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MECHANICS
THE MACMILLAN COMPANY
NEW YORK • BOSTON • CHICAGO • DALLAS
ATLANTA • SAN FRANCISCO
MACMILLAN & CO., LIMITED
LONDON • BOMBAY • CALCUTTA
MELBOURNE
THE MACMILLAN COMPANY
OF CANADA, LIMITED
TORONTO
MECHANICS
BY
WILLIAM FOGG OSGOOD, PH.D., LL.D.
PERKINS^ PROFESSOR OF MATHEMATICS, EMERITUS
IN HARVARD UNIVERSITY
NEW YORK
THE MACMILLAN COMPANY
1949
COPYRIGHT, 1937,
BY THE MACMILLAN COMPANY.
All rights reserved — no part of this book may be
reproduced in any form without permission in writing
from the publisher, except by a reviewer who wishes
to quote brief passages in connection with a review
written for inclusion in magazine or newspaper.
Published June, 1937.
Reprints! Nov. 1946.
Reprinted, May, 1948.
Reprinted November, 1949
S»T UP AND ELECTROTYPED BY J. S. GUSHING CO.
• PRINTED IN THE UNITED STATES OF AMERICA •
PREFACE
Mechanics is a natural science, and like any natural science
requires for its comprehension the observation and knowledge of a
vast fund of individual cases. Arid so the solution of problems is
of prime importance throughout all the study of this subject.
But Mechanics is not an empirical subject in the sense in which
physics and chemistry, when dealing with the border region of tUe
human knowledge of the day are empirical. The latter take
cognizance of a great number of isolated facts, which it is not as
yet possible to arrange under a few laws, or postulates. The laws
of Mechanics, like the laws of Geometry, so far as first approxima-
tions go — the laws that explain the motion of the golf ball or the
gyroscope or the skidding automobile, and which make possible
the calculation of lunar tables and the prediction of eclipses —
these laws are known, and will bo as new arid important two
thousand years hence, as in the recent past of science when first
they emerged into the light of day.
Here, then, is the problem of training the student in Mechanics
— to provide him with a vast fund of case material and to develop
in him the habits of thought which refer a new problem back to the
few fundamental laws of the subject. The physicist is keenly
alive to the first requirement and tries to meet it both by simple
laboratory experiments and by problems in the part of a general
course on physics which is especially devoted to "Mechanics."
The interest of the mathematician too often begins with virtual
velocities and d'Alembert's Principle, and the variational principles,
of which Hamilton's Principle is the most important. Both arc
right, in the sense that they are dping nothing that is wrong ; but
each takes such a fragmentary view of the whole subject, that his
work is ineffectual.
The world in which the boy and girl have lived is the true
laboratory of elementary mechanics. The tennis ball, the golf
ball, the shell on the river ; the automobile — good old Model T,
in its day, and the home-made autos and motor boats which
vi PREFACE
youngsters construct and will continue to construct — the amateur
printing press ; the games in which the mechanics of the body is a
part ; all these things go to provide the student with rich laboratory
experience before he begins a systematic study of mechanics. It
is this experience on which the teacher of Mechanics can draw, and
draw, and draw again.
The Cambridge Tripos of fifty years and more ago has been
discredited in recent years, and the criticism was not without
foundation. It was a method which turned out problem solvers —
so said its opponents. But it turned out a Clerk Maxwell and it
vitally influenced the training of the whole group of English
physicists, whose work became so illustrious. In his interesting
autobiography, From Emigrant to Inventor, Pupin acknowledges in
no uncertain terms the debt he owes to just this training, and to
Arthur Gordon Webster, through whom he first came to know
this method — a method which Benjamin Osgood Peirce also
prized highly in his work as a physicist. And so we make no
apologies for availing ourselves to the fullest extent of that which
the old Tripos Papers contributed to training in Mechanics. But
we do not stop there. After all, it is the laws of Mechanics, their
comprehension, their passing over into the flesh and blood of our
scientific thought, and the mathematical technique and theory,
that is our ultimate goal. To attain to this goal the mathematical
theory, absurdly simple as it is at the start, must be systematically
inculcated into the student from the beginning. In this respect
the physicists fail us. Because the mathematics is simple, they
do not think it important to insist on it. Any way to get an
answer is good enough for them. But a day of reckoning comes.
The physicist of to-day is in desperate need of mathematics, and
at best all he can do is to grope, trying one mathematical expedient
after another and holding to no one of these long enough to test
it mathematically. Nor is he to be blamed. It is the old (and
most useful) method of trial and error he is employing, and must
continue to employ for the present.
Is the writer on Mechanics, per contra, to accept the challenge
of preparing the physicist to solve these problems? That is too
large a task. Rather, it is the wisdom of Pasteur who said :
"Fortune favors the prepared mind" that may well be a guide for
us now and in the future. What can be done, and what we havo
attempted in the present work, is to unite a broad and deep knowv
PREFACE vii
edge of the most elementary physical phenomena in the field of
Mechanics with the best mathematical methods of the present
day, treating with completeness, clarity, and rigor the beginnings
of the subject ; in scope not restricted, in detail not involved, in
spirit scientific.
The book is adapted to the needs of a first course in Mechanics,
given for sophomores, and culminating in a thorough study of the
dynamics of a rigid body in two dimensions. This course may be
followed by a half-course or a full course which begins with the
kinematics and kinetics of a rigid body in three dimensions and
proceeds to Lagrange's Equations and the variational principles.
So important are Hamilton's Equations and their solution by
means of Jacobi's Equation, that this subject has also been in-
cluded. It appears that there is a special need for treating this
theory, for although it is exceedingly simple, the current text-
books are unsatisfactory. They assume an undefined knowledge
of the theory of partial differential equations of the first order, but
they do not show how the theory is applied. As a matter of fact,
no theory of these equations at all is required for understanding
the solution just mentioned. What is needed is the fact that
Hamilton's Equations are invariant of a contact transformation.
A simple proof is given in Chapter XIV, in which the method
most important for the physicist, namely, the method of separating
the variables, is set forth with no involved preliminaries. But
even this proof may be omitted or postponed, and the student may
strike in at once with Chapter XV.
The concept of the vector is essential throughout Mechanics,
but intricate vector analysis is wholly unnecessary. A certain
minute amount of the latter is however helpful, and has been set
forth in Appendix A.
Appendix D contains a definitive formulation of a class of
problems which is most important in physics, and shows how
d'Alembert's Principle and Lagrange's Equations apply. It ties
together the various detailed studies of the text and gives the
reader a comprehensive view of the subject as a whole.
The book is designed as a careful and thorough introduction to
Mechanics, but not of course, in this brief compass, as a treatise.
With the principles of Mechanics once firmly established and
clearly illustrated by numerous examples the student is well
equipped for further study in the current text-books, of which may
viii PREFACE
be mentioned : Routh : An Elementary Treatise on Rigid Dynamics
and also Advanced Dynamics, by the same author; particularly
valuable for its many problems. Webster, Dynamics — good
material, and excellent for the student who is well trained in the
rudiments, but hard reading for the beginner through poor pre-
sentation and lacunae in the theory ; Appell, Mecanique rationelley
vols. i and ii — a charming book, which the student may open
at any chapter for supplementary reading and examples. Jeans,
Mechanics, may also be mentioned for supplementary exercises;
as a text it is unnecessarily hard mathematically for the Sopho-
more, and it does not go far enough physically for the upper-
classmari. It is unnecessary to emphasize the importance of
further study by the problem method of more advanced and
difficult exercises, such as are found in these books. But to go
further in incorporating these problems into the present work
would increase its size unduly.
It is not merely a formal tribute, but one of deep appreciation,
which I wish to pay to The Macmillan Company and to The
Norwood Press for their hearty cooperation in all the many dif-
ficult details of the typography. Good composition is a distinct
aid in setting forth the thought which the formulas are designed
to express. Its beauty is its own reward.
To his teacher, Benjamin Osgood Peirce, who first blazed the
trail in his course, Mathematics 4, given at Harvard in the middle
of the eighties the Author wishes to acknowledge his profound
gratitude. Out of these beginnings the book has grown, developed
through the Author's courses at Harvard, extending over more
than forty years, and out of courses given later at The National
University of Peking. May it prove a help to the beginner in
his first approach to the subject of Mechanics.
WILLIAM FOGG OSGOOD
May 1937
CONTENTS
CHAPTER I
STATICS OF A PARTICLE
PAGE
1. Parallelogram of Forces 1
2. Analytic Treatment by Trigonometry 3
3. Equilibrium. The Triangle of Forces. Addition of Vectors . 4
4. The Polygon of Forces 7
5. Friction 9
6. Solution of a Trigonometric Equation. Problem 12
Exercises on Chapter I 15
CHAPTER II
STATICS OF A RIGID BODY
1. Parallel Forces in a Plane 21
2. Analytic Formulation ; n Forces 23
3. Centre of Gravity 26
4. Moment of a Force 28
5. Couples in a Plane 29
6. Resultant of Forces in a Plane. Equilibrium 32
7. Couples in Space 34
8. Resultant of Forces in Space. Equilibrium 36
9. Moment of a Vector. Couples 37
10. Vector Representation of Resultant Force and Couple. Resul-
tant Axis. Wrench 38
11. Moment of a Vector about a Line 40
12. Equilibrium 41
13. Centre of Gravity of n Particles 42
14. Three Forces 43
Exercises on Chapter II 46
CHAPTER III
MOTION OF A PARTICLE
1. Rectilinear Motion 49
2. Newton's Laws of Motion 50
3. Absolute Units of Force 55
4. Elastic Strings 58
ix
x CONTENTS
PA OB
5. A Problem of Motion 60
6. Continuation; the Time 63
7. Simple Harmonic Motion 64
8. Motion under the Attraction of Gravitation 69
9. Work Done by a Variable Force 72
10. Kinetic Energy and Work 75
11. Change of Units in Physics 76
12. The Check of Dimensions 79
13. Motion in a Resisting Medium 81
14. Graph of the Resistance ; 84
15. Motion in a Plane and in Space 86
16. Vector Acceleration 90
17. Newton's Second Law 92
18. Motion of a Projectile 93
19. Constrained Motion 95
20. Simple Pendulum Motion 97
21. Motion on a Smooth Curve 99
22. Centrifugal Force 101
23. The Centrifugal Oil Cup 105
24. The Centrifugal Field of Force 106
25. Central Force 108
26. The Two Body Problem 114
27. The Inverse Problem — to Determine the Force 114
28. Kepler's Laws 115
29. On the Notion of Mass 118
CHAPTER IV
DYNAMICS OF A RIGID BODY
1. Motion of the Centre of Gravity 120
2. Applications 123
3. The Equation of Moments 126
4. Rotation about a Fixed Axis under Gravity 127
5. The Compound Pendulum 130
6. Continuation. Discussion of the Point of Support .... 132
7. Kater's Pendulum 133
8. Atwood's Machine 134
9. The General Case of Rotation about a Point 136
10. Moments of Inertia 137
11. The Torsion Pendulum 139
12. Rotation of a Plane Lamina, No Point Fixed 139
13. Examples 141
CONTENTS xi
PAGE
14. Billiard Ball, Struck Full 143
15. Continuation. The Subsequent Motion ........ 145
16. Further Examples 146
Exercises on Chapter IV 151
CHAPTER V
KINEMATICS IN TWO DIMENSIONS
1. The Rolling Wheel ' 154
2. The Instantaneous Centre 155
3. Rotation about the Instantaneous Centre 157
4. The Centrodes 159
5. Continuation. Proof of the Fundamental Theorem .... 162
6. The Dancing Tea Cup 165
7. The Kinetic Energy of a Rigid System 166
8. Motion of Space with One Point Fixed 168
9. Vector Angular Velocity 170
10. Moving Axes. Proof of the Theorem of § 8 172
11. Space Centrode and Body Centrode 174
12. Motion of Space. General Case 175
13. The Ruled Surfaces 176
14. Relative Velocities 177
15. Proof of the Theorem of § 12 179
16. Lissajou's Curves 182
17. Continuation. The General Case. The Commensurable Case.
Periodicity 186
Professor Sabine's Tracings of Lissajou's Curves
between pages 190-191
CHAPTER VI
ROTATION
1. Moments of Inertia 191
2. Principal Axes of a Central Quadric 194
3. Continuation. Determination of the Axes 196
4. Moment of Momentum. Moment of a Localized Vector . . 197
5. The Fundamental Theorem of Moments 199
6. Vector Form for the Motion of the Centre of Mass .... 201
7. The Invariable Line and Plane 201
8. Transformation of 0 202
9. Moments about the Centre of Mass 204
10. Moments about an Arbitrary Point 205
xii CONTENTS
PAGE
11. Moments about the Instantaneous Centre 207
12. Evaluation of a for a Rigid System ; One Point Fixed ... 208
13. Euler's Dynamical Equations 210
14. Motion about a Fixed Point 212
15. Euler's Geometrical Equations 214
16. Continuation. The Direction Cosines of the Moving Axes . . 216
17. The Gyroscope 217
18. The Top 220
19. Continuation. Discussion of the Motion 222
20. Intrinsic Treatment of the Gyroscope 225
21. The Relations Connecting v, F, and K 229
22. Discussion of the Intrinsic Equations 231
23. Billiard Ball 237
24. Cartwheels 241
25. R&ume* 245
CHAPTER VII
WORK AND ENERGY
1. Work 248
2. Continuation : Curved Paths 250
3. Field of Force. Force Function. Potential 253
4. Conservation of Energy 256
5. Vanishing of the Internal Work for a Rigid System .... 258
6. Kinetic Energy of a Rigid Body 260
7. Final Definition of Work 261
8. Work Done by a Moving Stairway 264
9. Other Cases in Which the Internal Work Vanishes .... 266
10. Work and Energy for a Rigid Body 266
CHAPTER VIII
IMPACT
1. Impact of Particles 270
2. Continuation. Oblique Impact 274
3. Rigid Bodies 277
4. Proof of the Theorem 279
5. Tennis Ball, Returned with a Lawford 282
CHAPTER IX
RELATIVE MOTION AND MOVING AXES
1. Relative Velocity 285
2. Linear Velocity in Terms of Angular Velocity 285
CONTENTS xiii
PAGE
3. Acceleration 287
4. The Dynamical Equations 290
5. The Centrifugal Field 291
6. Foucault Pendulum 292
CHAPTER X
LAGRANGE'S EQUATIONS AND VIRTUAL VELOCITIES
1. The Problem 297
2. Lagrange's Equations in the Simplest Case 299
3. Continuation. Particle on a Fixed or Moving Surface . . . 303
4. The Spherical Pendulum 306
5. Geodesies 308
6. Lemma 310
7. Lagrange's Equations in the General Case 312
8. Discussion of the Equations. Holonomic and Non-Holonomic
Systems 313
9. Continuation. The Forces 315
10. Conclusion. Lagrange's Multipliers 316
11. Virtual Velocities and Virtual Work 318
12. Computation of Qr 320
13. Virtual Velocities, an Aid in the Choice of the TTT 321
14. On the Number m of the QT 322
15. Forces of Constraint 325
16. Euler's Equations, Deduced from Lagrange's Equations . . . 325
17. Solution of Lagrange's Equations 326
18. Equilibrium 330
19. Small Oscillations 333
Exercises on Chapter X 336
CHAPTER XI
HAMILTON'S CANONICAL EQUATIONS
1. The Problem 338
2. A General Theorem 339
3. Proof of Hamilton's Equations 342
CHAPTER XII
D'ALEMBERT'S PRINCIPLE
1. The Problem 345
2. Lagrange's Equations for a System of Particles, Deduced from
d'Alembert's Principle 348
3. The Six Equations for a System of Particles, Deduced from
d'Alembert's Principle 349
riv CONTENTS
PAGE
4. Lagrange's Equations in the General Case, and d'Alembert's
Principle 350
5. Application : Euler's Dynamical Equations 352
6. Examples 353
CHAPTER XIII
HAMILTON'S PRINCIPLE AND THE PRINCIPLE OF
LEAST ACTION
1. Definition of 5 356
2. The Integral of Rational Mechanics 360
3. Application to the Integral of Kinetic Energy 362
4. Virtual Work 364
5. The Fundamental Equation 364
6. The Variational Principle 370
7. Hamilton's Principle 370
8. Lagrange's Principle of Least Action 372
9. Jacobi's Principle of Least Action 377
10. Critique of the Methods. Retrospect and Prospect .... 379
11. Applications 379
12. Hamilton's Integral a Minimum in a Restricted Region . . . 381
13. Jacobi's Integral a Minimum in a Restricted Region .... 386
CHAPTER XIV
CONTACT TRANSFORMATIONS
1. Purpose of the Chapter 389
2. Integral Invariants 392
3. Consequences of the Theorem 395
4. Transformation of Hamilton's Equations by Contact Trans-
formations 400
5. Particular Contact Transformations 403
6. Theft-Relations 407
CHAPTER XV
SOLUTION OF HAMILTON'S EQUATIONS
1. The Problem and Its Treatment 410
2. Reduction to the Equilibrium Problem 413
3. Example. Simple Harmonic Motion 415
4. H, Independent of t. Reduction to the Form, H' = Pi . . . 420
5. Examples. Projectile in vacuo 424
CONTENTS xv
PAGE
6. Comparison of the Two Methods 429
7. Cyclic Coordinates 430
8. Continuation. The General Case 433
9. Examples. The Two-Body Problem 434
10. Continuation. The Top 438
11. Perturbations. Variation of Constants 440
12. Continuation. A Second Method 444
APPENDIX
A. Vector Analysis 447
B. The Differential Equation : (du/dt)z = f(u) 456
C. Characteristics of Jacobi's Equation 466
D. The General Problem of Rational Mechanics 476
INDEX . . . . 491
MECHANICS
CHAPTER I
STATICS OF A PARTICLE
1. Parallelogram of Forces. By a force is meant a push or a
pull. A stretched elastic band exerts a force. A spiral spring,
like those used in the upholstered seats of automobiles, when
compressed by a load, exerts a force. The earth exerts a force
of attraction on a falling rain drop.
The effect of a force acting at a given point, 0, depends not
merely on the magnitude, or intensity, of the force, but also on
the direction in which it acts. Lay off a right line from
O in the direction of the force, and mpke the length of ^ — ^
the line proportional to the intensity of the force ; for pJG 1
example, if F is 10 Ibs., the length may be taken as
10 in., or 10 cm., or more generally, ten times the length which
represents the unit force. Then this directed right line, or vector,
gives a complete geometric picture of the force. Thus if a barrel
of flour is suspended by a rope (and is at rest), the attraction of
gravity — the pull of the earth — will be represented by a vector
pointing downward and of length W, the weight of the barrel.
On the other hand, the force which the rope exerts on the barrel
will be represented by an equal and opposite vector, pointing
upward. For, action and reaction are equal and opposite.
When two forces act at a point, they are
equivalent to a single force, which is found as
follows. Lay off from the point the two vec-
tors, P and Q, which represent the given forces,
and construct the parallelogram, of which the
Fia 2 right line segments determined by P and Q are
two adjacent sides. The diagonal of the paral-
lelogram drawn from 0 determines a vector, R, which represents
1
2 MECHANICS
the combined effect of P and Q. This force, R, is called the
resultant of P and Q, and the figure just described is known as the
parallelogram offerees.
Example 1. Two forces of 20 pounds each make an angle of
60° with each other. To find their resultant.
Here, it is obvious from the geometry of the
figure that the parallelogram is a rhombus, and
that _the length of the diagonal in question is
20 N/3 = 34.64. Hence the resultant is a force of
34.64 pounds, its line of action bisecting the angle between the
given forces.
Example 2. Two forces of 7 pounds and 9 pounds act at a
point and make an angle of 70° with each other. To find their
resultant.
Graphical Solution. Draw the forces to scale, constructing the
angle by means of a protractor. Then complete the parallelo-
gram and measure the diagonal. Find its direction with the
protractor.
Example 3. A picture weighing 15 Ibs. hangs from a nail in
the wall by a wire, the two segments of which make angles of 30°
with the horizon. Find the tension in the wire.
Here, the resultant, 15, of the two unknown tensions, T and T9
is given, and the angles are known. It is evident from the figure
that T also has the value 15. So the answer is : 15 Ibs.
Decomposition of Forces. Conversely, a given force can be de-
composed along any two directions whatever. All that is needed
is, to construct the parallelogram, of which the given force is
the diagonal and whose sides lie along the given lines.
FBlTKf,
FIG. 4 FIG. 5
If, in particular, the lines are perpendicular to each other,
the components will evidently be :
F cos <p, F sin <p.
STATICS OF A PARTICLE 3
EXERCISES
1. Two forces of 5 Ibs. and 12 Ibs. make a right angle with
each other. Show that the resultant force is 13 Ibs. and that it
makes an angle of 22° 37' with the larger force.
2. Forces of 5 Ibs. and 7 Ibs. make an angle of 100° with each
other. Determine the resultant force graphically.
3. If the forces in Question 1 make an angle of 60° with each
other, find the resultant.
Give first a graphical solution. Then obtain an analytical
solution, using however no trigonometry beyond a table of natural
sines, cosines, and tangents.
4. If two forces of 12 Ibs. and 16 Ibs. have a resultant of 20 Ibs.,
what angle must they make with each other and with the resultant?
5. A force of 100 Ibs. acts north. Resolve it into an easterly
and a north-westerly component.
6. A force of 50 Ibs. acts east north-east. Resolve it into an
easterly and a northerly component. Ans. 46.20 Ibs. ; 19.14 Ibs.
7. A force «of 12 Ibs. acts in a given direction. Resolve it into
two forces that make angles of 30° and 40° with its line of action.
Only a graphical solution is required.
2. Analytic Treatment by Trigonometry. The problem of
finding the resultant calls for the determination of one side of a
triangle when the other two sides and the included angle arc
known; and also of finding the remaining angles. The first
problem is solved by the Law of Cosines in Trigonometry :
(1) c2 = a2 + 62 - 2ab cos C.
Here, a = P, b = Q, c = R,
C = 180° - w
and hence c
(2) 722 = P2 + Q2 + 2PQ cos co. FIG. 0
Example. Forces of 5 Ibs. and 8 Ibs. make an angle of 120°
with each other. Find their resultant. Here,
R2 = 25 + 64 - 2 X 5 X 8 X £ = 49 ;
R = 7 Ibs.
To complete the solution and find the remaining angles we
can use the Law of Sines :
4 MECHANICS
~ I,
(3)
1
(4)
sin A sin B sin C
Thus
sm ^> sm
There is no difficulty here about the sign when the adjacent
angle is used, since sin (180 — to) = sin co.
In the numerical example above, Equation (4) becomes :
8 7
sin <p |\/3
Thus
sin <p = 4-V3, cos <p = | , <t> = 81° 47'.
The third angle is found from the fact that the sum of the
angles of a triangle is two right angles :
A + B + C = 180°.
To sum up, then : Compute the resultant by the Law of Cosines
and complete the solution by the Law of Sines.
EXERCISES
Give both a graphical and an analytical solution each time.
1. Forces of 2 Ibs. and 3 Ibs. act at right angles to each other.
Find their resultant in magnitude and direction.*
2. Forces of 4 Ibs. and 5 Ibs. make an angle of 70° with each
other. Find their resultant.
3. Equilibrium. The Triangle of Forces. Addition of Vec-
tors. In order that three forces be in equilibrium, it is clearly
necessary and sufficient that any one of them be equal and opposite
to the resultant of the other two. The condition can be expressed
conveniently by aid of the idea of the addition of vectors.
First of all, two vectors are defined as equal if they have the
same magnitude, direction, and sense, no matter where in the
plane (or in space) they may lie.
* Observe that in this case it is easier to determine the angle from its tangent.
Square roots should be computed from a Table of Square Roots. Huntington's
Four-Place Tables are convenient, and are adequate for the ordinary cases that
arise in practice. But cases not infrequently arise in which more elaborate tables
are needed, and Barlow's will be found useful.
STATICS OF A PARTICLE 5
Vector Addition. Let A and B be any two vectors. Construct
A with any point, 0, as its initial point. Then, with the terminal
point of A as its initial point, construct B. The vector, C, whose
initial point is the initial point of A and whose terminal point
is the terminal point of B, is defined as the vector
sum, or, simply, the sum of A and B :
C = A + B.
It is obvious that
B + A = A + B.
Any number of vectors can be added by applying the definition
successively. It is easily seen that
(A + B) + C = A + (B + C).
Consequently the sum
A! + A2 + - - - + An
is independent of the order in which the terms are added.
For accuracy and completeness it is necessary to introduce the
nil vector. Suppose, for example, that A and B are equal and
opposite. Then their sum is not a vector in any sense as yet
considered, for the terminal point coincides with the initial point.
When this situation occurs, we say that we have a nil vector, and
denote it by 0 :
A + B = 0.
We write, furthermore,*
B =- A.
Equilibrium. The condition, necessary and sufficient, that
three forces be in equilibrium is that their vector sum be 0. Geo-
metrically this is equivalent to saying that the vectors which
represent the forces can be drawn so that the fig-
ure will close and form a triangle. From the Law
of Sines we have :
P Q _ E
sin p sin q sin e
FIG. 8 " p + q + e = 180°.
* It is not necessary for the present to go further into vector analysis than the
above definitions imply. Later, the two forms of product will be needed, and
the student may be interested even at this stage in reading Chapter XIII of the
author's Advanced Calculus, or Appendix A.
6
MECHANICS
FIG. 9
Since sin (180° — A) = sin 4, we can state the result in the
following form. Let three forces, P, Q, and E, acting on a par-
ticle, be in equilibrium. Denote the angles between
the forces, as indicated, by p, q, e. Then Equa-
tion (1) represents a necessary condition for equilib-
rium. Conversely, this condition is sufficient.
We thus obtain a convenient solution in all cases
except the one in which the magnitudes of the forces,
but no angles, are given. Here, the Law of Cosines *
gives one angle, and then a second angle can be
computed by the Law of Sines.
Example 1. Forces of 4, 5, and 6 are in equilibrium. Find
the angles between them.
First, solve the problem graphically, measuring the angles.
Next, apply the Law of Cosines :
42 = 52 + 62 _ 2 X 5 X 6 X cos p,
cos <p = f , <p = 41° 23'.
A second angle is now computed by the Law of Sines :
5 4
sin i
5V7
16 '
sin <p
55° 46'.
The third angle is 82° 51'.
Example 2. A 40 Ib. weight rests on a smooth horizontal
cylinder and is kept from slipping by a cord that passes over the
cylinder and carries a 10 Ib. weight at its other
end. Find the position of equilibrium.
The cord is assumed weightless, and since it
passes over a smooth surface, the tension in
it is the same at all points. The surface of
the cylinder is smooth, hence its reaction is
normal to its surface. Let 6 be the unknown
angle that the radius drawn to the weight FIG. 10
* If we had a large number of numerical problems to solve, it would pay to use
the more elaborate theorems of Trigonometry (e.g. Law of Tangents). But for
ordinary household purposes the more familiar law is enough.
STATICS OF A PARTICLE
makes with the vertical. Then from inspection of the figure
we see that 10 40
or
sin 6 sin 90°'
sin 6 = •£, 0 = 14° 29'.
EXERCISES
Find
1. Forces of 7, 8, and 9 pounds keep a particle at rest,
the angles they make with one another.
2. Forces of 51.42, 63.81, and 71.93 grs. keep a particle at rest.
What angle do the first two forces make with each other? Find
the other angles.
3. A weightless string passes over two smooth pegs at the same
level and carries weights of P and P at its ends. In the middle,
there is knotted a weight W. What
angle do the segments of the string p
make with the vertical, when the
system is at rest ? w
Ans. sin 6 — -^
4. A boat is prevented from drifting down stream by two ropes
tied to the bow of the boat, and to stakes at opposite points on the
banks. One rope is 125 ft. long; the other, 150 ft.; and the
stream is 200 ft. broad. If the tension in the shorter rope is
20 Ibs., what is the tension in the other rope?
6. Two smooth inclined planes, back to
back, meet along a horizontal straight
line, and make angles of 30° arid 45° with
the horizon. A weight of 10 Ibs. placed
on the first plane is held by a cord that
passes over the top of the planes and
carries a weight W, resting on the other plane and attached to
the end of the cord. Find what value W must have.
4. The Polygon of Forces. From the
case of three forces the generalization to
the case of n forces acting at a point pre-
sents no difficulty. Add the forces geometri-
cally, i.e. by the vector law. The vector
sum represents the resultant of all n forces.
Thus, in the figure, the resultant is given by FIG. 13
FIG. 12
8 MECHANICS
the vector whose initial point is the point 0 and whoso terminal
point is P.
The condition for equilibrium is clearly that the resultant be a
nil vector, or that the broken line close
and form a polygon — but not necessarily
a polygon in the sense of elementary
geometry, since its sides may intersect, as
in Figure 14.
This condition will obviously be ful-
filled if and only if the sum of the projections of the forces along
each of two lines that intersect is zero.*
Analytically, the resultant can be represented as follows.
Let a Cartesian system of coordinates be assumed, and let the
components of the force Fk along the axes be Xk and Yk. If,
now, the components of the resultant are denoted by X and Y,
we have : v _ v , v , , v
A — A! -\- A2 ~r * • • ~r An,
Y = Yt + F2 + • • • + Yn.
Such sums are written as
(1) X*, or Xk> or Z*>
t=l I
depending on how elaborate the notation should be to insure
clearness.
The forces will be in equilibrium if, and only if, the resultant
force is nil, and this will be the case if
(2) 2) Xk = 0, 2 Yk = 0.
t-i t=i
EXERCISE
A 4 Ib. weight is acted on by three forces, all of which lie in the
same vertical plane : a force of 10 Ibs. making an angle of 30° with
the vertical, and forces of 8 Ibs. and 12 Ibs. on the other side of
the vertical and making angles of 20° with the upward vertical
and 15° with the downward vertical respectively. Find the force
that will keep the system at rest.
Space of Three Dimensions. If more than two forces act at a
point, they need not lie in a plane. But they can be added two
* Cf. Osgood and Graustein, Analytic Geometry, pp. 1-6.
STATICS OF A PARTICLE 9
at a time by the parallelogram law, the first two thus being
replaced by a single force — their resultant — and this force in
turn compounded with the third force; etc. The broken line
that represents the addition of the vectors no longer lies in a plane,
but becomes a skew broken line in space, and the polygon of forces
becomes a skew polygon. The components of the resultant
force along the three axes are :
n n^ n
(3) X = 2^ X^ Y = 2^ Yk, Z = 2^ Zk.
The condition for equilibrium is :
In all of these formulas, Xk, Yk, Zk are algebraic quantities,
being positive when the component has the sense of the positive
axis of coordinates, and negative when tho sense is the opposite.
In solving problems in equilibrium it is frequently simpler to single
out the components that have one sense along the line in question
and equate their sum, each being taken as positive, to the sum
of the components in the opposite direction, each of these being
taken as positive, also. The method will be illustrated by the
examples in friction of the next paragraph.
5. Friction. Let a brick be placed on a table ; let a string be
fastened to the brick, and let the string be pulled horizontally with
a force F just sufficient to move the brick. Then the law of
physics is that A
F = pR, R\
r— J— -p£^
where R (here, the weight of the brick) is p 15
the normal* pressure of the table on the
brick, arid ju (the coefficient of friction) is a constant for the two
surfaces in contact. Thus, if a second brick were placed on top
of the first, R would be doubled, and so would F.
We can state the law of friction generally by saying: When
two surfaces are in contact and one is just on the point of slipping
* Normal means, at right angles to the surface in question. The normal to a
surface at a point is the line perpendicular to the tangent plane of the surface at
the point in question.
10 MECHANICS
over the other, the tangential force F due to friction is propor-
tional to the normal pressure R between the surfaces, or
where /*, the coefficient of friction, is independent of F and R,
and depends only on the substances in contact, but not on the
area of the surfaces which touch each other. For metals on metals
n usually lies between 0.15 and 0.25 in the case of statical fric-
tion. For sliding friction /* is about 0.15; cf. Rankine, Applied
Mechanics.
A simple experiment often performed in the laboratory for
determining /x is the following. Let one of the surfaces be repre-
sented by an inclined plane, the angle of
which can be varied. Let the other surface
be represented by a rider, or small block of
the substance in question, placed on the
plane. If the plane is gradually tilted from
a horizontal position, the rider will not slip
for a time. Finally, a position will be reached
for which the rider just slips. This angle of the plane is known as
the angle of friction and is usually denoted by X. Let us show that
n — tan X.
Resolve the force of gravity, W, into its two components along
the plane and normal to the plane. These are :
W sin X, W cos X.
And now the forces acting up the plane (i.e. the components
directed up the plane) must equal the forces down the plane, or
F = W sin X ; •
and the forces normal to the .plane and upward must equal the
forces normal to the plane and downward, or
R = W cos X.
Hence
F sin X , x
— - = £an ^
R cosX
But
F = /J2.
Consequently
n — tan X.
STATICS OF A PARTICLE 11
Example. A 50 Ib. weight is placed on a rough inclined plane,
angle of elevation, 30°. A cord attached to the weight passes
over a smooth pulley at the top of the plane and carries a weight W
at its lower end. For what values of W will the system be in
equilibrium if jut = -J ?
Here, X < 30°, and so, if W is very small, the 50 Ib. weight will
slip down the plane. Suppose W is just large enough to prevent
slipping. Then friction acts up the plane, and the forces which
produce equilibrium are those indicated. Hence
'F+W
50 sin 30°
W
50 cos 30°
Fio. 17 Fia. 18
F + W = 50 sin 30° = 25,
R = 50 cos 30° = 25 V3,
and F = %R.
It follows, then, that
W = 6~ 25 = 17.8 Ibs.
o
If, now, W is slightly increased, the 50 Ib. weight will obviously
still be in equilibrium, and this will continue to be the case until
the 50 Ib. weight is just on the point of slipping up the plane.
This will occur when W = 32.2 Ibs. as the student can now prove
for himself. Consequently, the values of W for which there is
equilibrium are those for which
17.8 ^ W g 32.2.
EXERCISES
1. If the cylinder of Example 2, § 3, is rough, /z = £, find the
total range of equilibrium. Ans. 4° 49' g 0 g 23° 44'.
2. Consider the inclined planes of Exercise 5, § 3. If the one
on which the 10 Ib. weight rests is rough, /z = T^, find the range
of values for W that will yield equilibrium.
12
MECHANICS
3. Prove the formula
H = tan X
by means of the triangle of forces, i.e. the Law of Sines, § 3, (1).
6. Solution of a Trigonometric Equation. Problem. A 50 Ib.
weight rests on a rough horizontal plane, ju = •£-. A cord is
fastened to the weight, passes over a smooth pulley 2 ft. above
the plane, and carries a weight of 25 Ibs. which hangs freely at
its other end. To find all the positions of equilibrium.
Resolving the forces horizontally and vertically, we find :
25 cos 6 = ±R,
25 sin 0 + R = 50.
Hence, eliminating R, we obtain
the equation :
FIG. 19 (1)6 cos 6 + sin 0 = 2.
This equation is of the form :
(2) a cos 0 + b sin 0 = c,
and is solved as follows. * Divide through by Va2 + b2 :
50
\OJ /— COS 0 -1 7=
Vo2 _|_ ^2 Va2
and then set
fA\ a
(4; — = — = = cos a,
Va2 + b2
Thus
— SHI V — —jr-- ~~- '
b
• ,- — - - — sin a.
Va2 + b2
cos a cos 0 + sin a sin 6 =
Va2 + 62'
or
(5)
cos (0 — a) =
Va2 + b2
* The student should observe carefully the trigonometric technique set forth
in this paragraph, not merely because equations of this type are important in them-
selves, but because the practical value of a working knowledge of trigonometry
is not confined to solving numerical triangles. Of far greater scope and importance
in practice are the purely analytical reductions to other trigonometric foi-ms, and
the solution of trigonometric equations. That is one of the reasons why the
harder examples at the end of the chapter are valuable. They not only give
needed practice in formulating mathematically physical data ; they require also
the ability to handle analytical trigonometry according to the demands of practice.
STATICS OF A PARTICLE 13
The angle a is most easily determined from the equation :
(6)
Thus a is seen to be one of two angles — which one is rendered
clear by plotting the point on the unit circle :
z2 + 2/2 = 1,
whose coordinates are
= __JL__ 7 b __
x ' y ~
The angle from the positive axis of x to the radius drawn to this
point is a. Thus we have a graphical determination of a. It is
not necessary to compute the coordinates accurately, but merely
to observe in which quadrant the point lies, so as to know which
root of the equation for tan a to take. Thus if b > 0, a must be
an angle of the first or second quadrant. Finally, if c/Va2 + b2
is numerically greater than 1, the equation has no solution.
In defining a, it would, of course, Jiave answered just as well if
sin a and cos a had been interchanged, and if either or both the
ratios in (4) had been replaced by their negative values.*
Returning now to the numerical equation above, we see that
tan a = £, sin a > 0, a = 9° 28' ;
cos (6 - 9° 280 -
0 - 9° 28' = ± 70° 48'.
Since 6 in the problem before us must be an angle of the first
quadrant, the lower sign is impossible, and
0 = 80° 16'.
We have determined the point of the plane at which all the
friction is called into play and the 50 Ib. weight is just on the
point of slipping. For other positions, F will not equal pR.
Such a position will be one of equilibrium if the amount of friction
actually called into play, or F, is less than the amount that could
* Equation (2) might also have been solved by transposing one term from the
left- to the right-hand side and squaring. On using the Pythagorean Identity :
sin20 -f cos20 = 1,
we should be led to a quadratic equation in the sine or cosine. This equation will,
in general, have four roots between 0° and 360°, and three of them must be excluded.
Moreover, the actual computation by this method is more laborious.
14 MECHANICS
be called into play, or pR. It seems plausible that such points lie
to the right of the critical point ; but this conclusion is not im-
mediately justified, for, although the amount of friction required,
01 F = 25 cos B,
is less for a larger 0, — still, the amount available, or
n 50-25 sin 0
Mfi- - g - ,
is also less. We must prove, therefore, that
F < nR
or
OK ^50-25 sin 0
25 cos 0 < - 5 --
b
This will obviously be so if
6 cos 6 < 2 - sin 0, 80° 16' < 0 < 90°,
or if
6 cos 0 + sin 0 < 2,
°rif 6 1 2
or if . •
cos (0 - 9° 287) <
-='
As 0, starting with the value 80° 16', increases, 0 — 9° 28' also
increases, and consequently cos (0 — 9° 28') decreases. Conse-
quently our guess is borne out by the facts, and the 50 Ib. weight
will be in equilibrium at all points on the table within a circle of
radius .343 ft., or a little over 4 in., whose centre is directly under
the pulley.
EXERCISES
1. Solve the same problem if the plane is inclined at an angle
of 15° with the horizon, and the vertical plane through the weight
and the pulley is at right angles to the rough plane.
Ana. 43° 33' g0g 69° 24'.
2. At what angle should the plane be tilted, in order that the
region of equilibrium may just extend indefinitely down the plane?
3. Find the angle of the third quadrant determined by the
equation : n . „
M 3 sin 6 - 2 cos 0 = 1.
STATICS OF A PARTICLE 15
4. Show that there are in all eight ways of solving Equation (2),
given by setting the right-hand sides of Equations (4) equal to
± cos a, + sin a and ± sin a, ± cos a, where the ± signs are
independent of each other.
5. Evaluate the integral :
dx
s-.
a cos x + b sin x
6. Solve the equation :
2 cos2 <p — 4 cos <p sin <p — 3 sin2 <p = — 5.
Suggestion. Introduce the double angle, 2<p.
EXERCISES ON CHAPTER I*
1. A rope runs through a block, to which another rope is
attached. The tension in the first rope is
120 Ibs., and the angle it includes is 70°.
What is the tension in the second rope ?
2. A man weighing 160 Ibs. is lying in a
hammock. The rope at his head makes an
angle of 30° with the horizon, and the rope
at his feet, an angle of 15°. Find the tensions in the two ropes.
3. A load of furniture is being moved. The rope that binds
it passes over the round of a chair. The tension on one side of the
round is 40 Ibs. and on the other side, 50 Ibs. ; and the angle is
100°. What force does the round have to withstand ?
4. A canal boat is being towed by a hawser pulled by horses
on the bank. The tension in the hawser is 400 Ibs. and it makes
4 an angle of 15° with the bank. What is
the effective pull on the boat in the direc-
tion of the canal?
5. A crane supports a weight of a ton
as shown in the figure. What arc the
forces in the horizontal and in the oblique
FlG-21 member?
6. Three smooth pulleys can be set at pleasure on a horizon-
tal circular wire. Three strings, knotted together, pass over the
* The student should begin each time by drawing an adequate figure, illustrating
the physical objects involved, and he should put in the forces with colored ink or
pencil. A bottle of red ink, used sparingly, contributes tremendously to clear
thinking.
16
MECHANICS
pulleys and carry weights of 7, 8, and 9 Ibs. at their free ends.
How must the pulleys be set, in order that the knot may be at
rest at the centre of the circle ?
7. A telegraph pole at two cross-
roads supports a cable, the tension
in which is a ton. The cable lies in
a horizontal plane and is turned
through a right angle at the pole.
The pole is kept from tipping by a
FIG. 22
FIG. 23
stay from its top to the ground, the stay
making an angle of 45° with the verti-
cal. What is the tension in the stay?
8. The figure suggests a stake of a
circus tent, with a tension of 500 Ibs.
to be held. What is the tension in
the stay, if the stake could turn freely?
9. Two men are raising a weight of 150 Ibs. by a rope that
passes over two smooth pulleys and is
knotted at A. How hard are they
pulling?
10. If, in the preceding question, in-
stead of being knotted at A, the two
ropes the men have hold of passed over
pulleys at A and were vertical above A,
how hard would the men then have to
pull?
11. A weight W is placed in a smooth hemispherical bowl ; a
string, attached to the weight, passes over the edge of the bowl
and carries a weight P at its other end. Find the position of
equilibrium.
12. Solve the same problem for a parabolic bowl, the rim being
at the level of the focus.
13. One end of a string is made fast to a peg at A. The string
passes over a smooth peg at J?, at the same level as A, and carries a
weight P at its free end. A smooth heavy bead, of weight W, can
slide on the string. Find the position of equilibrium and the
pressure on the peg at B.
14. A bead weighing W Ibs. can slide on a smooth vertical
circle of radius a. To the bead is attached a string that passes
FIG. 24
STATICS OF A PARTICLE 17
over a smooth peg situated at a distance %a above the centre of
the circle, and has attached to its other end a weight P. Find
all the positions of equilibrium.
16. A 50 Ib. weight rests on a smooth inclined plane (angle
with the horizontal, 20°) and is kept from slipping by a cord which
passes over a smooth peg 1 ft. above the top of the plane, and
which carries a weight of 25 Ibs. at its other end. Find the
position of equilibrium.
16. A heavy bead can slide on a smooth wire in the form of a
parabola with vertical axis and vertex at the highest point, A
string attached to the bead passes over a smooth peg at the focus
of the parabola and carries a weight at its other end. Show that
in general there is only one position of equilibrium ; but some-
times all positions are positions of equilibrium.
17. A weightless bead * can slide on a smooth wire in the form
of an ellipse whose plane is vertical. A string is knotted to the
bead and passes over two smooth pegs at the foci, which are at
the same horizontal height. Weights of P and Q «firo attached
to the two ends of the string. Find the positions of equilibrium.
18. An inextensible flexible string has its ends made fast at two
points and carries a weightless smooth bead. Another string is
fastened to the bead and drawn taut. Show that every position
of the bead is one of equilibrium, if the second string is properly
directed.
19. Give a mechanical proof, based on the preceding question,
that the focal radii of an ellipse make equal angles with the tangent.
20. A bead of weight P can slide on a smooth, vertical rod.
To the bead is attached an inextensible string of length 2a,
carrying at its middle point a weight W and having its other end
made fast to a peg at a horizontal distance a from the rod. Show
that the position of equilibrium is given by the equations :
P tan <p = (P + W} tan 6, sin 0 + sin <p = 1,
where 0, <p are the angles the segments of the string make with
the vertical.
* Questions of this type may be objected to on the ground that a force must act
on mass, and so there is no sense in speaking of forces which act on a masslcss ring.
But if the ring has minute mass, the difficulty is removed. The problem may be
thought of, then, as referring to a heavy bead, whose weight is just supported by
a vertical string. Since the weight of the bead now has no influence on the position
of equilibrium, the mass of the bead may be taken as very small, and so, physically
negligible.
18 MECHANICS
21. If, in the preceding question, P = Wy show that 0 = 21° 55',
V = 38° 49'. Determine these angles when W = 2P.
22. A flexible inextensible string in the form of a loop 60 in.
long is laid over two smooth pegs 20 in. apart and carries two
smooth beads of weight P and W. Find the position of equi-
librium, if the beads cannot come together.
Ans. W sin 6 = P sin ^>,
cos 6 + cos (f> = 3 cos 6 cos p ;
cos4 e - | cos3 e - I (i - ~) cos2 e
2
hence
9 cos4 6 - 6 cos3 6 - 8\ cos2 8 + 6X cos 6 - X = 0,
where _ (W2 - P2)
X~ TP
23. Show that if, in the preceding question, P = 5 and
TF = 10, 0 = 25° 8', and find the reaction on the peg.
24. One end of an inextensible string a in. long is made fast to
a peg A and at the other end is knotted a weight W. A second
string, attached to W, passes over a smooth peg at B, distant
b in. from A and at the same level, and carries a weight P at its
other end. Find the position of equilibrium.
If P = Wj how far below the level of the pegs will the first
^ - 25.* Observe the braces that stiffen the
frame of a railroad car. Formulate a
reasonable problem suggested by what you sec, and solve it.
26. A bridge of simple type is suggested by the figure. In
designing such a structure, the stiffness of the members at a point
A B
J_
FIG. 26
* The following four problems are given only in outline, and the student thus
has the opportunity of filling in reasonable numerical data and formulating a clean-
cut question. It is not necessary that he respond to all the problems; but he
should demand of himself that he develop a number of them and supplement these
by others of like kind which he finds of his own initiative in everyday life. For,
imagination is one of the highest of the intellectual gifts, and too much effort
cannot be spent in developing it.
STATICS OF A PARTICLE 19
where these come together is not to be utilized, but the frame is
planned as if the members were all pivoted there. Draw such a
bridge to scale and find what the tensions and thrusts will be if
it is to support a weight of 20 tons at each of the points A, B.
Make a reasonable assumption about the weight of the road bed,
but neglect the weight of the tie rods, etc. j 100 lbs<
27. The tension in each of the traces attached to _j_ Jo,
o
100 Ibs.
a whiffle-tree 3 ft. long is 100 Ibs. The distance from 1(j
the ring to the whifflc-tree is 10 in. What is the ten-
sion in the chains ? Fm- 27
28. Have you ever seen a funicular — a small passenger car,
hauled up a steep mountain by a cable? How is the tension in
tho cable related to the weight of the car? When the direction
of the cable is changed by a friction pin, or roller, over which the
cable passes, what is the pressure on the pin?
FRICTION
29. Consider the inclined planes of Question 5, § 3. If both
.ire rough and /z = -fa for each, what is the range of values for W
consistent with equilibrium?
30. A weightless bead can slide on a rough horizontal wire,
ju = 0.1. A cord is attached to the bead and carries a weight at
its other end, thus forming a simple pendulum. Through what
angle can the pendulum swing without causing the bead to slip?
31. A water main 5 ft. in diameter is filled with water to a
depth of 1 ft. A mouse tumbles in and swims to the nearest point
on the wall. If the coefficient of friction between her feet and the
pipe is -J-, can she clamber up, or will she be drowned?
32. A heavy bead is placed on a rough vertical circle, the
coefficient of friction being f . If the angle between the radius
drawn to the bead and the vertical is 16°, find whether the bead
will slip when released.
33. A rope is fastened to a weight that rests on a rough hori-
zontal plane, and pulled until the weight just moves. Find the
tension in the rope, and show that it will be least when the rope
makes, with the horizontal, the angle of friction.
34. The same question for an inclined plane.
20 MECHANICS
35. A 50 Ib. weight is placed on a rough inclined plane, M = t,
angle of inclination, 10°. A string tied to the weight passes over
a smooth peg at the same level as the weight and carries a weight
of 7 Ibs. at its lower end. When the system is released from rest,
will it slip ?
36. A weight is placed on a rough inclined plane and is attached
to a cord, the other end of which is made fast to a peg in the plane.
Find all positions of equilibrium.
37. If the parabolic wire described in Question 16 is rough, and
the weights are P and W, find all positions of equilibrium.
Ans. When P ^ W, the limiting position is given by one or
the other of the equations :
i - P ~ W 1 ? - W ~ P 1
tan 2 ~" P + W ' /z' tan 2 " W + P ' n'
Find the other positions of equilibrium, and discuss the
case P = W.
38. Cast iron rings weighing 1 Ib. each can slide on a rough
horizontal rod, M = -J. A string 6 ft. long is attached to each of
these beads and carries a smooth bead weighing 5 Ibs. How far
apart can the two beads on the rod be placed, if the system is to
remain at rest when released ?
39. An elastic string 6 ft. long, obeying Hooke's Law, is stretched
to a length of 6 ft. 6 in. by a force of 20 Ibs. The ends of the
string are made fast at two points 6 ft. apart and on the same
level. A weight of 4 Ibs. is attached to the mid-point of the string
and carefully lowered. Find the position of equilibrium, neglect-
ing the weight of the string. Ans. 0 is given by the equation :
cot 0 = 120 (1 - cos 0).
40. Solve the preceding equation for 0, to one-tenth of a degree.
Ans. 0 = 14|°.
41. The mast of a derrick is 40 ft. high, and a stay is fastened
^20 to a block of stone weighing
4 tons and resting on a pave-
ment, M = f • The boom is
35 ft. long, and its end is dis-
™ oo tant 20 ft. from the top of the
riG. &Q . 11-
mast. Is it possible to raise
a 5 ton weight, without the derrick's being pulled over, the dis-
tance from the stone to the derrick being 120 ft.?
CHAPTER II
STATICS OF A RIGID BODY
1. Parallel Forces in a Plane. Lot two parallel forces, P and
Q, act on a body at A and B, and let them have the same sense.
Introduce two equal and opposite forces, S and S', at A and B as
shown in the figure, and, com-
pounding them with P and Q
respectively, carry the resulting
forces back to the point D in
which their lines of action meet.
These latter forces are now seen
to have a resultant,
(1) R = P + Q,
parallel to the given forces and
having the same sense, its line
of action dividing the line AB
into two segments, AC and CB. Let the lengths of the segments
be denoted as follows: AC = a, CB = 6, AB = c, DC = h.
From similar triangles it is seen that
frI(3 29
Hence
P
S
h
a'
Q h
AS' 6'
aP = bQ.
a + 6 = c.
(2)
Moreover,
(3)
To sum up, then : The original forces, P and Q, have a resultant
determined by the equations (1), (2), and (3).
Example. The familiar gravity balance, in which one arm, a,
from which the weight P to be determined
is suspended, is short, and the other arm, b,
from which the rider Q hangs is long, is a
FIG. 30 case in point.
21
22
MECHANICS
Q
•1
Opposite Forces. If P and Q are opposite in direction, and
unequal (Q > P, say), they also have a resultant. Introduce a
force E (Equilibriant) parallel to P and Q and having the sense
of P, determining it so that Q will be equal and opposite to the
resultant of P and E. Then
Q = P + E,
cP - bE,
a = b + c.
Thus P and Q are seen to have
a resultant,
FlG- 31 (4) R = Q - P,
having the sense of Q, its line of action cutting AC produced in
the point B determined by the equations :
(5) aP = bQ,
(6) a = b + c.
If P and Q are equal, they form a couple and, as we shall show
later, cannot be balanced by a single force; i.e. they have no
resultant (force).
Example. Consider a pair of nut crackers. The forces that
act on one of the members are i) P, the pull of the hinge ; if) Q,
the pressure of the nut; and Hi) the
force E the hand exerts, balancing the
resultant, R, of P and Q.
We have here made use of the so-called
Principle of the Transmissibilily of Force,
which says that the effect of a force on a
body is the same, no matter at what
point in its line it acts. Thus a service
truck will tow a mired car as effectively
(but no more effectively) when the tow-rope is long, as when
it is short, provided that in each case the rope is parallel to the
road bed.
Moreover, it is not necessary to think of the point of application
as lying in the material body. It might be the centre of a ring.
For we can always imagine a rigid weightless truss attached to the
body and extending to the desired point. But we always think
of a body, i.e. mass, on which the system of forces in question acts.
FIG. 32
STATICS OF A RIGID BODY 23
EXERCISES
1. A 10 ton truck passes over a bridge that is 450 ft. long.
When the truck is one-third of the way over, how much of the
load goes to one end of the bridge, and how much to the other
end? Ans. 6f tons to the nearer end.
2. A weight of 200 Ibs. is to be raised by a lever 6 ft. long, the
fulcrum being at one end of the lever, and the weight distant 9 in.
from the fulcrum. What force at the other end is needed, if the
weight of the lever is negligible?
3. A coolie carries two baskets of pottery by a pole 6 ft. long.
If one basket weighs 50 pounds and the other, 70 pounds, how far
arc the ends of the pole from his shoulder?
2. Analytic Formulation ; n Forces. Suppose that n parallel
forces act. Then two, which are not equal and opposite, can be
replaced by their resultant, and this in turn combined with a third
one of the given forces, until the number has been reduced to two.
These will in general have a resultant, but, in particular, may form
a couple or be in equilibrium. Thus the problem could be solved
piecemeal in any given case.
An explicit analytic solution can be obtained as follows. Begin
with n = 2 and denote the forces by Pl and P2. Moreover, let P1
arid P2 be taken as algebraic quan-
tities, being positive if they act in
one direction ; negative, if they act p\
in the opposite direction. — • - • - 1 -
AT ±1 T VI* XssQ 3J=1 ** *2
Next draw a line perpendicular* Fl(J 33
to the lines of action of P1 and
P2, and regard this line as the scale of (positive and negative)
numbers, like the axis of x. Let xlf x2 be the coordinates of the
points in which P1? P2 cut the line. We proceed to prove the
following theorem.
The forces Pl and P2 have a resultant,
(l) R = P, + P»
provided Pl + P2 ^ 0. Its line of action has the coordinate:
* An oblique direction could be used, but in the absence of any need for such a
generalization, the orthogonal direction is more concrete.
24 MECHANICS
Suppose, first, that Pt and P2 are both positive. Then, by § 1,
R = P, + P2
,
where
a = x — x1? o = x2 — x,
provided Xj < x2 (algebraically). Hence
(x - x,) Pl = (x2 - x) P2,
and from this equation, the relation (2) follows at once.
The case that x2 < xl is dealt with in a similar manner, as is also
the case that Pl and P2 are both negative.
Next, suppose P1 and P2 have opposite senses, but
Pl + P, * 0.
Let P1 < 0, P2 > 0, | P1 | < P2,
where | x | means the numerical or absolute value of x. Thus
| - 3 | = 3, | 3 | = 3. Moreover, let xl < x2. Then, by § 1,
Pl and P2 have a resultant,
12 = -?! + P2,
and the coordinate, x, corresponding to it is obtained as follows :
a = x — xlt b = x — xtl
and hence, from § 1, (5) :
(x - x{) (- Pt) = (x - x2) P2.
So again we arrive at the same formulas, (1) and (2), as the
solution of the problem.
It remains merely to treat the remaining cases in like manner.
The final result will always be expressed by formulas (1) and (2).
We are now ready to proceed to the general case.
THEOREM 1. Let n parallel forces, Plt • • • , Pn, act. They will
have a resultant,
R = P! ^ ---- + Pn,
provided this sum ^ 0, and its line of action will correspond to x,
where
/o\ ;= _ P\ Xl i" * ' ' "T Pn Xn
(3) x -
STATICS OF A RIGID BODY 25
The proof can be given by the method of mathematical induc-
tion. The theorem is known to be true for n = 2. Suppose
it were not true for all values of n. Let m be the smallest value
of n for which it is false. We now proceed to deduce a con-
tradiction.
Suppose, then, that PD - • • , Pm is a system of parallel forces,
for which the theorem is false, although it is true for n — 2,
3, • • • , m — - 1. By hypothesis,
Pi + • • ' + Pm * 0.
Now, it is possible to find m — 1 of the P, 's whose sum is not 0 :
Pi + • ' • + Pm-, * 0,
let us say. These m — I forces have, by hypothesis, a resultant :
R' = Pl + • - - + Pw-lf
and its x has the value :
_/ /I 3*i r * ' " 'T~ -* w— i #w~i
x ~ />! + ••• -TP^; '
since the theorem holds by hypothesis for all values of n < m.
Next, combine this force with Pw. Since
R' + Pm = 1\ + - - - + Pm * 0,
the two forces have a resultant,
R = R' + Pm = P1 + - - • + Pw,
and its line of action is given by the equation
_ R'x' + Pmxm Plgl + • • • +Pwa,»
X #' + PM A + --+P™
But this result contradicts the assumption that the theorem is
false for n = m. Hence the theorem is true for all values of n.
Couples. Let
(4) P, + - - - + Pn = 0, Pn * 0.
Thei1 Pl + - • • + Pn-! ^ 0,
and the forces Plf • • • , Pn_1 have a resultant,
ft' = P 4- ... 4- P
•^ — •» i r n^ •*• n— 1>
whose line of action is given by the equation :
-/ = P! %l H~ ' ' • "T Pn-i ^n~i
26 MECHANICS
If, in particular, x' = xn, this resultant, 72', will have the same
line of action as Pn ; and since
R' + Pn = 0, or Rf = - Pn,
the n forces will be in equilibrium. We then have :
_ PI X\ I ' ' • ~T 1 n-i Xn-i
Xn r> y
f n
(5) Pl*l + •'• + PnXn = 0.
And conversely, if this condition holds, we can retrace our stops
and infer equilibrium. But, in general, x' ^ xn. Hence
_^ P, Xl H ----- h Pn-i gn-i
~
(6) Pl*l + • ' ' +PnXn^Q.
We then have a couple. And conversely, if (4) and (6) hold, we
can retrace our steps and infer that we have a couple. We have
thus proved the following theorem.
THEOREM 2. The n parallel forces Plt • • • , Pn form a couple if,
arulonlyif PI + . . . + P. = 0.
Equilibrium. The case of equilibrium includes not only the
case above considered (Pn ^ 0), but also the case in which all
n forces vanish. • We thus have the following theorem.
THEOREM 3. The n parallel forces Ply • • • , Pn are in equilibrium
if , and only if PI + . . . + Pn = 0^
Pi Xl + ---- h Pn Xn = 0.
3. Centre of Gravity. Let n particles, of masses Wi, • • • , mn,
be fastened to a rigid rod, the weight of which may be neglected,
and let them be acted on by the force of gravity. If the rod is
supported at a suitable point, (?, and is at rest, there will be no
tendency to turn in any direction. This point is called the centre
of gravity of the n particles, and its position is determined by the
equation :
mlxl + • • - + mnxn
-
ml + - • • + mn
STATICS OF A RIGID BODY 27
If the particles lie anywhere in a plane, being rigidly connected
by a truss work of weightless rods, and if we denote the coordinates
of mk by (xk, yk)1 the centre of gravity is defined in a similar manner
(sec below) and its coordinates, (x, ?/), are given by Equations
(1) and
/9x fl _ *KI y\ H ---- + ™>n yn
(2) y ~ n^ + .-.+m. '
For, let the plane of the particles be vertical, the axis of x being
horizontal. Then the system is acted on by n parallel forces,
whose lines of action cut the axis of x at right angles in the points
%!,•'', xny and their resultant is determined in position by Equa-
tion (1). On rotating the plane through a right angle and repeat-
ing the reasoning, Equation (2) is obtained.
The centre of gravity of any material system, made up of par-
ticles and line, surface, and volume distributions, is defined as a
point, (7, such that, if the parts of the system be rigidly connected
by weightless rods, and if G be supported, there will be no tendency
of the system to rotate, no matter how it be oriented. We have
proved tho existence of such a point in the case of n particles
lying on a line. For n particles in a plane we have assumed that
a centre of gravity exists and lies in the plane, and then we have
computed its coordinates. We shall prove later that n particles
always have a centre of gravity, and that its coordinates are given
by Equations (1), (2), and
. .
ml + • • • + mn
In the case of a continuous distribution of matter, like a tri-
angular lamina or a solid hemisphere, the methods of the Calculus
lead to the solution. It is the definite integral, defined as the
limit of a sum, that is here employed, and Duhamel's Principle
is essential in the formulation. In the simpler cases, simple
integrals suffice. But even in some of these cases, surface and
volume integrals simplify the computation.
The following centres of gravity are given for reference.
fl) Solid hemisphere : x = fa.
b) Hemispherical surface : x — £a.
c) Solid cone : x = f h.
d) Conical surface : x — f A.
e) Triangle : Intersection of the medians.
28
MECHANICS
4. Moment of a Force. Let F be a force lying in a given
plane, and let 0 be a point of the plane. By the moment of F about
0 is meant the product of the force by the distance from 0 of its
line of action, or hF. A moment may furthermore be defined as
an algebraic quantity, being taken as positive when it tends to turn
the body in one direction (chosen arbitrarily as the positive
direction), and negative in the other case. Finally, if 0 lies on
the line of action of the force, the moment is defined as 0.
Let a force F act at a point (x, y), and let the components of F
along the axes be denoted by X, Y. Then the moment (taken
algebraically) of F about the origin, 0, is :
(1) xY-yX.
Proof. Let the equation of the line of action of F be written
in Hesse's Normal Form :
x cos a + y sin a = h.
/
i FIG. 34
O
FIG. 35
Suppose, first, that the moment is positive. Then it will be
hF = x (F cos a) + y (F sin a).
Here, 2ir — a is the complement of 6, Fig. 34 :
Hence
and since
cos a = sin 6,
a = 0 - + 2rr.
sin a = — cos 0,
X = F cos 6, Y = F sin 6,
the proof is complete.
STATICS OF A RIGID BODY 29
If, however, the moment is negative, a. and 6 will be connected
by the relation, Fig. 35 :
,-a-«--Bt «-« + |.
Hence
cos a = — sin 9, sin a = cos 6.
The moment will now be repressed as
— hF = x (— F cos a) + y (— F sin a),
and thus we arrive at the same expression, (1), as before. The same
is true for a nil moment. Hence the formula (1) holds in all cases.
From (1) we prove at once that the moment of the resultant of
two forces acting at a point is the sum of the moments of the two
given forces. Let the latter be Fj, F2, with the moments
xY1 — yXl and xY2 — yX2.
The components of the resultant force are seen to take the form :
Xl + X2 and Yl + F2, and its moment is
From this expression the truth of the theorem is at once obvious.
Finally, the moment of a force about an arbitrary point, (x0, ?/0),
is seen to be :
(2) (x-x0)Y-(y-y,}X.
The physical meaning of the moment of a force about a point
is a measure of the turning effect of the force. Suppose the body
were pivoted at 0. Then the tendency to turn about 0, due to
the force F, is expressed quantitatively by the momenta And a
set of forces augment or reduce one another in their combined
turning effect according to the magnitude and sense of the sum
of the moments of the individual forces. From this point of view
a moment is often described in physics and engineering as a torque.
6. Couples in a Plane. A couple has already been defined as
a system of two equal and opposite parallel forces. A couple
cannot be balanced by a single force, but is an independent me-
chanical entity ; the proof is given below.
By the moment of a couple, taken numerically, is meant the
product of either force by the distance between the lines of action
of the forces.
30 MECHANICS
THEOREM. Two couples having the same moment and sense are
equivalent.
Suppose first that the forces of the one couple are parallel to the
forces of the other couple. Then, by proper choice of the axis
of x, we can represent the couples as indicated, where
J P. + P.-O. 0<P,;
P.I » «, I l •. P3 + P4 = 0, 0<P3;
p*\ " P, x^xt + h, 0<h;
Fxo.36 J *4 = *3 + J, °<l>
Now consider the system of four forces, Plf P2, — P3t — P4.
These are in equilibrium. For,
P, + P2 - P3 - P4 = 0
and
Pr _|_p-r _ P ,». _ /:> /r —
1 ^l I •* 2 **/2 •*• 3 "^S •* 4 **/4
(P -4- P W — P It — (P -\- P } r 4- P 7 = fl
^JTj -f X 2/ ^2 -I i « V.^3 I -* 4/ •*'4 ' ^3 ^ — VJ>
Hence the first couple is balanced by the negative of the second
couple, and thus the theorem is proved for the case that all the
forces are parallel.
If the forces of the two couples are oblique to each other, let
OA and OB be two lines at right angles to the forces of the first
couple and to those of the second
couple respectively. Lay off t wo equal
distances, OA = h and OB = h, on
these lines. Then by the theorem just
proved the first couple can be repre-
sented as indicated by the forces P
and P.* Furthermore, the second
couple, reversed in sense, can be rep-
resented by the forces Q and Q. Let
the lines of action of P at A and Q at B meet in C, and carry
these forces forward so that each acts at C. Then the four
* It might seem that there are two cases to be considered, for cannot the vectors
that represent the forces of the first couple be opposite in sense? True. But
then we can begin with the second couple. Its forces will be represented by the
Q and Q of the diagram ; and the forces of the first couple, reversed in sense, will
now appear as P and P.
STATICS OF A RIGID BODY
31
A:(i,o)
forces obviously are in equilibrium, for all four are equal in magni-
tude, since by hypothesis the moments of the two given couples
are equal ; and the forces make equal angles with the indefinite
line OC, in such a manner that the resultant of one pair is equal
and opposite to that of the other pair, the lines of action of these
resultants coinciding.
To sum up, then, the effect of a couple in a given plane is the
same, no matter whore its forces act, and no matter how large or
small the forces may be, provided only that the moment of the
couple is preserved both in magnitude and in sense.
Composition of Couples. From the foregoing it appears that two
couples in a plane can be compounded into a single couple, whose
moment is the sum of the moments
of the constituent couples ; all mo-
ments being taken algebraically.
For, assume a system of Cartesian
axes in the plane, and mark the
point A : (1,0). The first couple
can be realized by a force P1 at A
parallel to the axis of y (and either
positive or negative) and an equal
and opposite force —Pi at the w
origin, acting along the axis of y.
The moment of this couple, taken algebraically, is obviously P,.
Dealing with the second couple in a similar manner, we now
have as the result two forces, P1 and P2, at A parallel to the axis
of y', and two equal and opposite forces at 0 along the axis of
y. These forces constitute a resultant couple, whose moment is
the sum of the moments of the given couple.
This last statement is at fault in one particular. It may happen
that the second couple is equal and opposite to the first, and then
the resultant forces both vanish. In order that this case may not
cause an exception, we extend the notion of couple to include a
nil couple : i.e. a couple whose forces are both zero, or whose forces
lie in the same straight line ; and we define its moment to be 0.
We are thus led to the following theorem.
THEOREM. // n couples act in a plane, their combined effect is
equivalent to a single couple, whose moment is the sum of the moments
of the given couples.
32 MECHANICS
Remark. The moment of a couple is equal to the sum of the
moments of its forces about an arbitrary point of the plane. This
is seen directly geometrically from the definition of a moment. In
particular, let a point 0 be chosen at pleasure. The couple can
be realized by two forces, one of which passes through 0. The
moment of the couple is then equal to the moment of the other
force about 0.
6. Resultant of Forces in a Plane. Equilibrium. Let any
forces act in a plane. Then they are equivalent f) to a single
force, or ii) to a single couple ; or, finally, Hi) they are in equi-
librium. Let 0 be an arbitrary point of the plane. Beginning
with the force FD let us introduce at 0 two
forces equal and opposite to Fj. The two forces
checked form a couple, and the remaining
* force is the original force Fx, transferred to
the point 0.
Proceeding in this manner with each of the
, ' remaining forces, F2, • • • , Fn, we arrive at
a new system of forces and couples equiv-
alent to the original system of forces and consisting of those n
forces, all acting at 0, plus n couples. . These n forces are equiva-
lent to a single force, R, at 0; or are in equilibrium. And the
n couples are equivalent to a single couple, or are in equilibrium.
In general, the resultant force, R, will not vanish, nor will the
resultant couple disappear. The latter can, in particular, be
realized as a force equal and opposite to R and acting at 0, and a
second force equal to R, but having a different line of action.
Thus the resultant of all n forces is here a single force. Inciden-
tally we have shown that a non-vanishing couple can not be
balanced by a non- vanishing force ; for, the effect of such a force
and such a couple is a force equal to the given force, but trans-
ferred to a new line of action, parallel to the old line.
It may happen that the resultant force vanishes, but the result-
ant couple does not. For equilibrium, it is necessary and suffi-
cient that both the resultant force and the resultant couple vanish.
This condition can, with the help of the Remark at the close of
§ 5, be expressed in the following form.
EQUILIBRIUM. A system of n forces in a plane will be in equi-
librium if, and only if
STATICS OF A RIGID BODY
33
i) they are such as would keep a particle at rest if they all acted at
a point; and
ii) the sum of the moments of the forces about a point (one point
is enoughj and it may be chosen anywhere) of the plane is zero.
Analytically, the condition can be formulated as follows. Let
the point about which moments are to be taken, be chosen as
the origin, and let the force Fr act at the point (xr, yr). Then
Xr = 0,
r=l
2) (XrYr- yrXr) = 0.
FIG. 40
Example. A ladder rests against a wall, the coefficient of
friction for both ladder and wall being the same, M- If the ladder
is just on the point of slipping when inclined at
an angle of 60° with the horizontal, what is the
value of ju ?
Since all the friction is called into play, the
forces are as indicated in the figure, R and S being
unknown, and JJL also unknown.
Condition ?) tells us that the sum of the verti-
cal components upward must equal the sum of the
vertical components downward, or
R + »S = W.
Furthermore, the sum of the horizontal components to the right
must equal the sum of the horizontal components to the left, or
S = nR.
Finally, the moments about a point, 0, of the plane must
balance. It is convenient to choose as 0 a point through which a
number of unknown forces pass; for example, one end of the
ladder, say the upper end. Thus
2a cos 60° R = 2a sin 60° »R + a cos 60° TF,
or
R =V3/
Hence
W
2(1-
S
2 (1 - MV3)
34 MECHANICS
We can now eliminate R and S. The resulting equation is
Thus
/i = 2-V3 = 0.27.
The other root, being negative, has no physical meaning.
EXERCISES
1. If in the example just discussed the wall is smooth, but the
floor is rough, and if /* = £, find all positions of equilibrium.
2. If in the example of the text /* could be as great as 1, show
that all positions would be positions of equilibrium.
3. A ladder 12 ft. long and weighing 30 Ibs. rests at an angle
of 60° with the horizontal against a smooth wall, the floor being
rough, /u = £. A man weighing 160 Ibs. goes up the ladder. How
far will he get before the ladder slips?
4. In the last question, how rough must the floor be to enable
the man to reach the top?
6. Show that a necessary and sufficient condition for equilibrium
is that the sum of the moments about each of three points, A, B,
and C, not lying in a line, shall vanish for each point separately.
7. Couples in Space. THEOREM I. A couple may be transferred
to a parallel plane without altering its effect} provided merely that its
moment and sense are preserved.
It is sufficient to consider two couples in parallel planes, whose
moments are equal and opposite, and to show that their forces are
in equilibrium. Construct a cube
with two of its faces in the planes
of. the couples. Then one couple can
be represented by the forces marked
P and Pj and the reversed couple,
by the forces Q and Q (P = Q).
Consider the resultant of P at
A and Q at B. It is a force of
p + Q (= 2P) parallel to P and
having the same sense, and passing
through the centre, 0, of the cube. Turn next to P at C and
Q at D. The resultant of these forces is obviously equal and
STATICS OF A RIGID BODY 35
opposite to the resultant just considered and having the same
line of action. The four forces are, then, in equilibrium. This
completes the proof.
Example. In a certain type of auto (Buick 45-6-23) the last bolt
in the engine head was so near the cowl that a flat wrench could
not be used. The garage man immediately bent a flat
wrench through a right angle, applied one end of the
wrench to the nut and, passing a screw driver through
the opening in the other end, turned the nut. Thus the
applied couple was transferred from the horizontal
plane through the screw driver to the plane of the riut. IQ'
Vector Representation of Couples. A couple can be represented
by a vector as follows. Construct a vector perpendicular to the
plane of the couple and of length equal to the moment of the
couple. As regards the sense of the vector, either convention is
permissible. Let us think of ourselves as standing upright on the
plane of the couple and looking down on the plane. If we are
on the proper side of the plane, we shall see the couple tending to
produce rotation in the clock-wise sense. And now the direction
from our feet to our head may be taken as the positive sense of
the vector — or, equally well, the opposite direction.
THEOREM II. The combined effect of two couples is the same as
that of a single couple represented by the vector obtained by adding
geometrically the two vectors which represent respectively the given
couples.
The theorem has already been proved for the case that the
planes of the given couples are parallel or coincident. If they
intersect, lay off a line seg-
ment of unit length, AB,
on their line of intersection,
and take the forces of the
couples so that they act at
FlQ 43 A and B perpendicularly to
the line AB. Then it is
easily seen that the resultant of the two forces at A and the result-
ant of the two forces at B form a new couple.
Finally, the vector representations of these three couples are
three vectors perpendicular respectively to the three planes of
the couples, equal in length to the forces of the couples, and so
\R
QX^/
36 MECHANICS
oriented as to give the same figure yielded by three of the forces,
properly chosen, only turned through 90°.
8. Resultant of Forces in Space. Equilibrium. Let any n
forces act on a body in space. Let them be represented by the
vectors F,, • • • , Fn. Let 0 be an arbitrary point of space. Intro-
duce at 0 two forces that are equal and opposite to the force Fk.
Then the n forces Flf • • • , Fn at 0 have a resultant :
(1) R = F! + • • • + Fn,
acting at 0, or are in equilibrium. And the remaining forces,
combined suitably in pairs, yield n couples, Clt • • • , Cn, whose
resultant couple, C, is :
(2) C = C, + - - - + Cn,
or, in particular, vanishes ; the couples being then in equilibrium.
In general, neither R nor C will vanish. Thus the n given forces
reduce to a force and a couple. The plane of the resultant couple,
C, will in general be oblique to the line of action of the resultant
force, and hence the vector C oblique to the vector R. Let
C = C, + C2,
where C^ is collinear with R, and C2 is perpendicular to R. The
couple represented by C2 can be realized by two forces in a plane
containing the resultant force, R ; and its forces can be combined
with R, thus yielding a single force R, whose line of action, how-
ever, has been displaced. This leaves only the couple Cj. We
have, therefore, obtained the following theorem.
THEOREM. Any system of forces in space is in general equiva-
lent to a single force whose line of action is uniquely determined,
and to a single couple, whose plane is perpendicular to the line of
action of the resultant force.
In particular, the resultant force may vanish, or the resultant couple
may vanish, or both may vanish.
Equilibrium. The given forces are said to be in equilibrium if
and only if both the resultant force and the resultant couple
vanish.
For completeness it is necessary to show that the resultant
force, R, together with its line of action, and the resultant couple,
Ct, are uniquely determined. For it is conceivable that a differ-
STATICS OF A RIGID BODY
37
ent choice, O', of the point O might have led to a different result.
Now, the vector R is uniquely determined by (1), and so is the
same in each case ; but C depends on the choice of 0', and so Cl
might conceivably be different from CJ, though each would be
collinear with R. This is, however, not the case. For, reverse R
in the second case, and also the couple C[. Then the reversed
force and couple must balance the first force and couple. But
this situation leads to a contradiction, as the reader will at once
perceive.
9. Moment of a Vector. Couples. Given a force, F, acting
along a line //, and any point 0 in space. By the vector moment
of F with respect to (or about) 0 is meant the vector
product * /\F
(1) M - r X F,
where r is a vector drawn from 0 to a point of L.
It is a vector at right angles to the plane of 0 and
L, and its length is numerically equal to the
moment of F about 0 in that plane. Its sense
depends on whether we are using a right-handed or a left-handed
system. Referred to Cartesian axes
FIG. 44
(2)
(3)
M =
x — a y — b z — c
X Y Z
L = (y-V)Z- (z-e)Y
M = (z -c)X - (x- a)Z
N = (x-a)Y - (y-
P:(x,y,z)
r = r' - rj,
r' = x i + y j + z k,
FIG. 45
* The student should read § 3 of Appendix A. This, together with the mere
definitions that have gone before, is all of Vector Analysis which he will need for
the present.
38 MECHANICS
Vector Representation of a Couple. Let a couple consist of two
forces, F! and F2 :
F, + F2 = 0,
acting respectively along two lines L± and L2. The vector C
which represents the couple is seen from the definition of the
vector product to be :
(4) C = r X Plf
where r represents any vector drawn from a point of L2 to a point
of Lj. Let 0 be any point of space. Then the sum of the vector
moments of Fx and F2 with respect to 0 yields the vector couple :
(5) C = r, X P! + r2 X F2,
where r^ r2 are any vectors drawn from 0 to L1 and L2 respec-
tively. For
hence
r X F! = T! X F! - r2 X Fj = T! X Fj + r2 X F2.
10. Vector Representation of Resultant Force and Couple.
Resultant Axis. Wrench. Let Pi, • • • , Fn be any system of
forces in space. Let P be any point of space, and let equal and
opposite forces, F* and — FA, k = 1, • • • , n, be applied at P.
Consider the n forces F^ - • - , Fn which act at P. Their re-
sultant is
R = F! + - • • + Fn.
The remaining forces yield n couples, consisting each of F* at
PA', (%k, Vk, Zk) and — F& at P: (x, y, z). Let rk, r be the vec-
tors drawn from the origin of coordinates (chosen arbitrarily) to
Pk and P respectively. Then the Jfc-th couple, C*, is represented
by the equation :
C* = r* X F* - r X F*.
We are thus led to the following theorem.
THEOREM. The given forces Flf • • • , Fn are equivalent to a single
force,
(6) R = F, + • • • + Fn,
acting at P; and to a couple,
(7) C = 5) r* X F* - r X R.
fc-i
STATICS OF A RIGID BODY
39
Resultant Axis. The resultant axis is the locus of points P,
for which C lies along R ; i.e. is collinear with R. The condition
for this is obviously the vanishing of the vector product :
(8) R X C = 0, R * 0.
Let
(9) iXF^Li
(10)
Thus
(11) L =
R =
Fj + Zk.
Then the condition (8) becomes, by virtue of (7) :
R X (Li + M j + tf k) = R X (r X R),
i
j
k
i J k
X
Y
Z
=
X Y Z
yZ-zY
zX-xZ
xY-yX
L M N
or:
(12)
This is the equation of the resultant axis in vector form. To
reduce to ordinary Cartesian form, equate the coefficients of i, j,
k respectively. Thus we find :
Y + zZ = YN - ZM
(13)
- Z (xX
zZ) = ZL - XN
zZ) = XM - YL
One of these equations may become illusory through the vanish-
ing of all the coefficients ; but some two always define intersecting
planes, for the rank of the determinant is 2, since R > 0 ; and
between the three equations there exists an identical relation.
Let (£ , ij, f ) be the coordinates of the nearest point of the line to
the origin. Then
Hence
(15) $
YN -ZM
>? =
ZL-XN
ZM- YL
ft2
40 MECHANICS
Thus we have found one point of tt*e resultant axis, and the
direction of the axis is that of R. The resultant couple is given
by (7), where
(16) r = £i + Tjj + fk.
Wrench. A wrench is defined as two forces, acting at arbitrary
points; moreover, neither force shall vanish, and their lines of
action shall be skew.
Let the forces be F*, acting at (xk, y^ Zk), k = 1, 2. The reader
will do well to compute the resultant force, axis, and couple.
Suppose, in particular, that Fx is a unit force along the positive
axis of Z, and F2 is a force of 2, parallel to the axis of y and acting
at the point (1, 0, 0).
EXERCISE
Let F! and F2 be two forces, the sum of whose moments about
a point 0 is 0. Show that Ft, F2, and 0 lie in a plane.
11. Moment of a Vector about a Line. Let a line, L, and a
vector, F, be given. Let L' be the line of F, and let 0, Of be the
points of L and L' nearest together. Let r be the vector from
0 to 0', and let | r | = h. Let a be
a unit vector along L. Assume co-
z ordinate axes as shown. Then
° r L^r ° By the moment of F about L is meant :
FlG- 46 M = hY a,
where a = k. The moment M can be expressed in invariant
form as follows. Since
r = hi,
we have :
r X F =-
k • (r X F) = hY.
Hence
(1) M = {a-(rXF))a.
More generally, r may be any vector drawn from a point of L
to a point of L'. We have thus arrived at the following result.
STATICS OF A RIGID BODY 41
The moment of a vector F about a line L is given by the formula :
M = {a- (r X F)}a,
where r is any vector drawn from a point of L to a point of the line
of F, and a is a unit vector collinear with L and having the sign
attributed to L.
In particular, the moments of F about the three axes are re-
spectively :
yZ - zY, zX - xZ, xY - yX.
EXERCISE
A force of 12 kgs. acts at the point (— 1, 3, — 2), and its direc-
tion cosines arc (—3, 4, — 12). Find its moment about the
principal diagonal of the unit cube; i.e. the line through the
origin, making equal angles with the positive axes.
12. Equilibrium. In § 8 we havo obtained a necessary and
sufficient condition for the equilibrium of n forces, F,, • • • , Fn,
in terms of the vanishing of the resultant force and the resultant
couple. By means of Equations (6) and (7) of § 10 we can formu-
late these conditions analytically. The first, namely, R = 0,
gives :
and now the second, namely, C = 0, reduces Equation (7) to the
vanishing of the first term on the right, or
(2) 2 (yk Zk - zk Yk) =0, 2 (zk Xk - xk Zk) = 0,
This last condition, which was obtained from the vanishing of
a couple, admits two further interpretations in terms of the
vanishing of vector moments, namely :
i) The sum of the vector moments of the given forces with
respect to an arbitrary point 0 of space is 0.
ii) The sum of the vector moments of the given forces about
an arbitrary line of space is 0.
The condition ii) is equivalent to the following :
iif) The sum of the vector moments of the given forces about
each of three particular non-complanar lines is 0.
42 MECHANICS
Necessary and Sufficient Conditions. It is important for clear-
ness to analyse these conditions further, as to whether they are
necessary or sufficient or both.
Condition i), regarded as a necessary condition, is broadest when
0 is taken as any point of space. But Condition ii) is sufficient
if it holds for the lines through just one particular point 0, the
condition (1) being fulfilled.
Condition ii), regarded as a necessary condition, is broadest
when the line is taken as any line in space. But as a sufficient
condition, though true as formulated, it is less general than
(1) and Condition ii'), which may, therefore, be taken as the
broadest formulation of the sufficient condition.
EXERCISES
1. Show that Condition i) is sufficient for equilibrium.
2. Show that Condition ii) is sufficient for equilibrium.
13. Centre of Gravity of n Particles. Let the n particles
^i, • • • , wn be acted on by gravity. Thus n parallel forces arise,
and since they have the same sense, they have a resultant not 0.
Let the axis of z be vertical and directed downward. Then
the resultant is a force directed downward and of magnitude
n
(1) R = Wj g + • • • + mn g = g 5) mk,
the resultant axis being vertical. Furthermore, Xk = 0, Yk = 0,
Zk = mkg. Thus, §10, (11):
n n
(2) L = g V mk yk, M = - g V mk xk, N = 0.
t-l *-i
The nearest point of the resultant axis to the origin has the coordi-
nates given by (15), § 10 :
(3) * = *=5-, „.,_*£*> .-f-0.
If any point of this line is sustained, the system of particles
(thought of as rigidly connected) will be supported, and the
system will remain at rest. In particular, one point on this line
has the coordinates :
STATICS OF A RIGID BODY 43
If, secondly, we allow gravity to act parallel to the axis of x,
the resultant axis now becomes parallel to that axis, and the nearest
point to the origin is found by advancing the letters cyclically
in Equations (3). Again the point whose coordinates are given
by (4) lies on this axis. And, similarly, when gravity acts parallel
to the axis of y. It seems plausible, then, that if this point be
supported, the system will be at rest, no matter in what direction
gravity acts. This is, in fact, the case. To prove the statement,
let the point P of § 10 be taken as (x, y, 0). Then C = 0. For
mka mkp mk y
i j k
X ij 2
S nik y
since the coefficient of each of the unit vectors i, j, k is seen at
once to vanish, no matter what values a, 0, 7 may have.
Thus the existence of a centre of gravity for n particles is
established. It is a point such that, no matter how the system
be oriented, the resultant couple due to gravity is nil.
14. Three Forces. If three non-vanishing forces, acting on
a rigid body, are in equilibrium, they lie in a plane and either pass
through a point or are parallel.
Proof. If two forces in space are in equilibrium, they must be
equal and opposite, and have the same line of action ; or else each
must vanish. Exclude the latter case as trivial. Take vector
moments about an arbitrary point, O, in the line of action of one
of the forces. Then the vector moment of the other force must
vanish by § 12. Thus the second force either vanishes or passes
through 0 ; i.e. through every point of the line of action of the
first force. Finally, they must be equal arid opposite.
