that, if the angular velocity of OA be maintainedconstant, AB will oscillate relatively to OA with a period T s/l/a, where Tis the time of revolution of OA about 0, and a and I are the lengths of OAand AB respectively. [M. T. 1927]

6. One end of a uniform rod rests on a smooth horizontal plane and theother end is attached by a string of length I to a point whose height abovethe table is greater than I. Shew that, if small oscillations take place in avertical plane, the length of the equivalent simple pendulum is 21.

7. A uniform plank, length 2a and thickness 2h, rests in equilibriumon the top of a fixed rough cylinder of radius r whose axis is horizontal.Prove that, if r is greater than h, the equilibrium is stable; and that, if theplank is slightly disturbed, the period of an oscillation is that of a simplependulum of length

(a2+4A2)/3 (?•-£). [Coll. Exam. 1913]

8. Three equal uniform rods AB, BC, CD of length a are smoothlyjointed at B and C and suspended from points A, Din the same horizontalline at a distance a apart. Shew that, when the rods move in a verticalplane, the length of the equivalent simple pendulum is 5a/6.

9. Two masses rn and m', which are free to move on a given circularwire of radius a, are connected by a light rod subtending an angle 2/3 atthe centre of the wire; find the position of stable equilibrium when the

EXAMPLES 259

plane of the wire is vertical; and shew that, if the masses are slightlydisturbed, the length of the equivalent simple pendulum is

a (m+m')/</(m? + m"2 + 2mm' cos 2/3).

10. A solid uniform sphere of radius 6 makes small oscillations at thebottom of a fixed hollow sphere of internal radius a, the surfaces beingsufficiently rough to prevent sliding and the motion being in a verticalplane. Shew that the length of the equivalent simple pendulum is -| (a — b).

11. A circular cylinder of radius a surrounds another cylinder of smallerradius b. The former rolls on the latter, which is fixed. Shew that theplane through the axes moves like a simple pendulum of length

k being the radius of gyration of the moving cylinder about its axis.[M. T. 1910]

12. A cylindrical shell of mass M and radius b is free to turn about itsaxis, which is horizontal, and another rough cylindrical shell of mass mand radius a is placed inside i t ; prove that the plane through the axeswill oscillate like a simple pendulum of length

(b - a) (2M+ m)j(M+ TO).

13. A body of uniform density, having the form of a sector of a circleof radius 2a, performs rolling oscillations of small amplitude with itscircular edge in contact with a rough horizontal plane. Its centre ofgravity is half-way between this edge and its centre. Prove that thelength of the simple equivalent pendulum is 2a. [M. T. 1928]

14. A particle can move in a smooth circular tube which revolves abouta fixed vertical tangent with uniform velocity a>. Find the position ofrelative equilibrium of the particle, and shew that the time of a smalloscillation about that position is

2n I sin a 1»~^~' (.1+sin3 of '

where a is the angle of inclination to the vertical of the radius to theparticle when in relative equilibrium. [S. 1917]

15. A conical pendulum executes small oscillations about a state ofsteady motion in which the string of length I is inclined to the horizon atan angle a : shew that the period of its oscillation is

. Exam. 1907]

A N S W E E S

3. The period of the balance-wheel is 2 sj{ljl+i) sees.

14. a»2 (1 +sin a) =g tan a, where a is the radius.