In the case of three forces, no one of which vanishes, and no
two of which have the same line of action, take vector moments
about a point 0 in the line of action of the first force, but of no
other force. The sum of the second and third vector moments
about 0 must be zero. Hence the second and third forces lie in
a plane through 0. They are, therefore, equivalent to a single
force — a couple is impossible, since it could not be balanced by the
first force. Thus the first force reduces to the resultant, reversed
in sense, of the second and third forces, and the theorem is proved.
44
MECHANICS
A Trigonometric Theorem. The following trigonometric theorem
is useful in many problems of the equilibrium of a body acted on
by three forces. Let a line be drawn from the vertex of a triangle,
dividing the opposite side into two segments of lengths m and n,
and making angles 6 and <p with these sides. Then
(m + n) cot if/ — m cot 6 — n cot <p,
where \l/ is the angle this line makes with the
segment n.
The proof is immediate. Project the sides
of the triangle on this line, produced :
(m + n) cos \(/ = a cos 0 — 6 cos <pt
and then apply the law of sines :
m b n
FIG. 47
—
sin i
sin0'
sn
sn <p
Example 1. A uniform rod of length 2a is held by a string
of length 21 attached to one end of the rod arid to a peg in a
smooth vertical wall, the other end of the rod
resting against the wall. Find all the positions
of equilibrium.
The three forces of W, T, and R must pass
through a point, and this must be the mid-point
of the string. Hence, applying the above trigo-
nometric theorem to either of the triangles ABO
or ABC, we have :
(1) 2 tan 6 = tan <p.
A second relation is obtained from purely geometrical consider-
ations, namely : *
(2) I cos 0 = 2a cos <p.
It remains to solve these equations. Squaring (1) and reducing,
we have :
4 sec2 6 = 3 + sec2 <p,
or:
4 cos2 <p
cos2 0 =
1+3 cos2 <p
* It would be possible to use the geometric relation
/ sin 0 = a sin <p.
But the further computation of the solution would be less simple.
STATICS OF A RIGID BODY 45
Combining with (2), we get
4Z2 cos2 <p
= 4a2 cos2 t
1 + 3 cos2 <p
Since cos <p cannot vanish, it follows that
cos <p =
But a and I are not unrestricted, for 0 < cos ? < 1. Hence
and so a < I < 2a,
or, the string must be longer than the rod, but not twice as long.
Furthermore, there are always two positions of equilibrium, in
which the rod is vertical, regardless of I and a.
Remark. What the trigonometric theorem has done for us is
to eliminate the forces. Without it, we should have been obliged
to write down two or three equations involving T and R, and then
eliminate these unknowns, with which we have no concern so
far as the position of equilibrium goes.
Example 2. Suppose that, in the last example, the wall is
rough. Then there is, in addition, an upward force of friction,
F — /xft, making four forces in all, — when the rod is just on
the point of slipping down the wall. But the
forces R and F can be compounded into a single
force S making an angle \ with the normal to
the wall, and so the problem is reduced to a
three-force problem. Applying the trigonometric
theorem to the triangle ABC we find :
2a cot <p = a cot 6 — a tan \,
(3) 2 cot <p = cot 0 - /z.
It is better here to take the geometric relation in the form :
(4) I sin 8 = a sin (p.
From (3) we now have :
esc2 0 = 4 cot2 <p + 4/i cot <p + M2 + 1.
Hence
72
Z2 sin2 0 = -A — -0 - :— : - : - ; — r-rr = «2 sin2
4 cot2 99 + 4/i cot <p + M 2 + 1
46
MECHANICS
This last equation can be given the form :
Z2
4 cos2 <p + 4/i cos <p sin <p + (1 + /j2) sin2 <p = —
This equation, in turn, could be reduced to a quartic in sin <p
or cos <p ; but such procedure would be bad technique. Rather,
let
2 cos2 <p = 1 + cos 2y?, 2 cos <p sin v? = sin 2<p,
2 sin2 p = 1 — cos 2<p.
The equation is thus reduced to an equation of the form :
A cos 2<p + B sin 2^ = C,
and now can be solved by the method of Chapter I, § 6.
EXERCISE
1. Complete the study of Example 2, i) computing A, B, C,
•and ii) finding when the rod is just on the point of slipping up.
(K2
- J - 5 - M2 ;
ii) the same equation with the sign of M reversed.
2. If a = 1, I = If, p, = 0.1, find all positions of equilibrium.
EXERCISES ON CHAPTER II*
1. Show that, in a tackle and fall which has n pulleys in each
block, the power, P, exerted is 2n + 1 times
the tension in the rope.
2. What force applied horizontally to the
hub of a wheel (at rest) will just cause the wheel
to surmount an obstacle of height h?
3. Two heavy beads of the same weight can
slide on a rough horizontal rod. To the bead
is attached a string that carries a smooth heavy
bead. How far apart can the beads on the
rod be placed if they are to remain at rest when
FIG. 50 released ?
4. A gate is raised on its hinges and does not fall back. How
rough are the hinges?
* Begin each problem by drawing a figure showing the forces, and the lengths
and angles which enter.
STATICS OF A RIGID BODY 47
5. There has been a light fall of snow on the gate. A cat
weighing 5 Ibs. walks along the top of the gate, and the gate
drops. The disconcerted cat springs off. It is observed from
her tracks in the snow that she reached a point 2 ft. from the
end of the gate. The distance between the hinges is 2^- ft.,
and the centre of gravity of the gate is 5 ft. from the vertical
line through the hinges. If the gate weighs 100 Ibs., what is
the value of /z?
6. A rod rests in a smooth hemispherical bowl, one end inside
the bowl and the rim of the bowl in contact with the rod. Find
the position of equilibrium. a -f V32r2 + a2
Ans. cos 6 = ,
or
where the radius of the bowl is r, the distance of the centre
of gravity of the rod from its lower end is a, and the inclina-
tion of the rod to the horizon, 0, provided a < 2r.
7. A uniform rod rests with one end on a rough floor and the
other end on a smooth plane inclined to the horizon at an angle a.
Find all positions of equilibrium.
8. The same problem where both floor and plane are rough.
9. A picture hangs on a wall. Formulate the problem of
equilibrium when the wall is smooth, and solve it.
10. The same question where the wall is rough.
11. A smooth rod rests with one end against a vertical wall,
a peg distant h from the wall supporting the rod. Find the
position of equilibrium. [ft
Ans. cos 0 = \ —
* a
12. The same problem where the wall is rough, the peg being
smooth. Find all positions of equilibrium.
13. A barrel is lying on its side. A board is laid on the barrel,
with its lower end resting on the floor. Find all positions of
equilibrium. (Barrel, floor, and board are all rough.)
14. A plank 8 ft. long is stood up against a carpenter's work-
bench, which is 2 ft. 8 in. high. The coefficient of friction between
either the floor or the bench and the plank is £. If the plank
makes an angle of 15° with the vertical, will it slip down when
let go?
48 MECHANICS
15. A smooth uniform rod rests in a test-tube. Find the
position of equilibrium.
Ans. The solution is given by the equations :
2 tan 0 = cot ^>, r sin ^ + r = 2a cos 0.
16. A uniform rod 2 ft. long rests with one end on a rough
table. To the other end of the rod is attached a string 1 ft. long,
made fast to a peg 2 ft. above the table. Find all positions of
equilibrium.
Ans. One system of limiting positions is given for ju = 2
by the equations :
cot <p = 2 + 2 cot 0, 2 cos 6 + cos <p = 2.
Solve these equations by means of the Method of Successive
Approximations.
17. A water tower is 100 ft. high and 100 ft. in diameter. Find
approximately the tension in the plates near the base.
18. Water is gradually poured into a tumbler. Show that the
centre of gravity of the glass and the water is lowest when it is in
the surface of the water.
19. If one attempts to pull out a two-handled drawer by one
handle, what is the condition that the drawer will stick fast ?
CHAPTER III
MOTION OF A PARTICLE
1. Rectilinear Motion.* Tube simplest case of motion of mat-
ter under the action of force isHhat in which a rigid body moves
without rotation, each point of the body describing a right line,
and the forces that act being resolved along that line. Consider,
for example, a train of cars, and neglect the rotation of the wheels
and axles. The train is moved by the draw-bar pull of the loco-
motive, and the motion is resisted by the friction of the tracks
and the wind pressure. Obviously, it is only the components of
the forces parallel to the tracks that count, and the problem of
Dynamics, or Kinetics, as it is more specifically called, is to
determine the relation between the forces and the motion; or,
if one will: — Given the forces, to find the distance traversed
as a function of the time.
A more conventional example, coming nearer to possible experi-
mentation in the laboratory, would be that of a block of iron
* The student must not feel obliged to finish this chapter before going on. What
is needed is a thorough drill in the treatment of the early problems by the present
methods, for these are the general methods of Mechanics, to inculcate which is
a prime object of this book. Elementary text-books in Physics sometimes write
down three equations :
* = ^at2, v — at, t?2 = 2as,
and give an unconscionable number of problems to be solved by this device. The
pedagogy of this procedure is totally wrong, since it replaces ideas by a rule of
thumb, and even this rule is badly chosen, since it disguises, instead of revealing,
the mechanical intuition. Now, a feeling for Mechanics is the great object to
be obtained, and the habits of thought which promote such intuition are, fortu-
nately, cultivated by just the same mathematical treatment which applies in the
more advanced parts of Mechanics. It is a happy circumstance that here there
is no conflict, but the closest union, between the physics of the subject and the
mathematical analysis.
A thorough study of §§ 1-12 through working each problem by the present general
methods is most important. Moreover, § 22 should here be included with, of
course, the definition of vector acceleration given in § 16 and the statement of
Newton's Second Law in § 17. The student should then turn to Chapter IV, the
most revealing chapter in the whole elementary part of the book, and study it ic
all detail. The remaining sections of the present chapter should be read casually
at an early stage, so as not to impede progress. Ultimately, they are important ;
but they are most useful when the student comes to recognize their importance
through his experience gathered from the later work above referred to.
49
50 MECHANICS
placed on a table and drawn along by cords, so applied that the
block does not rotate and that each point of it describes a right
line — with varying velocity.
It is clear that a block of platinum having the same mass,
i.e. containing the same amount of matter, if acted on by the
same forces, would move just like the block of iron, if the two were
started side by side from rest or with the same initial velocities.
We can conceive physical substances of still greater density, and
the same would be true. On compressing the given amount of
matter into smaller and ever smaller volume, we are led to the
idea of a particle, or material point, i.e. a geometrical point, to
which the property of mass is attached. This conception has
the advantage that such a particle would move exactly as the
actual body does if acted on by the same forces; but we need
say nothing about rotation, since this idea does not enter when
we consider only particles. Moreover, there is no doubt about
where the forces are applied — they must be applied at the one
point, the particle.
2. Newton's Laws of Motion. Sir Isaac Newton (1642-1727),
who was one of the chief founders of the Calculus, stated three
laws governing the motion of a body.
FIRST LAW. A body at rest remains at rest and a body in motion
moves in a straight line with unchanging velocity, unless some external
force acts on it.
SECOND LAW. The rate of change of the momentum of a body
is proportional to the resultant external force that acts on the body.
THIRD LAW. Action and reaction are equal and opposite.
The meaning of the First Law is clear enough, if we restrict
ourselves for the present to bodies and particles as described and
moving in § 1.* The Third Law, too, is self-explanatory. Con-
sider, for example, two particles of unequal mass, connected by
a spring, the mass of which is negligible. Then the pull (or push)
of the spring on the one particle is equal and opposite to its pull
(or push) on the other particle.
The Second Law is expressed in terms of momentum, and the
momentum of a particle is defined as the product of its mass by
*We might consider, furthermore, such material distributions as laminae, i.e.
material surfaces; and also wires, or material curves. Finally, rigid combinations
of all these bodies.
MOTION OF A PARTICLE 51
its velocity, or mv. Here, v is not an essentially positive quan-
tity — the mere speed. We must think of the position of the
body as described by a suitable coordinate, s. The latter may
be the distance actually traversed by the particle; or we may
think of the path of the particle as the axis of x, and s as the
coordinate of the particle. The velocity, v, will then be defined
as ds/dt :
and is positive when s is increasing ; negative, when s is decreasing.
The Second Law can now be stated in the form :
(2) ««t.
Here, / denotes the resultant force, and is positive when it tends
to increase s ; negative, when it tends to decrease s.
Ordinarily, m is constant — always, in the case of the bodies
cited above — and so
d (mv) _ dv
~dT ~m'dt
The quantity dv/dt is defined as the acceleration, and is often
represented by a :
It is positive when v is increasing, negative when v is decreasing.
Newton's Second Law can now be stated in the form : The mass
times the acceleration is proportional to the force :
(4) ma oc /.
From the proportion we now pass to an equation :
(5) ma = X/,
where X is a physical constant. The value of X depends on the
units used. If these are the English units, the pound being the
unit of mass, the foot the unit of length, the second the unit of time,
and the pound the gravitational unit of force, then X has the
value 32 (or, more precisely, 32.2), and Newton's Second Law of
Motion becomes here :
A,) m
52 MECHANICS
In the decimal system, the gramme being the unit of mass, the
centimetre the unit of length, the second the unit of time, and the
gramme the gravitational unit of force, X = 981, and Newton's
Second Law of Motion becomes here :
A2) mft = 981 /.
In § 3 we shall discuss the absolute units. In particular, the
units of mass, length, and time having been chosen arbitrarily,
as in Physics, the so-called "absolute unit of force " is that unit
which makes X = 1 in Newton's Equation, so that here :
A \ dv .
A,) mdi=f-
Three Forms for the Acceleration. The acceleration is defined
as dv/dt, and since v = ds/dt, we have :
x/2«
/£!\ U *
(6) a = ^
A third form is obtained by starting with the equation :
(7\ dv _ dsdv
( ' dt ~ di ds
and then replacing ds/dt by its value, v. Thus
(8) . - .*.
These three forms for the acceleration :
dv dzs dv
connect the three letters s, t, v in pairs in all possible ways. Which
form it is better to use in a given case, will become clear from
practice in solving problems.
Example 1. A freight train weighing 200 tons is drawn by
a locomotive that exerts a draw-
jt PP-I /==8000t bar pull of 9 tons. 5 tons of this
s force are expended in overcoming
frictional resistances. How much
speed will the train have acquired at the end of a minute, if it
starts from rest?
MOTION OF A PARTICLE 63
Here we have
m - 200 X 2000 = 400,000 Ibs.,
/ = 9 X 2000 - 5 X 2000 = 8000 Ibs.*
and hence Equation A,) becomes :
400,000^ = 32X8000,
or — = — •
dt 25
Integrating with respect to t, we find :
V == Tjngt ~f~ C .
Since v = 0 when t = 0, we must have C — 0, and hence
At the end of a minute, t = 60, and so
^ = M X 60 = 38-4 ft- Per sec.
To reduce feet per second to miles per hour it is convenient
to notice that 30 miles an hour is equivalent to 44 ft. a second,
as the student can readily verify; or roughly, 2 miles an hour
corresponds to 3 ft. a second. Hence the speed in the present
case is about two-thirds of 38.4, or 26 miles an hour.
Example 2. A stone is sent gliding over the ice with an initial
velocity of 30 ft. a sec. If the coefficient of friction between the
stone and the ice is -fa, how far will the stone go ?
Here, the only force that we take account of is the retarding
force of friction, and this amounts to one-tenth of a pound of force
for every pound of mass there is m
in the stone. Hence, if there are — 4 \ 10-q
m pounds of mass in the stone s
the force will be ^m lbs.,f and FlG* 52
since it tends to decrease s, it is to be taken as negative:
* The student must distinguish carefully between the two meanings of the word
pound, namely (a) a mass, and (6) a force — two totally different physical objects.
Thus a pound of lead is a certain quantity of matter. If it is hung up by a string,
the tension in the string is a pound of force.
t The student should notice that m is neither a mass nor a force, but a number,
like all the other letters of Algebra, the Calculus, and Physics.
54 MECHANICS
Now what we want is a relation between v and s, for the ques-
tion is: How far (s = ?), when the stone stops (v = 0)? So we
use the value (8) of a and thus obtain the equation :
dv 16
Vds=~-5>
or v dv = — ±/-ds.
rr "2 16 . n
Hence T> = — ^- s + C.
To determine C we have the data that, when s = 0, v = 30.
Since in particular the equation must hold for these values,
^! = 0 + C, C = 450,
and so v2 = 900 - ^s.
When the stone stops, v = 0, and we have
0 = 900 - 3£s, s = 141 ft.
EXERCISES *
1. If the train of Example 1 was moving at the rate of 4 m.
an hour when we began to take notice, how fast would it be mov-
ing half a minute later? Give a complete solution, beginning with
drawing the figure. Ans. About 17 m. an h.
2. A small boy sees a slide on the ice ahead, and runs for it.
He reaches it with a speed of 8 miles an hour and slides 15 feet.
How rough are his shoes? Ans. M = .15.
3. Show that, if the coefficient of friction between a sprinter's
shoes and the track is TV> n^ ^cs^ possible record in a hundred-
yard dash cannot be less than 1 5 seconds.
4. An electric car weighing 12 tons gets up a speed of 15 miles
an hour in 10 seconds. Find the average force that acts on it,
* It is important that the student should work these exercises by the method
set forth in th4e text, beginning each time by drawing a figure and marking (t) the
force, by means of a directed right line, or vector, drawn preferably in red ink ;
and (t'i) the coordinate used, as s or x, etc. He should not try to adapt such formulas
of Elementary Physics as
v = at, a = £a£2, vz = 2as
to present purposes. For, tho object of these simple exercises is to prepare the way
for applications in which the force ia not constant, and here the formulas just cited
do not hold.
MOTION OF A PARTICLE 55
i.e. the constant force which would produce the same velocity in
the same time.
6. In the preceding problem, assume that the given speed is
acquired after running 200 feet. Find the time required and
the average force.
6. A train weighing 500 tons and running at the rate of 30 miles
an hour is brought to rest by the brakes after running 600 feet.
While it is being stopped it passes over a bridge. Find the force
with which the bridge pulls on its anchorage. Ans. 25.2 tons.
7. An electric car is starting on an icy track. The wheels
skid and it takes the car 15 seconds to get up a speed of two miles
an hour. Compute the coefficient of friction between the wheels
and the track.
3. Absolute Units of Force. The units in terms of which we
measure mass, space, time, and force arc arbitrary, as was pointed
out in § 2. If we change one of them, we thereby change the
value of X in Newton's Second Law. Consequently, by changing
the unit of force properly, the units of mass, space, and time being
held fast, we can make X = 1. Hence the definition above given:
DEFINITION. The absolute unit of force _is that unit which
makes X = 1 in Newton's_ Second Law of Motion ij*
(1) moT^f.
In order to determine experimentally the absolute unit of force,
we may allow a body to fall freely and observe how far it goes in
a known time. It is a physical law that the force with which
gravity attracts any body is proportional to the mass of that body.
Let the number g be the number of absolute units of force with
* We have already met a precisely similar question twice in the Calculus. In
differentiating the function sin x we obtain the formula
Dx sin x = cos x
only when we measure angles in radians. Otherwise the formula reads:
Dx sin x — X cos x.
In particular, if the unit is a degree, X = Tr/180. We may, therefore, define a radian
as follows : The absolute unit of angle (the radian) is that unit which makes X = 1
in the above equation.
Again, in differentiating the logarithm, we found ;
X
This multiplier reduces to unity when we take a = e. Hence the definition :
The absolute (natural) base of logarithms is that base which makes the multiplier
logo e in the above equation equal to unity.
56 MECHANICS
which gravity attracts the unit of mass. Then the force, measured
in absolute units, with which gravity attracts a body of m units
of mass will be mg. Newton's Second Law A3) gives for this case :
dv , dv
ds
s = %gt* + K, K = 0,
and we have the law for freely falling bodies deduced directly
from Newton's Second Law of Motion, the hypothesis being
merely that the force of gravity is constant. Substituting in
the last equation the observed values s = S, t = T, we get :
28
9 = ?*'
If we use English units for mass, space, and time, g has, to
two significant figures, the value 32, i.e. the absolute unit of
force in this system, a poundal, is equal nearly to half an ounce.
If we use c.g.s. units, g ranges from 978 to 983 at different parts
of the earth, and has in Cambridge the value 980. The absolute
unit of force in this system is called the dyne.
Since g is equal to the acceleration with which a body falls
freely under the attraction of gravity, g is called the acceleration
of gravity. But this is not our definition of g ; it is a theorem
about g that follows from Newton's Second Law of Motion.
The student can now readily prove the following theorem,
which is often taken as the definition of the absolute unit of
force in elementary physics : The absolute unit of force is that
force which, acting on the unit of mass for the unit of time, gener-
ates the unit of velocity.
Incidentally we have obtained two of the equations for a freely
falling body :
v = gt, s = %gt2.
The third is found by setting a = v dv/ds and integrating :
dv
2gs.
MOTION OF A PARTICLE
57
Example. A body is projected down a rough inclined plane
with an initial velocity of VQ feet per second. Determine the
motion completely.
The forces which act are : the component of gravity, mg sin 7
absolute units, down the plane, and the force of friction, pR =
nmg cos 7 up the plane. Hence
ma = mg sin 7 — nmg cos 7,
dv
-IT = g sm 7 —
cos 7.
Integrating this equation, we
get
v = g (sin 7 — /z cos 7) t + C,
»o = 0 + <?>
— = g (sin 7 -
cos 7) £ + v0.
A second integration gives
B) s = \g (sin 7 - ^ cos 7) P + VQ t,
the constant of integration here being 0.
To find v in terms of s we may eliminate t between A) and
B). Or we can begin by using formula (8), § 2, for the acceler-
ation :
dv f . ,
v-r = g (sin 7 - /x cos 7),
do
%v2 = g (sin 7 — M cos 7) s + K,
^ = 0 + X,
t;2 = 20 (sin 7 — /z cos 7) s + #o-
EXERCISES
1. If, in the example discussed in the text, the body is pro-
jected up the plane, find how far it will go up.
2. Determine the time it takes the body in Question 1 to
reach the highest point.
3. Obtain the usual formulas for the motion of a body pro-
jected vertically :
v2 = 2gs + vl or = — 2gs + vl ;
v = gt + VQ or = - gt + VQ ;
8 = ot* + v0t or =-tf* + M.
58 MECHANICS
4. On the surface of the moon a pound weighs only one-sixth
as much as on the surface of the earth. If a mouse can jump
up 1 foot on the surface of the earth, how high could she jump
on the surface of the moon? Compare the time she is in the air
in the two cases.
6. A block of iron weighing 100 pounds rests on a smooth
table. A cord, attached to the iron, runs over a smooth pulley
at the edge of the table and carries a weight of 15 pounds, which
hangs vertically. The system is released with the iron 10 feet
from the pulley. How long will it be before the iron reaches the
pulley, and how fast will it be moving?
Ans. 2.19 sec. ; 9.1 ft. a sec.
6. Solve the same problem on the assumption that the table is
rough, n = ^, and that the pulley exerts a constant retarding
force of 4 ounces.
7. Regarding the big locomotive exhibited at the World's
Fair in 1905 by the Baltimore and Ohio Railroad the Scientific
American said : "Previous to sending the engine to St. Louis, the
engine was tested at Schenectady, where she took a 63-car train
weighing 3150 tons up a one-per-cent. grade."
Find how long it would take the engine to develop a speed
of 15 m. per h. in the same train on the level, starting from rest,
the draw-bar pull being assumed to be the same as on the grade.
8. If Sir Isaac Newton registered 170 pounds on a spring
balance in an elevator at rest, and if, when the elevator was
moving, he weighed only 169 pounds, what inference would he
draw about the motion of the elevator?
9. What does a man whose weight is 180 pounds weigh in an
elevator that is descending with an acceleration of 2 feet per
second per second ?
4. Elastic Strings. When an elastic string is stretched by a
moderate amount, the tension, T, in the string is proportional
to the stretching, i.e. to the difference, s, between the stretched
and the unstretched length of the string :
(1) T oc s, or T = ks,
where fc is a physical constant, whose value depends both on the
particular string and on the units used.
MOTION OF A PARTICLE 59
Suppose, for example, that a string is stretched 6 in. by a force
of 12 Ibs. ; to determine k. If we measure the force in gravitar
tional units, i.e. pounds, then
T = 12 when s = £.
Hence, substituting these values in equation (1), we have:
12 = &£, or k = 24,
(2) T = 24s.
If we had chosen to measure the force in absolute units, i.e.
poundals, then, since it takes (nearly) 32 of these units to make
a pound, the given force of 12 pounds would be expressed as
(nearly) 12 X 32, or precisely 120, poundals. Hence, substitut-
ing the present value of the force in (1), which, to avoid con-
fusion, we will now write in the form :
T = k's,
we have : 120 = k' \ or k' = 240,
(3) T' = 240s.
When the string is stretched 1 in., s = ^ and the tension
as given by (2) is T = 2, i.e. 2 pounds. Formula (3), on the
other hand, gives 20, or 64 (nearly) as the value of the ten-
sion, expressed in terms of poundals, and this is right; for it
takes 64 half-ounces to make 2 pounds, and so we should have
T' = 20.*
The law of strings stated above is familiar to the student in the
form of Hooke's Law:
rr\ __.
I '
where I is the natural, or unstretched, length of the string, and
lf, the stretched length; the coefficient E being Young's Mod-
ulus. For a given string, E/l = k is constant, and V — I = s is
variable.
* It is easy to check an answer in any numerical case. The student has only
to ask himself the question: "Have I expressed my force in pounds, or have I
expressed it in terms of half-ounces?" Just as five dollars is expressed by the
number 5 when we use the dollar as the unit, but by the number 500 when we
use the cent, so, generally, the smaller the unit, the larger the number which expresses
a given quantity.
60 MECHANICS
EXERCISES
1. An clastic string is stretched 2 in. by a force of .3 Ibs. Find
the tension (a) in pounds; (b) in poundals, when it is stretched
s ft. Ans. (a) T = 18s ; (6) T = 180s.
2. When the string of Question 1 is stretched 4 in., what is
the tension (a) in terms of gravitational units; (b) in terms of
absolute units? Ans. (a) 6 pounds; (6) 192 poundals.
3. An elastic string is stretched 1 cm. by a force of 100 grs.
Find the tension (a) in grs. ; (6) in dynes, when it is stretched
s cm. Ans. (a) 100s; (b) 98,000s.
4. One end of an elastic string 3 ft. long is fastened to a peg
at A, and a 2-pound weight is attached to the other end. The
weight is gradually lowered till it is just supported by the string,
and it is found that the length of the string has thus been doubled.
Find the tension in the string when it is stretched s ft.
Ans. f s Ibs. ; ^s poundals.
5. A Problem of Motion. One end of the string considered
in the text of § 4 is fastened to a peg at a point 0 of a smooth
horizontal table ; a weight of 3 Ibs. is attached to the other end
of the string and released from rest on the table with the string
.stretched one foot. How fast will the weight be moving when
the string becomes slack?
The weight evidently describes a straight line from the starting
point, A, toward the peg 0, and we wish to know its velocity
when it has reached a point B, one foot from A.
The solution is based on Newton's Second Law of Motion.
It is convenient here to take as the coordinate, not the distance
AP that the particle has travelled at
- - • - < - ' - -j any instant, but its distance s from B.
T-, KA The force which acts is the tension
riG. 54 i.i
of the string; measured in absolute
units it is 240rs. Since it tends to decrease s, it is negative.
Hence Newton's Law becomes :
(1)
(2)
___ fJ2 a tit)
To integrate this equation, replace -^ by its value v -=- :
MOTION OF A PARTICLE 61
Hence v dv = —
(3)
I v dv = — 8g I sds,
To determine C, observe that initially, i.e. when the particle
was released at A, v = 0 and s = I. Hence
0 = - 4(7 + C, C = 40,
and (3) becomes
(4) t;2 = 80 (1 - s2).
We have now determined the velocity of the particle at an
arbitrary point of its path, and thus are in a position to find its
velocity at the one point specified in the question proposed,
namely, at B. Here, s = 0, and
v2Uo = 8g = 8 X 32, v\s=sQ = 16 (ft. per sec.)
EXERCISES *
1. The weight in the problem just discussed is projected from
B along the table in the direction of OB produced with a velocity
of 8 ft. per sec. Find how far it will go before it begins to return.
Ans. Newton's equation is the same as before, and the
integral, (3), is the same ; but initially s = 0 and v = 8.
Hence C = 32, and the answer is 6 inches.
2. If, in the example worked in the text, the table is rough
and the coefficient of friction, /i, has the value ^, how fast will
the body be moving when it reaches B ?
Ans. Newton's equation now becomes :
3^=-24g8 + i-3g,
and the answer is : 4 Vl5 = 15.49 ft. per sec.
3. Solve the problem of Question 1, for a rough table, M = T-
Ans. The required distance is the positive root of the
equation 16s2 + s — 4 = 0, or s = .4698 ft., or about
5fin.
* In the following exercises and examples, it will be convenient to take the value
of g as exactly 32 when English units are used. Begin each exercise by drawing a
figure showing the coordinate used, and mark the forces in red ink.
62 MECHANICS
4. Find where the weight in Question 2 will come to rest
if the string, after becoming slack, does not get in the way.
6. The 2 Ib. weight of Question 4, § 4, is released from rest
at a point B directly under the peg .4 and at a distance of 3 ft.
from A ; the string thus being taut, but not stretched. Find
how far it will fall before it begins to rise. Ans. 6 ft.
6. If, in the last question, the weight is dropped from the
peg at A, find how far it descends before it begins to rise.
Ans. To a distance of 6 + 3\/3 = 11.196 ft. below A.
7. If the weight in the last two questions is carried to a point
7 ft. below A and released, show that it will rise to a distance of
5 ft. below A before beginning to fall.
8. If, in the last question, the weight is released from a point
10 ft. below A, show; that it will rise to a height of 1 ft. and 10 in.
below A.
9. The string of the example studied in the text of § 4 is
placed on a smooth inclined plane making an angle of 30° with
the horizon, and one end is made fast to a peg at A in the plane.
If a weight of 1£ Ibs. be attached to the other end of the string
and released from rest at A, find how far down the plane it will
slide. Assume the unstretched length of the string to be 4 ft.
10. The same question if the plane is rough, /* = ^V3.
11. A cylindrical spar buoy (specific gravity ^) is anchored
so that it is just submerged at high water. If the cable should
break at high tide, show that the spar would jump entirely out
of the water.
Assume that the buoyancy of the water is always just equal
to the weight of water displaced.
12. A particle of mass 2 Ibs. lies on a rough horizontal table,
and is fastened to a post by an elastic band whose unstretched
length is 10 inches. The coefficient of friction is -£, and the band
is doubled in length by hanging it vertically with the weight at
its lower end. If the particle be drawn out to a distance of
15 inches from the post and then projected directly away from
the post with an initial velocity of 5 ft. a sec., find where it will
stop for good.
MOTION OF A PARTICLE 63
6. Continuation ; the Time. The time required by the body
whose motion was studied in § 5 to reach the point B can be
found as follows. From equation (4) we have :
(5) v = -^ = ± V8g Vl — s2.
Since s decreases as t increases, ds/dt is negative, and the lower
sign holds. Replacing V8g by its value, 16, we see that
(6)
This differential equation is readily solved by separating the
variables, i.e. by transforming the equation so that only the
variable s occurs on one side of the new equation, and only t on
the other ; thus
(7) I6dt = -
Hence 16* = - f , ds = - sin-1 s + C.
= - f , ds = - si
J Vl - s2
If we measure the time from the instant when the body was
released at A, then t = 0 arid s = 1 are the initial values which
determine C :
0 = - sin-1 1 + C, C = ~
Thus l&t = - sin-1 s.
The right-hand side of this equation has the value cos""a«.
Hence we have, as the final result,*
(8) 16t = cos"1 s, or s = cos 16$.
* In evaluating the above integral we might equally well have used the formula
Vl - «2
We should then have had :
16* = cos-1 8 - C'.
Substituting the initial values t - 0, * = 1 in this equation, we find :
0 = cos'1 1 - C", or C" - 0,
and the final result is the same as before.
64 MECHANICS
This equation gives the time it takes the body to reach an
arbitrary point of its path. In particular, the time from A to
B is found by putting s = 0 :
(9) W = cos-1 0 = |, t = J2 = .09818 sec.
EXERCISES
1. Show that if the body, in the case just discussed, had been
released from rest at any other distance from the peg, the string
being stretched, the time to the point at which the string becomes
slack would have been the same.
2. Show that it takes the body twice as long to cover the first
half of its total path as it does to cover the remainder.
Find the time required to cover the entire path in the case
of the following exercises at the close of § 5.
3. Exercise 1. Ans. ^ = .09818.
OA
4. Exercise 5.
Ans. t = \^K / / -', total time, TrA/^ = .9618 sec.
* 04 J v 6s — s2 ^
5. Exercise 6. Ans. t = \— ^ + sin-1 77= =?
6. Exercise 7. Ans. .9618 sec.
7. Exercise 9. 8. Exercise 10. 9. Exercise 8.
7. Simple Harmonic Motion. The simplest and most important
ease of oscillatory motion which occurs in nature is that known
as Simple, Harmonic Motion. It is illustrated with the least
amount of technical detail by the following example, or by the
first Exercise below.
Example. A hole is bored through the centre of the earth, a
stone is inserted, the air is exhausted, and the stone
is released from rest at the surface of the earth.
To determine the motion.
The earth is here considered as a homogeneous
sphere, at rest in space. Its attraction, F, on
^ the stone diminishes as the stone nears the cen-
FIG. 55 tre, and it can be shown to be proportional, at
MOTION OF A PARTICLE 65
any point of the hole, to the distance of the stone from the
centre: „ ~ ,
F oc r, or F = kr.
To determine the constant fc, observe that, at the surface,
r = R (the radius of the earth), and, if we measure F in absolute
units, F = wgr, where m denotes the mass of the stone. Hence
mg = kR or k = ~^,
/£
and F = ^r
R
As the coordinate of the stone we will take its distance, r,
from the centre of the earth. Then Newton's Second Law gives
us:
/i\ d*r mg
(1) m^=-Rr-
For, when r Ls positive, the force tends to decrease r, and so is
negative. When r is negative, the force tends to increase r alge-
braically, and so is positive. Hence (1) is right in all cases.
In order to integrate Equation (1), which can be written in
the form :
(2) <*L=-Lr
w dp Rr>
we employ the device of multiplying through by 2dr/dt:
cydrd^r = _ 2g dr
dtdt*~ RTdt
d /rfr\2
The left-hand side thus becomes -77 ( -n ) • Hence each side
at \dt/
can be integrated with respect to t : *
* This method can be applied to any differential equation of the form :
Multiply through by 2 dy/dx :
a4?
<&c cte2
The left-hand side thus becomes — ( ~ J • Hence
Integrating, we have
66 MECHANICS
--^.Cr^
~ Rrdt
Cd(
Jdt\dt
or
_ - -
dt ~ R - R
To determine (7, observe that initially, i.e. when the stone was
at Ay r = R and the velocity, dr/dt, = 0. Hence
0=~|ft2 + c, or C = !#2.
7£ it
Thus finally :
<3> (I)'- 1 <«'-">•
At the centre of the earth, r = 0, and (dr/dt)2 = gR. If we
take the radius of the earth as 4000 miles, then R = 4000 X 5280,
g = 32, and the velocity is about 26,000 ft. a sec., or approxi-
mately 5 miles a second.
The stone keeps on with diminishing speed and comes to rest
for an instant when r = — J?, i.e. it just reaches the other side
of the earth, and then falls back. Thus it oscillates throughout
the whole length of the hole, reaching the surface at the end of
each excursion, and continuing this motion forever. The result
is not unreasonable, for there is no damping of any sort, — no
friction or air resistance.
The Time. To find the time we proceed as in §6. From
Equation (3) it follows that
Hence, separating the variables, we have :
'" dr
dt = - \— —F====^.
9 VR* - r
or t = ^\|— cctfh1 ~ + C.
I
Initially, t = 0 and r = 72 ; thus (7 = 0, and
(4) t = \— cos-1 ~, or r = R cos (^ VB) '
* <7 /t \ ¥A//
MOTION OF A PARTICLE 67
The time from A to 0 is found by putting r = 0 :
*
-'
On computing the value of this expression it is seen to be 21 min.
and 16 sec. The time from A to B is twice the above. Hence
the time of a complete excursion, from A to B and back to A is
0
This time is known as the period of the oscillation.*
The General Case. Simple Harmonic Motion is always dom-
inated by the differential equation
A\ ** X — _ „% ~
' ~dfi ~ '
where the coordinate x characterizes the displacement from the
position of no force. This equation can be integrated as in the
special case above, and it is found that
B)
where h denotes the value of x which corresponds to the extreme
displacement. The velocity when x = 0 is numerically nh, and
thus is proportional both to n and to h. A second integration
gives
C) x = h cos ntj
provided the time is measured from an instant when x = h.
The period, T, is inversely proportional to n :
^ „ 27T
and the amplitude is 2h. Thus the period is independent of
the amplitude.
The motion represented by Equation C) is known as Simple
Harmonic Motion. The graph of the function is obtained from
* In the first equation (4) the principal value of the anti-cosine holds during
the first passage of the stone from A to B. The second equation (4) holds with-
out restriction.
68 MECHANICS
the graph of the cosine curve by plotting the latter to one scale
on the axis of t, and to another scale on the axis of x.
FIG. 56
EXERCISES
1. Two strings like the one described in the text of § 4 are
fastened, one end of each, to two pegs, A and B, on a smooth
horizontal table, the distance AB being double the length of
either string, and the other end of each string is made fast to
a 3 Ib. weight, which is placed at 0, the mid-point of AB. Thus
each string is taut, but not stretched. The weight being moved
to a point C between 0 and A and then released from rest, show
that it oscillates with simple harmonic motion. Find the velocity
with which it passes 0 and the period of the oscillation. It is
assumed that the string which is slack in no wise interferes with
or influences the motion.
Ans. The differential equation which dominates the motion
d2x
is —ffi = — 256z, where x denotes the displacement of the
at
3 Ib. weight ; hence the motion is simple harmonic motion.
The required velocity is numerically 16h, where h denotes
the maximum displacement. The period is 27T/16 =
.3927 sec.
2. Work the same problem for two strings like the one of
Question 4, § 4, and a 2 Ib. weight.
3. Show that the motion of Example 7, § 5, is simple harmonic
motion, and find the period.
4. If a straight hole were bored through the earth from Boston
to London, a smooth tube containing a letter inserted, the air
exhausted from the tube, and the letter released at Boston, how
long would it take the letter to reach London?
MOTION OF A PARTICLE 69
6. If in the problem of Question 9, § 5, the weight were re-
leased with the string taut, but not stretched, and directed straight
down the plane, show that the weight would execute simple
harmonic motion. Determine the amplitude and the period.
6. Work the problem of the text for the moon; cf. the data
in §8.
7. A steel wire of one square millimeter cross-section is hung
up in Bunker Hill Monument, and a weight of 25 kilogrammes is
fastened to the lower end of the wire and carefully brought to
rest. The weight is then given a slight vertical displacement.
Determine the period of the oscillation.
Given that the force required to double the length of the wire
is 21,000 kilogrammes, and that the length of tho wire is 210 feet.
Ans. A little over half a second.
8. A number of iron weights are attached to one end of a long
round wooden spar, so that, when left to itself, the spar floats
vertically in water. A ten-kilogramme weight having become
accidentally detached, the spar is seen to oscillate with a period
of 4 seconds. The radius of the spar is 10 centimetres. Find
the sum of the weights of the spar and attached iron. Through
what distance docs the spar oscillate ?
Ans. (a) About 125 kilogrammes ; (6) 0.64 metre.
8. Motion under the Attraction of Gravitation. Problem. To
find the velocity which a stone acquires in falling to the earth
from interstellar space.
Assume the earth to be at rest and consider only the \A
force which the earth exerts. Let the stone be re- j
leased from rest at A, and let r be its distance from
the centre 0 of the earth at any subsequent instant.
Then the force, F, acting on it is, by the law of
gravitation, inversely proportional to r :
El A
*-•?
Since F = mg when r = JR, the radius of the earth,
X , r mgR*
mq = -F^ and v = —^ —
70 MECHANICS
Newton's Second Law of Motion here takes on the form :
dzr _ mgR2
m~dT* ~ ~7*~'
Hence
To integrate this equation, we employ the method of § 7 and
multiply by 2 dr/dt :
drd2r 2gR2dr d/dr\* = 2gR*dr
dt dt2 r2 dt' °r di\dt) r2 dt
Integrating with respect to t we find :
Initially dr/dt = 0 and T = Z ; hence
o.' + c, c~
Since dr/dt is numerically equal to the velocity, the velocity
V at the surface of the earth is given by the equation :
If I is very great, the last term in the parenthesis is small, and
so, no matter how great I is, V can never quite equal V2gR.
Here g = 32, R = 4000 X 5280, and hence the velocity in ques-
tion is about 36,000 feet, or 7 miles, a second.
This solution neglects the retarding effect of the atmosphere;
but as the atmosphere is very rare at a height of 50 miles from
the earth's surface, the result is reliable down to a point com-
paratively near the earth.
In qrder to find the time it would take the stone to fall, con-
sider the equation derived from (2) :
Hence
— 2
,
and
MOTION OF A PARTICLE 71
Vl r rdr
Turning to Peirce's Tables, No. 169, we find :
dr
Vlr - r2
= — V lr - r2 + ~ sin-
Thus t = fji
Initially t = 0 and r = I :
Finally, then,
fr-H + s s-sm-
For purposes of computation, a better form of this equation
is the following :
(3)
EXERCISES*
1. If the earth had no atmosphere, with what velocity would
a stone have to be projected from the earth's surface, in order
not to come back?
2. If the moon were stopped in its course, how long would
it take it to fall to the earth? Regard the earth as stationary.
Ans. 4 days, 18 hrs., 10 min.
* In working these exercises, the following data may be used :
Radius of the moon, ^ that of the earth.
Mass of moon, ^T that of earth.
Mean distance of moon from earth, 237,000 miles.
Acceleration of gravity on the surface of the moon, £ that on the surface of the
earth.
Diameter of sun, 860,000 miles.
Mass of sun, 333,000 that of the earth.
Mean distance of earth from sun, 93,000,000 miles.
Acceleration of gravity on the surface of the sun, 905 ft. per sec. per sec.
72 MECHANICS
3. Solve the preceding problem accurately, assuming that the
earth and the moon are released from rest in interstellar space
at their present mean distance apart. Their common centre of
gravity will then remain stationary.
4. The same problem for the earth and the sun.
6. If the earth and the moon were held at rest at their present
mean distance apart, with what velocity would a projectile have
to be shot from the surface of the moon, in order to reach the
earth?
6. If the earth and the moon were held at rest at their present
mean distance apart, and a stone were placed between them at
the point of no force and then slightly displaced toward the earth,
with what velocity would it reach the earth ?
7. If a hole were bored through the centre of the moon, as-
sumed spherical, homogeneous, and at rest in interstellar space,
and a stone dropped in, how long would it take the stone to reach
the other side ?
8. Show that if two spheres, each one foot in diameter and of
density equal to the earth's mean density (specific gravity, 5.6)
were placed with their surfaces £ of an inch apart and were acted
on by no other forces than their mutual attractions, they would
come together in about five minutes and a half. Given that the
spheres attract as if all their mass were concentrated at their
centres.
9. Work Done by a Variable Force. If a force, F, constant
in magnitude and always acting along a fixed line AB in the
same sense, be applied to a particle,* and if the particle be dis-
placed along the line in the direction of the force, the work done
by the force on the particle is defined in elementary physics as
FI W = Fl,
A p B where I denotes the distance through which
•p eo C3
the particle has been displaced.
Suppose, however, that the force is variable, but varying con-
tinuously and always acting along the same fixed line. How
shall the work now be defined ?
*Or, more generally, to one and the same point P of a rigid or deformable
material body.
MOTION OF A PARTICLE 73
Let a coordinate be assumed on the line ; i.e. think of the line
as the axis of x. Let the particle be displaced from A: x = a
to B: x = 6, and let a < b. Let F, to begin with, always act
in the direction of the positive sense along the axis. Then
F=f(x),
where f(x) denotes a positive continuous function of x.
Divide the interval (a, b) up into n parts by the points xl9
xz, • • • , zn_i, and let XQ = a, xn = b. Then, if
xt+i — Xk = Azfc,
the work, ATF*, done by the force in displacing the particle through
the fc-th interval ought, in order to correspond to the general
physical conception of work, to lie between the quantities
FiAx and Fi'Az,
where FJ and F'k' denote respectively the smallest and the largest
values of f(x) in this interval.* We have, then :
(1) FiAx ^ ATFt ^ Fi'te.
On writing out the double inequality (1) for k = 0, I, • • • ,
n — 1 and adding the n relations thus resulting together, we find
that W = 2 AWk lies between the two sums :
(2) F'^x + F(Ax + • • • + n_, Az,
(3) F'Jte + F('&x + • • • + K'^Az.
Each of these sums suggests the sum
(4) /(*0) Az0 + f(Xl) A*! + - • • + /(*„_,) Axn,
whose limit is the definite integral,
ft
(5) lim [/(x0) A* + /(x,) Ax + • • • + /(*._0 Ax] = f /(*) As.
n-oo J
a
That W is in fact equal to this integral :
h
(6) W = Jf(x)dx,
a
follows from DuhamePs Theorem.
* This statement is pure physics. It is the physical axiom on which the general-
ization of the definition of work is based. More precisely, it is one of two physical
axioms, the other being that the total work, W, for the complete interval is the sum
of the partial works, ATT*, for the subintervals.
74 MECHANICS
If the force F acts in the direction opposite to that in which
the point of application is moved, we extend the definition and
say that negative work is done. For the case that F is constant,
the work is now defined as follows :
(7) W = F(b - a).
Here, F is to be taken as a negative number equal numerically
to the intensity of the force.
Thus (7) is seen to hold in whichever direction the force acts,
provided that a < 6. Will (7) still hold if 6 < a? It will.
There are in all four possible cases :
i) + + ii) Hi) H — «0 — h
In cases i) and ii) the force overcomes resistance, and positive
work is done. In cases in) and iv) the force is overcome, and
negative work is done. Hence (7) holds in all cases.
It is now easy to see how the definition of work should be
laid down when F varies in any continuous manner. The consider-
ations are precisely similar to those which led to Equation (6),
and that same equation is the final result in this, the most general,
case : &
W = Cf(x) dx.
a
Example. To find the work done in stretching a wire. Let the
natural (or unstretched) length of the wire be I, the stretched
length, V. Then the tension, T, is
__, r-^-i T> i given by Hooke's Law :
O A P B
FIG. 59 T = \ ~~
where X is independent of I and V, and is known as Young's Mod-
ulus.
Let the wire, in its natural state, lie along the line OA, and
let it, when stretched, lie along OB, OP being an arbitrary inter-
mediate position. Let x be measured from A, and let x = h at
B. Then
T = \~
, TT7 /\ x , x r ,
and W = I \jdx = j I xdx =
-==-
MOTION OF A PARTICLE 75
This is the work done on the wire by the force that stretches it.
If the wire contracts, the work done by the wire on the body to
which its end P is attached will be
/(->!)*<"
Xa2
21
21
EXERCISES
1. In the problem of § 7 compute the work done by the earth
on the particle when the latter reaches the centre.
2. A particle of mass m moves down an inclined plane. Show
that the work done on it by the component of gravity down the
plane is the same as the work done by gravity on the particle
when it descends vertically a distance equal to the change in
level which the particle undergoes.
3. A particle is attracted toward a point 0 by a force which
is inversely proportional to the square of the distance from 0.
How much work is done on the particle when it moves from a
distance a to a distance b along a right line through 01
4. If the earth and the moon were stopped in their courses
and allowed to come together by their mutual attraction, how
much work would the earth have done on the moon when they
meet?
5. Find the work done by the sun on a meteor when the latter
moves along a straight line passing through the centre of the
sun, from an initial distance R to a final distance r.
10. Kinetic Energy and Work. Let a particle of mass m
describe a right line with velocity v = ds/dt. Its kinetic energy
is defined as the quantity :
mv2
2 '
Let the particle move under the action of any force F which
varies continuously: F = /(s). Then Newton's Second Law
can be written in the form :
dv f/ ^
fi»5 -/(«).
Hence
mvdv = f(s)ds.
76 MECHANICS
Integrate this equation between the limits a and 6, denoting
the corresponding values of v by vl and v2 :
6
/ mv dv = I f(s) ds.
The left-hand side has the value :
mv2 Pa
~2~ n
The right-hand side is, by definition, the work W done on the
particle by the force F. Hence
(1)
i
and we infer the result :
THEOREM. The change in the kinetic energy of a particle is
equal to the work done on it by the force which acts on it.
This theorem expresses, in this the simplest case imaginable,
the Principle of Work and Energy in Mechanics. By means of
it a first integral of the equation arising from Newton's Second
Law can be found in the case of a particle, when the force is
known as a function of the position, and the student will do well
to go back over the foregoing problems and exercises, and ex-
amine their solution from this new point of view ; e.g. Equation
(4) in § 5, Equation (3) in § 7, and Equation (2) in § 8 are, save
for the factor m/2, the Equation of Energy, as (1) is often called.
EXERCISES
Work the Exercises of §§5, 7, 8, so far as possible, by the
Method of Work and Energy.
11. Change of Units in Physics.* To measure a quantity
is to determine how many times a certain amount of that sub-
stance, chosen arbitrarily as the unity is contained in a given
* The introduction of this paragraph and the next at this stage seems to require
justification. If these two purely physical subjects are sufficiently important to be
taken up here, then why not, at the beginning of this chapter, the first time they
are needed ? But if they are merely for reference, why break the unity, coherence,
of the presentation by placing them here rather than at the end of the chapter?
The Author feels that this is about the time when the beginner in Mechanics should
turn his attention systematically to these subjects, for until he has some knowledge
of the problems studied in this chapter, ho can hardly be expected to recognize the
importance of Change of Units and of the Check of Dimensions.
MOTION OF A PARTICLE 77
amount of the substance.* Thus to measure the length of right
lines is to find how many times a right line chosen arbitrarily as
the unit of length — a foot or a centimetre or a cubit — is con-
tained in a given right-line segment. The number, s, thus result-
ing is called the length of the line. It depends on two things —
the particular line and the unit chosen. If a different unit of
length be chosen, the same line will have a different number,
s', assigned to it, and its length then becomes s'.f Now, for all
lines, s' will be proportional to s :
(1) s' oc s or s' = c s,
where c is a constant depending on the units. It is determined
in any given case by substituting particular values for s and s',
known to correspond. Thus if we wish to transform from feet
to yards, consider in particular a line which is a yard long. Here,
s' will equal 1 and s will equal 3, so
1 = 3c, c = i,
and,t
(2) «' = $8.
Example. If a yard is the unit of length, a minute the unit
of time, a ton the unit of mass, and a kilogramme the unit of
force, find X in Newton's Second Law.
We will start with Newton's Equation in the English units :
^2<?
(3) *<y = 32/,
* The word substance here may be too narrow in its connotations, for we want
a word that will include every measurable quantity, from the length of a light-
wave to the wheat crop of the world. Such a woid obviously does not exist,
and so we agree to use substance in this sense as a terminus tcchnicus.
t It would seem paradoxical to say that the same line has a length of 6 when
the foot is the unit, and a length of 2 when the yard is the unit. But it must bo
remembered that the length is a function of two variables, the unit being one of
them. The attempt is sometimes made to meet the apparent difficulty by saying
"3 ft. = 1 yd." But this makes confusion worse confounded; for 3 = 1 is not
true, while on the other hand to try to introduce "concrete numbers," like 3 ft.,
10 Ibs., 5 sees., into mathematics, is not feasible. To try to change units in this
way leads to blunders and wrong numerical results. There is only one kind of
number in elementary mathematics. To attempt to qualify it as abstract, is to
qualify that which is unique. The denominate attribute (3 ft., 10 Ibs., etc.) is part
of the physical thing conceived ; it does not pertain to the mathematical counterpart,
which is purely arithmetical.
t Compare this equation with the attempted form of statement mentioned in
the last footnote : " 1 yd. = 3 ft." It would seem to follow from that state-
ment that «' yds. = 3s ft. But «' = £«. What a cheerful prospect for getting
the right answer by that method 1
78 MECHANICS
and write the transformed equation in the form :
Then the problem is to determine X'. Here, from (2) :
Next,
mf = km,
i _ j. v 9000 lc — m' — m •
I - ft X ZOOO, lc - 20Q(), m - 2QO()
Similarly,
i' _ ± ? = $
60' J 2.20'
Thus
,d2s' _ 602 d2s
m "^ - 2000*X3m^
XT = __L_\'/
*j 220 A;-
The left-hand sides of these equations are equal by (4). On
equating the right-hand sides and dividing by (3) we find :
602 X'
2000 X 3 2.20 X 32'
X' = 422.4.*
On dropping the primes, Newton's Second Law, written in the
new units, appears in the form :
EXERCISES
1. If the units of length, time, and mass are respectively a mile,
a day, and a ton, compute the absolute unit of force in pounds.
2. If the acceleration of gravity is 981 in the c.g.s. system,
compute g in the English system.
3. If the acceleration of gravity is 32.2 in the English system,
compute g in the c.g.s. system.
* More precisely, the result should bo tabulated as :
X' = 4.2 X 102,
since the data, namely, X - 32, are correct only to two significant figures.
MOTION OF A PARTICLE 79
4. If the unit of force be a pound, the unit of time a second,
and the unit of length a foot, explain what is meant by the absolute
unit of mass, and show that it is equal (nearly) to 32 Ibs.
6. Formulate and solve the same problem in the decimal
system.
6. If the unit of mass is a pound, the unit of length, a foot, and
the unit of force, a pound, find the absolute unit of time.
Arts. .176 sees.
12. The Check of Dimensions. The physical quantities that
enter in Mechanics can be expressed in terms of the units of
Mass [Af], Length [L], and Time [7"]. Thus velocity is of the
dimension length/time, or L/T = LT~l. Acceleration has the
dimension LT~2, and force, the dimension ML/T~*.
When, an equation is written in literal form, as
each term must have the same dimension. For, such an equation
remains true, no matter what the units of mass, length, and time
may be; and if two terms had different dimensions in any one
of the fundamental quantities (mass, length, time), a change
of units would lead to a new equation not in general equivalent
to the old one.
This principle affords a useful check on computation. Thus,
if an ellipse is given by the equation :
all the quantities x, y, a, b are of dimension one in length, or L.
The dimension of its area must be L2 ; and it is, for A = irab.
The volume of the ellipsoid of revolution corresponding to rota-
tion about the axis of x should be of dimension L3, and it is :
V = £7ra62.
This principle affords a useful check on putting in or leaving
out g, when problems are formulated literally. Thus in the
Example of § 3, if we had forgotten our g in writing down the
right-hand side, the check of dimensions would immediately
have shown up the oversight. For, the left-hand member is of
dimension ML/T~~*\ hence every term on the right must have
80 MECHANICS
this same dimension. It does, in the correct equation of the text.
It is, of course, only when all the quantities which enter are in
literal form, that the check can be used. If some are replaced
by numbers, the check does not apply.
Observe that in computing the dimension of a derivative, like
d2s/dt2, we may think of the latter as a quotient, the numerator
being a difference, and hence of the dimension of the dependent
variable, while the denominator is thought of as a power.
EXERCISES
Determine the dimension of each of the following quantities:
1. Kinetic energy. Ans. ML2T~2.
2. Work. Ans. ML2T~2.
3. Moment of inertia. Ans. ML2.
4. Momentum. Ans. MLT~l.
6. Couples. Ans. ML2T'2.
6. Volume density. Ans. ML~Z.
7. Surface density. Ans. ML*2.
8. Line density. Ans. ML"1.
9. The acceleration of gravity. Ans. LT~2.
10. The wind resistance can often be assumed proportional
to the square of the velocity. If it is written as cv2, what is
the dimension of c? Ans. ML~l.
11. In Question 10, what is the answer when the wind re-
sistance is taken per square foot of surface exposed ?
12. Check the dimensions in each equation occurring in § 3.
13. In § 4, Equation (3), the check fails. Explain why.
14. What are the dimensions of Young's Modulus?
15. In the Example treated in § 8, we wished to find the velocity
of the stone at the centre of the earth in miles per second. But
if we substituted for R, in the formula (dr/df)2 = gR, the value
of R in miles (i.e. 4000), we obtained a wrong answer, even though
the dimensions of both sides of this equation are the same, namely,
L2/T~2. Explain why, and show how Formula (2), which is
one hundred per cent literal, can be used to yield a correct result,
when R = 4000.
16. Examine each equation in § 8 as to whether the Check
of Dimensions is applicable.
MOTION OF A PARTICLE 81
13. Motion in a Resisting Medium. When a body moves
through the air or through the water, these media oppose re-
sistance, the magnitude of which depends on the velocity, but
does not follow any simple mathematical law. For low velocities
up to 5 or 10 miles per hour, the resistance R can be expressed
approximately by the formula :
(1) R = av,
where a is a constant depending both on the medium and on the
size and shape of the body, but not on its mass. For higher
velocities up to the velocity of sound (1082 ft. a sec.) the formula
(2) R = cv*
gives a sufficient approximation for many of the cases that arise
in practice. We shall speak of other formulas in the next para-
graph.
Problem 1. A man is rowing in still water at the rate of 3 miles
an hour, when he ships his oars. Determine the subsequent
motion of the boat.
Here Newton's Second Law gives us :
/o\ dv
(3) m^=-av.
TT j* m dv
Hence at = ,
a v
/A\ A m i VQ
(4) t = - log-^
where VQ is the initial velocity, nearj^^Bft. a sec.
To solve (4) for v, observe that
a* i *>o
— = log--, or
m v v
Hence
_at
(5) 9 v = VQ e ™.
It might appear from (5) that the boat would never come to
rest, but would move more and more slowly, since
_at
lim e~™ = 0.
« = 00
We warn the student, however, against such a conclusion. For
the approximation we are using, R = av, holds only for a limited
82 MECHANICS
time, and even for that time is at best an approximation. It
will probably not be many minutes before the boat is drifting
sidewise, and the value of a for this aspect of the boat would
be quite different, — if indeed the approximation K = av could
be used at all.
To determine the distance travelled, we have from (3) :
dv
mv~r = — av,
ds
and consequently :
(6) » = »o-^.
Hence, even if the above law of resistance held up to the limit,
the boat would not travel an infinite distance, but would ap-
proach a point distant
feet from the starting point, the distance traversed thus being
proportional to the initial momentum.
Finally, to get a relation between s and t, integrate (5) :
ds -?
/>7\ ffiVn /1 ~m\
(7) 8 = — -(l - e m).
From this result is also evident that the boat will never cover
a distance of S ft. while the above approximation lasts.
EXERCISE
If the man and the boat together weigh 300 Ibs. and if a steady
force of 3 Ibs. is just sufficient to maintain a speed of 3 miles
an hour in still water, show that when the boat has gone 20 ft.,
the speed has fallen off by a little less than a mile an hour.
Problem 2. A drop of rain falls from a cloud with an initial
velocity of v0 ft. a sec. Determine the motion.
We assume that the drop is already of its final size, — not
gathering further moisture as it proceeds, — and take as the
law of resistance :
R = ct;2.
MOTION OF A PARTICLE 83
The forces which act are i) the force of gravity, mg, downward,
and if) the resistance of the air, cv2, upward. As the coordinate
of the particle we will take the distance AP, Figure 60, which
it has fallen. Then, Newton's Second Law becomes :
dv , A
m-jj = mg — cv2.
TT dv mg — cv2
Hence v~r = — ,
as m
cv*
mv dv
mg — cv*'
s = - log (mg - cv2) + <7,
and thus finally FIG. 60
/ON W i m<7 — £00
(8) s = 5- log — - -t'
v 7 2c m# — cv2
Solving for v we have
mg — cv2
(9) ^gg-^-^g
v y
When s increases indefinitely, the last term approaches 0 as
its limit, arid_hence the velocity v can never exceed (or quite
equal) v — Vmg/c ft. a sec. This is known as the limiting velocity.
It is independent of the height and also of the initial velocity, and
is practically attained by the rain as it falls, for a rain drop is
not moving sensibly faster when it reaches the ground than it was
at the top of a high building.
EXERCISES
1. Work Problem 2, taking as the coordinate of the rain drop
its height above the ground.
2. Find the time in terms of the velocity and the velocity in
terms of the time in Problem 2.
3. Show that, if a charge of shot be fired vertically upward,
it will return with a velocity about 3£ times that of rain drops
84 MECHANICS
of the same size ; and that if it be fired directly downward from
a balloon two miles high, the velocity will not be appreciably
greater.
4. Determine the height to which the shot will rise in Question
3, and show that the time to the highest point is
where v0 is the initial velocity.
14. Graph of the Resistance. The resistance which the at-
mosphere or water opposes to a body of a given size and shape
can in many cases be determined experimentally with a reason-
able degree of precision and thus the graph
of the resistance :
JL can be plotted. The mathematical problem
^ '2 then presents itself of representing the curve
with sufficient accuracy by means of a simple
function of v. In the problem of vertical motion in the atmos-
phere, Problem 2, § 13,
dV ^ f/ \
m -^ = mg ± f(v),
according as the body is going up or coming down, s being meas-
ured positively downward. Now if we approximate to f(v) by
means of a quadratic polynomial or a fractional linear function,
or
we can integrate the resulting equation readily. And it is obvi-
ous that we can so approximate, — at least, for a restricted range
of values for v.
Another case of interest is that in which the resistance of the
medium is the only force that acts, as in Problem 1 :
dv ff ,
m _«-/(„).
A convenient approximation for the purposes of integration is
/(») = avb.
MOTION OF A PARTICLE 85
Here a and b are merely arbitrary constants, enabling us to im-
pose two arbitrary conditions on the curve, — for example, to
make it go through two given points, — and are to be determined
so as to yield a good approximation to the physical law. Some-
times the simple values 6 = 1, 2, 3 can be used with advantage.
But we must not confuse these approximate formulas with simi-
larly appearing formulas that represent exact physical laws.
Thus, in geometry, the areas of similar surfaces and the volumes
of similar solids are proportional to the squares or cubes of cor-
responding linear dimensions. This law expresses a fact that
holds to the finest degree of accuracy of which physical measure-
ments have shown themselves to be capable and with no restric-
tion whatever on the size of the bodies. But the law R = av2
or R = cv* ceases to hold, i.e. to interpret nature within the limits
of precision of physical measurements, when v transcends certain
restricted limits, and the student must be careful to bear this
fact in mind.
EXERCISES
Work out the relations between v and s, and those between
v and I, if the only force acting is the resistance of the medium,
which is represented by the formula :
1. R = a + bv + cv\ 2. R = ~-~V-- 3. R = av*.
7 + dv
4. Show that it would be feasible mathematically to use the
formulas of Questions 1 and 2 in the case of the falling rain drop.
5. A train weighing 300 tons, inclusive of the locomotive, can
just be kept in motion on a level track by a force of 3 pounds
to the ton. The locomotive is able to maintain a speed of 60
miles an hour, the horse power developed being reckoned as 1300.
Assuming that the frictional resistances arc the same at high
speeds as at low ones and that the resistance of the air is pro-
portional to the square of the velocity, find by how much the
speed of the train will have dropped off in running half a mile
if the steam is cut off with the train at full speed.
6. A man and a parachute weigh 150 pounds. How large
must the parachute be that the man may trust himself to it at
any height, if 25 ft. a sec. is a safe velocity with which to reach
the ground? Given that the resistance of the air is as the square
86 MECHANICS
of the velocity and is equal to 2 pounds per square foot of oppos-
ing surface for a velocity of 30 ft. a sec.
Ans. About 12 ft. in diameter.
7. A toboggan slide of constant slope is a quarter of a mile
long and has a fall of 200 ft. Assuming that the coefficient
of friction is T§Q, that the resistance of the air is proportional
to the square of the velocity and is equal to 2 pounds per square
foot of opposing surface for a velocity of 30 ft. a sec., and that
a loaded toboggan weighs 300 pounds and presents a surface
of 3 sq. ft. to the resistance of the air ; find the velocity acquired
during the descent and the time required to reach the bottom.
Find the limit of velocity that could be acquired by a tobog-
gan under the given conditions if the hill were of infinite length.
Ans. (a) 68 ft. a sec. ; (b) 30 sees. ; (c) 74 ft. a sec.
8. The ropes of an elevator break and the elevator falls with-
out obstruction till it enters an air chamber at the bottom of
the shaft. The elevator weighs 2 tons and it falls from a height
of 50 ft. The cross-section of the well is 6 X 6 ft. and its depth
is 12 ft. If no air escaped from the well, how far would the
elevator sink in? What would be the maximum weight of a
man of 170 pounds? Given that the pressure and the volume
of air when compressed without gain or loss of heat follow the
law : pvl -4l = const.,
and that the atmospheric pressure is 14 pounds to tho square
inch.
9. In the early days of modern ballistics the resistance of
the atmosphere to a common ball was determined as follows.
A number of parallel vertical screens were set up at equal dis-
tances, the ball was shot through them (with a practically hori-
zontal trajectory), and the time recorded (through the breaking
of an electric circuit) at which it cut each screen. Explain the
theory of the experiment, and show how points on tho graph of
the resistance as a function of the velocity could be obtained.
16. Motion in a Plane and in Space. Vector Velocity. When
a point P moves in a plane or in space, its position at any instant
can be represented by its Cartesian coordinates :
(1) *-/(*), 2/
MOTION OF A PARTICLE
87
where the functions are continuous, together with any deriva-
tives we shall have occasion to use.
The velocity of P has been defined as ds/dt. For, hitherto,
we have regarded the path as given, and it was a question merely
of the speed and sense of description of the path. But now we
need more. We need to put into evidence the direction and
sense of the motion, and so we extend the idea, defining velocity
more broadly as a vector. Lay off on the tangent to the path,
in the sense of the motion, a directed line segment whose length
is the speed of the point, and let the vector thus determined be
defined as the vector velocity of the point P.
Composition and Resolution of Velocities. A mouse runs across
the floor of a freight car. To determine the velocity of the mouse
in space, if the velocity of the car is u,
and the velocity of the mouse relative
to the car is v.
Let the mouse start from a point P
on one side of the car and run across
the floor in a straight line with constant
velocity, v, relative to the car. Let Q
be tho point she has reached at the end
of t seconds. Then p~ _
Let the velocity, u, of the train be constant, and let 0 be the
initial position of P. Then
OP = ut.
In Figure 62, the line OA represents the vector velocity u of
the train, and AB represents the vector velocity v of the mouse
relative to the freight car. Their geometric, or vector, sum is
represented by 07?. From similar triangles
it appears that the path of the mouse in
space is the right line through 0 and 5,
and that her velocity in space is the vector
OB, or u + v.
Thus her velocity in space may be de-
scribed, from analogy with the parallelo-
gram of forces, as the resultant of the two component velocities,
u along the direction of OA and v along the direction OC through
O parallel to A£.
ut
FIG. 62
FIG. 63
88 MECHANICS
Similarly, any vector velocity may be resolved into two com-
ponent velocities along any two directions complanar with the
given velocity ; Fig. 63.
The extension to space is obvious. Any three non-complanar
vector velocities can be composed into a single velocity by the
parallelepiped law. And conversely any given vector velocity
can be decomposed into three component vector velocities along
any three non-complanar directions.
The General Case. Returning now to the general case of motion
in a plane or in space, we may define the average vector velocity for
the At seconds succeeding a given instant as the vector * (PP')
divided by A£, or the vector (PQ) :
When At approaches 0 as its limit, the length of this vector,
namely, the chord PP , divided by At, approaches the speed of
the point at P ; or,
r ~P**' r As n • n
hm -— = hm — = Dts = v, numerically.
A« = O At AJ-=O At
Moreover the direction of the variable vector (PQ) approaches
a fixed direction as its limit. And so the variable vector (PQ)
approaches a fixed vector, v, as its limit, or
lim --V- = lim (PQ) = v.
A* = 0 At AZ = 0
This vector, v, is defined as the vector velocity of the point P.
Cartesian Coordinates. To prove that the above limit actually
exists, consider the components of (PP') and (PQ) along the.
axes. These arc :
i A# Ay Az
Ax, Ay, A* and -, -, -•
The last three variables approach limits :
lim -7 = Dix, lim --7 = Dty, lim ~ = Dtz.
A«-0 At Af-0 At A/=*0 At
Hence (PQ) approaches a limit, v, and the components of v along
the axes are :
* When it is not feasible to represent vectors by bold face type, the ( )
notation may be used, as: (PP') or, later, (a). — The student should draw the
figure which represents the vectors (PP') and (PQ)>
MOTION OF A PARTICLE 89
dx __ dy __ dz
Vx ~Tt' Vv~ ~dt' Vz " «'
These equations admit the following physical interpretation.
Consider the projections, L, Af, TV, of the point P on the axes of
coordinates. The velocities with which these points are moving
along the axes are precisely dx/dt, dy/dt, and dz/dt. And so we
can say : The projections of the vector velocity v along the axes are
equal respectively to the velocities of the projections.
Finally, observe that, just as the average vector velocity ap-
proaches the actual vector velocity as its limit, so the projections
of the average vector velocity approach the projections of the
actual vector velocity as their limits.
Remark. The student may raise the question : If vx, vy, and
vz are the components of the vector velocity, v, are they not,
therefore, themselves vectors, and should they not be written
as such, vz, vy, vz? Yes, this is correct. But it does not con-
flict with the other view of vx, vy, and vz as directed line segments
on the axes of x, y, and z. For, a system of vectors whose direc-
tion (but not sense) is fixed, constitute a system of one-dimen-
sional vectors, and these are equivalent to directed line segments,
since the two systems stand in a one-to-one relation to each
other. One-dimensional vectors can be represented arithmeti-
cally by the ordinary real numbers, positive, negative, and zero.
EXERCISES
1. Show that, if polar coordinates in the plane are used, the
component velocities along and orthogonal to the radius vector
are respectively :
dr dB
2. A point moves on the surface of a sphere. Show that
dB . A d<p
where 6 and <p denote respectively the co-latitude and the longi-
tude.
3. A point moves in space. Show that
dr dB
where r, 6, <p are the spherical coordinates of the point.
90 MECHANICS
16. Vector Acceleration. Let a point describe a path, as
in § 15. By the vector change in its velocity is meant the vector
(1) Av = v' - v,
cf. Fig. 65, p. 96. The average vector acceleration is defined as the
vector
Av
AT
When A£ approaches 0, the average vector acceleration approaches
a limiting value, and this limiting vector is defined as the vector
acceleration of the point :
/ \ r Av
(a) = hm — •
Ar=o At
Cartesian Coordinates. The components of the vector acceler-
ation along the Cartesian axes, atx, a^, and az, arc readily com-
puted. For, the components of the vector (1) along the axes
are respectively :
v'x — vx v'y — vy v'z — vz
AJ ' AJ ' AZ
As in the case of velocities, the components of the limiting vector
and the limits approached by the components of the variable
vector are respectively equal.* Hence
,. At>z ~ ,. Avy ~ .. Avz ~
ax = hm — = Dtvx, av = hm — = Dtvv, az = hm — - = Dtvt,
A/ = 0 *** A/ = 0 &t A/ = 0 At
or:
d'2z
Osculating Plane and Principal Normal.] Let a vector r be
drawn from an arbitrary fixed point 0 of space to the variable
point P that is tracing out the curve (1), § 15. Then
dr .
v--a = *-
Let s be the arc, measured in the sense of the motion ; and let
.
ds
* This theorem is true of any vector which approaches a limit, as the student
can readily verify.
f Cf. the Author's Advanced Calculus, p. 304, § 8.
MOTION OF A PARTICLE 91
Then r' is a unit vector lying along the tangent and directed
in the sense of the motion. Furthermore,
= r
ds
is a vector drawn along the principal normal, toward the con-
cave side of the projection of the curve on the osculating plane,
and its length is the curvature, K, at P.
On the other hand, the acceleration
dv , dsdr
=¥ and *-__
Hence
EXERCISES
1. A point describes a circle with constant velocity. Show
that the vector acceleration is normal to the path and directed
toward the centre of the circle, and that its magnitude is
"2 2
— , or or r.
2. Show that, when a point is describing an arbitrary plane
path, the components of the vector acceleration along the tangent
and normal are : <j2s v*
at = w' an = ?
whoro p denotes the radius of curvature, and the component <xn
is directed toward the concave side of the curve.
3. A point describes a cycloid, the rolling circle moving forward
with constant velocity. Show that the acceleration is constant
in magnitude and always directed toward the centre of the circle.
4. Prove by vector methods that, in the case of motion in
where ar, a0 denote the components of the acceleration along
and perpendicular to the radius vector.
Use the system of -ordinary complex numbers, a + bi, where
i = V— 1, and set „ .
' r = re6\
5. Obtain the same results by geometric methods.
92 MECHANICS
17. Newton's Second Law. Let a particle move under the
action of any forces, and let F be their resultant. Let (a) be
its vector acceleration. Then Newton's Second Law of Motion
asserts that the mass times the vector acceleration is proportional to
the vector force , or, if the absolute unit of force is adopted,
(1) m(a) = F.
In Cartesian coordinates the law becomes :
W>-J7Z = X,
(2)
td?z
If, in particular, X, Y, Z are continuous functions of x, y, z,
dx/dt, dy/dt, dz/dt, and ty it then follows from the theory
of differential equations that the path is uniquely determined
by the initial conditions; i.e. if the particle is projected from a
point (z0, 2/o> ZG) with a velocity whose components along the axes
are (u^ v0, w0), the path is completely determined. This remark
is striking when one considers that the corresponding theorem
is not true if one determines the motion by means of the principle
of Work and Energy; cf. the Author's Advanced Calculus, p. 351,
Singular Solutions. The essential point here is that Equa-
tions (2) never admit a singular solution, whereas the equations
of Work and Energy do.
In the more general cases it is also seen that the path is uniquely
determined by the initial conditions. This statement is con-
firmed in the case of each of the examples considered below.
For a general treatment, cf. Appendix A.
Osculating Plane. The force, F, always lies in the osculating
plane of the path. For, from § 16, and Equation (1) above,
Hence we can resolve F into a component T along the path and a
component N along the principal normal, and we shall then have :
y"72 o tYllfi
m—— = T = N
where p = l//c.
MOTION OF A PARTICLE
93
EXERCISE
Show that (r' x r") • F = 0.
Hence, in Cartesian coordinates,
-(z'x")Y+(x'y")Z = 0,
where
and
y'z"-z'y",
etc.
/ dx fr
ff1 — f —
JU — ~; . JU —
ds
etc.
18. Motion of a Projectile. Problem. To find the path of a
projectile acted on only by the force of gravity.
The degree of accuracy of the approximation to the true motion
obtained in the following solution depends on the projectile and
on the velocity with which it moves. For a cannon ball it is
crude, though suggestive, whereas for the 16 Ib. shot, used in
putting the shot, it is decidedly good.
Hitherto we have known the path of the body; here we do
not. The path will obviously be a plane curve, and so Newton's
Second Law of Motion becomes :
m
dt2
where X , Y are the components of the resultant force along the
axes, measured in absolute units.
In the present case X = 0, Y = — mg, and we have
(2)
FIG. 64
If we suppose the body projected from 0 with velocity VQ at
an angle a with the horizontal, the integration of these equations
«ives: dx
dt
-jj = (J = VQ cos a, x = v0 1 cos a ;
•^ = vQ sin a - 0tf , y = V0 / sin a -
94 MECHANICS
Eliminating t we get :
OX"
(3) y = x tan a - -^- — 5- •
2^ cos2 a
The curve has a maximum at the point A : (z,, t/J,
t$ sin a cos a #?, sin2 a
/M _ U /%• U
*> ff ' *> - ~2T~
Transforming to a set of parallel axes through A, we have :
x = x1 + a?!, y = y' + ylt
V ' * 2«2 COS2 n,
&UQ LUo Cc
This curve is a parabola with its vertex at A. The height
of its directrix above A is v\ cos2 a/2</, and hence the height
above 0 of the directrix of the parabola represented by (3) is
vl sin2 a . v% cos2 a _ v%
~~2g~ + 2i~ " 27'
The result is independent of the angle of elevation a, and so
it appears that all the parabolas traced out by projectiles leaving
0 with the same velocity have their directrices at the same level,
the distance of this level above 0 being the height to which the
projectile would rise if shot perpendicularly upward.
EXERCISES
1. Show that the range on the horizontal is
R = — sin 2a,
j/
and that the maximum range R is attained when a = 45° :
~ g'
The height of the directrix above 0 is half this latter range.
2. A projectile is launched with a velocity of VQ ft. a sec. and
is to hit a mark at the same level and within range. Show that
there are two possible angles of elevation and that one is as much
greater than 45° as the other is less.
3. Find the range on a plane inclined at an angle j3 to the
horizon and show that the maximum range is
* ~ ~g 1 + sin 0
MOTION OF A PARTICLE 95
4. A small boy can throw a stone 100 ft. on the level. He
is on top of a house 40 ft. high. Show that he can throw the
stone 134 ft. from the house. Neglect the height of his hand
above the levels in question.
6. The best collegiate record for putting the shot was, at
one time, 46 ft. and the amateur and world's record was 49 ft.
Gin.
If a man puts the shot 46 ft. and the shot leaves his hand at
a height of 6 ft. 3 in. above the ground, find the velocity with
which he launches it, assuming that the angle of elevation a is
the most advantageous one. Am. v0 = 35.87.
6. How much better record can the man of the preceding
question make than a shorter man of equal strength and skill,
the shot leaving the latter's hand at a height of 5 ft. 3 in. ?
7. Show that it is possible to hit a mark B : (x6, 2/&), provided
8. A revolver can give a bullet a muzzle velocity of 200 ft.
a sec. Is it possible to hit the vane on a church spire a quarter
of a mile away, the height of the spire being 100 ft. ?
9. It has been assumed that the path of the projectile is a
piano curve. Prove this assumption to be correct by using all
three Equations (2), § 17.
19. Constrained Motion. Let a particle be constrained to
move in a given curve, like a smooth bead that slides on a wire.
Consider first the case of a plane curve. Let the component
of the resultant of all the forces along the tangent be T and along
the normal be N. Then Newton's Second Law of Motion, § 17,
gives the following equations :
(1)
mv
2 = N.
P
The proof given in § 17 was based on vector analysis. We will
give one for the plane case without the use of vector methods.
Geometric Proof. Compute the components of the vector
acceleration along the tangent and along the normal. Let <p
96
MECHANICS
be the angle which the tangent has turned through in passing
from P to P' . Then the component of Av along the tangent
will be
vf cos tp — v = (t; + Av) cos <p — v
= Av cos <p — v (1 — cos <p).
By the definition of curvature,
K = lim~r, p = lim
PP'
PP'
Now, the component of the average ac-
celeration along the tangent is
vf cos & — v Av 1 — cos (p
TT = -r- COS <f> — V — -•
At At At
Let At approach 0. Then <p approaches 0, and the limit of the
first term on the right is
/,. AZA/V \ ~
f lim — Mlimcos^J = Dtv.
To evaluate the limit of the second term, write
1 — cos y? __ 1 — cos <p jp_ As
At <p As At
The first factor approaches 0, and the second and third factors
remain finite, since each approaches a limit. Hence the limit
of the right hand side is 0.
We have proved, then, that
,. vf cos <p — v ~
lim — = Dtv,
and thus the first of Equations (1) is established.
To obtain the second of Equations (1), consider the component
of the average acceleration along the normal, or
This can be written as
v sm <p
At
;sin <p (?_ As
v
_
<p As At'
MOTION OF A PARTICLE 97
where s is assumed to increase with t. The limit of this product
is seen to be : i 2
v X 1 X - X v = -,
P P
and this proves the theorem.
The component N measures the reaction of the curve. It is
the centripetal force due to the motion.
The foregoing analysis yields the first of Equations (1) for
twisted curves.
EXERCISE
Use the present geometric method to obtain the formulas :
1 d
where ar, ote denote respectively the components of the vector
acceleration along and perpendicular to the radius vector.
20. Simple Pendulum Motion. Consider the simple pendu-
lum. Here ^s
m -772 = - mg sin 0,
at
and since s = 1 0,
This differential equation is characteristic for Simple Pendulum
Motion. We can obtain a first integral by the method of § 7 :
2g . de
~~T: 77^ — ~T~ sin u -jr.
dt dt2 I at
0 = cos « + C,
where a is the initial angle ; hence
(2) ^ = ^(cos0-cos«).
The velocity in the path at the lowest point
is I times the angular velocity for 0 = 0, or
V20Z (1 — cos a), and is the same that would have been acquired
if the bob had fallen freely under the force of gravity through the
MECHANICS
same difference in level. Equation (2) is virtually the Integral
of Energy.
If we attempt to obtain the time by integrating Equation (2),
we are led to the equation :
de
20 J Vcos 0 — cos a
This integral cannot be expressed in terms of the functions at
present at our disposal. It is an Elliptic Integral.* When 0,
however, is small, sin 6 differs from 6 by only a small percentage
of either quantity, and hence we may expect to obtain a good
approximation to the actual motion if we replace sin 6 in (1) by 0 :
(3)
_ g
-~
This latter equation is of the type of the differential equation
of Simple Harmonic Motion, § 7, A), n2- having here the value
g/l. Hence, when a simple pendulum swings through a small
amplitude, its motion is approximately harmonic and its period
is approximately
9
The Tautochrone. A question that interested the mathema-
ticians of the eighteenth century was this : In what curve should
a pendulum swing in order that the period of oscillation may bo
absolutely independent of the amplitude? It turns out that
the cycloid has this property. For, the differential equation of
motion is
dzs
FIG. 67
where $ is measured from the lowest
•' point, and since
s = 4a sin r,
, d*s g
we have -j-z = — -~ s.
at2 4a
* Cf. the author's Advanced Calculus, Chapter IX, page 195, where this integral
is reduced to the normal form.
MOTION OF A PARTICLE
99
This is the differential equation of Simple Harmonic Motion,
§ 7, A), and hence the period of the oscillation,
9
is independent of the amplitude.
A cycloidal pendulum may be constructed by causing the cord
of the pendulum to wind on the evolute of the path. The resist-
ances due to the stiffness of the cord as it winds up and unwinds
would thus be slight ; but in time they would become appreciable.
21. Motion on a Smooth Curve. Let a bead slide on a
smooth wire under the force of gravity. Consider first the plane
case. Choosing the axes as indicated, we have :
(1)
Hence
dt*
dx
ds
_ 9
2~ g
~ds~dt
Integrating this equation with respect
to /, we find :
If we suppose the bead to start from
rest at A, then
0 = 2gxQ + C,
\A:(xQ,y0)
FIG. 68
(2)
—
But the velocity that a body falling freely from rest a distance of
x — x0 attains is expressed by precisely the same formula.
In the more general case that the bead passes the point A
with a velocity v0 we have :
(3)
eg = 2gx<> + C,
t>2 - eg = 2g (x - z0).
Thus it is seen that the velocity at P is the same that the bead
would have acquired at the second level if it had been projected
vertically from the first with velocity »0.
100 MECHANICS
The theorem also asserts that the change in kinetic energy is
equal to the work done on the bead ; cf. § 10.
If the bead starts from rest at A, it will continue to slide till
it reaches the end of the wire or comes to a point A' at the same
v* >^L level as A In the latter case it will in gen-
eral just rise to the point A1 and then retrace
its path back to A. But if the tangent to the
FIG. 69 curve at A ' is horizontal, the bead may
approach A1 as a limiting position without ever reaching it.
EXERCISES
1. A bead slides on a smooth vertical circle. It is projected
from the lowest point with a velocity equal to that which it would
acquire in falling from rest from the highest point. Show that
it will approach the highest point as a limit which it will never
reach.
2. From the general theorem (2) deduce the first integral
(2) of the differential equation (1), § 20.
Space Curves. The same treatment applies to space of three
dimensions. It is interesting, however, to give a solution based
on Cartesian coordinates. Choose the axis of x as before positive
downward. Then we have :
(4)
. D
~dT* = "% + R*>
d*y _ D
~* ~ "'
where Rx, Rv, Rt are the components of the reaction R of the wire
along the axes. Since R is normal to the curve, we have :
<» s-S+*-8+ *•!-»•
To integrate Equations (4) multiply through respectively by
dx/dty dy/dt, dz/dt and add. We thus find, with the aid
of (5):
MOTION OF A PARTICLE 101
But
2 /efo\2 . (dy\2 /dz\*
"2 = (&) + U) + b) •
Hence (6) reduces to
(7) ^ m d (v2) = mg dx.
On integrating this equation, we find :
/o\ mvZ mva / \
(8) -7j --- 2^ = "V (* - *<>)•
This is precisely the Equation of Energy. It could have been
written down at the start from the Principle of Work and Energy.
It is the generalization of (2) for space curves.
EXERCISE
A bead slides on a smooth wire in the form of a helix, axis
vertical. Determine the reaction of the wire in magnitude and
direction.
22. Centrifugal Force. When a particle of mass m describes
a circle with constant velocity, the acceleration is directed toward
the centre, and its magnitude is
The force which holds the particle in its path is,
therefore, normal to the path and directed inward.
Its magnitude is
,r mv2 9 FIG. 70
N = - = mco2r.
r
Why, then, the term " centri/u^aif force" — the force that "flees
the centre"? The explanation is a confusion of ideas. If the
mass is held in its path by a string fastened to a peg at the centre,
0, does not the string tug at 0 in the direction OP away from the
centre and is not this force exerted by the particle in its attempt,
or tendency, to fly away from the centre? The answer to the
first question is, of course, "Yes." Now one of the standard
methods of the sophists is to begin with a question on a non-
controversial point, conceded without opposition in their favor,
and then to confuse the issue in their second question — "and
is not this force exerted by the particle?"
102 MECHANICS
Matter cannot exert force, for a force is a push or a pull, and
matter can neither push nor pull ; it is inert. The particle does
not pull on the string, the string pulls on the particle. But even
this statement will be accepted only half-heartedly, if at all,
by people who have not yet grasped the basic idea of the science
of Mechanics — the study of the motion of matter under the
action of forces. What comes first is a material system — solid
bodies, particles, laminae and material surfaces, wires, any combi-
nation of these things, including even deformable media (hydro-
dynamics, elasticity) — and then this system is acted on by
forces.
ISOLATE THE SYSTEM
The man who first uttered these words deserves a monumentum
aere. In the present case there are two systems, each of which
can be isolated : (1) the particle ; (2) whatever the peg is attached
to — think of a smooth table, the particle going round and round
in a horizontal circle and being held in its path by a string whose
other end is attached to a peg at a point 0 of the table. In the
case of the first system, the force that acts is the pull of the string
toward the centre, and this force is what is now-a-days described
as "centripetal" force — the force that " seeks the centre."
The second system has nothing to do with the particle. In
particular, this system may be the table. In that case, the floor,
as well as gravity, exerts certain forces, and under the action of
all the forces, the table stays at rest. The force of the string,
varying in direction, causes the forces of the floor to vary.
And now, after all is said and done, comes the rejoinder: "But
the particle did pull on the string, for otherwise the string would
not have pulled on the peg." There is no answer to these people.
Some of them are good citizens. They vote the ticket of the
party that is responsible for the prosperity of the country ; they
belong to the only true church ; they subscribe to the Red Cross
drive — but they have no place in the Temple of Science ; they
profane it.
Example 1. A bullet weighing 1 oz. is shot into a sling, con-
sisting of a string 5 ft. long with one end fastened at 0, the other
end carrying a leather cup. If the velocity of the bullet is 600 ft.
a sec., how strong must the string be, not to break?
MOTION OF A PARTICLE 103
The tension in the string will be o 5 ^
mv2 I
FIG. 71
where m = ^, v = 600, r = 5 ; or
6002
16X5
4500 ;
4500 what? pounds? No, for the force is measured in absolute
units, or poundals, and so, to get the answer in pounds, we must
divide by 32. The tension, then, that the string must be able
to withstand is 141 Ibs.
Example 2. A railroad train rounds a curve of 1000 ft. radius
at 30 m. an hour. How high should the outer rail be raised,
fc if the flanges of the wheels are not to
press against either track? Standard
gauge, 4 ft. 8| in.
If a plumb bob is hung up in a car,
and does not oscillate, then it should be
at right angles to the axles of the wheels.
It will describe its circular path in space
under the action of two forces, namely,
gravity, mg, downward, and the tension,
\ T, of the string. Let the string make
mg „ 7 an angle a. with the vertical. Then the
vertical component of T just balances
gravity, for there is no vertical motion of the bob. Hence
T cos a. ~ mg.
The horizontal component of T yields the normal force N which
keeps the bob in its circular path, or
Hence
v2 442
Since the distance between the rails is 4 ft. 8£ in., it follows that
the outer rail must be raised 3.42 in.
104 MECHANICS
EXERCISES
1. A particle weighing 4 oz. is attached to a string which
passes through a small hole, 0, in a smooth table and carries
a weight W at its other end. If the first weight is projected
along the table from a point P at a distance of 2 ft. from 0 with
a velocity of 50 ft. a second in a direction at right angles to OP,
the string being taut and the part below the table vertical, how
great must W be, that the 4 oz. weight may describe a circular
path? Ans. 9 Ibs. 12 oz.
2. A boy on a bicycle rounds a corner on a curve of 60 ft.
radius at the rate of 10 m. an hour, and his bicycle slips out from
under him. What is the greatest value p, could have had ?
Ans. Not quite |.
3. A conical pendulum is like a simple pendulum, only it is
projected so that it moves in a horizontal circle instead of in a
vertical one. Show that
Zo>2 = g sec a.
4. If the earth were gradually to stop rotating, how much
would Bunker Hill Monument be out of plumb? Given, that
the height of the monument is 225 ft. and the latitude of Charles-
town is 42° 22'. Ans. About 4£ in.
5. An ocean liner of 80,000 tons is steaming east on the equator
at the rate of 30 knots an hour. If she puts about and steams
west at the same rate, what is the increase in her apparent weight ?
6. If the earth were held in her course by steel wires attached
to the surface on the side toward the sun and evenly distributed
as regards a cross-section by a plane at right angles to them,
show that they would have to be as close together as blades of
grass. It is assumed that their other ends are guided near the
earth's surface.
7. Show that a steel wire one end of which is made fast to
the sun and which rotates in a plane with constant velocity,
making one rotation in a year, could just about reach to the
earth without breaking. Neglect the heat of the sun and all
forces of gravitation.
8. A steel wire 1 sq. mm. in cross-section, breaking strength
70 kgs., is strung round the earth along the equator. Show that,
if the earth gradually stopped rotating, the wire would snap.
MOTION OF A PARTICLE
105
9. What is the smallest latitude such that the wire described
in the preceding question, if strung round the earth on that par-
allel, would not break ?
10. If the earth had a satellite close by, how often would
the latter rise and set in a day? Ans. About 18 times.
11. A boy swings a bucket of water around in a vertical circle
without spilling any. Does not the bucket exert a pull on the
boy's hand ?
Explain the situation by isolating a suitable system, namely:
i) the bucket of water ; ii) the boy.
23. The Centrifugal Oil Cup. A device once used for deter-
mining the speed of a locomotive consisted of a cylindrical cup
containing oil and caused to rotate about its axis, which was
vertical, with an angular velocity proportional to the speed of
the train. Let us sec how it worked.
Suppose the oil to be rotating like a rigid body, with no cross
currents or other internal disturbances. What will be the form
of the free surface ? Imagine a small par-
ticle floating on the oil. It will be acted
on by the force of gravity, mg, downward
and the buoyancy, B, of the oil normal to
the surface. The resultant of these two
forces must just yield the centripetal
force N required to keep the particle in
its path. Now
N = mco2x.
FIG. 73
On the other hand, the slope of the curve is determined by the
fact that the tangent is normal to B. Thus
B cos T = mg, B sin r = N.
Hence
U'X
or
dx " g X'
It follows, then, that
0) v-$*.
Thus it appears that the free surface is a paraboloid of revolution
106 MECHANICS
To Graduate the Cup. It is easily shown that the volume of
a segment of a paraboloid of revolution is always half the volume
of the circumscribing cylinder. If, then, we mark the level of
the oil when it is at rest, the height, h, to which it rises above this
level when it is in motion will just equal the depth, h, of the
lowest point of the surface below this point. From (1) it follows,
then, that if a denotes the radius of the cup,
or
EXERCISES
1. A tomato can 4 in. in diameter is filled with water and
sealed up. It is placed on a revolving table and caused to rotate
about its axis, which is vertical, at the rate of 30 rotations a sec.
Find the pressure on the top of the can.
Ans. The weight of a column of water 4 ft. high (nearly)
and standing on top of the can.
2. How great is the tendency of the can to rip along the seam?
24. The Centrifugal Field of Force. It is possible to view the
mechanical situation in the oil cup from a statical standpoint.
Imagine very tiny insects crawling slowly round on tho surface
of the oil. To them the oil and all they could sec of the walls
and top of the cup would appear stationary, and they would
refer their motion to the rotating space as if it were at rest.
We can reproduce the situation, so far as statical problems
are concerned, in a space that is actually at rest by creating a
field of force, in which the force which acts on a particle of mass m
distant r from a fixed vertical axis is the resultant of the force
of gravity, mg, vertical and downward, and a force mcoV directed
away from the axis, where co is a constant. Thus the magnitude of
the force would be
+ (mco2r)2 =
and it would make an angle <p with the downward vertical, where
wV
tan <p = — •
0
MOTION OF A PARTICLE 107
To bring the mechanical situation nearer to our human intui-
tion, we might think of a large round cup, 500 ft. across at the
top, constructed with the flooring in the form of the paraboloid
in question and rotating with the suitable angular velocity.
There would be a small opening at the vertex, through which
observers could enter and leave. The view of all surrounding
objects would be cut off, and the mechanical construction would
be so nearly perfect that, when we were inside the cup, we should
not perceive its motion. Suppose, for example, that the slope
of the floor along the rim were 45°. Then, since
uPx
tan T = — ,
g
it follows that
~ 32 '
co = T4T (nearly), or .36.
The time, T, of a complete revolution is given by the equation :
27T
«- jT,
or
T = — = 17 sees.
0>
Thus the cup would make nearly four revolutions a minute.
Since co2/gr = ^^, the intensity of the field would be
(0.004r)2,
and upon the rim of the cup, this would amount to mgV2, or
41 per cent greater than gravity on the fixed surface of the earth
— roughly, two-fifths more. A movie actress who was main-
taining her weight in Hollywood, would tip the scales at, —
well, how much ?
What we have said applies, however, only to bodies that are
at rest in the field. When a body moves, still other forces enter,
and these will be considered in the chapter on Relative Motion.
Nevertheless, we can describe the motion of a projectile directly,
since it would be a parabola in the fixed space we started with.
Imagine a tennis court laid out with its centre at the lowest
point of the bowl. Lob the ball from the back line to the back
line, and watch the slice !
108 MECHANICS
One may reasonably inquire concerning the engineering prob-
lems of the construction. There will be a tendency of the cup
to burst — to fly apart, due to the "centrifugal force." Can it
be held together by reinforcing it with steel bands round the
outer rim, or will these have all they can do to hold themselves
together? It turns out that only one-seventieth of the break-
ing strength will be needed to hold the band together, thus leav-
ing sixty-nine seventieth^ for reinforcing.
But since at the rim the "centrifugal force" is as great as the
force of gravity, any unbalanced load will cause the cup to tug
on its anchorage unmercifully. A hundred men weigh approxi-
mately 8 tons, and if they were bunched at a point of the rim,
the reaction on the anchorage would be 8 tons. The student
will find it interesting to compute the reaction in case a racing
car were driven along the rim at 100 miles an hour.
25. Central Force. Let a particle be acted on by a force
directed toward a fixed point, O, and depending only on the
distance from O, not on the direction. Newton's Second Law
of Motion, § 17, then becomes :
(1>
™d.(r*—\ = 0
rdt\ dt) '
where R is a continuous function of r.
Law of Areas. The second equation admits a first integral :
(2) r^==h'
This equation admits a striking interpretation. Consider the
area, A, swept out by the radius
vector drawn from 0 to the par-
ticle. Then
A = / r*dd,
FIG. 74 dt ~~ dt
MOTION OF A PARTICLE 109
and hence
(3) A=ft(«-g,
or, equal areas are swept out in equal times.
We have tacitly assumed that h 5* 0. If h = 0, then (2)
reduces to dd = 0, and the path is a straight line through 0.
Work and Energy. The kinetic energy of the particle is
nw^_m(W W
W 2 " 2 W + r <
By virtue of (2) this becomes :
= i
2 ~ 2 2 2
On the other hand, the work, cf. Chap. VII, § 3.
r
(6) TF = Cfidr.
Hence
I _
~ J?
1 _ 2 /
7> ~ mh*J
This is a differential equation of the first order connecting
r and 6, and its integral gives the form of the path.
The Law of Nature. Newton discovered the Law of Universal
Gravitation, which says that any two particles in the universe
attract each other with a force proportional to their masses and
inversely proportional to the square of the distance between
them. This law is often referred to as the Law of Nature.
In the present case, then, the particle is attracted toward 0
with a force proportional to 1/r2, and so
(8) l«l«^, «=-£
Thus
(9) ^
The Law of Energy, as expressed in the form of Equation (7),
here becomes :
110 MECHANICS
where X = m/x, and C is a constant depending on the initial condi-
tions.
The form of this equation suggests a simplification consisting
in substituting for r its reciprocal :
(11) u = J-
Thus (10) becomes :
/io\ i 9 2/4 . ~
(12) ^5 + «« = -S;« + C'.
This equation admits further reduction. Write :
Since the left-hand side can never be negative, the right-hand
side can be written as 52, and B itself may be chosen as either
one of the square roots. Finally, set
* - u ~ ?»•
Then (13) goes over into :
dr2
(14) * + ^ - B".
The general integral of this differential equation can be written
in the form :
(15) x = B cos (0 - 7),
where y is the constant of integration. When 5 = 0, the truth
of this statement is obvious, for then (14) reduces to
— + *» - 0
de* + x u'
and the only solution of this differential equation is *
x = 0.
If S2 * 0, then (14) yields :
d8 =
* We have here an example of a differential equation of the first order, handed
to us by physics, whose general integral does not depend on an arbitrary constant,
but consists of a unique function of 0 alone.
MOTION OF A PARTICLE 111
where, however, the two ± signs are not necessarily the same.
But in all cases this last equation leads to (15).*
We set out to integrate Equation (12), and we have arrived
at the result :
(16) u = ~ + B cos (e - 7).
This equation can be thrown into familiar form by taking B
as the negative radical and setting
where e now is the constant of integration. Thus (16) yields :
(\7\ r = — -
1 } M 1 - ecos(0 - 7)
The Orbit. The path of the particle is given by Equation (17).
This is the equation of a conic referred to a focus as pole and
having the eccentricity e.
The Case e < 1. If e < 1, the conic is an ellipse, and the
length of the transverse axis is
M (1 - e*)
Denoting the length of the semi-axes by a and 6, we have :
h* . A2
- Md-e2)' MVf=T'
The distance between the foci is
(19) c =
The area of the ellipse is
(20) TTOfc =
Th
tion:
A = pr.
Hence
(21) T2 = 47r2--
* It is worth the student's while to follow through these multiple-valued func-
tions, that he may secure a firmer hold on the Calculus, even though the final
result — Equation (15) — is simple.
(1 - e2)1
The periodic time T is connected with the area A by the rela-
112 MECHANICS
Determination of the Constants of Integration. Let the body be
projected from the point (r, 0) = (a, 0) with an initial velocity
VQ in a direction making an angle ft with the prime direction
0 = o. To determine the orbit.
We will mention first a general formula. Let \l/ be the angle
from the radius vector produced to the tangent. Then
since each side represents the component ve of the vector velocity,
v, perpendicular to the radius vector. By virtue of (2) this
becomes :
(22) h = vr sin ^,
and this is the formula we had in mind.
To determine the constants in (17), then, write the equation
in the form :
/f>O\ _ r* / '-j //j \\
Hence
(24) - = -^ (1 — e cos 7), e cos 7 = 1
ah* an
Furthermore,
du fie .
— = — - sin (Q — -y).
dO h'2
Since
du __ 1 dr __ i dr __ v cos \f/
d6 r2 dO h dt h '
we have initially :
/rt/»\ M^ • ^n COS p . Vi\ il COS p
(26) — T^ sin 7 = —, — -. e sin 7 = —
hi h M
From (22),
(27) h = VQ a sin j8, cos2 ft = 1 ^—-
U i / i - 2 y»^
Squaring the second equation in (24) and (26), and adding,
we find by the aid of (27) :
(28) e
MOTION OF A PARTICLE 113
The evaluation is now complete. By means of (27), h is deter-
mined ; (28) then gives e, and (24) and (26) yield 7.
From (28) we infer that
and this equation contains the interesting result that the orbit
will be the following conic :
i) ellipse, if v% < — ;
ii) parabola, if v\ = — ;
Hi) hyperbola, if v% > — ,
irrespective of the direction, 0, in which the body is launched.
For motion in a circle, e = 0. From (24)
(30) - = 4 h2 = MO.
an2
Moreover, from (26) we see that £ = ± 7r/2, and so we infer from
(27) that
A2 = via*.
Hence, by the aid of (30),
(31) v* = Jj.
Conversely, conditions (30) and (31) are sufficient, that the
path be a circle. For from (27) follows that cos2/3 = 0, and
(29) gives 6 = 0. The result checks with the fact that the numer-
ical value of R, or w/*/a2, is equal to the centripetal force, or
EXERCISES
1. Show that if
then
2 , ,
r2-rr = h and u = -,
dt r'
114 MECHANICS
2. It has been assumed that the orbit is a plane curve. Prove
this to be the case by means of a constraint, consisting of a smooth
plane through 0, the point of projection, and the tangent to the
path at that point. Use Newton's Equations, § 17, (2), and
show that the force of the constraint is 0.
26. The Two Body Problem. If two particles of masses w,
m'j attracting each other according to the law of nature, and
acted on by no other forces, be projected in any manner, their
centre of gravity, G, will describe a right line, with constant
velocity, or remain permanently at rest ; cf. Chapter IV, § 1. Con-
sider the latter case. Let G be the fixed point 0, and let the
distances of the particles from 0 be r, r'. Then the force of their
mutual attraction is
/ = K (r + r')2>
where K is the gravitational constant. On the other hand,
mr = m'r'.
Hence
, _ m + m'
r ~m' r'
and so
' \2
f m \2 ,
= ( - - - ,) m'.
\m + ml
Thus the particle m is attracted toward 0 with the force that
would be exerted by a mass M fixed at 0, and so the orbit of m
is determined by the work of § 25. In particular, if m describes
an ellipse, mf will describe a similar ellipse with the same focus,
being turned through an angle of 180°.
27. The Inverse Problem — to Determine the Force. Let
a particle move in a plane according to the Law of Areas. Then
r^ = h
T dt ft'
and the component of the force perpendicular to the radius vector,
0, is nil. Hence the particle is acted on by a central force, R,
either attractive or repulsive. From Exercise 1, § 25, we have :
MOTION OF A PARTICLE 115
Example. Let the path be an ellipse (or, more generally, any
conic) with the centre of force at a focus. Then
1 — e cos (0—7) ,
u = — , p = const.,
P
d*u . 1
W* + U = p>
and
mh*l
K~ p r*
The force is, therefore, an attractive force, inversely proportional
to the square of the distance from the centre, when r lies between
its extreme values for this ellipse. But an arbitrary range of
values, 0 < a < 7 < 0, can be included in such an ellipse, and so
the result is general.
EXERCISE
Show that if the path is an ellipse with the centre of force at
the centre, the force is proportional to the distance from the
centre.
28. Kepler's Laws. From observations made by Tycho
Brahe, Kepler deduced the laws which govern the motion of
the planets.
1. The planets describe plane curves about the sun according
to the law of areas;
2. The curves are ellipses with the sun at a focus;
3. The squares of the periodic times of revolution are proportional
to the cubes of the major axes of the ellipses.
Newtonys Inferences. From Kepler's laws Newton drew the
following inferences. Consider a particular planet. From the
first law it follows that the force acting on it is a central force,
since the component © at right angles to the radius vector is nil.
From the second law, combined with the first, it follows from
§ 27 that the force is inversely proportional to the square of the
distance from the centre, or
116 MECHANICS
It has been shown in § 25, (21) that
T2 = 47T2-,
M
where T denotes the periodic time, and a is the semi-axis major.
For a second planet,
Kepler's third law gives, then, that p! = p, or that /* is the same
for all the planets.
To sum up, then, Newton inferred that the planets are at-
tracted toward the sun with a force proportional to their masses
and inversely proportional to the square of their distances from
the sun.
From here it is but a step to the Law of Universal Gravita-
tion. If the sun attracts the planets, so must, by the principal
of action and reaction, the planets attract the sun. Let M denote
the mass of the sun, thought of as at rest.* Then
Thus the Law of Universal Gravitation is evolved : Any two
bodies (particles) in the universe attract each other with a force
proportional to their masses and inversely proportional to the
square of the distance between them, or
win
The factor K is called the gravitational constant. Its value
in c.g.s. units is
# = 6.5X 10-*;
cf. Appell, l.c., pp. 390-405.
EXERCISES
1. Show that the first of the equations (1), § 25 :
dt* dt2
* For a more detailed treatment cf. Appell, Mecanique rationnette, vol. 1, 3d ed.,
1909, § 229 et seq.
MOTION OF A PARTICLE 117
on making the transformation (11) :
.
r* —
r dt
and employing (2) :
goes over into the equation :
Hence obtain (16) :
u = ~ + B cos (B — 7).
2. Prove that
h
v — -,
P
where p denotes the distance from 0 to the tangent to the path.
3. Show that the earth's orbit, assumed circular, would
become parabolic if half the sun's mass were suddenly annihilated,
the sun being assumed to be at rest.
4. A smooth tube revolves around one end in a fixed plane
with constant angular velocity. A particle is free to move in
the tube. Determine the motion.
5. If, in the preceding question, an elastic string is made
fast to the particle and attached to the end of the tube, deter-
mine the motion.
6. A particle is attracted toward a fixed centre with a force
proportional to the distance. Show that the path is a plane
curve, and that it can be represented by the equations :
x = A cos (nt + a), y = B sin (nt + a).
Is it an ellipse?
7. Show that a comet describing a parabolic path cannot
remain within the earth's orbit, assumed circular, for more than
(2 \
«- 1-th part of a year, or nearly 76 days.
8. A shell is describing an elliptical orbit under a central
attractive force. Prove that, if it explodes, all the pieces will
meet again at the same moment ; and that after half the interval
between the explosion and the collision, each piece will be moving
118 MECHANICS
with the same velocity as at the instant of explosion, but in the
opposite direction.
9. Show that a particle, moving under the action of a central
force, cannot have more than two apsidal distances ; cf . Appendix B.
10. Find the law of force when a particle describes a circle,
the centre of force being situated on the circumference.
Ans. The inverse fifth power.
11. If two spheres, each one foot in diameter and of density
equal to the mean density of the earth (5.6) were released from
rest in interstellar space with their surfaces -^ inches apart, how
long would it take them to come together?
How great would the error be if their mutual attraction were
taken as constant?
12. A cannon ball is fired vertically upward from the Equator
with a muzzle velocity of 1500 ft. a sec. How far west of the
cannon would it fall, if the earth had no atmosphere ?
13. Show that a particle acted on by a central repulsive force
varying according to the inverse square, will in general describe
a branch of a hyperbola with the centre of force at that focus
which lies on the convex side of the branch. What is the excep-
tional case?
29. On the Notion of Mass. Matter is inert. It cannot exert
a force; it cannot push or pull. It yields to force, acquiring
velocity in the direction in which the force acts — we are think-
ing of a particle. By virtue of its inertness it possesses mass,
which may be described as the quantity of matter which a body
contains.
Mass is measured by the effect which force produces on the
motion of a body. We assume that force may be measured by
a spring balance. If a force, constant in magnitude and direc-
tion, be applied to a body initially at rest, the body will acquire
a certain velocity in a given time. If the same force be applied
to another body, and if the second body acquire the same veloc-
ity in the same time, the two bodies shall be said to have the
same mass. Thus different substances can be compared as to
their masses and on adopting an arbitrary mass as the unit in
the case of one substance, the unit can be determined in the
case of other substances.
MOTION OF A PARTICLE 119
It was proved experimentally by Newton that the forces with
which gravity attracts two masses equal according to the above
definition, are equal. And so one is led to infer the physical
law that the weight of a body is proportional to its mass. This
law affords a convenient means of measuring masses, namely,
by weighing.
In abstract dynamics, however (to quote from Maxwell),
matter is considered under no other aspect than that under which
it can have its motion changed by the application of force. Hence
any two bodies are of equal mass if equal forces applied to these
bodies produce, in equal times, equal changes of velocity. This
is the only definition of equal masses which can be admitted in
dynamics, and it is applicable to all material bodies, whatever
they may be made of.*
In Engineering it has become customary to define masses as
equal when their weights are equal. We have here a question
of a sense of values, and Maxwell has gone on record as declaring
unequivocally for the inertia property. To use weight to define
mass is like saying that two lengths are equal when the rods by
which we measure them have the same weight. Just as space
and time stand above mass and force, so, in its elementary impor-
tance, the inertia property towers above the law of gravitation.
* Maxwell, Matter and Motion, Art. XL VI.
CHAPTER IV
DYNAMICS OF A RIGID BODY
1. Motion of the Centre of Gravity. Let a system of particles
be acted on by any forces whatever. The latter may be divided
into two classes : i) the internal forces ; ii) the external forces.
By i) we mean that the particle mz exerts on m1 a force F12
which may have any magnitude and any direction whatever,
or in particular not be present at all, F12 = 0. The particle
ml exerts a force on w2, which is denoted by F2l. And now we
assume the physical law that action and reaction are equal and
opposite; i.e. that the vector F21
>2/2) is equal and opposite to the vec-
tor F12, or
Fi2 + F21 = 0.
FIG. 75 For convenience we will think of
the particles and forces as lying in
a plane. The transition to space of three dimensions is immediate.
Denote the components of a vector force F along the axes of
coordinates by X, Y. Then
*i2 + X21 = 0, F12 + 721 = 0.
Suppose there are three particles. Then Newton's Second Law
of Motion gives for the first of them the equations :
j. __
= X1 + Xn + X 1
There are in all three such pairs of equations, those in x being the
following :
d*x d2x
d2x
~dfi = -^3 +
120
DYNAMICS OF A RIGID BODY
121
On adding these three equations together, the components
Xu on the right, arising from the internal forces, annul one an-
other in pairs, and only the sum of the Xi remains :
a X\ . a Xn , Ci X-i -*r . *rr . •»•
m*~dP + m*~W + m*^W = Xl + X>2 + *3*
In a similar manner we infer, by writing down the three equa-
tions in y and adding, that
m* "eft2^ "^ "*2 ^ft2^ "^ "*3 ~eft^ ~~ * * ^~ * ? ~"~ * 3'
Precisely the same reasoning shows that if, instead of three,
we have any number, n, of particles, the internal forces annul
one another in pairs, and thus we obtain the result :
Coordinates of the Centre of Mass. The left-hand sides of
these equations admit a simple interpretation in terms of the
motion of the centre of mass of the system. The coordinates,
(xy y), of the centre of mass are given by the equations :
(2)
x =
V =
mn xn 2
mn
mnyn
m1 + • • • + wn
If we denote the total mass by M, then
! rnk xk =
Hence we have :
d*xk
a'Xk _ TUT-— V
~M ~ M ~jfLi 2*
fc l*l/ H/w j^
and thus Equations (1) can be written in the form :
(3)
122 MECHANICS
These equations are precisely Newton's Second Law of Motion
for a particle of mass Af, acted on by the given external forces,
each transferred to the particle. We can state the result as
follows.
THEOREM. The centre of mass of any system of particles moves
as if all the mass were concentrated there and all the external forces
acted there.
In the case of particles in space, there is a third equation,
(3) being superseded now by
(4) * = ^ , .
Remark. There is one detail in the statement of the theorem
that requires explicit consideration. We have written down
tbe differential equations of the motion, but we have not inte-
grated them. If we do not start the particle of mass M in coin-
cidence with the initial position of the centre of mass, it obvi-
ously cannot describe the same path. More than this, we must
give it the same initial velocity (i.e. vector velocity). Is this
enough to insure its always remaining in coincidence with the
centre of mass? The answer to this question is a categorical
Yes ; cf. Chapter III, § 17 and Appendix B.
Generalized Theorem. We have proved the theorem of the
motion of the centre of mass for a system of particles. In the
case of a rigid body, we can think of the body as divided up into
a large number of cells, each of small maximum diameter; the
mass of each cell as then concentrated at one of its points, and
the n particles thus resulting as connected by masslcss rods,
after the manner of a truss.* To this auxiliary system of par-
ticles the theorem as above developed applies. Arid now it is
intuitionally evident, or plausible, that the system of particles
will move in a manner closely similar to that of the rigid body,
when the cells are taken very small. One is tempted to say
* It is often necessary to use a truss, at some of whose vertices there are no
masses. We may think of minute masses attached at these points and acted on
by gravity or by no external forces at all. The effect of these small masses is to
modify slightly the value of M in Equations (4). And now it follows from the
theory of differential equations that the integrals of (4) are thereby also modified
only slightly. Hence the physical assumption is made, that Equations (4) hold
even when there are no masses at the vertices in question.
DYNAMICS OF A RIGID BODY 123
that the motion of the actual body is the limit approached by
the motion of the system of particles as n grows large and the
cells small. And this is, in fact, true. But this is not a mathe-
matical inference — far from it — it is a new physical postulate.
We thus extend the theorem and elevate it to a Principle.*
PRINCIPLE OF THE MOTION OF THE CENTRE OF MASS. The
centre of mass of any material system whatsoever moves as if all the
mass were concentrated there, and all the external forces acted there :
A)
2. Applications. The Glass of Water. Suppose a glass of
water is thrown out of a third-story window. As the water falls,
it takes on most irregular forms, breaking first into large pieces,
and these into smaller ones. The forces that act are gravity
and the resistance of the atmosphere, the latter spread out all
over the surfaces of the pieces. And now the Principle of the
last paragraph tells us that the centre of gravity moves as if
all the mass were concentrated there and all these forces trans-
ferred bodily (i.e. as vectors) to that point.
The Falling Chain. Let a chain hang at rest, the lower end
just touching a tablo, and let it be released. To determine the
pressure, F, on the table.
We idealize the chain as a uniform flexible string, of length /
and density p (hence of mass M = pi), and think of it as im-
pinging always at the same fixed point, 0, of the table. Let s
be the distance the chain has fallen and let x be the height of
the centre of gravity above the table. Then the Principle of
the Motion of the Centre of Mass gives the equation :
* A "Principle" in Mechanics is well described in the words of Professor Koop-
nrian (of. the Author's Advanced Calculus, p. 430): "According to the usage of
the present day the word principle in physics has lost its metaphysical implication,
and now denotes a physical truth of a certain importance and generality. Like
all physical truths, it rests ultimately on experiment ; but whether it is taken as a
physical law, or appears as a consequence of physical laws already laid down,
does not matter."
124
MECHANICS
(1)
Now,
~ a '
s - ld*s I
G *
' Opl
I dt*'
Moreover, from the laws of freely falling
bodies,
Fia. 76
On substituting those values in (1), we
obtain :
(2)
Hence
(3)
OP (s - 1) +
F - gpl.
or
F = gps + pv2,
F = 3gps.
This means that the pressure of the chain on the table is always
just three times the weight of that part of the chain which has
already come to rest on the table.
It appears, then, that F is made up of two parts, i) the pres-
sure gps on the table, of that part of the chain already at rest;
and ii) a pressure
(4) P = pv*,
due to the impact of the chain against the table.
A Stream of Water, Impinging on a Wall. Suppose a hose is
turned on a wall (or a convict !). To determine the pressure.
We idealize the motion by thinking of the stream as hitting
the wall at right angles, the water spattering in all directions
along the wall and thus giving up all its velocity in the line of
motion of the stream.
Dynamically, this is precisely the same
case as that of the chain falling on the table,
so far as the impact is concerned, and hence
the pressure is given by (4) : FIG. 77
PA
== ov •*
DYNAMICS OF A RIGID BODY 125
Example. A fire engine is able to send a 2 in. stream to a
vertical height of 200 ft. Find the pressure if the stream is
played directly on a door. Ans. 541 Ibs.
The Crew on the River. The crew is out for practice. Ob-
serve the cut-water of the shell and describe how it moves, and
why it moves as it does. What system do you decide to isolate ?
the shell? or the shell, oars, and crew?
EXERCISES
1. If a man were placed on a perfectly smooth table, how
could he get off?
2. If a shell were fired from a gun on the moon and exploded
in its flight, what could yo.u say about the motion of the pieces ?
3. A goose is nailed up in an air tight box which rests on plat-
form scales. The goose flies up. Will the scales register more or
less or the same?
4. A pail filled with water is placed on some scales. A cork
is held submerged by a string tied to the bottom of the pail.
The string breaks. Do the scales register more or less or the
same?
6. A man, standing in the stern of a row boat at rest, walks
forward to the prow. What can you say about the motion of
the boat?
6. When the man stops at the prow of the boat, boat and man
will be moving forward with a small velocity. Explain why.
7. A uniform flexible heavy string is laid over a smooth
cylinder, axis horizontal, and kept from slipping by holding
one end, A, fast, the part of the string from A up to the cylinder
being vertical. The part of the string on the other side of the
cylinder is, of course, also vertical, its lower end, /?, being below
the level of A, and the whole string lies in a vertical plane per-
pendicular to the axis of the cylinder. The string is released
from rest. Determine the motion, there being a smooth guard
which prevents the string from leaving the upper side of the
cylinder.
8. If, in Question 7, the difference in level between A and
R is 2 ft., and if the distance from A up to the cylinder is 8 ft.,
126
MECHANICS
compute the velocity of the string when the upper end reaches
the cylinder, correct to three significant figures.
9. Find how long it takes the upper end of the string to reach
the cylinder.
10. The sporting editor of a leading newspaper recently re-
ported a new stroke which a certain coach had developed, the
advantage of which was that it gave an even motion to the shell
and avoided the jerkiness 01 the old-fashioned strokes. Examine
this news item.
3. The Equation of Moments. Recall the formula for the
moment of a force F about the origin,
namely,
(1) xY-yX.
Consider a system of particles acted
— on by any external forces whatever,
and interacting on one another by
forces that are equal and opposite, but
are now assumed each time to lie in the lino joining the two par-
ticles in question. Moreover, the particles shall lie in a fixed
plane. Begin with the case of
three particles, as in § 1, and
write down the six equations that
express Newton's Second Law
of Motion for these particles.*
Next, form the expression :
FIG. 78
FIG. 79
ml
r
and compute its value from the equations in question, namely,
(x, Yl - Vl X,) + (x, Y12 - Vl X12) '+ (x, Yn - y, Xn).
The parentheses represent respectively the moments of FD F12,
F13 about the origin.
Now, do the same thing for the particle ra2, and finally, for ra3.
On adding these three equations together, it is seen that the
* It is important that the student do this, and do it neatly, and not merely gaze
at the three equations printed in § 1 and try to imagine the three not printed.
He should write out the full equation derived below from these, neatly on a single
line, and then write the other two under this one.
DYNAMICS OF A RIGID BODY
127
moments of the internal forces about the origin destroy one
another, and there remains on the right-hand side only the sum
of the moments of the applied forces.
If there are n > 3 particles, mlt w2, • • • , mn, the procedure is
the same, and we are thus led to the
THEOREM OF MOMENTS:
B)
mk
d*Xk
- yA).
We refrain from writing down the corresponding theorem in
three dimensions, because we shall have no need of it for the
present.
4. Rotation about a Fixed Axis under Gravity. Let the
system of particles of § 3 be rigidly connected, and let one point,
O, of the trass-work be at rest, so that the system rotates about
0 as a pivot. For example, take the case of a uniform rod, one
end of which is held fast, and which is released from rest under
gravity. Divide the rod into n equal parts, x
and concentrate the mass of each part, for defi-
niteness, at its most remote point. We thus
have a system of n particles, and we connect
them rigidly by a massless truss-work as shown
in the figure.*
We are now ready to compute each side of
Equation B) for the auxiliary system of n par-
ticles. Let r be the distance from 0 to any
point fixed in the rod. Then
(1) x «= r cos 0, y = r sin 0,
where 0 varies with the time, t, but r is constant with respect
to t. Hence
Fia. 80
(2)
dx . _ dO
-rr = — T Sill 0 -7T,
dt dt
$-»-•£
We observe next that, in all generality, by mere differentia-
tion, i.e. purely mathematically,
(3)
<L ( d-i — 4*\ - d*y _
di\x~di ydt)~xd?
* Cf. the footnote, § I.
128 MECHANICS
and we proceed to compute the parenthesis by means of Equa-
tions (1) and (2). We find :
dy dx ~dO
In the present case we have :
d ( dyk dxk\ _ d*8
dt\Xk dt yk~dt) -~rt~dT»
for rk does not change with the time, and so drk/dt = 0. Hence
The sum which here appears is the moment of inertia* of the
system about 0 :
Thus the left-hand side of the Equation of Moments reduces to
the expression :
Turning now to the right-hand side of B) we see that the
kth particle, m/t, yields a moment about 0 equal to the quantity
— mkgrk sin 0, and so the sum in question becomes
]£ - mkgrk sin 0, or - (2) mkrk)g sin 0.
t t
But mkrk = Mhy
where h is the distance from 0 to the centre of gravity, (?, of the
system of particles. Hence, finally,
(6) I^2=-MghsmO.
This is substantially the equation of Simple Pendulum Motion,
Chapter III, § 20 :
^\ d26 g . .
* Moments of inertia for such bodies as interest us here are treated in the
Author's Introduction to the Calculus, p. 323.
DYNAMICS OF A RIGID BODY 129
Hence the system of n particles oscillates like a simple pendulum
of length
l = m
or
(82) I = j, where 7 = M W,
k denoting the radius of gyration.
More precisely, what we mean by the last statement is this.
Let a simple pendulum be supported at 0, let its length be k*/h,
and let it be placed alongside the rod, the bob being at a point
distant Z from O. If now both be released from rest at the same
instant, they will oscillate side by side, though not touching each
other.
The Actual Rod. As n grows larger and larger, the massless
rod weighted with the n particles comes nearer and nearer to
the actual rod, dynamically. This is not a mathematical state-
ment. It expresses our feeling from physics for the situation —
our intuition. And so whon we say that the motion of the actual
rod is the limit approached by the motion of the auxiliary rod,
we are stating a new physical postulate. The result is, that the
actual rod oscillates like a simple pendulum of length
fc_2-iL2-?/
h # 3
EXERCISES
Apply the method set forth in the text, introducing each time
an auxiliary set of particles, and proceeding to the limit. Do
not try short cuts by attempting to use in part the result of the
exercise worked in the text.
1. A rod 10 ft. long and weighing 30 Ibs. carries a 20 Ib. weight
at one end and a 30 Ib. weight at the other. It is supported
at its middle point. Find the length of the equivalent simple
pendulum. Ans. 30 ft.
2. Equal masses are fixed at the vertices of an equilateral
triangle and the latter is supported at one of the vertices. If it
be allowed to oscillate in a vertical plane, find the length of the
equivalent simple pendulum.
130
MECHANICS
3. A rigid uniform circular wire * 6 in. in diameter and weigh-
ing 12 Ibs. has a 4 Ib. weight fastened at one of its points and
is free to oscillate about its centre in its own plane. Find the
length of the equivalent simple pendulum.
4. Equal particles are placed at the vertices of a regular hexa-
gon and connected rigidly by a weightless truss. The system
is pivoted at one of the particles and allowed to oscillate in a
vertical plane under gravity. Find the length of the equivalent
simple pendulum.
6. Generalize to the case of n equal particles placed at the
vertices of a regular n-gon.
6. The Compound Pendulum. Consider an arbitrary lamina,
or plane plate of variable density. Let it be supported at a
point 0 arid allowed to oscillate freely in its own plane, assumed
vertical, under gravity. This is essentially the most general
compound pendulum. To determine the motion.
Divide the lamina up in any convenient manner into small
pieces and concentrate the mass of each piece at one of its points.
Connect these particles with one
another and with the support at 0
by a truss-work. The auxiliary sys-
tem can be dealt with by the Prin-
ciple of Moments. Set
%k — ^k COS 6k,
Then
yk = rk sin 0*.
FIG. 81
Now, draw a line in the lamina, —
for example, the line through 0 and
the centre of gravity, G, of the
particles, — and denote the angle it makes with the axis of x by 0.
Then
where ctk varies with fc, but is constant as regards the time. Hence
d0k = de d2Bk = d*0
dt " dt' dP ~~ dt*
* By a wire is always meant a material curve.
DYNAMICS OF A RIGID BODY 131
Thus the left-hand side of the Equation of Moments, § 3,
becomes
n\ V 1&® Tdze
(1) ?m*r*^ = /^>
where 7 denotes the moment of inertia of the system of particles
about 0.
The right-hand side of B), § 3, can be written
(2) 2) •" mkgyk = -02) mky*'
The last sum has the value My, whore the coordinates of G
are denoted by (x, y). Let the distance from 0 to G be h. Then
y = h sin 0
and (2) becomes
(3) - M0/i sin 6.
On equating (1) and (3) to each other, we have
d~n
(4) /5y[=- JfffAsinfl.
This is the Equation of Simple Pendulum Motion, and
It appears, then, that the auxiliary system of particles oscillates
like a simple pendulum. As we allow n to increase without limit,
the maximum diameter of the little pieces approaching 0, it seems
plausible that the motion will approximate more and more closely
to that of the actual compound pendulum, and this consider-
ation leads us to lay down the physical law, or postulate, that
Equation (4) holds for the compound pendulum, where 7 and
h now refer to the latter body.
Remark. We have thought of the mass of the compound
pendulum as two-dimensional, or lying in a plane. But this is
obviously an unnecessary restriction. Conceive a block of
granite, blasted from the quarry — as irregular and jagged as
you please. Mount it on two knife-edges, so it can swing about
a horizontal axis. Now this block will obviously oscillate exactly
as a plane lamina perpendicular to the axis would, if the mass
of the actual block were projected parallel to the axis on a plane
at right angles to the axis.
132 MECHANICS
The above "obviously" is not to be taken mathematically,
but is a new physical law, or postulate. It is true that when
we come to treat the general case of motion in three dimensions,
this postulate will be merged in more general ones.
EXERCISES
Find the length of the equivalent simple pendulum when the
compound pendulum is one of the following.
1. A uniform circular disc, free to rotate in its own plane
about a point in its circumference. Ans. I = f r.
2. A circular wire, about a point of the wire. Ans. I = 2r.
3. Question 1, when the axis is tangent to the disc.
Ans. I = fr.
4. Question 2, when the axis is tangent to the wire.
Ans. I = -Jr.
6. A rectangular lamina, about a side.
6. A square lamina, about a vertex.
7. A triangle, about a vertex.
6. Continuation. Discussion of the Point of Support. Let
70 = Mfc2
be the moment of inertia of the compound pendulum about a
parallel axis through the centre of gravity, (?. By the theorem
of § 10 the moment of inertia about the actual axis will be :
and the length of the equivalent simple pendulum is soon from
(5), § 5, to be :
(6) I = ^±A2-
The question arises: What other points of support, 0 (i.e.
what other parallel axes), yield the same period of oscillation?
Clearly they are those, and only those, whose distance, x,
from 0 satisfies the equation,
,_*• + *
x
(7) x* - Ix + Jk2 = 0,
DYNAMICS OF A RIGID BODY
133
where k and I are given, and where, more-
over, (6) is true, or
(8) h* - Ih + k2 = 0.
One root of Equation (7) is xl = h.
The other is seen to be
_i j, _ *2
^ - I - ft _ ~
We can state the result as a theorem.
FIG. 82
THEOREM. The locus of the points 0, for which the time of oscil-
lation is the samey consists of two concentric circles with their centre
at G, their radii being
, , k2
h and -r-
fi
EXERCISES
1. Draw two concentric circles about G, of radii h and k*/h.
Show that the length, Z, of the equivalent simple pendulum cor-
responding to an axis through a point 0 on one of these circles
is obtained by drawing a line from 0 through G, and terminating
it whore it meets the other circle.
This theorem is due to Huygens.
2. Show that the locus of the points of support, for which
the time of oscillation is least, form a circle with G as centre and
of radius k.
7. Kater's Pendulum. The experiment for determining the
value of g, the acceleration of gravity, by means of a simple
pendulum and the formula
T =
9
is familiar to all students of physics and mathematics. The
chief error in the result arises from the error in determining I.
The bob is not sensibly a particle and the string stretches.
To attain greater accuracy, Kater made use of Huygens's
Theorem, §6, Ex.1, constructing a compound pendulum that
could be reversed. It consists essentially of a massive rod,
or bar, provided with two sets of adjustable knife-edges. These
edges lie in two parallel lines, and the centre of gravity, (?, is
134 MECHANICS
situated in their plane, at unequal distances, h and A', from them.
The knife-edges are now so adjusted experimentally that the
period when the pendulum oscillates about the one pair is the
same as when it is reversed and allowed to oscillate about the
other pair. Since
I = h + h',
the determination of the length of the equivalent simple pendulum
can now be made with great accuracy by measuring the distance
between the knife-edges. Indeed, the accuracy in thus deter-
mining g is now so great that very small errors, like those due
to the buoyancy of the air, the changes in the pendulum due to
changes in temperature, and the give of the supports have to
be considered. For an elaborate and interesting account, cf.
Routh, Elementary Rigid Dynamics, § 98 et seq.
8. Atwood's Machine. An Atwood's Machine consists of a
pulley free to rotate about a horizontal axis, and a string passing
over the pulley and carrying weights, M and M + m, at its two
ends. It may be used to measure the acceleration of gravity.
Our problem is to determine the motion of the system. The
"system" which we choose to isolate is the complete system
of pulley and weights, the mass of the string being assumed
negligible. This is not a rigid system, but still, if we replace
the pulley by a system of particles rigidly connected, the internal
forces of the complete auxiliary system will satisfy the hypothesis *
of § 3, and thus the Equation of Moments will hold.
For the auxiliary system of particles due to the wheel the
contribution to the left-hand side of the Equation of Moments,
B), § 3, becomes as in the case of the compound pendulum :
where I denotes the moment of inertia of this system about the
axis, and B is the angle through which the wheel has rotated.
* Consider a short interval of time in the duration of the motion. In the
auxiliary system, let each vertical segment of the string be fastened to a particle
near the point of taiigericy of the string in the actual case. Then it is plausible
physically that the motion of the auxiliary system during this short interval differs
but slightly from that of the actual system. Hence we may assume that the force
of the string always acts at the points of tangency with the wheel, and neglect the
rest of the string which is in contact with the wheel. But this is a new physical law.
DYNAMICS OF A RIGID BODY
135
Let the radius of the wheel (more precisely, of the groove in
which the string lies) be a. Observe, too, that
yl = const. + a0,
= const. — aO.
Thus the remaining contributions to the left-
hand side of B), § 3, will be
(2) (M + m) a2 -^ + Ma2 -^
The right-hand side of B) reduces to
(3) (M + m) ga — Mga = mga.
Thus B) becomes :
dt2
This, for the auxiliary system of particles. And now we assume,
physically, that the limit approached by the motion of the auxiliary
system is the motion of the actual system ; i.e. that Equation (4)
holds for the actual system.
Let s denote the distance the weight and rider have descended.
Then s = aO, and from (4) it follows that
(5)
mga'
dt2 I + (2M + m) a2
On integrating this equation we have, in particular, that
(6)
s =
I + (2M + m) a2
Corresponding values of s and t can be observed experimentally.
Thus Equation (6) is equivalent to a linear equation in the two
unknowns, //a2 and g :
~
II
0.
If M is held fast and m is given different values, it is clear that the
coefficient of g will take on different values, and so we shall have
two independent linear equations for determining the unknown
physical constants, 7/a2 and g.
136 MECHANICS
EXERCISES
In working these exercises use the method, not the result, of
the text. Begin each time by drawing a figure.
1. Suppose that the wheel is a uniform circular disc weighing
10 Ibs., and that 5 Ib. weights are fastened to the two ends of
the string. What will be the acceleration due to a 1 oz. rider?
2. Work the case in which the wheel is a hoop, i.e. a uniform
circular wire, the masses of the spokes being negligible; and
show that the acceleration of the rider does not depend on the
radius, but only on the mass of the hoop, and M and m.
3. Determine the tensions in the string in the general case.
4. Find the reaction on the axis.
6. Prove the assertion in the text about the coefficient of g's
taking on different values when m is varied.
6.* How rough must the string be in the general case, in order
not to slip?
9. The General Case of Rotation about a Point. Consider
an arbitrary rigid body in two dimensions, acted on by any forces
in its plane, and free to rotate about a point 0, i.e. about an
axis through 0 perpendicular to the plane. Then, I say, its
motion is determined by the Principle of Moments,
B) / -jT2 = 5J Moments about 0.
The Principle is rendered plausible by dividing the actual
distribution into small pieces, as in the example of the corn-
pound pendulum and the Atwood's machine, and observing that
the Principle is true for the auxiliary system. The limit ap-
proached by the motion of the auxiliary system is the motion
defined by Equation B) of the present paragraph. And thus we
are led to lay down the physical postulate that this is the motion
of the actual system. Equation B), then, is an independent
physical law, made plausible by the mathematical considerations
set forth above, but not following mathematically from them.
The Effect of Gravity. Whenever gravity acts, the contribu-
tion of this force to the right-hand side of Equation B) can
* This problem is more difficult than the others, and is essentially a problem in
the Calculus ; cf. the author's Advanced Calculus, Chapter 14, § 8.
DYNAMICS OF A RIGID BODY 137
always be written as the moment of a single force, that force
being the attraction of gravity on a single particle of mass equal
to the mass of the entire body and situated at the centre of gravity
of the body. This is true in the most general case of motion,
when no point of the body is permanently at rest. Here, again,
we have a new physical postulate.
EXERCISES
1. A turn table consisting of a uniform circular disc is free
to rotate without friction about its centre. A man walks along
the rim of the table. Find the ratio of the angle turned through
by the table to the angle described by the man, if man and table
start from rest.
2. The same problem when the man walks in along a radius
of the table, — the system not being, however, initially at rest.
10. Moments of Inertia. The moment of inertia of the
simpler and more importanjb distributions of matter are deter-
mined by the methods of the Integral Calculus ; cf. for example
the author's Introduction to the Calculus, p. 323, and the Advanced
Calculus, pp. 58, 79, 88.
Ml2
1. A uniform * rod of length I about one end : — 5—
o
TI/T 2
2. A rod of length 2a about its midpoint : —$ —
3. A circular disc about its centre : —5 —
&
Mr2
4. A circular disc about a diameter : ~T~~'
5. A square about its centre : f M a2.
6. A square about a side; cf. Example 1.
7. A scalene triangle about a side : — —,
where h denotes the altitude.
o A i i * r *
8. A sphere about a diameter :
9. A cube about a line through the centre parallel to an edge ;
cf. Example 5.
* It will henceforth be understood that the distribution is uniform unless the
contrary is stated.
138 MECHANICS
A GENERAL THEOREM. The moment of inertia of any distribu-
tion of matter whatever, about an arbitrary axis, is equal to the mo-
ment of inertia about a parallel axis through the centre of gravity,
increased by Mh2:
where h denotes the distance between the axes.
We will begin by proving the theorem for a system of particles.
Let the first axis be taken as the axis of z in a system of Cartesian
coordinates, (x, y, z) ; and let the second axis be the axis of z'
in a system of parallel axes. Then
/ = 2 mk (x,? + 2/*2), 70 = 2 w* W + yi2).
Since
x = x' + x, y = y' + y,
it follows that
) = 2
k .
2x mk xi + 2y ™<k yi-
The last two terms vanish because 0' is the centre of gravity,
and hence
2 mk x't = o, 2
It remains merely to interpret the terms that are left, and
thus the theorem is proved for a system of particles.
If we have a body consisting of a continuous distribution of
matter, we divide it up into small pieces, concentrate the mass
of each piece at its centre of gravity, form the above sums,
and take their limits. We shall have as before 2mkXrt = 0,
S mk yi = 0, and hence
lim 2 ™*(**2 + 2/t2) = Km
n-oo j7 n = oc
or
since these limits are by definition the moments of inertia for the
continuous distribution.
DYNAMICS OF A RIGID BODY 139
Example. To find the moment of inertia of a uniform cir-
cular disc about a point in its circumference. Here, 70 = %Mr2
and h = r. Hence T Q ,. , 0
/ = f Mr2.
11. The Torsion Pendulum. Let a rod be clamped at its
mid-point to a steel wire and suspended, the rod horizontal and
the wire vertical. Let the rod be displaced slightly in its hori-
zontal plane, the wire remaining vertical, arid then released. To
determine the motion.
The forces acting on the rod amount to a couple, due to the
torsion of the wire, and the moment of the couple is propor-
tional to the angle through which the rod is displaced — such
is the law of elasticity. Thus the Principle of Moments, § 9,
yields in this case the differential equation,
Ma2
where / = — ^— is the moment of inertia of the rod, and K is the
constant of the wire.
Equation (1) is the equation of Simple Harmonic Motion,
and thus the period of oscillation,
(2) T = 2*
is the same, no matter what the initial displacement may have
been, provided merely that the distortion of the wire is not so
great as to impair the physical law above stated, and provided
damping is neglected.
12. Rotation of a Plane Lamina, No Point Fixed. Let a rigid
plane lamina be acted on by any forces in its plane, and let it
move in its plane. To determine the motion.
The centre of gravity will move as if all the mass were con-
centrated there and all the forces were transferred to that point ;
§ 1. It remains to consider the rotation.
PRINCIPLE OF MOMENTS. The lamina rotates as if the centre
of gravity were held fast and the same forces acted on the lamina as
those applied in the actual casey
(1) ^"77/2 = S Moments about (?,
140 MECHANICS
where I denotes the moment of inertia about the centre of gravity, G;
6 is the angle that a line fixed in the lamina makes with a line fixed
in the plane, and the right-hand side is the sum of the moments of
the forces about G.
Proof. Consider first a system of particles rigidly connected.
Let (x, y) be axes fixed in the plane, and (£ , 77) parallel axes whose
origin is at G. Then
(2) x = £ + a, y = 77 + y,
and
the omitted terms vanishing for the reason that
>k £* = 0, ]£ mk rjk = 0,
and hence, too,
Remembering that
dt\dt
we see that Equation B), § 3, here becomes :
Because
xk = & + x, yk = rik + y,
the right-hand side of Equation (4) becomes :
2 fen - rjkXk) +
7
Since
DYNAMICS OF A RIGID BODY 141
it follows that
On subtracting this equation from (4), there remains :
(5) *
In this equation is contained the proof of the theorem for a
system of n particles. For, the left-hand side reduces to the
left-hand side of (1), since the distance of the point (&, ?/*) from
the centre of gravity, G, does not change with t; and the right-
hand side expresses precisely the sum of the moments of the
applied forces about G.
Finally, we pass to a continuous distribution of matter in
the usual way, laying down a new physical postulate to the effect
that Equation (1) shall hold for all rigid distributions of matter
in a plane.
13. Examples. A hoop* rolls down a rough inclined plane
without slipping. Determine the motion.
The forces are: the force of gravity and the reaction of the
plane. Let the latter force be resolved into a normal com-
ponent, R, and the tangential force of fric-
tion, F, acting up the plane. Then, for
the motion of the centre of gravity, we shall
have:
-. FIQ 84
The second equation for the motion of the centre of gravity
merely tells us that
(2) R = M g cos a,
a fact that we could have guessed, since the centre of gravity
always remains at the same distance from the plane. However,
let us formulate the second equation, and prove our guess right.
Let y denote the distance of the centre of gravity from the plane.
* A pipe, the thickness of which is negligible, when placed on the plane with its
axis horizontal, would move in the same way. The two problems are dynamically
identical.
142 MECHANICS
Then
But y = a, the radius of the hoop, and so the left-hand side of
this equation is 0.
Turning now to the rotation of the hoop, we write down Equa-
tion (1) of the Theorem, § 12 :
(4) /-^ = aF, I = Ma2.
Since there is no slipping,
(5) 5 = a0,
where, for convenience, we take as 6 the angle that the radius
drawn to the point of contact with the plane at the start has
turned through, 5 being also 0 at the start.
• Equations (1) and (4) can now be written in the form :
Ma -JTJ = Mg sin a — F,
(6)
Ma*<^ = aF.
On eliminating F between these equations, we find :
or
(8) ^ = |sina.
Hence it appears that the centre of the hoop moves down the
plane with just half the acceleration it would have if the plane
were smooth.
Equation (2) appears to have played no part in the solution.
But we have assumed that there is no slipping, and so F cannot
be greater than pR :
(9) F ^ »R.
To ascertain what this condition means for the coefficient
of friction, ju, and the steepness of the plane, a, solve Equations
(6) for F and substitute :
F — ^y s*n a
2 '
DYNAMICS OF A RIGID BODY 143
Mg sin a . .
— 2-5 ^ M^ cos a,
(10) tan a ^ 2M.
Hence it appears that a may not exceed tan"1 2ju.
EXERCISES
1. Show that, if the hoop be released from rest,
gt . at* .
v = ~ sm a, s = ~- sm a,
v2 = 0s sin a.
2. Show furthermore that
at . at2 .
« = TJ- sin a, ^ = T- sin a,
<&& 4<z
W2 — *L gjn a
a
3. Solve the problem studied in the text for a sphere. Show
that
d*6 5g . d*s 50 .
^5 = ^ am a, —^J^Bina.
4. Prove that the sphere will slip unless
tan a g JJLI.
6. Make a complete study of a disc, or solid cylinder.
14. Billiard Ball, Struck Full. A billiard ball is struck full
by the cue. To determine the motion.
The forces are: the force of gravity, acting downward at the
centre of gravity, and the reaction of the billiard table, which
yields a vertical component, R, and a horizontal component, F.
Let s be the space described by the centre of the ball, and 6, the
angle through which the ball has turned.*
The Principle of the Motion of the Centre of Gravity, § 1,
yields the equations :
(1)
__
dt*
F
R = Mg FIG. 85
* It is of prime importance that the student begin each new problem, as here,
by drawing a figure showing the forces and the coordinates used in setting up the
differential equations of the motion. It is well, too, to note at the same time any
auxiliary relations, as in the present instance, F =» pR.
144 MECHANICS
The Principle of Rotation about the Centre of Mass, § 12,
yields the equation :
tv\ rd*e nw T
(2) 7 = °^ I
Finally, so long as there is slipping,
(3) F = MB.
From Equations (1), (2), and (3) it appears that
/A \ &S
(4.) is=-«r'
d<2 ~ 2a
The integrals of these equations are as follows :
,g . f v = v0 - ngt, s = v0t-
\ t>2 = vl -
and
Thus as the ball advances, its centre moves more and more
slowly, while the speed of rotation steadily increases. Finally,
pure rolling will set in. This takes place when the velocity of
the point of the ball in contact with the table is nil. Now, the
velocity of this point of the ball is made up of two velocities,
namely, i) the velocity of translation, or the velocity the point
would have if the ball were not rotating, i.e. v, as given by (50;
and ii) the velocity due to rotation, or the velocity the point
would have if the ball were spinning about its centre, thought of
as at rest. The latter is a velocity of ao> in the direction opposite
to the motion of the centre, and is given by (52). Thus the ve-
locity forward of the point of the ball in contact with the table is
(6) v — ow.
Slipping continues so long as this expression is positive, and
ceases when it vanishes :
(7) v — oo> = 0.
DYNAMICS OF A RIGID BODY 145
The time is given by the equation
or
The corresponding value of s is seen to be :
The angle through which the ball turns is
Finally,
(11) »1 = |>0, «! = T2'
7 la
EXERCISES
1. Solve the same problem in case the table is slightly tipped
and the ball is projected straight down the plane.
2. Work the last problem with the modification that the ball
is projected straight up the plane.
15. Continuation. The Subsequent Motion. At the end of
the stage of the motion just discussed, the ball has both a mo-
tion of translation and one of rotation, the point of the ball
in contact with the table being at rest. If from now on the force
exerted by the table on the ball consists solely of an upward
component R and a tangential component F, the latter force will
vanish, and the ball will continue to roll without slipping. For,
suppose the table is rough enough to prevent slipping. Then
s = oQ, and since equations (1) and (2) still hold, we have :
Hence F vanishes, and the angular and linear accelerations are
both 0, too.
But in practice the ball will slow up. How is this to be ac-
counted for, if the resistance of the air is negligible? The answer
is, that the reaction of the table is not merely a force, with com-
ponents R and F. but, in addition, a couplej the moment of which
146 MECHANICS
we will denote by C. This couple has no influence on the motion
of the centre of gravity; thus Equations (1), §14, remain as
before. But Equation (2) now becomes
O
(13)
FIG. 86 Furthermore,
(14)
Hence
dP 7 Ma' dt* 7 Ma*' la
Since C is small, the ball slows up gradually.
EXERCISES
1. If the centre of the ball was moving initially at the rate of
6 ft. a sec. and if the ball stops after rolling 18 ft., show that
C = IMa.
2. If the initial velocity of the centre was VQ and if the ball
rolled I ft., show that C is proportional to the initial kinetic energy
and inversely proportional to the distance rolled.
16. Further Examples, i) HOOP ON ROUGH STEEPLY IN-
CLINED PLANE. Suppose, in the Example studied in the text
of § 13, that a does exceed tan"1 2/*. What will the motion then
be, the hoop being released from rest?
Equations (1), (2), and (4) will be as before. But now (5) is
replaced by the equation :
(i) p = &
all the friction now being called into play. On eliminating P
and R, we find :
(2)
dzs
- = <7(sin<* -
-£= = — cos a.
dt2 a
The integrals of these differential equations can be written
down at once. In particular, it is seen that the ratio of s to 6
is constant, if the hoop starts from rest :
s a (sin a — u cos a) ,, , % -,
- = — - - - '- = o(tan a cot X - 1).
0 p cos a
DYNAMICS OF A RIGID BODY
147
FIG. 87
The last parenthesis has the value 1 when tan a = 2/z, and
is > 1 when a is larger. Thus the motion is one in which a cir-
cle of radius , „_
r = a (tan a cot X — 1)
and centre at the centre of the hoop rolls
without slipping on a line parallel to the
plane and beneath it. We have here an
illustration of the general theorem that any
motion of a lamina in its own plane can be
realized by the rolling without slipping of a curve drawn in the
lamina on a curve drawn in the plane ; cf. Chapter V, § 4.
ii) LADDER SLIDING DOWN A SMOOTH WALL. First, draw a
figure representing the forces and the coordinates.
The three equations of motion thus become :
(3)
FIG. 88
"77/2 = a^ s*n 0 "~ a^ cos ^ ^ =
U/t"
With these three Dynamical Equations are associated two
Geometrical Equations :
(4) x = a cos 0, y = a sin 6.
These five equations determine the five unknown functions x, y,
0, R, S, the time being the independent variable ; or they deter-
mine five of the variables x, y, 0, R, /S, t as functions of the
sixth. Eliminate 72, S between the first three equations:
<5> '¥.-
> -^ — M a cos 0 —~ — Af gra cos 6.
From the Geometrical Equations follows :
dx . „ dO
-~~ = a cos 6 -jr,
dt dt
d2x
148 MECHANICS
Combining these with (5) and reducing we obtain :
This differential equation can be integrated by the device of
multiplying through by 2d0/dt and then integrating each side
with respect to t :
30 0dO
-
dt
Since
d
it follows that
fdO\2 30 .
(dt) =-2Hs
The constant of integration, C, is determined by the initial
conditions. If the ladder is released from rest, making an angle a
with the horizontal, then dB/dt = 0 and 6 = a initially, and so
0 = - jj£ sin a + C.
Zd
Hence, finally,
*> ©'-£(-..-*.)•
To find where the ladder will leave the wall. This question is
answered by computing R and setting it = 0 :
E» ™d2x ™ - />d20 „ */de\2
R = M --trr = — Ma smO-jrz — Ma cos 6 [ -77 ) ,
at1 at2 \at/
(8) R = f Mg cos 6 (3 sin 6 - 2 sin a).
Hence R = 0 when
3 sin 6 — 2 sin a = 0.
Let ft be the root of this equation :
ft = sin-1 (f sin a).
Observe that cos 8 cannot vanish when 0 g 6 < £•
DYNAMICS OF A RIGID BODY 149
The intuitional evidence is here complete : — the ladder leaves
the wall and slides along with the lower end in contact with the
floor. But suppose a person is unwilling to trust his intuition
and says: — "Ah, you have not proven your point in merely
showing that R = 0 for a certain value of 0. The ladder might
still remain in contact with the wall, R increasing as the ladder
continues to slide." The logic of this objection is valid. The
objection can be met as follows.
Think of the upper end of the ladder as provided with a ring
that slides on a smooth vertical rod. Then the ladder will not
leave the wall. How about R in this case? Formula (8) now
holds clear down to the floor ; but R < 0 when 6 < sin"1 (| sin a),
and so the vertical rod has to pull on the ladder instead of push-
ing. This proves that our intuition was correct.
The Time. From Equation (7) it appears that
(9)
Vsin a — sin 6
This integral cannot be evaluated in terms of the elementary
functions. On making the substitution
x = sin 6,
the integral goes over into an Elliptic Integral of the First Kind,
and can be treated by well-known methods; cf. the Author's
Advanced CakuluSj Chapter IX.
iii) COIN ON SMOOTH TABLE. A coin is released from rest
with one point of the rim touching a smooth horizontal table.
To determine the motion.
The forces acting are: Gravity, Mg, down, and the reaction,
R, of the table upward. Thus the centre of gravity of the coin
descends in a right line. Let its height above the table be de-
noted by y. Then the further Dynamical Equations become :
(10)
=-aRcosO.
at*
The Geometrical Equation is :
(11) y — osinfl.
150 MECHANICS
On eliminating R and y we find :
(12) (fc2 + a2 cos2 0) ^ - a2 sin 0 cos (9 (~f)2 = - ag cos 0.
(Zf \ttf /
This differential equation comes under a general class, namely,
those in which one of the variables fails to appear explicitly.
The general plan of solution in such cases is to introduce a new
variable,
And this can be done here. But in the present case there is a
short cut, due to the special form of the differential equation.
It is observed that, on multiplying the equation through by
2dQ/dty the left-hand side becomes the derivative of a certain
function with respect to ty so that the equation takes on the form :
(13)
On integrating each side of this equation with respect to t, w°
find:
(F + o2 cos2 0) =-2ag sin 9 + C.
\Gfv /
To determine C make use of the initial conditions, dO/dt = 0
and 0 = a. Thus
(14) (A;2 + a2 cos2 ^{^) = 2a0 (sin a - sin 0).
The angular velocity, w, of the coin when it falls flat on the
table is given by the equation :
_ 2ag sin a
Wl ~ » »
But here is an assumption, namely, that the point of the coin
initially in contact with the table remains in contact till 0 = 0.
This is plausible enough physically; but in this gunws, is there
not an appreciable admixture of unimaginativeness and the
question which the moron so frequently asks: "Why shouldn't
it?" The angular velocity dd/dt of the coin is steadily increas-
ing, as we see both intuitionally and from (14). May it not
increase to such an extent that the lowest point in the coin may
DYNAMICS OF A RIGID BODY 151
kick up and leave the table before the centre comes clear down?
The moron certainly cannot answer this objection by physical
intuition.
It is here that mathematics sits as judge over the situation.
Replace the actual problem by one equivalent during the early
stage of the motion, and see whether this stage lasts through to
the end. Let the lowest point of the coin be provided with a
ring that slides on a smooth horizontal rod. Then the coin will
fall as we guessed. Compute now the reaction, R. The test
is: Does R remain positive throughout the motion? We leave
it to the student to find out.
EXERCISES ON CHAPTER IV
1. A homogeneous solid cylinder is placed on a rough inclined
plane and released from rest. Will it slip as it rolls, or will it
roll without slipping?
Ans. It will slip if the angle of inclination of the plane
is greater than tan"1 (|M)-
2. The same problem for a homogeneous spherical shell
(material surface).
3. A billiard ball is set spinning about a horizontal axis and
is released, just touching the cloth of the billiard table. How
far will it go before pure rolling sets in ?
4. A circular disc has a string wound round its circumference.
The free end of tho string is fastened to a peg, A, and the disc is
released from rest in a vertical plane with its centre below the
level of A, and the string taut and vertical. Show that the
centre of the disc will descend in a vertical right line with two-
thirds the acceleration of gravity.
6. The disc of the preceding problem is laid flat on a smooth
horizontal table ; the string is carried over a smooth pulley at the
edge of the table, and a weight equal to the weight of the disc
is attached to the end of the string. The system is released from
rest, the string being taut and the weight hanging straight down.
Show that the acceleration of the weight is three-fourths that
of gravity.
6. Find the tension of the string in the last question.
7. Solve the problem of Question 3 with the modification
that the table is inclined at an angle a to the horizon.
152 MECHANICS
Discuss in full the case that the rotation of the ball is in such
a sense that the ball moves down the plane faster than it would
if it had not been rotating.
8. Study the problem of the last question when the rotation
is in the opposite sense.
9. A billiard ball is placed on a billiard table inclined to the
horizontal at an angle a, and is struck full by the cue, so that
it starts off straight down the plane without any initial rotation.
Study the motion.
10. The same problem when the ball is so struck that it starts
straight up the plane.
11. If a man were placed on a perfectly smooth table, how
could he turn round ?
. 12. A plank can rotate about one end, on a smooth horizontal
table. A man, starting from the other end, walks toward the
pivot. Determine the motion.
13. A smooth tube, the weight of which may be neglected,
can turn freely about one end. A rod is placed in the tube and
the system is released from rest with the rod horizontal. Deter-
mine the motion.
14. A spindle consists of two equal discs connected rigidly
with an axle, which is a solid cylinder. The spindle is placed
on a rough horizontal table, and a string is wound round the axle
and carried over a smooth pulley above the edge of the table.
A weight is attached to the lower end of the string and the system
is released from rest. Determine the motion.
Consider first the case in which the string leaves the axle from
the top ; then, the case that the string leaves the axle from the
bottom. In each case, the segment of the string between the
axle and the pulley shall be horizontal and at right angles to the
axis, and the part below the pulley, vertical.
15. The centre of gravity of a four-wheeled freight car is
5 ft. above the track and midway between the axles, which are
8 ft. apart. The coefficient of friction between the wheels (when
they are locked) and the track is ^-. If the car is running at the
rate of 30 m. an h., in how short a distance can it be stopped by
applying the brakes to the rear wheels only? How far, if the
brakes are applied to the fronfr wheels only?
DYNAMICS OF A RIGID BODY 153
16. A uniform rod is suspended in a horizontal position by
two vertical strings attached to its ends. One string is cut.
Find the initial tension in the other one.
17. A hoop is hung up on a peg and released. Find whether
it will slip.
18. A uniform circular disc, of radius 1 ft. and weight 10 Ibs.,
can rotate freely about its centre, its plane being vertical. There
is a particle weighing 1 Ib. fixed in the rim, and a fine inextensible
weightless string, wound round the rim of the disc, has a weight
of P Ibs. fastened to it. The system is released from rest with
the 1 Ib. weight at the lowest point and the other weight hanging
freely at the same level. How great may P be, if the 1 Ib. weight
is not to be pulled over the top?
19. A billiard ball rolls in a punch bowl. Determine the
motion.
20. A solid sphere is placed on top of a rough cylinder of revolu-
tion, axis horizontal, and slightly displaced, under the action of
gravity. Find where it will leave the cylinder.
21. A uniform rod is released from rest, inclined at an angle,
with its lower end in contact with a rough horizontal plane.
Will it slip at the start? Determine the motion.
22. A packing box is sliding over an icy side walk. It comes
to bare ground. Will it tip up?
CHAPTER V
KINEMATICS IN TWO DIMENSIONS
1. The Rolling Wheel. When a wheel rolls over a level road
without slipping, the nature of the motion is particularly acces-
sible to our intuition, for the points of the wheel low down move
slowly, the point in contact with the ground actually being at
rest for the instant, and it is much as if the whole wheel were
pivoted at this point and rotating about it as an axis. This is,
in fact, precisely the case, the velocity of each point of the wheel
at the instant being the same as if the wheel were rotating per-
manently about that point.
If the wheel is skidding, it is not so easy to see that a similar
situation exists, — and yet it does. No matter how the wheel is
moving, provided it is rotating at all, there is at each instant a
definite point (far away it may be), about which the wheel rotates
at that instant. This point is called the instantaneous centre.
To prove this assertion, we will begin by giving a general formu-
lation of the problem of the motion of any plane lamina in its
plane. It makes the problem more concrete to think of an actual
lamina, like a disc or a triangle or a finite surface, S. But we
are really dealing with the motion of the whole plane, thought
of as rigid.
The motion may be described mathematically as follows.
Draw a pair of Cartesian axes in the moving plane; i.e. think
X
FIG. 90
of this plane as a sheet of paper, and draw the (£, ?/)-axes in red
ink on the paper. Assume further a system of axes fixed in
154
KINEMATICS IN TWO DIMENSIONS 155
space — the (x, £/)-axes. Then the (£, ^-coordinates of an arbi-
trary point P of the moving plane are connected with the (x, y)-
coordinates of the same point by the relations :
x = XQ + £ cos 6 — 77 sin 0,
y — 2/0 + ? sin 0 + y c°s 0-
The position of the moving plane is known when one point,
as 0', is known and the orientation, as given by 0, is known.
The motion may, therefore, be completely described by stating
how XQ, 2/0) 0 vary with the time; i.e. by saying what func-
tions x0, 2/o> 0 are °f £ :
Wo shall assume at the outset that these functions are continuous
and possess continuous derivatives of the first order. Later, it
will bo dosirable to restrict them further by requiring that they
have continuous derivatives of the second order.
EXERCISE
Express £ and rj in terms of x and yy i) geometrically, by reading
the result off from the figure ; ii) analytically, by solving Equa-
tions A) for £, 77. The formulas are :
f f = (x - x0) cos 0 + (y - 2/0) sin 0,
I rj = — (x — XQ) sin 0 + (y — yQ) cos 0.
2. The Instantaneous Centre. Let P be a point fixed in the
moving plane — mark it with a dot of red ink on the sheet of
paper. Lot the coordinates of P be (x, y). Then they are deter-
mined as functions of t by Equations A), (£, rj) being the coordi-
nates of P with reference to the moving axes. Of course, £ and
TJ are constants with respect to the time, for the red ink dot does
not move in tho paper — it moves in space.
The vector velocity, v, of P in space can be determined by
moans of its components along the axes of x and yy which are fixed
in space, Chapter III, § 15 :
« -^ -, -^-
156
MECHANICS
These derivatives can be computed in terms of the known func-
tions (1), namely, x0, yQ, 0, and of their derivatives, by means
of Equations A). Thus
(2)
dx ___ dxQ ,
dy _ dy* , f
dt dt
The parentheses that here enter are seen from Equations A)
to have the values :
Hence
-(y-
dx _
dt ~
X - XQ.
~dt
dy __ dy, , _
dt~~dt+(X
dB
These equations express the components of the vector velocity
v of the point P along the axes fixed in space, in terms of the
coordinates (x, y) of P and the known functions (1).
New Notation. Since derivatives with respect to the time
occur frequently in the work which follows, the Newtonian nota-
tion with the dot is expedient :
(4)
. _dx
X ~ '
x =
dt*'
etc.
Thus the formula for the components of the velocity assumes tho
final form :
B)
* = z0 - (y - 2/0) 0>
y = 2/0 + (* - #o) *•
The Instantaneous Centre. We now inquire what point or
points (if any) of the body are at rest at a given instant. A
point is "at rest" if its velocity is 0. Hence the condition is,
that x = 0 and y = 0, or :
(5)
0 = x0 - (y - j/0) 6,
0 = y« + (* - zn) 6.
KINEMATICS IN TWO DIMENSIONS 157
These equations yield a unique solution for the unknown x
and y when, and only when, 6 ^ 0 :
C)
+ Xn
- - - - A
THEOREM. At any instant at which d6/dt = 6 is not 0, there is
one and only one point of the body at rest.
This point is called the instantaneous centre, and its coordinates^
(xu !/i)i are given by Equations C).
If 6 = 0, no point of the body is at rest, or else all points are ;
there is never a single point at rest, to the exclusion of all
others.
When 6 = 0, XQ and yQ not both vanishing, all points of the
body are moving in the same direction with the same speed,
and we have a motion of translation.
EXERCISES
1. Show that the coordinates (£j, 77 j) of the instantaneous
centre, referred to the moving axes, arc given by the equations :
— ^o sin 6 — ?yn cos &
"
a
xQ cos 0 + 2/0 sin 6
1/1 = ^
2. A circle rolls on a line without slipping. Show that the
point of contact is at rest.
3. A billiard ball is struck full by the cue. Find the instan-
taneous centre during the subsequent motion.
3. Rotation about the Instantaneous Centre. The very
name " instantaneous centre7' implies that the motion of the body
is one of rotation about that point. Let us make this state-
ment precise.
Suppose the body is rotating about the origin, 0, with angular
velocity 6 = o>. What will be the vector velocity of an arbitrary
point P:(x, y)? The answer is given by Equations B), where
0* is taken at O, and thus x0 = yQ = ±0 = yQ = 0. Hence
158 MECHANICS
(1) f *"""*
I y = xu.
The result checks, for these are the components of a vector at
right angles with the radius vector r drawn from 0 to P and hav-
ing the sense of the increasing angle 6, its length being
Vx2 + y2 ft = ru, 0 < 6.
If 6 < 0, its sense is reversed.
It is the form of Equations (1) that is important. We say
that any motion of the points of the (x, 2/)-plane such that, at
a given instant, the velocity of each point is given by (1), is one
of rotation of the plane as a rigid body about 0. The velocities
of the points in the actual motion before and after the instant in
question may be different from those of the rigid body that is
rotating permanently about 0. But for a short space of time
before and after the instant, the discrepancy will be small because
of the continuity of the motion, and at the one instant, the veloci-
ties will all tally exactly.
If the point about which the body is permanently rotating
had been the point (a, 6) instead of the origin, Equations (1)
would have been the following :
(2)
f *=-&-&)*,
1 y = (x - a)6.
We are now ready to state and prove the following theorem.
THEOREM. The motion of the actual body at an arbitrary instant
t, at which 6 ?* 0, is one of rotation about the instantaneous centre.
To prove the theorem we have to show that, at the instant 2,
where (xlt y^ are given by Equations C), § 2, and (#, y} are given
by Equations B) of the same paragraph. To do this, eliminate
XQ and 2/0 between Equations B) and C). This can be done most
conveniently by writing Equations C) in the form (5) of § 2 :
0 = *0 - (Vi - 2/oH
0 = 2/0 + (xl - x0)6,
KINEMATICS IN TWO DIMENSIONS 159
and then, in this form, subtracting them respectively from Equa-
tions B). The result is Equations (3) of this paragraph, and the
theorem is proved.
Translation and Rotation. From the foregoing result a new
theorem about the motion of the plane can be derived at once.
Let A be an arbitrary point, and let its vector velocity be denoted
by V. Impress on each point of the plane, as it moves under
the given law, a vector velocity equal and opposite to V. Then
A is reduced to rest, and the new motion is one of rotation about
A with the same angular velocity as before. We thus have the
THEOREM. The field of vector velocities is the vector sum of the
fields consisting i) of the translation field due to the vector velocity
of an arbitrary point, A ; and ii) of the rotational field with A as
centre.
In other words, the given motion consists of rotation about an
arbitrary point, A, plus the translation of A.
The theorem also follows immediately from Equations B), if
we take the point 0' at A.
4. The Centrodes. We are familiar with the motion of a
circular disc when it rolls without slipping on a right line or a
curve — a wheel rolling on the ground. Con-
sider, more generally, the motion of a lamina
when an arbitrary curve drawn in it rolls
without slipping on an arbitrary curve fixed
in space. Wo may think of a brass cylinder,
or cam, as cut with its face corresponding to
the first curve, and attached to the body;
a second such cam, with its face corresponding to the second
curve, being fixed in space. And now the first cam is allowed
to roll without slipping on the second cam.
Thus a great variety of motions of the lamina can be realized,
and now the remarkable fact is that all motions can be generated
in this way, with the single exception of the translations, —
provided that the functions (1) of § 1 have continuous deriva-
tives of the second order, and the space centrode is traced out
by the instantaneous centre with non-vanishing velocity.
A necessary condition for the truth of this statement is evident
from intuition, namely : — the point of contact of the two cams
160 MECHANICS
must be the instantaneous centre of the actual motion. This fact
suggests the proof — the faces of the cams, i.e. the curves, must
be the loci of the instantaneous centres in the body and in space.
Definition. The locus of the instantaneous centre in the body
is called the body centrode, and the locus of the instantaneous
centre in space is called the space centrode.
THEOREM. Any motion of a rigid lamina which is not transla-
tion can be generated by the rolling of the body centrode (without
slipping) on the space centrodc, provided the space centrode is traced
out by the instantaneous centre with non-vanishing velocity; the
functions (1) of § 1 having continuous derivatives of the second order.
Before wo can prove the theorem, we must make clear to our-
selves how to formulate mathematically the rolling of one curve
without slipping on a second curve. As the independent variable,
the *timo most naturally suggests itself; but it is better at the
outset not to choose it, but to take, rather, a variable X which
merely corresponds to tho fact that, for an arbitrary (i.e. variable)
value of X, the curves meet in a (variable) point P. And now
we shall demand further :
i) that the curves be tangent to each other at P ;
ii) that the arc of the one curve corresponding to any two
different values of X, namely, Xj and X2 ; and the arc of the other
curve corresponding to the same values of X, have the same length.
Thus, in particular, both curves may be moving — a more
general case than the one that interests us here.
Let the equation of the one curve, C, referred to a system of
Cartesian axes, (x, y), be :
(1) * = ff(A), » = A(X),
where the functions #(X), h(\) are continuous together with
their first derivatives, and the latter do not vanish simultane-
ously :
(2) 0 <<7'(X)2 + /i'(X)2.
Let the second curve, F, referred to a second system of Cartesian
axes, (£, r;), be represented by similar equations,
(3) £
(4) 0
KINEMATICS IN TWO DIMENSIONS
161
The coordinates of any point of the plane, referred to the one
set of axes, are connected with the coordinates of the same point,
referred to the other set of axes, by the equations :
x = x0 + % cos 6 — TI sin 0,
y = 2/o + £ sin 9 + rj cos 0.
And now we require that XQ, yQy 6 be functions of X which have
continuous first derivatives :
(5)
(6) *o=/00, 2/o=* 00, 0 =
where /'(X), ^'(X), ^'(X) exist and are continuous.
Since C and F always meet in a point P, whose coordinates
are expressed by the equations (1) and (3), it follows that Equa-
tions (5) will hold identically in X if the values of x} y from (1),
and those of £, 77 from (3), be substituted therein.
The vector v whose components are
_ dx __ dy
Vx ~ d\' Vy ~~ d\
is tangent to C at P and its length is
The vector u whose components arc
is tangent to F at P and its length is
The requirements i) and ii) demand that these two vectors
be identical. This condition is both necessary and sufficient.
The analytical formulation of tho condition is as follows :
(7)
vx = u$ cos B — Uy sin 0,
vy = u% sin B + u^ cos B.
We now have all the material
out of which to construct the
proof. From Equations (5) it
follow? that
FIG. 92
162 MECHANICS
dx
The first line in these equations is nothing more or less than
the first of Equations (7), and the latter equations we have set
out to prove. Hence the second line must vanish, if the equation
is to be true, and so, by the aid of (5), we obtain the first of Equa-
tions (8) :
<8)
The second equation is obtained in a similar manner from the
second of the above equations.
Equations (8) represent a new form of necessary and suffi-
cient condition for the fulfilment of Conditions i) and ii).
5. Continuation. Proof of the Fundamental Theorem. It
is now easy to prove the theorem of § 4. The two curves, C
and F, are here the space centrode and the body centrode, and
we will now take as our parameter X, the time t. Equations (1),
§ 4, thus represent the coordinates, xl and ylt of the instanta-
neous centre in space, and in Equations (8), the (x, y) are the
coordinates of this same point, (xl9 yj. The other quantities
that enter into (8) are the functions (6) that determine the posi-
tion of the moving body; and X = t. Thus Equations (8) go
over into the following :
f *o - (2/i - 2/o) * = 0,
1 2/o + (xl - z0) 6 = 0.
But these are precisely Equations (5) of § 2, which determine the
instantaneous centre. Equations (8) are thus shown to be true.
KINEMATICS IN TWO DIMENSIONS
163
Discussion of the Result. From Equations (8) § 4, it appears
that a necessary condition for the truth of the theorem is, that
the coordinates of a point P of C satisfy the equations :
(10)
X = Xn -
dy0 /d0
dX / dX'
, dxa I dB
y = 2/o + -3r / 3T-
But these conditions are not sufficient, since the functions x and
y thus defined will not in general admit derivatives.
To meet this latter requirement we demand, therefore, that
the functions (6) possess, furthermore, continuous second deriv-
atives. But this is not enough, even if the case that x and y
reduce to constants is excluded (rotation about a fixed point).
It is, however, sufficient when we add the hypothesis of (2),
§ 4, and so demand that
dy° /de\
~dx/dx/'
d
de
be not both 0 (dO/d\ being, of course, 5^0). In other words,
the equations
(12)
t
d0 d2x0 _ d^cteo _, ^!^o = 0
dX dX2 dX2 dX "*" dX2 dX U|
dX'dX2" "" dX*"d\ "~ dX2 dX = °
shall never hold simultaneously. This excluded case includes
the case in which an ordinary cusp occurs; but it also includes
more complicated singularities.
If, in particular, the functions (6) are analytic in the neighbor-
hood of a point, X = X0, and if the case of permanent rotation
about a fixed point be excluded, the curve C will at most have
a cusp and otherwise be smooth in the neighborhood of the point ;
and the same will be true of P.
Acceleration of the Point of Contact. Let the point (XQ, y0),
at a given instant, t, be taken at the point of contact of C and T.
Then it follows from (10) — or, more simply, from (8) — since
x = xQ and y = yQ> that
dX
164 MECHANICS
Let the origin, furthermore, be taken at this point (x0, y0), and
let C be tangent to the x-axis here. Now, the derivatives of x
and y in (10) cannot both vanish. On computing them it is
seen that they reduce respectively to
d\2 ~d\' d\2 d\
The second, dy/d\, has the value 0, since C is tangent to the axis
of x at the origin. Hence we infer that
If X is the time, t, these derivatives become the components
along the axes of the acceleration of the point of contact, thought
of as a point fixed in the moving body. From (13) it appears that
this acceleration is never 0, but is a vector orthogonal to the
centrodes at their point of contact. The reader can verify this
result in the case of the cycloid.
Example. A billiard ball is projected along a smooth hori-
zontal table with an initial spin about the horizontal diameter
which is perpendicular to the line of motion of the centre. Deter-
mine the two centrodes.
Take the path described by the centre of the ball as the axis of x,
and the centre of the ball as (x0, y0). Then
Equations (10) give :
FIG. 93 x = XQ, y =--.
Hence the space centrode is a horizontal straight line at a distance
c/co below the centre of the ball, and the instantaneous centre
is always beneath the centre of the ball. This means that the
ball rolls without slipping on a right line distant c/w below the
centre. Hence the body centrode is a circle of radius c/co about
the centre of the ball.
EXERCISE
A billiard ball is struck full by the cue. Determine the
space centrode and the body centrode during the stage of slipping ;
cf. Chapter IV, § 14.
KINEMATICS IN TWO DIMENSIONS
165
The coordinates being chosen as in the Example, the equations
of the space centrode are :
SC ~~ *C rt ~~ Cv """"
_ 2a _
V ~ 5
2acl
5/ t'
where a denotes the radius of the ball, and c the initial velocity
of its centre. The time that elapses during the stage of slipping
is 2c/7/A<7 seconds. The space centrode meets the billiard table at
the angle
The equations of the body centrode, referred to suitable polar
coordinates, are :
2ac 1 2a
P =-
t
5 '
6. The Dancing Tea Cup. When an empty tea cup is set
down on a saucer, the cup sometimes will dance for a long time
before coming to rest. Two features of this phenomenon attract
attention; first, that the energy, obviously slight, is not earlier
dissipated by damping, and secondly, that we can hear a noise
in which so little energy is involved. The second point can be
disposed of easily because of the physical fact that the energy
of sound waves is surprisingly small.
To examine the first critically we need more light on the nature
of the motion. The results which wo have obtained in this
chapter furnish the clue. The following example is highly sug-
gestive.
Consider the motion of a lamina, in which
the body centrode is a right line making a
small (variable) angle with the horizontal.
For the space centrode take a curve suggested
by the figure. Such a curve can be defined
suggestively as follows. Begin with the curve
(1)
y = sin -•
166 MECHANICS
This curve gives satisfactorily the part of the figure not too near
the lines y = ± 1, but it is tangent to these lines, whereas it
should have cusps on them.
The desired modification is simple. For example, to convert
the curve
y=f(X)=X*
from one which is tangent to the axis of x into one which has a
cusp on the axis, it is enough to replace f(x) by [/(z)]* :
Apply this idea to the curve (1). It will suffice to set
(2) y
as the reader can easily verify.
Now allow the body centrodo — the tangent line — to descend
according to a reasonable law. We have here a picture of what
goes on as the tea cup dances. The line oscillates through smaller
and smaller angles as its point of intersection with the axis of x
descends.
Tyndall,* in his popular lectures, showed an experiment with
a coal shovel illustrating the same phenomenon. The all-metal
shovel was heated near its centre of gravity and laid across two
thin lead plates clamped in a vise, with their edges horizontal.
As the shovel bore more heavily on one of the plates, the latter
expanded with the heat, throwing the shovel onto the other
plate. Then the process was reversed. Thus vibrations like
those of the tea cup arose, and died down.
7. The Kinetic Energy of a Rigid System. The kinetic en-
ergy of any system of particles is defined as
T = i 2) m^\ vk2 = xk2 + yi2 + zt2.
We restrict ourselves to two dimensions, and thus
vi? = xj? + yj?.
Suppose, now, that the particles are rigidly connected. In
Equations A), § 1, let the point (x0, y0) be taken at the centre
of gravity, (x, y). Thus Equations (2), § 2, become :
* Tyndall, Heat Considered as a Mode of Motion, Lecture IV.
KINEMATICS IN TWO DIMENSIONS 167
±k = x — (& sin 0 + *;* cos 0) 0,
Vk = £ + (& cos 0 - 77* sin 0) 0.
On squaring and adding, multiplying by mk, and then adding
with respect to k, we find :
For, each of the remaining terms involves as a factor one of the
quantities
2)
and each of these is 0, since the centre of gravity is at the origin
of the (£, 7y)-axes. Hence it follows that
(1) T = |MF2 + i/fl2,
where V denotes the velocity of the centre of gravity and I is the
moment of inertia about the centre of gravity, 12 being the angular
velocity.
Second Proof. The result may also be obtained by means of
the instantaneous centre, 0. For the motion, so far as the
velocities that enter into the definition of T are concerned, is
one of rotation about 0. Hence
(2) T = i/'ft2,
where /' denotes the moment of inertia about 0. Now,
/' = I + Mh\
where h is the distance from 0 to the centre of gravity, and
(3) V = Aft.
On substituting this value of /' in (2) and then making u$e of (3),
T takes on the form (1), and this completes the proof.
Generalization. The most general rigid bodies with which we
are concerned are made up of particles and material distribu-
tions spread out continuously along curves, over surfaces, and
throughout regions of space. When such a body rotates about
an. axis, the kinetic energy is defined by Equation (1). We
can state the result in the form :
The kinetic energy of any rigid material system which is rotating
about an axis, is given by the formula:
T
168 MECHANICS
Remark. The formula holds even for the most general case of
motion of any rigid distribution of matter in space. For, such
motion is helical, i.e. due to the composition of two vector fields
of velocity, i) a field corresponding to rotation about an axis;
and ii) a field of translation along that axis ; cf . § 12 below.
EXERCISES
1. A ball rolls down a rough plane without slipping. Deter-
mine the kinetic energy in terms of the velocity of its centre.
2. A ladder slides down a wall, the lower end sliding on the
floor. Find the kinetic energy in terms of the angular velocity.
3. A uniform lamina in the form of an ellipse is rotating in
its plane about a focus. Compute the kinetic energy.
4. A homogeneous cube is rotating about one edge. Determine
the kinetic energy.
8. Motion of Space with One Point Fixed. Consider any
motion of rigid space, one point, 0, being fixed. We shall show
that there is an instantaneous axis, i.e. a line through 0, the
velocity of each point of which is 0; and that the velocities of
all the points of the moving space, considered at an arbitrary
instant, form a vector field which coincides with the vector field
arising from the permanent rotation of space about this axis.
We give first a geometrical proof which appeals strongly to
the intuition. The refinements which a critical examination
of the details calls for are best given through a new proof by
vector methods.
Let Q be a point of the fixed space, distinct from O. If its
velocity is 0, then the velocity of every point of the indefinite
right line through 0 and Q is 0, since a variable right line is evi-
dently at rest if two of its points are at rest.
If, on the other hand, Q is moving, pass a sphere, with centre
at 0, through Q and consider the field of vector velocities cor-
responding to the points of this sphere. The vectors are evidently
all tangent to the sphere, and they vary continuously, together
with their first derivatives, for we are not concerned with dis-
continuous motions.
Pass a great circle, (7, through Q perpendicular to the vector
velocity of Q. Let P be a point of C near Q. Then the vector
KINEMATICS IN TWO DIMENSIONS 169
velocity of P will also be at right angles to the plane of C and on
the same side of C as the vector at Q. For, since the vector
velocity of Q is at right angles to the chord QP, the vector velocity
of P must lie in the plane through P perpendicular to QP. But
it also lies in the tangent plane to the sphere at P. And now I
say, there, must be a point A of C (and hence two points) whose
velocity is 0. For, otherwise, all the vectors that represent the
velocities of the points of C would be directed toward the same
side of C. In particular, then, the point Q' diametrically opposite
Q would have such a vector velocity. But that would mean
that the mid-point of the diameter Q'Q, i.e. the centre 0 of the
sphere, is not at rest. From this contradiction follows the truth
of the assertion that there is a point A of C which is at rest. Hence
the whole indefinite line through 0 and A is at rest, and the exist-
ence of an instantaneous axis, 7, is established.
Rotation about the Instantaneous Axis. It remains to prove
that the vector field of the actual velocities coincides with the
field of the vector velocities due to a rotation about I. Con-
sider an arbitrary point, P, not on /. Then P cannot be at
rest, unless all space is at rest. For, if three points, not in a line,
of moving space are at rest, all points must be at rest. Pass
a plane, M, through P and the axis. Then the vector velocity
of P must be perpendicular to M. For let Q be any point of /.
Since Q is at rest, the vector velocity of P must lie in a plane
through P perpendicular to QP.
Consider next the circle, C, through P with / as its axis. The
vector velocity of P is tangent to C. For it is perpendicular to
any line joining P with a point of /. Moreover, the vector
velocities of all points of C are of the same length. For other-
wise two points of C would be approaching each other, or reced-
ing from each other.*
Lastly, the magnitude of the vector velocity of P is propor-
tional to its distance from /. Let M be the plane determined
by 7 and P. Consider two points, P1 and P2, in M but not on
7, distant A, and h2 respectively from 7. Let uhl be the magni-
tude of the vector velocity of Pt. Then coA2 is the magnitude of
the vector velocity of P2. For otherwise P2 would issue from
the rigid plane M.f This completes the proof.
* Exercise 4 below. t Exerciso 5
170
MECHANICS
EXERCISES
1. Give a rigorous analytic proof that if two points of a moving
straight line are at rest, every point of the line is at rest.
2. A point Q is moving in any manner, and a second point, P,
is so moving that, at a given instant, it is neither approaching Q
nor receding from Q. Give a rigorous analytic proof that the
vector velocity of P is orthogonal to the line QP, if the vector
velocity of Q is orthogonal to that line.
3. If three points of space are at rest, and if these points do
not lie on a line, all space is at rest. Prove rigorously analyt-
ically.
4. Prove analytically the statement of the text which refers
to this Exercise.
5. The same for this statement.
9. Vector Angular Velocity. Let space rotate as a rigid body
about a fixed axis, L, with angular velocity o>. Let P be an
arbitrary point fixed in the moving space.
Then the velocity of P will be represented by
a vector v perpendicular to the plane deter-
mined by P and the line L, and of length
hw, where h is the distance of P from L.
Let 0 be any point of L. Lay off from 0
along L a vector of length w and denote this
vector by (co). Let r be the vector drawn from O to P. Then
the vector velocity v of P is represented by the vector product
of (w) and r :
FIG. 95
(1)
Xr;
cf. Appendix A.
Let a system of Cartesian axes (or, y, z) be assumed with 0 as
origin, and let i, j, k be unit vectors along these axes. Write
(2)
Then
(3)
co2k.
i J k
x y z
KINEMATICS IN TWO DIMENSIONS 171
The components of v along the axes are thus seen to be :
vx =
Vy = XUg — ZO)
Vg =
(4)
Composition of Angular Velocities. Consider two rotations about
axes which pass through 0. Let them be represented by the
vectors (o>) and (a/)- An arbitrary point P of space has a vector
velocity v given by (1) :
v = («) X r,
due to the first rotation, and a vector velocity v' :
V = («') X r,
due to the second rotation.
Let these vectors, v and v', be added. Then a third vector
field results — one in which to the point P is assigned the vector
v + v'. It is not obvious that this third vector field can be
realized by a motion of rigid space — far less, then, that it is
precisely the field of velocities due to the angular velocity repre-
sented by the vector
(5) (0) = («) + («0-
That this is in fact true — that is the Law of the Composition of
Angular Velocities.
The proof is immediate. We have :
(6) v + V = («) X r + («') x r.
Now, the vector, or outer, product is distributive :
(7) {(«) + («')} X r = («) X r + («') X r.
Hence
(8) v + v' = {(«) + (a/)} X r = (12) X r,
and we are through. The result can be formulated as the fol-
lowing theorem.
THEOREM. Angular velocities can be compounded by the Law
of Vector Addition.
EXERCISE
Prove the law of composition for angular velocities by means
of Equations (4).
172
MECHANICS
10. Moving Axes. Proof of the Theorem of § 8. Let space
be moving as a rigid body with one point, 0, fixed. Let i, j, k
be three mutually orthogonal unit vectors drawn from 0 and fixed
in space, and let a, 0, 7 be a second set of such vectors fixed in
the body. The scheme of their direction cosines shall be the
following :
a j8 7 £ rj f
(1)
Thus
'l '2
a =
with similar expressions for /3, 7, where the direction cosines
are any functions of the time, t, continuous with their first (and
for later purposes their second or even third) derivatives, and
satisfying the familiar identities ; cf. Appendix A. Observe that
(2)
187 =
aa = 0, etc.
07 + 7/3 = 0, etc.
We are now in a position to prove analytically the existence of
an instantaneous axis. Let P be an arbitrary point fixed in the
body, and let r be the vector drawn from the fixed point 0 to P.
Then
(3) r = f a + vp + f 7.
Since P is fixed in the body, £, 77, f are constant with respect
to the time, and so
(4) t = $a + 40 + fy.
A necessary and sufficient condition that P be at rest is, that
the projections of f on three non-complanar axes all vanish.
Hence, in particular, the condition that P be at rest can be ex-
pressed in the form :
(5) at = 0, fit = 0, 7* = 0.
Applying this condition to the vector (4), we find the three
ordinary equations :
KINEMATICS IN TWO DIMENSIONS 173
+ f ay = 0
(6) tfa+ f 07 = 0
=0
Let
(7) a = yp, b = ay, c = pa.
From (2) it follows that
a = — pjy b = — 7«, c = — aft.
Equations (6) arc now seen to admit the particular solution:
£ = a, 77 = b, f = c.
These cannot all be 0 unless the body is at rest, since the vanishing
of the above scalar products would mean that
pa = 0, ya = 0 ;
and of course aa = 0. Thus the vector a would be at rest, and
likewise, each of the other vectors, P and 7.
The general solution of the equations (6) is given by the equa-
tions :
(8) £ = Xa, ij = X6, f = Xc, - oo < X < oo .
These points, and these only, are at rest. They form the instan-
taneous axis, and it remains to show that the* latter deserves
its name.
Instantaneous Axis. Let a vector (co) be defined as follows:
(9) o>£ = 7/3, a, = ay, co$ = fta ;
(10) (CO) = C0£0! + ^0 + W^7.
Then (co) is collinear with the instantaneous axis, whose equa-
tions (8) can now be written in the form :
(11) 1 = JL = L.
CO^ COr, CO^
We have seen that the vector velocity v of an arbitrary point
fixed in the body is given by (4). The components of v along
the (£, 77, f )-axes can be written in the form :
V{ = at = %aa + yap + f ay
= yt = %ya + rjyp + £yy
174 MECHANICS
Hence
ff = f «„ —
From (12) it follows that
(13) V = COf W, CO^ = (CO) X T,
r
and so we see that the actual vector velocity v of P is the same
as the vector velocity which P would have if rigid space were
rotating about the instantaneous axis with angular velocity co.
Thus the actual field of vector velocities of the points P coin-
cides with the field of vector velocities due to rotation about the
instantaneous axis represented by the vector angular velocity
(co), and the proof is complete.
11. Space Centrode and Body Centrode. The locus of the
instantaneous axis in fixed space is called the space centrode, and
its locus in the moving space, the body centrode. The actual
motion consists of the rolling of the one cone (the body centrode)
without slipping on the other cone (the space centrode).
To prove this statement consider the path traced out by a
specified point in the instantaneous axis. Take, for instance,
the terminal point of the vector (co), the initial point being at 0.
The locus of this point is a certain curve C on the space centrode :
(co) = coxi + coj + cosk,
and a certain curve T of the body centrode :
It is sufficient to show that these curves are tangent and that
corresponding arcs are equal. This will surely be the case if
d(u>)/dt for C is equal to d(co)/cft for F. Now, the first vector
has the value
The value of the second vector is :
+ a*,/? + co$
KINEMATICS IN TWO DIMENSIONS 175
The last line vanishes because it represents the velocity of
the point (o>) fixed in the body, this point lying on the instan-
taneous axis. The first line is the vector (co). This completes
the proof.
EXERCISE
Treat the motion of the plane by analogous vector methods.
Let _
be the vector drawn from the fixed to the moving origin, and
let p, a- be unit vectors drawn along the positive axes of £ and rj.
Let r be the vector from the fixed origin to an arbitrary point P.
Then
r = f + £p + r?<7.
The vector velocity in space of a point P fixed in the plane is
given by the vector equation :
* = f + f P + ^-
The instantaneous centre is given by setting t = 0.
On the other hand,
p = e", a = «C+D'.
The complete treatment can now be worked out without diffi-
culty.
12. Motion of Space. General Case. Let rigid space be
moving in any manner, subject to the ordinary assumptions
about continuity. Reduce a point A to rest by impressing on
all space a motion of translation whose vector is equal and oppo-
site to the vector velocity of A. The vector field of the velocities
in the original motion is compounded by the parallelogram
law of vector addition out of the two vector fields i) of translation
and if) of rotation about the instantaneous axis, 7.
Let the vector that represents the translation be resolved into
two vectors, one, T, collinear with 7, the
other, A, at right angles to 7. The velocity ~H
of any point, P, distant h from the axis, is,
in the case of pure rotation, hu ; its direc-
tion is at right angles to the plane through P and the axis, and its
sense is a definite one of the two possible senses. Hence it is seen
176 MECHANICS
that it is possible to find a point, B, whose vector velocity due to
the rotation is equal and opposite to the vector velocity A. (Draw
a line from a point 0 of the axis, perpendicular to / and A, and
measure off on it, in the proper direction, a distance h = A/u.)
All points in the line L through B parallel to / will also be at
rest. It thus appears that the original motion is one of rotation
about L compounded by the law of vector addition with a motion
of translation parallel to L tiad represented by the vector T.
This vector field is, in general, the same as that of the vector
velocities of the points of a nut which moves along a fixed machine
screw (or of the points of a machine screw which moves through
a fixed nut). The two exceptional cases are those of rotation,
corresponding to a pitch 0 of the threads, and translation, the
limiting case, as the pitch becomes infinite.
13. The Ruled Surfaces. We have seen in § 12 that the
vector field of velocities, in the general case of the motion of rigid
space, is the sum of two vector fields — one, rotation about an
axis, L ; the other, translation parallel to L. The locus of L in
space is a ruled surface S, the space centrode, and the locus of L
in the moving space is also a ruled surface, S, the body centrode.
From analogy with the rolling cones we should anticipate Jhe
theorem governing the present case.
THEOREM. The surface £ is tangent to S along L, and it rolls
and slides on S.
An intuitional proof can be given as follows. First of all, it
is clear from the very definition of L that S slides on S along L.
So it is necessary to prove only the tangency of the two surfaces.
Let L0 be the line L at time t = £0, and let P0 be a point of L0.
Pass a plane through P0 perpendicular to L0, cutting S in the
curve Cj and let P be the point in which L at time t = tQ + At
cuts C. Let Q be the point fixed in S, which will coincide with
P at time tQ + A£. The vector velocity of Q at the instant t —
t0 has a component, c, parallel to L0 and a component hu at right
angles to the plane through L0 and Q. Obviously h is infinitesimal
with AJ. In time M the point Q will, then, have been displaced,
save as to an infinitesimal of higher order, parallel to L0 by a
distance cAt. But it will have reached P.
The proof is now clear. The plane through L0 and Q makes
an infinitesimal angle with the tangent plane to S at PQ because
KINEMATICS IN TWO DIMENSIONS 177
it contains a point Q of S infinitely near to P0, but not on L0.
The plane through L0 and P makes an infinitesimal angle with
the tangent plane to S at PQ because it contains a point P of S
infinitely near to P0, but not on Z/0. And these two planes make
an infinitesimal angle with each other, because when Q is dis-
placed parallel to L0 by a distance cA£, its distance from P is an
infinitesimal of higher order than the distance of P from P0.
Instead of developing the details needed to make the intuitive
proof rigorous, we will treat the whole question by vector methods.
First, however, a digression on relative velocities.
14. Relative Velocities. Let a point P move in any manner
in space, and let its motion be referred to a system of moving
axes.
Consider, first, the case that the moving axes have a fixed
origin, 0. Let a system of axes (x, y, z), fixed in space, with
origin at 0 be chosen ; let the moving axes be denoted by (£, 77, f),
and referred to the fixed axes by the scheme of direction cosines
of § 10. Let r be the vector drawn from 0 to P :
0) r = f« + 4/9 + [ 7.
Then
or
(3) v = vr + v«,
where the terms on the right have the following meanings. The
vector,
,.. d£ , drj _ , df
<4> ^-dr + s'-1-*"
represents the velocity of P relative to the moving axes; i.e.
what its absolute velocity would be if the (£, t;, f)-axes were fixed
and P moved relative to them just as it actually does move.
Secondly, the vector
(5) ve = ** + it + r 7
represents the velocity in space of that point fixed in the body,
which at the instant t coincides with P. To say the same thing
in other words : — Let us consider the point P at an arbitrary
instant of time, t = t. Let Q be the point fixed in the body,
178
MECHANICS
which at this one instant coincides with P. Then ve is the vector
velocity of Q. It is the vitesse d'entrainement, the velocity with
which the point Q is being transported by the body at the instant t.
The analytic expression for ve we know all about. In vector
form it is :
(6) ve = («) X r
or
(7)
Its components along the axes, if we write v' = vej are :
(8)
Thus we have as the final solution of our problem this: The
components of the vector velocity of P along the axes of % , ry, f are :
(9)
= +
General Case. Let the axes of (x, y, z) be fixed in space.
Let (£, TJ, f) be the moving axes, whose origin, 0', has the coor-
dinates (XQ, i/o, 20). Then
(10) r = r0 + r'.
Hence
(11) v = v0 + v'.
Here,
FIG. 97
(12)
dxQ .
and v' is given by (9). The v' of (11) is, of course, not the V
of (8).
KINEMATICS IN TWO DIMENSIONS
EXERCISE
179
Denoting the components of v0 along the (£ , 17, f )-axes by
> v*y vl> show that the components of v along these axes are :
(13)
Here,
dy
'dt+^~
4 +
= 7*0.
16. Proof of the Theorem of § 12. Let a system of Cartesian
axes fixed in space, (x, y, 2), with origin in 0 be assumed. Let
O' : (x0, y^ ZQ) be a point fixed in the body, the motion of 0'
being known :
(1) *o=/(0, 2/o = v(0, *o = lKO.
Finally, let P be any point fixed in the body. Then
(2) r = rc + r',
cf. § 14, (10) with the specialization that here
(3)
dt
Then the components of the absolute velocity of P (i.e. its velocity
in fixed space) along the axes of (£ , 77, f ) are given by the formulas
of §§ 14, 13 :
(4)
We can formulate the problem as follows : To find a point
^•(£i> i/D fi) fixed in the moving space whose absolute vector
velocity is collinear with the vector (o>), or is 0 :
(5) *, = £(<•>).
Here, (co) is the vector angular velocity of the moving space,
whose rotation is defined by the direction cosines of § 10.
180
MECHANICS
By virtue of (4) the vector equation
three ordinary equations :
(6)
(5) is equivalent to the
Since
(7)
0 —
0 -
= 0,
a further necessary condition is :
(8) utat0 + co^f0
We can dispose at once of the case o> = 0 ; for then the space
in which 0' is at rest, is stationary, and so the motion of the given
space is translation (unless it be at rest). Thus all lines parallel
to the vector that represents the translation are axes such as
we seek.
If o) ^ 0, we obtain from (8) a unique determination of k.
On substituting this value in (6), two of these equations, suitably
chosen, determine uniquely two of the three unknowns £lf rjl} ft
as linear functions of the third, and then the remaining equation
(6) is true because of (8).
£i = «i> i7i = 61, fi = ci
be a particular solution of (6), then an arbitrary solution,
£i» 'Ju Ti> will satisfy the equations :
(£1 - fli)
+ (fi - ciK = 0
- (f i - c^cu* = 0
Hence
(9)
and thus &, T^, ft is seen to be any point of the line through
(a,, bi, Cj) collinear with (<*>). This line we define as L. These con-
ditions are sufficient as well as necessary.
KINEMATICS IN TWO DIMENSIONS 181
The locus of L in space is the ruled surface S ; its locus in the
body (i.e. the moving space) is the surface S. These surfaces
have the line L in common. We wish to show that they are
tangent along L, and that S slides over S in the direction of L.
The last fact is clear from the definition of L.
The point (&, rjlf ft) is not uniquely determined by the time,
but may be any point of L. We will, for our purposes, select it
as follows. Let Z/0 be a particular L, and let P0 be an arbitrary
point of L0, once chosen and then held fast. Pass a plane M
through P0 orthogonal to L0. Then (xlf ylt zj shall be the inter-
section of the variable line L with M , and its locus shall be denoted
by C. The point (ft, yly fj shall be the point of S which coin-
cides with (xu yi, 2i) at time t = t. Its locus in S shall be denoted
by F. This curve can be represented in the form :
F: ti=F(t), 77! = $(0, fi=*(0-
Its tangent vector at an arbitrary point is
£ka + *?i* + #,
dt + dt1* + dt 7'
provided this vector 7* 0.
To show that two surfaces which intersect at a point P0 are
tangent it is sufficient to show i) that they have a common tangent
vector, t ; and ii) that a tangent vector tj to the one surface and
a tangent vector t2 to the other surface, neither collincar with t,
are complanar with t, all three vectors, emanating from P0.
The surfaces S and 2 satisfy i) because they are both tangent
to L. Secondly, consider the vector ^ drawn from 0 to the point
of intersection of C and F at time t — t. Its derivative is a vector
tangent to C, provided it 5^ 0.
On the other hand, consider the point (|t, 77^ J\) of F, for which
t = t. Let the vector drawn from 0' to this point be denoted
by r[. Then
*i = TO + i{,
where r0 is given by (1), and
182 MECHANICS
Hence
^ -^tt-L*'!* j-*j,y
dt " dta^'Mft^ dty
4-fo + ^d + 7/^ + ^7.
This last line is precisely the vector velocity of that point fixed
in S, which at the instant in question, t = t, coincides with
(x\> V\y zi)- This vector, t, let us call it, lies along L because of
(5), — unless it be 0.
The other vector on the right of (11) is the vector (10); i.e..
a vector t2 tangent to T at (£, rjly f,) or (xl9 ylt zj. Equation
(11) thus says that
ti = t, + 1.
Now, the vector drjdt = tx ^ 0 will not lie along L. Hence t.2
will not, either. Consequently Condition ii) is satisfied, arid the
surfaces S and F are tangent along L. The case t = 0 is included ;
it does not lead to an exception.
EXERCISE
Let a cylinder of revolution roll and slide on a second cylinder
which is fixed, the first cylinder always being tangent along an
element, and there being no slipping oblique to the element.
Choose the point (x0, i/0, z0) in the axis of the moving cylinder,
and discuss the whole problem by the method of this paragraph.
16. Lissajou's Curves. In one dimension, or with one degree
of freedom, the most important periodic motion is Simple Har-
monic Motion. It can be represented analytically in the form :
(1) x = a cos (nt + 7),
where
(2) T = -
' n
is the period, where a is the amplitude, and where 7 is determined
by the phase.
In two dimensions, or with two degrees of freedom, an impor-
tant case of oscillatory motion about a fixed point is that in
which the projections of the moving point on two fixed axes
KINEMATICS IN TWO DIMENSIONS 183
at right angles to each other, execute, each by itself, simple har-
monic motion :
I x = a cos (nt + 7)
\u) 1
I y = b cos (mi + e)
It is possible to generalize at once to n dimensions :
(4) xk = ak cos (nkt + yk), fc = Z, • • • , n.
Let us study first the two-dimensional rase, beginning with
some simple examples. We may set 7 = 0 if, as usually hap-
pens, the instant from which the time is measured is unimpor-
tant.
Example 1 : m = n. Dynamically, this case can bo realized
approximately by the small oscillations of a spherical pendulum.
Let 7 = 0,
<p = nt.
Then
mt + e = <f> + €,
(x = a cos <p
y = A cos <p — B sin <p
A = b cos e, B = b sin c.
Assume that neither a nor b vanishes, since otherwise we should
be thrown back on right line motion along one of the axes. We
will take a > 0, b > 0.
In general, B ^ 0. The path of the moving point is then
an ellipse with its centre at the origin. For,
(6) cos <p = ^, sin <p = -^ x - ^ y.
On squaring and adding we find :
(7) B2x2 + (Ax - ay)2 = a2B2,
and this equation represents a central conic which does not reach
to infinity, i.e. an ellipse.
184 MECHANICS
The axes can be found by the methods of analytic geometry,
or computed directly by making the function
COS2 ^ __ 2AB cos <p sin ^ + B* sin2 ^
We have omitted the special case : B = 0. Here, e = 0 or
TT, and the motion is rectilinear, along the line :
In all cases, the path is confined within the rectangle :
x = ± a, y =±b,
and continually touches all four sides, — sometimes being a
diagonal, but, in general, an ellipse inscribed in the rectangle.
Example 2. m = n + h, where h is small. If we write the
equations in the form :
x = a cos nt
(9)
y = b cos (n£ + A/ + e)
then, for the duration of time T — 2w/n,
ht + € is nearly constant, and the path is
FIG. 98 nearly an ellipse — which, however, does
not quite close. And now, in the next in-
terval of time, the path again will be a near-ellipse, but in a
slightly different orientation — its points of tangency with the
circumscribing rectangle will be slightly advanced or retarded,
depending on whether h is positive or negative.
Thus a succession of near-ellipses will be described, all inscribed
in the same fixed rectangle x = ± a, y = ± 6. The motion can
be realized approximately experimentally as follows.
Blackburn's Pendulum. By this is meant the mechanical sys-
tem that consists of an ordinary pendulum, the upper end of
* Their directions are determined, in either way, by the formula :
cos 2* = or cos 2y
.
— 62 sin 2e 2ab cos c
where y denotes the angle from the axis of x to an axis of the conic. The length*
of the axes are found to be :
where
A8 = a4 -f 2a*&* cos 2« + b*.
KINEMATICS IN TWO DIMENSIONS 185
which is made fast at the mid-point of an inextensible string whose
two ends are fastened at the same level. When the bob oscillates
in the vertical plane through the supports, the second string
remains at rest, and we have simple pendulum motion, the length
of the pendulum being Z, the length of the first string.
Secondly, let the bob oscillate in a vertical plane at right angles
to the line through the points of support, and mid-way between
these. Again, we have simple pendulum motion ; but the length
is now I' = I + d, where d denotes the sag in the second string.
For small oscillations, the coordinates of the bob will evidently
be given approximately by Equations (3), and by suitably choos-
ing I and d, we can realize an arbitrary choice of m and n.
The Sand Tunnel.* If the bob of the pendulum is a tunnel
of small opening, filled with fine sand, the sand, as it issues from
the tunnel, will trace out a curve on the floor which shows ad-
mirably the whole phenomenon of the Lissajou's Curves. In
particular, if the second string is drawn as taut as is feasible,
so that d is small, the two periods will be nearly, but not quite,
equal ; and it is possible to observe the near-ellipses steadily
advancing, flashing through near-right lines (the diagonals of
the fixed rectangle).
Example 3. m — 2n. Begin with the case 7 = 0* e = 0, and set
<p — nt:
(10) x = a cos <p, y = b cos 2<p.
Hence
(11) */ = I** - 6
and the curve is an arc of a parabola, passing through the vertices
(a, 6), (—a, 6) of the circumscribing rectangle and tangent to
the opposite side at the mid-point. The sand pendulum may
be released from rest at the point (a, 6), and it then traces re-
peatedly the parabolic arc. In the general case,
(12) x = a cos <p, y = A cos 2<p — B sin 2<p ;
A = b cos 6, B = b sin e.
* This experiment should be shown in the course. It is not necessary to have
a physical laboratory. A tunnel can be bought at the Five and Ten, and string
is still available, even in this age of cellophane and gummed paper.
186
MECHANICS
When e is small, the curve runs along near to the parabola; cf.
Fig. 100. It is symmetric in the axis of y, since ^ and <p' — <p + TT
give x' = — x, y' = y. It is tangent once to each of the sides
x = a, x = — a of the circumscribing rectangle, and twice
to each of the other two. sides. When e has increased to ?r/2,
A = 0 and
(13) x = a cos <p, y = — b sin 2p
or
FIG. 99
This curve is obtained at once by affine
transformations from the curve
(15) yz
which is readily plotted.
When c has reached the value TT, we have again an arc of a
parabola — the former arc, turned upside down. As e continues
to increase, the new curves are the mirrored images of the old in
the axis of x, for e' = e + TT reverses the signs of A and B. All
these curves except the arcs of parabolas are quartics, inscribed in
the fixed rectangle, and having symmetry in the axis of y.
Example 4. m = 2n + h, where h is small. Here,
f x = a cos nt
(16) _
I y = b cos [2n* + ht + e]
and for a single excursion, M is nearly 0.
Thus the new curve runs along close to an
old curve for a suitable fixed c, but as time
elapses, the suitable e advances, too. FIG. 100
The student can readily trace these curves
with the sand tunnel. If he does his best to make d = J, there
will be enough discrepancy to provide for a small h.
17. Continuation. The General Case. The Commensurable
Case. Periodicity. Let m and n be commensurable,
where p and q are natural numbers prime to each other. Then
n = ap, m = aq.
KINEMATICS IN TWO DIMENSIONS 187
(x = a<
,
y = b(
Let <p = at, 7 = 0. Then
= a cos p<p,
y = 6 cos (£«? + e).
These functions are periodic with the primitive periods 2ir/p
and 27r/<7, and evidently have the common period 2ir. The
smallest positive value of co for which
a cos p(<p + co) = a cos p^>
6 cos {(/(p + co) + e} = fe cos {</p + e}
is co = 27T. For, if to is to be a period of the first function, then
x2?r
CO = A
P
And if co is to be a period of the second function, then
27T
CO = M —
Hence
X /* .
- = -, Xg = «,,
and the smallest values of X, n in natural numbers which satisfy
this equation arc X = p, n = q.
From the periodicity of the functions it appears that the curve
is closed, arid thus, as t increases, the curve is traced out re-
peatedly. For a non-specialized value of e, the curve is tan-
gent to each of the sides x = a, — a of the circumscribing rec-
tangle p times, corresponding to the solutions of the equations
cos p<p — 1, — 1 ; and q times to each of the sides y — 6, —6.
A line x = x', — a < x' < a, cuts the curve in 2p points ; a
line y = y'j — b < y' < 6, in 2q points.
These curves are all algebraic, and rational, or unicursal.
For, on setting £ = tan -J-p, the variables x and y appear, by
de Moivre's Theorem, as rational functions of £. The curves
are all symmetric in the axis of y.
The Incommensurable Case. Aperiodic. If on the other hand
n/m is incommensurable, the curve never closes. It courses
every region contained within the rectangle. If P be an arbi-
trary point of the rectangle, the curve will not in general pass
through P ; but it will come indefinitely near to P — not merely
once, but infinitely often ; possibly, occasionally passing through P.
188 MECHANICS
The proof can be given as follows. Consider a circle and the
angle <p at the centre. Let <f> = £ = 2wa be an angle which is
incommensurable with 2?r; i.e. let a be irrational. Then the
points of the circle which correspond to £, 2£, 3£, • • • (denote
them by Pt, P2, • • •) are all distinct. Hence they must have at
least one point of condensation, P. But from this follows that
every point must be a point of condensation. For, let Pn and
Pm be two points near P. Then the point corresponding to
n£ — w£ must be near the point corresponding to <p = 0. Hav-
ing thus obtained an arc of arbitrarily small length, we have but
to take multiples of it, i.e. to construct the points P*(n-m)i
k = 1, 2, 3, • • • , to come arbitrarily near to any point on the
circumference.
Turning now to the equations of the curve, let
m
<p = nt, — = a.
n
Then
x = a cos <f>,
1 y = b cos (cup + 17).
Let — a g x' ^ a, and let <p = <p' be a root of the equation
xf — a cos <p.
The curve cuts the line x = x' in the points for which
y = b cos {<*(<?' + 2kw) + r?}, 6 cos {«(- <?' + 2kw) + y}.
And now, since the angles 2kair lead to points on the circle which
are everywhere dense, the corresponding values of the cosine
factor are also everywhere dense between — 1 and + 1.
It is of interest to study the multiple points of the curve. These
occur when
t) <f>' = kw + 1^, I ^0;
it) ^ = )br + ^~a' fc?£0;
provided
(19) -n * (10 + k0a)ir.
When the inequality (19) holds, there is a one-to-one cor-
respondence between the values of <p and the points of the curve,
KINEMATICS IN TWO DIMENSIONS 189
provided the multiple points (which are always double points
with distinct tangents) are counted multiply. If, however,
(20) 77 = (10 + kQa)7r,
then
a<p + f\ = a (<p + fc07r) + i07r.
Set
(21) 0 = <f> + kQw.
Then the Equations (18) become :
x = a! cos 0
<22> I „=»•
where a! = a or — a, and likewise &' = 6 or — 6. Let
0 ^ 0 < oo.
Then there is a one-to-one correspondence between the values
of 6 arid the points on the curve. The point for which 9 = 0 :
x = a', y = &',
is an end-point of the curve. It is simple, no other branch going
through it. The double points correspond to the values
0' = kw + - > 0,
a
where
/CTT - - > 0, I ^ 0 ;
a.
or where
- far + - > 0, k ^ 0.
and n-Dimensions. In the case of motion with three
degrees of freedom, the equations can be reduced to the form :
(23)
x — a cos <p
y = b cos (ay + ??)
z = c cos ($v + f )
The case that a, /3 are both commensurable can be discussed as
before. The curve closes, the motion is periodic. When a
and ft are both irrational, and their ratio is also irrational, it can
190 MECHANICS
happen that the curve courses every region, however small, of
the parallelepiped :
— a^xga, —6^2/^6, — c ^ z ^ c,
and has no multiple points, the correspondence between the
points of the curve and the values of <p when — oo < <p < oo
being one-to-one without exception Whether the former prop-
erty is present for all such values of a and 0, provided further-
more that or, /3, and /3/a are not connected by a linear non-homo-
geneous equation with integral coefficients, and that 77, J" are not
specialized, I cannot say, though I surmise it to be. The latter
property, however, can be established.
The same statements hold in the general case,
xk = ak cos (oLk<p + rik), k = 1, - - • , n.
If; may happen, in a dynamical system with n degrees of free-
dom and coordinates qlt • • • , qn, that only a sub-set, <ft, • • • ,
qm, 1 g m < n, execute a Lissajou's motion. Thus a Black-
burn's Pendulum suspended in a moving elevator will have its
projection on a horizontal plane executing a Lissajou's motion,
whereas the vertical motion is not periodic at all.
The late Professor Wallace Clement Sabine drew mechanically
some very beautiful curves, which are here reproduced in half-tone.
I still have the half-tone which Dr. Sabine gave me. So far as I
have been able to ascertain, the curves were never published. The
figures here shown were made from lantern slides in possession of
the Jefferson Physical Laboratory, and it is through the courtesy of
the Laboratory that I have been enabled to reproduce them here.
CHAPTER VI
ROTATION
1. Moments of Inertia. The moment of inertia of n particles,
w,-, with respect to an axis is defined as the sum :
(1) / = i><r<«,
<=i
whore r» denotes the distance of wt- from the axis; cf. Chapter
IV, § 10.
Let 0 be an arbitrary point of space, and let Cartesian axes
with 0 as origin be assumed. Let the moments of inertia about
the axes be denoted as follows :
(2)
A =
The products of inertia are defined as the sums :
(3) -0 = 5 mi Vi z<> E =
These definitions are extended in the usual way by the methods
of the calculus to continuous distributions.
In terms of the above six constants it is possible to express
the moment of inertia about an arbitrary axis through 0. Let
the direction cosines of the axis be a, 0, y and let P : (x, y, z)
be an arbitrary point in space. Then
r2 = p2 _ <^
or
r2 = x2 + tf + z2 - (ax + py + yz)2.
Since o p
of + P + 72 = 1, FlG-
the last expression for r2 can be written in the form :
(x2 + 2/2 + *2)(«2 + P + 72) - (ox + fry + yz)*.
191
192 MECHANICS
Hence
- 2yazx -
Thus
/ = a2 5) m;(^2 + z>2) + 02 5J mt-(*i2 + a*2) + etc.
or:
(4) 7 = 4a2 + 502 + CV - 2D07 ~ 2#y« - 2Fa/3.
This is the desired result. The meaning of the formula can
be illustrated by the Ellipsoid of Inertia. Consider the quadric
surface,
(5) Ax2 + By2 + Cz* - 2Dyz - 2Ezx - 2Fxy = 1.
It is known as the Ellipsoid of Inertia, and its use is as follows.
Let an arbitrary line through 0 with the direction cosines «, 0, 7
meet the surface in the point (X, Y, Z), and let p be the length
of the segment of the axis included between the centre of the
ellipsoid and its surface. Then
X = ± ap, Y = ± 0p, Z = ± yp.
Since X, Y, Z satisfy (5), it follows that
(6) p2(^la2 + Bp + C72 - 2D/37 - 2#ya - 2Fa/3) = 1.
On combining (4) and (6) we find :
(7) P2/ = 1, / = ^
and / is seen to be the square of the reciprocal of p. From this
property it appears that the Ellipsoid of Inertia is invariant of
the choice of the coordinate axes.
If all n particles mz lie on a line, Equation (5) no longer repre-
sents an ellipsoid. Let the axis of z be taken along this line.
Then A = B and all the other coefficients vanish. Thus (4) be-
comes 7 = 4 («' + £*).
Here, A ^ 0 except in the single case that i = 1 and ml lies at
the origin. In all cases but this one, the quadric surface (5)
still exists, being the cylinder of revolution
(8) A(*2 + 2/2) = 1,
and the theorem embodied in (7) is still true.
ROTATION 193
Suppose, conversely, that (5) fails to represent a true (i.e.
non-degenerate) ellipsoid. If all the coefficients A, B, • • • ,
F are 0, the system of particles evidently reduces to a single
particle situated at 0. In all other cases, (5) represents a central
quadric surface, S. If this is not a true ellipsoid, then there is
a line, L, which meets 8 at infinity; i.e. which does not meet S
in any proper point, but is such that a suitably chosen variable
line L' always meets S, the points of intersection receding in-
definitely as Lf approaches L. The moment of inertia about
L' is given by (7) and approaches 0 as L' approaches L. Hence
the moment of inertia about L is 0. But if the moment of inertia
of a system of particles about a given axis is 0, it is obvious that
all the particles must lie on this axis.
We see, then, that (5) represents a true ellipsoid in all cases
except the one in which the particles lie on a line, and that (7)
holds in all the latter cases, too, except the one in which the
system reduces to a single particle situated at 0.
Parallel Axes. We recall, finally, the theorem relative to par-
allel axes ; Chapter IV, § 10 :
THEOREM. The moment of inertia, I, about any axis, L, is
equal to the moment of inertia, 70, about a parallel axis, L0, through
the centre of gravity, plus the total mass times the square of the dis-
tance, h, between the axes:
I = J0 + MW.
EXERCISE
Show that the moment of inertia about any line, L, in space
is given by the formula :
7 = {A + M(y\ + *?)} a2 + {B + M(z\ + x$\ /32
+ {C + M(x\ + tf)} 72 - 2(Z> + MVlzJ 0y
- 2(E + MZ.X,) ya - 2(F + Mx.y,) aft
where the origin of coordinates is at the centre of gravity and
x\> y\j z\ arc ^ie coordinates of any point on L, and where a, ft y
are the direction cosines of L.
For an arbitrary system of axes, replace xlt yly zl respectively by
*i ~ *> y\ - y> *i - *>
194 MECHANICS
where #, ?/, 2 are the coordinates of the centre of gravity, and
xu Vn z\ are the coordinates of any point on L — all referred to
the new axes.
2. Principal Axes of a Central Quadric. Let a quadric surface
be given by the equation :
(1) Ax2 + By* + Cz* + 2/)7/* + 2Ezx + 2Fxy = 1,
where the coefficients are arbitiary subject to the sole restriction
that they shall not all vanish. The problem is, so to rotate the
axes that the new equation contains only the square terms. Let
(2) F(x, y, z) = Ax2 + Eif + Cz* + 2Dyz + 2Ezx + 2Fxy,
(3) *(*, P,s) = x* + y* + z*.
Consider the value of the function F(x, y, z) on the surface
of the sphere
(4) ' x* + ?/2 + z2 = a2, or *(x, y, z) = a2.
Since F(x, y, z) is continuous and the sphere is a closed surface,
the function must attain a maximum value there, and also a
minimum.
Let the axes be so rotated that the maximum value is assumed
on the axis of z} in the point (0, 0, f), where f = ± a. We think
of Equations (2) and (3) now as referring to the new axes.
In accordance with the Method of Lagrange * we form the
function
F + \$,
the independent variables being x, y, z, with X as a parameter ;
and we then sot each of tho first partial derivatives equal to 0 :
(5) b\ + X*, = 0, F2 + X$2 = 0, t\ + X$3 = 0.
These three equations, combined with (4), form a necessary
condition on the four unknowns x, y> Zj X for a maximum :
' Ax + Fy + Ez + \x = 0
(6) Fx + By + Dz + \y = 0
Ex + Dy + Cz + \z = 0
But we know that the point (0, 0, f), f ^ 0, yields a maximum.
Hence D = 0, E = 0,
* Lagrange's Multipliers, Advanced Calculus, Chapter VII, § 5.
ROTATION 195
and the new F(x, y, z) has the form :
. F(x, y, z) = Ax* + 2Fxy + By* + Cz\
If the coefficient of the term in xy does not vanish, it can be
made to do so by a suitable rotation of the axes about the axis
of z; cf. Analytic Geometry, Chap. XII, §2. Thus F(x,y,z) is
reduced finally by at most two rotations (these may be combined
into a single rotation, but that is unessential) to the desired form :
(7) F(x, y, z) = Ax* + By* + Cz\
Here, A, B, C may be any three numbers, — positive, negative,
or 0, — except that we have excluded as trivial the case that
all three vanish. The original equation (1) will obviously repre-
sent an ellipsoid if and only if the new coefficients A, B, C in (7)
are all positive. We have thus established the following theorem.
THEOREM. An arbitrary homogeneous quadratic function
F(x, y, z) can be reduced by a suitable rotation of the axes of coordi-
nates to a sum of squares. The new coefficients of xr, y', z1 may
be any numbers, positive, negative, or 0.
EXERCISE
Show, by the method of mathematical induction and La-
grange's Multipliers, that an arbitrary homogeneous quadratic
function in n variables,
can be reduced to a sum of squares by a suitable rotation.
By a rotation is meant a linear transformation :
x{ = anx1 + • • • + ainxn
x'n = ani#i + • • • + annxn
such that, for any two corresponding points (xlt • • • , xn) and
(x(9 • • • , x'n), the relation holds :
r'2 _L . . . 4. r'2 — r 2 J_ . . . 4. r 2
*l \ \ *n ~ *\ \ \ An )
and the determinant of the transformation,
A = S ± an • • • ann,
which necessarily has the value ± 1, is equal to + 1.
196 MECHANICS
It is easy to write down the conditions that must hold between
the coefficients of the transformation, but these conditions are
not needed for our present purpose. Obviously, the result of
any two rotations is a rotation.
3. Continuation. Determination of the Axes. In the fore-
going paragraph we have been content to show the existence of
at least one rotation, whereby the given function is reduced
to a sum of squares. We have not computed the values of the
new coefficients, nor have we determined the lengths of tho axes.
Now, any rotation of the axes carries the second function, $, over
into itself :
*'(*', */', z') s *(*', »', z') = *(x, y, z).
The function :
fl = F + \3>,
goes over into the function :
F' + X*',
where X remains unchanged. Now, the condition :
an _ 00
~ ' ~ '
a? ~ '
is equivalent to the condition :
since the determinant of the linear transformation does not
vanish. Hence Equations (6) of the preceding paragraph will
be of the same form for the transformed functions. When
F'(x', y', *') = A'x'* + B'y'* + C'z">,
the equation for determining X reduces to the following :
(A' + X)(B' + X)(C" + X) = 0.
Thus the three roots of the determinant of Equations (6),
A +\ F E
(8) As F B + \ D
E D C +
ROTATION 197
arc seen to be the negatives of the coefficients A', B't C', and so
the axes of the quadric are found. If the roots of the deter-
minant (8) are denoted by \19 X2, X3, the lengths of the semi-
axes are
In case a X» = 0, the quadric reduces to a cylinder, or more spe-
cially, to two planes. All three X»'s will vanish if and only if the
original F(x, y, z) vanishes identically.
When tho Xt have once been determined, Equations (6) give
the equations of the axes of the quadric. In general, the three
\i are distinct, and Equations (6) then represent a right line
for each X».
4. Moment of Momentum. Moment of a Localized Vector.
Let A be a vector whose initial point, P, is given, and let 0 be
any point of space. Let r be the vector drawn from 0 to P.
By the moment of A with respect to 0 is meant the vector, or outer
product : *
(1) r X A.
We have met this idea in Statics, where the moment of a force
F, acting at a point P, with respect to a point 0 was defined as
the vector
M = r X F.
The moment of momentum of a particle with respect to a point is
defined as the vector |
(2) ff = r X mv,
whore r is the vector drawn from the point to
the particle, and v is the vector velocity of the FIG. 102
particle.
The moment of momentum of a system of particles with respect
to a point is defined as the vector
n
(3) ff = J) rk X mkvk,
1=1
* Cf. Appendix A.
f Contrary to tho usual notation of writing vectors in boldface, as a, x, i,
etc., or by parentheses, as (co), it seems here expedient to denote the vector moment
of momentum by a, the vector momentum by p, and the vector angular velocity
by o>.
198
MECHANICS
where rk is drawn from the point in question to nik, and v^ is tin
vector velocity of m,k.
In the case of a continuous distribution of matter the extensioi
of the definition is made in the usual way by definite integrals.
In Cartesian form a has the value, for a single particle :
(4)
J 1
y -
dy
'dt mdt mdt
dx
dz
dz di
dx
dz
dx
the origin being at 0. And so, for a system of particles, th(
components of a along the axes are :
(5)
Rate of Change of <r. Since
(&} *L ( ^y. _ di
dt \ dt dt
it is seen from Equations (5) that
(7)
dxk dz
dx
dzx
- y-^~v
These equations, in vector form, become :
(8) ljt=^ mkTk X a*'
where a^ denotes the vector acceleration of the fc-th particle.
ROTATION 199
The result, Equation (8), could have been obtained at once
from (3). If we differentiate Equation (2), we find :
da ^, dv . di ^,
_ = mrx__ + w._*xv.
Now,
-77 = v and v X v = 0.
at
Hence
do- . y dv
dt = mr X dt = ™r * a>
where a denotes the vector acceleration. Similarly, from (3)
we derive (8).
5. The Fundamental Theorem of Moments. In Chap. IV,
§ 3, it was shown that, in the case of any system of particles in a
plane such that the internal forces between any two particles
are equal and opposite and lie along the lino through the particles,
the moments of the internal forces annul each other, and the
equation of rotation becomes :
(1) I; mk (xk ^ - yk dji*) = 2 (Xk Y
k
fc= 1
The theorem and its proof can be generalized at once to space
of three dimensions. Newton's Second Law of Motion is ex-
pressed for the particle mk by the equations
d*Xk _ v _i_
'<~W~Xk +
J
= Yk + X Y«
where F*/ denotes the internal force which is exerted on the
particle w* from the particle my. Multiplying the third of these
equations by y^ the second by — zk and adding, and observing
that the moments of the internal force cancel in pairs, since the
forces F/jfc and F*/ are equal and opposite and have the same line
of action, the first of the following three equations is obtained.
The other two are deduced in a similar manner.
200 MECHANICS
(2)
These equations express the Fundamental Theorem of Mo-
ments. In vector form it is :
(3) S-i>XF,
or:
^7 = 2) (Moments of the Applied Forces about 0).
Equation (3) can be deduced more simply by vector methods.
Write Newton's Second Law in the vector form :
(4) mkak = F* + 2) Fw.
Next, form the vector product,
(5) mkTk X a* = rfc X F* + 2) r* X F*/,
and add. The sum on the left is equal to d<r/dt by § 4, (8). On
the right, the vector moments of the internal forces cancel in
pairs, and there remains the right-hand side of (3).
FUNDAMENTAL THEOREM OF MOMENTS. The rate of change
of the vector moment of momentum of any system of particles is
equal to the vector moment of the applied forces, provided that the
internal forces between each pair of particles are equal and opposite
and in the line through the particles :
or, in Cartesian formt Equations (2).
The foregoing result applies to the most general system of
particles, subject merely to internal forces of the very general
nature indicated. By the usual physical postulate of continuity
we extend the theorem to the case of continuous distributions
ROTATION 201
of matter, or to any material point set. For example, our solar
system is a case in point, and we will speak of it in detail in § 7.
6. Vector Form for the Motion of the Centre of Mass. Let 0
be an arbitrary fixed point in space, and let f be the vector drawn
from 0 to the centre of gravity, (?, of a material system. Let
Fi, • • • , Fn be the forces that act; i.e. the applied, or external,
forces. Then the Principle of the Motion of the Centre of Mass
is expressed by the equation :
(1) M -
.,
where v = df/dt. Equation (1) is merely the vector form of
Equations A), Chapter IV, § 1. It can be derived by vector
methods, by adding Equations (4), § 5, and observing that
Mi = 2)
= 1
Let p denote the momentum,
p = M v.
Equation (1) now takes on the form :
(2) £
Thus we have for any system of particles, rigid or deformable,
and even for rigid bodies and fluids, the two equations of momen-
tum:
THE EQUATION OF LINEAR MOMENTUM :
A) \
THE EQUATION OF MOMENT OF MOMENTUM :
7. The Invariable Line and Plane. In case no external forces
act,
<» a - "•
202 MECHANICS
and the vector cr remains constant. The line through 0 collinear
with <T is called the invariable line with respect to 0, and a plane
perpendicular to it, the invariable plane with respect to 0.
The solar system is a case in point, if we may neglect any force
the stars may exert. We may consider the actual distribution
of matter and velocities, and then, on choosing a fixed point,
0, the corresponding value of a will be constant.
Or we may replace the sun and each planet by an equal mass
concentrated at its centre of gravity, and consider this system.
Again, the vector a- corresponding to a given point 0 will be
constant, and obviously nearly equal to the former a.
Let us choose one of these cases arbitrarily and discuss it further.
The vector a depends on the choice of 0. Can we normalize
this choice? The centre of mass of the solar system is not at
rest, and so, since we are neglecting any force exerted by the
stars*, the momentum of the system, p = Mv, is constant and
5^ 0. The direction of this vector, p or v, does not depend on
the choice of 0. The point 0' can be so chosen that a' is collinear
with p.
We shall show in the next paragraph, Equation (5), that
(2) a = a' + Mr0 X V,
where a, a' are referred to 0, 0' respectively. If a is not already
collinear with p, let <r be resolved into two components; one,
collinear with p, the other, cr0, at right angles. We wish, then,
so to determine r0 that
(3) MrQ X v = (70, <TO * 0.
Since o-0 and v arc perpendicular to each other, this can be done.
The point Of will be any point of a line collinear with p. This
is known as the invariable line of the solar system. For a further
discussion, cf. Routh, Rigid Dynamics, Vol. I, p. 242.
8. Transformation of <r. Let 0 be a point fixed in space,
p and let O' be a second point, moving or fixed. Let
P be the position of a particle of the system. Then
o r CD f r = r' + r0;
Fia. 103 I V = v' + V0,
where v' expresses the velocity of P relative to 0', and V0 is the
velocity of 0' .
ROTATION 203
For a single particle, the moment of momentum with respect
to 0 is the vector
<r = r X rav = mr' X (v' + v0) + mr0 X v,
or
(2) <r = mr' X v; + mr0 X v + mr' X v0.
The first term on the right has the value
a'r = r' X wv',
or the relative moment of momentum, referred to 0' as a moving
point.
For a system of particles we infer that
* x v*
or
(3) <r = <r'r + Mr0 X v + Mi' X v0,
where o> is the relative moment of momentum referred to 0' as
a moving point ; v is the velocity of the centre of mass ; and f '
is the vector drawn from 0' to the centre of mass.
The second term on the right,
Mr0 X v = r0 X Mv,
can be interpreted as the moment of momentum, relative to O,
of the total momentum, Mv, of the system, thought of as a mass,
M , concentrated at O1 and moving with the velocity v.
The third term,
Mr' X V0 = f' X Mv0,
is the moment of momentum, relative to 0', of the total mass,
M j concentrated at the centre of gravity and moving with ve-
locity V0.
If, in particular, 0' be taken at (?, then f' = 0, v0 = v, and
(4) <r = <r'r + Mr X v,
where o> denotes the relative moment of momentum, referred
to G as a moving point, and
Mr X v = f X Mv
is the moment of momentum, Mv, of the total mass, concentrated
at G and moving with the velocity of G, referred to 0.
204 MECHANICS
Let cr' denote the value of a referred to the point 0' as a fixed
point; i.e.
X vk.
Then
</ = 5J m^i X vi + ]£ mkri X v0
or
</ = v'r + Mr' X v0.
Thus Equation (3) goes over into :
(5) a = </ + Mr0 X v.
This amounts to setting v0 = 0 in (3).
9. Moments about the Centre of Mass. We have the Funda-
mental Equation of Moments, § 5 :
(I)' S
And we have the Equation of Transformation, § 8, (4) :
(2) a = *'r + Mf X v,
where o> is the relative moment of momentum, referred to G as
a moving point.
Differentiate this last equation, observing that
since df/dt = v is cither 0 or else collinear with v. Thus we find :
/ON da _ dff'r M d?
(6) dt ~ Ht + M X Tt
On the other hand,
r = r' + f ,
where r, f are drawn from 0 ; r' from 0. Thus
(4) 5) r* X F» = g r; X F* + f X 5) F*.
t k Ic
Substituting in Equation (1) the values found in Equations (3)
and (4), we obtain the result :
ROTATION 205
Recall the Equation of the Motion of the Centre of Mass, § 6 :
(6) "£
From it follows that
Mr X ^ = f X £ F*.
fc
Thus these terms cancel in (5) and there remains :
In this equation is embodied the result which may be described
as the
PRINCIPLE OF MOMENTS WITH RESPECT TO THE CENTRE OF
MASS. The rate of change of the relative vector moment of momen-
tum, referred to the centre of mass G regarded as a moving point,
is equal to the sum of the vector moments of the applied forces with
respect to G:
EXERCISE
Show that a rigid body is dynamically equivalent, in general,
to a pair of equal masses on the axis of or, a second pair on the
axis of y, and a third pair on the axis of z, each pair being situ-
ated symmetrically with respect to the origin, and all six dis-
tances from the origin being the same ; it being assumed that the
principal axes of inertia lie along the coordinate axes. Discuss
the exceptional cases. Use the results of § 12.
10. Moments about an Arbitrary Point. Consider the most
general transformation, § 8, (3) :
(1) * = <j'T + Mr0 X v + Mr' X v0,
and differentiate :
On the other hand,
(3) 2 r" x F* = 2 ri x F*
t I
206 MECHANICS
From the Equation of Linear Momentum, § 6, (1) follows that
(4) MIO X % - 2 ro x F*-
G/v ^^T
Moreover,
-jj£ V -f- V0 - .
For,
hence
v0 X v = v0 X v' + v0 X v0 = v0 X v',
and
v0 X v' + v' X v0 = 0.
Substituting, then, in the Equation of Moment of Momentum,
§6, B):
g = 2r*XF*,
t
and reducing, we find :
This equation is general, covering all cases of taking moments
about a moving point 0', relative to that point. When, however,
one uses the expression: " taking moments about a point O'"
the meaning ordinarily attached to these words is, that the equa-
tion
(6) f = ?ri><F*
shall be true. Hence we must have
(7) f'X§ = °
for every value of t.
Let t = T be an arbitrary instant. Let 0' be a point which
describes a certain path,
(8) r0 - r0ft r).
Consider this as the vector r0 of the foregoing treatment. Then
ROTATION 207
At the instant t = r,
/dv*\ /a2f<A
(-5-),., = (IF;,.;
Then Equation (7) is to hold for this vector r0 at the one instant
t = T.
Thus we have in general, not a single curve traced out by 0'
and (7) considered for a variable point of that curve, as in the
case of § 8, where f' = 0, — but a one-parameter family of
curves, and Equation (7) considered for one point of each curve ;
cf. for example, the next paragraph.
11. Moments about the Instantaneous Centre. Consider the
motion of a lamina, i.e. a rigid plane system, in its own plane.
Let Q be the instantaneous centre at a given instant, t = T.
Then <r, referred to the point Q, is a vector perpendicular to the
plane, and its length is
Tde
1 dt'
where / is the moment of inertia of the lamina about Q.
What does it mean to "take moments about Q"? From the
foregoing it means to take moments about a point 0' describing
a curve
TO = *o& T)
which at the instant t = r passes through Q.
There is an unlimited set of such curves. Let us select, in par-
ticular, the curve C which is the path of that point fixed in the
lamina, which passes through Q at the instant t = r. Observe
that this is an arbitrary choice of C. This curve C is known in
terms of the rolling of the body centrode on the space centrode.
The velocity of 0' at Q is 0, but its acceleration, if Q is an
ordinary point, is normal to the centrodes at Q and does not vanish ;
Chapter V, § 5. If, then, Equation (7), § 10 is to be satisfied, the
centre of gravity, G, must lie in the normal to the centrodes. In
particular, the normal to the body centrode must pass through
the centre of gravity. Hence the body centrode must be a circle
with the centre of gravity at the centre, if the condition is to be
permanently satisfied. The equation of moments now becomes :
I-JJ2 ~ S (Moments about Inst. Centre).
208 MECHANICS
The only case, then, of motion in a plane, in which we may
permanently take moments about the instantaneous centre,
thought of as a point fixed in the moving body, is that in which
a circle rolls on an arbitrary curve, the centre of gravity being
at the centre of the circle; and the limiting case, namely, that
the point Q is permanently at rest. This last case corresponds
to the identical vanishing of dvjdt.
Moments about an Arbitrary Point. Consider now an arbi-
trary point 0' fixed in the body. Let it be at Q at the instant
t = r, and let C be the curve,
TO = r0(J, T),
which it is describing. Take moments about Q with reference
to this point, 0'. Then
da[ = d?0
dt dt2'
If, furthermore, Equation (7), § 10 is satisfied, the equation of
moments becomes :
/-72/)
^77/2" = 2} (Moments about Q).
Equation (7) here means, in general, that the acceleration of 0'
is collinear with the line determined by Q and G. In particular,
the equation is satisfied if the acceleration of O' is 0 at Q ; or if
Q coincides with G. *
EXERCISE
A billiard ball is struck full by the cue. Consider the motion
while there is slipping. Show that it is not possible to take
moments about the instantaneous centre.
Find the points of zero acceleration and verify the fact that
it is possible to take moments about them, explaining carefully
what you mean by these words.
Show that the points whose acceleration passes through the
centre of the ball lie on a circle through the centre of the ball,
of radius one-fifth that of the ball, the centre being directly above
the centre of the ball.
12. Evaluation of <r for a Rigid System; One Point Fixed.
Consider a rigid system of particles with one point, 0, fixed.
* Ed ward V. Huntington has discussed this question, Amer. Math. Monthly,
vol. XXI (1914) p. 315.
ROTATION
209
The motion is then one of rotation about an axis passing through
0 ] cf . Chapter V, § 8. Let the vector angular velocity be
denoted by w, and let a, p, 7 be a system of mutually perpen-
dicular unit vectors lying along Cartesian axes with the origin
at 0. When we wish these axes to be fixed, we shall use the
coordinates (x, yy z) and replace QJ, j3, 7 by i, j, k. In the general
case, the coordinates shall be £, 17, f .
Let P be any point fixed in the body, and let r be the vector
drawn from 0 to P :
a)
The velocity of P,
(2)
r = $ « + lift + f y.
"£
is expressed in terms of the vector co as follows (Chapter V, § 9) :
(3) v = co X r.
In Cartesian form,
(4) v =
or
(5)
For a single particle, then, a has the value :
(6) a = r X rav,
(7)
Hence
(8)
a = m
y
r
210 MECHANICS
These formulas lead in turn to the following :
<T£ = m [(rj* + f 2) C0£ - &CO,, -
(9)
= m [ - rf»t + (f
= m [ - f£cof - ftw, + (£2 + i;2
For a system of particles they become :
and so, finally,
(10)
or = -
These are the formulas which give the components of a along
the axes of £, 77, f when the origin is fixed. It is obviously im-
material whether the axes are fixed or moving.
13. Euler's Dynamical Equations. Consider the case of a
rigid body, one point 0 of which is fixed. The Equation of
Moments,
(1) -rr = 2 ( Moments about 0 ),
referred to this point, admits a simple expression in terms of the
angular velocity, &, of the body. Let the (£, r/, f)-axes be fixed
in the body, and let P be a point which moves according to any
law. Let
r = £<* + T70 + f%
where r is the vector drawn from 0 to P. Then we have seen
(Chapter V, § 14) :
dr =
dt
This result applies to the vector :
o- = cr$ a + (Ty 13 + <TS 7,
and thus gives us the left-hand side of (1).
ROTATION
On the right-hand side of (1) let
X F* = La + M/3 + Ny.
211
Thus we have :
(2)
= L,
dt
dt
dfft -LJ
~ _j." | C0£ O"TJ WTJ (7A == xV.
at
On substituting for <T£, o^,, a^ their values from (10), § 12, the
equations known as Euler's Dynamical Equations result. In
particular, if the (£, 77, f)-axes are laid along the principal axes
of inertia, then
and Equations (2) assume the form :
dp
/j -
(3)
where
P =
r =
When the axes of coordinates do not coincide with the prin-
cipal axes of inertia, Euler's Equations take the general form :
(4)
1
dt
dt
dt
- (En, -
(C - B) w^ = L,
and two others obtained by advancing the letters cyclically.
Eider's Dynamical Equations also apply to the rotation of
a rigid body about its centre of mass ; § 9. Here, there is no
restriction whatsoever on the motion.
212
MECHANICS
14. Motion about a Fixed Point. Let the body move under
the action of no forces, save the reaction at 0. Then Euler's
Equations become the following :
A first integral is obtained by multiplying the equations re-
spectively by p, q, and r, and adding :
(2)
Ap* + Bq* + Cr* = h.
This is the Equation of Energy, Chapter VII, §§ 5, 6.
A second integral is found by multiplying Equations (1) re-
spectively by Ap, Bq, and Cr, and adding :
dr „
(3) A2p2 + B2q2 + C2r2 = I
This equation corresponds to the fact that
dt
= 0,
and so
a = Apa + Bqf$ + Cry
is constant.
From Equations (2) and (3), two of the variables, as p2 and
#2, can, in general, be determined in terms of the third, and then,
on substituting in the third equation (1), a differential equation
for r alone is found. It is seen that t is expressed as an elliptic
integral of the first kind in r. Thus, p, qt and r are found as
functions of t.
Exercise. Let A = 3, B = 2, C = 1 ; and let p, q, r all have
the initial value 1. Work out the value of t in terms of the in-
tegral.
ROTATION 213
The Body Cone. On multiplying (2) by Z, (3) by A, and sub-
tracting, we find :
(4) A(l- Ah) p* + B(l- Bh) g2 + C(l - Ch) r2 = 0.
The equations of the instantaneous axis are :
(5) -* = ? = £,
' p q r'
when p, q, r are the above functions of t. Hence the locus of
the instantaneous axis in the body is the quadric cone :
(6) A(l- Ah) ? + B(l- Bh) T,2 + C(l - Ch) f2 = 0.
More explicitly, let p, q, r satisfy (2) and (3) and hence (4).
Then any point (£, ry, f) of (5) satisfies (6), and hence lies on
the quadric cone. Conversely, let (£, 77, f) be a point of the
quadric cone, (6). If (£, TJ, f) ^ (0, 0, 0), determine p, q, r, p
by the four equations
P = /*£> Q = M, r = /if,
Thus (3) is satisfied. And (4) holds, too. Hence (2) is true.
Consequently, (£, r/, f ) lies on an instantaneous axis.
The Space Cone. Poinsot obtained an elegant determination
of the space centrode. Consider the ellipsoid of inertia, 2. It
is a surface fixed in the body. Let m
be the point in which the ray drawn
from 0 and collinear with co cuts S.
Then the tangent plane M to S at m is
a plane fixed in space, the same for all
points, m. The motion is seen to be one
of rolling of the surface S on the plane
M without slipping. pIG ^4
To prove the first statement, it is
sufficient to show that the tangent plane at m is perpendicular to
cr, and that its distance from O does not depend on m. The
equation of S is :
S: 4£2 + 5T;2 + Cf2 = 1.
The coordinates of m are :
PP, PQ, pr, where
214 MECHANICS
Hence the equation of M is
M : PAp^ + pBqv + pCrf = 1.
The direction components of its normal are Ap, Bq, Cr. But
these are precisely the projections of <r on the axes. Hence M
is perpendicular to a. Moreover, the distance of the plane M
from 0 is
1 = 1 = J h
P^l * l'
and so is constant. This completes the proof of the first state-
ment.
To prove the second statement; consider so much of the body
cone, (6), as lies in S. Let F be the curve on S which marks
the intersection of these two surfaces. Then F rolls on M with-
oujb slipping, and the curve of contact, C, can servo as a directrix
of the space centrode. For the body cone, (6), rolls without
slipping on the space cone, and the curves F, C are two curves
on these cones, which curves aro always tangent at the point M
of the instantaneous axis. The angular velocity, w, is propor-
tional to the distance Om ; for
15. Euler's Geometrical Equations. Euler introduced as co-
ordinates describing the position of a rigid body, ono point
of which is fixed at 0, the three angles, 6, p, and ^ represented
in the figure. Between the components of the angular velocity
w about the instantaneous axis,
and these coordinates and their derivatives, exist the following
relations :
d$ , • dd
V = — Sin 0 COS <p —rr + SHI (p -r.
dt at
... <ty , de
q = sin 0 sin ip-j- + cos tp-r
(it Ctt
(1)
dt ' dt
These are known as Enter' s Geometrical Equations.
ROTATION
215
A geometrical proof can be given by computing the vector
velocities of certain suitably chosen points in two ways. Begin
FIG. 105
with the point C in which the positive axis of f pierces the surface
of the unit sphere. As we look down on the sphere from above
this point, it is evident from the figure that
dO
dt
= p sin <p + q cos <p
sin 6 -— = — p cos <p + q sin <p.
(it
FIG. 106
These equations yield the first two of Equations (1).
To obtain the third equation, consider the motion of E. Its
velocity is made up of a velocity o>£ tangent to the arc EA, and
two velocities perpendicular to this arc. On the other hand,
its velocity is composed of the velocity tp tangent to the arc EA ;
the velocity \ft cos 0, also tangent to EA ; and 0 perpendicular
to EA. Hence
and this is the third Equation (1).
216
MECHANICS
If p, <7, r have onc6" been determined as functions of the time,
Equations (1) yield a system of three simultaneous differential
equations of the first order for determining 0, <p, $ as functions
of the time. Thus in the problem of § 14, — the motion of a
rigid body under no forces, or acted on by the one force of con-
straint that holds the point O fixed, — p, q, r were determined
explicitly as functions of the time, and the further study of the
problem is based on the above Equations (1).
16. Continuation. The Direction Cosines of the Moving Axes.
The moving axes are related to the fixed axes by the scheme,
£ V f
y
z
and the question is, to express the nine direction cosines in terms
of the Eulerian angles. This can be done conveniently by vector
methods, if we effect the displacement one step at a time. Let
i, j, k be unit vectors along the original axes, and a, /ft, 7 unit
vectors along the displaced axes. Let the first displacement
be a rotation about the axis of z through the angle ^, and let
i, j go over into i1; jj. Then
it = i cos ^ + j sin ^
J! = — i sin ^ + j cos ^
k — k
*V| — JEL.
Next, rotate about the axis of jt through an angle 6, whereby
ix goes into i2, and kx into k2 = 7 :
12 = it cos 0 — kt sin 0
J2 = Ji
k2 = ij sin 0 + kj cos 6.
Finally, rotate about k2 through an angle ^, whereby i2 goes
over into i3 = a and J2 goes into j3 = 0 :
13 = i2 cos <p + J2 sin <p
j3 = — i2 sin <p + j2 cos v
k, = k,.
ROTATION
217
From these equations it appears that
Zj = cos 0 cos <p cos ^ — sin <p sin ^
ml = cos 6 cos (f> sin ^ + sin ^ cos ^
n! = — sin 0 cos v?
Z2 = — cos 6 sin <p cos ^ — cos <p sin ^
m2 = — cos 6 sin p sin ^ + cos p cos ^
n2 = sin 0 sin v?
Z3 = sin 0 cos ^
w3 = sin 6 sin ^
n3 = cos 0.
17. The Gyroscope. It is now possible to set forth in simplest
terms the essential characteristics of the motion of a rotating
rigid body, which is the basis of gyroscopic action. By a gyro-
scope is meant a rigid body spinning at high velocity about an
axis passing through the centre of gravity, which is at rest, and
acted on by a couple whose representative vector is perpendicular
to the axis.
Consider, in particular, the following motion. Let A = J?,
C 7* 0, and let the axis of f be caused to rotate with constant
angular velocity, c, in the plane ^ = 0. What will be the couple?
Here, d\l//dt = 0 and Euler's Geometrical Equations give
P
dO
de
3~T7* T =
where dd/dt = c. The components L and M are unknown,
but N = 0. The third of the Dynamical Equations becomes :
~
dt
= 0; hence r = v,
and P is a large positive constant. Since here
d<p .
Hence, from the first two equations,
p = c sin vtj q = c cos vt.
218 MECHANICS
On substituting these values in the Dynamical Equations, we
find:
L = Ccv cos vt, M = — Ccv sin vt.
We may think of the couple as made up of a force F acting
at the point C : (£, 77, f) = (0, 0, 1) in Fig. 105, and an equal and
opposite force at 0. Then F will be tangent to the sphere at
C. Let it be resolved into two components, one perpendicular
to the (£, f )-plane ; the other, in that plane. The first will have
the value L, taken positive in the sense of the negative ry-axis ;
the second will equal M, taken positive in the sense of the £-axis.
When t = 0,
L = Ccv, M = 0,
and at any later time, the result is the same. This can be seen
directly from the nature of the problem, since the motion of the
axis of f in the plane \f/ = 0 is uniform, and hence the force which
the constraint exerts will be the same force relative to the body
at one instant as at any other instant.
It is easy to verify analytically the truth of the last statement.
For, the force normal to the plane \l/ = 0 will always be
L cos <p — M sin <p = Ccv ;
and the force in that plane will always be
L sin if> + M cos p = 0.
This result brings out in the simplest form imaginable the
essential phenomenon in gyroscopic action, namely, this: To
cause the axis to move in a plane with constant angular velocity,
a couple must be applied whose forces act on the axis in a direction
at right angles to that plane.
Finally observe that if one thinks of onesolf as
standing on the gyroscope and moving with it, one's
e exerted body along the positive axis of f and facing in the
direction of the motion of the axis, the force L ap-
plied to the gyroscope will be directed toward one's
left, and hence the reaction of the gyroscope on the
FIG. 107 constraint will be directed toward the right, the gyro-
scope spinning in the clockwise sense as ono looks
down on it. Of course, if the sense of the rotation were reversed,
the sense of the reaction would be reversed also.
ROTATION 219
EXERCISES
1. Show that, if no assumption regarding 6 is made, but ^ = 0
and r = v, then
T A - 4d*e . n 4de
L = A sin vt -rz + Cv cos vt -T:
at* at
TM A *d*e n • <dd
M = A COS vt -T7T ~ Cv Sin vt -rr-
at1 at
2. If the axis presses against a rough plane, \l/ = 0, the tan-
gential force being p, times the normal force, then *
dB
provided dO/dt > 0 and furthermore the point of the axis in
contact with the plane moves backward, i.e. in the sense of the
decreasing 6.
On the other hand, the axis must have a sufficiently large
radius so that the requirement below relating to the motion of
the point of contact can be fulfilled. Hence
dO c^t , Q cA ^t
37 = ce A and 9 = -/Y — e A
at CIJLV
whore c denotes the initial value of dO/dt, and initially 0 = cA/Cp,v.
Moreover,
O
VSS-T8'
where v = dO/dt and s = 0 refer to the point in which the sphere
(of radius 1) is cut by the axis.
3. Prove that, in the problem of the preceding question, the
normal reaction of the constraint is
4. Show that, if a (small) constant couple, of moment €, acts
on the gyroscope, the vector that represents the couple being
at right angles to the plane \l/ = 0 and directed in the proper
* If we think of the material axis as a cylinder of small radius, there will be a
small couple about the axis, tending to reduce r. But as this couple approaches 0
when the radius of the cylinder approaches 0, we may consider the ideal case of an
axis that is a material wire of nil cross section, the couple now vanishing.
220 MECHANICS
sense, and if the axis of the gyroscope be constrained to move
in the plane \l/ = 0, the acceleration of 6 is constant :
This last equation is true, even when e varies with the time.
6. Prove that, no matter how 8 varies, the axis of the gyro-
scope always being constrained to move in the plane ^ = 0,
the reaction on the constraining plane \l/ = 0 is numerically
its sense being that of the increasing \l/ when dO/dt > 0, but the
opposite when dd/dt < 0.
6. It has been shown that a rigid body is equivalent dynami-
cally to three pairs of particles situated at the six extremities of
a three dimensional cross ; § 9.
Let the equivalent system move as the gyroscope did in the
text, i.e. with \l/ = 0 and dO/dt = c. Consider, in particular,
an instant, at which the moving axes are flashing through the
fixed axes ; i.e. 0 = <p — \l/ — 0. Show, by aid of the expres-
sions for a, 0, 7, that the vector acceleration of each of the four
particles on the £- and the f-axes passes through 0 ; but, in the
case of each of the other two particles, is parallel to the axis of
f . Hence explain the reaction of the gyroscope on the constraint.
7. Discuss the problem of Question 2 for the case that the
point of contact is allowed to slip forward. Consider also all
cases in which dB/dt < 0 initially.
18. The Top. The top is a rigid body having an axis of
material symmetry and, in the case of a fixed peg, supported at
a point 0 of the axis. Let the positive axis of f pass through the
centre of gravity, (7, distant h from 0.
The third of Euler's Dynamical Equations becomes, since the
applied forces — gravity and the reaction of the peg — both pass
through the axis of f ,
(1) C% = 0.
Hence r = v (constant).
The equation of energy here becomes :
(2) A (p2 + £2) + CV> = H - 2Mgh cos 6.
ROTATION
221
Furthermore, the vertical component of the vector a is constant.
For, the applied forces giving a vector moment at 0 reduce to
gravity, which is vertical, and so its vector moment with respect
to O is horizontal. Now, the components of o- along the moving
axes are Ap, Bq, and Cr. Hence the vertical component of a
is (§ 16) :
Bqn2
On substituting for nlt n2, nz their values from § 16 we have :
(3) — Ap sin 6 cos <p + Aq sin 6 sin <p + Cv cos 0 = K.
Turning now to Euler's Geometrical Equations, we find :
(4)
dt , - dO
p = — sin 0 cos <p-~ + sin <p-jr
...
q = sin 0 sin <p
cos
dO
-=r
at
On substituting these values of p and q in (2) and (3) we find :
(5)
(6)
sin2 e -- = 0 - bi> cos
6 =
C
the constants a and 0 depending on the initial conditions of the
motion; i.e. they are constants of integration; whereas a and
b are constants of the body.
The third Equation (4) determines <p after 0 and ^ have been
found from (5) as functions of t :
(7)
<p = vt — I cos 6
dt
dt.
Returning now to Equations (5) and eliminating d\l//dt, we
obtain :
(8) sin2 6 (~)2 = sin2 e(a - a cos 6) - (J8 - 6i> cos 0)2.
222 MECHANICS
The result is a differential equation for the single dependent vari-
able, 6. It can be improved in form by the substitution
(9) u = cos 9 :
(10) 2 = (1 - u*)(a - au) - CJ - bvuY = f(u).
Thus f(u) is seen to be a cubic polynomial, which we will
presently discuss in detail. But first observe that the second
Equation (5) gives :
d* P "" bvu
Hence \f/ is given by a quadrature after u has once been found
as a function of t.
Retrospect and Prospect. To sum up, then, we have reduced
the problem to the solution of Equation (10) for u as a function
of t.' Equation (9) gives 0; Equation (11) gives \f/; and Equa-
tion (7) gives <p. We may concentrate, then, on the solution of
Equation (10).
19. Continuation. Discussion of the Motion. The Polynomial
(1) fM = (1 - u*)(a - au) - (0 - bmY
becomes positively infinite for u = + oo. It is negative or 0
for u =+ 1, — 1. Hence in general the graph will be as indi-
cated, or
0 </(M), u, <u<u2]
fM = /(iO = o.
Moreover, — 1 < u^ < u2 < 1, and
f(u) has one root, u' > 1. The roots
w^ u2 will, therefore, be simple roots.
The differential equation
» (a)'-™
comes under the class discussed in Appendix B. In particular,
the solution is a function
(3) u = <*>(0
single-valued and continuous for all values of t and having the
period T7, where
ROTATION
223
(4)
or
(5) <p(t +
Furthermore, if
(6) *! =
then
And similarly, if
//w\ /, \
then
/*7'\ /* \ /j I \
\t ) ^C*2 — T) ~ ^(^2 ~T~ Ty.
Physical Interpretation. Let a sphere /S be placed about 0
as centre, and let P be the point of intersection of the positive
axis of f with S. Lot C be the curve that P describes on S. The
results just obtained show that C lies between the two parallels
of latitude corresponding to
(8) u = ult u = u2.
For convenience let t be measured from a point on the upper
parallel, u = u2. Then there are three cases according as initially
III. < 0.
CASE I. Since
FIG. 109
— bvu
eft 1 - u2
is positive when u has its greatest value, u2j d^/dt will remain
positive, and so ^ will steadily increase with t. Let ^ = 0 when
t = 0. As t increases to iT, \!/ will increase to
224 MECHANICS
where u = ^(0, Equation (3). When t = T, \p will have in-
creased by ^, and one complete arch of the curve C will have
been described. The arch is symmetric in the plane ^ = -J-^.
The rest of C is obtained by rotating this arch about the polar
axis of S through angles that are multipla of ^.
CASE II. Here, d\f//dt is 0 at the start, and hence
|8 - bvu2 = 0.
Since u decreases, it follows that in the further course of the
motion
0 < 0 — bvu,
and so \[/ steadily increases. The curve C has cusps on the upper
parallel of latitude.
CASE III. Here
0 - bvu < 0
at the start, and it is conceivable that this relation should persist
forever. But even if this were not the case, it is still conceivable
that the value of \l/ when P reaches the lower parallel of latitude
should be less than or equal to the initial value, ^ = 0. That
neither of these cases is possible — that the value of \l/ corre-
sponding to the first return of P to the upper circle is positive —
has been shown by Haclarnard.* The curve C has double points
in this case, but it proceeds with increasing t in the sense of the
advancing ^, as indicated.
Special Cases. There is still a variety of special cases to be
discussed, one of which is that in which f(u) has equal roots
lying within the interval :
— 1 < MI = tig < 1.
Since
/(I) £ 0
in all cases, and since
there must be a third root u1 ^ 1. Thus u± is a double root and
f(u) = (w - u,
where
x(u) < 0, - 1 < u < 1.
* Butt, des Sci. math. 1895, p. 228.
ROTATION 225
The only solution of Equation (2) in this case, which takes
on the value u^ when t = 0, is
u = wt.
The curve C reduces to a parallel of latitude.
When u = 1 is a root, various cases can arise. The point P
may pass through the north pole with a velocity ± 0 ; or it may
gradually climb, approaching the north pole as a limit; or the
top may permanently rotate about the polar axis. Similarly,
when u = — 1 is a root.
There is a great wealth of literature on the gyroscope and
the top. The reader can refer to the article on the Gyroscope
in the Encyclopaedia Britannica; to Webster, Dynamics; to
Routh, Elementary Rigid Dynamics; and to Appell, Mecanique
rationelle, vol. II.
EXERCISE
Treat the case of a top on a smooth table. Assume that the
peg is a surface of revolution. The distance, then, from the
centre of gravity to the vertical through the point of contact
with the table will be a function of the angle of inclination of
the axis.
Assume axes fixed in the body with the origin at the centre
of gravity.
Write down i) the equation of energy; ii) the equation that
says that the vertical component of or is constant.
From this point on the procedure is precisely as before, and
the result is again a differential equation of the type treated in
Appendix B. Discuss all cases, and show that in general the
axis oscillates between two inclinations, both oblique to the
vertical.
Begin with the special case that the peg is a point. Having
studied this case in detail, proceed to the general case and study
it in detail, also. Then derive the special case as a particular
case under the general case.
20. Intrinsic Treatment of the Gyroscope.* The most general
case of motion of a gyroscope reduces to one in which a single
couple acts on the body, and this couple can be broken up into
*The results of this paragraph arc contained in a paper by the Author: "On
the Gyroscope," Trans. Amer. Math. Soc., vol. 23, April, 1922, p. 240.
226 MECHANICS
two couples — one, represented by a vector at right angles to
the axis of the gyroscope ; the other, by a vector collinear with
the axis. In the most important applications that arise in prac-
tice, the latter couple vanishes. But in the general case, it gives
rise to the third of the Dynamical Equations in the form :
The former couple can be realized by a single force F per-
pendicular to the axis and acting at the point P in which the
positive f-axis cuts the unit sphere, — the other force of the
couple and the resultant force acting at 0.*
Definition of the Bending, K. Let C be the curve described
on the unit sphere by P, and let S be the cone which is the locus
of the axis of the gyroscope, and of which C is the directrix. Con-
sider the rate at which the tangent plane to S is turning when P
describes C with unit velocity. This quantity shall be denoted
as the bending of the cone and represented by the number K.
It is also the rate at which the terminal point of a unit vector
drawn from 0 at right angles to the tangent plane traces out its
path on the unit sphere. K shall be taken positive when an ob-
server, walking along C, sees C to the left of the tangent plane,
and negative, when C is to his right.
It is easy to compute K. Let V be the angle from the parallel
of latitude through P with the sense of the increasing ^ to the
tangent to C with the sense of the increasing s. Then it appears
form an infinitesimal treatment that
«__ .
v ' ds ds
Since
tan V = , . . — -, or V = tan"1
, . . — -, -77-; — -,
d\f/ sin B $' sin 0'
where accents denote differentiation with respect to s, and since
ds2 - d6* + dj* sin2 0, or 0/2 + V* sin2 0 = 1,
it follows that
(3) jc = WB" - 0' iH sin 0 - (1 + 0/2) V cos 0.
* The point O need not be the centre of gravity in the following treatment.
It may be any point fixed in the axis of material symmetry.
ROTATION 227
From the definition it follows at once that the bending of a
cone of revolution must be constant. To find its value, let the
coordinates be so chosen that the equation of the cone is 0 = a.
Then the length of the arc of C is
5 = ^ sin a and so \l/' sin a = 1.
From (3) it now is seen that
(4) K = — cot oc.
The cone lies to the right of the observer, as he travels along C.
If he reverses his sense, the sign of K will be changed. But both
cases are embraced in the single formula (4), the second corre-
sponding to a cone whose angle is TT — a, or
again for which s is replaced by — s.
Conversely, if K is constant, C is a circular
cone. For, the equation (3) can, by elimi-
nating ^',
U -Ax • ±1 f
be written in the form :
ntt
(5) ± K =
the — sign holding whenever ^' < 0. Hence
If K is constant, set K = — cot a. Then Equation (6) admits
one solution, 0 = a, or 0 = 7r — a; and, as is shown in the
theory of differential equations, this is the only solution which,
at a point s = s0, takes on the value a, or TT — a, and whose
derivative vanishes there.
Further Formulas for K.* From (3) it follows further that
-7-77: sin B — 2 -J-T^ cos 6 — sin2 6 cos 0
,~ ,
(7) ± * =
where the — sign holds whenever \f/' < 0.
* These results are inserted for completeness. They will not be used in what
follows, and the student may pass on without studying them. They are chiefly
of interest to the student of Differential Geometry.
228
MECHANICS
If K is known, or given, as a function of s, then Equation (6)
determines 0 as a function of s, and \l/ is then found by a quad-
rature :
(8)
The bending, K, is connected with the curvature, K, of C, re-
garded as a space curve, by the formula
(9) K* = K2 + 1.
Furthermore, cf. Fig. Ill below:
( i j k
(10) n = a X t
x y z
x' y' z'
(11)
hence
(12)
Since I K\ = 1 1' I and
= yz" -
zx" - xz"
KZ' = xy» _ yX".
(13) ^2 = x"* + y"* + z"*,
formula (9) follows at once from (12) and (13). Moreover, from
(12) it follows that
(14)
*=-
x y z
x' y' z'
x" y" z"
Finally, the torsion, T, of C is connected with /c by the relation :
vl*=± T,
the result obtained by Professor Haskins.*
* For the proof of this formula cf. the Author's paper cited above.
d
Ts(
ROTATION 229
21. The Relations Connecting v, F, and *. The physical phe-
nomenon which it is most important to bring home to one's
intuition is the effect of the force F on the motion of the gyro-
scope. Any such explanation must take account of all three
quantities, v, F, and K. But many popular explanations — claim-
ing correctly to be "non-mathematical," but incorrectly to be
accurate in their mechanics — fail because they are unaware
of K. Thus, for example, the statement often made that " when
a couple is applied to a rotating gyroscope, the forces of the
couple intersecting the axis of the gyroscope at right angles, the
axis will move in a plane perpendicular to the plane of the forces
of the couple" is false. In fact, the axis will begin to move
tangentially to this plane, if it starts from rest, and all inter-
mediate cases are possible, according to the initial motion of the
axis.
A simple and accurate explanation, in terms of v, F, and K,
can be given as follows.* First of all, however, the third of
Euler's Dynamical Equations, which here becomes :
(1)
and requires no further comment than i) that it is perfectly
general, applying to the motion of the gyroscope under any
forces whatever; and ii) that in the case which most interests
us, namely that in which there is only the force F (and the reac-
tion at 0) we have : N = 0, and so r = *>, a constant.
Let F, then, be resolved, in the tangent plane, into a component
T along the positive tangent, and a component Q, taken positive
when directed toward the left of the observer; i.e. Q is posi-
tive when K is positive. Then
(2)
AKV* + Crv = Q,
where v = ds/dt and s increases in the sense of the motion of P,
r being given by Equation (1).
*Cf. the Author's paper "On the Gyroscope" cited above, p. 240.
230
MECHANICS
Proof. Let the unit vector from 0 to P be denoted by a (it
is the vector 7 of the coordinate system) ; let t be a unit vector
along the positive tangent to C at P; and let n be a unit
vector normal to a and t and so oriented
'a with regard to them as ft is with regard
\p X to 7 and a. These are principal axes of
t \ inertia, and the moments of inertia about
them are :
L = A In = A, Ia = C.
The components of the angular velocity
& about them are :
FIG. Ill
o>t = 0, wn = I),
Now, (T has the value :
a = In C0n n + It CO* t + Ia C0a a.
Hence
(3)
From this equation we can compute -rr :
do-
dv
dn
da
It is clear that
(4)
da
dt
vt.
Furthermore, from the definition of the bending, it appears that
/^x dn
(5) -j-r = KVt
Hence, finally,
d<r
(6)
= Av-n + (Aw* + C»)t + C
Let the vector M which represents the resultant moment of
all the applied forces about 0 be written in the form :
M = Mnn + Mtt + Maa.
Since
da -.
ROTATION 231
we have :
(7) Av^- = Mn, AKv* + Crv = Mt, C % = Ma.
us dt
Turning now to the case in which we are most interested,
namely, that in which a force F acts at P in a direction at right
angles to OP :
F=-Qn+Tt,
we see that Mn = T, M t = Q, and thus Equations (2) are
established. Equation (1) is the third of Equations (7).
We have thus obtained Euler's Dynamical Equations in the
form:
dv
(8)
Av^- = T
ds
AKV* + Crv = Q
22. Discussion of the Intrinsic Equations. The first of Equa-
tions (8), § 21,
AV£ = r,
ds '
admits a simple interpretation. It shows that the point P de-
scribes the curve C exactly as a smooth bead of mass m = A
would move along a wire in the form of C if it were acted on by
a tangential force T.
The third equation,
C~ = N
^ dt *'
shows that the component r of the angular velocity & about the
axis of the gyroscope varies exactly as it would if the axis were
permanently at rest and the same couple N relative to the axis
acted.
The second equation,
A) AKV* + Crv = Q,
expresses the sole relation which holds between the four variables
K, v, r, and Q. In the applications, however, r is constant, r = v,
and so the equation
A') AKV* + Cw = Q
expresses the sole relation between K, v, and Q.
232 MECHANICS
The Case F = 0. Let us begin with the case that F vanishes,
but the axis is not at rest. Here, Q = 0, T = 0. Equation A)
gives
i) AKV + Cr = 0,
or, on introducing the radius of bending, p = 1/| K |, and choos-
ing r > 0 :
Av
p== 'Cr'
If r is a positive constant, r = v > 0, then
Cv
and since v is constant, for
A __ n
FIG. 112 ds~ '
K is also constant, and negative. The axis of the gyroscope is
describing a cone of semi-vertical angle a, where
cot a = | K |, or tan a = p,
and the sense of the description is such
that the observer, walking along C in the
positive sense, has the cone on his right.
The Case K = 0. Here, the path of P
is an arc of a great circle, and
& ' FIG. 113
Q = Crv, or Q = Cw,
no matter what T and the motion of P along its path may be.
The pressure of the axis against the constraint, in a normal direc-
tion, is to the right, and is proportional to r and to v ; — or, if
r is constant, to v, the coefficient then being Cv. Thus we obtain
anew, and with the minimum of effort, the main result of § 17.
General Interpretation of Equation A). We can now give a
simple physical interpretation to Equation A) :
Am* + Crv = Q.
The left-hand side is the sum of two terms. The second term
expresses the force,
Q2 = Crv,
ROTATION
233
that would be required to cause P to describe a great circle on
the sphere; i.e. to make the axis move in the plane through 0
tangent to C. This force, Q2, is always directed toward the
left, for Q2 > 0.
The first term,
accounts for the bending.
~ , 9
Ql = AKV*,
Its numerical value,
can be interpreted as the centripetal force exerted on a particle,
of mass m = A, to make it describe a circle of radius p with
velocity v. Wlien K is positive, this force is positive, and so is
directed toward the left ; and vice versa.
Consider now the force Qf along the normal — n at P, which
(combined with the smooth constraint of the surface of the sphere)
would be required to hold a particle of mass m = A in the path C.
Let the vector a be written in the form :
Then
a = xi
yj + zk.
v = xi + y] + zk = vt,
where x' = dx/dst x = dx/dt, etc. Furthermore,
n = a X t = (yzf - zyf) i + (zz' - xzf) j + (**/' - yx')k.
The acceleration, (a), of P in space is, of course :
(«) = zi + « j + ^k.
Now, the component of the acceleration along the normal n
to the plane of a and t is n-(a), which can be written in the
form :
x y z
x' yf z'
x y z
Since x = vx', it follows that
x = v*x" + vx', etc.,
and so
x y z
x' y' z'
x y z
x y
x" y" z"
234 MECHANICS
Thus mw2 is equal to the force Q' tangent to the sphere and
normal to (7, which would be required to hold a particle of mass
w, describing (7, in its path; the component along t being
mvdv/ds, and the third component, along a, being the reaction
normal to the sphere, in which we are not interested.
It is natural to think of the point pn on the line through P
along n as the centre of bending. If we draw the osculating cone
of revolution through P, this is the point Q in which
that line meets the axis of the cone. An obvious
interpretation for this force of mw2 is the centripetal
force of a particle describing a circle of radius p, with
centre at Q, tangent to C at P, the velocity being v.
_, . The force Q2 can be realized physically as follows.
Let an electro-magnetic field of force be generated by
a north-pole situated at 0, and let the particle m carry a charge,
e, of electricity. The force exerted on e by the field will be at
right angles to the path and tangent to the sphere, — and,
finally, proportional to the velocity, v, of m. Hence e can be so
chosen that this force will be precisely equal to Q2 = CW.
In the more general, but less interesting, case that r is variable,
the physical interpretation can still be adapted by using a variable
charge.*
Summary of the Results. To sum up, then, we can say : The
point P, in which the axis of the gyroscope meets the unit sphere
about 0, moves like a particle of mass m = A constrained to
lie on the sphere and carrying a charge of electricity, e. The
forces that act on m are supplied by the electromagnetic force
of the field, Q2 = Cw, acting on ey and a force F acting on m,
the components of F along the tangent and normal at P being
T and Ql respectively. The case of a variable r can be met by
a variable charge, e.
As regards the physical realization of the condition that the
particle lie on the surface of the sphere, we may think of a mass-
less rod of unit length, free to turn about one end which is pivoted
at 0, and carrying the particle at the other end.
* The idea of using the above electro-magnetic field to obtain #2 was suggested
to me by my colleague, Professor Kemble, to whom I had just communicated the
results of the text, down to this point. (Note of Jan. 23, 1933.)
ROTATION 235
EXERCISES
1. Suppose the axle P of the gyroscope is caused to move in
a smooth slot in the form of a meridian circle, which is made to
rotate in any manner. The force F will then be normal to the
meridian, or tangent to the parallel of latitude. Show that
d26 .
Suggestion : Combine Euler's Geometrical Equations with
Euler's Dynamical Equations.
2. Let the components of F along the meridian in the sense
of the increasing 6 and along the parallel of latitude in the sense
of the increasing ^ be denoted respectively by ® and ^ Show
that
If ® and ^ are known as functions of 6, \f/} t, these equations
suffice to determine the path of P.
3. Consider small oscillations of the axis of the gyroscope in
the neighborhood of the axis 6 = v/2, $ = 0. Let
Show that the equations of Question 1 lead to the approximate
equations :
4. Generalize the equations of Question 2 to the case that A,
, C are all distinct.
236 MECHANICS
5. Intrinsic Equations. From the equations :
Av% = T,
ds '
AKV* + Cvv = Q,
K =
the path can be determined if T, Q are known as functions of
s and v.
6. . Ship's Stabilizer. The gyroscope can be used to reduce the
rolling of a ship. A massive gyroscope is mounted in a cage,
or frame, its axis being fixed with reference to the frame, and
vertical. The frame is mounted on trunnions, with axis hori-
zontal and at right angles to the keel, and it is provided with
a brake to dampen its oscillations about this axis. Thus the
axis of the gyroscope has two degrees of freedom; it can rotate
in the plane through the keel and the masts, and this plane rotates
with the rolling of the ship.* Isolate the following systems :
i) The ship, exclusive of the gyroscope and frame ;
ii) The frame ;
Hi) The gyroscope.
The rolling of the ship is governed by the equation :
where the first term on the right is due to the damping of the
water; the second, to the righting moment produced by the
buoyancy ; and the third, to the force exerted by the trunnions.
The frame may be thought of as rotating about the point, 0,
regarded as fixed, in which the axis of the gyroscope cuts the
*A picture and an account of the ship's gyroscope is found in the article on
the "Gyroscope" in the Encyclopaedia Britannica and in Klein-Sommerfeld,
Theorie des Kreisels, vol. iv., p. 797. For the discussion which follows the reader
also needs, however, the theory and practice of Oscillatory Motion with Damping ;
cf. the Author's Advanced Calculus, Chap. XV.
ROTATION 237
axis of the trunnions. Let Euler's Angles be so chosen that the
axis of the sphere : 6 = 0, ^ = 0, is parallel to the keel, the
plane ^ = 0 being vertical. Moreover, let 0 be replaced by $,
where
The motion of the gyroscope about its centre of gravity, the
point (9, will be governed by the approximate equations of Ques-
tion 3.
Finally, the motion of the frame is governed by the equations
called for in Question 4 above. These equations are modified by
the condition <p = 0, and then reduced still further by setting
sin & = 0, cos # = 1. Thus
dt
where the first term on the right is due to the brake and other
damping, and the second, to gravity, since the frame is so con-
structed that its centre of gravity is appreciably below O.
On combining these five equations and neglecting. A + A' in
comparison with 7 we find :
These are the equations which govern the motion. They are
discussed at length in Klein-Sommerfcld, I.e.
23. Billiard Ball. Let a billiard ball be projected along the
table, with an arbitrary initial velocity of the centre, 0, and an
arbitrary initial velocity of rotation. To determine the motion.
Let the (x, y)-plane of the axes fixed in space be horizontal.
Let moving axes of (£, 17, f) be chosen parallel to (#, y, z), but
with the origin at the centre of the ball.
The point of the ball P, in contact with the table, shall be
slipping, and the angle from the positive direction of the axis
of x or £ to the direction of its motion shall be ^.
238
MECHANICS
The forces acting are : gravity, or M gt downward at 0 • R = M g
upward at P ; and the force of friction, nMg, at P in the sense
opposite to that of slipping. Hence, for the motion of the centre
of gravity,
M -jp = — pM g cos \f/
CD rf2.
dt2
The vector momentum a-, referred to the centre of gravity,
has for its components along the moving axes :
where
The moment equation,
thus gives :
f Ma2.
d<r
(2)
Hence
(3)
/ -—• = — nMga sin
u-6
7 --— = nMga cos ^
'if-0-
= const.
5 dt
The angle ^ is unknown. Eliminate it by combining Equa-
tions (1) and (2) :
(4)
Hence
(5)
dt*
^y.
dt2
5 dt
2a
2a
where A, B are constants of integration depending on the initial
conditions. They may have any values whatever.
ROTATION 239
Let V be the velocity of the lowest point of the ball. Then
(6)
Vx = V cos $ = -7- — aco,
Vv = V sin ^ = -£ + au(.
Combining these equations with (5) we get :
where
2\dt ^
A' = -
B' =- fB.
Equations (1) now take on the following form. For abbrevi-
ation let
dt
A',
Then
(8)
Hence
and consequently
du
dt
dv
du dv ^
v -7T — u -77 = 0,
dt dt y
— av,
where a, /3 are constants not both 0. Moreover, u and v are
not both 0.
Suppose u > 0, a > 0. Then
av
av
u
The proof of this last equation requires the consideration of the
two cases : i) 0 7* 0 ; ii) 0 = 0. These formulas are general,
holding in all cases in which w ^ 0.
240
MECHANICS
It thus appears that
(9)
cos
sn =
ft
Hence d?x/dt2 and d2y/dt* are constants, and consequently the
centre of the ball describes in general a parabola; in particular,
a straight line. The direction, however, in which the point P
is slipping, is always the same; cf. Equations (9).
This result comprises the main interest of the problem, so long
as there is slipping. Slipping ceases when V = 0, or
(10)
The Subsequent Motion. From this instant on the motion is
pure rolling — we are, of course, neglecting rolling friction and
all other damping. For, at the instant in question, V — 0, and
the angular velocity is related to the linear velocity of the centre
of gravity as follows. Let the centre of the ball be at the origin
and let its velocity be directed along the positive axis of x. Then
(11)
dx
Tt
t-O
c #
c> Jt
0;
au( | t-o = 0, aco, \t-o = c, wf | t-o — y,
where 7 can have any value, positive, negative, or 0.
Let us consider the motion which consists in pure rolling and
pivoting, and see what force at P is necessary. First, we have
(12)
Mw =
-
dt*
- Y
~ Y>
where X, Y are the components of the unknown reaction at P.
Next, taking moments about the centre of gravity, we find :
(13)
UOi) f UWf) -w*
___. — i ___. — _ ax
dt dt
dt dt
ROTATION 241
Finally,
(14)
Tr A
y. - - - a*, - 0
These seven equations, (12), (13), (14), together with the
initial conditions (11), formulate the problem completely, and
determine the seven unknown functions, x, y, co$, w^, o^, X , F,
as we will now show.
From the second equation (13) it appears that
Subtracting this equation from the first equation (12), we find :
But the left-hand side of this equation vanishes because the first
equation (14) is an identity in t. Hence X = 0. Similar con-
siderations show that Y = 0.
On substituting these values in (12) and (13), these five equa-
tions can be solved subject to the five initial conditions (11), and
the other condition, that initially x = 0, y = 0. The centre of
the ball describes the positive axis of x with constant velocity, c.
The angular velocity w is also constant, its components along
the axes being given by their initial values (11). Since 7 is
arbitrary, w may be any vector whatever in the (77, f)-plane,
whose component along the r;-axis is c/a.
The foregoing discussion may be abbreviated by means of the
Principle of Work and Energy, Chapter VII.
The motion of pure rolling with pivoting requires, then, no
force to be exerted by the table. It is uniquely determined by
the initial conditions, and hence it coincides with the actual
motion of the billiard ball.
24. Cart Wheels. Consider the forewheels of a cart. Ideal-
ized they form two equal discs connected by an axle about which
each can turn freely. To determine the motion on a rough
inclined Diane.
242
MECHANICS
We will begin with a still simpler case — that of a single wheel,
or disc, mounted so that it can turn and roll freely, but will always
have its plane perpendicular to the plane on which it rolls. The
frame which guides it may be thought of as smooth. Its mass
can be taken into account, but we will disregard it, in order not
to obscure the main points of the problem.
We will choose the coordinates as indicated, the axis of £ being
in the disc and always parallel to the plane ; the axis of 77, being
the axis of the disc, is also
parallel to the plane. The
axis of y lies in the plane and
is horizontal. The axis of x is
directed down the plane. Let
v? be the angle through which
the disc has turned about the
axis of rj ; let s be the arc de-
scribed by the point of con-
tact, P ; and let 0 be the angle
from the positive axis of x to the positive tangent at P. Let the re-
action of the plane be
F = Xa + Yj9 + ZT,
where a, 0, y are unit vectors along the moving axes.
Z = Mg cos e, where e is the inclination of the plane, and
Then
(1)
d2x
M -;-£ — X cos 6 — Y sin 6 + Mg sin 6
= Xsin0 + Ycos0
The angular velocity,
has the value
Moreover,
0 = - da, 7 = 0.
Take moments about the centre of gravity :
(2)
dt
ROTATION
243
Since
and A = C =
Hence
or
(3)
In computing the right-hand side of Equation (2), the couple
which keeps the axis of the disc parallel to the plane must be
taken into account. The vector which represents it is collinear
with the axis of £ . Hence the couple may be realized by the two
forces :
= 0, co,, = <p, o>£
a2, # = pf a2, we have :
= - Beta. + 5^j3 +
Thus
at r2=-0.
r* X F* = j8 X Fl7 + (- 18) X (- F,y) + (- ay) X F;
r* X F, = (2F, + aY) a - aX0.
or, finally :
(4)
Equating, then, the vectors (3) and (4) we find :
(5)
Finally, the condition of rolling without slipping can be written
in the form :
dx
~dt
= v cos 0,
dy =
dt
v sin ^,
244 MECHANICS
where
and so
/n\ dx d<p n dy d<p . „
(6) -TT = a-—- cos 0, -~ = a-— sin 0.
v ' dt dt dt dt
The formulation is now complete. There are seven unknown
functions, namely : x, y, 0, p, X, Y, Fl9 and seven equations to
determine them, namely, Equations (1), (5), (6).
To solve these equations, begin by determining 6 from (5) :
3/\
(7) ft = X, 9 = \t + M.
Next, eliminate Y in (1) :
cos<9 + sin^ = x + Mg sin e cos 6.
at1 diL J
And now X can be eliminated by (5), and xy y by (6). Thus
Ma -—• = — i Ma ~ + Mg sin e cos 0,
or
(8)
where
Hence
(9) ^ = *(sin»-slnM) + *o,
and
fc , M . / . fc sin M\ .
v? = ^5 (cos M - cos 0) + ^0 -- — Jt + v?o-
From (6), x and y can now be found as functions of t; and finally
X, Y, Fu can be determined from (1) and (5).
The system of Equations (1), (5), (6) is an example of equa-
tions called non-holonomic by Hertz because some of them, —
namely (6), — involve time-derivatives of the first order only
and cannot be replaced by geometric equations between the
coordinates.
An interesting case of a non-holonomic problem is that of a
coin rolling on a rough table. It is studied in detail by Appell,
ROTATION 245
Mecanique raiionelle, Vol. I, p. 242, of the 1904 edition, and an
explicit solution is obtained in terms of the hypergeometric
function.
EXERCISES
The student should first, without reference to the book, repro-
duce the treatment just given in the text, arranging in his mind
the procedure: i) figure, forces, coordinates; ii) motion of the
centre of gravity; Hi) moments about the centre of gravity;
iv) conditions of constraint ; v) the solution of the equations.
1. Solve the problem of the two wheels mentioned in the
text.
2. Coin rolling on a rough table. Read casually Appell,
adopting his system of coordinates. Then construct independ-
ently the solution, following the method used in the problem
of the text.
3. The problem of the text, when the mass of the frame is
taken into account. Begin with the case that the bottom of the
frame is smooth and its centre of mass is at the centre of the disc.
4. Study the motion of the centre of gravity of the disc treated
in the text, by means of the explicit solution of x, y in terms of t.
25. Resume. In dealing with the motion of a rigid body,
there are the two vector equations :
equivalent to six ordinary equations.
It is always possible to take moments about the centre of
gravity.
The Principle of Work and Energy frequently gives a useful
integral of the equations of motion.
If the right-hand side of the Moment Equation is a vector
lying in a fixed plane, the component of a normal to this plane is
constant, and thus an integral of the equations of motion is
obtained.
Sometimes there are conditions which are expressed by equa-
tions between time-derivatives of the first order, t = t, but which
cannot be expressed by equations between the coordinates only.
246 MECHANICS
The first step in solving a problem is to draw the figure, mark
the forces, and pass in review each of the items just mentioned ;
reflecting, in case these are not adequate, on considerations of
like nature, which may be germane to the problem.
With the forces and the geometry of the problem in mind,
next choose a suitable
Coordinate System. If it is desirable to refer a to the centre
of gravity, a Cartesian system with its origin there is usually
the solution. These axes may be fixed in the body, coinciding
with the principal axes of inertia. Or their directions may be
fixed in space. Or they may move in the body and in space
subject to some condition peculiar to the problem in hand.
Final Formulation. It remains to write down the equations
arising from each of the above considerations. They must be
in number equal to the number of unknown functions. Besides the
differential equations of the second order, these may also include
differential equations of the first order, not reducible to equations
between the coordinates.
The solution of these equations is a purely mathematical prob-
lem. Go back frequently over familiar problems and recall the
mathematical technique, writing the equations down on paper,
neatly, and carrying through all details of the solution. In this
way, analytical consciousness is developed; it is composed of
experience and common sense.
Further Study. There is a vast fund of interesting problems
in Rigid Dynamics, of all orders of difficulty, and two invaluable
treatises are Appell, Mecanique rationelle, Vols. I and II, and
Routh, Rigid Dynamics, Vols. I, II. Routh's exposition of the
theory is execrable, but his lists of problems, garnered from the
old Cambridge Tripos Papers, are capital.
The earth is a top, and the study of the precession and nuta-
tion of the polar axis is a good subject for the student to take
up next.
Webster's Dynamics is also useful in the important applica-
tions it contains. The text is hard reading; but the student
who once dominates the method as set forth, for example, in the
foregoing treatment, can and should construct his own solution
of the problem in hand.
Finally, Klein-Sommerfeld, Theorie des Kreiselsy in four volumes.
This is a classic treatment of the subject. The first three vol-
ROTATION 247
umes treat the theory of the top by modern mathematical methods.
The fourth volume, devoted to the applications in engineering,
can be studied directly through the theory which we have de-
veloped above, without reference to the earlier volumes. There
is a detailed study of the gyroscopic effect in the case of rail-
road wheels, the Whitehead torpedo, the ship's stabilizer, the
stability of the bicycle, the gyro-compass, the turbine of Leval,
and a large number of further topics.
CHAPTER VII
WORK AND ENERGY
1. Work. In Elementary Physics work is defined as the
product, force by distance :
(1) W = Fl,
the understanding being that a force F, constant in magnitude
and direction, acts on a particle, P, or at a point P fixed in a
rigid or elastic body, and displaces P a distance I in the direction
of tho force.
The definition shall now be extended to the case of a variable
force, still acting on a particle or at a fixed point of a material
body. Let
a ^ x g 6
be the interval of displacement. Let
F=f(x)
be the force, where f(x) is a continuous function. Divide the
interval into n parts by the points XQ = a, xl9 • • • , xn-\, xn = b,
Fk and consider the fc-th sub-interval :
- _, .
FlG 116 3*-i ^ x ^ xk, Az* = xk - «»_!.
And now we demand that the extended definition of work shall
be so laid down that
i) the total work shall be equal to the sum of the partial works :
ii) the work for any interval shall lie between the work cor-
responding to the maximum value of the force in that interval,
and the work corresponding to the minimum force :
g AWk ^
248
WORK AND ENERGY 249
where
Fi £ F £ Fi'
in the interval in question.
Now, since f(x) is a continuous function, it takes on its mini-
mum value, Fi, in the interval :
and similarly, its maximum value :
FZ =/(*;'), **-
Hence W lies between the two sums :
But each of these sums approaches a limit as n increases, the
longest Axjb approaching 0, and this limit is- the definite integral :
Hm
i>
= f
J
Hence the requirements, i.e. physical postulates i) and ii) are
sufficient to determine the definition of the work in this case : *
b
(2) W=Jf(x)dx.
a
The foregoing definition applies to a negative force, and also to
the case that 6 < a ; the work now being considered as an alge-
braic quantity. Thus if a force, instead of overcoming resistance,
is itself overcome ; i.e. yields, it does negative work.
The work which corresponds to a variable displacement, x,
where a ^ x ^ 6, is by definition :
X
(3) W=Jf(x)dx.
a
Hence
(O f - ,.
* Strictly speaking, we have shown that (2) is a necessary condition for the
definition of work according to the postulates i) and ii) . It is seen at once, however,
that conversely Equation (2) affords a sufficient condition, also.
250 MECHANICS
EXERCISES
1. Show that the work done in stretching an elastic string is
proportional to the square of the stretching.
2. Find the work done by the sun on a meteor which falls
directly into it.
3. The work corresponding to a variable displacement from
x to 6, where a ^ x ^ 6, is by definition :
&
(5) W=ff(x)dx.
X
What is the value of dW/dxl
2. Continuation : Curved Paths. Suppose the particle describes
a curved path C in a plane, and that the force, F, varies in mag-
nitude and direction in any continuous manner. What will be
the work done in this case?
Suppose the path C is a right line and the force, though oblique
to the line, is constant in magnitude and direction; Fig. 117.
Resolve the force into its
two components along the line
and normal to it. Surely, we
tett >f< j*a must lay down our definition
r^— - — -~- i of work so that the work done
\ii \
by F is equal to the sum of
FlG 117 the works Of the component
forces. Now, the work done
by the component along the line has already been defined, namely,
Fl cos ^, where F = | F | is the intensity of the force.
It is an essential part of the idea of work that the force over-
comes resistance through distance (or is overcome through dis-
tance). Now, the normal component does neither; it merely
sidles off and sidesteps the whole question. It is natural, there-
fore, to define it as doing no work. Thus we arrive at our final
definition: The work done by F in the particular case in hand
shall be
(6) W = Fl cos ^.
A second form of the expression on the right is as follows. Let
X and Y be the components of F along the axes. Let T be the
angle that the path AB makes with the positive axis of x. Then
WORK AND ENERGY
251
the projection of F on A B is equal to the sum of the projections
of X and Y on AB, or
F cos ^ = X cos T + Y sin T.
On the other hand,
x2 — xl = I cos T, 2/2 — 2/i = Z sin r.
Hence
(7) TF = X(xz — arO H
General Case. If C be any regular curve, divide it into
n arcs by the points s0 = 0, st, • • • , sn-i, sn = J. Let F£ be
the value of F at an arbitrary point of
the /b-th arc, and let ^i be the angle
from the chord (st-i, sk) to the vector
F£. Then the sum
Jb-l
FIG. 118
where Zfc denotes the length of the
chord, gives us approximately what we should wish to understand
by the work, in view of our physical feeling for this quantity.
The limit of this sum, when the longest h approaches 0, shall be
defined as the work, or
(8)
W
Since
~^- = 1,
As* '
it is clear that the above limit is the same as *
/
n /*
lim V Fk cos fa Asfc —IF cos \l/ ds.
— A J
We are thus led to the following definition of work in the case
of a curved path :
iF cos ^ ds.
o
(9)
W
* Cf. the author's Advanced Calculus, p. 217. It is imperative that the student
learn thoroughly what is meant by a line integral.
252 MECHANICS
A second formula for the work is obtained by means of (7) :
< i
(10) W= Ax COST + 7 SUITES = C(x^ + Y-
J J ^ as t
0 0
or
(11) W = Cxdx + Ydy.
The extension to three dimensions is immediate. The defini-
tion (9) applies at once without even a formal change. Formula
(11) is replaced by the following :
(12) W = Cxdx + Ydy + Zdz
or
(«', 6'. c')
Cxdx+Ydy + Zdz.
(a.b.c)
Example. To find the work done by gravity on a particle
of mass m which moves from an initial point (XQ, yQ, ZQ) to a final
point (xlt ylt zj along an arbitrary twisted curve, C.
Let the axis of z be vertical and positive downwards. Then
X = 0, Y = 0, Z = mg ;
W = jXdx + Ydy + Zdz = jmgdz = mg(zl - z0).
C ZQ
Hence the work done is equal to the product of the force by the
difference in level (taken algebraically), and depends only on
the initial and final points, but not on the path joining them.
EXERCISES
1. A well is pumped out by a force pump which delivers the
water at the mouth of a pipe which is fixed. Show that the
work done is equal to the weight of the water initially in the well,
multiplied by the vertical distance of the centre of gravity be-
low the mouth of the pipe.
2. The components of the force which acts or? a particle are :
X = 2x — 3y + 4z — 5, Y = z — x + 8, Z = x + y + z + l2.
WORK AND ENERGY 253
Find the work done when the particle describes the arc of the
helix
x = cos 0, y = sin 0, z = 70,
for which 0 ^ 0 g 2*.
3. If the curve C is represented parametrically :
C: x= /(X), y = *(X), 0 = f (X), X0 S X ^ Xi,
show that the work is given by the integral :
3. Field of Force. Force Function. Potential. A particle
in the neighborhood of the solar system is attracted by all the
other particles of the system with a force F that varies in magni-
tude and direction from point to point. Thus F is a vector point-
function throughout the region of space just mentioned. Its
components along Cartesian axes, namely, X, Yy Z, are ordinary
functions of the space coordinates, x, y, z, of the particle. In
vector form :
(1) F = Xi + Y j + Zk.
The example serves to illustrate the general idea of a field
of force. We may have an electro-magnetic field, as when a
straight wire carries a current. If the north pole, P, of a magnet
is brought into the neighborhood of the wire, it will be acted on
by a force F at right angles to any line drawn from P to the wire
and of intensity inversely proportional to the distance of P from
the wire, the sense of the force depending on the sense of the
current.
If the axis of z be taken along the wire, then
Z-0,
where (r, 0, z) are the cylindrical coordinates of P, and C is a
positive or negative constant. Thus in vector form
and
254 MECHANICS
Force Function. It may happen that there is a function
(4) u = <p(x, y,z)
such that
, v Y _ du v _ du 7 __ du
(5) X~te> W ~d~z
Such a function, w, is called a force function. In vector form :
F can be written in symbolic vector form as follows. Let
V be a symbolic vector operator, namely :
(7) V = i— +'— + k— -
Then Vu is defined as :
Hence
(9) F = Vu.
Gravitational Field. In the case of the field generated by a
single particle of attracting matter, there is a force function :
(10) « = £
where r is the distance from the given fixed particle to the variable
particle, and X is a positive constant.
In the case of n particles,
(n) « - 2 Tk>
provided the units are properly chosen.
Ekctro-Magnetic Field. For the electro-magnetic field above
described,
(12) 11 = CO.
We may also write :
(13) u = C tan~l -;
v / X9
but this formula is treacherous, since only certain values of the
multiple-valued function are admissible. However, since the
wrong values differ from the right ones only by additive con-
WORK AND ENERGY 255
stants, we can use the formula for purposes of differentiation,
and we shall have :
(14) X = |H = C-=g-v Y = f* - C^jhi, Z = 0.
' dx x2 + 2/2' £?/ x2 + y2
Work. When a particle describes an arbitrary path in a field
of force, the work done on the particle by the field is given by
Equation (12) of § 2. If there is a force function, this formula
becomes :
i i &u 7 . du , \
i.e. the change which u experiences along the curve C. If the
region in which C lies is simply connected, or if u is a single-
valued function, then
(16) W = u + const.
Thus W is independent of the path by which the particle arrived
at its final destination, and depends only on the starting point
and the terminal point :
(17) W = u(x, y, z) - u(a, 6, c).
For any closed path, W = 0.
Such a field of force is called conservative. It is true conversely
that if the field represented by the vector (1) is conservative,
then there always is a force function, u. For then the integral :
(*.*.«
(18) u= IXdx + Ydy + Zdz
(a.b.c)
is independent of the path and so defines a function u(x,y,z).
Moreover,
(19) **, *y, *z.
dx dy dz
Potential Energy. When a field of force has a force function,
u, the negative of u, plus a constant, is defined as the potential
energy :
(20) * = - u + C.
In case, then, a potential <p exists,
256 MECHANICS
EXERCISES
1. Show that the field of force defined by the vector (3) is
not conservative. But if R be any region of space such that an
arbitrary closed curve in R can be drawn together continuously
to a point not on the axis, without ever meeting the axis, though
passing out of R, then the field of force defined in R by (3) is
conservative.
2. A meteor, which may be regarded as a particle, is attracted
by the sun (considered at rest) and by all the rest of the matter
in the solar system. It moves from a point A to a point B.
Show that the work done on it by the sun is
W = Km(- -
where r0 and rx represent the distances of A and B, respectively,
from the sun, and K is the gravitational constant.
4. Conservation of Energy. Let a particle be acted on by
any force whatever. The motion is determined by Newton's
Second Law :
(i) '
U/l/ U>C/ U/l/
Multiply these equations respectively by dx/dt, dy/dt, dz/dt,
and add :
dzd^z\ _ ydx , ydy -dz
} m \dt~dt2 ~r~dt~dt2 ^ dt ~dfi) ~ dt dt + dt
The left-hand side of this equation has the value :
~2dtv*'
W + ~dP + ~dt2 '
where
v2 =
Hence
m d 2 Y dx , y, dy t ^ dz
2 dt dt dt dt
Each side of this equation is a function of t, and the two func-
tions are, of course, identical in value. If, then, we integrate
WORK AND ENERGY 257
each side between any two limits, <„ and tlt the results must
tally :
f™*.vtdt= (
J 2dtVdt J \
«o 'o
The left-hand side of this equation has the value :
The right-hand side is nothing more or less than
Cxdx + Ydy + Zdz,
taken over the path of the particle ; § 2, (13). But this is pre-
cisely the work done on the particle by the force that acts. Hence
The quantity
mv
is defined as the kinetic energy of the particle. We have, then,
in Equation (3) the following theorem.
THEOREM. The change in the kinetic energy of a particle is equal
to the work done on the particle.
If instead of a single particle we have a system of particles,
the same result is true. For, from the equations of motion of
the individual particles :
,.^ d2xk v
(4) mt-^r-Xt, ™*-- = ,
we infer that
^ ( v dxk . dyk , dzk
2(X
xk . v dyk , 7 dzk\
+^
The kinetic energy of the system is defined as
258 MECHANICS
On integrating, then, between any limits <„ and <1; we have
89 r.-'.-t
The right-hand side represents the sum of the works done on
the individual particles, or the total work done on the system.
The result is the Law of Work and Energy in its most general
form for a system of particles.
THEOREM. The change in the kinetic energy of any system of
particles is equal to the total work done on the system.
Conservative Systems. In case the forces are conservative;
i.e. if there exists a force function U such that
the right-hand side of Equation (5) becomes C/j — Ug, and so
(7) 1\ -T0 = U, - U»
The potential energy, <l>, is defined as :
(8) $ = - U + const.
Hence (7) can be written :
(9) 71! + *t = T0 + *o-
Lett the total energy be defined as
(10) E = T + *.
We have, then :
(11) E, = Em
or the total energy remains constant. This is the Law of the Con-
servation of Energy in its most general form for a system of
particles.
6. Vanishing of the Internal Work for a Rigid System. Con-
sider a set of particles which form a rigid system. Let them be
held together by massless rods connecting them in pairs. Thus
the internal forces with which any two particles, rat- and m/,
react on each other are equal arid opposite :
(1) Fty + Fn = 0,
WORK AND ENERGY 259
and lie along the line joining the particles, and furthermore the
distance between the particles is constant ; i.e.
(2) rl = (Xi - xtf + (yi - 2/y)2 + (Zi - ztf
is independent of the time, or
In general, however, if each particle is connected by these
rods with all the others, there will be redundant members, so
that the stresses in the individual rods will be indeterminate. In
that case, let the superfluous rods be suppressed.
Consider the work done on the particle mi by the rod con-
necting it with mj. It is :
/
+ Y^ dyi + Z»/ dzi
and can be expressed by means of the parameter t in the form :
C( Y —{ -L- v ^Mi -u 7 ^A ,//
By the same token, the work done on rn,j by m» is
Since
Xu + Xn = 0, Ya + Y,-t = 0, Za + Za = 0,
the sum of these two works can be written in the form :
«i
/(
i v (dyi dy\ „ (dz{ dz
ij' ~~ + Yii\dt ~ ~ " ~
This last integral vanishes. For, the force Fi; is collinear with
the line segment connecting wt- with mjf or :
Hence the integrand vanishes identically by (3).
260 MECHANICS
We have thus obtained the result that the work done by the
internal forces of a rigid system of particles is nil. It follows,
then, that the change in the kinetic energy of such a system is
equal to the work done by the external, or applied, forces. Look-
ing backward and also forward we can now state the general
THEOREM. The change in the kinetic energy of any rigid system
whatever is equal to the work done by the applied forces.
For a system of particles the proof has been given. Before
we can extend it to rigid bodies, we must generalize the defini-
tions of kinetic energy and work.
6. Kinetic Energy of a Rigid Body. Consider a rigid body.
Let the volume density, p, be a continuous function. Denote
by v the velocity of a variable point P of the body. Then the
kinetiQ energy is defined as
(i) T =
extended throughout the region T of space, occupied by the body.
The vector velocity v of P is the vector sum i) of the velocity
V along the axis of rotation and ii) the velocity v' at right angles
to that axis. Hence
(2) v* = V* + r*a>*
where r denotes the distance of P from the axis, and co is the
angular velocity about the axis. Substituting this value in (1) we
find:
Hence
(3) i-Tr+TT
Let v denote the velocity of the centre of gravity, G ; and lot
h be the distance of G from the axis of rotation. Then
0t = F2 + A2co2.
Moreover,
I = 70 + Mh\
where 70 is the moment of inertia about a parallel axis through
G. Hence
(4) r
WORK AND ENERGY 261
In equations (3) and (4) is contained the following general
theorem.
THEOREM. The kinetic energy of a rigid body is the sum of
the kinetic energy of translation along the instantaneous axis and the
kinetic energy of rotation about the instantaneous axis.
It can also be expressed as the sum of the kinetic energy of a particle
of like mass, moving with the velocity of the centre of gravity, and the
kinetic energy of rotation about an axis through the centre of gravity,
parallel to the instantaneous axis.
One Point Fixed. Let a point 0 of the body be at rest. Let
the (£, 17, f)-axes lie along the principal axes of inertia, 0 being
the origin. Then the components of the vector velocity v of any
point fixed in the body are :
= 770^ — fa?,/
— f w£ ~~ £w£
Vt
Hence
(5) T = %(Aw$2 + BuJ + Cco^2).
If the axes of coordinates are not the principal axes of inertia,
then
(6) T = i
The General Case. From (4) and (5) we infer that
(7) T = ±Mv* + ±(Ap* + Bq* + Cr2),
where A, B, C are the moments of inertia about the principal
axes of inertia through the centre of gravity, and p, q, r are the
components of the vector angular velocity w along these axes.
7. Final Definition of Work. We have hitherto assumed that
the point of application, P, of the force is fixed in the body. Sup-
pose P describes a curve C either in the body or in space. How
shall the work now be defined?
Take the time as a parameter. Divide the interval TO ^ t ^ rt
into n parts by the points £0 = r0 < tv < • - • < tn-\ < tn = rv
Let Qk be the point fixed in the body, which at time t = tk will
262
MECHANICS
reach C ; let Tk be its path in space, and let Vk be its velocity in
space when it reaches C ; cf . Fig. 120. For the interval of time
we may take the force as constant, F = FA, the value of F at
the intersection of I\ with C, and let F*
act on the point Qk throughout the in-
terval Atffc. Then J?k will do work equal
approximately to
(1)
Fk vk cos \{/k A^,
FIG. 119
where \l/k is the angle from Tk to F& at
Pk. If Qk is displaced along the tangent
to Tfc a distance vkktk, the expression (1) represents the work
precisely.
We will now define the work as
F
Km
cos
or, dropping the r-notation and expressing
the interval of time as tQ g t ^ /t :
FIG. 120
(2)
cf. Fig. 119. In vector form the work is
W = I Fvcostdt;
(3)
!
= /Fvttt,
?r
where v is the vector velocity of Q at P, and Fv is the scalar
product of these vectors.
We have used the time as the independent variable, or the
parameter, in terms of which to define the displacement. But
the result is in no wise dependent on the time in which the dis-
placement takes place. Any other parameter, X, would have
done equally well, provided d\/dt is continuous and positive (or
negative) throughout. For
j. ds ,. ds,x
v at = -TT at = -=- aA.
WORK AND ENERGY
263
This formulation of the definition of work in the general case
is due to Professor E. C. Kemble.
Example 1. A billiard ball rolls down a rough inclined plane
without slipping. Find the work done by ^
the plane.
Here, C is either the straight line or the
circle; each curve T is a cycloid with
cusp at P and tangent normal to C ; and
v = 0. Hence
W = 0.
FIG. 121
Example 2. The same, except that the ball slips.
The curve C shall be taken along the plane. The normal
component R = M g cos a does no work ;
the component along the plane,
F =
cos
does. Let s be the distance travelled
by the centre of the ball ; 0, the angle
through which the ball has turned.
The curve T is a trochoid tangent to C at P. Hence \l/ = 0 or *,
FIG. 122
ds
dd
and
' cos «{(«! — s0) - <* (0i - 0o)l-
Observe that in the definition, Equation (2), v is positive or 0.
It would not, therefore, be right in this example to write
_ ds d8
v~Tt adt
EXERCISES
1. Check the result in Example 2 by determining the motion
of the ball and computing the change in kinetic energy.
2. A train is running at the rate of 40 m. an h. The baggage
car is empty, and the small son of the baggage master is disport-
264
MECHANICS
ing himself on the floor. He runs forward, then slides. If he was
running at the rate of 6 m. an h. when he began to slide, and slid
5 ft., how much work did he do on the car?
Compute by the definition and check your work by solving
for the motion.
3. A rope is frozen to the deck of a ship. The free end is
^ — I™ haaled , over a smooth pulley at P.
, •**! — ^~^~"^ , It takes a vertical component of
B R = 20 Ibs. to free the frozen part.
How much work is done ?
Take the frozen part as straight, and P in the vertical plane
through it.
4. Extend the definition of work to a body force, F, where F
is a continuous vector, defined at each point of the body :
5. Show that the internal work due to the rope in an At wood 's
machine is nil. Would this be the case if the rope stretched?
6. A number of rigid bodies are connected by inextensible
cords that can wind and unwind on them in any manner without
slipping. Show that the sum of the works done by the; cords
on the system and the system on the cords is nil. First, extend
the definition of work so as to include the case of the work done on
the system by the part of a cord which is in contact with a body.
8. Work Done by a Moving Stairway. Consider the work
which an escalator, or moving stairway, does on a man as he
walks up. The forces that act on the man are /£, S, and Mg,
where R, S are the components of the
force which the escalator exerts on his
foot, and Mg acts at his centre of
gravity. The curve T is always a
right line lying in the inclined plane,
and
ds
dt
FIG. 124
where s denotes the distance the escalator has moved since the
man came aboard
WORK AND ENERGY 265
The force R does no work, since for it ^ = ir/2. The whole
work is due to S = F cos ^, and is :
0
The speed of the escalator is constant ; denote it by c. Thus
(2) W
And
(3) I = c*t,
where I is the distance the escalator has moved while the man
is running up.
On the other hand, consider the motion of the centre of grav-
ity of the man. Let the axis of x be taken up the plane. Then
M --f = S — Mg sin a,
(4) Mu^ — Mu0 = I S dt —
sn a,
where u — dx/dt.
It follows, then, from (2) and (4) that
(5) W = c(Mul — MV,Q) + Mgct^ sin a.
If the man steps off with the same velocity with which he
stepped on, u^ = w0, then, with the help of (3),
(6) W = Mgl sin a.
Now
h = I sin a
is the vertical distance by which the man would have been raised
in tho time he was on the escalator if he had not run, but stood
still. Hence, finally,
(7) W = Mgh.
It makes no difference, then, whether the man runs fast or
slowly, up or down. The one thing that counts is how long he
is on the escalator. Thus when small boys play on the escalator,
running up and down, the work the escalator does increases in
266 MECHANICS
proportion to the time they are on it, provided they arrive and
leave with the same velocity.
9. Other Cases in Which the Internal Work Vanishes.
i) Two Rigid Bodies, Rolling without Slipping. Here, the
action and reaction are equal and opposite, though not in
general normal to the surfaces. Moreover,
the vector velocity of the point of contact,
regarded as a point fixed in the one body,
- - ^p^f — - must be the same as the vector velocity of the
--- ^[ >^ point of contact, regarded as a point fixed in
Vi'v2 the other body.
The works done by the two forces Fj, F2 on
FIG. 125 the two bodies are :
*i «i
W1 = CF.V, cos ft dt, W2 = (*F2v2 cos ft dt.
to
But Fl = F2, i\ = v2, ft + ft = TT. Hence
W, + W2 = 0.
ii) Two Smooth Rigid Bodies, Rolling and Slipping. Here
the forces Fl and F2 are equal and opposite, and normal to the
surfaces at the point of contact. The ve-
locities Vj and v2 are not equal when there is
slipping ; but their projections on the normal
are equal :
vl cos ft + vz cos ft = 0,
Since furthermore Fl = F2) we have :
W1 + W2 = 0. Fio. 126
We have already mentioned the case of rigid bodies on which
inextensible massless strings wind and unwind, § 7, Exercise 6:
and massless rods were shown in § 5 to do no work. Thus syj?
terns of rigid bodies connected by inextensible strings and rods,
even though the point of application of the force exerted by the
string or rod be variable, show no internal work.
10. Work and Energy for a Rigid Body. THEOREM. The
change in the kinetic energy of a rigid body, acted on by any forces,
is equal to the work done by these forces.
WORK AND ENERGY 267
We prove the theorem first for two special cases.
CASE I. No Rotation. Here, the change in kinetic energy is
m ^i2 _ -fl^o2
w 2 2 '
i.e. the change in the kinetic energy of a particle of mass M,
moving as the centre of gravity is moving.
On the other hand, consider the work done by one of the
forces, F :
(2) W =
Since there is no rotation, v = v, ^ = ^,
and
(3) w = C
Fvcos$dt.
Hence W is the work done on a particle at the centre of gravity
by the same force, and the theorem is true by § 4.
CASE II. One Point Fixed. Here, Eulcr's Dynamical Equa-
tions, Chapter VI, § 13, determine the motion. Consider a force
F which acts on the body at P. Let r be the vector drawn from
0 to P. Then the v of the definition of work, § 7, (3) is
v = & X r.
Hence
(4) Fv cos ^ = F - v = F • (w X r).
On the other hand the vector moment of F about 0 is
From Euler's Equations, I.e., we have :
A dp . n dq . ~ dr T , ,, , ,r
^TT/ "•" '5/ •" /it ~ P ' ^ ' '
Hence
(5) $(Ap* + Bq* +
The left-hand side is the change in kinetic energy. Now
(6) Lp + Mq + Nr = M - w = co - (r X F),
268
MECHANICS
and so
(7) Fv cos ^ = Lp + M q + Nr.
For it is true of any three vectors that
a - (b X c) + c • (b X a) = 0,
since
a - (b X c) =
a1 a
Moreover,
Hence
w X r =- (r X o>).
F-(w Xr) = «-(r X F).
From (5) it follows, then, that for a single force, the change
in kinetic energy is equal to the work done. For the case of
n forces the proof is now obvious. The extension to the case of
body forces and forces spread out continuously over surfaces or
along curves, presents no difficulty.
Remark. We have shown incidentally that the work done
on a rigid body with one point fixed is
h
J o> t,
where
M = La + Mp + Nj
is the resultant couple.
The General Case. Consider first a single force, F. The work
it does is
W = /Fvd*.
Here,
V = V + V',
where v is the velocity of the centre of gravity and v' is the ve-
locity of the point Q relative to the centre of gravity, as it flashes
through P. Hence
W =
WORK AND ENERGY 269
The first integral has the value
Mv*
The second integral is equal to the right-hand side of Equation (5).
Thus the theorem is proved for one force. For a number of
forces the proof is now obvious.
EXERCISES
1. A ball is placed on a rough fixed sphere of the same size
and slightly displaced near the highest point. Find where it
will leave the sphere. Let p, have any value.
2. A weightless tube can turn freely about one end. A smooth
rod is inserted in the tube and the system is released from rest
with the tube horizontal. How fast will it be turning when it
is vertical ?
3. A cylindrical can is filled with water and sealed up. It is
mounted so that it can rotate freely about an element of the
cylinder. Show that it oscillates like a simple pendulum, pro-
vided the can is smooth.
4. In the preceding problem, the height of the can is equal
to its diameter, and the can weighs 5 Ibs. The water weighs
31 Ibs. Find the length of the equivalent simple pendulum.
5. A circular tube, smooth inside, plane vertical, is partly
filled with water. The tube is held fast and the water is dis-
placed, then released from rest. Show that it oscillates like
a simple pendulum, and determine the length of the latter.
6. A bent tube in the form of an L is mounted so that it can
slide freely on a smooth table. The vertical arm is filled with
water, and the system is released from rest. How fast will it
be moving where the vertical arm has just been emptied?
Assume the tube smooth inside; and also take the weight of
the tube with its mount equal to the weight of the water.
7. The can of Question 4 is allowed to roll down a rough in-
clined plane, starting from rest. Find the acceleration of the
centre of gravity.
CHAPTER VIII
IMPACT
1. Impact of Particles. Let two particles, of masses ml and
W2, be moving in the same straight line with velocities Ui and
w2, and let them impinge on each other. To find their velocities
after the impact.
Isolate the system consisting of the two particles. Then no
j*1 w2 external forces act, and so the
/$|Q xgfr
^f u2 momentum remains unchanged.
FIG. 128 Hence
(1) m^ — m^! + m2u2 — w2w2 = 0.
As yet, nothing has been said about the elasticity of the par-
ticles. The extreme cases are: perfect elasticity (like two bil-
liard balls) and perfect inelasticity (like two balls of putty).
In each case there is deformation of the bodies — for now we
will no longer think of particles, but, say, of spheres, and the
velocities ult w2, etc. refer to their centres of gravity.
During the deformation the mutual pressures mount high,
and even if other (ordinary) forces act, their effect is negligible,
compared with the pressures in question. In the case of perfect
inelasticity, there is no tendency toward a restitution of shape,
and so, when the maximum deformation has been reached, the
mutual pressures drop to nothing at all. At this point, the
velocities of the two centres of gravity are the same,
u( = u2,
and hence this common velocity, which we will denote by f7, is
given by the formula :
(2) u = miUl "*" m*u*
ml + m2
Thus the problem is solved for perfect inelasticity. For partial
elasticity, it is helpful to picture the impact as follows. The
270
IMPACT
271
motion of the centre of gravity of each ball is given by the
equation :
For the first stage of the impact, i.e. up to the time of greatest
deformation, t = T, we have, on integrating each side of each
equation between the limits 0 and T :
T T
(4)
mlul
<-T /»
i-o J
t=T /»
= I J
/ = 0 e/
Rdt.
The integral is called an impulse,* and is denoted by P :
T
(5) P = CR dt.
o
Hence
(6)
= — P
= P.
m2U —
The second stage of the impact now begins, as the balls are
kicked apart by their mutual pressures. On integrating the
equations (3) between the limits T and 7V, we have :
T'
(7) m^
fftdt',
CR'dt'.
Now it is easily intelligible physically if we assume that, in
the case of partial elasticity, the value of R' stands in a constant
ratio to the value of R at corresponding instants of time, or that
(8)
when
R' = eR,
T - t = t' - T.
Here the physical constant e is
called the coefficient of restitution.
It lies between 0 and 1 :
t T t'
FIG. 129
(9)
0 < e < 1,
* Sometimes spoken of as an impulsive force ; but this nomenclature is unfor-
tunate, since P is not of the nature of a force, which is a push or a pull, but rather
is expressed by a change of momentum. Moreover, the dimensions of impact are
ML/T, not ML/T*.
272 MECHANICS
being 0 in the case of perfect inelasticity and 1 for perfect elas-
ticity. Hence
7" T
(10) P' = /V dtf = e CR dt = eP.
Equations (7) thus yield the following :
(U) { ^-Ilf/ = df
The four equations, (6) and (11), contain the solution of the
problem. Between them, U and P can be eliminated, and the
resulting equations can then be solved for u[, u2. The result is :
0
Uf\ — :
* vn.. 4-
(12)'
U2
ml + m,2
e) m1u1 + (ra2
The Case m2 = oo. If in Equations (12) we allow w2 to increase
without limit, we obtain the equations :
U2 = U2.
These equations do not prove that, when the mass m2 is held
fast, or is moving with unchanging velocity u2, the velocity of the
mass ml after the impact will be given by the first equation (13),
but they suggest it. The proof is given by means of the first
of the equations (6) and (11), resulting as they do respectively
from the first of the equations (4) and (7), combined with (10);
U having here the known value u2.
If, in particular, u2 = 0, we have :
(14) u{ = -««,.
Perfect Elasticity, e *= 1. Equation (14) becomes in this case
u[ = — -MI, and the ball recedes with the same velocity as that
with which it impinged.
If the masses are equal, ml = w2, Equations (12) become :
IMPACT 273
and the balls interchange their velocities. This latter phe-
nomenon can be illustrated suggestively by two equal ivory
balls suspended side by side from strings
of equal length, after the manner of two
pendulums. If one ball hangs vertically
at rest, and the other is released from
an angle with the vertical, the second
ball will be reduced to rest by the im-
pact, and the first will rise to the same
height on its side of the vertical as that F
from which the second ball was released.
Thus the velocities will be successively interchanged at the lowest
point of the circular arc.
Critique of the Hypothesis (8). In this hypothesis we have
taken for granted an amount of detail in the phenomenon before
us far in excess of what the physicist will admit as reasonable
in viewing the actual situation, and he may easily be repelled by
so dogmatic an assumption in a case that cannot bo submitted to
direct physical experiment and which, after all, is far less simple
than we have led the reader to suppose, since the problem is
essentially one in the elasticity of three-dimensional distribu-
tions of matter. The objection, however, is easily met. We
may take Equation (10) :
P' = eP,
as the physical postulate governing impact.
EXERCISES
1. A ball of 6 Ibs. mass, moving at the rate of 10 m. an h.
overtakes a ball of 4 Ibs. mass moving at the rate of 5 m. an h.
Determine their velocities after impact, assuming that the coeffi-
cient of restitution is £. Ans. 7 and 9.5 m. an h.
2. The same problem, when the balls are moving in opposite
directions.
3. A perfectly elastic sphere impinges on a second perfectly
elastic sphere of twice the mass. Find the velocity of each after
the impact.
4. Newton found that the coefficient of restitution for glass
is ^f . If a glass marble is dropped from a height of two feet
on a glass slab, how high will it rise?
274 MECHANICS
5. In the last question, what will be the height of the second
rebound? What will be the total distance covered by the marble
before it comes to rest? *
6. Find the time it takes the marble to come to rest.
7. In the experiment with the pendulums described in the
text, the impinging ball will not be quite reduced to rest, because
no two material substances are quite elastic. If, for given balls,
e — 0.9, show that the ball which is at rest should be about 11
per cent heavier than the other one, in order to attain complete
rest for the latter. What per cent larger should its radius be ?
8. If, in the last question, the pendulum bobs are of glass.
e = ^f , find the ratio of their diameters.
9. If two perfectly elastic balls impinge on each other with
equal velocities, show that one of them will be brought to rest
if it is three times as heavy as the other.
10. Determine the coefficient of restitution for a tennis ball
by dropping it and comparing the height of riie rebound with
the height from.which it was dropped.
11. Some pitchers used to deliver a slow ball to Babe Ruth,
believing that he could not make a home run so easily as on a fast
ball. Discuss the mechanics of the situation.
J 2. Continuation. Oblique Impact. Let two spheres impinge
at an angle, and suppose them to be perfectly smooth. To de-
termine the velocities after the
impact.
Let the line of centres be taken
as the axis of x. The deforma-
tion of each sphere is slight, and
Fia 131 the force exerted by the other
sphere, spread out as it ^s over
a very small area and acting normally at each point of this area,
will yield a resultant force, K, nearly parallel to the axis of x.
For the first sphere we have :
* The physics of the second part of this problem (and of the next) is altogether
phantastic. After a few rebounds we pass beyond the domain within which the
physical hypothesis of the text applies, and the further motion becomes a purely
mathematical fiction. It is amusing for those who have a sense of humor in science.
But for the literal-minded person, be he physicist or mathematician, it is dangerous.
IMPACT 275
(15)
where
X = # cos 6, F = 72 sin c,
e being numerically small and xlt yl referring to the centre of
gravity, and for the second sphere,
(16) rn^X, *,— F.
On integrating (15) we obtain :
,.,,_>.
(17)
= - /-Yd*, m,^1 ^ = /Yrf*.
*-o J dl * = o J
And now we denote the first impulse by P, and lay down the
postulate that the second impulse is 0 :
(18)
0 0
Thus the integrals of (15) and (16) load to the equations :
m^U — mlul =— P [ mtF — m^^ — 0
(19)
1 = P
which hold for the first epoch of the impact, the equations for
the second epoch being, as in the corresponding case of § 1, the
following :
m} u\ — m, U = — eP { m* v\ — m} V = 0
1.
4 — m2U = eP I m^ — m2F = 0.
For we assume as there the physical postulate :
(21) P' = eP.
The result at which we have arrived is seen to be the following.
The component of the velocity of each sphere perpendicular to
the line of centres has been unchanged by the impact,
(22) v[ = vl9 v'2 = v2.
276 MECHANICS
The components of the velocity along the line of centres are
changed precisely as in the case of direct impact, § 1, Equa-
tions (12) :
u\ = — - — *--
mi
(23)
Mj =
ml
Kinetic Energy. When e — lt i.e. when the spheres are per-
fectly elastic, the total kinetic energy is unchanged by the impact,
for then
, 2 _ >\ ,
r ~~ ~ " ~~ '
as is shown by direct computation from (23), and the equa-
tions (22) hold in all cases, whether e = 1 or e < 1.
When e = 0, i.e. when the spheres are totally inelastic, an easy
computation shows that the kinetic energy has been diminished.
The intermediate case, 0 < e < 1, is treated in the same way.
It follows from direct computation that the left-hand side of
Equation (24) has the value :
(m^ + w2M2)2 + mlmz(ul — u2)ze2
m2)
and this is at once shown to be less than the right-hand side.
The terms arising from Equations (22) do not, of course, affect
the result.
EXERCISES
1. A smooth ball travelling south-east strikes an equal ball
travelling north-east with one-quarter the velocity, their line
of centres at the time of impact being east and west. If e = ^,
find the velocities of the balls after impact.
2. A smooth ball strikes a horizontal pavement at an angle
of 45°. Find the angle of rebound if the coefficient of restitution
is f.
3. Show that the kinetic energy of the balls of Question 1 is
diminished in the ratio of 245/272 by the impact.
4. The corresponding question for the ball of Question 2.
IMPACT
277
3. Rigid Bodies. Let a rigid body be acted on by a single
impulse. By that is meant the postulates about to be laid
down, suggested by the following physical picture. A force F
acts at a point (x, y) for a
short time, mounting high in
intensity. Ordinary forces, if
present, produce in this inter-
val of time, 0 ^ t ^ T, only
slight results, and in the ulti-
mate postulates do not appear,
so they are not considered in ^ FrG 132
the present picture.
The three equations which govern the motion are :
- = Y
dt* *'
On integrating with respect to the time we find •
(2)
dt
M(u' - u) = Cxdt, M(v' - v) = CY
0 U
T T
7(0' - 0) = Ax - x) Ydt - f(y - y)Xdt.
Concerning F we will assume that the vector changes con-
tinuously in magnitude and direction during the interval of
time in question, and that the point of application, (x, T/), also
moves continuously, remaining near a fixed point (a, b) through-
out the interval. Let
(3) x = a + £, y = 6 + 17.
Then £ , rj are infinitesimal with T. Let
(4)
P = Cxdt, Q = CYdt.
278 MECHANICS
The last Equation (2) now becomes :
(5) /(«' - 0) = (a - *) Q - (b - y) P
T T
+ CtYdt - Cr,Xdt.
0 0
We should like to infer mathematically that from the hypoth-
esis that the integrals (4) approach limits when T approaches 0,
the integrals in the last line of (5) converge toward 0; for then
we should have the equations :
M(u'-u)=P,
(6) ' 7(12' - $2) = (a - x) Q - (b - y) P,
P and Q here denoting the limiting values of the integrals (4).
This inference can in fact be drawn, provided the angle through
which the vector F ranges is less than 180°. Equations (6)
then hold, and, in particular, it follows, on eliminating P and
Q between them, that
(7) 7(Q; - fl) = M(a - x)(v' - v) - M(b - y)(u' - u)
or
(8) fc2(12' - Q) = (a - x)(v' -v)- (b- y)(u' - u).
An Example. A rod is rotating about one end, and it strikes
an obstruction, which brings it suddenly to rest without any
reaction on the support. What point of the rod comes into
contact with the obstruction ?
{Let the distance from the stationary end
be h, and let I be the length of the rod.
O k ' Let
FIG. 133 v = C; then ft = - C.
Since
M/2
u = 0, fi' = 0, ^ = 0, 12' = 0, / = ^-,
we have :
Hence
The point is called the centre of percussion.
IMPACT
279
4. Proof of the Theorem. The proof is given by means
of the Law of the Mean, which is as follows. Let f(x), <p(x)
be two functions which are continuous in the closed interval
a ^ x g b, and let <p (x) not change sign there. Then
ft 0
//*
f(x) <p(x) dx = f(x') I <p(x) dx, a < a:' < 6.
e/
(9)
In the present case the axes can be so chosen that
0 g Y.
Hence
(10)
C^Ydt = £' f*Ydt,
where £' is the value of £ at a suitable point, t = £', in the interval
0 g £ g 77. Now, by hypothesis, £ and T? approach 0 uniformly,
i.e. the largest numerical value that either has in the interval
0 ^ t g T approaches 0; and furthermore, also by hypothesis,
the integral on the right approaches a limit, Q. Hence the inte-
gral on the left approaches 0.
If the range of the angle of F does not exceed 90°, the axes of
coordinates can be so chosen that neither X nor Y changes sign
in the interval 0 ^ t ^ 77, and then it can be shown as above
that both integrals in the second line of (5) approach 0.
In the more general case that F is contained merely within
an angle less than TT, the axes can be chosen in more ways than
one so that Y will not change sign.
If (x, y) refer to one such choice and
(x', y') to a second, then
(11)
where
x' = ax + by
yr = ex + dy
FIG. 134
a = cos 7, b = sin 7,
c = — sin 7, d = cos 7.
The same transformation holds * with respect to the vector F :
* It is in such a case as the present one that the scientific importance of the
proper definition of (-artesian coordinates, laid down in Analytic Geometry, ia
revealed. That definition begins with directed line segments on a line, proceeds
to the theorem that the sum of the projections of two broken lines having the same
280 MECHANICS
f X' = aX + bY
(12> I r-T + fl'
and also with respect to (£, jj) :
r = of + 6,
Hence
r r T T T
(14) A'*" dt = ac Ax A + 6c Ax dt + bd Cr,Y dt + ad ft-Y dt.
0 0000
The integral on the left, and the last two integrals on the right,
approach the limit 0 with 77, as has been shown above. We
will show that this is true also of the other integrals, and hence
in particular of the last integral in (5). To do this, write down
Equation (14) for two choices of axes (xf, y') subject to the
above conditions and characterized by two values of y: yl and
72, where yl ^ 0, y2 ^ 0, yl ^ 72, and solve the resulting equa-
tions for the two integrals in question. The determinant of
the equations,
has the value
sin yl sin y2 sin (yz - 7^,
and so does not vanish. Thus the integrals for which we are
solving are seen to be linear functions of integrals that are known
to approach 0, and this completes the proof.
The Restriction. The theorem is not true when F is required
merely to vary continuously with t in the in-
terval 0 ^ t ^ T, as the following example
shows. Let
X = F cos <?j Y = F sin <p ;
£ = p cos ^, 77 = p sin \l/.
FIG. 135 Then
extremities, on an arbitrary line, is the same for 'both lines, and ends by declaring
the coordinates of a point as the projections on the axes of the vector whose
initial point is the origin and whose terminal point is the point in question. With
that definition, Equations (12) and (13) are merely particular cases of Equations
(11), since both sets of equations express the projections of a vector on the coordinate
IMPACT 281
Let F and p be constants, and let
_ 2wt TT _ 2wt
Then
r T
P-fx* — Ffa™*-0,
0
0
71
7*
Q=fYdt = Fj
fc
0 0
r
yv - ,,2
•)'
0
If, now, we set
__ /TT r» __ •*•_
P * > f rpy
the integrals (4), being always 0, each approach limits, and so
the P and Q of Formulas (6) have each the value 0. But the
third Equation (6) does not hold.
But may it not still be sufficient, in order to secure the vanish-
ing of the limit of the integral
T
/(*r-
0
to demand that Y ^ 0? That this is not enough, is shown by
modifying the above example as follows. Let
X = F cos ?, Y = 0.
The integral then has half the value it had before; hence, etc.
EXERCISES
1. A uniform rod at rest is struck a blow at one end, at right
angles to the rod. About what point will it begin to rotate?
2. A packing box is sliding over an asphalt pavement, when
it strikes the curbstone. Find the speed at which it begins to
rotate.
3. If, in the preceding question, the pavement is icy, and if
the box, before it reaches the curb, comes to a bare spot, M = 1>
what is the condition that it should not tip?
282 MECHANICS
4. If the box tips, find whether it will slide, or rotate about
a fixed line.
6. Show that greater braking power is available when the
brakes are applied to the wheels of the forward truck of a rail-
road car.
6. If all four wheels of an automobile are locked, compare the
pressure of the forward wheels on the ground with that of the
rear wheels.
7. A lamina is rotating in its own plane about a point 0, when
it is suddenly brought to rest by an obstruction at a point P
situated in the line OG produced. Show that OP is equal to the
length of the equivalent simple pendulum, when the lamina
is supported at 0 and allowed to oscillate under gravity in a
vertical plane.
5. Tennis Ball, Returned with a Lawford. Consider a tennis
ball, returned over the net with flat trajectory arid rotation such
that the lowest point of the ball is moving backward. The
ground thus exerts a forward force, and we will assume that this
state of affairs holds throughout the impact. We shall have,
then, the following formulation of the problem :
M(U - M0) = »Q Mfa - {/) '== e»Q
M(V -VQ) = Q M(v, - F) = eQ
where
^^^^ no
F=pR
Fid. 136
is the impulse. First of all,
V = 0,
for the point of greatest deformation is marked by the centre
of gravity of the ball's ceasing to descend. Thus we have seven
equations for the seven unknowns, u{, vlt cot, [/, V, 12, Q.
It is now easy to solve. Observe that
VQ < 0, co0 < 0, u0 > 0.
IMPACT 283
We have, then :
Q = M(- t>0), vl = e(- VQ),
The value of c^ is not interesting. What we do want to know is
the slope of the trajectory at the end of the impact ; i.e.
v± __ _ e( — VQ) _ __ _ e\ _
^ " u0 + (1 + e)M(-t,0) ~ 1 + (1 + e)/i\f
where X = (— VQ)/UQ is the numerical value of the slope before
the impact.
As the ball has a drop due to the cut, X will be considerably
larger numerically than the slope in the part of the trajectory
just preceding the last ten feet or so before touching the ground.
It might conceivably have a value as great as £. The value of
e is about 0.8. ju varies considerably and might be as high as £.
Thus
& = .15,
*i
as against X = .20, or the steepness of the rebound is only three-
fourths the steepness of the incident path.
Not only does the ball rise at a smaller angle, but the horizontal
velocity is increased by nearly 10 per cent ; for
u, = w0[l + (1 + <0/*A] = 1.09t*0.
For this discussion to be correct it is essential that the ball
maintain its spin throughout the whole impact. This explains
the nature of the stroke. The racquet has a high upward velocity
while the ball is on the guts.
The ball loses spin during the flight before the impact, due to
the air resistance causing the drop, and this loss may easily be
comparable with the loss during the impact. It would be inter-
esting to take motion pictures of the ball, showing the trajectory
just before and just after the impact.
EXERCISES
1. A billiard ball, rotating about a horizontal axis, falls on
a partially elastic table. Find the direction of the rebound if
ju = % and e = .9.
284 MECHANICS
2. A rod, moving in a vertical plane, strikes a partially elastic
smooth table. Determine the subsequent motion.
3. The preceding question, with the change that the table is
rough, M = i-
4. Question 2 for a table that is wholly inelastic and infinitely
rough.
5. A rigid lamina is oscillating in a vertical plane about a
point 0 when it strikes an obstacle at P whose distance from 0
is equal to the length of the equivalent simple pendulum. Show
that it will be brought to rest without any reaction on the axis.
For this reason P is called the centre of percussion.
6. A rigid lamina, at rest, is struck a blow at a point 0. Find
the point about which it will begin to rotate.
7% A solid cone, at rest, is struck a blow at the vertex in a
direction at right angles to the axis. About what line will it
begin to rotate ?
CHAPTER IX
RELATIVE MOTION AND MOVING AXES
1. Relative Velocity. It is sometimes convenient to refer the
motion of a system to moving axes. Let O be a point fixed
in space. Let 0' be a point moving in any manner, like the
centre of gravity of a material body, or the centre of geometric
symmetry of a body whose centre of gravity is not at 0'; it is
a point whose motion is known, or on which we wish particularly
to focus our attention. Finally, let P be any point of the system
whose motion we are studying. Then
(1) r = r0 + r',
dr _ dr0 di^
dt " dt ^ dt'
or
(2) v = v0 + v'.
, • • i • , i FlG- 137
The choice of notation is here particularly
important — boldface letters denote as usual vectors — because
we have two analyses to emphasize, namely, i) the breaking up
of the velocity v into the two velocities v0 and v'; and if) the
breaking up of v' into the two velocities :
(3) v' = vr + v.,
where vr, the relative velocity, and ve the vitesse d'entrainement are
presently to be defined. For this purpose we must first recall
the results of an earlier study.
2. Linear Velocity in Terms of Angular Velocity. In Chap-
ter V, § 8, we have studied the motion of a system referred to
moving axes (£, 17, f ) with fixed origin 0. Here,
(4) r = £a + iff + f y
and
+ r/4 + f 7-
285
286
MECHANICS
This equation represents an analysis of the velocity
(6) v = £
of the point P into two velocities, namely,
(7) v = vr + ve,
where
(%} v=— 4-— #4-^
is the relative velocity of P with respect to the moving axes ; i.e. the
absolute velocity which P would have if the (£ , r;, f )-axes were
at rest and the point P moved relatively to them just as it does :
(9) £ = f(t), ?? = <p(t), f = \l/(i).
Secondly,
(10) Ve = £a + rift + f 7
is the vitesse d'entrainement, the Schleppgeschwitidigkeit, the
velocity with which that point Q fixed in the moving space and
flashing through P at the one instant, t, is moving in space. Let
(w) be the vector angular velocity of the moving axes :
(11) (w) = pa + qp + ry,
where
Then
(12) ve = («) X r,
FIG. 138
or
(13)
+ (&• - f p) j8 + (77?? -
y.
The final result is as follows : The components of v, or dr/dt,
along the axes are
(14)
RELATIVE MOTION AND MOVING AXES 287
I repeat : These are the formulas when the moving axes
have their origin, 0', fixed : r0 = 0, v0 = 0, v = v'.
3. Acceleration. Returning now to the point P of § 1 and
Equations (1) and (2), we define its acceleration as the vector:
Hence
(16) a = — ° + — ,
or
(17) a = a0 + a'.
The first term on the right, a0, requires no further comment.
It is merely the acceleration in fixed space of the known point 0'.
The second term, a', relates to the rotation and admits of a num-
ber of important evaluations.
First Evaluation. The first of these is as follows. Let a' be
denoted by a. Then
<•» -T
We may identify the variable vector v' with the variable vector
r of § 2, Formula (4) ; for, of course, r was any vector, moving
according to any law we wish. Now, we have evaluated the
right-hand side of (18) by means of Equations (14). Hence the
components of the right-hand side of (18) are obtained by sub-
stituting in the right-hand side of (14) for £, rjf f respectively
v£> v^ v£-
On the other hand, write
(19) a = a$ a + a^ ft ~f~ fl<7-
Thus wo arrive at the final determination of a in terms of known
functions :
__ dv* * __
(20)
dVrt
= ~
a$ — — jT- + pity — qv$
These are the formulas referred to as the First Evaluation.
288 MECHANICS
Second Evaluation. The Theorem of Coriolis. Another form
for the vector a can be obtained by differentiating (5), § 2 :
+ f « + 77/3 + f y.
Thus
o | _
y* dt ! c»^ l <ft e&y
+ £<* + r/j3 + f 7.
This result is due to Coriolis.
The first and third lines admit immediate interpretations. For,
is the relative acceleration, or the acceleration of P referred to the
(£> Vy f)-axes as fixed. Next,
(23) ae==^ + 77d^ + f W~
is the acceleration d'entrainementj or the SchleppbesMeunigung,
the acceleration with which the point Q, fixed in the moving
space and coinciding at the instant t with P, is being carried
along in fixed space.
The vector (23) :
ae = & + tip + fy,
can be computed as follows. Since — as is geometrically, or
kinematically, immediately obvious —
(24) a = r/3 - qy, $ = py - ra, y = qa - p/3,
we have :
(25) -J»-}»
+ rp - qy.
The last line has the value :
RELATIVE MOTION AND MOVING AXES 289
where
(co) = pa + q@ + ry.
Hence
/oc\ / \dQ dr\ f.dr >.dp\ / dp
(26) *• - r - ' a + - f ' + " -
- w2(£« + i?/3 + r 7) + (p$ + qr> + rf)(pa + g/8 + ry),
or
(27) ae = («0 X r - «2r + ((«)r) («),
where
This vector (w') is the velocity relative to the fixed axes (£, 77, f),
with which the terminal point of (co) is moving when the initial
point is at 0' ; it is the relative angular acceleration, referred to the
(£> *?> f )-axes as fixed.
Finally, the vector
a =
' dtdt^ dtdt^ dt dtj
can be expressed in the form :
(30) a» = (co) X vr,
or:
For, on recurring to Formula (10) of § 2 and taking, as the arbi-
trary vector r, the vector vr, which is given by (8), the right-
hand side of (10) comes to coincide with the right-hand side of (29).
With the aid of (12), this vector can be written in the form of the
right-hand side of (30), and this completes the proof.
To sum up, then : — From (17),
(31) a = a0 + a',
where a' = a, and a is given by (20). A second evaluation of a
is given by (21),
(32) a = ar + 2at + ae,
where ar is given by (22) and ac by (23) ; the latter, in a differ-
ent form, by (26) or (27). Finally, at is given by (29) or (30).
290 MECHANICS
4. The Dynamical Equations. From Newton's Second Law of
Motion, written in the form :
(1) ma = F,
it follows that
(2) ma0 + ma = F,
where a0 is the acceleration of the moving origin, 0', and
a = ar + 2a, + a*.
The vector ar is the relative acceleration and is given by For-
mula (22), § 3. The vector ac is the acceleration d' entrainement
and is defined by (23) ; it is represented by (26) or (27). Finally,
a* is defined by (29) and is represented by (30).
If the motion of the moving axes is regarded as known, then
QO is'a known function of tt and (o>), i.e. p, q, r are known from
(11), § 2. Equation (2) can now be written in the form :
(3) mar = F — maQ — 2ma» — mae.
On substituting for a* its value from (30) and for ae its value
from (26), a system of differential equations is found for deter-
mining £, 77, f :
rf£ \dt dt dt
(4)
where the functions /, <p, \l/ can be written down explicitly from
the above formulas.
More generally, Equation (1) can be thrown into the form
required in a given problem by using a suitable form for a, a0,
ar, a^ a« as pointed out at the end of § 3. Each one of these
accelerations must be studied in the particular case. There is
no single choice of sufficient importance to justify writing down
the long formulas. But the student will do well to make hi^
own syllabus, writing down the value of ar and each form foi
a,, a{.
RELATIVE MOTION AND MOVING AXES 291
EXERCISE
Obtain the dynamical equations in explicit form from La-
grange's Equations, Chapter X. Observe that
•f 6? + & — f p + tfop)2
where
_ __
V»y ~ V0 ' 7 - *3 -ft- + ™* -fa + "I -5-
5. The Centrifugal Field. Let space rotate with constant
angular velocity about a fixed line through 0, the (£, rj, f)-axes
being fixed in the moving space. Then the vector angular
velocity (o>) is constant, and a0 = 0. The vector ae, § 3, (27)
reduces to :
(1) a.=-«*r+((«).r)(«),
and is easily interpreted. Kinematically, it is, of course, the
centripetal acceleration ; geometrically, it is a vector drawn from
the point P toward the axis and of length o>2p, where p is the
distance from P to the axis.
Newton's Law takes the form :
ma = F,
where a is given by § 3, (32), and thus
(2) a = + 2q - 2r - ««« + pfa + 9, + rf )
292
MECHANICS
Axis of f, the Axis of Rotation. In this case,
r = w, and the equations reduce to the following :
(3)
p = q = 0,
Fia. 139
Thus the motion along the axis of f is the same as it would
be if space were not rotating. The projection of the path on
the (£, 77)-plane is the same as the path of a particle in fixed, or
stationary, space, when acted on i) by the applied force F ; ii) by
a force rao>2/> directed away from 0 ;
and Hi) by a force at right angles to
the path, equal in magnitude to 2ma)V,
and so oriented to the vector velocity
v as the positive axis of £ is, with re-
spect to the positive axis of 17.
This third force is known as the
Coriolis force. In the case of the
Centrifugal Oil Cup, and the corre-
sponding revolving tennis court, Chapter III, § 23 it was enough,
for problems in statics, to take into account the u centrifugal
force," or the force ii) above. But for problems in motion, this
is not sufficient. There is the Coriolis force Hi) at right angles
to the path, like the force an electro-magnetic field exerts on a
moving charge of electricity.
6. Foucault Pendulum. Consider the motion of a pendulum
when the rotation of the earth is taken into account. We may
think, then, of the earth as rotating about a fixed axis through
the poles, which we will take as the axis of z, the axes of x and y
lying in the plane of the equator.
Let P be a point of the northern hemisphere, and let its dis-
tance from the axis be p. By the vertical through P is meant
the line in which a plumb bob hangs at rest, or, more precisely,
the normal to a level surface. Let f be taken along the vertical,
directed upward; let £ be tangent, as shown, to the meridian
RELATIVE MOTION AND MOVING AXES 293
through the point of support of the pendulum; then 77 will be
tangent to the parallel of latitude through the point of support,
and directed west. Let X be the latitude of P; i.e. the angle
that f makes with the plane of the
equator.
The earth rotates about its axis
from west to east, and so the
vector angular velocity, (co), is
directed downward. Thus
p = co cos X, q = 0,
r = — w sin X,
2rr
co =
24 • 60 - 60
= .000727.
FIG. 140
We can now write down the differential equations that govern
the motion. These are contained in the single vector equation
of § 4 :
ma0 + ma = F.
Let P be the point of support, and let (f, 1?, f ) be the coordinates
of the pendulum ; Z, its length,
First, compute F :
F = G + N,
where
- au , w.c>u
is the force due to gravity, or the attraction of the earth ; and
N = - j^a-Jtffl- j-#7
is the tension of the string.
Next, a0 is the centripetal acceleration, or :
a0 = — co2p(« sin X + 7 cos X).
Finally, a is given by the formulas (2) of §5. For, although
these were written down for the particular case a0 = 0, they apply
generally, where a0 is arbitrary, provided the vector angular
velocity of the moving space is constant.
Thus we can write down explicitly the three equations of mo-
tion. These we replace by approximate equations obtained as
294 MECHANICS
follows. Approximate, first, to the field of force by the gravity
field.
field,
U = -
U = - mg£ .
Next, suppress those terms which contain co2 as a factor, or are
of the corresponding order of small quantities. Thus
Finally,
? + T? + f 2 =
= -I + r? + terms of higher order.
We introduce the further approximations which consist in sup-
pressing the term in d£/dt in the second equation, and setting
N = mg. The first two equations thus become :
A)
Discussion of the Equations. Multiply the first equation A)
through by di/dt, the second by dq/dt, and add. The resulting
equation,
,
dtdt*~^dtdt*
integrates into the equation of energy :
2~ Z 2
or, on introducing polar coordinates,
RELATIVE MOTION AND MOVING AXES 295
Next, multiply Equations A) by »j and £ respectively, and
subtract :
This integrates into
drj d£ _ , 2 r
%-17^-cor + C'
or
(2) fJ = «'r» + Cf
where co' = « sin X.
4 Special Case. Let the pendulum be projected with a small
velocity from the point of equilibrium. Then initially r = 0 ;
hence C = 0 and
dB
di = "'
It follows, then, that
0 = w't.
This means that, if the motion be referred to moving axes, so
chosen that f' coincides with f, but £' makes an angle u't with £,
the pendulum will swing in the (£', f ')-plane. It is now easy to
determine r as a function of t from (1) ; r executes simple harmonic
motion.
The General Case. Returning now to the general case, let
the motion be referred to a moving plane through 0 (the point
of equilibrium of the pendulum), perpendicular to the f-axis,
and rotating with constant angular velocity a/ about 0. Then
(3) <f> = B - u't
is the angular coordinate in the new plane. Equation (2) now
becomes :
(4) r-l-C,
and this is the equation of areas in its usual form.
Equation (1) goes over into :
or
(5) ^ + r«^ + 2»'C + rV = - |r2 + .
296 MECHANICS
On suppressing the term r2«'2 because of its smallness, we find :
« £+-¥-!-+*••
But this is precisely the equation of energy corresponding to an
attracting central force of intensity -y^r. Hence the motion
I
is elliptic with 0 at the centre ; i.e. the pendulum, once released,
describes a fixed ellipse in the moving plane. The axes of this
ellipse rotate in the positive sense, i.e. the clockwise sense, as one
looks down on the earth. But the pendulum describes the ellipse
in either sense, according to the initial conditions, the degenerate
case of the straight line lying between the description in positive
sense and that in negative sense. In the Foucault experiment
in the Pantheon the pendulum was slightly displaced from the
position of equilibrium and released from rest relative to the
earth. It then described the ellipse in the negative sense. For
initially dr/dt was 0, so that it started from the extremity of
an axis (obviously the major axis) and its initial motion rela-
tive to the moving plane was in the negative sense of rotation ; i.e.
counter clockwise. At the end of twenty-four hours, t has the
value : t = 24 X 60 X 60, and hence
0 = u't = 27rsinX.
The result checks, for at the equator 0 should be 0, and at the
North Pole, 2w.
EXERCISE
Obtain the equations of motion A), directly from Lagrange's
Equations, Chapter X.
CHAPTER X
LAGRANGE'S EQUATIONS AND VIRTUAL VELOCITIES
INTRODUCTION
In the preceding chapters, the treatment has been based on
Newton's Second Law of Motion. Work and Energy have
entered as derived concepts. It is true that certain general
theorems have been established, whereby some of the forces of
constraint have been eliminated, like the theorem relating to
the motion of the centre of gravity, and the theorem of rotation of
a rigid body. But in the last analysis, when there have been forces
of constraint which have not annulled one another in pairs, the
setting up of the problem has involved explicitly any unknown
forces of constraint, as well as the known forces, and the former
have then been eliminated analytically, anew in each new problem.
We turn now to methods whereby, in certain important cases,
the forces of constraint can be eliminated once for all, so that
they will not even enter in setting up the equations on whose
solution the problem depends. Moreover, we introduce intrinsic
coordinates and intrinsic functions. The intrinsic coordinates are
a minimum number of independent variables whose values locate
completely the system. They are often called generalized coordi-
nates, and are denoted by qlt • • • , qm. The intrinsic functions are
the kinetic energy, the work function or its negative, the potential
energy, and the Lagrangean function L. These we have called
intrinsic because they do not depend on any special coordinate
system, or on any special choice of the q's. Later, we shall consider
intermediate cases in which the number of q's, though highly re-
stricted, is not a minimum, and in which, moreover, the unknown
forces, or constraints, have not been wholly eliminated.
1. The Problem. A material system may be determined in
its position by one or more coordinates, qlt - • • , </„,* and the
* We choose, in general, the letter m to denote the number of the q's. But we
replace it by n in these early examples to avoid confusion with the m that refers
to the mass of the particle.
297
298 MECHANICS
time, t. For example, let a bead of mass m slide freely on a
smooth circular wire, which rotates in a horizontal plane about
. one of its points, 0, with constant angular
velocity. The angle <p that the radius
drawn from 0 makes with a fixed hori-
zontal line is given explicitly,
Let 9 be the angle from OQ produced to
the radius, QP9 drawn to the bead. Then
the position of m is fully determined by 6 and t. Thus if we
set 0 = q,
x = f(q, t), y = \(/(q, t).
The problem of determining the motion is that of finding q
as a function of t.
More generally, let a smooth wire, carrying a bead, move
according to any law, and let the bead be acted on by any forces.
To determine the motion. We will treat this prob-
lem in detail presently.
As the second illustrative example, consider n
masses, ml9 • • • , mnj fastened to a weightless in-
extensible flexible vstring, one end, 0, of which is
held fast, and let the system be slightly displaced
from the position of equilibrium. To determine the m j
oscillation.
Finally, we may think of a rigid body, acted on
by any forces. If there are no constraints, it will
require six coordinates, qlt qz, • • • , qB9 to determine
the position. These may be the three coordinates
of one of the points of the body, as the centre of
gravity, x, ?/, 2; and the three Eulerian angles,
0, tp9 \f/9 which determine the orientation of the
body.
We may, however, also introduce constraints. If
one point is fixed, there are three degrees of free-
dom, and so three coordinates, ql9 </2, qZ9 — for ex- /
ample, the Eulerian angles — are required. Or,
again, the body might rotate about a fixed axis.
Then n = 1, and ql = q9 — a single coordinate — would be
sufficient. Or, finally, the body might be free to slide along
LAGRANGE'S EQUATIONS. VIRTUAL VELOCITIES
a fixed line and rotate about it. Here n = 2 and qlf q« are the
coordinates.
Each of the last two examples may be varied by causing the
line to move in an altogether specified manner. Then, beside
ql = q, or q1 and q2, the time, t, would enter explicitly.
In all such cases, the motion is determined by Lagrange's
Equations, which, when there is a force function U, take the
form:
A\ 1(1T_^-^ _ i
A; dtdb dq~,~dqr' r-1, ...,m,
where T denotes the kinetic energy, and qr — dqr/dt. We turn
now to the establishment of these equations, beginning with the
simplest cases.
2. Lagrange's Equations in the Simplest Case. Let a bead
slide on a smooth wire whose form as well as position varies with
the time :
(1) x=f(q,f), */ = *>(<?, 0, * = *(</,*),
where the functions f(q, t), <p(q, t), \f/(q, t) are continuous to-
gether with whatever derivatives we need to use, and where
fa 8y dz
V 'dq' dq
do not all vanish simultaneously. The motion is determined by
the equations :
"
(2)
where X, F, Z refer to the given, or applied forces, i.e. forces
other than the reaction, (X, Y, Z), of the wire, and
(3) XX + /*Y + v1 = 0,
where (X, /z, v) are the direction components of the tangent to
the wire — for the reaction of the wire is normal to the wire,
though otherwise unknown.
300 MECHANICS
Multiply these equations through by dx/dq, Sy/9q, 8z/8q re-
spectively and add :
8y Pz8z\ _
^ + dt* dq) ~y>
the remaining terms, namely :
y dx . y dy 7 dz
A TT~ i • "o~ i *- ~^r~i
dq dq dq
vanishing because of (3), since dx/dq, dy/dq, dz/dq are the di-
rection components of the tangent to the wire.
The left-hand side of Equation (4) can be transformed as
follows. We write :
tYI
(6) T = ~ (x* + y* + z*)9
where the dot notation means a time derivative :
dx . dq .
x = Tt> ^Tv ctc-
From (1)
dx _
dt ~ dqdt " dV
or
/«\ . dx . . dx
(7) ^ITqO + Jt'
with similar expressions for y and z. On substituting these
values in (6), T becomes a function of q, q, t. And now the left-
hand side of Equation (4) turns out to be expressible in the form :
<W <LdL-?L
W dt dq dq
For, first, we have :
dT f.dx..dy..ai\
-z-r = m[x— + y-~ + z~).
dq \ dq dq dq/
From (7) it follows that
dx = dx
dq ~ dq
LAGRANGE'S EQUATIONS. VIRTUAL VELOCITIES 301
with similar expressions for dy/dq and dz/dq. Thus
dT
Next, differentiate with respect to the time :
.V d dT __ /d2x dx d2y dy d2
' dtdq \dt2 dq dt2 dq dt2 dq,
. / . d dx . . d dy . d <
On the other hand,
dT _ / . dx _j_ . dy L . dz
_ _ m^
Now,
,1AN
(10) — = ^-.
dq \ dq dq dq/
dx _ d dx
'dq ~ dt~dq'
For, from (7),
f)'V s)^ IF /ft IT
UJU (/ JU . . C/ JU
~dq ~ ~difq dqdt'
and, of course,
d dx d2xdq d2x
Jt~dq = d^~dt dtdq'
Substituting the value given by (11) and the corresponding
values for dy/dq, dz/dq in the right-hand side of (10), we see
that dT/dq is equal to the last half of the right-hand side of (9),
and thus the proof is complete : — the left-hand side of (4) has
the value (8). We arrive, then, at the final result :
-
dt^j 8q ~ '
This is precisely Lagrange's Equation for the present case.
The case that the applied forces have a force function, U, is of
prime importance in practice. Here
dV v dU „ 8U
~> Y = '' ~'
and thus Q becomes :
= 4. 4. .
dx dq dy ~dq "*" dz dq
302 MECHANICS
Lagrange's Equation now takes the form :
dt dq dq dq
Example. Consider the problem stated at the opening of the
paragraph. Here, the applied forces are absent, and so U = const.
Furthermore,
x = a cos ut + a cos (6 + coJ)
y = a sin ut + a sin (0 + ui)
where q = 0 ; and
x = — aco sin ut — a (0 + co) sin (0 + co£)
2/ = aco cos to£ + a (^ + co) cos (^ + co<)
T = ^- ((^ + a?)2 + 2co(0 + co) cos 0 + co2)
orn /^T1
— r = ma2(^ -f co + co cos 0), — = —ma2co(0 + co) sin 0,
d dT 3T „ /d*B , o .
5ay"-8» a"ma"(3i+w8i
or
g=-..dn,
This last is the equation of Simple Pendulum Motion,
d*Q g • ^
-& — }**'•
Thus the bead oscillates about the moving line OQ as a simple
pendulum of length
Z -£
^~co2
would oscillate about the vertical.
EXERCISES
1. A bead slides on a smooth circular wire which is rotating
with constant angular velocity about a fixed vertical diameter.
Show that
tl^fi
-^ = co2 sin 6 cos B + 2 sin 0.
at* a
LAGRANGE'S EQUATIONS. VIRTUAL VELOCITIES 303
2. If the bead is released with no vertical velocity from a
point on the level of the centre of the circle, show that it will
not reach the lowest point if
3. A bead slides on a smooth rod which is rotating about
one end in a vertical plane with uniform angular velocity. Show
that d2r
— = w2r + g sin wt.
4. Integrate the differential equation of the preceding ques-
tion.
6. A bead slides on a smooth rod, one end of which is fixed,
and the inclination of which does not change. Determine the
motion, if the vertical plane through the rod rotates with constant
angular velocity.
6. In the problem discussed in the text, determine the reac-
tion, N, of the wire. Ans. N = ma2 [w2 cos 6 + (0 + w)2].
7. A smooth circular wire rotates with constant angular
velocity about a vertical axis which lies in the plane of the circle.
A bead slides on the wire. Determine the motion.
8. Show that if, in Question 1, the axis is a horizontal di-
ameter, the motion is given by the equation :
~~ — f w2 cos (p + ~ cos coH sin <p = 0,
where <p is the angle which the radius drawn to the bead makes
with the radius perpendicular to the axis.
9. A bead can slide on a smooth circular wire which is expand-
ing, always remaining in a fixed plane. One point of the wire is
fixed, and the centre describes a right line with constant velocity.
Determine the motion of the bead. . „ , 8
Ans. 6 = a + -•
t
10. The same problem if the centre is at rest and the radius
increases at an arbitrary rate.
3. Continuation. Particle on a Fixed or Moving Surface. Let
a particle, of mass w, be constrained to move on a smooth sur-
face, which can vary in size and shape,
(1) x = f(ql9 q2, 0, V
304 MECHANICS
where the functions on the right are continuous, together with
whatever derivatives we wish to use, and the Jacobians
do not vanish simultaneously. The motion is given as before by
Equations (2) of § 2, where now, however, the reaction, due to
the surface, is known in direction completely.
Multiply these equations through respectively by Sx/dql9
dy/d<li, dz/d<li> and add. On the right-hand side there remains
only
since dx/dqlf etc. are the direction components of a certain line
in the surface, drawn from (x, y, z), and X, Y, Z are the com-
ponents of a force normal to the surface. Hence
* dq, dt* dq, *
The reduction of the left-hand side is similar to the reduction
in the earlier case. It is, however, just as easy to carry this
reduction through for a system with n degrees of freedom, and
this is done in the following paragraph. Thus we see that
dt dqt dql l '
and, similarly :
dt dq% dq%
These are Lagrange's Equations for the case of two variables;
i.e. the case of (ql9 qz). If, in particular, a force function, [/,
exists, then Lagrange's Equations take the form :
\ d (/ 2 u L uLJ d u L u L uU
dt dqi dql dq± dt dq2 ^2 ^2
Example. A particle is constrained to move, without fric-
tion, in a plane which is rotating with constant angular velocity
about a horizontal axis. Determine the motion.
LAGRANGE'S EQUATIONS. VIRTUAL VELOCITIES 305
Let the axis of rotation be taken as the axis of x, and let r
denote the distance of the particle from the axis. The coordi-
nates of the particle are (z, y, z),
where
y = r cos 0, z = r sin 0, 0 = co£,
and we take ql = x, q2 = r. Then
y — r cos w£ — 7*w sin co£,
z — r sin o>2 + rw cos ait,
a result that may be read off directly, without the intervention
of y, z. Furthermore,
U = —mgz = —mgr sin ut.
Lagrange's Equations now become :
dT . dT n dU n
-r-r = mx, ~^~ = 0, -— = 0;
dx dx J dx
a result immediately obvious. Next,
dT . dT
or
— = — mg sin <
ttr* — mco2r = — ing sin co/,
^2 - o> r =-^sm^.
A special solution of this equation is found, either by the method
of the variation of constants or, more simply, by inspection, to be :
sni
r
" 2co
Hence the general solution is
Since 0 = w^, the equation can be written in the form :
306 MECHANICS
A particular case of interest is that in which A = B = 0.
This corresponds to the initial conditions of launching the particle
from a point in the axis with a velocity whose projection along
the axis is arbitrary, the projection normal to the axis being g/2w.
The path is then a helix.
EXERCISES
1. A cylinder of revolution is rotating with constant angular
velocity about a vertical axis, exterior to the cylinder, the axis
of the cylinder being always vertical. A particle is projected
along the inner surface, which is smooth. Determine the motion.
2. Use Lagrange's Equations to determine the motion of a
particle in a plane, referred to polar coordinates :
d*r dd*\ md
4. The Spherical Pendulum. Consider the spherical pendu-
lum ; i.e. a particle moving under gravity and constrained to lie
on a smooth sphere. Take as coordinates the colatitude, 6, and
the longtitude, <p, the north pole being the point of unstable
equilibrium . Then
fY)f]1
T = ^- (02 + p Sin2 0)^ jj = _mfif2 = _ mga cos e .
O/TI PT7 %TT
— - = ma26, — = ma2<f>2 sin 6 cos 0, — - = mga sin 0,
00 Cv Cv
and the first of Lagrange's Equations becomes :
ma2S — ma2(f>2 sin 6 cos 6 = mga sin 0,
or
a
Proceeding now to the second equation, we have :
Hence
LAGRANGE'S EQUATIONS. VIRTUAL VELOCITIES 307
This equation integrates into
(2) ^sin*0 = h,
where the constant h is of dimension — 1 in the time, [T~1].
Combining Equations (1) and (2), we obtain:
» -
This is the Equation of Spherical Pendulum Motion. In case
the motion is to be studied for small oscillations near the lowest
point of the sphere, it is well to replace 6 by its supplement,
6' — TT — 6. Equation (3) then becomes :
This last equation reduces to the Equation of Simple Pendulum
Motion when h = 0. Any differential equation of the form :
fA\ d26 A cos0 D .
(4) —r = A --T-T + B sm 6,
at2 sm3 6
where A > 0 and B > 0 are arbitrary constants, can obviously
be interpreted in terms of spherical pendulum motion.
A first integral of (3) can be obtained in the usual manner :
2A2cosgdfl 2g . dfi
dtdP sin3 0 dt "*" a *m dt'
Integrating each side with respect to 0, we obtain :
<» £--;£-?-• + *
For a further discussion of the problem, cf. Appell, Mecanique
rationelk, vol. i, § 277.
EXERCISES
1. Give an approximate solution for small oscillations near
the point of stable equilibrium, 0, using Cartesian coordinates.
Here,
and the approximate path is an ellipse with 0 as centre.
308 MECHANICS
2. Treat the motion of a particle constrained to move on a
smooth surface, 0 2
22 = a2£2 + 022/2,
for small oscillations near the origin.
3. Show that a top which is not spinning moves like a spherical
pendulum. More precisely, we mean the body of Chapter VI,
§ 18, when v = 0.
6. Geodesies. Let a particle be constrained to move on a
smooth surface under no applied forces. The path is a geodesic.*
Let the surface be given by the equations :
x = f(u, i;), y = <p(u, 0), z = t(u, v),
where these functions are continuous together with their first
derivatives, and not all the Jacobians
d(u,vy d(u,v)' d(u,v)
vanish. The element of arc is given by the formula :
ds* = Edu* + 2Fdudv + Gdv\
where the coefficients arc easily computed,
The kinetic energy has the value :
T = % (Eu* + 2Fui> +
Lagrange's Equations now become, since U = 0 :
(6)
(Eu + Fit) - i (Eu w2 + 2FU uv + Gu t-2) = 0,
™ (Fu + Gv) - -£ (Ev u2 + 2Fvuv + Gvv2) = 0.
On the other hand, the geodesies, in their capacity of being
the shortest lines on the surface, are given as extremals of the
integral A,
= fVEu'2
L = / VEu'2 + 2Fu'v' + Gv'2 d\
*0
* By a geodesic is meant a line of minimum length on a surface — minimum, at
least, if the points it connects are not too far apart ; cf. Advanced Calculus, p. 411.
LAGRANGE'S EQUATIONS. VIRTUAL VELOCITIES 309
in the Calculus of Variations (cf. the Author's Advanced Calculus,
p. 411) by the equations :
Eu' + Fv' Eu u'2 + 2FU u'v' + Gu v'2 „
- = 0
d Fu' + Gv' Ev u'* + 2FV u'v' + G, v'* _
= u
(7)
d\ VEu'^'2Fu'vr+ Gv'2 2V'Eu'2 + ^'u'i/^W2
The parameter X can be replaced by any other parameter, /u :
M = /(A),
provided that/(X) is continuous, together with its first derivative,
and/'(X) 9* 0. In particular, then, the choice t = /z is a possible
one. But then, because of the equation of energy, T = h, or :
(8) (#w'2 + 2*W + Gv'*) = h, u' = u, v' = *,
£i
it follows that Equations (7) reduce to Equations (6).
Since the velocity along the path is constant, the only force
being normal to the path, t is proportional to 5. In fact, (8)
says that
mds *
Thus the transformation of the parameter from X to t amounts
in substance to a transformation to s; i.e. Equations (6) are
virtually the intrinsic differential equations of the geodesic :
(9),
~ (Eu' + Fv') - £ (E. w'2 + 2FU u'v' + Gu v"*) = 0
as
J- (Fuf + Gv') - $ (E, w'2 + 2FV u'v' + G, v'2) = 0
as
. du . dv
U* = -J-, V' = -7-
rl n ' ft n
EXERCISES
1. Obtain the geodesies on a cylinder of revolution. Observe
that, when the cylinder is rolled out on a plane, the geodesies
must go over into straight lines.
2. The same problem for a cone of revolution.
310
MECHANICS
3. Show that the geodesies on an anchor ring, or torus, are
given by the differential equations :
d20 . h2 sin 6
(10)
a(b + acosfl)3
"
where a and & > a are the constants of the anchor ring, and A is a
constant of integration.
4. Show that if, in the preceding question, initially 6 = ir/2,
= or,
= A then
(ID
a2 (6 + a cos0)2 a2
dt (b + a cos 0)2
6. Lemma. We have seen in § 2 that, in the case of a single #,
dx (Py dy
dq dt2 dq
d^z
dt2
dt dq dq
It is important to recognize this equation as a purely analytic
identity, irrespective of any physical meaning to be attached to
T. It says that, if
T = ~(x2 + y2 + z2)
and
* = f(q, 0, y = v(<7, 0, ^ = ^(<7, 0,
where these functions arc any functions subject merely to the
ordinary requirements of continuity, then Equation (1) is true.
We turn now to the general case of n particles, m» with the
coordinates (a?», yt,Zi)9 i = 1, 2, • • • , n. Let the position of
the system be determined by m parameters, or generalized coordi-
nates, ?!,•••, qm, and the time, t :
*i = /ifoi, ' ' ', ?m, 0
yi= <Pi(qly • • -, ?m, 0
2i = ti(qi, - • •, gm, 0
where the functions /,-, (p», ^t, are continuous together with their
partial derivatives of the first two orders, and where the rank of
the matrix a) is m :
A)
LAGRANGE'S EQUATIONS. VIRTUAL VELOCITIES 311
a)
dxl
dqm
dql
The kinetic energy,
can be expressed in terms of glt - • • , qm, qly • • • , qm, t, for
with similar formulas for yiy z^
Our aim is to establish the general fundamental formula cor-
responding to (1) :
i \
-^~' - — "
r = 1,
The independent variables in the partial differentiations are
(<7i> • ' " » (7m, (7i> • • • > qm, 0> and x^ y^ Zi are given by (3). We
have :
From (3) :
Hence
_ _
dqr 'dqr
- etc.
Differentiate with respect to ^ along a given curve :
(A\ <L?L = V (tfxjdxj d^jjdyi d2Zj dz
W *^r V miV"^2 0?r ^2 ^r ^2 07
d dxi d^dy/i . d[ dzj\
2'
312
MECHANICS
On the other hand,
(5) I =
Now,
For, from (3),
\ * dqr * dqr l dqr<
^
dt
while
d
dqrdt'
dtdqr
Similar relations hold for diji/dqr and dZi/dqr. On substituting
these values in (5), it is seen that the last sum in (4) has precisely
the value dT/dqr, and thus the relation I. is established.
7. Lagrange's Equations in the General Case. Let a system
of particles nti with the coordinates (xi, y^ Zi) be acted on by any
forces whatever, X», Ft, Zi. By Newton's Second Law of Motion
(i)
v
= Xi
Let the position of the system be determined, as in § 6, by m
coordinates ql9 • • • , qm and t :
A)
where the functions /»-, ^,-,
derivatives of the first two orders, the matrix
ifei, • ' • , <7«»» 0
(<7i, • • • , ?w, 0
i(^li ' ' ' i (7m, 0
are continuous together with their
dql
LAGRANGE'S EQUATIONS. VIRTUAL VELOCITIES 313
being of rank m. Multiply the first of Equations (1) by 8xi/dqr,
the second by dyt/dqr, the third by dzi/dqr, and add :
/0, v (x* x<
(2) 2, *< VdP" ft
dx< 4. v ^< j. 7 ^A
^ ^ ^'
r = 1, • • • , ra.
The left-hand side has the value expressed by the Fundamental
Equation I. of § 6. Let the right-hand side be denoted by Qr :
(3) «"
It thus appears that
d dT
These are known as Lagrange's Equations.
We have deduced Lagrange's Equations from Newton's Second
Law of Motion. They include Newton's Law as a particular case.
For if we set
t =
then T7 becomes :
and Q3<, Q3<+i, Qa<+2 are now the components of the force which
acts on mt-. Thus Newton's Equations result at once.
8. Discussion of the Equations. Holonomic and Non-
Holonomic Systems. We have before us the most general case.
No restrictions have been made on the forces. These may, then,
comprise dissipative forces, like those of friction or air resistance.
On the other hand, there may be one or more equations of the
form:
(4)
where the function F does not depend on the initial conditions.
314 MECHANICS
Moreover it may happen, whether there are relations of the
form (4) or not, that the <?/s and their time derivatives are bound
together by one or more equations :
(5) $(?!, • ' • ,?m, % • • • , ?m, 0 = 0.
An airplane, rising at a given angle, would be an example.
The case which is most important in practice, is that in which
<f> is linear in the qr :
(6) AM + -- +Amqm + A = 0,
where the A's are functions of (qlt • • • , qm, C), independent of
the initial conditions.
It may happen that a relation of the form (6) is equivalent to
one of the form (4). Thus if
' ds dO
this relation is equivalent to the equation :
(8) s - ad = c,
which is essentially of the form (4).
If no relations (5) or (6) are present ; or if such relations (5) or
(6) as may have entered in the formulation of the problem are
all capable of being replaced by equations of the form (4), the
system is said to be holonomic. Examples of non-holonomic systems
are the Cart Wheels of § 24 infra, and the Billiard Ball on the
rough table, rolling and pivoting without slipping, p. 240 ; also the
coin on the rough table, and the bicycle.* But when the Billiard
Ball slips, p. 237, the system is holonomic, for the unknown reac-
tion of the table can be computed explicitly, as the reader can
easily verify, in terms of the velocity of the point of contact,
and thus its components are expressed in terms of the time deriv-
atives of the generalized coordinates.
We are still leaving in abeyance the question of whether La-
grange's Equations admit a unique solution. Our conditions are
necessary for a solution of the mechanical problem, but not always
sufficient. The study of sufficient conditions will be taken up in
§ 17 and in Appendix D.
* Appell, Micaniqae rationelle, Vol. II, Chaps. XXI, XXII.
LAGRANGE'S EQUATIONS. VIRTUAL VELOCITIES 315
9. Continuation. The Forces. The question of holonomic or
non-holonomic has to do with the left-hand side of Lagrange's
Equations, i.e. with conditions on the qr, qr, t which do not involve
the forces or contain the constants of the initial conditions.
The forces appear on the right, and it is to these that we
now turn our attention. It may happen that the total force
Xi, Yij Zi can be decomposed into two forces :
(9) x, = xi + xi Y< = Yi + 17, zt = z( + z;
in such a manner that the X{, Y^ Z\ will be essentially simpler
than the X», Yi, Ziy and that the X*, F*, Z* disappear altogether
from Lagrange's Equations. For example, the X{, F{, Z\ may be
expressible in terms of a force function :
v, _ till v, _ d(J r/r _ dU
Xi ~ ~ Yi " ' "
where U is known explicitly in terms of X*, 2/», «,-, t.
As regards the disappearance of the A7, Y*, Z* the problems
discussed in §§ 1-6 have afforded ample illustration. These
were the so-called forces of constraint, and they did not appear in
Lagrange's Equations.
Returning now to the general case, we observe that it may
happen that the X*, F*, Z* fulfil the condition :
r — 1, • • • , m. When this is true, the Qr on the right of La-
grange's Equations take on the simpler form :
This case is important in practice because it enables us to get
rid of some or all of the unknown forces of the problem arising
from constraints ; cf . § 1 5. But even when all of the latter forces
cannot be eliminated in this way, their number can be reduced to
a minimum ; and then the method of multipliers set forth in the
next paragraph leads to the final elimination.
316
MECHANICS
10. Conclusion. Lagrange's Multipliers. Consider a system,
the motion of which is given by Lagrange's Equations :
ddT_dT^ = Q
It can happen that the Qr's can be split in two :
(14) Qr = Qr + &*, r = 1, - • • , m,
in such a manner that the Q' are essentially simpler than the Qr
— known functions, for example — whereas the Q* have the
property that
(15) Q*TT, H + Qiirm = 0,
where irlf • • • , wm are any m numbers which satisfy the equations :
(16) disTTi + • • • + amsTrm = 0, s = 1, ••-,/*< m.
Let the rank of the matrix
(17)
be u.
an
ami
From Equations (13) it follows that
d dT dT ~
no matter what numbers the TT, may be. If, in particular, the
Qr and the jr, are subject to the condition expressed in (14),
(15), and (16), then
(18)
d dT
Multiply the /z equations (16) respectively by arbitrary num-
bers, Xi, • • • , X^, and subtract from (18) :
Of the m numbers vlt • • • 9vm it is possible to choose some
set of m — p, arbitrarily, and then the rest are determined by
(16). For definiteness, suppose
LAGRANGE'S EQUATIONS. VIRTUAL VELOCITIES 317
(20)
Then ir^+i, • • • ,irm are arbitrary, and iru • • • , TT> are determined.
Now let the Xj, • • • , V be so chosen that
(21)
This is possible because the determinant of these /z linear equa-
tions in X,, • • • , XM is not zero. For these values of X Equa-
tions (19) reduce to the following m — ju equations :
(22)
The determination of the X«'s by (21) is independent of any
choice of the TT/S. The numbers TJ>+I, • • • , TTOT are wholly arbi-
trary. Hence each coefficient in (22) must vanish. We have
thus established the following
THEOREM. When Qr can be written in the form:
(23)
w/ie
(24)
r = 1,
(25) , als*-! +
the rank of the matrix
(26)
+ ama7rm = 0,
= 1,
am\
Ctl/x ' ' '
iwgf /z, </ien i< zs possible to find p numbers \19 • • • , X^ swcA
(27)
dT
numbers are determined by & of the Equations (27), and the
values thus obtained are then substituted in the remaining m — p
equations.
318 MECHANICS
Applications of this theorem occur in practice in a variety of
problems in which the qr, qr, and t are connected by relations of
the form (6), § 7 :
(28) ai,^ + • • • + am»qm + a, = 0, « = 1, • ••,/*,
where the ar,, a, depend on the qr and £, but not on the ini-
tial conditions. As a matter of fact, in a number of such
cases the coefficients ara in (28) do lead to a system of equations
(16) which control a set of numbers TT^ • • • , wm for which an
analysis (14) of Qr with the resulting relation (15) is possible.
In other problems, however, the ars of equations (16) have noth-
ing to do with any such equations as (28), if indeed the latter
exist, but may even themselves depend on qr, as well as qr and t.
For a complete discussion cf . Appendix D.
We turn now to a direct determination of the Qr from purely
mechanical considerations.
11. Virtual Velocities and Virtual Work. Let a system of
particles ra» with the coordinates (x%1 ?/t, zt) be given (i — 1,
• • • , ft). Let dxi, dyt, dzi be any 3n numbers, and let m» be
carried to the point (xi + 8xl, yi + fyi, Zi + dZi). Then tha
system is said to experience a virtual displacement (5xi, tyi, dZifl
the word " virtual" expressing the fact that the actual system
may not be capable of such a displacement, even approximately.
Thus a particle constrained to move on a curve or a surface would
in general be taken off its constraint, and not lie even in the tan-
gent line or plane.
If forces (Xi, Y^ Z$ act on the system, the quantity
(1) W* = J (X, dxi + Yi dyi + Zi dzi)
t = i
is defined as the virtual work due to the virtual displacement.
It is convenient in many applications to restrict the virtual
displacements admitted to consideration by linear homogene-
ous equations between the 5xt-, 6?/t, 5zi. Consider, in particular,
a system of particles whose coordinates are given by Equations
A), §7:
^ = fi(qi, - • - , qm, t)
A)
, 0
*, 0
LAGRANGE'S EQUATIONS. VIRTUAL VELOCITIES 319
where the rank of the matrix
a)
is m. Let
(2)
fat t dxl
09i dqm
^ dZn
dql dqm
where dq}, • • • , 3qm are m arbitrary quantities. If m < 3n, the
toi, 8yi} dzi are subject to one or more linear homogeneous equa-
tions. Thus only a limited number of them can now be chosen
arbitrarily, the rest being then determined.
Consider the actual displacement (Axt-, A?/*, Az») which the
system experiences in time At as it describes its natural path.
It is :
AXt = fi (ql + Ag,, • • • , qm + Agm, t + At) — /»• (qly • • • , qm, t)
(3) A?/t = <pi (ql + At/!, • • • , qm + A^m, t + A^) — pi (qlf • • • , gm, ^)
Since the 5#r are arbitrary, it is possible to choose them equal to
the Aqr, or dqr = Aqr. It does not follow, however, that the cor-
responding tot, diji, dzi will differ from Axi} Ayiy Az» by infinites-
imals of higher order with respect to A£. This will, in fact, be
the case if the functions /t-, <pt, \f/i do not contain the time, i.e.
if dfi/dt = 0, etc. But otherwise in general not.
The virtual work has the value :
(4)
,
where all m of the numbers 8qr are arbitrary. If, in particular,
the forces can be broken up as in (9) § 9 :
(5) Xi = XI + XI Ft = Y( + F?, Zt = Z\ + Zt,
320 MECHANICS
so that (11) holds:
(6) jjj(ir* +17* + *
r = 1, • • • , m, then (4) takes on the simpler form :
(7) r,-
In either case,
(8) W5 = Q.dq, + ...+Qm 8qm.
Consider the actual displacement (Ax», Ayi, AZi) of the system
in time At as it describes its natural path. If /,-, ^?t-, \f/i do not
contain t, the virtual work Ws will differ from the actual work,
ATT, by an infinitesimal of higher order than A£; otherwise, this
will not in general be the case.
12. Computation of Qr. In Equation (8), § 11 the dqr are
m arbitrary numbers. We may, then, set dqk — 0, k ^ r;
5qr 7* 0, and compute the corresponding value of W6. We shall
then have :
Consider, for example, a particle that is constrained to lie
on a moving surface. Its coordinates are subject to the condi-
tions :
(10) x = f(qly q2, 0, y = <p(qi, q» 0, 2 = iKfc, q* t).
A virtual displacement means that we fix our attention on an
arbitrary instant of time, t, and consider the surface represented
by (10) for this value of t. Next, consider a point (x, y, z) of
this surface. Then a virtual displacement (&r, dy, dz) of this
point means an arbitrary displacement in the tangent plane to
the surface at the point in question. In particular, if we set
dq2 = 0 and take dq1 ^ 0, then the virtual displacement takes
place along the tangent to that curve in the surface whose coor-
dinates are represented by (10) when </2 and t are held fast.
Now, the natural path of the particle under the forces that
act does not in general lie in the surface just considered, nor is
it tangent to the surface. If the surface is smooth, the reaction
on the particle will be normal to the surface, and so the virtual
LAGRANGE'S EQUATIONS. VIRTUAL VELOCITIES 321
work Ws due to the reaction will be 0. But the actual work
done by the reaction in A£ seconds along the natural path will
in general be an infinitesimal of the same order as At.
It is now easy to see how to compute Ql and Qz m case the
surface is smooth. The virtual work of the reaction of the surface
is nil, and so we need consider only the other forces. The virtual
displacement takes place in the tangent plane to the surface,
and we can compute directly the virtual work corresponding
to the successive virtual displacements given by 8ql ^ 0, 8q2 = 0
and 5ft = 0, 8qz ^ 0 ; cf. further § 16 infra.
13. Virtual Velocities, an Aid in the Choice of the irr. In
the general theorem of § 10 there was no indication as to how
the 7rr may bo chosen. In certain cases which arise in practice,
the motion being subject to Lagrange's Equations :
/ d 3T dT
it happens that there are geometric or kinematical relations
between the qr's of the form (28), § 10 :
(2) ali(fft + ' • ' + dmsqm + CL8 = 0, 8 = 1, ••-,/*< 1»,
where the ar8, a8 are known functions of qlt • • • , qm, t,* which
do not depend on the initial conditions, and where the rank of
the matrix
an
(3)
IS fJL.
If, now, the possible virtual displacements corresponding to
an arbitrary choice of 5ft, • • • , 8qm are so restricted that
(4) als 8qi + - - • + am8 8qm = 0, 5 = 1, • • • , /*,
it turns out that the virtual work of certain forces ("forces of
constraint ") will vanish. Hence by identifying these "forces
of constraint " with the Q* and setting 8qr = 7rr, the hypotheses
of that theorem arc fulfilled.
* It may happen that some or all of these equations may be integrated in the
form: F (qi, • • -, </m, t) = 0, where F does not depend on the initial conditions;
but it is not important to distinguish this case.
322 MECHANICS
Example. Consider the disc of Chapter VI, § 24 as free to
roll without slipping on a rough horizontal plane which is moving
in its own plane according to any given law ; for example, rotat-
ing about a fixed point with constant velocity. The force which
the plane exerts on the disc at the point of contact will do work
on the disc. But the virtual work of this force, when the virtual
displacement is restricted as above, is nil. Thus we are led to
a suitable set of multipliers 7rr, namely, the 5qr thus restricted.
14. On the Number m of the qr. For a system of particles
the qr's, as has already been pointed out, can always be identified
with the coordinates :
Here, m = 3w, and Lagrange's Equations become identical with
Newton's Equations.
In 'theory, then, there is no difference between the two systems.
In practice, Lagrange's Equations provide in many cases an
elimination of forces in which we arc not interested.
Consider a ladder sliding down a wall. If the wall and floor
are smooth, we may take m = 1, q = 0, and all the forces in
which we are not interested will be eliminated.
s Here,
(2) x = a cos 0, y = a sin 0.
MgeMK Hence
FIG. 144 (3) r = ^(02 + £1)0.
Lagrange's Equation becomes :
where
U = — Mg a sin 0.
Thus we find as the equation governing the motion :
M (a2 + A;2) -^ = - Mga cos 0
or
fK\ d?0 ag
LAGRANGE'S EQUATIONS. VIRTUAL VELOCITIES 323
In this example, the maximum of elimination has been at-
tained at one blow. If we think of the ladder as made up of a
huge number of particles connected by weightless rods, the forces
in the rods have been eliminated, and also the forces exerted by
the floor and the wall.
Suppose, however, that the floor and the wall
are rough. We can still write down a single
Lagrangean equation,
the left-hand side being as before. But now
W
(7) Q = _J = - Mga cos 0 + 2a cos Op,S + 2a sin 0 »R.
dq
We have not equations enough to solve the problem.
The difficulty can be met by taking m = 3 and setting
(8) 0i = x, 02 = 27, 03 = 0-
T is given by (1). And now
Q! = S - M#, Q2 = R + »S - Mg,
(9)
1 = a(sin 0 + » cos 0) S + a(/i sin 0 - cos 6) R.
Lagrange;s three equations become :
(10)
= a(sin 0 + n cos 0) S + aO» sin 0 - cos 0)
The first two of these are the equations of motion of the centre
of gravity ; the third, the equation of moments about the centre
of gravity.
What Lagrange's method here has done, is first to eliminate
the internal forces between the particles, just as we did in Chapter
IV, § 1, when we proved the theorem about the motion of the
centre of gravity; and similarly, when we proved the theorem
of moments, Chapter IV, § 3 and § 9.
324 MECHANICS
Let
Qr = ® + Qf,
where
Qr~s-MB, Q; = /Z + /*S,
Q* = a(sin 0 + /x cos 0) S + a(p, sin 0 - cos 6) R.
We wish to find three multipliers, TT^ ?r2, 7r3, such that
(11) Ql^ + Q*T2 + Q,*T, = 0.
This can easily be done algebraically, with the result that
TTI + /i7T2 + a(sin 6 + /x cos 0) 7r3 = 0
(12) ,
+ 7r2 + «(M sin 0 — cos 0) 7r3 = 0
a solution of these equations being :
TJ = — sin (0 + 2X), ?r2 = cos (0 + 2X), 7r3 = I/a.
But a mechanical derivation is easy, too. Consider the result-
ant of the forces R and p,R. Draw a perpendicular to it through
the lower end of the rod, and displace this end along this line.
Do the same thing at the upper end of the rod, and displace the
upper end along this line. The result is, that
f Sql =-asin (9 + 2\) 8q,
' lo) 1
I dq2 = a cos (6 + 2X) S#3
Corresponding to such a displacement the virtual work of the
"constraints" must vanish. And now it is merely a question of
trigonometry to show that our expectation is fulfilled :
(14) Qj>i + Ql fy, + «,*«?, = 0.
Equation (11) corresponds to Equation (15) of § 10, and Equa-
tions (12) are the Equations (16) of that paragraph. But Equa-
tions (13), though corresponding to Equations (4), § 13, do not
have their origin in Equations (2), § 13. The latter would arise
from differentiating (2).
Returning now to Equations (10), we see that the unknown
reactions R and S are eliminated by (14), where 5qlf 8q.2, 8qs sat-
isfy (13). Hence
(15) Jflfift + (M§ + Mg) dq, + Mk*0dqz = 0,
or
(16) - MX a sin (0 + 2X) + (My + Mg) a cos (0 + 2X)
+ Mk* '6 = 0.
LAGRANGE'S EQUATIONS. VIRTUAL VELOCITIES 325
This equation, combined with Equations (2), leads at once
to the solution of the problem :
(17) (/c2 + a2 cos 2X) ~~ + a2 sin 2X ^~ + ag cos (0 + 2X) = 0.
15. Forces of Constraint. A definition of "forces of con-
straint" from the point of view of physics, which shall be both
accurate and comprehensive, has, so far as the author knows,
never been given. They would be included in such forces as
the X*, ¥*, Z* of §9, which disappear from the Qr; I.e., Equa-
tion (11). Arid still again, the Q* of § 10 arise from unknown
forces and are eliminated by the method of multipliers.
Perhaps these two cases are comprehensive in Rational Me-
chanics. Are there problems in this science not included here?
If not, the asterisk forces could be defined as the forces of con-
straint ; cf . Appendix D.
16. Euler's Equations, Deduced from Lagrange's Equations.
When a rigid body rotates about a fixed point, the kinetic energy is
(1) T = i(4p2 + #?2 + O2).
Let p, q, r be expressed in terms of Euler's angles, Chapter VI,
§15:
p = 6 sin (p — \}/ sin 0 cos <p
(2) q — 6 cos <p + \l/ sin Q sin <p
r — <j> -{- \l/ cos 0
The second of Lagrange's Equations is readily computed :
dT A dp . ,, dq . „ dr „
— = Ap~ + Bq~~ + CV — = Cr;
v<P O(D V(D (/<p
TT- = 8 cos <p + ^ sin 6 sin <p = g,
— = — 6 sin v> -\- $ sin 0 cos ^> = — p,
Hence
326
MECHANICS
To compute <$>, observe that, no matter what forces may act,
they can be replaced by a force at 0 and a couple. The latter
can be realized by means of three forces :
o
ii)
Hi)
a force L y acting at the point * r =
" Ma " " r =
T = a.
A virtual displacement 80 = 0, 5<p ^ 0, d\l/ = 0 gives as the
virtual work
N5<p
and this is equal to &8<p. Hence <t> = N and we have :
Cj|- (A -B)pq = N.
This is the third of Euler's Dynamical Equations. The other
two follow from this one by symmetry, and are obtained by
advancing the letters cyclically.
EXERCISE
Obtain the six equations of motion of a rigid body by means of
Lagrange's Equations.
17. Solution of Lagrange's Equations. We have seen in
§ 10 that if the Qr satisfy the conditions of the Theorem of that
paragraph, then
(i)
where the matrix :
(2)
-
dt dqr dqr
y* ^
= Qr + 2) ar8\9, r = 1, • • • , m,
is of rank ju < m. In the cases which arise most frequently in
practice, there are n equations of the form :
* By the " point r " is meant the terminal point of the vector r when the initial
point is at 0.
LAGRANGE'S EQUATIONS. VIRTUAL VELOCITIES 327
(3) a,isqi + • • - + amsqm = aa, s = 1, • • • , ju,
where the a's are functions of glt • • • , qm, t.
The kinetic energy T is a positive definite quadratic form in
the ql9 • • • , qm, but not necessarily homogeneous :
T = T2 + T,+ T0,
where
and T19 T0 are homogeneous of degree 1 or 0 in the qr, or van-
ish identically. The coefficients are functions of qlf - - • , qm, t.
The form T2, or
(4) 2 </*<**
*,./
is a positive definite homogeneous quadratic form. For T can be
written in the form :
/tu, •••, juw, (7 are functions of #i, ••-, ^m, t.
Finally, let Q'r be a known function of qr, qr, t.
THEOREM. The m + /* Equations:
d dT 8T / A
i H ---- + o«.g« = a., s = 1, • • • , /z,
determine uniquely the m + p, functions (ft, • • • , </„», Xx, • • • , XM.
Proo/. The first m of these equations have the form :
(5) Airqi + • • • + Amrqm — arlXj — • • • — ar/AXM = Br,
r = 1, • • • , m, where £r is a function of the </,., qr, and £. The
remaining /* equations give, on differentiating :
(6) ai,qi -\ ---- + amsqm = Ca, s = 1, • • • , /*,
where C. is likewise a function of the gr, qr, and <. Thus we have
m + M linear equations in the m + n unknowns: qlt • • • , #m,
Xj, • • • , X^. Their determinant :
328
MECHANICS
_
"" an • • • a«ii 0 • • • 0
Oi,A • • • OmM 0 • • • 0
does not vanish. For otherwise the m + n linear homogeneous
equations :
uti + • • • + AmiZm + an*?! + • • • + ai^ = 0
lmti + • * ' + Ammtm + Oml^ + ' ' ' + Omplfc = 0
= 0
= 0
would admit a solution (£, - • • , £w, 77^ • • • , T/M) not the identity.
Moreover, not all the fi, • • • , £m in this solution could vanish ;
for then we should have :
+
+
= 0.
But the rank of the matrix of these equations, namely the matrix
(2), is /x. Hence all the i)l9 • • • , ^ of the solution vanish — a
contradiction.
Next, multiply the r-th equation (8) by £ r, r = 1, • • • , m, and
add. The terms in i^, • • • , 77^ drop out because of the last p. of
the equations (8), and so there results the equation :
where not all the £j, • • • , £m are 0. This is impossible, since (4)
is a positive definite quadratic form.
Equations (5) and (6) admit, therefore, a solution :
(9)
0,
0
, m ;
LAGRANGE'S EQUATIONS. VIRTUAL VELOCITIES 329
Assuming for definiteness that the determinant
an • • • <v
I/*'** Uft.fi.
wo see that the first m of the Equations (9) admit the integral
given by (3), or :
where, in particular, /, is linear in <fo+i, • • • , qm- This is a par-
ticular integral which is independent of the initial conditions of
the mechanical problem. Let
(11) qa = Ka, a = » + 1, • • - , m.
The system of m differential equations (9) is now seen to be equiv-
alent, under the restrictions of the dynamical problem, to the
system :
dq8 f , j\ 1
' ' ' ' ' * • - - • '
(12)
dKa
, Km, 0, tt = /* + 1,
Here is a system of 2m — p differential equations of the first order
for determining the 2m — M unknown functions qr, Ka. Their
solution yields the m desired functions, qlt - - - , qm. The X8 are
now uniquely determined as functions of t and the initial condi-
tions. As regards the freedom of the initial conditions, on which
of course the determination of the constants of integration depends,
the initial values </r°, gr° of qr, qr are restricted by the equation
fl\sf]\ H I * * * I drnsqm ~ $.s-
It is important here, as in so many problems of the kind dis-
cussed in this chapter, to distinguish between constants that are
connected with the choice of coordinates and constants that arise
from the initial conditions of the mechanical problem. Thus in the
problem of ('hap. IV, § 13, p. 141, Fig. 84, s might equally well
have been measured from a different level, and then the relation
would have been :
s = ad + c.
330
MECHANICS
18. Equilibrium. Let a dynamical system be given, with n
particles m,, the motion being subject to Newton's Law :
(1)
The forces are said to be in equilibrium if
Zit
1,
n.
(2)
= 0, Yt = 0,
0,
A necessary and sufficient condition for equilibrium is, that
(3)
= 0,
= 0,
= 0.
We are not interested in the general case, which, in accord with
the definition just given, relates to a single instant of time, the
forces not in general being in equilibrium at any other instant.
We have concern rather with a permanent state of rest of a system
capable of certain motions which are subject to geometric condi-
tions. We are thinking primarily of such problems in the statics
of particles and rigid bodies as were studied in Chapters I and II ;
but also of more general problems, like the following : — A uni-
form circular disc has a particle attached to its rim. The disc
rests on a smooth ellipsoid and a rough table which contains
two axes of the ellipsoid. Find the positions of equilibrium.
More precisely, the system shall be capable of assuming the
positions defined by the equations :
(4)
= fi(q\, •••,?*)
where the functions /», ^, \f/i do not depend on t, and where the
rank of the matrix
(5)
s ra.
Observe that this last requirement does not imply that m has
the least value for which the xiy y^ Zi can be represented by equa-
tions of the form (4), satisfying the above requirements. It is
LAGRANGE'S EQUATIONS. VIRTUAL VELOCITIES 331
still possible that the q19 • • • , qm may be connected by relations
of the form :
(6) Ft (ft, • • • , qm) = 0, j = 1, • • • , p < m.
On the other hand it does imply that if the x^ yiy 2», all vanish,
then this is true of the qr, and conversely; and if, furthermore,
both the ±ij #», z», and the xiy #», z» all vanish, then this is true
of the qr and the qr, and conversely.
The motion of the system is, first of all, subject to the equa-
tions :
(7) [71, = Qr, r = 1, • • • , M < m,
where by definition
\T\ =<L2L-VL.
1 lr dt dqr dqr
To these may be added further equations :
(8) alaqi + • • • + am*qm = 0, a = 1, •••,/*< m,
where the rank of the matrix
(9)
is (1. It is possible that some or all of these equations can be
expressed in the form (6), but this is unimportant.
A first necessary and sufficient condition for equilibrium is that
(10) Qr = 0, r = 1, - - - , m.
For, a necessary and sufficient condition for the vanishing of the
left-hand side of (7) for ft = 0, •••,&, = 0 is that ft = 0,
• - • , qm = 0.
If there are relations of the form (8), it may happen that the
Qr can be split up as follows :
(11) Qr = Qr + Q?>
where
(12) Q*«ft + • • • + Q*m*qm = 0,
provided the dqr are so chosen that
(13) 01,30! + • • • + Om^qm = 0, s = 1, • • • , /*.
332 MECHANICS
Under these circumstances a necessary and sufficient condi-
tion for equilibrium is that
(14) <?>/! + •• - +Q'mdqm = 0,
provided the 8qr satisfy (13).
That the condition is necessary appears from the fact that
(10) is true, and hence
(15) (Q[ + <#) «& + ••• + (<& + Qi) «?* = 0
for all values of the dqr. If the 5qr satisfy (13), it follows that
(12) is true, and (14) now follows.
Suppose conversely that (12) and (14) hold when the 8qr are
subject to (13). Then the system is in equilibrium. Suppose
the statement false. From (12) and (14) it follows that (15) holds,
provided (13) is true, and hence from (7) it follows that
(16) i)[r]r5<yr = 0,
T~.\
provided (13) holds. Lot
(17) qr = cr, r = 1, • • • , m,
initially. Then not all the cr are 0. Now,
(18) dqr = cr, r = 1, • • • , m,
is a system of values satisfying (13). For, on differentiating (8)
with respect to t and then setting t = /0, qr = 0, these relations
follow, namely :
0>lsC\ + ' ' ' + OmsCm = 0, S = 1, • • • , (JL.
Consequently (16) holds for those values of cr. Now,
(19) ' T = ^A^qaqft.
a, ft
Hence
f)T
— = Air(ji + ' ' ' + Amrqmj
v(]r
d BT .... , A .. . . . ,. . ,
^ -QT = Air q\ + - - • + Amr qm + terms in (ql9 • - • , qm),
and so initially
m m
^ [T]rdqr = ^ MirCj H + Amrcn)cr = ^ Aapcacft.
»•-! r=l a,/3
LAGRANGE'S EQUATIONS. VIRTUAL VELOCITIES 333
But hero is a contradiction, since (19), being a positive definite
quadratic form, can vanish only when all the arguments are 0.
We can state the result as a
THEOREM. A necessary and sufficient condition that a dynamical
system, the motion of which is governed by the equations :
_rf d_T __ 8T
(20) dt'dqr Wr~V" r-1'"->m;
«i«7i + • • • + Om. (7m = 0, « = 1, •••,/*< m,
where the rank of the matrix of these last p, equations is /*, be in
equilibrium and at rest, is that qr = 0, r = 1, • • •, m, and that
(21) Qi«ffi + • • • +Qmdqm = 0,
where. 5q}, - - • , dqm are subject to the condition:
(22) a\»8qi + • • • + am5<$tfw = 0, » = 1, ••-,/*.
7T is a homogeneous positive definite quadratic form in
//, in particular,
(23) Qr = Qr + Qr*, f = 1, • • • , m,
and ?/ # is known that
(24) Q*fy, + • • • + Q*ndqm = 0,
where dql9 • • • , 5gm are subject to (22), ttew (21) can 6e replaced by
(25) QlSft + • - • + Q'mdqm = 0.
19. Small Oscillations. Two equal masses are knotted to a
string, one end of which is made fast to a peg at 0. Determine
the motion in the case of small vibrations. Here,
T = ^ (202 + ^2 + 2^ cos fc, - 6)),
U = mga(2 cos 0 + cos <p).
Since 6, d, <p, <p arc small, these functions can be re-
placed by the approximations :
. 9X FIG. 146
const.
U = - mga(Q* + ^) +
334 MECHANICS
Equations (1) arc typical for an important class of problems
in small oscillations of a system about a position of stable equi-
librium, the applied forces being derived from a force function,
(7. Let T and U both be independent of t\ let qr = 0 for
r = 1, • • • , m, be the position of equilibrium, and let T, U be re-
placed by their approximate values when qr, qr are all small. Then
(2) T = V dra qr<ls
r.s
r, s = 1, - • • , ra, where the coefficients or, = aar, br, = b,r are
constants and each of the quadratic forms is definite.
Lagrange's Equations now take the form :
(3) Orltfi + ' ' ' + Clrmqm = ~ (&rl<7l + • • • + &rm?m),
r = 1, - - - , m.
*To integrate these equations, it is convenient to introduce new
variables as follows. It is a theorem of algebra * that by means
of a suitable linear transformation with constant coefficients :
(4) qr = »riq( + • • • + Mmrf,, r = 1, - • • , ro,
the quadratic forms (2) can each be reduced to a sum of squares :
T = q(* + • • • + q'm\
U = - nt2 ft'2 - • - - - nm2 ^2.
Lagrange's Equations now become :
«g $-V* r-l,...,».
Their integrals take the form :
(7) q'r = Cr cos (nrt + 7r), r = 1, • • • , m.
Returning to the original variables qr, we find as the general
solution of Equations (3) the following :
(8) qr = CiMri cos (nj + TI) + • • • + Cmfirm cos (nmt + 7m),
r = 1, • - • , m,
where the Cr, yr are the 2m constants of integration.
* Bocher, Higher Algebra, Chap. 13.
LAGRANGE'S EQUATIONS. VIRTUAL VELOCITIES 335
In this result, complete as it is in theory, there appear, how-
ever, the coefficients /*r, of the linear transformation (4). These
can be determined by the following consideration.
Let Cr = 0 in (8) when r 9* s; and let C8 = 1. Thus we have
a special solution :
(9) q? = \r cos (nt + 7),
where Xr = /ir« and n = na, 7 = y8. Substitute q? in (3) :
(10) (bn - n*arl) Xi H ----- h (bm - n2arm) \m = 0,
r = 1, • - - , m.
A necessary condition that (9) be a solution is, that the m linear
Equations (10) admit a solution in which the X/s are not all 0.
Hence the determinant of these equations must vanish :
(ii)
n2an
=o.
6mm ~ n2a
If the nr2 are all distinct, they form precisely the m roots of
this equation in n2. Moreover, each n? leads to a unique deter-
mination of the ratios of the X's through the m Equations (10),
and our problem is solved.
It may happen that k of the roots n2 of (11) coincide. In that
case, k of the X's can be chosen arbitrarily, and so we still have k
linearly independent solutions (9) corresponding to such a root
n2. More precisely, let n2 be a multiple root of order fcj ; n22,
a multiple root of order fc2; etc. Let n2 be set = n^ in Equa-
tions (10). Then, of the unknown \lt • • • , Xm, it is possible to
choose a certain set of ^ arbitrarily, and then the rest will be
uniquely determined. Let all but one of these A^ X's be set
= 1. Thus we get ^ sets of (Xx, • - - , Xw), and each set gives a
solution of Equations (3). Moreover these solutions are obvi-
ously linearly independent. Proceeding to n22 we determine
in the same manner fc2 further sets of (Xn .• • • , Xm), each set giv-
ing a solution of (3) ; these solutions are likewise linearly inde-
pendent of one another and also of the earlier solutions. And
so on, to the end. Thus in all cases the roots of (11) lead to m
linearly independent solutions (9).
The variables q'r are known as the normal coordinates of the
problem. Each is uniquely determined, save as to a factor of
336 MECHANICS
proportionality, when the roots of (11) are distinct. But in
the case of equal roots, an infinite number of different choices
are possible.
EXERCISE
Carry through the example given at the beginning of the para-
graph.
Ans. Two sets of linearly independent solutions are the
following :
f 0j = cos fat + 7j), f 02 = cos fat + 72) ;
i ^ = A/2 cos fat + 7i), I 0»2 = ~ ^2 cos fat + 72),
where
n* = (2 - V2) 2 n22 = (2 + \/2) ?-
Cv C*
The general solution is :
0 = <?! cos (n^ + TJ) + C2 cos fat + 72),
tf? = C^A/2 cos (n^ + TI) - C2V2 cos fat + 72).
EXERCISES ON CHAPTER X
1. A smooth wedge rests on a table. A block is placed on
the wedge, and the system is released from rest. Determine
the motion.
2. Two billiard balls are placed one on top of the other, on
a rough table, and released from rest, slightly displaced from the
position of equilibrium. Determine the motion.
3. A uniform rod is pivoted at one end, and is acted on by
gravity. Will it move like a spherical pendulum ?
4. A uniform rod of length 2a and mass 3m can turn freely
about its mid-point. A mass m is attached to one end of the
rod. If the rod is_rotating about a vertical axis with an angular
velocity of \/2ng/a, and so released, show that the heavy end
will dip till the rod makes an angle of cos~1(Vn2 + 1 — ri) with
the vertical, and then rise again to the horizontal.
5. Obtain the equations of the top from Lagrange's Equations.
6. Determine the motion of a top whose peg, considered as
a point, slides on a smooth horizontal plane.
7. The same question when the size of the peg is taken into
account.
LAGRANGE'S EQUATIONS. VIRTUAL VELOCITIES 337
8. The ladder of p. 322, the initial position being oblique
to the line of intersection of the wall and the floor.*
9. Two equal rods are hinged at their ends and project
over a smooth horizontal plane. Determine the motion.
SUGGESTION. Take as coordinates (1) the x, y of the centre
of gravity ; (2) the inclination 6 of the line through the centre of
gravity and the hinge ; (3) the angle a between this line and either
of the rods.
Two of Lagrange's Equations control the motion of the centre
of gravity. A third expresses the fact that the total moment
of momentum with respect to the centre of gravity, is constant.
And fourthly there is the equation of energy, f
10. A rough table is rotating about a vertical axis. Study
the motion of a billiard ball on the table, assuming that there is
no slipping.
11. The same problem with slipping.
12. Work the problem of § 19, p. 333, when the particles are
not required to move in a vertical plane.
13. Two equal uniform rods arc hinged at one of their ends,
and the other end of one rod is pivoted. Find the motion for
small oscillations in a vertical plane.
14. The same problem when the rods are not restricted to
lying in a plane.
16. A uniform rod is supported by two strings of equal length,
attached to its ends, their other ends being made fast at two
points on the same level, whose distance apart is equal to the
length of the rod. A smooth vertical wire passes through a
small hole at the middle of the rod and bisects the line joining
the fixed points. Determine the motion.
16. If in the preceding question the wire is absent, study the
small oscillations of the rod about the position of equilibrium.
17. A bead can slide on a circular wire, no external forces
acting. Determine the motion in two and in three dimensions.
Begin with the case of no friction.
* Routh, Elementary Rigid Dynamics, p. 329.
t Appell, Mtcanique rationelle, vol. ii, chap. 24, § 446. Many other problems
of the present kind are found in this chapter.
CHAPTER XI
HAMILTON'S CANONICAL EQUATIONS
1. The Problem. The problem of this chapter is the deduc-
tion of Hamilton's Canonical Equations :
dqr_m dpr__W i ...
dt " dpr' dt dqr' ' ' '
from Lagrange's Equations :
d dL dL n t
:r; Q~ -- ^"~ = 0, r = 1, • • • , m.
cftd</r dqr
The transition is purely analytical, involving no physical con-
cepts whatever, and for that reason it is well to set the theorem
and proof apart in a separate chapter.
The problem can be stated as follows. We start out with a
Lagrangean System. Such a system is defined as a material
system which can be located by means of m generalized coordi-
nates qly • • • , qm and whose motion is determined by Lagrange's
Equations. If we set qr = Kr, these go over into the 2m equa-
tions :
A)
d ML _
dt dKr dqr
r = 1, < • • , m, where
(1) L = L(ql9 • • ' , tfm, *i, • ' • , Km, 0 = i(0r, *r, 0
is a function of the 2m + 1 independent variables #/., *r, <.
The function L is called the Lagrangean Function. In case
there is a work function,
L is given by the equation :
(2) L = T + U,
338
HAMILTON'S CANONICAL EQUATIONS 339
where KT = qr and
T = T(qr, ?r, 0
is the kinetic energy. In any case, the Hessian Determinant,
the Jacobian :
/«\ d(L1? • - • , Lm)
(6) 0(*i,---,O'
where
shall not vanish.
The (2m + l)-dimensional space S2m+i of the variables (qr, *r, t)
shall be transformed on the (2m + l)-dimensional space R2m+i
of the variables (qr, pr, t) by means of the transformation :
/ M \ 8L t
(4) Pr = ^-, r = 1, • • • , m,
the f/r going over individually into themselves. The system of
2m differential equations of the first order in the 2m dependent
variables qr, Kr, namely, Equations A), thereby goes over into
a system of 2m like equations in the 2m dependent variables
qr, pr. These last equations are the following:
fr. dqr 811 dpr 8H -
(5) -*=& -*=-*? r-1'-"-^
where II = H (qr, pr, t) is defined by the equation :
(6) H = I) prKr ~ L,
r=l
the Kr being functions of (qr, pr, 0 defined by (4).
This is the theorem which is the subject of this chapter and to
the proof of which we now turn. The converse is true under
suitable restrictions.
2. A General Theorem. Let F(xl9 • • •, xn) be any function,
continuous together with its derivatives of the first two orders,
and such that its Hessian Determinant, the Jacobian :
<"
Let a transformation, T, be defined by the equations :
T: r-£ r
340 MECHANICS
Let G (ylt • • • , yn) be defined by the relation :
(2) G(yi, • • • , j/n) = J£ xryr - F(xl> ••-,xm)>
r=l
where xl9 • • • , xn are the functions of ylt • • • , yn defined by
1. Then
(3) xr-%, r- !,...,».
For, differentiate (2), regarded as an identity in the independent
variables ylt • • • , yn :
dxs
The right-hand side, by the equation^ defining T, reduces to xr,
and the proof is complete. Furthermore,
For, on performing first the transformation T, then the transfor-
mation T~~lj the result is the identical transformation. Hence
d(xu - - • , xn) d(yl9 • • • , yn) '
or:
• • • , G.) 1
3(*i, • • • , *n)
In particular, then,
/rv d(Gi. • • • , Gn) r$(ri, • • • , /n)"T~
f^i r= I — I •
These results may be stated in the following theorem.
THEOREM I. Let F(xly • • • , xn) be a function satisfying the
condition:
Perform the transformation :
dF
T • 11 — r = 1 • • • w
^ . 2/r - , r 1, , n.
HAMILTON'S CANONICAL EQUATIONS 341
Let G(ylt • • • , yn) be defined by the equation:
n
(6) G(yly • • • , y») = 5) xryr - F(xl9 • • • , xn),
r=l
where xr is determined as a function of (ylt • • • , yn) by the inverse,
T~l, of T. Then the inverse of T is represented as follows :
T~l • x - — r = 1 • • • n
1 . xr- ^, r l, , n.
Moreover,
d(Gi, ' ' ' > Gn) , Q
^(2/1, • • -,2/n)
In particular,
The identical relation can be written in the symmetric form :
(7) F(xlt •••,xn)
r-1
where
3F dG
and the Hessian Determinants of F, G are ^ 0.
We now proceed to a second theorem, which is of importance
in the applications of the results of this chapter in mechanics.
THEOREM II. // F, G are defined as before, and if each depends
on a parameter, % , the relation
(8) J'ft; a?,,..-, x^ + Oft; yu • • • , 2/n) = J xryr
T=l
being an identity, because of T or T~l, either in the n + 1 arguments
(£ ; #!,•••, Xn) or in the n + 1 arguments (£ ; yly • • • , yn
Let (£ ; «!,•••, XB) be the independent variables in (8). Then
=
yr ft & ft'
But
and the proof is complete.
342 MECHANICS
3. Proof of Hamilton's Equations. We start out with the
Lagrangean Function L(qr, KT, t), which fulfils the condition :
'"' L
w n(if • • • if } 9
V \Kli ) Km/
and make the transformation :
(2) pr = £-, r = 1, • • • , m.
The Hamiltonian Function // (qr, pr, f) is then defined by the
equation :
(3) L + H = % prKr.
T
If, then, we set xr — Kr, yr = pr, and regard the qr and t as
parameters, all the conditions of the theorems of § 2 will be met.
It follows, then, that the inverse of (2) is given by the equation :
and furthermore that
(5; — + g^ ~ t r —,-•-, m.
It is also true that
dt dt '
although this relation is not important for our present purposes.
Turn now to Lagrange's Equations :
dq,
~dt
d <9L _ cuu
dt d*r dqr
The first of these, combined with (4), gives :
m dqr-m r-\ ••• m
\OJ • , — ~ , / — 1,, III.
From the second, combined with (2) and (5), we infer that
(9) -37- = — -r-— , T = 1, • • • , m.
HAMILTON'S CANONICAL EQUATIONS 343
But these are precisely the Hamiltonian Equations (5) of § 1 :
dqr_9H dpr__3H
dt ~ dpr> dt - dqr> T ~ A' ' ™'
which we set out to establish.
The mathematical converse is simple. Given Equations (10)
with the condition
Equations (4) define a transformation, and then L is defined
by (3). Then (2) arid (5) follow from the theorems of § 2. And
now the first of Equations (10), combined with (4), gives
dqr .
-£ = «„ r = l,...,m.
The second equation (10), combined with (2) and (5), leads to
the equation :
d dL _ dL _ 1
~7i ~n — ~^ t f — 1> " " " ) W"
dt 8Kr dqr
Thus wo arrive at Lagrange's Equations (7).
We see, then, that a knowledge of the function H is sufficient
for a complete mathematical formulation of the motion. But
what can we say of the physical meaning of // in the general
case? There is an important restricted class of cases in which
the definition is simple. Suppose there is a force function U
depending on qr, t alone, and furthermore that the kinetic en-
ergy T is a positive definite homogeneous quadratic form in the
(/!,-•-, qm> Let L be defined by the equation :
(11) L = T + U.
Let Kr = qr> Then
The transformation :
8L
*-aTr
now becomes :
344 MECHANICS
Thus
§?,«,- 2 «,j£-2T.
r = l r c/"r
Hence
(12) ff = J) pr*r -L = T-U.
T
Still more specially, if neither T nor [7 depends on t, then //
becomes the total energy of the system. That H is here constant
along any given path appears as follows. We have in the general
case the relation :
dH m
as is seen at once by differentiating :
*H = VM^: + T?dH_dpr , BH
dt $, d(Ir dt ^ dpr dt "*" dt '
and then making use of Hamilton's Equations. But in the gen-
eral case dPI/dt ^ 0, and so H is not constant along an arbitrary
path. If, however, H does not contain t, then 3H/dt = 0 and
since now dH/dt = 0, we have :
(14) H = h.
CHAPTER XII
D'ALEMBERT'S PRINCIPLE
1. The Problem. The general problem of Rational Mechanics,
so far as it relates to a system of particles, can be formulated :
i) in terms of the 3n equations given by Newton's Second Law
of Motion :
A) miXi = Xi, rmi/i = Yiy w»z» = Zit i = 1, • • - ,n ;
ii) in terms of further conditions equivalent to 3n relations
between the 6n + 1 variables (xi} yi} ziy Xif Yi, Zi, t). A postu-
lational treatment of these conditions will be found below in
Appendix D.
Two extreme cases may be mentioned at the outset. First,
each variable Xi, Yiy Zif may be given as an explicit function
of the Xi, yi, Zi and their first derivatives with respect to the
time, and t :
X, = */ fa, yt, zif Xi, yiy z{, t)
YJ = */ fa, yiy Zi, ±i, yi} Zi, t)
Thus A) reduces to a system of simultaneous differential equa-
tions for determining Xi, yiy Zi as functions of the time, and with
the solution of this problem the determination of the Xi, Yi, Zi
is given by substitution.
Secondly, at the other extreme, the path of each particle and
the velocity of the particle in its path may be given. Thus
#ij y\> Zi become known functions of t, and again the Xi, Ft, Zi
are found by substitution.
Between these two extremes there is a class of problems in
which constraints occur which can be eliminated by a general
principle due to d'Alembert. We shall not attempt to give
a general definition of " constraints/' for no such definition exists ;
but we can formulate a requirement which embraces the ordinary
cases that arise in practice. Let 8xit dyi, 6zt be any 3n quantities
345
346 MECHANICS
whatsoever. Then it is seen at once from A) that the following
equation is true :
(1) 2) (m& ~ x*> *x* + (m*y* ~ r») fy< + (m<*< - z*> dZi = °-
«-i
This equation is sometimes referred to as the General Equation
of Dynamics.
Now it may happen that the force X^ F», Zi can be broken
up into two forces :
(2) Xi = XI + X!, Yi = Y( + Yl Zi = Z( + ZI
where the two new forces, namely, the X'it F«, Z\ and the
X*t F|*, Z*, are simpler than the old for the following reasons.
i) The X't, Yi, Z{ are either explicit functions of the x^ ?/», Zi,
±i, yif Zi, t or they involve in addition a restricted set of un-
known functions arising from forces which are not given as func-
tions of these &n + 1 variables.
ii) The Xf, Yf, Zf have the property that
(3) 2) XfdXi + Yfdyi + ZfdZi = 0
!--=!
for all values of the 6xi, 5t/», dzt which satisfy the \i equations :
n
(4) ^/Aiadxi + Biadyi + CieSz^ 0, a = 1, • • • , M,
<=1
where the coefficients are given functions of the 6n + 1 variables
Xiy yij Zi, Xif j)it Zi, t, and the rank of the matrix :
•"•11 " * " -"nl
(5) '
is /x ; and conversely.
By means of- Equations (4) the 3n quantities X*, Yf, Z* can
be expressed in terms of /x unknowns as follows. Multiply the
a-th Equation (4) by an arbitrary number \a, and subtract the
new equation from (3). Thus
(6) 2) (ft -
i-l a-l
D'ALEMBERTS PRINCIPLE 347
Now, in Equations (4), a certain set of 3n — n of the &c», By*, dZi
can be chosen at pleasure and then the remaining /* of these
quantities will be uniquely determined, for at least one At-rowed
determinant from the matrix (5) does not vanish. It follows,
then, that \, • • • , X^ can be determined uniquely from a suitable
set of ju equations chosen from the 3n equations :
H M f-
(7) X* — 2^ Aia\a, Yi
a= 1
And now the remaining 3n — /z Equations (7) will be satisfied
by these values of the X's. For Equation (6) has become an
equation in which only those terms appear for which 6x,, 8yi, 6z»
are arbitrary, and hence their coefficients must each vanish.
Equations A) can now be written in the form :
(8) mix* = X( + 5) Aia\a, nnyi = Y( + J fiiaX«,
where the Xt, • • • , X^ have come to us as linear combinations
of /z suitably chosen X*, Y*9 Z*. On the other hand, they appear
in Equations (8) merely as /z unknown functions, which can be
determined from /* of these equations and then eliminated from
the remainder.
Virtual Work. To put first things first was never more impor-
tant than in the statement of d'Alembert's Principle. The 3n
quantities 8xi, Syiy dZi are to begin with 3n arbitrary numbers,
and we then proceed to restrict them by the equations (4). Never-
theless, whatever values they may have, they determine by defi-
nition a virtual displacement of the system of points (Xi, t/», 2,-),
and the quantity :
Ws - 2) X{ dxi + Yi By* + Z, dZi
«=i
is by definition the virtual work corresponding to this virtual
displacement. Thus Equation (3) says that the force X*t Y*, Z*
is such that it does no virtual work when the virtual displacement
is subject to the conditions (4).
348 MECHANICS
D'ALEMBERT'S PRINCIPLE FOR A SYSTEM OF PARTICLES. Given
a system of particles, the motion of which is determined in part by
Equations A). The discovery of an analysis of Xiy Yi, #»• by (2)
and of the most general virtual displacement 5Xi, dy^ 6zif whereby the
virtual work of the force Xf, Yf, Z* vanishes, this virtual displace-
ment being expressed by (4); finally the elimination of dXi, dyif dzir
and X*j Yf, Z*, as above set forth, whereby 3n — n equations free
from these unknowns result; — this is the spirit and content of
d 'A lembcrt's Principle .
This enunciation of the Principle does not represent its historic
origin, but rather its interpretation in the science today ; cf .
Appendix 1).
2. Lagrange's Equations for a System of Particles, Deduced
from d'Alembert's Principle. Let the coordinates x», yi, z» of the
system of particles considered in § 1 be expressible in terms of m
parameters and the time :
(D
i?m, 0
, <im, 0
i tfm, 0
where (ql} • • • , qm) is an arbitrary point of a certain region of
the (</!, • • • , gm)-spacc and the rank of the matrix is m. Let
Then by the purely mathematical process of differentiation and
substitution Equation (1) of § 1 yields :
dT 8T
ZdtWr'Wr
where
and
(4)
Now, the m quantities 6ql9 • • • , dqm are wholly arbitrary. Hence
the coefficient of each term in (3) must vanish, and so we arrive
D'ALEMBERT'S PRINCIPLE 349
at Lagrange's Equations in their most general form for a system
of particles :
,,, d 8T 8T
Here, no restriction whatever is placed on the forces X», Ft,
nor is the number of qr'$ required to be a minimum.
In an important class of cases which arise in practice,
(6) x, = x'< + xi F, = F; + 17, Zi = z; + z«*,
where
Hence
r = 1, • • • , TW.
Equations (5) now become Lagrange's Equations for this restricted
case. The cases of constraints that do no virtual work are here
included. Cf. further Appendix D.
3. The Six Equations for a System of Particles, Deduced from
d'Alembert's Principle. Let the system of particles of § I be
subject to internal forces such that the action and reaction between
any two particles are equal and opposite and in the line through
the particles :
X*
ij __
~
And let any other forces X'tj Y[, Z( act. Let
dXi = a + pzt - yiji
fyi = b + yxi — aZi
dZi = C + ayi - pXi
where a, 6, c, a, /?, 7 are six arbitrary quantities. Since the
internal forces destroy one another in pairs, and likewise, their
moments, Equation (1) goes over into the following:
350 MECHANICS
0 = a J (m,£< - X'() + b 2 (»*<#< - F|) + c 2 (»»<*< - Zf)
18 2) (
x
Now, set any five of the six quantities a, 6, c, a, #, 7 equal to 0,
and the sixth equal to 1. Thus the six equations of motion,
from which the internal reactions have been eliminated, emerge :
(1)
n n
) m*g» = 2J
In vector form these equations appear as the Equation of Linear
Momentum :
^ = F
and as the Equation of Moment of Momentum :
dt
We have used d' Alembert's Principle to deduce a set of necessary
conditions. These are not in general sufficient, because the fore-
going choice of dxif 8yiy dzi is not in general the most general one.
4. Lagrange's Equations in the General Case, and d'Alembert's
Principle. Consider an arbitrary system of masses, to which
Lagrange's Equations, on the basis of suitable postulates, apply :
/i\ d dT dT n i
(1) 5 a£ -«£-«" r-i,..-,*.
If we set by way of abbreviation :
(2) ddT _dT _
(2) ~ [ ^
D'ALEMBERT'S PRINCIPLE 351
then
(3) 2 ([71 - Qr) Sqr = 0,
where dql9 • • • , dqm are any m quantities whatever. It may
happen that Qr can be expressed in the form :
Qr = Q'r + Q],
where Q'r is for some reason simpler than Qr and where, moreover,
(4) Qffy, + • • • + Q;s?m = 0,
provided
(5) daidqi + ' • ' + a>am8qm = 0, a = 1, • • • , /I,
the rank of the matrix :
being /*. By reasoning precisely similar to that used in § 1, it is
seen that the Q* can be represented in the form :
M
Qr — 2L a<*r^a, r = 1, • • • , W,
where the X« can be interpreted physically as certain linear com-
binations of a suitable set of /* of the quantities Q* .
Moreover, Lagrange's Equations take on the form :
d dT dT , A _
where now the Xa are thought of as unknown functions, which
can be determined by /* of these equations and then eliminated
from the remaining m — n equations.
Virtual Work. In all cases the expression
can be interpreted as the virtual work done on the system by the
forces which correspond to the Qr. In particular, then, the
condition (4) means that the virtual work of the forces which
lead to the Q* is nil, provided that the virtual displacement cor-
responds to the condition expressed by Equations (5).
352 MECHANICS
5. Application: Euler's Dynamical Equations. Consider a
rigid body, one point, 0, of which is fixed, and which is acted on
by any forces. Its position may be described geometrically in
terms of Euler's Angles, Chapter VI, § 15 :
(1) Qi = 0, q2 = t, q* = <p.
Its kinetic energy is, by Chapter VII, § 6 :
(2) T = -i (Ap* + Bq* + Cr2),
where
p = — \j/ sin 6 cos <p + 6 sin <p
(3) q = \l/ sin 6 sin (p + 6 cos <p
r = ^ cos 0 + ^
By d'Alembert's Principle, § 4 :
where all three 5^r are arbitrary. Let 5q{ — 0, dq2 = 0,
Compute
The value is seen at once to be :
On the other hand, Q3 can be computed as follows. Denote
the vector moment of the applied forces, referred to 0, as
M = La + Mf3 + Ny.
Then the virtual work corresponding to the virtual displacement
(5qly 8q2) 5g3) = (0, 0, dqz) is seen to be :
Hence Q3 = N, and we find :
Thus one of Euler's Dynamical Equations is obtained, and
the other two follow by symmetry, through advancing the letters
cyclically.
D'ALEMBERT'S PRINCIPLE 353
The reader will say : " But this is precisely the same solution
as that given earlier by Lagrange's Equations, Chap. X, § 16."
True, so far as the analytic details of the solution go ; and this is
usually the case with applications of d'Alembert's Principle. It
is the approach to the problem through the General Equation
of Dynamics, § 1, which here yields (4), and the concept and use
of virtual work, that brings the treatment under d'Alembert's
Principle.
6. Examples. Consider the problem of the ladder sliding
down a smooth wall; cf. Fig. 88, p. 147. Let us regard this
problem as the motion of a lamina, moving in its own plane.
AH the generalized coordinates of the lamina we may take the
coordinates of the centre of gravity :
?i = x, q2 = y, qz = 8.
Then
(1) T = %M(fr + p) + pf F 02,
where k is the radius of gyration about the centre of gravity.
By d'Alembert's Principle,
(2) 2 ([T\r - Qr) 8qr = 0.
r-l V '
In the present case,
Q^ = S5x, Q26</2 = (R - Mg) by,
Q3 5<73 = a (S sin 0 - R cos 6) d6,
and thus Qr is determined. Let
Qr = Q; + QM
where
Q* = s, Ql = R, Q3* = a OS sin 9 - R cos 8).
Now, x, y, 6 are connected by the relations:
(3) x = a cos 0, y = a sin 6.
If, then, we subject dx, dy, SO to the corresponding relations:
8x = — a sin 6 60, 6?y = a cos 0 60,
we see that
(4) Qr^i + Ql 5<72 + Q,*«ff, = 0.
354 MECHANICS
Thus the virtual work of the forces Q?, corresponding to such
a displacement, is seen to be nil, and so Equation (2) is replaced
by the simpler equation :
(5) i; (m, - $W == o.
r-l V '
Hence
M^j(- a sin 0) 66 + (^^f + Mg) a cos 6 69 + MW — 66 = 0.
On replacing these second derivatives of x and y by their values
from (3) a differential equation in the single dependent variable
0 is obtained :
and this determines the motion.
Rough Wall. Suppose, however, the wall is rough; Fig. 145,
p. 323. Equations (1) and (2) still hold. But now
Qi*fc = (S ~ »R) «x, Q2 6q2 = (R + »S - Mg) 8ff,
Cs ^3 = a [S (sin 6 + /z cos 6) + R (/x sin 6 - cos 0)] 66.
Let
Qr = Q; + Q;,
where
Qr = S - /»«, Q* = B + MS,
Q* = aS(sin 0 + M cos 0) + aR(n sin 0 — cos 0).
The values of 5^D 6q2, 8qz which make the virtual work of the
force Q* vanish :
QI^ + c;»?i + cr«?i = o,
are found by making the coefficients of B and S zero in this last
equation :
—/z&7i + 6q2 + a (M sin 6 — cos 0) 6qz = 0
. .. 4- a (sin 0 + M cos 0) 6q3 = 0
Hence
f (1 + M2) fyi = a [— (1 - /i2) sin 0 - 2M cos 0] 6q^
I (1 + M2) 5^2 = a [- 2M sin 0 + (1 - M2) cos 0] 6q3
D'ALEMBERT'S PRINCIPLE 355
Equation (5) now becomes :
M^jSq, + (M^JL + Mg) Sq, + Mk^dq, = 0,
and it remains only to substitute the values of 5qlf 8q2 from (7),
and the values of x, y from (3), and reduce. The result is :
((1 - M2) a* + (1 + M2) *») JJ +
ag [2/x sin 6 - (1 - /*2) cos 0].
The virtual displacement (dqlt 5g2, dq^) which here led to the
elimination of the unknown reactions R, S was not one which
in any wise conformed to the " constraints " in the sense of the
floor and the wall. If we replace R and /z# by their resultant
and draw a line L± through the bottom of the ladder perpendicular
to it, and then do the same thing at the top of the ladder, thus
obtaining a line L2, the above virtual displacement corresponds
to an actual displacement in which the bottom of the ladder is
moved along Lt and the top along L2.
CHAPTER XIII
HAMILTON'S PRINCIPLE AND THE PRINCIPLE
OF LEAST ACTION
1. Definition of 8. A new and independent foundation for
Mechanics is given by Hamilton's Principle and certain other
Principles of like nature. An integral, for which Hamilton's
Integral :
«j
V + t/) dt,
is typical, is extended along the natural path of the system, and
then its value is considered for a neighboring, or varied, path.
The Principle asserts that the integral is a minimum for the
natural path, or at least that the integral is stationary for this
path, i.e. that its variation vanishes :
(T + U) dt = 0.
It is to the treatment of this subject that we now turn. Obvi-
ously we must begin by defining what is meant by a varied path
and by a variation 8.
Let F(xl9 • • • , xn> x[, • - • , x'n, u) be a function of the 2n + I
variables indicated. Here, (xl9 • • • , xn) shall lie in a certain
region R of the n-dimensional space of the variables (xly • • • , xn) ;
the variables x[9 • • • , x'n shall be wholly unrestricted ; and u
shall lie in the interval : a g u g b. The function F shall
be continuous, together with its partial derivatives of the first
and second orders.* Let
*As regards assumptions of continuity, we lay down on GO and for all the
requirement that whatever arbitrary functions are introduced shall be continuous,
together with whatever derivatives we may wish to use, unless the contrary is
stated.
For an introductory treatment of the Calculus of Variations cf. the author's
Advanced Calculus, Chap. XVII.
356
HAMILTON'S PRINCIPLE. LEAST ACTION 357
C : Xi = Xi(u), a ^ u g 6, i = 1, • • • , n,
be a path lying in R. Let
, _ dxj(u)
Xl~ du '
Thus a path T in the (2n + l)-dimensional space of the arguments
of F is determined.
By a varied path, F', is meant the following. Let Cf be a curve
in R defined by the equations :
C' : Xi = Xi(u, e), a ^ u g 6, f = 1, • • • , n,
where
a?t(w, 0) = Xi(u),
and € is considered only in a region for which | e | is small. De-
note partial differentiation with respect to u by d. Let
, e) // \ -i
• •
' ~~ du ' ' , , ,
be chosen as the values of the #(,••-, #£. The curve F' in the
(2n + l)-dimcnsional space is what is meant by a varied curve.
The variation of #,-, or &c,-, is defined by the equation :
Since xl(u, e) is any function that conforms merely to the general
requirements of continuity, we see that
dxi = IH(U), i = 1, • • • , n,
is a wholly arbitrary function, restricted only by the above re-
quirements of continuity.
The variation of x't, or bx( is not, however, arbitrary, but is
defined by the equation :
dude >o
Thus
Hence
. A<
-r--oa;» = o-j —
du du
It is now natural to lay down the further definition :
(4)
358 MECHANICS
Definition of SF. By the variation of F (a;,-, x't, w) is meant :
8» "=(£)..,
where Xi and xj on the right-hand side are set equal to x» (u,
and ZI'(M, e). Hence
(6) ^-|(J>' + I
It is obvious that
8(F + $) = 8F + d$;
and also that
where
, •••,«*,«,•••, x'n, u), k = 1, • • • , m.
Finally, the definition :
(7) ddF = ddF,
corresponding to the theorem :
/ON *dF d8F
(8) a5?-"5T
And similarly,
6 b
(9) J FSd* = *
a a a
The dependent variables x^u), • • • , xn(u) play a r61e in the
foregoing definitions analogous to that of the independent variables
in partial differentiation. But the analogy holds only up to a
certain point, and to assume it beyond theorems like the above
formulas which we can prove, has led to confusion and error in
physics.
Variation of an Integral. Consider the integral :
b
/f
F(Xi, u ' ' > XH9 X19 • • • , Xn, U) du,
HAMILTON'S PRINCIPLE. LEAST ACTION 359
taken along the path F. By 8J is meant the following : — Extend
the integral along the path F'. Thus a function «/(e) is defined.
and now, by definition :
(10) ^
It follows at once as a theorem that
(ID
The integral is said to be stationary for a particular path F if
b
8 CFdu = 0,
a
no matter what functions 8x* = tm(u) may be. The condition
is readily obtained in case dxi is restricted to vanish for u = a
and f or u = 6 :
(12) dxi | „.« = in (a) = 0; &e< | Ussb = ^(6) = 0.
For:
d ( $F ^ \ „ dF f i d ^ *
d^ V^J 5X V " ~d$ 8Xi + dH W< 8Xi'
Hence
03)
If, now, the integral in question is to vanish for an arbitrary
choice of &C;, it is easily seen that each parenthesis in the inte-
grand of the last integral must vanish, or :
These are known as Euler's Equations.
It is clear that
6 b
(15) d jFdu= j8Fdu.
360 MECHANICS
The limits of integration may be varied, too. Let
a' = ^(a,€), 6' = *(&,€),
where ^?(a, 0) = a, \f/(b, 0) = 6. Let
6'
J (c) = IF [Xi (u, e), x'i (u, c), u\ du.
a'
Then the variation of the integral is defined as before, by (10).
It follows that
6 b
(16) dCFdu = C6Fdu + F(Bi9Bfi9b
a a
where
. A* = z«(o), A'< = x£(a), £<
EXERCISE
Since
l/, 6)
it follows (under the ordinary hypotheses of continuity), on let-
ting e approach 0, that the right-hand side approaches
f».
du
The left-hand side approaches 8&. Hence
,eM> d „
5 -j- = -7- 6$.
aw aw
Thus Equation (7) is obtained as a theorem, and no new definition
is necessary. Explain the error.
2. The Integral of Rational Mechanics. All that has gone
before merely leads up to the definition of the variation of the
following integral :
ti
(1) j>F(xlt---,xn,xl,-",xn,()dt,
HAMILTON'S PRINCIPLE. LEAST ACTION 361
where the limits of integration may be constant or variable. The
answer would seem to be simple, since t is the variable of integra-
tion and hence the independent variable in each of the functions
Xi = Xi(t).
But the symbol written down as the integral (1) is taken in Physics
to mean something totally different. Let
(2) t = t(u), 0 ^ u
be any function of u such that, in the closed interval (0, 1),
0<f -f.
du
Then the symbol (1) is taken to mean the integral :
(3) / = JF (xlt •••,*„, |f, ..-, |?) t'du,
0
where x[ = dxi/du, and the "variation of the integral (1)" is
understood to be the variation of this last integral. Thus
(4) dJ = Cd(Ft')du,
o
where u is the variable of integration, and the independent vari-
able in each of the functions Xi = Xi(u), t = t (u).
This last variation, (4) : &/, comes under the earlier definition
of the variation of an integral. In particular, Equation (4) may
be written in the form :
8J =
0
In each of these integrals the variable of integration may be
changed back from u to t, and thus
«t <t
(5) SJ = CdFdt+ foddt.
fbFt'du+ CF8tfdu.
362 MECHANICS
The variation of t, namely St, is an arbitrary function of u :
U = T(M),
and
dSt T'(U)
dt t'(u)'
where u is the inverse function defined by (2). Moreover, by
6F in (5) is meant the following :
m » -•
where
/z,'\ t'5x't - xW , d .., d
8 (7) = - ?5 — . «*, = Tu $x<, st = ^ a.
We see, then, that
'
Now, when 2 is the independent variable,
/o\ *- d dXi
(8) 5^* = -ir-
The two formulas, (7) and (8), show that 5i» is not invariant
of the independent variable. Why should it be? Similarly,
the variation of an integral is not invariant of the variable of
integration. Much of the confusion in the literature arises from
losing sight of this fact. The " variation of the independent
variable" is supposed to cover this case. It does so when and
only when it becomes identical in substance with the above
analysis.
3. Application to the Integral of Kinetic Energy. By defini-
tion, the kinetic energy
Hence
n
(1) &T = 5} mi(±idXi + yifoji
i-i
no matter what the independent variable and the dependent
functions may be. If the former is u} then
HAMILTON'S PRINCIPLE. LEAST ACTION 363
(2) 8T = «,[*, 5 () + fr S () + « 6
By Formula (7) of § 2 and the corresponding formulas involving
yiy Ziy this equation becomes :
a, w.
Variation of the Integral :
(4) J =
Let the natural path in space be represented parametrically by
the equations :
Xi = Xi(u), 0 ^ u ^ 1,
and let
The variation of this integral has, by § 2, (5) the value :
(5)
By the aid of (3),
f*Tdt= (*8Tdt+ CTdd
*o to t0
The first term on the right can be transformed by integration
by parts, the integrand obviously having the value :
Hence
*t <i
(7) fdTdt=-Cj? rm (Xi dXi + Hi dyi + 2< 62^) d<
V V <"1
'l
+ 5) wiifofa, + yi% + ^fe{) tl ~ 2 TlTd^.
< . i <o J
364 MECHANICS
Finally, then :
/i ^
(8) 5 CT dt = - / 5) mt (ft fa, +
i:sl
The variations dxi, dy^ bz^ dt are 3n + 1 arbitrary functions,
subject merely to the ordinary conditions of continuity. If,
in particular, we impose on &c»-, dyiy 5zt- the condition that they
vanish at the extremities of the interval of integration, i.e. for
t =• tQj tly then
*, /t <i
(9) Ardt = - f 5) mi(ft fa* + »*«»< + 2*820 * ~ 2 J
/o 'o ~ *o
and
'' /! n
(10) d (*Tdt =- f T m< (ft fa,- + y< fyi + 5f ««0 -
«/ «/ i-i
/O h 0
4. Virtual Work. By the virtual work of the forces Xi, F,-, Zt-,
considered along the natural path :
(1) Xi = Xi(u), yi = y»(w), 2. = ti(u), UQ-^U ^ uly
is meant the quantity :
(2) Ws =
where fai, 5i/i, 5^^ are 3n arbitrary functions of u, subject merely
to the ordinary conditions of continuity.
This quantity is often denoted by dW ; but it is not, in general,
the variation of any function, and so it is better to avoid this
confusion, writing 5W only when W8 is the variation of a function.
6. The Fundamental Equation. Combining Equation (9) of
§ 3 with Equation (2) of § 4 we have :
HAMILTON'S PRINCIPLE. LEAST ACTION 365
(1)
Here the 3n + 1 variations &rt-, 6?yt, 5zt-, 5£ are arbitrary except
that the first 3n of these vanish for t — tQ, ^ ; and the n forces
Xi, Y{, Zi are any forces whatever. If these are the total forces
acting on the particles, the right-hand side of (1) will vanish since
each parenthesis vanishes by Newton's Law, and we shall have :
t\
(2) J(ST + 2T<~ + Wt) dt = 0.
h
Let the total force be broken up into two forces :
(3) Xi = xi + xi Y> = y; + Y;, z, = z; + zi
Then
(4) Ws = Wv + Ws*.
Suppose that W&* vanishes :
(5) W8* = X;txt + Yt8y{ + Ztdz< = 0,
t=i
when the variations dX{, 8yi, dzi are chosen subject to certain
conditions. Then Equation (2) takes the form :
(6)
where now dxt, 8yi, dzi satisfy these conditions, dt being still wholly
arbitrary, and
(7) W9 = 2) XJ
t=i
These conditions usually take the form :
(8)
366
MECHANICS
where Ata, Bta, Cia are functions of x,-, «/<, Zi, xit yf, zit t, and the
rank of the matrix :
An • • • Cni
(9)
* 1 I/A * * * v> n/A
is M. This case includes both the holonomic and the non-holo-
nomic cases. But it must be observed that 8T is in general no
longer, or not yet, the variation of a function.*
Generalized Coordinates. Suppose that the coordinates xt-, yi} Zi
of each mass mi can be expressed in terms of m parameters
q.u ' ' ' i q™ an<i the time :
Xi = fi(q\j • • • > qm, f)
Vi ~ <?i(qu ' • • i q*n, 0
Zi = ^i(q^ • • • , qmj t)
where the rank of the matrix
(10)
is m. Suppose further that Equation (5) is satisfied when
dfi s , , dfi ,
OXi == 7 — OQ'i ~i~ " * * H O^m
^^j ^Q'm
*„. _ ^» ^ . . fa *„
(11)
the 5^, • • • , 8qm being arbitrary. Let
/1O\ /^ ''O ( V ^i I V/ ^2/» I ^
(12) Qr = > ( A< -r— + Yi -r— + /
+ I*-'**.
r = 1, • • • , m.
* The definition of the variation of a function, it will be recalled, is based on
the dependence of the latter on certain arbitrary functions, whose variations may
also be taken as arbitrary. These arbitrary functions are analogous, let us repeat,
to the independent variables in the case of partial differentiation. And so further
assumptions (i.e. postulates or definitions) are needed before dT can again mean a
variation.
HAMILTON'S PRINCIPLE. LEAST ACTION 367
Then
(13) W? = Q! «& + ••• +QmSqm.
On the other hand,
(14) T = T(ql9 • ••,«.,*, •• •,&,,<)
and 5T7 as given by § 3, (2), becomes the variation of this latter
function, where qi(u), •••, qm(u), t(u) are the independent
functions. Equation (6) of the present paragraph now takes the
form:
where dT means what it says — the variation of T — and where
Ws is the Wy of (13).
We will denote this equation as the Fundamental Equation.
It embraces Equation (2) above, for the Xi, t/», z» can always be
taken as m = 3n generalized coordinates.
This equation is sometimes written in the form :
ti
X) C(dT + 5W) dt + 2T 8dt = 0,
where
(15) dW = Qldql + -- +Qmdqm.
Let us see just what this means. First of all, the equation is
true under the hypotheses which led to Equation I. These were,
that the path is the natural path of the system, given by the
equations :
(16) qr = ffr(tO, t = t(u),
where £(0) = t0, t(l) = tlt and the variations
8qr = ir(u), 8t = »/0(w),
are arbitrary functions subject merely to the conditions :
7?r(0)=0, r = 0, 1, ...,m; r?r(l) = 0, r = 1, • • • , m,
368 MECHANICS
and possibly to a further restriction :
I rjr(u) | < A, | jr(u) | < h, r = 0, 1, • • • , m,
where h is a definite positive constant.
Furthermore, in Equation X), 8W = W$ is not in general the
variation of any function of qlt • • • , qm, t. The Qr have definite
values at each point of the natural path, and so are definite func-
tions of u; but they do not in general have any meaning at a
point (qr, t) not on the natural path, nor does dW.
Finally,
(17) ^SfV
where
^
\du du du du \du
The last term in the integral has the value :
/27
0
And now the meaning of Equation X) is this : — If qr and t
are set equal to the functions (16) which define the natural path
and if dqr, 8t are chosen arbitrarily, subject merely to the general
conditions above imposed, Equation X) will be fulfilled. Thus
Equation X) expresses a necessary condition for the motion of
the system — and this in all cases, be they holonomic or non-
holonomic.
Since Equation X) is true for all variations 8qr, 8t, it still repre-
sents a necessary condition when these functions are subject to
any special restrictions we may choose to impose on them. For
example, it may happen that the Qr can be broken up into two
functions :
Qr = Qr + Qr*, r = 1, • • • , m,
such that
QiiQi + • • • + Qlfym = 0,
provided that
a«*i8qi + • • • + Oamfyw = 0, a = 1, • • • , /i,
HAMILTON'S PRINCIPLE. LEAST ACTION 369
where aar = aar (ql9 • • • , qm, qlf • • • , qm, 0> and the rank of the matrix :
is IJL. Here, dW is replaced by
(18)
5W =
+
but only m — /i of the &/r, and dt, can now be chosen arbitrarily.
In what sense is 5T now a "variation"? Emphatically, in no
sense ; for no definition has been laid down which reaches out
to this case, and it is only from a definition that 5T can derive
its meaning. Nevertheless, Equations (17) and (18) continue
to define the values of the terms dT, bW that appear in Equa-
tion X), and thus this equation continues to have a meaning, and
to hold when a certain set of m — /* variations 8qr, and dt, are
chosen arbitrarily.
Force Function. Finally, there may be a force function, U :
(19)
where U is a function of the #,-, y^ z, and t. Thus the Funda-
mental Equation (2) becomes in this case :
II.
where the (Xi, F<, Zi) of (19) is the total force acting on m,-,
provided (/ does not depend on t ; otherwise we must understand
by d(J the virtual variation of U, or :
Again, there may be a function U(ql, • • • , qm> <) such that
in (15)
„ SU
r — 1,
m.
Then
&W = BU
370 MECHANICS
and the Fundamental Equation takes on the same form, II.,
provided U does not depend on t ; otherwise,
'U~W «* + ~'+Wt*"
oq\ oqm
6. The Variational Principle. The variational principle as ex-
pressed by the Fundamental Equation I. of § 5, or even by
Equation II., does not assert that the integral of some function,
or physical quantity, is a minimum, or even stationary :
5 I (something) =0 or Id (something) = 0.
For the integrand is not a variation, in the sense of the Calculus
of Variations; nor are the forces of the problem varied; they
are considered only along the natural path of the system.* The
Principle expresses a necessary condition for the motion of the
system. In the non-holonomic case, the condition cannot be
sufficient, since the first-order differential equations have not
been incorporated into the formulation of the problem.
We turn now to certain further restrictions whereby Hamil-
ton's Integral or an analogous integral does become stationary,
and in fact, in a restricted region, a minimum.
7. Hamilton's Principle. If we set :
t(u, c) = U, tQ^U^ tlt
then
8t s 0
and the Fundamental Equation I. becomes :
(1)
We can now suppress the parameter u since the time is not to be
varied.
* In its leading ideas this treatment was given by Holder, Gottinger Naehrickten,
1896, p. 122. Unfortunately Holder felt impelled to defer to the primitive view of
variations as " infinitely small quantities" in the sense of little zeros, i.e. infinitely
small constants or functions of Xi, ylt z», t. In the foot-notes on pp. 130, 131 the
"neglect of infinitesimals of higher order" renders obscure — in fact, vitiates —
the treatment, so far as clean-cut definitions go. The writer cannot but feel that
the inner Holder would have preferred such a treatment as that of the text, but
that he did not have the courage to break with the unsound traditions of the little
zeros, for fear of losing his clientele.
HAMILTON'S PRINCIPLE. LEAST ACTION 371
Suppose that a force function U exists, which depends only
on the Xij 2/», z» and t, or on the qr and t :
(2) U=U (xif y<, ziy f) or U = U (qr, t).
Since t is here the independent variable with respect to variation
5, we have
W8 = 6U
in the sense of the Calculus of Variations, and (1) becomes :
*i
(3) C(ST
= 0.
It must be remembered, however, that the variations &r,, fly,-,
bZi or 5qr satisfy the condition of vanishing when t = t0 and when
t = t1.
Here we meet our first example of an integral,
(4)
to
the variation of which vanishes :
ti
(5) * f(T + U) dt = 0.
h
This equation embodies Hamilton's Principle, which we may
formulate as follows.
HAMILTON'S PRINCIPLE. Let T be the kinetic energy of a system
of particles f and let a force function U = U(qr, t) exist. The
natural path of the system is that for which Hamilton's Integral:
(6) l(T + U)dt,
is stationary:
(7) d I (T + U) dt = 0.
Here t is the independent variable, and the variations of the dependent
variables are such as vanish when t = tQ and when t = tv
372 MECHANICS
A necessary and sufficient condition that (7) be true is afforded
by Euler's Equations, § 1, which here become :
d dT dT dU ,
_ __ — ______ rp — . 1 , , . i*jj
dtdqr dqr~ dqr' ' '
But these are precisely Lagrange's Equations for the system.
Incidentally we have a new proof of Lagrange's Equations, in
case we make Hamilton's Principle our point of departure.
We have proved the Principle for systems of particles with m
degrees of freedom, and it can be established in certain more
general cases, e.g. for systems of rigid bodies; provided each time
that a force function exists. The case is also included, in which
relations of the form :
*>«(?!> ' ' ' > ?•», 0 = 0, a = 1, • • • , Ml
exist ; cf . Bolza, Variationsrechnung, p. 554. The most general
case is that of a system having a Lagrangean Function, or kinetic
potential, L. In the above cases,
L = T + U.
When it is not possible to establish it without special postulates —
consider, for example, the motion of a perfect fluid or of an elastic
body — it is taken as itself the postulate governing the motion
of the system. The Principle consists, then, in requiring that
the Lagrangean Integral :
d
(8) fldt
to
be stationary ; or that
«i
b Cldt =• 0.
V
It will be shown in § 14 that Hamilton's Integral (8) is actually
a minimum for a path lying within a suitably restricted region ;
but the minimum property does not necessarily hold for un-
restricted paths.
8. Lagrange's Principle of Least Action. Our point of departure
is the Fundamental Equation II., § 5, in which U now does not
depend on t :
HAMILTON'S PRINCIPLE. LEAST ACTION 373
Moreover, T does not depend on t :
T = T(qi, • • • , qm, qiy • • • ,?m).
Thus we have :
ti
(1) C(ST + SU) dt + ZTddt = 0.
<0
Here each 5 represents a variation in the sense of the Calculus
of Variations, the independent functions being ql9 • - - , </m, t ;
but the integrand is not the variation of some function, nor is
the integral the variation of some integral. Nevertheless, the
equation is true when all m + 1 variations, qlt • • - , qm< t, are
chosen arbitrarily. Let us examine more minutely the meaning
of this last statement. These variations are defined by arbitrary
functions :
(2) 0r(u,e), «(u,€),
such that
qr(u,0) =gr(u), t(u,0) = t(u).
Moreover :
?r(0,€) = <?r(0), ?r(l, 0 =gr(l)j
j(0, c) = <(0, 0) = <0 = const. ; i(l, 0) = ^.
But in general ^(1, c) ^ tv Thus
5«|«-o = 0; ^|tt=1 ^ 0.
In particular, then, the functions (2) may be restricted by any
further conditions which are compatible merely with the general
conditions of continuity. Such a condition is the one that, not
merely for the natural path corresponding to € = 0, but also for
all varied paths :
(3) T = U + ft,
or, more explicitly :
(3') T [?r(u, «), ^|] = U [?r(u, c)] + A,
where fe is a constant. Since T is here a homogeneous quadratic
polynomial in qlf - - - , qm, it is clear that t (uy e) is obtained by a
quadrature when the qr(u,€), r = 1, • • • , m, are chosen arbi-
trarily.
374 MECHANICS
Let
be any function. By dF we shall now mean the following :
(4) SF = -
u
where qr(u, e), t(u,e) are restricted by the relation (3), i.e. (3').
And similarly :
(5) b \Fdu = -j- iFdu ,
J 0*J e-O
0 0
where the integrand on the left is formed for the arguments
qr(u), etc., and the integrand on the right, for qr(u, c), etc.;
Equation (3') still holding. Thus it follows, in particular, that
(6.) bT = 8U.
Although these definitions are in form identical with the earlier
ones, where the m + 1 functions (2) were arbitrary, they are in
substance distinct, since these m + 1 functions are now related
by (3')-
Equation (1) now becomes, on suppressing the factor 2 :
r,
(7) CdTdt + Tddt = 0.
Since obviously, under our new definition of 5,
5 /TT/'\ xT7 tf i rn jj^'
0(1 1 ) = 01 'I -f- 7 Ot,
and since 5t' = dbt/du, Equation (7) takes the form:
i
(8) I d(Tt')du = 0.
o
Hence, finally :
(9) lj*Tdt = 0.
We are thus led to the following Principle.
LAGRANGE'S PRINCIPLE OF LEAST ACTION. Let a system of
particles have the kinetic energy T and a force function U, where U
depends only on the position of the system, not on its velocity or the
HAMILTON'S PRINCIPLE. LEAST ACTION 375
time, and where T is independent of t. Then a necessary and suffi-
cient condition for the natural path of the system is, that
(10)
subject to the hypothesis that all varied paths fulfil the requirement
that
(11) T = U + h.
In addition^ the variations of the coordinates shall vanish for t = t0
and t = Jj.
The Principle thus formulated presents a Lagrangean problem
in the Calculus of Variations with variable end points and one
auxiliary condition :
(12)
CTM = o,
The method of solution developed in that theory* employs
Lagrange's Method of Multipliers. Briefly outlined it is as
follows. Set
F = T + \v,
where X is a function of t, and let qr(t), X($), be determined by
the m + 1 equations
/*o\ *fl? d &F f\ t
<13> Wr'dtWr^0' '-I.'".*.
and the second equation (12). From Equation (13) it follows
that
^ + X^_^r^ + X
tyr dqr dt Ldqr d
or
These equations, combined with the second equation (12), give :
(14) X = - -J-.
* Cf. Bolza, Variationarechnung, p. 586, where the case is considered that there
are, in addition, relations between the coordinates, not involving the time.
376 MECHANICS
Hence the qr (t) are determined from the resulting equations,
8T . dU d dT „
The latter are Lagrange's Equations. Incidentally we have
a new deduction of them, based on Lagrange's Principle of Least
Action.
As in the case of Hamilton's Principle, so here we can give a
direct proof of Lagrange's Principle of Least Action by means
of the Calculus of Variations. For, as above pointed out, the
Principle is equivalent to the Lagrangean problem represented
by (12).
Recurring to the condition (3) we see that the functions ql (u, e),
' m ' 9 Qm(u, e) may be chosen arbitrarily, and the function t (w, e)
then determined by (3'). If the function t(u, e) thus determined
be substituted in the integral :
i
(16) fft'du,
0
then t is completely eliminated from that integral. For
(17) T=%Ar.qrq., AT. = A.r,
r. s
the coefficients Ar8l depending only on the ql9 - • • , qm. Now,
/™ = /2r'
to 0
(18)
Let
(19)
where, as usual, q'r = dqr(u)/du. Then
or
(20)
From (20) and (3) it follows that
Tt' =
HAMILTON'S PRINCIPLE. LEAST ACTION 377
and thus t is eliminated, the integral (18) taking the form :
f-
0
We now have before us a problem in the Calculus of Variations,
of much simpler type — the simplest type of all, considered at
the outset. It is the integral (21), formed for the functions
qr(u), that is to be stationary, and these functions are all arbi-
trary. After this problem has been solved, t is determined from
(20), or
(22) t-t, = f-j-JL
J Vu
0
This is Jacobi's Principle of Least Action, which we will treat
in the next paragraph as an independent Principle. But it is
interesting to see how it can be derived from the Fundamental
Equation of § 5, and proved as a particular case under Lagrange's
Principle of Least Action.
EXERCISE
Show that Equation (1) under the restrictions named can be
thrown into the form :
<i
dT , dU d dT\ „ ,,
-x- — h o 17 TT-) &Qr dt = 0.
O(lr 0qr (It G(j[r'
Hence deduce Lagrangc's Equations.
9. Jacobi's Principle of Least Action. Let a system of particles
have the kinetic energy T and a force function U, where U depends
only on the position of the system, not on its velocity or the time, and
where the conditions imposed on the coordinates do not contain the
time explicitly. Then a necessary and sufficient condition for the
natural path of the system is, that the integral :
(1) fVU + hVTdt
be stationary:
(2)
378
MECHANICS
The time is given by the equation:
(3) T = U -
or
(4)
where
(5)
r
vir+T
=Vrdt.
where <h(u
stationary :
We can give a direct proof as follows. The integral (1) has
the value :
i
(6) JVu + hVSdu,
0
, qm(u) are arbitrary functions. It is to be
i
(7) d I VU + h VS du = 0.
0
Hence Euler's Equations must hold, or :
(8) — (V(7 + h V5) — 7 (V[J + h VS) = 0,
T = 1, • • • , m.
d /vTT+1
Hence
.
S d /
du\
'
Equations (9) determine the path ; t has not yet entered in
the solution. Equations (3) and (5) now determine t; it is
given by (4).
It follows furthermore that
dqr \du/ dqry dq'r du dqr
Combining these equations with (4) and (9) we find :
/i i \ .__ — „.
\ ' J4 O^, O^. Q
at oqr oqr oqr
HAMILTON'S PRINCIPLE. LEAST ACTION 379
Thus we arrive at Lagrange's Equations. If we assume them,
then we have a proof of Jacobi's Principle. Conversely, if we
assume Jacobi's Principle, we have a new proof of Lagrange's
Equations.
10. Critique of the Methods. Retrospect and Prospect The
symbol d is treacherous. It can and does mean many things, and
writers on Mechanics are not careful to say what they mean
by it. In d'Alembert's Principle the dxt, 8yif dZi began life by
being 3n arbitrary numbers. In their youth they were dis-
ciplined to conform to certain linear homogeneous equations.
Thus still a number of them were arbitrary quantities; the rest
had no choice, they were uniquely determined.
Enter, the Calculus of Variations. And now the dxij dy^ dzi,
and 8t become the variations of functions of a parameter, or inde-
pendent variable, u. From now on these <$'s must be dealt with
under the sanctions of the Calculus of Variations — at least,
if tho findings of that branch of mathematics are to be adopted.
The Future. As the physicist fares forth over the uncharted
ocean of his ever-expanding science, his compass is the Principles.
He seeks an integral which in the new domain will do for him what
Hamilton's Principle achieved in classical mechanics. There is
mysticism about this integral. Imagination must guide him,
and ho will try many guesses. But he will not be helped by an
undefined d. He must make a cloan-cut postulate defining the
integral, and then lay down a clean-cut definition of what he
moans by tho variation. There is no short cut. A thorough-
going knowledge of the rudiments of the Calculus of Variations
is as essential in Mechanics as perspective is in art.
11. Applications. Lot a particle be acted on by a central
attracting force inversely proportional to the square of the dis-
tance. Then
(1) r =
where the pole is at the centre of force, and it is assumed that the
motion takes place in a plane (cf. Exercise 4, below). Then the
integral :
(2) f\ £
r*0'*du
380 MECHANICS
must be made a minimum. Set
F (r, B, r', 0') = >(? + h)(r'2 + r2*'2)'
Then
A ^ - ^ = n
dw 00' 20
Since dF/dO = 0, it follows that
™ ^F - ^/M . . r*0' _ „
(6) W'-Vr + h Vr'* + r*0'* "
If c = 0, then
0 = const.
and the motion takes place in a right line. But if c 7* 0, 0 may
be taken as the variable of integration : * u = 0, and (3) becomes :
Hence
e , . cdr
*/;
rVhr2 + »r - c2
Change the variable of integration :
= 1
"" r
Then
*=+/
/T i 22
v Al ~| /it/* C 1L
Performing the integration, we find :
_ 1 — e cos (0 — y) = K
EXERCISES
1. Discuss in detail the case c = 0.
2. In the general case, determine the constants e, K, y in terms
of the initial conditions.
* It is true that the interval for u was (0, 1) ; but it might equally well have
been an arbitrary interval : a ^ u ^ b .
HAMILTON'S PRINCIPLE. LEAST ACTION 381
3. Obtain the time.
4. Allowing the particle free motion in space, show that its
path is a plane orbit.
Suggestion : — Use Cartesian coordinates.
5. Discuss the motion of a particle in vacuo under the force of
gravity. Assume the path to lie in a plane.
6. In Question 5, prove that the path must lie in a plane.
7. Explain the case of motion in a circle under the solution
given in the text.
12. Hamilton's Integral a Minimum in a Restricted Region.*
THEOREM. The integral
f
(1) ldt
to
is a minimum for the natural path, provided tQ and ^ are not too far
apart.
The Lagrangean function :
• • • , ?m, ft,
has the properties :
(3) 2
1,1
is a positive definite quadratic form. Moreover,
(4) H + L = J Prqr,
T
where
(5) Pr = J| r - 1, • • • , m,
and the Hamiltonian function
H(qly •••,?«,?!,•• -,pm, 0
has the properties :
* Oarathfodory has given a proof of this theorem : Riemann-Weber, Partielle
Differenlialgleichungen der mathematischen Physik, 8. cd. 1930, Vol. I, Chap. V.
382 MECHANICS
A necessary condition that the integral (1) be a minimum is, that
h
8 fL
8 I L dt = 0.
The extremals are given by Kuler's equations :
which are precisely Lagrange's equations.
By the transformation (5), the inverse of which is given by
(7), Lagrange's equations (8) go over into Hamilton's canonical
equations, Chap. XI :
dqr^M dpr___M r-i ... m
(9) dt ~ dpr' dt * dqr' r~1' 'm'
The latter can be solved by means of Jacobi's equation ; cf. Chap.
XV and Appendix C :
/im W . uf dv dv
(10) _ + ff(,1,...,,m,_ ...,_
as follows. Let (qrQ, pr°, t0) be a point in the neighborhood of
which Equations (9) are to be solved. A solution of (10) :
(11) V = S(ql9 • • - , qm, «!,•••, «», t),
can be found * such that
* *
(12) - f ^ 0
^(«ll ' ' ' , «»)
* The existence theorem in question follows at onoo from the theory of character^
istics as applied to Equation (10). That theory tells us thut there exists a solution
of (10) :
V -Stai, • • -, Qm, 0,
such that, when t = to, S reduces to a given function ^ (71, • • •, Qm) :
S (qi, - • -, qn, to) - <f> (qi, • • •, qm).
Here, <p (q\, • • •, qm) is any function which, together with its first derivatives, is
continuous in the neighborhood of the point (qi°, • • •, qm°). Such a function is:
<P (Ql, ' ' '. Qm) = S«r7r,
r
where the oti, •••,«»» are m arbitrary constants, or parameters. The function *S
thus resulting is the function required in the text.
If, as we may assume, the function // (qr, pr, 0 is analytic in the point (qr°,
pr°, Jo), and if, as is here the case, <f> (qr) is analytic in the point (gr°), then the
fundamental existence theorem of the classical Cauchy Problem, formulated for the
simplest case, applies at once, and the theory of characteristics is not needed.
HAMILTON'S PRINCIPLE. LEAST ACTION 383
in the point (qrQ, <*r°, tQ) and furthermore the equations :
/-<o\ SS ~ 8S t
(13) pr = Wf, fir — -^, r = l,.-.,m
are satisfied — the first set, when (qv°, pr°, J0) are given, by the
values ar = ar° ; and then the second set determines /3r°.
By means of this function S Equations (9) are solved. The
solution is contained in (13) and is obtained explicitly by solving
(13) for qr, pr :
( . f ?r = /r(«i, • • • , «m, ft, ' • • , ftn,
I pr = grfai, ' ' ' , OW, ft, ' ' ' , j8m,
t)
Properties of the Extremals. If r/r = </r(0 represents an
extremal, and if qr = dqr/dt, then by (5) and (13) :
05) ^r = ^f, r = l, .-.,n.
Moreover :
(16) 2) 'M' + & = £•
r
For, since S is a solution of (10), it follows, by the aid of (13), that
(17) St + H(ql9 •••,?»,?!, •••,p*,0 =0.
On substituting this value of H in (4), and replacing pr in the
resulting equation by its value from (13), Equation (16) results.
The Function E(qr, q'T, qr, t). Consider the function
V = L(qr,q'r,t),
where (</r, </J, t) are 2n + 1 independent variables. Let (qr, qr, 0
be an arbitrary point, and develop L' about this point by Taylor's
Theorem with a Remainder. We have :
(18) V = L + 2) Lir(q'r - qr) + E(qr, & gr, 0,
r
where L, Lqr are formed for the arguments (qri qr, t), and
(19) E(qr, Qr, <jr, 0 = i 2 liri, (q'r - qrM - ?.)
r, s
the coefficient I^rQ8 being the value of L<jr^ for a mean value of
the arguments qr, namely, qr + 0(q'r — qr), where 0 < 8 < 1.
The quadratic form (3) is positive definite. Hence
384 MECHANICS
(20) 0<E(qr,q'r,qr,t)
if (q'i, • • • > £m) is distinct from (ft, • • • , gm).
Proof of the Minimum Property. Consider an arbitrary extremal
(OQ through the point P0 : (qr°, t<>), represented by (14) :
£0: Qr = qr(t), r = 1, • - • , m.
Let P! : (qrl, t\) be a second point on <£~0 near by. Connect P0
and P! by an arbitrary curve
C: qr = qr(t), r = 1, • • • , m,
and let q'r (t) = dqr/dt. The curve C shall, however, be a weak
variation,
l?r(0-tfr«| <>7,
Let
L = L(qr,q'r,t).
Let (<?r, 0 be an arbitrary point on C. Through this point
there passes an m-parameter family of extremals, (13) or (14).
We select one of them as follows. Let alt • • • , am retain the
values they have for <£~0; but let ft, • • • , ftn have new values,
namely, those given by the second of the equations (13), when
qr = qr) i = t. The corresponding value of qr will be given
by (7). It is the value found, for the ar, $r in question, by dif-
ferentiating the first of the equations (14) with respect to t. These
values of «r, qn qr, t satisfy Equations (15) and (16); the pr do
not enter explicitly in these equations, and so the fact that they
depend on t does not complicate the equations.
We now apply Equation (18), setting qr = qr, giving to qr
the value just found, and letting q'r refer to C, <fr = q'r. Thus
(21) L = L(qr, qr, t) + % L^r(qrj qry t)(q'r - qr) + E(qr, q'r, qr, 0-
r
The first two terms on the right of (21) can be modified as
follows. First,
(22) ^^ = £ S,r(qr, f)q'r + St(qr, t).
Next, from (16) :
(23) 2 s«, W" 0 9r + S, ($„ 0 - L (gr, (/„ 0 = 0.
r
HAMILTON'S PRINCIPLE. LEAST ACTION 385
Subtracting (23) from (22) we have :
^jj^ = 2) S,r(qr, t)(q'r - qj + L(qr, qr> f).
Finally, since from (15)
the first two terms on the right of (21) have the value dS(qr, t)/dt,
and (21) can be written :
(24) L = - + E(qr, q'n qr, f).
We now proceed to integrate this equation from t0 to tv Ob-
serve that
to
has precisely the value of the integral :
ti
CL <u,
taken along the natural path of the system. For, along <§~0 Equa-
tion (16) says that
r> Q
~
and so
ti
t.
/Lett = 5far,0
to
But in the end points, qr(t) = gr(0- We thus arrive at the final
result :
(25)
i i
CL dt = /L dt + CE(qr, $, qr, t) dt.
to to «o
If, then, C differs from c£F0, there will be points of C at which
E > 0, and so the integral of L over C (i.e. the integral on the
left) will be greater than the integral of L over (£"0 (i.e. the first
integral on the right) and our theorem is proved.
386 MECHANICS
13. Jacobi's Integral a Minimum in a Restricted Region. In
Jacobi's Integral :
(1)
the functions T and 'U do not contain t explicitly, and T is homo-
geneous in the qr :
(2) T=
The varied functions, qr(u, «), are arbitrary, subject merely to
the condition that dqr — 0 in each end-point, t — t0, t^ It is
obvious that the integral (1) lias the same value as the integral :
rt
(3) • JTdt,
subject to the restriction :
(4) T = U + h.
This condition shall hold for the varied paths, too. Thus qr(u, c)
is still arbitrary; but t(u, c) is determined by (4). To prove,
then, that the integral (1) is a minimum for the natural path, it
is sufficient to show that the integral (3) has this property, if
(4) holds for the varied paths.
In the present case,
(5) L = T + U.
(6) H = T - U.
From (4) and (5),
(7) L = 2T - h.
Let (nQ be the path defined in § 12, and let C' :
(8) §r = 7r(w, 0> ?=<(",€),
be a varied path. Consider the varied integral. From (7)
t\ *i
(9) /2? dl = fl dt + h(tl- *0),
HAMILTON'S PRINCIPLE. LEAST ACTION 387
where
The right-hand side of Equation (9) can be computed as follows.
From the analysis used in § 12, Equation (24), we see that
Li'
du
Hence
Ldt = S(qr, I) "' + CEdl
<b
Since qr = qr for u = w0, MJ, the first term on the right has the
value :
StorSZD-StorVo)-
Hence
*« *i
(10) /2T df = S fer1, ?i) - S far0, «o) + h (t, - t0) + CE dl
to to
Since // is independent of t :
it follows as in Chap. XIV, § 4, that a function S of the form :
S = - ht + W(ql9 ••-,?*, A, «2, ••-, a*)
can be found, where h is to be identified with av Using this
function S in (10), we have :
(11) ArdT = I^to,1) - Ftor°) + Csdl
<0 <0
If we allow C" to coincide with <§"0, then J? ^ 0, and
t\
J*2T<lt = W(qr1) - W(qr°).
388 MECHANICS
Thus (11) becomes:
C2Tdt = C2Tdt
to t0
This proves the theorem. For, if C" is distinct from co'o, then
J5, which is never negative, will be positive for some parts of the
interval of integration, and hence the integral (3), extended over
C', will exceed in value the same integral extended over the
natural path, as was to be proved.
The case U = const, leads to the geodesies on a manifold for
which the differential of arc is given by the equation :
ds2 = 5} A rs dqt dqy.
r, s
Thiis we have a proof that a geodesic on a manifold obtained ar
above is the shortest line connecting two points which are not too
far apart.
CHAPTER XIV
CONTACT TRANSFORMATIONS
1. Purpose of the Chapter.* The final problem before us is
the integration of Hamilton's Canonical Equations :
.. dqr 8H dpr 8H ,
A) df = Wr' dT=-W,' r=l'-'m-
The method consists in finding a large and important class of
transformations of the variables (qr, pry t) into new variables
(q'r, p'r, t'), such that Equations A) are carried over into a new
system of like form :
A/, dq'r 3H' dpr 3H' t
A) W = W dr=~W r-1'-'w'
or, as we say, transformations with respect to which Hamilton's
Equations remain invariant.
The most general class of such transformations we shall con-
sider, are the so-called Canonical Transformations. A one-to-one
transformation :
<lr = qr(qly • ' -,7m,
Pi, • '
• , Pm, 0
Pr = P' (ft, ' ' ' , (7m,
Pli ' '
' , Pm, 0
f = «' (ft, ••-,?»,
Pi, * *
' , Pm, 0
» (ft', ' "^m. Pi', "
* , Pm,
O^n
01, ' ' ' , tfm, Pi, ' ' ' , Pm, 0
is said to be canonical if there exist two functions,
H(<li, ' ' • , 7m, Pi, • • • , Pm, 0 and H' (q(, - • • , q'm, p(, • • • , Pm, t')
(not in general equal to each other) such that
* This introductory paragraph is designed to give an outline of the treatment
contained in the following chapter. The student should read it carefully, not,
however, expecting to comprehend its full meaning, but rather regarding it as a
guide, to which, in his study of the detailed developments, he will turn back time
and again for purposes of orientation.
389
390 MECHANICS
(1) /(2 V'M - H'df) = J(2 Prdqr - Hdt),
1" 1'
where F is an arbitrary closed curve of the (2m + l)-dimensional
(gv, ?>r, 0-sPa('0> ail(l V' *s its image in the transformed (#J, pj, <')-
space, these spaces being thought of as simply connected.
To a canonical transformation there corresponds a function
' ' ' i <7m, Pi, - • • , Pm, 0 such that
(2) 2 ?' M * H'dt' = ^ Prdqr - Hdt
And conversely, when three functions H', H, ^ exist, for which
the latter relation is true, the transformation is canonical.
Contact Transformations. An important sub-set of these ca-
nonical transformations consists in those for which the last Equa-
tion I. is
(3) f = t.
On equating the coefficients of dt on the two sides of Equation (2)
we find :
'
Since /' = /, we may say that the variable t is not transformed,
and treat it as a parameter. Equations I. thus take on the form :
with
<?' = q'rfal, ' ' ' iQmiPi, ' • ' , Pm, t)
Pr = Pr(q\y ' ' ' , Qm, Pi, ' • • , Pm, t)
^(<7?> • • • »<?m, P'}, ' '• 9 Pm) _^0
d(9l9 ' ' ' , (7m, Pi, ' ' ' , Pm)
And now comes an important modification of Equation (2).
Since we now are regarding the (qrj pr), and not the (qr, pr, t),
as the independent variables, (2) can be written by the aid of
(4) in the form :
(5) %(p'dq'r- prdqr) =d*.
r
Of course, d$ has different meanings in (2) and (5). In (2),
^\ ^T,
(6) l»
CONTACT TRANSFORMATIONS 391
since here the independent variables are qr, pr, t, whereas in (5),
/*\ JT
(7) *
since here the independent variables are gr, pr ; a similar remark
applying to the other differentials, dq'r. This is not an exception,
or contradiction, in principle, but only in practice, since the
differential of any function, *(«i, • • • , xn), depends on the inde-
pendent variables :
__ ^ <Nf
and it is not until we have said what these shall be — i.e. defined
our function — that we can speak of its differential.
A transformation — we will henceforth change the notation
from m to n :
( , f Qr = ?'(<?!, ' ' ' , 9n, Pi, ' ' ' , Pn)
I Pr = Pifei, ' ' ' , tfn, Pi, ' ' ' , Pn)
3 (ft', ' ' ' , 0i> Pl'» ' ' ' , Pn) , n
#ft> ' ' ' , 0n, Pi, ' ' ' , P»
such that
(9) Jpfdtf = jprdqr,
r' r
where F is an arbitrary closed curve of the (QV, pr)-space, thought
of as simply connected, and F' is. the curve into which it is trans-
formed, shall be called a contact transformation. There cor-
responds to such a transformation a function ^(qlt • • • , qnj
Pi> ' • ' > Pn) for which
(10) 2,(prdq'r-prdqr) = d*.
r
And conversely, a transformation (8) for which (10) is true satis-
fies (9) and so is a contact transformation.
A contact transformation may, of course, depend on certain
parameters, p', q'r and ¥ thus becoming functions of these para-
meters as well. The transformation II. above is a case in point.
Finally, the canonical transformations form a group; i.e. the
result of applying first one and then a second such transfor-
mation may itself be expressed as a canonical transformation.
392 MECHANICS
The contact transformations also form a group. The group of
contact transformations II. is a subgroup of the group of canonical
transformations I.
The approach to the contact transformations is through the
Integral Invariants of Poincarg.
The contact transformations, as defined generally by (8) and
(9), are of especial importance in Mechanics because any such
transformation carries an arbitrary system A) of Hamiltonian
Equations over into a second such system, A') ; cf. infra, § 4.
We shall treat the application of these transformations to the
integration of Hamilton's Equations at length in Chapter XV.
If the student is willing to take this one property of contact
transformations for granted, he can turn at once to Chapter XV,
and he will find no other assumptions needed in the study of
thajt chapter.
2. Integral Invariants. Consider the action integral :
*i
(1) j*L(qr,q'n{)dt,
to
and the extremals, which are the path curves, given by Lagrange's
Equations :
/o\ ^ **L ^ _ n — 1
(2) diWr Wr~ ' r-V-.,n,
where L is the Lagrangean function, or the kinetic potential.
The general solution can be written in the form :
(3) qr = qr(t ; qf, • • • , qJ, qf, • • • , tfn°), r = 1, • • • , n,
where gr°, qr° are the initial values of qr, qr, i.e. their values
when t = fo. In the (2n + l)-dimensional space of the variables
(<7u * ' * 9 q*> <ii> * ' ' > q*9 0 these equations, together with the n
further equations :
(4) <?r = <?r('; «1°, •••,?»°, ?1°, •••,<7n°)>
represent a curve C, — or more properly, a 2n-parameter family
of curves C. Let a closed curve, F0 :
(5) ?r° = ?r°(X), tfr° = gr°(X), r = l, • • • , n, X0 S X g X,,
be drawn in the plane t = £0. The curves C which pass through
the points of F0 form a tube of solutions, which we will denote by S.
CONTACT TRANSFORMATIONS
393
Let the action integral, (1), be extended along the curves C
which form S. Its value is a function of X :
(6)
where qr, qr are given by (3)
and (4) and qr°, qr° by (5).
Differentiate / (X) :
FIG. 147
- C
" J
On integrating by parts, observing that
we have :
Hence
d dqr
di~d\'
C'^^LM =3L<^- C^^^SLM
J d</r 3\ d(jr 3\ J dtdqr 3\
t\
'(\\ - C V (^J d 3L\ fyr,, | y dL8qr
() ~ J $ \Wr dtWr' 8\ CU + -f 2qr d\
The integral vanishes, because qr is by hypothesis a solution
of (2).
We now make the transformation, Chapter XI, § 3 :
(7)
Thus
(8)
dL
Since F0 is a closed curve, qr°(\) = ^r°(
and (6) gives :
(9)
Hence
X
0,
394 MECHANICS
and so from (8) :
*i
/» k P/y <l
eJX = 0.
Let t0 be thought of as constant, but tl9 which is also arbitrary,
as variable ; denote the latter by t. Then
This equation represents the theorem in which the whole in-
vestigation of this paragraph culminates. In substance it can
be stated as follows. We may regard Equation (7), along with
the n + 1 further identical equations, qr = qr, t = t, as repre-
senting a transformation of the (qr, qr, £)-space on the (qr, prj 0"
space. Observe that the Jacobian
• - » ?j
•••
Chapter XI, § 3. Thus the curves C of the first space go over
into curves C" of the second space, and F0 goes over into a curve
TO, S being transformed into a tube S'.
Let us now make the second space — the space of the vari-
ables (<?r, pr, t) — our point of departure and, dropping the primes,
consider a closed curve in the plane t = tQ of that space :
Consider furthermore curves C through its points, which are
obtained by transforming the curves C of the earlier space. The
integrals (11) now become line integrals in the present space.
If we change the notation, setting
(13) qrQ = «r, Pr° = Pr,
then (11) assumes the form :
(14)
/5J Prdqr = / ^0rdar,
r *J r
where F is the curve of intersection of the arbitrary plane / = t
with the tube S determined by T0, an arbitrary closed curve of
CONTACT TRANSFORMATIONS
395
the plane t = tQ. But this is precisely the definition of a contact
transformation, t being thought of as a parameter : *
, .
Pr(«i,
, «„, ft,
' i «n, ft,
, fti, 0
3. Consequences of the Theorem, a) Hamilton's Canonical
Equations. Lagrange's Equations (2), § 2 form a system of n
simultaneous total differential equations of the second order.
By means of the transformation (7) these are carried over into
a simultaneous system of 2n total differential equations of the
first order in the (qr, prj 0-space. Let these be written in the
form :
(16) = Qrfop,0, Pr(?,P,0,
Since the right-hand side of (14) is independent of t, the deriva-
tive of the left-hand side with respect to t must vanish. Hence '
or
*1
d C ^-\ dqr j\ _ (\
dij 2,P'^dX-u'
?1T a\r + praTax)dx = °'
Integrate by parts :
d\
n
pr
Since T0 is a closed curve,
= 0,
*The geometric picture is here slightly different from the earlier one, since
the variables (ari /3r) and (qrt pr) are interpreted in different planes. But of course
one may think of a cylinder on T0 as directrix, with its elements parallel to the
t-axis. On cutting this cylinder with the plane t = t, we have a curve F0 lying in
the same plane with F. Or, to look at the situation from another angle, t is only
a parameter, and it is the spaces of (ar, Pr) and (qr, pr) which concern us.
396 MECHANICS
and we have :
Here,
dqr __ dqr _ n tyr __ dpr _ p
8t ~ dt ~ Wr> dt ~ dt "
Thus Equation (17) may be written in the form :
(Prdr-Qrdr) = 0.
But r may be any closed curve of the plane t = /, since to any
such curve in that plane corresponds a F0 in the plane t = t0.
It follows, then, that we can define a function // by moans of
the integral :
,
(19) H = f £ -Prdqr
'
where the fixed point (alt • • • , an, blf • • • , &„, £) of the plane t = t
is connected with the variable point (qly • • • , qni Pi, • • • , pn, 0
of this same plane by a curve lying in the plane. Because of
(18) the value of the integral does not depend on the path,
and thus // is defined as a function of (qr, pr) for the particular
value of t.
Let the point (a, &, t), for dofinitcness, lie on the extremal through
the point (a, 0', 20). Then H becomes a function of (qrj pr, 0-
If (c/, 0', J0) is replaced by a different point (a", /3", tQ), the
new H will differ from the old H by an additive term which is a
function of t, but not of (qr, pr).
More generally, let // be defined by the equation :
(20) II = li+f(i),
where H is a specific one of the functions 77 just defined, and
f(t) is an arbitrary function of / alone.
From (19) it follows that
dn
CONTACT TRANSFORMATIONS 397
Thus the system of equations (16) is seen to have the form :
, , dgr _SH dpr _ 8H
(22) Hf-Wr' ~dt~~Wr' r=l,. ..,n.
First fruits of our theorem. The Hamiltonian Function H
grows naturally out of Equation (14) ; for (18) is but another
form of (14), and (18) at once suggests the definition of // by
(J 9) and (20). Thus if we had never heard of H through the trans-
formations of Chapter XI, we should still be led to it by the
theorem of this paragraph.
The Function V and Its Relation to H. Equation (14) can be
written in the form :
f
To
lr- Prdctr) =0,
where gr, pr are given by (15), the curve F0 being as before any
closed curve in the plane t = t0. It follows, then, that the integral :
(<*,$)
(23)
extended over an arbitrary path in the plane t = t0 joining the
points (a', jft'), (a, /?), is independent of the path and thus
defines a function of (a, 0), t entering as a parameter :
(«J5)
(24) g (Prdqr - Prdctr) = V (<*, ft t).
f
(«'>) '
Differentiate this equation with respect to / :
dqr\ __ dV
'~
where the italic d means differentiation along a curve (15). Trans-
forming through integration by parts we have :
/ocx rj r , • dv
(25)
r
_
398 MECHANICS
The integral^ on the left is precisely the negative of the integral
(19), or — H(qr, pr, f). Hence
(26) H = 2 PrQr ~ ^,
where H is given by (20), and
On the other hand, the Lagrangean Function L (qr, qr, t) is
connected with H (qr, pr, t) by the relation (cf . Chapter XI, § 3) :
(27) L + H = ]T prqr.
r
Hence it appears that
<*> £ - -
Just as H was defined only save as to an additive function of t,
so V can be modified by adding any function ty and the same
is true of L. But it is convenient to restrict these additive func-
tions so that (26) and (27) will hold.
From the foregoing reasoning we can draw a more general con-
clusion, and then supplement it with a converse.
THEOREM I. Let
r = l,---, n,
be an arbitrary system of simultaneous differential equationst and let
qr = <pr(t ; «!,-•-, orn, ft, • • • , j9»)
be the solution, where ar, Pr mean the initial values of qr, pr cor-
responding to t — t0. Let TQ be an arbitrary closed curve lying
in the plane t = tQ of the (2n + l)-dimensional space of the vari-
ables (qr, pr, t). Let S be a tube consisting of the curves ii) which
pass through points of F0; and let T be the section of S by the plane
-«. //
Pr dqr
\ 5)
CONTACT TRANSFORMATIONS 399
is an integral invariant of Equations i) ; i.e. if
iv) I ^Prdqr = I ^LfPrdotr,
r ' r0
then Equations i) form a Hamiltonian System:
. dqr dH dpr dH ,
»» i -/*_ = . r . = — r = 1 • • • ti
"' dt 8pr' dt 8qr> T lf >n'
Conversely, if Equations i) form a Hamiltonian System y), then
Hi) will be an integral invariant, or iv) mil be satisfied.
Observe, however, that Theorem I. is more general than its
origin from the action integral (1) and the transformation (7)
would indicate. It applies to any functions Qr, Pr for which
Hi) is an integral invariant ; or, in the converse, to any function
H, provided that the determinant
02 TI
Hu".Hnn* 0, Hi3 = -—
But a system of Equations v) may conceivably not lead to a
mechanical problem — why should it ?
b) Contact Transformations. The content of Theorem L can
be restated in terms of contact transformations.
THEOREM II. Let
qr — 9r(aD ' ' ' , <*n, ft, ' * ' , ft, t)
a)
hr(otlt ' ' ' , «n, ft, ' ' ' , fti, 0
where r = 1, • • • , n,
Pi,
0^, ••-,«., ft, --^ftr '
a?id
f ar = gfrC^, • • • , an, ft, • • • , ft, <0)
I ^r = Ar(a!, • • • , an, ft, • ' • , ft, O
fee a transformation of the 2n-dimensional (<xr, pr) -space on the
(qr, pr)-space; and let
400 MECHANICS
be the system of differential equations corresponding to a) ; i.e.
defined by a). // a) is a contact transformation; i.e. if
I ^Prdqr = j ^Prdar,
or
2J (prdqr - Prdoir) = dV (a, /3, /),
r
then b) z's a Hamiltonian System:
C' ~dt ^ ~dp~r' ~dt = '"~ ~dqr' r== >'">n>
and conversely.
4. Transformation of Hamilton's Equations by Contact Trans-
formations. If we start out with a given system of Hamiltonian
Equations :
, , dqr _ dH dpr dH_ __ 1
() ~dt ~ 8p~r' ~dt " ~dq~r' r-I,---,n,
and make an arbitrary transformation :
/0, f 9r = /rfe, ' ' ' , ^n, Pi, • • ' , Pn, 0
W 1 , , A
I Pr = Vr(q\y ' ' ' , ?n, Pi, ' ' ' , Pn, 0
the transformed equations :
(3)
r = 1, • • • , n, will not in general be of the form (1) ; i.e. they
will not have the form :
where /f ' = H' (q'T, p'r t) is some function of the arguments q'T, p'r L
A sufficient condition that (3) be Hamiltonian, i.e. of the form
(4), is that (2) be a contact transformation.
The proof is based on Theorem II., § 3 and the fact that the
contact transformations form a group. Let (2), then, be a contact
transformation. Denote it by T. Let (aj, #) be the initial val-
CONTACT TRANSFORMATIONS 401
ues of (?', p'r) for t = tQ. They arise from (ar, 0r) by T, formed for
* = *o J ^o> let us write it. Thus, symbolically,
(*;, #) = T0(ar, 0r), or («r, 0r) = TQ-*(cl, ft).
Again, we may write symbolically :
(#, PJ) = T (gr, pr).
Finally, consider the solution of (1), whereby the space of the
(ar, pr) is carried over into the space of the (gr, pr). This
transformation is the Transformation a) of Theorem II. , § 3, and
so because of (1) is a contact transformation. Denote it by D :
D («r, 0r) = (qr, pr).
On the other hand, the effect of the transformation defined by
the differential equations (3) is to carry the space of the (a'Ty ft)
over into the space of the (q'r, pr). Denote it by A :
And now we see that this result — this transformation A — can
be obtained as follows : Perform first the contact transformation*
TQI on the (aj, $)-space, thus obtaining the (ar, /3r)-space :
(ar, 0J = 57 («;, #)•
Next, perform the contact transformation D on the (ar, fir)-
space, thus obtaining the (qr, pr)-space :
Finally, perform the contact transformation T on the latter space,
thus obtaining the (q'r, p'J
fer', Pr) = T (qr, pr)
We have in this way obtained A as the result of three contact
transformations :
A = TDT?.
Hence A is itself a contact transformation, and so the system (3)
is Hamiltonian, by Theorem II., § 3.
This is the result on which the developments of Chapter XV
depend. It may be stated as follows.
* The inverse of a contact transformation is obviously itself a contact transfor-
mation.
402 MECHANICS
THEOREM. // a system of Hamiltonian Equations (1) be trans-
formed by a contact transformation (2), the result is a Hamiltonian
system (4). The condition is sufficient, but not necessary.
Computation of H'. The original system of Hamiltonian
Equations (1) leads to the contact transformation Z), for which
the relation :
(5) ^ pr dqr - 2 ^ dar = dV (ari 0r, t),
r r
is characteristic, where
JT — "V /n A __
dt
(6) 77 = 2 Prtfr - ~ •
The transformed Hamiltonian Equations (4) lead likewise to
a contact transformation D' = A, for which the relation
(7) 2 P' M - 2 # da' = dV> («'. £'> 0
r r
is characteristic, where
j jr/
Let
r r
be the characteristic relation of the contact transformation T.
Then
(10) 2 # da' " S ^ dc*r = dW (<*" &r> *o)
r r
will be the characteristic relation corresponding to T0.
Each of the differentials on the right is taken on the supposi-
tion that t is a parameter, and so a constant. Moreover, (qr> pr)
are given in terms of (ar, Pr) by equations of the type a), § 3.
From (5), (9), (10) we infer that
r r
d[- W(ctr, ftr, <0) + V(ar, 0r, 0 H
CONTACT TRANSFORMATIONS 403
Hence
-
dt ~ dt dt'
provided V (ar, $, f) and W(qr, pr, 0, which arc determined only
save as to additive functions of t, are chosen properly. From
(6) and (8) we now infer, by means of (12), that
(13) H' = H - + £ (p'^ - prqr).
Each of the functions H', H, dW/dt was originally defined
only save as to an additive function of t, and it is only when
these additive functions are suitably restricted, that (13) holds.
6. Particular Contact Transformations. In applying the theory
we have developed it will be convenient to denote the trans-
formed variables by Qrj Pr instead of by q'n p'r. Thus a trans-
formation :
f Qr = /rfoi, • • • , <7», Pi, • • • , p», 0
I Pr = <7r(<7i, - • ' , Qn, Piy ' • ' , Pn, 0
where
' • , (jft, 1 l)'*'? •* n) ^ Q
' " * ' '
is a contact transformation if
(3) 2 (Pr dQr - Pr.rf7r) = ^^ (<7r, Pr, 0,
r
where £ is regarded as a parameter and the differentials are taken
with respect to (qr, pr) as the independent variables.
If such a transformation be applied to the Hamiltonian system :
(4\ dqr__<M ^Pr__^ff -_!...„
W dt dpr' dt dqr' lj 'n'
these equations go over into a new Hamiltonian system :
dQ,_£ff' dPr__W _j ... _
W <tt ~ »P,' rf< ~ 3Q,' ' ' '
where H' (Qr, Pr) 0 is connected with H (qr, pr, t) by Equation
(13), §4, or:
(6) H' = H - + (PrQr - prqr).
404 MECHANICS
The (qr, Qr, t) as Independent Variables. Equations (1) repre-
sent 2n relations between the 4n variables (gr, prj Qr, Pr), and
when (qr, pr) are chosen as the independent variables, (2) and
(3) hold. It may be possible to choose the 2n variables (qrj Qr)
as the independent variables, t always being regarded as a para-
meter in (3). Write
(7) W(qr,pr,t) = W'(qr,Qr,t).
Thus (3) becomes :
(8) 5 (PrdQr - PrdQr) = dW (qr, Qr, t) .
On equating the coefficients of dQr, dq, in (8) we find:
Pr =
(9)
Equation (6) can now be transformed as follows : Since
dW dW dW'dQr SW'dqr 8W
'
dt dt 'dQr dt •? 8qr dt dt '
we have :
Hence (6) becomes :
(10) ff ' - ff -
OO OO
The Transformation: pr = -^-, Pr = — ^T' We can write
Oqr Glflr
down a particular contact transformation, in which (qr, Qr) can
be taken as the independent variables. Let
• ' • , qn, «!, - • • , any t)
be a function of the 2n + 1 arguments such that
CONTACT TRANSFORMATIONS 405
Set ar = Qr and make the transformation :
/irk\ &S r> &S i
(12) Pr = Wr, Pr=~Wr, r = l,...,n.
The first n of these equations can be solved for the Qr in terms
of the (qr, pr) because of (11), and then the Pr are given by the
last n equations. Thus a transformation (1) results, the Jacobian
(2) not vanishing.*
The transformation will be a contact transformation, for
(PrdQr ~ Prdqr) = ~ £ (|| dQr + J dqr) = ~ d3,
and we may set W = — S, since Tf and hence FT' is determined
only save as to an additive function of t. Equation (10) now
becomes :
(13) H' = H + -
How such a function S can be found, which will enable us to
solve Hamilton's equations explicitly, will be shown in Chapter
XV.
Conversely, the most general contact transformation (1) which
can be written in the form :
is given by (12). For, Equations (9) must be true, and it remains
only to set S — — W. It is seen at once that the Wr of (9) must
satisfy (11), since otherwise there would be a relation between
the Pr.
* The proof is as follows. If
Vr = fr (Xl, • • -, Xn), T = 1, • • •, U,
be a transformation having an inverse
Xr = V?r (l/Ii ' • ', 2/n), f = 1, • • •, n,
where fr, <f>r are all functions having continuous first derivatives, then
d (y\, - • -, yn) . d (x\, • • •, xn) _ j
d(xi, - • -,*„) d(yi,--,yn)
Consequently neither Jacobian can vanish.
In the present case, the qr, Pr can be expressed in terms of the Qr, Pr, since the
value of the determinant (11) is unchanged if the qr, ar are interchanged.
406 MECHANICS
EXERCISES
OO ^Cf
1. The Transformation: pr = T— , Qr = ^p- Study the
oqr Ofr
analogous case, in which (qr, Pr) can be taken as the indepen-
dent variables, t being, as usual, a parameter. Show that, if
£(#u • ' • > Qn> <*i> ' * * > <*n, 0 be chosen as before, and if we set
Pr = ar, then
/-. ^\ v*S s^ O& -
(14) p, = w, Qr=W, r = l,..-,n,
will give a contact transformation. Observe that (3) can be
transformed by means of the identity
d(PrQr) = PrdQr + QrdPr,
so that it takes on the equivalent form :
rdPr + prdqr) = d(~ W" + Pr Qr).
Choose W(qr, pr, 0 = W"(qr, I\, t), therefore, -so that
S=- W"
Compute dW"/dt and show by the aid of (14) that (6) yields :
(15) ff' = // + —.
State also, and prove, the converse.
2. Computation of //' in the General Case. Let TT^ • • • , 7T2ri
be any set of 2n variables, chosen from the 4n variables
far, Pry Qr, Pr), in terms of which the remaining 2n variables
can be expressed. Show that
06) «-,«
where qr, Qr, and W are expressed as functions of (TT*, 0-
/)^f OC|
3. The Transformation : qr = -5 — , Pr = -^r- If (pr, Qr) can
Gpr #Vr
be taken as the independent variables, and if we set
S = W + Pr?r,
CONTACT TRANSFORMATIONS
407
where qr, W, and S are now functions of (pr, Qr, t), then the
transformation takes the form :
(17)
and (6) yields :
(18)
= _
^ dPr' '
H' = H -
Conversely, if S(qlt • • • , qn, alt • • • , an) be chosen as before, and
if we set Qr = QLr, then (17) will define a contact transformation.
6. The fi-Relations. There is one case of importance still
to be considered, namely, that in which W is a function of
(qr, Qr, t), but the (qrj Qr, t) cannot be chosen as the independ-
ent variables. The extreme case would be that in which
Qr = Wr
T =
n,
Tho general case is that in which 0 < m ^ n independent rela-
tions between the (qr, Qr, t) exist, and no more :
where the rank of the matrix :
X^ Xli
(2)
Wn
is m. Thus m of the Q*'s can be expressed as functions of the
remaining ju = n — m Q/s and q^ • • • , qn, t. As a matter of
notation let the above m Qfc's be QD • • • , Qm :
/O\ /") ___ / ' f\ f\ n ff /\ __ 1 /yyj
Then the determinant whose matrix consists of the first m columns
of (2) will not vanish. Among the 2n (pr, Pr) it shall be possible
to choose m variables, TTJ, • • • , wm such that (irlt • • • , 7rm, Qm+i,
* ' ' > Qn, <7i, • • • , ^n, 0 can serve as the 2n + 1 independent
408 MECHANICS
variables. But the function W(qr, pr, t), when expressed in
terms of the new variables, does not depend on irly • • • , irm :
(4) W(qf,Pr,t) = IF* for, Or, fl.
It is not, of course, unique, because of the Q-relations, (1).
Equation (3), § 5 now takes on the form :
(5) 2) ( Pr dQr - pr cfyr) = dW* (qr, Qr, 0.
We will rewrite it in the form :
I (^-CK -?(*•
It is not, however, in general true that the coefficients of the
differentials vanish.
By means of the m equations (1) the first m differentials
dQu • • • , dQm can be eliminated, the resulting equation being
of the form :
(7) Xm+ldQm+t + • • • + XndQn + Y.dq, + - - • + Yndqn = 0.
The differentials in (7) are independent variables, and so we can
infer that
Xm+l = 0, - . • , Xn = 0, Y, - 0, • • • , Yn = 0.
The actual elimination can be conveniently performed by
means of Lagrange's multipliers. From Equations (1) we infer
that
n
n = o
(8)
Multiply the fc-th of these equations by X& and add to (6). Then
determine the X/t's so that the coefficients of dQ}, • • • , dQm vanish.
The resulting equation is of the form (7), and so its coefficients
vanish automatically. We thus arrive at the 2n equations :
(9)
_
' ~ Wr + ' Wr
r = l ... m
CONTACT TRANSFORMATIONS 409
The first m of these equations determine the \k's. The remainder
are satisfied as shown above. The result is symmetric and holds,
no matter what set of m Qk's is determined by (1) ; i.e. no matter
what ?n-rowed determinant out of the matrix (2) is different
from 0.
It is now easy to determine Hf by means of (13), § 4 :
On replacing Pr, pr here by their values from (9) and observing
that
dW*
=
dt dQr dt , dQr d{ dt >
dttr ^ dttr dQr . ^ d&r ^r , ^r ^
dt "" 2? 3Qr dt ^ ^ dqr dt ^ dt '
we find the following result :
-- — -J— » —
If, in particular, the 12's do not contain t explicitly, this equa-
tion reduces to
(11) H' = H
CHAPTER XV
SOLUTION OF HAMILTON'S EQUATIONS
1. The Problem and Its Treatment. We have considered a
great variety of problems in mechanics, the solution of which
depends, or can be made to depend, on Hamilton's Canonical
Equations :
m dqr - dH dpr --m r-1 ..- n
(i) ~dt~~Wr' ~dt ~ Wr ' ' '
where H is a function of (qr, pr, £). The object of this chapter is
to solve these equations explicitly in the important cases which
arise in practice.
The method is that of transformation. By means of a suitably
chosen transformation :
Qr = Fr(q19 • • • , qn, Pi, • • • , P», 0
Pr = Gr(qly ' - - , qn, Pi, • ' ' , Pn, 0
Equations (1) are carried over into equations of the same type :
dt ~ d/V dt
'
but more easily solved. Here, H' is a function of (QT, Pr, <)> 11(>t
in general equal to H .
The determination of a convenient transformation (2) depends
on a partial differential equation of the first order, due to Jaeobi * :
(A\ dV _i_
(4) -- +
It is not the theory of this equation, however, but the practice,
that concerns us, for all we need is a single explicit solution,
(5) V = V(qlf • • • , g», «i, • • • , «n, 0,
depending in a suitable manner on n arbitrary constants, or
parameters, «!,••-, «n. Such a solution is found in practice by
means of simple devices, notably that of separating the variables.
* Hamilton came upon this equation ; but its use as here set forth is due to Jaeobi.
110
SOLUTION OF HAMILTON'S EQUATIONS 411
The function (5) once found, the further work consists merely
in differentiation and the solution of equations defining the
qr, pr implicitly. Two cases are especially important, namely :
a) Reduction to the Equilibrium Problem. Here, a solution
(5) of (4) enables us so to choose (2) that the transformed H
vanishes identically, //' = 0. Equations (3) can now be inte-
grated at sight :
where ar, ($r are arbitrary constants. On substituting these val-
ues in (2), the inverse transformation,
f qr = /r(Q!, ••-,««, PI, •••,P»,0
yields the desired solution :
qr = /r(«i, ' ' ' , «n, ft, ' ' • i fti, 0
(8)
Pr = 0r(«i, ' ' ' , <*», ft, ' ' ' , ft», 0
The transformation (2) in this case, as will be shown in § 2, is
given by the equations :
,0. _ dV p _ dV _ t
^ Pr~Wr r~~Wr' r-l,..-,^
where F is written for the arguments qr, Qr :
Thus the solution (8) is obtained by solving the equations :
dV 0 dV
pr = Wr' &T=~^ r=l,..-,n,
where the present V has the form (5).
b) Constant Energy, H (qr, pr) — h. The second case is that
in which H does not contain the time explicitly :
H = H(ql9 • • • ,g», Pi, • • • ,pn).
It is here possible to find a transformation (2) in which Fr, G>
do not depend on t,
Qr = Fr(qlf • • • , gw, plf • • • , pn)
(10)
' ' ' , ^n, Pi, • • ' ,
412 MECHANICS
such that the new H will depend only on the Prj but not on
Qrj t. In particular,
#' = P^
Equations (3) now take on the form :
= 0, r=l,...,n.
Thus*
Qi = t + /3}, Q, =ft, 5 = 2,
Pr = ar, r = 1, • • • , n.
Let the inverse of (10) be written :
(12)
Then the solution of (1) is given by the formula :
(13)
1 ft, ft,
The transformation (2) in this case, as will be shown in § 4, is
given by the equations :
HA\ dW n
(14) pfSSWr9 Qr =
where W is a solution of the equation :
iff m dW
^•••••'-'•^•••••
or:
Tf = W(ql9 • • • ,?n, A, «2, • • • ,«»)•
Here TF depends on the arbitrary constant A, and also, in a suitable
manner, on n — 1 further constants, or parameters, a2» " * ' t <**•
These are set equal respectively to the Pr :
Pi = h ; P, = aa, s = 2, • • • , n.
* The change of notation whereby the a/s and the /Vs are interchanged is made
for the purpose of conforming to usage in the literature.
SOLUTION OF HAMILTON'S EQUATIONS 413
Equations (14), combined with (11), thus yield :
(15) ,w
The last n — 1 of these equations can be solved for q^ • • • , qn
in terms of qly as will be shown in § 4, thus giving the form of the
path ; and then q± can be found from the first equation (15) in
terms of t.
We have characterized this case by the caption: "Constant
Energy," but this is not a physical hypothesis. Our hypothesis
is, that H does not depend explicitly on 2, and this is all we need
for the mathematical development. That H then turns out to
be constant along the curves of the natural path, is an impor-
tant consequence; but our treatment does not depend on this
hypothesis.
Contact Transformations. The transformations used in a) and
b), namely, (9) and (14), are examples of contact transformations.
A transformation (2) with non-vanishing Jacobian was defined
in Chapter XIV, § 1, to be a contact transformation if
(16) 2} (PrdQr - Prdqr) = dW (qr, pr, t),
r
where the differentials are taken with respect to the (qr, pr) as
the independent variables, t being regarded as a parameter. Such
a transformation always carries a Hamiltonian System (1) into
a Hamiltonian System (3). That the transformations (9) and
(14) satisfy the condition (16) is seen at once by substituting
in (16), observing in the case of (14) that
d(PrQr) = PrdQr + QrdPr.
This is all the theory the student need know from Chapter XIV,
to enter on the study of the present chapter, and this amount
of theory was all developed in §§ 1-4 of that chapter.
2. Reduction to the Equilibrium Problem. We have seen in
Chapter XIV, § 5, that a transformation :
414 MECHANICS
where
S = 8(qif ••-, fr, «„••-, «„)
is any function such that
and where Qr is set = ar, will carry the Hamiltonian System (1)
of the last paragraph over into a Hamiltonian System (3), where
(2) /r-ff + f.
The transformed function H' can be made to vanish identically
if we can find a solution V of the partial differential equation :
which depends on n arbitrary constants, aly • • • , an :
V = V(qlJ • • • , ?„, «!,••-, an, 0>
and is such that
^.......vj^
3(«ll •••>«»)
On setting S equal to this function F, and making the trans-
formation (1), H' as now determined vanishes identically. Thus
the transformation :
(5) p, = g, Pr — jfc r=l,...,n,
where ar is replaced by Qr in F, transforms the Hamiltonian
System (1) to the Equilibrium Problem:
« t-o- T'-«- '-1'-'"'
The solution of these equations is the system of equations (6),
§ 1. These are the values of Qr, Pr to be substituted in the
transformation (1) ; i.e. in the present case, in (5) :
(7) p, = g, 0,=~g, r-l,...f«.
The last n of these equations can be solved for the qr's because
of (4), and then the first n equations give the pr.
SOLUTION OF HAMILTON'S EQUATIONS 415
There is, of course, a further requirement in the large, namely,
that the ar, $r can be so determined as to correspond to the initial
conditions : t = tQ, qr = qrQ, pr = Pr°. Thus the equations :
Pr° = V«r «, • - • , gn°, «!,•••, «n, Q, r = 1, • - • , n,
must admit a solution, ar = ar°, and V (ft, • • • , qn, al9 • • • , e*n, 0
must satisfy all the conditions of continuity, notably (4), in the
neighborhood of the point (qr, ar) = (<7r°, ar°).
EXERCISE
Pass to the Equilibrium Problem by means of the transforma-
tion studied in Chapter XIV, § 5, Exercise 1 :
dS n dS 1
*'~W Qr = Wr> " = V--,".
Here,
Let V = V(QI, • • • , qn, alf • • • , an, 0 be the same function as
that of tho text — a solution of Equation (3). If, then, we
replace ar by Pr and set S = V, the transformed H ' will vanish :
//' = 0, and Hamilton's Equations will take on the form of
the Equilibrium Problem :
dQr dPr n ,
~W = °> ~di- = 0' r = l,...,n.
If we write their solution in the form :
Qr = - Pr, Pr = Qfr, T = 1, • • • , tt,
we are led to the same solution of the original Hamiltonian
Equations as before — namely, that given by (7).
3. Example. Simple Harmonic Motion. Here the kinetic en-
orgy T and tho work function U are expressible respectively in
the form :
(1) T = ^q\ V -- \q*, 0<X.
Thus
(2) L = T+V-?-f
(3)
416 MECHANICS
(4) ff-rt-L.-Lp. + lj..
Hamilton's Equations assume the form :
al dq - p dp - - \n
a; di ~m' Tt~ A9'
We propose to solve them by the method of § 2. The equation
for determining V, § 2, (3), here becomes :
We wish to find a function :
(6) V
which satisfies this equation.* One such function is enough.
Let us see if we cannot find one in the form :
(8) F = fl + W,
where 12 = 12 (f) is a function of t alone, and W = W (q) is a func-
tion of q alone. If this be possible, we shall have :
' + V = o.
This equation can be written in the form :
X 0 d!2
The left-hand side of (9) depends on q alone, the right-hand side,
on t alone. Hence each is a constant — denote it by a ; it is
obvious that a ^ 0 :
* Let the student disembarasa himself of any fears due to his ignorance of tho
theory of partial differential equations. No such theory is needed in the kind of
application in Physics which we are about to consider ; it would not even be help-
ful in practice. The single function V(q, a) is obtained by a simple device fully
explained in the text.
There is, of course, a most intimate relation between the theory of Hamilton's
Equations arid the theory of this partial differential equation, as is indicated, for
example, by the " theory of characteristics" ; cf. Appendix C. The point is, that
this theory is not employed in such applications as those illustrated here. For the
latter purpose, a single solution V(q\, • • •, qnt ai, • • •, «„, 0 is all that is required,
and such a solution is obtained by ingenious devices of a homely kind, as set forth
in this Chapter.
SOLUTION OF HAMILTON'S EQUATIONS 417
J_/dTF\2 + X 2==a
The first equation gives :
Q = — aif
no constant of integration being added because we need only a
particular integral, and so choose the simplest. From the second
equation,
(dW\2
-j— ) = 2ma — m\q2.
One solution of this equation is :
W = I V2ma - m\q*dq.
Thus
/»
(10) V=-o* +
«/
0
Equations (7), § 2 here become :
dq
(11) __^_ ^_
fl_-^--<- m -— Xg2
u
This last equation gives :
and thus
(12) q
From the first Equation (11),
(13) p = V2m^ cos \- (« - 0).
Equations (12) and (13) constitute a solution of Hamilton's
Equations, which, however, is at present restricted ; for we have
not paid heed to Condition (7) on the one hand or, on the other,
418 MECHANICS
considered that the second equation (11) is restricted. Here
then is a difficulty.* Either we must follow the theory as hitherto
developed, using single-valued functions W and V ; then t is con-
fined between certain fixed values. Or else we must introduce
multiple-valued functions F, and then we must go back and
revise and supplement the general theory.
There is, however, a third choice — a way out, whereby we
can remain within the restrictions of the present theory. Accord-
ing to that theory the solution given by (12), (13) is valid so long
as
Now, from the general theory of differential equations, Equa-
tions a) admit a solution single-valued and analytic for the whole
range of values — oo<2<+oo. Equations (12), (13) yield
a solution for a part of this interval. Therefore, by analytic
continuation, the solution (12), (13) must hold for the whole
interval.
EXERCISES
1. Obtain the solution of Equations a) in the form :
(14)
p = — V2ma sin ^~ (t —
by choosing as W the function :
W = — I V2m<x — m\q2dq + C(a),
o
and suitably determining the constant of integration C(a).
2. Solve Equations a) directly, eliminating p and thus obtain-
ing the equation
* There is also a further difficulty, since the first equation (11) may not admit
a solution (suppose p0 < 0), but this difficulty can be met by choosing the negative
radical,
— V; 2ma —
SOLUTION OF HAMILTON'S EQUATIONS 419
the general solution of which can be written in the form :
-fi, 0 ^ A.
nit
3. The Simple Pendulum. Let q be the angle of displacement
from the downward vertical. Then
m ml2 .9 Tr j
T = - g2, U = mgl cos q ;
Obtain the equation for motion near the point of stable equilib-
rium:
f
dq
where t is restricted. Hence discuss the two cases : — a) oscil-
latory motion (libration) ; b) quasi-periodic motion, when the
pendulum describes continually complete circles (limitation).
Observe that, when t passes beyond the restricted interval,
the sign of the radical changes, and q changes from increasing to
decreasing, or vice versa.
4. Freely Falling Body, or vertical motion under gravity.
Here, q shall be measured downward from the initial position.
p = mq,
±p*-
420 MECHANICS
dW
Since
= ± V2ma + 2m2gq.
_ dV dW
P dq dq'
we have no option as to which radical shall be taken. If the body
is projected upward, q will bo negative for a while, and so we must
choose the negative radical for this stage of the motion. At the
turning point, (7) is not fulfilled, since d2 V/dqda does not exist.
We have now a new problem, as the body descends. The
choice of W must be made on the basis of the positive radical.
Nevertheless, both stages of the motion are covered by the solu-
tion for the first stage :
ft 1 2^y
/> — ft — o\ 2 •"%/_____ // ___ /9i
p = mg (t — /3) — V2om.
Why?
4. H, Independent of /. Reduction to the Form, H' = Pi.
We have seen in Chap. XIV, § 5, Ex. 1, that if S be an arbitrary
function of the qr, of n arbitrary constants, or parameters, the
ar, and of t :
where
*'*' •'"'*S)* o,
and if we set ar = Pr, then the equations :
fn\ dS ~ dS .,
(2) pr = ^, Qr-W, r = l,...,n,
define a contact transformation whereby Hamilton's Equations
(4), § 5, go over into (5), § 5, and
(3) H' = H + ft-
If S does not depend on t, this equation reduces to the following :
(4) H' = //.
Suppose, furthermore, that H is also independent of t :
H = H(qlf • - - , qn, plt - • - , pn).
SOLUTION OF HAMILTON'S EQUATIONS 421
Then
H' = H'(Ql,".,Q*,Pl,---,Pn).
We propose the problem of determining S so that Hr will depend
only on the Pr :
H' = H'(Plt •••,Pn\
and, in fact, that //' will be an arbitrarily preassigned function
of the Pr. Begin with the case :
(5) H'(P19 • - • , Pn) = Px.
To find such a function S(qlt • • • , qn, (xlt • • • , an), consider
the equation :
,Rv
(6)
Suppose it is possible to find a solution :
W = W(qly • - • , gn, A, «2, • ••,««)
depending on n — 1 arbitrary constants «2, • • • , an — and of
course on h, which is also arbitrary — such that*
(7)
(7)
It then follows, as we will show later, that
, 2, • • • , n
This is the function which we will choose as S :
(9) S(ql, - - • , qn, a,, • • • , a») = TT^, • • • , tfn, A, «2, • • • , «»)»
where at = A. If now we set :
(10) Pj = ai = A ; P. = a., 5 = 2, • • • , n,
then (6) becomes, because of (2), (9), and (10) :
(H) //(ft, • • • , <?n, Pi, • • • , Pn) = PI,
and hence (4) gives :
#' = PI,
as was desired.
* In practice this is done by writing down an explicit function of the nature
desired, obtained by such artifices as the separation of variables.
422 MECHANICS
Thus the transformed Hamiltonian Equations become :
dQ.
-dt^1'
(12)
0,
dt ' dt
— r = o, r = 1, • • • ,n.
2,
The solution of this system is obviously :
Ci = < + ft, Q> = P; s = 2, ••-,«;
Pr = ar, r = 1, • • • , n.
(13)
Returning, then, to the original transformation (2), which now
takes on the form :
{T dPr'
n,
(14) pr =
we liave :
(15)
The last n — I of these equations can be solved for q2, • • • , qn
as functions of q1 because of (7), thus determining the form of
the curves of the natural path of the system. And then the
first equation can be solved for ql in terms of t. This last state-
ment is conveniently substantiated indirectly. All n equa-
tions (15) can be solved for ql9 - - • , qn in terms of t because of (8).
These functions qr(t) satisfy the last n — 1 equations (15), and
so the earlier solution of these equations for q2, • • • , qn in terms
of ql become identities in t when qr is replaced by qr(t) given by
using all n equations.
Proof of Relation (8). Observe that Relation (6) is an identity
in the h, aa as well as in the qr. Hence on differentiating suc-
cessively with respect to h, a2, • • • , an, we find :
+W^ +
(16)
+ ••• +
SOLUTION OF HAMILTON'S EQUATIONS 423
The determinant of these equations is the Jacobian that appears
in (8). If it were 0, it would be possible to determine n multi-
pliers \i, • • • , Xn, not all 0, such that the n equations :
(17)
= 0
= 0
are true, and since (7) holds by hypothesis, \ may be chosen at
pleasure. Now multiply the fc-th equation (16) by X* and add.
The coefficient of each Hp vanishes, and so the whole left-hand
side reduces to 0. But the right-hand side is Xu which is arbi-
trary. This contradiction arises from supposing that (8) is not
true, and the proof is complete.
The Equation of Energy. When the kinetic energy T and the
work function U are both independent of t, H is also independent
of t, and H represents the total energy (sum of the kinetic energy T
and the potential energy — U). Hence* // is constant and we
may write :
h = H(qi9 • - - , qn, pl9 • • • , p»).
Thus this equation appears to be derived from the physics of
the problem. It is. But this derivation is not helpful in the
present theory. For we are dealing with contact transforma-
tions which reduce Hamilton's equations to a desired form, and
Equation (6) takes its systematic place in that theory. It ex-
presses a condition for the function W that will make the desired
transformation possible. Nevertheless, the physics of the situ-
ation throws a side light on the situation, which it is well to note.
The Symmetric Form. We have set, unsymmetrically, h = Pl
in Equations (10). We might equally well replace (10) by the
equations :
(100 *(Pi, • • • , Pn) = A, P* = «., « = 2, • • • , n,
where <i> (alt • • • , an) is any function such that cfa/d^ 5^ 0. The
above reasoning, with an obvious modification in detail, shows
that the determinant :
*That H is here constant along a natural path follows from Chap. XI, §3 :
dH dH
Tt - IT - °'
424 MECHANICS
d(Wq, • - - , WQ )
(8') ~^ — -^ * 0,
3(alf <*2, • • • , a«)
where W = W(qly • • • , q*9 h, alt • • • , «n) is determined as before
from (6), and h = $(«!, «2, - • • , «„). Thus the transforma-
tion (14) is justified and Equations (12) become :
(120
dQr
dt
-
dt
1, • • • , n.
The solution of these equations is obvious, and symmetric.
First,
Pr = art r = 1, • • • , n,
where the ar are n arbitrary constants. Next,
Qr = Urt + Pr, r = 1, • • • , H,
where
o>r = $r(<*i, • • • , an), r = 1, • - • , n,
and the f}r are n arbitrary constants. Thus we have, finally :
(19) , .
•ft,
a wholly symmetric solution of Hamilton's Equations.
If we should wish to use a function $(0^, • • • , an), for which
some other derivative, as d$/daz, is ^ 0, then we should need
a solution W(qlt • • • , qn, alt - • • , an) such that
5. Examples. Projectile in vacuo. Let a particle of mass
m be acted on solely by gravity, and let it be launched so that it
will rise for a time. Let qlt q2, </3 be its Cartesian coordinates,
with ql vertical and positive downward. Then
T = &2 + ft2 + </s2), I/
SOLUTION OF HAMILTON'S EQUATIONS 425
tt = ^Cpl2 + p* + P^
The equation for W becomes :
Let us try to find the desired function,
0(a,f a,) '
by setting
W. = W, + W 2 + W 3,
where Wr = Wr(qr) is a function of qr only. Thus
n
0.
"j i j
Vd^/ Vrf?2/ Vrf
Since it is only a particular function W that is needed, satisfy-
ing the Jacobian Relation of Inequality, § 4, (8), it will suffice
to set
^J2 = 2m(/l-a22-a32)-
Herc, h is determined by the initial conditions from the equation
H = A, and «2> <*3 are anv ^wo parameters such that initially
2m (h — c*22 — «32) + 2m2gql > 0.
We now may choose :
JTTf
* — -v/O TTZ A /O O Q
where «8 is positive, negative, or zero, subject merely to the rela-
tion of inequality. But, in the choice of Wlf it is the negative root,
- a,* - a,*)
that must be chosen, since
_ _ dWl
Pl J
426 MECHANICS
and p1 < 0 in the stage we are considering. We may take
0i
W1 = - CV2m(h - «22 - a32) + 2m
Cl
where cx is the initial value of qr Thus, finally,
Q\
W = - A/2m(A - <*22 ~ «3
c,
The condition (7), § 4, is satisfied.
We are now in a position to write down the solution of the
problem. It is given by Equations (15), § 4 :
t + ff =^=-m f _ ^i
dh J V2m(h - «22 - «32) + 2m*gql'
A = |^ = 2maa f
8«. J
along with the equations :
The first of the equations in each of these sets of three is in
substance identical with the one which governs the vertical
motion of a falling body, § 3, Exercise 4, where now
a = h - <*22 - «32, ]8 = - ft ;
and hence :
- §
- «22 - «32)
In the earlier case, (7 = 0 initially, and so ct mnst be set = 0.
The last two equations in the first set give :
ft)
SOLUTION OF HAMILTON'S EQUATIONS 427
and so, finally :
s = 2, 3.
The method we have employed gives the solution of the prob-
lem so long as the body is rising — no longer ; for when it is
descending, pv becomes positive, and dWl/dq1 = dW/dqt cannot
be expressed by the negative radical. This second stage of the
motion, in which the body is falling, could be dealt with by apply-
ing the method afresh with suitable modifications — in particular,
by taking the positive radical for dWt/dq^ But this step can
be eliminated if we observe that the equations we are integrating,
Hamilton's Equations, here become :
di ~mPr>
dt
r = 1,2,3;
di
2,3.
The solution of these equations is unique, and is expressed by
functions of t which are analytic for all values of L Hence the
analytic continuation of the restricted solution found above gives
the general solution, and the formulas found for qr, pr are true
generally.
EXERCISES
1. Central Force, two dimensions, attracting according to the
law of nature. Let ql = r, q2 = <p. Then :
= R
2m\dr
428 MECHANICS
/dR\* n , . 2wX a2
(W) =2mA + — -^
Thus
W=± 2mh + — -^dr + <*«>,
J T T
r*
where either the plus sign or the minus sign holds throughout
the first stage. Hence
r
dr
t +
= + m f
~ J
Vo z, , 2mX
2mh H
r
02 =
2mh
r r£
Discuss the case that the radicand vanishes for two distinct
positive values of r, expressing r as a periodic function of ^>, and
evaluate the integral that expresses t ; cf. § 9.
2. The same problem in space. Let ql = r, q% = 0, qz = <p ;
x — r cos 6 cos ^?, 2/ = r cos 6 sin ^?, 2 = r sin 0 ;
~2m
TT = fl + ® + *;
2m\
J VO
Complete the solution and discuss the cases that the radicands
have distinct roots.
SOLUTION OF HAMILTON'S EQUATIONS 429
3. Discuss the problem of § 4 when n = 1. Show that W is
given by solving the equation :
and integrating :
= ef" ^ dh
I
W
Then
ff(q,h)dq.
_8W
y dP
Thus
"- 8q-
6. Comparison of the Two Methods. We have studied two
methods of solving Hamilton's Equations, a) Reduction to the
Equilibrium Problem ; b), when // does not depend on t, Reduc-
tion to the Form, //' = Pv
The first method, being general, must apply to the second case.
It does. Let us treat this case by the first method, as set forth
in the Exercise of § 2. We will choose as V the function :
(1) V = - ht + W,
where W = W(qlt • • • , qn, h, aa> • • * > «n) is the function of § 4,
and h, as have been replaced by Plt P8. The transformation of
that Exercise,
r)V ?)V
(2) pr = — ~ Qr — r = 1 • • • n
yields an H ' that vanishes identically. The transformed Hamil-
tonian Equations thus take the form :
(Ti ^®r — n ^r — n r — 1 . . . n
W -dT-°' dt ~°' r~lj 'n*
430 MECHANICS
Departing from the notation of the Exercise, write their integrals
in the form :
... I Qr = Pr, r = !,-•-,«;
(4) 1
I P, = h, P. = a., s = 2, ••• ,n.
The solution of the original Hamiltonian Equations is now given
by substituting these values in (2) :
= 8V
(5)
* = 2, . . • , n.
But
3V_ = 3W W =_ 3W 3V =
^ * dqr ~ dqr' dh + dh ' da, ~ da,'
Hence Equations (5) agree not only in substance, but even in
form, save for one exception, with Equations (15), § 4. The
equation arising from differentiation with respect to h in the
earlier case read :
Here it is :
7. Cyclic Coordinates. It frequently happens that H, besides
being independent of t, contains fewer than n g's. Begin with
the case of one q,
(1) H = H(ql9plt---,pn).
From Hamilton's Equations,
(2) f-f-0. -«.-.•,
and hence
(3) ?>« = ««, 5 = 2, • • • , n.
It is not difficult to complete the solution by means of Hamil-
ton's Equations and the integral of energy,
(4) h = H(ql9p19 • • • ,pn);
SOLUTION OF HAMILTON'S EQUATIONS 431
but this is not the form of solution in which we are interested.
We desire a discussion by the methods of § 4 ; in particular, by
the transformation :
dW
/*\
(5)
where
(6) W = W(qi1 • • - , ?„, h, a» • • - , an)
is a solution of the equation :
m\
l, —,...,— J,
,,,...,
••-
and Pl = A, P, = aa, s = 2, • • • , n.
To find such a solution we turn to the Method of Separation
of Variables, which has rendered such good service in the past.
Let
'(9) . w=Wl + -- + Wn,
where Wr = Wr (<?r) is a function of qr alone — and of the n
parameters, A, a2, • • • , «„. From (5) and (3) we see that
a" « « 2, - - - , n,
and so we try :
TT, = ««g., s = 2, • - - , n.
Let T^! be denoted more simply by v :
(10) W, = ^(fc^afc-'-.aO =t>.
Then (7) becomes :
(11) flr(g1,^,a2,---,«n)=A.
If we assume that
fiff
(12) l^ffp.feuPi."*'--.*..)*0'
^Pl
and solve the equation :
(13)
432 MECHANICS
for p, :
(14) p, =*(?„ A, «2, ••',«»),
we have :
(15) -^ = *(?i, A, a,, ••-,«„).
Now choose as » :
«i
(16) v = J *(<?i> A, a2, • • • , a.) dqlt
C
where c is a numerical constant.
We are thus led to a function
(17) W = V + otf, + • • • + anqn
of the desired kind, provided the Jacobian relation (8) is satisfied,
The Jacobian here reduces to
d'2v
" ' °r '
dq.dh " dh' dh'
where p1 is determined by (13). On differentiating (13) -we find :
M!£!
dp, dh
and so the Jacobian does not vanish.
Solution of Hamilton's Equations. We can now apply the
general theory of § 4. The transformation (5) of the present
paragraph carries Hamilton's Equations over into the form:
= 0, r-l,-..,n,
the solution of which is :
ft =« + fc, Q. = A, s = 2, ..-,n;
Px = A, P, = aa, s = 2, - - - , n.
These values for Qr, Pr are to be substituted in (5), and the result-
inc eauations solved for qr, pr :
SOLUTION OF HAMILTON'S EQUATIONS 433
<7i
dh J dh l9
aw
dh
8W r , , .
Pi = - = *Wi, h, a2, • • • , an),
(18)
dW 0
p9 = -7T-- = a,, s = 2, • • • , n.
The equations of the second line determine g. as a function
of ft:
/' (W
~^~dqly s = 2, '--,n.
8
The first equation gives ft as a function of t.
EXERCISE
Obtain the final result (18) directly from Hamilton's Equations.
8. Continuation. The General Case. Let H depend on
1 < v < n arguments qk :
(1) H = #(ft, ••-,?„?!,••• ,pn).
The method of treatment is similar, though the solution cannot
in general be obtained by quadratures. Equations (3) of § 7
now become :
(2) p, = a., s = v + I, • • • , n.
By analogy we now seek to determine W in the form :
(3) W = v + a*+lqr+1 + ••• + anqn,
(4) v = *(?„ •••,<?„, h, «2, • • • , «„),
(5)
a^T
434 MECHANICS
Equation (7), § 7, for W now becomes :
/c\ L rj
(b) n = //^ft, • • - , q,, TT-, • • • , Tjjp a^+i, • • • ,
This is an equation of the same type as (6), § 4, but with v < n
variables qr. As in the earlier case, only a particular solution
is sought, and such a solution may be found by special devices,
notably the method of separation of variables.
A function v once found, the solution proceeds as before.
(7)
/Si = - + qi, I = v + 1, • • • , n.
From the equations of the second line g* can be found in
terms of ft, k = 2, • - • , p. From the first equation ft is now
found in terms of t. Finally, qi is given by the last line,
I = v + 1, • • • , n.
9. Examples. The Two-Body Problem. Consider the motion
of two bodies (particles) that attract each other according to
the law of nature and are acted on by no other forces. Their
centre of gravity travels in a right line with constant velocity,
or else remains permanently at rest. We will assume the latter
case. Then each of the bodies moves as if attracted by a force
at 0, the centre of gravity, which is inversely proportional to
the square of the distance of the body from 0.
We will first discuss the motion in a plane — later, in space.
Let the particle be referred to polar coordinates, ft = r,
g2 = (p. Then
" 2\dt2 dt2
Hence
Let
W = v + a2q2.
SOLUTION OF HAMILTON'S EQUATIONS 435
Then v is given by the equation :
1 f/^\2 , «221 ^ _
2m \\dql/ (7j2J ql '
or
rt 7 . 2mX <*22 ,
2mA H 1- dr.
r r2 '
where a definite one of the two signs holds for the first stage of
the motion.
Equations (18) of § 7 now give the solution of Hamilton's
Equations in the form :
. • ^ dv /•* Sv .
or
,=»/-
c ±
dr
2mh +
2m\
ft =
The directness of the result is particularly noteworthy. It
has not been necessary to make use of skillful devices or to effect
complicated eliminations. From the evaluation of the second
integral r can be expressed as a trigonometric function of <p. But
the discussion of r in terms of t is more complicated ; cf . below
the reference to Charlier.
The Orbit in Space. To treat the motion in three dimensions
let
x = r cos 0 cos <pt y = r cos 0 sin ^>, z = r sin 0.
Then
Let
0,
Then
Pi
Ps
436 MECHANICS
H = ±(v * + Si P»2
2m \I tf,2 ?,2 cos2 ?2/ ql
Since H = H(qv q^, p1( pa, p3), we see that p3 = a3 (const.)- Thus
W = v + ci^s,
where v is given by the equation :
J_[7-^Y + A. fi?Y 4. «*2 i _ x = ,
2m lA^/ qf\dqj q,2 cos2 q.2J ql
Hero, t-hcro arc only two independent variables, q^ = r and
g.2 = 6. The equation can be written in the form :
0.
cos
On setting
v = R +
the variables can be separated :
- r*(~^ + 2mhr* + 2m\r =
Hence
ft
®
y
ft
= r±v^ _ a.,2
where the signs are determined for a particular stage of the motion,
and c, T are arbitrary numerical constants. Adding the further
term a3g3, we have:
W = v + ag^, v = R + 0.
We are thus led to the solution of the problem in the form given
by (7), §8:
SOLUTION OF HAMILTON'S EQUATIONS
,+A-.A *
437
(2)
&=-
r
dr
J . 9^/0 T, 2raX a,2
c ± r2 \2rnh -\ \
* r r2
6
d6
- a32 sec2
± v a,22 — a32 sec2
The discussion of this solution on the hand of the explicit
evaluation of the integrals and the inverse functions thus arising
presents practical difficulties. The problem is of so great impor-
tance in Astronomy that it has been treated at length by Charlier,
Mcchanik des Himmels, Vol. I, Chap. 4, p. 167. On p. 171,
Equations (7) are identical with our solution, save as to notation.
Failure of the Method. There are cases in which the method
breaks down. Consider, for example, motion in a plane. Sup-
pose the body is projected from a point A, distant a from the
centre of force, 0, at right angles to the line OA and with a velocity
v(} such that
It will then describe a circle, r = a. But the Equations (1)
or (2) can obviously never yield this solution. Why?
The function v was determined from the equation :
= 2mA
2raX
In the present case,
h =-
2a'
mav0,
and hence
= 0.
438 MECHANICS
Thus the condition
is not fulfilled, and so, of course, there is no reason why the method
should apply, since the hypotheses on which it depends do not
hold.
10. Continuation. The Top. We take over from Chapter
VI, § 18, the expression for the kinetic energy,
By Euler's Geometrical Equations, this becomes :
T =
Let
tfl ^ Q> <?2 = <P>
Then, since
= dT
we have :
Pi = Ad,
ps = C<p cos 0 + ( A sin2 0 + C cos2
Thus T, expressed in terms of the p's and <?'s, becomes :
Furthermore,*
U = — Mgb cos 0.
Thus
Hence it appears that the problem comes under the case of
cyclic coordinates treated in § 7. First, then,
* It is necessary to change from the earlier notation h for the distance from
the peg to the centre of gravity, since h plays so important a rdle in the present
theory. Let the distance be denoted by 6.
SOLUTION OF HAMILTON'S EQUATIONS 439
To determine v we have :
Irl dv* , 1 9 . l/a2cosq1 - a3\21 . ,_ , L
2 Li 55? + C^ + l( sin gl ) J + M*6 COS * - *•
^2
sin2 ft ^—2 = (24 A - La22 - N cos ft) sin2 ql - («2 cos q1 - «3)2,
_
„ = r+ V(2^ A - La22 - JV cos ft) sin2 g, - («2 cos ?, - «^
t/ sin ft ?1>
c
where € is an arbitrary numerical constant, not a parameter,
and the sign of the radical must be chosen with respect to the
special stage of the motion under consideration. Moreover, for
brevity,
L = 4, N = 2AMgb.
v/
The solution of the problem, as given in § 7, now takes on the
form:
«+*-*
Thus
A sin ft (
4- C-
'~e/~H
± v (2Ah — L«22 ~ A^ cos gj sin2 gj — («2 cos gj — as)2
Let
u = cos #,.
Then this equation becomes :
9l = r ^
c
cose,
e/ +V/<W
where
This is the same result obtained by elementary methods,
Chap. VI, § 18. But compare the technique. With only Euler's
440 MECHANICS
Dynamical and Geometrical Equations to work with,* elimina-
tions had to be made by ingenious devices, whereas the present
advanced methods free the treatment from all artifice. The
fundamental equation in desired form is evolved naturally, directly,
from the general theory, not untangled from a snarl of equations.
Instead of having to solve three equations for 6, <j>, \j/ by more or
less ingenious methods of elimination, the functions 77, [7, and
hence H are obtained without the use of any artifice whatever,
and the method of § 7 yields ql — 0 at once as a function of t, the
further equations giving q3 = <p and </3 = \f/ immediately.
EXERCISE
Study the motion of a top with hemispherical peg, spinning
and sliding on a smooth table. Show that
where
F(u) = (2h - ^ -
11. Perturbations. Variation of Constants. In the problem
of perturbations the motion which the system would execute if
only the major forces acted is regarded as fundamental, and
then the variation from this motion due to the disturbing forces,
thought of as slight, is studied.
This analysis of the physical problem is mirrored mathematically
by writing down Hamilton's Equations for the actual motion :
___ -.-
~dt ~Wr dt" dqr' ' ''
and then setting the characteristic function H of the actual prob-
lem equal to the H0 of the problem due to the major forces, plus
a remainder, Hl :
(2) H = H, + H{.
* It is true that in the earlier treatment we had two integrals of the differen-
tial equations of motion to work with at the outset, namely ; the equation of energy,
T = U + h, and the equation arising from the fact that the vector moment of mo-
mentum <r is always horizontal. But even so there were three equations in 0, $, <f>
to integrate.
SOLUTION OF HAMILTON'S EQUATIONS 441
Transformation of the Major Problem to the Equilibrium Problem.
First, the major problem, represented by Hamilton's Equations
in the form :
is solved by reducing it, through a contact transformation, to
the Equilibrium Problem. The contact transformation is given
by the equations :
fA\
(4) Pr r
where
(5) F« = F»(<7i, • • • , q«, P,, • • • , Pn, t)
is obtained as follows. Write down Jacobi's Equation, cor-
responding to Hamilton's Equations (3) :
Let
V° = V°fe, ••-,?», alf •", an, 0
be a solution of this equation such that the Jacobian
O / v 7~ VJ.
0(al9 - • • , an)
In this function, replace ar by Pr. The resulting function is the
function (5). [In practise, the function F0^, • • - , qny alf • • • ,
an, t) is obtained, not from an elaborate theory of partial differ-
ential equations, but by means of simple devices, ad hoc.]
Let the transformation (4) be written in the explicit form :
,. f Qr = Fr(plt ' * ' 9Pn9qi, ' * * 9qn,
I Pr = Gr(plt • • ' jpniQi, ' " ,qn,
or
f 9r = fr(P» • • • , P«, Q» -^Qn,
1 pr = gr(Ply ' • • , Pn, Q19 • • - , On,
To say that the major problem is thereby transformed to the
Equilibrium Problem means that, when the variables qry pr that
442 MECHANICS
form the solution of Equations (3) are subjected to the transfor-
mation (4), the resulting Hamiltonian Equations become :
(7) f = 0, ^ = 0, r=l,..-,n.
The solution of these equations can be written in the form :
(8) Qr = 0r, Pr = «r, f = 1, • • • , ft,
where ar, Pr are constants. Now transform the variables Qr, Pr
that are the solution of Equations (7), namely, the functions
given by (8), back by means of the transformation (4"), and we
have the solution of Equations (3) in the form :
f Qr = /r(«i, ' ' ' , «n, ft, • • • , 0n, 0
1 Pr = 0r(alf " -,«», ft, •' ',0«, 0
Thus the transformations (4;) or (4"), and (9), identical
except in notation, represent two distinct things:
a) In the form (4") these equations represent the Contact
Transformation (4).
b) In the form (9) they represent the Solution of the Hamil-
tonian Equations of the Major Problem, or (3).
Transformation of the Actual Problem by the Same Contact Trans-
formation. We now proceed to apply the contact transformation
(4), not to the variables (qr, pr) which satisfy Equations (3),
but to the variables (qr, pr) of the original problem, which satisfy
Equations (1). Since this is a contact transformation, we know
that Equations (1) will go over into new equations of the same
form:
dQ,_8ir dPr__9ff' j.
(W) dt ~ dPr' dt ~ Wr ' '
Here H' = H'(Qr, Pr, <) has the value, cf. Chap. XIV, § 5, Ex. 1,
(15):
(11) H' = H +
But from (6) :
Hence
H' = H -
SOLUTION OF HAMILTON'S EQUATIONS 443
Finally, from (2) it follows that
(12) Hf = fft.
Thus Equations (10) take the form :
dQr^dH, dPr _ m,
The result may be stated as follows. When the variables qrj pr
which form the solution of the actual problem represented by Equa-
tions (1) are transformed by the contact transformation (4) or (4'),
the transformed equations take the form (13), where Hl is the given,
or known, function of Equation (2), now expressed through (4) or
(4") in terms of Qr, Pr, t.
The Final Solution. It is now but a step to the solution of
Equations (1), which represent the actual problem. Solve Equa-
tions (13), thus determining Qr, Pr as functions of t. Then
transform these functions, the solution of (13), back by means
of (4) or (4") to the variables qr, pr. The latter satisfy Equa-
tions (1).
The result can be expressed in the form :
(14)
Pr =flTr(P,, •••,Pn,Qi, ' ' ' , Qn, 0
where Qr, Pr arc determined by Equations (13).
Variation of Constants. The method above set forth has been
called the "variation of constants." This expression is a mathe-
matical pun. It is a pun on the letters ar, 0r. These, in Equa-
tions (9), are constants — the equations there representing the solu-
tion of the major problem, (3). On the other hand, they can be
identified with the variables Pr, Qr of (14), these variables being
determined by (13), and then Equations (14) represent the solu-
tion of the actual problem, (1).
We can attain complete confusion of ideas, as is done in the
literature, by changing the notation in (13) and (14) from
Qr, Pr to 0r, oLr. Thus (14) goes over into the form of (9), and
(13) is replaced by the equations :
dar _ 3(- g,) df)r _ B(- H,)
dt ~ d0r ' dt ~ dar ' r-L>'"'n>
444 MECHANICS
where H1 = H1(Qr, Prjt) is now written as Hl(ft1)a1)t)) the
Hamiltonian function now being — H1 instead of ff,.
Thus the pun is explained — but it is a poor pun that has to be
explained.
12. Continuation. A Second Method. It is possible to treat
the problem of perturbations in still another manner. Let
<f>(alf • • • , an) be any given function whose first partial deriva-
tives are not all 0. Let the Hamiltonian Equations for the undis-
turbed motion, namely, (3), be transformed by a new contact
transformation :
where S is defined as follows. Consider the equation :
/i/r\ / \ T
(16) *(«,, -..,«.) =
Let
• • • , qn, «!,-•-, an, t)
be a solution such that *
Now, make the contact transformation :
/1>7\ ^ r> ^
(17) Pr-Wr, Pr=~W,
where
5 = 8(qll • • • , q^ Q,, • • • , Qn, 0-
* In order to find such a solution, begin with the equation :
, „/ OS dS \ , dS
h=H(qi,...,qn,~,...,--,t) + -,
where h is an arbitrary constant, and seek a solution :
S = S (<?,, • • •, qn, h, «2, • • •, On, Of
such that
t gt, ' ' ', Qn
d (h, as, • • -, «„)
where as, • • •, On are arbitrary. Substitute
h = <p (ai, • • •, an)
in iS. If d<p/dai ^ 0, this will be the function desired.
SOLUTION OF HAMILTON'S EQUATIONS 445
This transformation, applied to Equations (1), carries these over
into equations of the same type :
dt dPr' dt 8Qr' ' ' '
where, by Chap. XIV, § 5 :
(19) H' = H + ?j-
VI
But, by (16) and (17) :
Hence, with the aid of (2) :
the arguments now being the Qry Pr into which gr, pr have been
transformed by (17). Thus Equations (18) take the form :
dQr = 0/7, dPr _ _ 3/7, _ 3jp
" r-1"'n-
Solve these equations and substitute the functions of t thus
obtained, namely, the Qr, Pr, in (17). The functions qr, pr of t
obtained from these equations are the solution of the actual prob-
lem, or Equations (1).
Carathgodory * treats Equations (20) as follows. He writes
X//! instead of H1 :
He then develops the solution into a power series in X :
f Qr = Ctr + \Clr + X2C2r + • • • ,
<22> ( Pr = 0r-vat+ Wlr + ' ' • ,
where C*r, Dkr are functions of t, vanishing when t = 0 (for sim-
plicity we have set t0 = 0). On substituting these values for
Qr, Pr in (21) and equating coefficients of like powers of X, the
coefficients C*n Dkr can then be obtained by quadratures.
* Cf. reference above, p. 381. The page in R.-W. is 211.
APPENDIX A
VECTOR ANALYSIS
In Rational Mechanics only a slight knowledge of Vector
Analysis is needed. It is important that this knowledge be
based on a postulational treatment of vectors. The system of
vectors is a set of elements, forming a logical class. Certain
functions of these elements are defined, whereby two elements
are transformed into a third element. These functions are called
addition, multiplication by a real number (here, only one element
enters as the independent variable), the inner product (scalar
multiplication), and the outer product (vector multiplication).
The functions obey certain functional, or formal, laws, which
happen to be a subset of the formal laws of algebra :
A + B = B + A
AB = BA
A(BC) = (AB)C
A (B + C) = AB + AC
(B + C)A = BA + CA
A brief, systematic treatment such as is here required is given
in the Author's Advanced Cakulus, Chap. XIII. For a first
approach to the subject the Hamiltonian notation of S and V for
the scalar and vector products has the great advantage of clear-
ness in emphasizing the functional idea — the concept : transfor-
mation. On the other hand the notation pretty generally adopted
at the present day is the designation of vectors by Clarendon or
boldface, the scalar product being written as a • b or ab (read :
a dot b), and the vector product as a X b (read : a cross b). It
is useful, therefore, to have a syllabus of definitions and essential
formulas in this notation.
447
448 APPENDIX A
1. Vectors and Their Addition. By a vector is meant a directed
line segment, situated anywhere in space. Vectors will usually
be denoted by boldface letters a, A, or by parentheses; thus
a vector angular velocity may be written (w).
Two vectors, A and B, are defined as equal if they are parallel
and have the same sense, and moreover are of equal length :
A = B.
By the absolute value of a vector A is meant its length ; it is
denoted by | A |, or by A.
Addition. By the sum of two vectors, A and B, is meant their
geometric sum, or the vector C obtained by the parallelogram
law:
A + B = C.
In order that this definition
may apply in all cases, it is
necessary to enlarge the system
of vectors above defined by a nul vector, represented by the
symbol 0.
If B is parallel to A and of the same length,
but opposite in sense, then
A + B = 0, or B = - A. B
FIG. 149
Moreover, we understand by wA, where m is
any real number, a vector parallel to A and m times as long ; its
sense being the same as that of A, or opposite, according as m
is positive or negative. If m = 0, then raA is a nul vector :
OA = 0. The notation Am means wA, and also
aA + 6B a . , 6 _
r- j— means . , A H r—r B.
a + o a + o a + o
Vector addition obeys the commutative and the associative law
of ordinary algebra :
A + B = B + A
A + (B + C) = (A + B) + C
Subtraction. By A — B is meant that vector, X, which added
to B will give A :
B + X = A, X = A - B.
VECTOR ANALYSIS 449
To obtain X geometrically, construct A and B with the same
initial point ; then A — B is the vector whose initial point is the
terminal point of B, and whose terminal point is the terminal
point of A ; Fig. 149.
Cartesian Representation of a Vector. Let a system of Cartesian
axes be chosen, and let i, j, k be three unit vectors lying along
these axes. Let A be an arbitrary vector, whose components
along the axes are Alt A2J Az. Then evidently
A = AJ + A2j + Azk.
E = B,i + B,j + B,kt
then
A + B = (A, + BJi + (A, + B,)j + (A, + £3)k.
Also : _
A ^
Resultant. If n forces, Fx, F2, • • • , Fn, act at a point, their
resultant, F, is equal to their vector sum :
F = Ft + F2 + • • • + Fn.
If n couples, Mu M2, • • • , Mn, act on a body, the resultant
couple, M, is equal to their vector sum :
M = MJ + M2 + - • • + Mn.
Two or more vectors are said to be collinear if there is a line
in space to which they are all parallel. In particular, a nul vector
is said to be collinear with any vector. Three or more vectors
arc said to be complanar if there is a plane in space to which they
are all parallel. In particular, a nul vector is said to be parallel
to any plane. If three vectors, A, B, and C, are non-complanar,
then no one of them can vanish (i.e. be a nul vector) and any
vector, X, can be expressed in the form :
X = ZA + mB + nC,
where Z, m, n are uniquely determined.
Differentiation. Velocity. Acceleration. Osculating Plane. A
variable vector can be expressed in the form :
A =
450 APPENDIX A
where i, j, k are three fixed vectors mutually perpendicular. If
/(Oi <p(t), ^(0 have derivatives, the vector A will have a de-
rivative defined as
lim-rr =
Its value is :
Moreover,
If m is a function of x and A is a vector depending on x> and if
each has a derivative, then mA will have a derivative, and
d(mA) dm. . of A
-~^j — - = -y~ A + m~T'
ax ax ax
If a point P move in any manner in space, its coordinates being
given by the equations :
where /, <p, $ are continuous functions of the time, having con-
tinuous derivatives, and if
r = xi + yj + zk,
the vector velocity of P is represented by
W /i <f>, t have continuous second derivatives, the vector
acceleration of P is given by
The plane determined by the vectors r and f drawn from P
(on the assumption that neither is a nul vector) is the osculating
plane. Thus the vector acceleration always lies in the osculating
plane.
2. The Scalar or Inner Product. The scalar or inner product
of two vectors, A and B, is defined as the product of their absolute
values by the cosine of the angle between them. It is denoted
by A-B or AB and is read: "AdotB." Thus
A - B = AB = | A | | B | cos c.
VECTOR ANALYSIS 451
If one of the factors is a nul vector, the scalar product is defined
asO.
The commutative and the distributive laws hold :
AB = BA
A(B + C) = AB + AC.
The associative law has no meaning.
The scalar product vanishes when either factor is a nul vector ;
otherwise when and only when the vectors are perpendicular to
each other. Furthermore :
often called the norm of the vector.
:2 — i ;2 — i k2 _ i
i—i, j — i, K — i,
jk = 0, ki = 0, ij = 0.
Cartesian Form of the Scalar Product :
= AB
Differentiation :
If a is a unit vector, i.e. if | a | = 1, then
a2 = 1, and aa' = 0.
3. The Vector or Outer Product. Let two vectors, A and B,
be drawn from the same initial point. Then they determine
a plane, M, and a parallelogram in that plane. The vector
or outer product is defined as a vector perpendicular to M and
of length equal to the area of the parallelogram. Its sense
is arbitrary. It is defined with ref- AXB
erence to the particular system of
Cartesian axes to be used later. It is
denoted by
AXB
and is read : " A cross B." FIG. 150
452
APPENDIX A
If one of these vectors is 0, or if the vectors are collinear, neither
being 0, the vector product is defined as 0, and these are the
only cases in which it is 0. Otherwise, let e be the angle between
the vectors. Then
| A X B | = | A | | B | sin c.
The commutative law does not hold in general, for
AXB=-BXA.
The associative law does not hold ; e.g. (i X j) X j 5^ i X (j X j).
But the distributive law is true :
and
C)=AXB
as c^n be proved geometrically, or still more simply, analytically,
by means of the Cartesian form ; cf . infra.
It is convenient to choose the sense of the vector product so
that
i X j = k, j X k = i, k X i = j.
In any case
A X A = 0,
and so, in particular,
i X i = 0, j X j = 0, k X k = 0.
Cartesian Form of the Vector Product :
A X B =
i J
**l -^-
Differentiation:
-
dx
dx
-
dx
4. General Properties. Let A, B, C be three non-complanar
vectors drawn from the same point. The volume of the paral-
lelepiped determined by these vectors is numerically
A • (B X C).
VECTOR ANALYSIS
453
A necessary and sufficient condition that three vectors A, B,
C be complanar is :
A - (B X C) = 0.
Linear Velocity in Terms of Angular Velocity. Let space be
rotating about an axis / with vector angular velocity (w). Then
the velocity v of an arbitrary
point P will be :
v = («) X r,
where r is the vector drawn
from any point O of the axis to
the point P. If the axis passes through the origin, then
Fia. 151
v =
J
x y
and
vx = zwy — ywz
Vy = Xtl)g — Z<l)X
If it passes through the point (a, 6, c), then
vx = (z - c) coy - (y - 6) wz
vz — (y — 6) o)x — (x — a) uy
In tho general case of motion of a rigid body (i.e. motion of
rigid space), lot 0' : (z0, y0, z0) be a point fixed in the body, and
let (f, ??, f ) be the coordinates of any point P fixed in the body,
the origin being at 0' ; but otherwise the (£, rj, f )-axes may
Mp move in any manner. Then
FIG. 152
O'
where
« P 7
COf CO, C0£
* i r
V = V0 + V',
v0 =
454 APPENDIX A
Localized Vectors. It is sometimes convenient to prescribe the
initial point of a vector, or the line in which the vector shall lie,
as in the case of a force acting on a particle, or a force acting on
a rigid body. It is with reference to such vectors that the follow-
ing definitions are framed.
By the moment of a vector F with respect to a point 0 is meant
the vector
M = r X F,
where r is the vector drawn from 0 to any point of the line in
which F lies. In practice, F may be a force acting on a rigid body,
or F may be the vector momentum, mv, of a particle.
The moment of a couple can be expressed as
TI X F! + r, X F2,
where FD F2 are the forces of the couple and rt, r2 are vectors
drawn from any point 0 of space to any points Plt P2 of the lines
of action of FD F2, respectively.
By the moment of a vector F about a directed line L is meant the
vector
M = Ma, M = a - (r X F),
where a is a unit vector having the direction and sense of L, and
r is the vector drawn from any point of L to any point of the line
in which F lies. Thus if F is a force acting on a rigid body, let
its point of application be transferred to the point P nearest to
L, and let 0 be the point of L nearest to F ; i.e. OP is the common
perpendicular of L and the line of action of F. Decompose F
at P into a force parallel to L and one perpendicular to L. The
vector moment of the latter component at P with respect to 0
is Ma.
5. Rotation of the Axes. Direction Cosines. A transforma-
tion from one set of Cartesian axes to a second having the same
origin (both systems being right-handed, or both left-handed)
is characterized by the scheme of direction cosines :
X
y
z n^ n% 7i3 fc HI w2
VECTOR ANALYSIS
455
Between the nine direction cosines there exist the following
relations :
mS + m22 + m32 = 1
n* + n22 + n32 = 1
IS + mS +
1
n
+ tn2w3 + n2na = 0
+ m^ml + n3nx = 0
-f w,m2 + nxn2 = 0
== 0
n^j + n2Z2 + n3i!3 = 0
JiW, + Z2m2 + J3ms = 0
— m3n2
ml = n2Z3 — n3Z2
n2 = Z3mI — Z^j
Z3 =
1.
APPENDIX B
(dij\ ^
~jj) = /(«)
Differential equations of the form :
where
I. f(u) = (t*-a)(6 -*)
or
II.' /(w) = (w - a) (6 - tO
and ^ (w) is continuous and positive in the interval
a ^ u ^ bj
play an important role in Mechanics. Let us study their integrals.
CASE I. A particular integral of (1) is found by extracting
the square root :
du
and separating the variables :
du
dt =
<2> '-Ajsr
a ^ u ^ b.
Geometrically, the function on the right of (2) can be inter-
preted as the area under the curve,
a b
(3)
\ /
w
u The graph of the function
FIG. 153 (4) y = V(u - a) (6 - u
456
A DIFFERENTIAL EQUATION
457
is represented by Fig. 153. The reciprocal of an ordinate of this
curve gives the corresponding ordinate of the graph of the func-
tion (3), Fig. 154 :
/r\ .. _ m
V(u-a)(b -u)t(u)
The area under the curve (5), shaded in
the figure, represents the integral (2), or: v
(6) t
r du_
~~ J V^-~Mb~^
Thus this area expresses t and brings out
the fact that t increases as u increases.
Conversely, u increases as t increases. °
Let A be defined by the equation :
a u b
FIG. 154
(7)
r du^
J V(1T^~ci)(6~-
Then the graph of u, regarded as a func-
tion of ty
(8)
u =
0
FIG. 155
is as shown in Fig. 155. Its slope is 0 at
each extremity and positive in between.
The definite integral, (2) or (6), has now served its purpose.
It has yielded for a restricted interval,
0 ^ t g A,
a particular solution of (1).
Continuation by Reflection. Reflect the graph of the func-
tion (8), Fig. 155, in the axis of ordinates, and let the curve thus
obtained define a continuation of the function ^(0 throughout
the interval — A ^ t rg 0. Analytically the reflection is repre-
sented by the transformation :
Thus
= *(- 0,
-A
458 APPENDIX B
The extended function :
is seen to satisfy the differential equation
fdu\
(w) =
Hence the function <p(t) thus defined in the interval (— A, A), or
u = <*(0, - A ^t ^ A,
is a solution of (1).
To complete the definition of <p(f) for all values of t, i.e.
— oo < t < oo, we could repeat the process of reflection, using
next the lines / = A and t = — A ;
anc* so on' ^ut ** *s s^mP^er to
! /i\ introduce the idea of periodicity.
\
v
Periodicity. Let the function
now be extended to all values
of t by the requirement of peri-
-2A -A
FIG. 156 odicity.
(9) <f>(t + 2A) = <p(f), - oo < t < oo.
Then we have one solution of Equation (1).
The General Solution. The general solution of Equation (1) in
the present case can now be written in the form :
(10) u = <p(t + 7),
where 7 is an arbitrary constant. Observe that
(U) *>(- 0 = <f>(i).
Hence
/i o\ '/— t\ ft\
To an arbitrary value u0 of u such that a < w0 < 6 there
correspond two and only two values of t in the interval (— A, A),
for which
(13) UQ = ^>(0,
namely
If w0 = «, there is only one value, namely, t = 0; and if
w0 = &, then t — Ay —A. But only one should be counted,
A DIFFERENTIAL EQUATION 459
since the fundamental interval of periodicity should be taken as
an open interval,
c <t g c + 2A or c g t < c + 2A,
where c is arbitrary. Moreover, du/dt has opposite signs in J0
and t'Q, because of (12).
We can now prove that there is a solution of the given differ-
ential equation, which corresponds to arbitrary initial condi-
tions : u = ul9 t = t19 provided merely that
a g H! ^ 6.
Suppose that it is known from the physics of the problem that
du/dt is negative initially. Now, set UQ = ^ and determine
tQ as above so that
w0 = ^(a *'(*o) <0.
Finally, define 7 by the equation :
*i + 7 = tQ, 7i = <o - 'i-
Thus 7 = 7i is uniquely determined and the function
(14) u «*(« + 7i)
is the solution we set out to obtain.
But is this solution unique, or are there still other solutions
which satisfy the same initial conditions? If a < u^ < 6, the
answer is affirmative for values of t near ^ ; but for remote values,
the question of singular solutions arises, to which we now turn.
Singular Solutions. The given differential equation admits,
furthermore, singular solutions. The functions
u = a, u = 6
are obviously solutions of the differential equation :
(15)
each being considered in any interval for t, finite or infinite. Such
a solution, moreover, may be combined with a solution (10) at
any point. The solution now may follow (10) indefinitely; or
it may switch off on a singular solution again.
These solutions do not, however, have any validity in the
problems of mechanics, for which the above study has been made.
The mechanical problems depend each time on differential equa-
460 APPENDIX B
tions of the second order, and these have unique solutions, depend-
ing on the initial or boundary conditions. Equation (14) repre-
sents an integral of these equations. But the converse is not
true, namely, that every integral of (15) is an integral of the
second order equations — why should it be ? We see, then,
that we may be on dangerous ground when we replace the latter
equations, in part, by the integral of energy, for example ; since the
modified system may have solutions other than that of the given
mechanical problem. Cf. the Author's Advanced Calculus, p. 349.
Does this remark not call into question the validity of the
treatment in Chap. XV, since the equation :
is essentially the integral of energy? Not if we apply that method
as set forth in the text. For in a suitably restricted region there
is only one solution yielded by those methods, and we were careful
to point out that it is the analytical continuation of this solution
that yields the solution of the mechanical problem beyond this
region. Thus the singular solutions are automatically eliminated.
CASE II. This case :
(16) (J~J = (u - a) (6 - u)*t(u),
is more easily dealt with. A particular integral of (16) is given
by the formula
(17) t = f-(b _ dM— === , a ^ u < b
a
The inverse function,
(18) u = *>(0, 0 ^ t < oo,
represents an integral of (16) in the interval indicated. And
now this solution can be completed by the definition :
(19) <p(- t) = <p(t).
Thus we have one solution :
u=b (20) u = «,(<),
— oo < t < oo.
It is now shown as before that
the general solution is
A DIFFERENTIAL EQUATION 461
(21) u = ?(« + 7).
A further case, namely :
(22)
can be treated in a similar manner; or, more simply, be thrown
back on the case just considered by a linear transformation.
Finally, the case (not mentioned above) :
(23)
breaks up into the two distinct equations :
a) -£ = + (u - a) (6 - t«)vV(tt);
b) ft=-(u-a)(b-u)Vt(d.
Each of these is solved at once by a quadrature.
FURTHER STUDY OF CASE I. There is another treatment of
Case I which brings out the important fact that the function <p(t)
is essentially a sine or cosine function :
(24) u = C cos 0 + C',
where 0, in the simplest case, is proportional to the time :
* = j<,
and in the general case is of the form :
where h (/) is periodic with the period 2A :
h(t + 2A) = h(t).
This method, moreover, may simplify the computation in case
it is desired to tabulate the function (p(t).
The given differential equation :
(25)
462
APPENDIX B
can be reduced by a linear transformation :
, 2u - a - 6
U= b-a '
to the form, after dropping the accent :
(26)
Make the substitution :
(27) u = cos 6, 0 < 0 < IT.
Equation (26) becomes, on suppressing the factor * sin2 0 :
(28)
This equation is equivalent to the two equations :
(29)
(30) =
The solution of (30) is obtained from the solution of (29) by
changing the sign of t.
A particular solution of (29) is given by the quadrature :
(3D
Write
(32)
where
(33)
9
"/:
de
^(cosfl)'
- oo <e < oo.
f . d9 - 24.
_J V^(cose)
Then g(6) is periodic with the period 2v. For,
(34)
* In so doing we suppress the singular solutions of (26).
A DIFFERENTIAL EQUATION 463
But the value of the integral, because of the periodicity of the
integrand, is 2A for all values of 0. Hence
(35) g(e + 2r)
Equation (31), or its equivalent,
(36) t = ^ +
defines 6 as a single- valued function of t, since the integral (31)
represents a monotonic function of 0. Let 0 be written in the
form:
(37) 0= J + /KO-
Then h(t) has the period 2 A :
(38) h(t + 2A) = h(t).
For, let 0 have an arbitrary value 00 in (36) and let the correspond-
ing value of t be t0 :
Let 0 = 00 + 27r, and let t' be the new value of t :
,,*('. + 20 +,(,. + ar).
By virtue of (35),
or t' = <0 + 24.
From (37) we now infer :
Hence
and the proof is complete.
If we multiply (36) by v and (37) by A and add, we find :
(39) »0(»)+AA(0 =0.
464 APPENDIX B
We are now ready to express u in terms of /. In Equation (27)
B was restricted. Now, set generally :
(40)
This function is seen by direct substitution to be a solution of (26).
If we denote it by <p(f)y the general solution of (26) will be :
(41) u = *>(* + 7).
The other equation, (30), leads to the same result.
If it is a question actually of computing h(t), then the integral
(31) can be tabulated for values of 0 from 0 to TT, the reckoning
being performed by the ordinary methods for evaluating definite
integrals — Simpson's Rule, etc.
Integral of a Periodic Function. Let f(x) be a continuous peri-
odic function :
f(x + A) = /(*), - oo < x < oo,
where A is a primitive period, corresponding to 2 A above. Let
A
c =
0
Then
J + A
Jf(x)dx.
0
L
f(x) dx = C,
where x is arbitrary. For,
X + A
£//(*) dx = f(x + A)- /(*) = 0.
Let
X
/£V
f(x)dx - jx.
c
Then <p(x) is periodic :
<p(x + A) = <p(x).
For,
x+A
v(x + A) - *(x) = ff(x)dx -£(x + A)+
A DIFFERENTIAL EQUATION 465
Hence
where X = C/A. The result may be stated as follows.
THEOREM. The integral of a periodic function is the sum of a
periodic function and a linear function :
where
and
A
C=ff(x)dx, X = |-
0
Instead of the linear function \x we may write
Xz + 7 or X(z - xn),
the function <p(x) being changed by an additive constant. In
particular, X = 0 if and only if
A
ff(x)dx = 0.
o
APPENDIX C
CHARACTERISTICS OF JACOBI'S EQUATION
Although Jacobi's partial differential equation of the first order :
AN
A)
has played an important role in the solution of Hamilton's Equa-
tions :
rn d(Jr _ dH dpr _ m
*> ~dt~Wr ~dt~~Wr' r-l,..-,m,
where // = H(qly - • • , qm, ply • • • , pm, t), we have not found it
necessary to refer to the theory of characteristics, partly because
we have sought certain explicit solutions by means of ingenious
devices (separation of variables, for example) ; partly because,
when we have needed an existence theorem, it was supplied at
once by reference to the Cauchy Problem. Nevertheless it is
of interest for completeness to connect the equation with its
characteristics.
1. The Analytic Theorem. Consider the general partial dif-
ferential equation of the first order :
T .. .
L l' ' n> > ' dx
Let F(xly - - • , xn, z, 2/i, • • • , 2/n), together with its partial deriva-
tives of the first two orders, be continuous for those values of the
arguments for which (xl9 • • • , xn, z) is an interior point of an
(n + l)-dimensional region * R of the space of the variables
(x19 • • • , xn, z), and the yk are wholly unrestricted. Use the
notation :
/•i\ y __ dF 7 __ dF v 3F
(1) Xk~^' Z-~te> Yk~^
At a given point A : (a,, • • • , an, c, b19 • • • , bn) = (a, c, b) of R
let the Yk not all vanish ; in particular, let Yn ^ 0.
* R shall not include any of its boundary points.
466
CHARACTERISTICS OF JACOBFS EQUATION 467
The characteristic strips are defined by the system of 2n ordinary
differential equations :
II dxk = dz = """ dyk k = l
Yk S yk Yk Xk + ykZ* ' '
The solution of II. shall go through the point (x°, 2°, yQ), which
shall lie in the neighborhood of A and moreover on the manifold
F = 0, or
F (x ® • • • x ® z® v ® • • • '2/^^=0
Although there are 2n + 1 initial values — the (2°, 2°, y°) —
there is only a 2n-parameter family of solutions of II., for we
may without loss of generality set xn° = an once for all. The
solution of II. can now be written in the form :
T. = f-(r • -r ° • . • r° 7° 11 ° • • • 11 °^
•^t ^ i \^n , -^i , , -^n— 1) * 9 Ml 9 9 if* /9
i = l,---,n-l;
(2)
Along any curve (2) the function F(xlt • • • , xn, z, y^ • • • , yn)
is constant, since
dF = % Xkdxk + Zcfe +5) y*d»*-
On subjecting dxk, dz, dyk to the conditions imposed by II. it
appears that dF = 0. Hence
(3) F(xl9 •••,xn,z,yl,.-,yn)=C
is an integral of the system of differential equations II.
Characteristic strips are curves (2) for which
(4) F(x», - - • , 4-i, On, 2°, 2/i°, • • • , »»°) = 0,
i.e. C = 0 in (3). This equation can be solved for ynQ since
Thus there is a (2n — l)-parameter family of characteristic strips.
Consider now the (n + l)-dimensional space of the variables
(&!,•••, xny z), in which a solution :
(5) z = *(*!, • - -,*»),
468 APPENDIX C
of the partial differential equation I. will lie. In the (hyper-)
plane xn = an of this space let a manifold be defined by the
equation :
(6) 2° = Cdfo0, ' ' • , Xn-l),
where o)(xlt • • • , xn-\), together with its first partial derivatives,
is continuous in the neighborhood of the point (a,, • • • , an~0,
and
Furthermore, let
and let yn° be given by (4) or (4'). If, now, we regard the
as n — 1 independent parameters and for symmetry in notation
set
Xn = Un,
the first n equations (2), combined with this last equation, will
represent a (hyper-) surface parametrically, the equation of which
can be thrown into the form (5) by eliminating the (ulf • • • , wn),
and this function (5) is a solution of the given partial differential
equation I. Moreover it is the most general solution; i.e. any
solution (5), such that V satisfies the above requirements of
continuity, can be obtained in this manner.*
This is the general theorem of the solution of I. by means of
characteristics. We proceed to apply the result to Jacobi's
Equation A).
2. Jacobi's Equation. Let
(8) ( Xr = q" y' = Pr> r == 1, • • • , m = n - 1 ;
1 z=F, xn = *.
As regards yn, we see from I. and A) that it is given by the equa-
tion:
(9) F ss H(xlt • • • , xm, * , Vi, - • - , 2/m, xn) + yn = 0.
Equations II. now take the form :
*For the proof cf. Goursat-Hedrick, Mathematical Analysis, or the Advanced
Calculus, Chap. XIV, p. 366.
CHARACTERISTICS OF JACOBFS EQUATION 469
(10)
m i
m
= - dpr — dyn
d# dff
dqr dt
The initial values are :
Xr — Qr 3
(H)
/7° ?/ 0 = 7i 0 r =s 1 . • • *n •
</r > Jf r — Pr , r — 1, • , 771 ,
^o> 2/w = "" " > ^0*
From (10) follow first Hamilton's Equations :
(12)
dqr
dt dpr' dt dqr'
Furthermore, by the aid of (9),
( . dtf^m.
\ / fit QJ J
at ot
and finally, since from (12)
m.
(14)
dV
dt
= Z PrQr -
Observe in passing that the right-hand side of (14) is the La-
grangean Function, L :
H + L = 5) prqr,
r
and so
(15)
dV -L
~-L-
But we have anticipated the results of the general theory and
although obtaining the facts of the case in Equations (12); (13),
and (14), we have not brought out the direct testimony of the
general theory in the present case. Let us turn back, then, to
Equations (2) and Condition (4) or (4'). It appears that the
solution of Equations (10) takes the form :
(16)
n __ / // .
Qr — Jr(J> ,
7) — • / (t •
Pr ~ Jn+r \f ,
-«<)
Pi
n ° V
> Q.m , ' 0>
, <7m°, F0)
Pm°,
470 APPENDIX C
where r = 1, • • • , m = n — 1 and
(17) y»° = - H (<?,», •••, qn°, Pl«, ••-, pm«, t0)
In the case before us the functions frt /w+r, r = 1, • • • , m,
arising as they do from the solution of Equations (12), do not
depend on F0, and ynQ is given by (17). Thus we have
(18)
V ' '
and also the further integral of (10), given by (9) :
(19) 2/n = - //(ft, • - • , g«, p,, • • • , ?m, 0-
But V in (16) does depend on V0. It is given by (14) :
(30)
or
t
V =
to
So much, then, for the discussion of the solution of Equations
II., i.e. (10). As regards now the solution of Equation I., i.e. A) :
(21) +
we choose w subject to the conditions under (6) :
(22) F0 = cofoo, • • • , qm«)
and set
n
=
r 's
3. Application. We have seen in Chap. XV, § 2, that Hamil-
ton's Equations can be solved by a contact transformation :
/f\n\ d$ n d$ -i
(23) Pr = w, Pr=~W, r = l,--,m,
which transforms the given dynamical problem into the Equilib-
rium Problem, the solution of which is
Qr = ar, Pr = 0r, r = 1, • • • , m,
CHARACTERISTICS OF JACOBFS EQUATION 471
where ar, @r are arbitrary constants, wholly unrestricted so far
as the transformed Hamiltonian Equations :
dQr_Q d^-n r-1 ••• m
"rfT"0' ~3T~°' r-l,---,m,
are concerned.
The demands that the function Sfa, - • • , qm, Qif • • • , Qm, 0
fulfil are the following. First, it must be possible to solve the
equations :
(24) br = Sqr(alt • • • , Om, alf • • • , ctm, t0)
for the ar :
(25) oj = aj0, - - - , am = am°,
where ar = #r°, 6r = pr° are an arbitrary set of initial values
of </r, pr.
Furthermore, S(ql9 - • • , </m, «1? • • • , am, 0 shall be continuous,
together with its derivatives :
dS dS d*S
dqr' dctr dqrdoL8'
in the neighborhood of the point (ar, ar°, tQ), and
«!, •••,«„, , • • • , m
Finally, the function V = S(qr, <xr, t) shall satisfy Jacobi's Equa-
tion A).
The proof of the existence of such a function S is given by the
theorem of § 2 by setting
(27) cofo0, • - • , qm°) = a^i0 + • • • + <*mqm°.
For now the corresponding solution of Jacobi's Equation A) :
(28) V = S(qly - -,qm, an • • • , aw, 0,
has the property that
Moreover, the Jacobian determinant (26) is seen to have the value
1, and we are through.
We have obtained this existence theorem for the function £
by means of the theorem of § 2, the proof of which is based on
472 APPENDIX C
characteristics. But it might equally well have been derived
directly from the existence theorem which is usually referred to
as Cauchy's Problem, § 5 below, provided we are willing to assume
that H(qr, pr, 0 is analytic in the point (gv°, pr°, tQ).
4. Jacobi's Equation : H, Independent of t. Consider the
case that H does not depend on t :
(1) H = H(ql9 • - - ,<?m, plf • - • ,pm).
Jacobi's Equation now takes the form :
A/, *
A') _ +
We seek the special solution
(2) V = S(ql9 ••-,?„,«„ ••-,ob,0
defnanded in § 3. • «
It is possible to obtain S as follows. A solution of A') can be
found by setting
(3) V = - ht + W,
where
w = w(qi, ..-,?»)
does not depend on t. Then W will satisfy the partial differential
equation :
r,x v( dW 8W\ ,
C) ff^,...,^-.,... ,__)-*.
The derivatives of H(ql} - • • , qm, pl9 • • • , pm) with respect to
the pr are not all 0.* Let
» £*«•
Then the equation
(5) ff fe, ' ' ' , ?m, Pi, ' ' • , Pm) = ft
can be solved for pl :
(6) Pi = x(7i, •",?», A, P2i '••» Pm),
and C) is equivalent to the equation :
^ dW . dW dW
c) =
* Either because of the hypotheses of Chap. XI, § 3 or because H is a positive
definite quadratic function of the p/s.
CHARACTERISTICS OF JACOBFS EQUATION 473
Let
(7) W = W(qi,*--,qm,h,a2, ••-,«„)
be that solution of C) which reduces to
(8) W0 = a>(<72°, • • • , fr°)
when ql = q^. If, now, we set h = «!, the desired function S
is given by the equation :
(9) S(ql9 • • • , qmy «!,-•-, a™, 0 =
- o^ + TF(ft, • • • , qm, al9 • • • , am).
For,
dW
Hence
pr° = ar, r = 2, • • • , m,
are a system of equations which can be solved for the ar, r = 2,
• • • , m, and c^ is given by (5).
It remains to examine the Jacobian,
Since
(J1S\ = { 0,
\dardqa/0 I 1,
we have only to show that
0, r ?£ s 0
r, s = 2, • • • , m,
r = s
Now,
(12)
x i
is given by the equation (5) :
Hence with the aid of (4)
and the proof is complete.
474
APPENDIX C
Summary of Results. To sum up, then : — the solution of
Hamilton's Equations B) is given by the equations :
98
dW
8S
(14)
or
(15)
The ar are determined in terms of the (q°, p°) by the equations :
r = 2, - - - , m ;
(16)
= Pr°,
The #r are now given by (15) on setting qr = qrQ and substituting
for ar the value given by (16).
The Function W. The total differential equations which deter-
mine the characteristics of C) are :
(17)
Since
we have
(18)
or
(19)
dqr dW -dpr
^L 2 — —
Spr Pr 8pr dqr
m_ .
dPr ~ 9"
dW
W
r-1,
m.
If ff is a homogeneous quadratic function of pM • • • , pm, then
(20)
W = 2 Hdt + WQ.
CHARACTERISTICS OF JACOBFS EQUATION 475
6. The Cauchy Problem. Let F (xi, • • •, xn, z, y\, • • •, yn)
be analytic in the point (a, c, b) = (a1? • • • , an, c, bl9 • • • , &n)
and let dF/dx^ ^ 0 there. Consider the partial differential
equation :
fa 2Z\ n
,...,xn,2,— ,..-,— )=o.
Let
be analytic in the point (a2, • • • , an) and let
^(«2, • • • ,an) = c,
^*(oa, • • • , o») = 6*, A; = 2, • • • , M.
Then there exists one and only one function,
Z = ^(Xj, - • - , Xn),
which is analytic in the point (a^ • • • , an), has
^(«i, • • • , On) = c,
^/(ai> ' ' ' > «n) == 6/, y = 1, • • • , n,
and satisfies the given differential equation in the neighborhood
of the point (o1; • • • , an).
This is the existence theorem known as the Cauchy Problem.
Cf. Goursat-Hedrick, Mathematical Analysis, Vol. II, § 446.
APPENDIX D
THE GENERAL PROBLEM OF RATIONAL MECHANICS
I
PATHS
Consider* a system of n particles mt : (x», y^ Zi) acted on by
forces (Xi, Yi, Zi). Their motion is governed by Newton's Law:
A) niiXi — Xi niiiji = Yi mi'Zi = Z,-
Here are Qn dependent variables, the x,, yit z,-, X^ Yi} Zt, con-
nected by 3n equations. The problem of motion is to find 3n
supplementary conditions whereby these 6n variables will be
determined as functions of the time, t, and suitable initial condi-
tions, and to solve for these functions. Each member of the
family which forms the solution, namely the curve :
Xi Xi \l)) y\ y i \'/j Zi Zi \tj
Xi = Xi(t), Yi = Yi(t), Zi = Zi(t)
determines a curve :
Xi = xt(t), yi = 2/t(0, *< = *;(0
(2)
• _ . __ . __ ,
Xi ~ ~dT' Vi ~ "df ' Zi " ~dt
in the (6n + l)-dimensional space of the (a;,-, 2/t-, 2», ft, ?/», «», 0>
and such a curve is called a path. Obviously the paths (2)
stand in a one-to-one relation to the curves (1).
The problem of motion as so formulated transcends the domain
of Rational Mechanics. In order to restrict our attention to the
latter field, we now lay down the further postulate — which, be
it noted, is not satisfied by certain systems which occur in nature,
viz., certain systems in which electro-magnetic phenomena are
present.
* The following treatment is the result of a joint study of the problem by
Professor Bernard Osgood Koopman and myself.
4.76
GENERAL PROBLEM OF RATIONAL MECHANICS 477
POSTULATE I. DYNAMICAL DETERMINATENESS. In a given
dynamical system, when 6n + 1 constants (x^, yi(), zt-°, Xi°, 7/,°, z,-°, J0)
are arbitrarily assigned, not more than one path (2) exists which
passes through this point :
THE DOMAIN ZX Those points (#;, yi} zi} x», #;, «,-, 0 of the
(6n + 1) -dimensional space, through which paths pass, con-
stitute the domain D. This domain may consist of the entire
space, or of a region of it ; but in general neither of those things
will be the case. It is a point set, concerning the constitution of
which we need make no hypothesis at the present moment. It
will be restricted by later postulates.
THEOREM I. The variables Xi} Ft, Zi are uniquely determined
in the points of D :
Xi = Xi fa, y,, Zj, xh yit Zj, t)
(4) Yi = Yi (x/, y,-, Zj, Xj, fa, Zj, t)
Zi = Zi (Xj, yh Z3; Xj, 7/y, Zj, /)
where (xj, yh zit Xj, y,-, z}, t) is any point of D.
For, through each point of D passes a path, unique in virtue
of Postulate I. Along a given path Xi, Yi, Zi are uniquely deter-
mined as functions of t by A). Hence Xi, Yi, Zi are uniquely
determined at the point of D in question, but not in general in
points not lying on Z).
THE DOMAIN R. In the (3n + l)-dimcnsional space of the
variables (x^ yi, z^ f) those points which participate in paths
form a point set R, which may be described as the orthogonal
projection on this space of the domain D. In particular R may
consist of the whole space, or of a (3n + l)-dimensional region
in it. But in general neither of these things will be the case.
Let P be a point of R. To P there corresponds at least one
path given by (2). The points (xiy y^ Zi, t) represented by the
first line of (2), namely :
(5) Xi = Xi (0, Vi = Vi (0, Zi = ^ (0,
all belong to R. Hence the curve (5) lies wholly in R.
478 APPENDIX D
Consider an arbitrary line through P, but not perpendicular
to the axis of t. Let its direction components be a,-, ft, 7,-, K,
where K ^ 0. There may be a path corresponding to P, such
that at this point
Xi : yi : Zi = <*< : ft : 7*.
When this is not the case, not all lines through P correspond
to paths, and so certain relations between the direction com-
ponents (ca, pi, 7,, K) must exist. Thus we are led to a second
postulate.
POSTULATE II. The direction components at points of R, to which
paths correspond, are given by the equations :
(6) 2} (A3iai + B8i ft + Cai 7t) + D8K = 0, s = 1, • • • , cr,
i»i
where A8i, B8i, C,», D8 are functions* of (Xi, yl, zt, t) such that the
rank of the matrix :
All ' ' ' Aln Bn ' • • Bin Cll • ' • Cln
(7)
is <r.
Since along a curve (5)
«i ~~ ft ~~ 7i ; ~~ K
at the point P, it follows that
n
B) ^ (Ati Xi + B8i yi + C«, Zi) + D8 = 0, s = 1, - • •, <r.
These equations form a necessary and sufficient condition for
(£», yi, Zi) if (Xi, yiy ziy xiy yit zit t) is to be a point of D.
It may happen that the system of Equations B) (a system of
Pfaffians) admits certain integrals :
where the rank of the matrix :
* Throughout the whole treatment, the continuity of the functions which enter,
and the existence and continuity of such derivatives as it may be convenient to use,
are assumed.
GENERAL PROBLEM OF RATIONAL MECHANICS 479
(8)
is I. Since the system B) may obviously be replaced by any
non-specialized linear combination of these equations, it is clear
that Equations B) may be so chosen that the last I of them are :
(9)
dt
= 0,
The constants Ck come to us as constants of integration in the
system of integrals C) of the Pfaffians B). They contribute
toward determining the particular dynamical system we are
defining, different choices of the Ck leading to separate dynamical
systems. They are not to be confused with constants of integra-
tion that are determined by the initial conditions within a par-
ticular dynamical system.
Holonomic and Non-Holonomic Systems. If, in particular,
I = or, Equations B) can be replaced by Equations C) and thus
become completely integrable. The dynamical system we are
in process of defining is then said to be holonomic. But if there
remain a — I = /* > 0 Equations B), which then are non-
integrable, the system is said to be non-holonomic. Equations B)
shall now be replaced by the first p of them, and Equations C) :
B')
C)
(Aai±i + Baiyi + CaiZi) + Da = 0, a = 1, •••,/*;
= /x + I.
II
THE FORCES. D'ALEMBERT'S PRINCIPLE
The force Xt, Y^ Zi which acts on m» is made up in general of
a force- X'iy F«', Z\ which is known in terms of zt-, y^ zt-, xiy fa, zit t,
and of further forces XJ/, F{,-, Z«/, where j = 1, 2, • • • , p», the
components of these latter forces being wholly or in part un-
480
APPENDIX D
known. Denote the unknown components by S19 • • • , SK.
Then our postulates must provide for enough known equations
between the S's and the Xi, y^ zt-, Xi, ?/,-, 2,-, t to make possible the
elimination of the $'s between these equations and Equations A),
with the result that the equations thus obtained, combined with
Equations B') and C), will just suffice to determine xif y*, z> as func-
tions of t and the initial conditions. We proceed to the details.
D'ALEMBERT'S PRINCIPLE
In practice the equations which the Si, • • • , SK satisfy are
usually linear. Our problem shall be restricted to systems which
obey the following postulate.
POSTULATE III. The force Xi, Yit Z{ is the sum of two forces:
(10) 'Xi = xi + A7, Yi = y; + 17; Zi = z; + z;,
where X\, FJ, Z\ are known in terms of the coordinates a:,-, y^ z,-,
Xi, ilij Zi, t °f an arbitrary point of D, ami where
(11) ijxrfc + rrih. + z; $-, = (>
i=l
for all fi, r/i, f,- such that
(12) 2J Abb + Shu + C'pift = 0, j8 = 1,- - •, i/.
i=l
Here, A'ai, B'ai, C^i are known functions of the above #,-, 7/t-, z^
Xi9 tit* *i) t> and the rank of the matrix :
(13)
v. Conversely, when Equations (12) are satisfied. Equation (11)
Turning now to Equations A), we have what is known as the
General Equation of Mechanics :
(14)
= 0,
GENERAL PROBLEM OF RATIONAL MECHANICS 481
where £,-, T/,-, f i are 3n arbitrary quantities. Under the sanction
of Postulate III. this equation can be replaced by the following :
(15) 2) (mt£i ~ XI) fc + (m<fr - Ffl* + (w«2« - Z«')f« - 0,
<-l
where £,-, ?7t, f ,- are any 3n quantities which satisfy the condition
(12).
Multiply the 0-th equation (12) by X^ and subtract the re-
sulting equation from (15) :
(16) 2) (mtfi - X,' - 5) A'fiiljh + (mtfr - Y'< - % B'^n
<=1 0-1 /3-1
+ (mizi-Z'i-^C'ei\p)t< = 0.
0=1
Suppose for definiteness that the determinant whose matrix
consists of the first v columns of the matrix (13) is ^ 0. Then
the X's can be so determined that the coefficients of the first v
of the quantities £1, • • • , £n, i?i, • - • , *7n, fi, • • • , fn in (16) will
vanish. Substitute these values of X1? • • • , \v in the remaining
coefficients of (16). Thus a new linear equation in the £,-, 17,-, f$-
arises, in which only the last 3n — v of these quantities appear.
But the latter are arbitrary. Hence each coefficient must vanish.
The 3n — v equations thus obtained express the result of elim-
inating the unknown Sl • • • , SK, i.e. the X*, Yf, Zf, from the
problem. They contain only a?«, yiy ziy Xi, #», z^ xit yiy zi} t, and
can be written in the form :
E) %(Eyi Xi + Fyi y< + Gyi 2,) + #* = 0, 7 = 1, . • • , 3n - v,
<-i
where the coefficients Eyil Fyi, Gyi, Hy are known functions of
Zt, IJi, Zi, Xi, y*, Zi, t at each point of D.
Equations E) and B) form a necessary condition for the func-
tions Xi(f), yi(t), Zi(t) which define a path (2). Hence if P0:
(xfj yfj ZiQ, XiQ, yfj zP, tQ) is an arbitrary point of Z), Equations E)
and B) admit a solution having as its initial values the coordinates
of JP0. Furthermore, by virtue of Postulate I., this solution is
unique. We have now arrived at a complete analytical formula-
tion of the problem, for we can retrace our steps. Let
482 APPENDIX D
be a curve lying on C) and satisfying B'). Then a) gives rise to a
curve T which lies on D. Consequently all the coefficients in B'),
(12), and E) are determined in the points of a). Let a) also
satisfy E).
Since E) holds, it follows that XD • • • , X, can be determined so
as to make each parenthesis in (16) vanish. Next, determine
X*9 Yif Z* from these X's by the equations :
X* = 2* Afii X/3, Y* = £j Bfii Xj3, Z* = £j Cfti X/3.
0=1 /3=i 0=1
These quantities satisfy (11) and (12). On substituting them in
(10), values of Xt, F», Z{ are obtained for which A) is true, because
each parenthesis in (16) vanishes, and so T is a path. But there
is only one path through an arbitrary point P0 of D. Hence a) is
unique.
Retrospect. These Postulates complete the formulation of the
class of problems in Rational Mechanics which we set out to
isolate. The role which d'Alembert's Principle * plays is two-
fold. First, it requires that the relations between the unknown
S\, • ' m 9 &* shall be linear. Secondly, it performs the elimination
by a technique such that the multipliers {,-, T/;, f» can always be
interpreted as virtual displacements of the system of particles m» :
(xt, yi, Zi) by setting
(17) bXi = fc byt = ??i, bZi = ft.
Remark. In general there is no relation between the coefficients
A9i, B9i, C8i of Equations B) and the Apit B^i, C'^ of Equa-
tions (12). Hence the virtual displacements to,-, 6r/», fai of (17)
will not coincide save as to infinitesimals of higher order with any
possible displacement A#t, Ayiy As* due to an actual motion of the
system in time A/.
In a sub-class of cases it happens, however, that the 4«», J3,t, C«i
in B) and the Apt, B'ftiy C'pi in (12) are respectively equal to each
* Historically d'Alembcrt's Principle took its start in the assumption of a con-
dition, necessary and sufficient, that a system of forces, acting on a system of
particles, be in equilibrium, namely, that the virtual work corresponding to a
virtual velocity be nil. When a system of forces not in equilibrium acts on a system
of particles, the former can be replaced by a system of forces in equilibrium
through the introduction of "counter effective forces" or "forces of inertia"
(sic), and thus d'Alembert arrived at the General Equation of Dynamics.
GENERAL PROBLEM OF RATIONAL MECHANICS 483
other. But even so, if the Da are not all 0, the virtual displace-
ment will not tally save as to infinitesimals of higher order with
any possible actual displacement.
Finally it can happen that, in addition, the D, are all 0.
Then the virtual displacement corresponds to a possible displace-
ment. But this is a very special, though highly important, case.
Let
(18)
Ill
LAGRANGE'S EQUATIONS
', 0m, 0
-, ffm, 0
-,ff«, 0
where the rank of the matrix :
(19)
dq,
is w, and where, moreover, the region of the (x», y^ z», 0-space
which corresponds to the points (qly • • • , qm, f) in which /», ??,•, ^i
are defined, at least includes the points of R.
Let T denote the kinetic energy :
Then, for an arbitrary choice of the qr, since
T goes over into a function of gr, qrt t :
T = T(qr,qr,t).
Conversely, if £», ^», £», are any set of numbers for which these
equations are true, the qr are uniquely determined.
484 APPENDIX D
Consider a path (2). Since the points #» = Xi(t), yi = yi(t),
Zi = Zi (0 all lie in R, a curve of the (qr, 0-space is thus defined :
(20) qr = qr(t), r = 1, ...,m.
For the path in question we have :
__0
dtdqr Sqr~Q" '
where
8xi 4- Y 8yi 4- 7 8z
<+ i+
Equations (21) are always true under the foregoing restrictions.
They will be sufficient to determine the motion if the system is
holonomic and if
(22)
r = 1, • • •, m. For then
' dXi - dyi - ' dzr
"
and thus Qr is known — first, in terms of x<, y^ zlt xiy yiy ziy t, and
so finally in terms of qr, qr, t. That Equations (21) can be solved
for ql9 • • • , qm follows from the fact that T is a positive definite
quadratic form in the qly • • • , qm.
This means in terms of the foregoing treatment that a suitable
choice of the multipliers £,, 7?t, f t in Equation (12) is :
\
vhere the 8qr are arbitrary. Equations C), if present, are all
satisfied identically when the Xi, y^ Zi are expressed in terms of
tte qr and t by Equation (18). " Equations B') are not present in the
problem. The system is, to be sure, holonomic, but it is not the
only case in which this is so.
The General Case. We assumed in Postulate III. that X't, F{, Z(
are flee from the /S's, and that the S's coincide with the
X*, Yf) Z*. We now divide the S's into two categories :
t) a sub-set, denoted anew by Xf, F*, Z*, which fulfil the
former requirements (11), (12), (13);
GENERAL PROBLEM OF RATIONAL MECHANICS 485
ii) a second sub-set, R19 • • • , RT> on which the XI, Y(9 Z\ shall
now depend linearly.
Thus Equation (21) holds, where Qr is given by (23). Let
(25) Qr = Qr + Q?, r = 1, - • -, m,
where Q'r is known in terms of such values of qr, qrj t as correspond
to points of D.
In the particular case before us, namely, Equations (18), it
can happen that the equation :
(26) Qi* *-! + •••+ Qi *•• = 0
is true for all values of the multipliers irr for which the following
equations hold :
(27) Oft! Tt + - - - + a'tmKm = 0, 0 = 1, •••, vl9
where n'pr depends on values of qr, qr, t, which correspond to points
of D, and the rank of the matrix :
/
(28)
is vl'9 and conversely, when Equations (27) hold, then (26) is true.
Equations B'), C) go over in the present case into :
Bq) ®al*ll 4" " " " ~T~ ttam^wt ~f" &a == 0, Oi == 1, • • ', /Zj 9
Cq) <&k (fir) 0 == T*k) k = 1, * * ', lu
where ^ ^ n ; ^ g Z, and where the rank of the matrix :
(29)
is /zt ; the rank of the matrix :
(30)
Fg~
being ^.
486 APPENDIX D
Finally it can happen that the Rv • • • , Rr can be eliminated
between these equations, thus leaving a system of equations be-
tween the qrj qr, qr- Such a system yields a unique solution, cor-
responding to each path of the dynamical system with which we
set out. Thus the dynamical problem is completely formulated by
means of Lagrange's Equations.
All of the foregoing assumptions are in tentative form — " It
may happen • • • " At the one extreme, the choice of the functions
fit 9* ti can always be made so that all these things do happen ;
for the qr can, in particular, be identified with the xi} yi} z< :
At the other extreme, m may be chosen so small that Equations
(21), though true, will contain unknown functions which cannot
be eliminated — namely, the Rlt • • • , RT. This means that, for
such' a choice of the functions (18), the £t-, rj f, ft- as given by
(24) are too restricted. The &, rjif f. of Equations (11) are quan-
tities which must be able to take on every set of values which
satisfy (12). The £.-, 77,-, {*» which here figure, given by (24), are
not free under the condition (24), but arc unwarrantably restricted
by (24).
In a given problem the desideratum usually is, to choose m as
small as possible, subject to the requirement that the same degree
of elimination of the'S's through (24) shall have been attained,
as if Equations (11) and (12) had been used.
IV
NOTES
Consider the dynamical system that consists of a bead sliding
on a fixed circular wire and acted on by no other forces than the
reaction of the wire. Equations A) take the form :
mx = X, my = Y, mz = Z.
Let the wire be a circle whose axis is the axis of 2. Then Equa-
tions B) become :
xx + yy = 0
B) • • o
This system of Pfaffians is completely integrable :
f x2 + 2/2 = a2
I z = c
GENERAL PROBLEM OF RATIONAL MECHANICS 487
Different values of the constants of integration, a and c, give
different systems of paths, (2) ; but a path of one such system
has no point in common with a path of a second system.
Proceeding to the forces we see that Z = 0, since 3 = 0, and
so we have a two-dimensional problem.
The Smooth Wire. Assume first that the wire is smooth. Then
the reaction is along the inner normal.
X = X*, Y = Y*,
and
X*£+ F% = 0
provided
^ £ + y t] = 0.
Turning to Lagrange's Equations we set m = 1 and take
x = a cos #, y = a sin q.
Then
= X* (- a sin ?) + 7* (a cos ?)
= ;T(-2/) + F** = 0.
Hence, finally :
and it remains merely to integrate this differential equation.
The Rough Wire. Suppose, however, the wire is rough. Let
q > 0. Then
X* = — R cos q + pR sin q
Y * = — R sin q — pR cos q.
Lagrange's Equation :
Jt ~dij ~ ~dq * Q'
is still true. But
488 APPENDIX D
(a result at once obvious) and Lagrange's Equation becomes :
We have not enough equations to solve the problem. This is
the case in which Lagrange's Equations are said to "fail" or be
"inapplicable." The failure lies, not in Lagrange's Equations,
but in a misuse of them. We should take m = 2. Let us first
treat the problem, however, by the methods of Parts I., II., before
Lagrange's Equations were introduced in Part III. Here, then,
r72rr
m ±± = X* = - R cos B + »R sin 0
at*
m-j% = F* =-juftcos0 - Rsin0.
ut
Equation (11) now takes the form :
x** + 7*77 = 0,
or
(— R cos 0 + p,R sin 6) £ + (- nR cos 6 - R sin 0) rj = 0,
or, finally,
(— x + »y) £ + (— MZ - y) v = 0,
and this is the form of Equation (12). Hence we may take
£ = IJLX + y, rj =- x + ny.
On substituting these values in the General Equation of
Dynamics we have :
/ • \ d2x , t . ^ d?y ~
(i*x + V)-fc + (-x + /iy) ~ = 0.
This equation and Equation C), namely :
x* + y* = a2,
provide us with two equations for determining x and y as functions
of I, and thus the problem is reduced to a purely mathematical
problem in differential equations. Observe, however, that the
virtual displacement used in this solution :
GENERAL PROBLEM OF RATIONAL MECHANICS 489
is not one which is compatible with the constraints, i.e. the circular
wire — even save as to infinitesimals of higher order than e. It
corresponds to a displacement along a line at right angles to the
resultant of R and pR.
Turning now to Lagrange's Equations let us choose ql and qz
as the polar coordinates of the mass m. Then Lagrange's Equa-
tions (21) become :
(31)
7<S
Now, Equations Cq) here become :
Cfl) r = a.
On the other hand Equation (26) :
here becomes :
0,
and thus Equation (27) takes the form :
*"l + M^2 = 0-
If, then, we set :
TTi = — /i, ^2
Equations (31) and Cq) yield :
dO* , d*0
-.ma_7ri + ma_
or
#8 <W
and it remains merely to integrate this equation.
As a further illustration of the use and abuse of Lagrange's
Equations may be mentioned the Ladder Problems of pages 322
and 323.
INDEX
Absolute unit of force, 52
of mass, 79
Absolute value, 24
Acceleration, 50, 52, 287
d'entrainement, 288
of gravity, 56
Vector, 90
Addition of vectors, 4
d'Alembert's Principle, 345, 480
Angle of friction, 10
Angular velocity, Vector, 170, 285
Appell, 225, 244, 246, 307, 337
Areas, Law of, 108
Atwood's machine, 134
Axes, Principal, of a central quadric,
194, 196
Rotation of the, 454
B
Bending, K, 226
Centre of, 234
Billiard ball, with slipping, 143, 237,
314
without slipping, 145, 240, 314
Blackburn's pendulum, 184
Bocher, 334
Bolza, 372, 375
Brah6, Tycho, 115
Centripetal force, 102
Centrodes, 159
Space and Body, 174
Change of units, 76
Characteristics of Jacobi's Equation,
466
Charlier, 437
Check of dimensions, 79
Coefficient of friction, 10
of restitution, 271
Component of force, 2
of velocity, 87
Compound pendulum, 130
Cone, Body, Space, 213
Conservation of energy, 256
Conservative field of force, 255, 258
Constrained motion, 95
Constraint, Forces of, 315, 325
Contact transformations, 390, 399
Particular, 403
Coordinates, Cyclic, 430
Generalized or intrinsic, 297
Normal, 335
Coriolis, 288
Couples, 25, 29, 34, 37
Composition of, 31
Nil, 31
Resultant of n, 31
Vector representation of, 38
Cyclic coordinates, 430
Canonical equations, 338, 395
transformations, 389
Carathe'odory, 381, 445
Cart wheels, 241, 314
Cauchy problem, 475
Central force, 108, 379, 427, 434
Centre of bending, 234
Centre of gravity, 26, 27, 42
Motion of the, 120
Centre of mass, Motion of the,
123
Centrifugal force, 101
field of force, 106, 291
oil cup, 105
5, Definition of, 356
Critique of, 379
Dancing tea cup, 165
Decomposition of force, 2
Dimensions, Check of, 79
Direction cosines of the moving axes,
216, 454
Dyne, 56
120, E
Elastic strings, 58
Elasticity, Perfect, 272
Electromagnetic field, 254
Ellipsoid of inertia, 192
491
492
INDEX
Energy, Kinetic, 75, 260
Conservation of, 256
Potential, 255
Work and, for a rigid body, 266
Equation, Solution of a trigonomet-
ric, 12
Fundamental, 367
of moments, cf . Moments
Equilibrium of couples, 31
of a dynamical system, 330
of forces in a plane, 32
of forces in space, 36, 41
of n forces, 9
of three forces, 5
of a rigid body, 26
Problem, 413
Escalator, 265
Euler's Angles, 214, 215
Dynamical Equations, 210, 325,
352
Equations, 359
Geometrical Equations, 214
Field of force, 253
Centrifugal, 291
Gravitational, 254
Electromagnetic, 254
Force, 1
Absolute unit of, 52, 55
Central, 108, 379, 427, 434
Centrifugal, 101, 291
Centrifugal field of, 291
Centripetal, 102
Component of, 2
of constraint, 315, 325
Equilibrium of three, 5, 43; cf.
Equilibrium
Field of, 253
function, 253
Moment of a, 28
Parallel, in a plane, 21 ; in space, 36
Parallelogram of, 2
Polygon of, 7
Triangle of, 4
Foucault Pendulum, 292
Friction, 9
Angle of, 10
Coefficient of, 10
Problems in, 19
Function, Lagrangean, 338
Hamiltonian, 342
Fundamental equation, 367
Generalized coordinates, 297
Geodesies, 308
Goursat, 468
Gravitation, Motion under the attrac-
tion of, 69
Law of universal, 116
Gravitational constant, 116
Gravity, Acceleration of, 56
Gyration, Radius of, 129
Gyroscope, 217
Intrinsic treatment of the, 225
H
Hadamard, 224
Hamilton's Canonical Equations, 338,
395
Proof of, 342
Solution of, 410, 432
Reduction of, to the Equilibrium
Problem, 411, 413
for constant energy, 411, 420
Hamiltonian Function, 342
Hamilton's Principle, 371
Integral, 356
Integral a minimum, 381
Harmonic Motion, Simple, 64, 415
Haskins, 228
Hedrick, 468
Helical motion, 168
Hertz, 244
Holder, 370
Holonomic, 313, 479
Hooke's Law, 59, 74
Huntington, 208
Huygens, 133
Impact of particles, 270
Oblique, 274
of rigid bodies, 277
Impulse, 271
Inertia, 118
Ellipsoid of, 192
Moment of, 128, 137, 191
Product of, 191
Instantaneous centre, 154, 157, 160
axis, 168, 173
Integral invariants, 392
Integral of kinetic energy, 362
of a periodic function, 464
of rational mechanics, 360
INDEX
493
Internal work, 258
Intrinsic treatment of the gyroscope,
225
coordinates, functions, 297
equations of the gyroscope, 236
Invariable line and plane, 201
Inverse problem, 114
Isolate the System, 102
Jacobi's Equation, 410, 468, 472
Characteristics of, 466
Integral a minimum, 386
Principle of Least Action, 377
K, bending, 226
Rater's pendulum, 133
Kemble, 234, 263
Kepler's laws, 115
Kinetic energy, 75
of a rigid system, 166, 260
Integral of, 362
Klein-Sommerfeld, 236, 246
Koopman, 123, 476
Kreisel, 236, 246
Ladder, 147, 322, 323, 353
Lagrange's Equations, 299, 304, 312,
348, 350, 482
multipliers, 194, 316, 375
Principle of Least Action, 374, 377
Solution of, Equations, 326
Lagrangean function, 338
Lagrangean integral, 372
Lagrangean integral a minimum, 381
Lagrangean system, 338
Law of areas, 108
of nature, 109, 116
of universal gravitation, 116
of work and energy, 258
Least Action, 374, 377
Leval, Turbine of, 247
Lissajou's curves, 182, 190
M
Mass, Absolute unit of, 79
Moments about centre of, 139, 205
Motion of the centre of, 120, 201
Notion of, 118
Material point, 50
Maxwell, 119
Moment of a force, 28
of a couple, 29
of a vector, 37
of a localized vector, 197
of momentum, 197, 205
of inertia, 128, 137, 191
Theorem of Moments, 127, 200
Moments about the centre of mass,
139, 205
Moments about the instantaneous
centre, 207
Moments about an arbitrary point,
205, 208
of a vector about a line, 40
Momentum, 50, 201, 350
Moment of, 197, 200, 350
Motion under the attraction of gravi-
tation, 69
Newton's Laws of, 50
Simple Harmonic, 64, 415
Constrained, 95
Simple Pendulum, 97
Spherical Pendulum, 306
in a resisting medium, 81
in a plane and in space, 86
of a projectile, 93, 424
on a smooth curve, 99
on a space curve, 100
of the centre of gravity, 120
of space, General case of, 175
about a fixed point, 212
Moving axes, 172, 216
curve, 299
surface, 303
N
Newton's Laws of Motion, 50
Second Law, 92, 290
Non-holonomic, 244, 313, 479
Normal, 9
Principal, 90
Normal coordinates, 335
Nul vector (or nil vector), 5, 447
couple, 31
Numerical value, 24
Operator, Symbolic vector, 254
Orbit of a planet, 111, 113, 435
Oscillations, Small, 333
Osculating plane, 90, 92
494
INDEX
Parabolic motion, 93
Parallel forces in a plane, 21, 23
in space, 36
Parallelogram of forces, 2
Particle, 50
Pendulum, Blackburn's, 184
Compound, 130
Foucault, 292
Rater's, 133
Simple, 97, 419
Spherical, 306
Torsion, 139
Periodic time, 111
Perturbations, 440
Poincar6, 392
Poinsot, 213
Potential, 253
energy, 255
Poundal, 56
Principal axes of a central quadric,
194
Principle of the motion of the centre
of mass, 123
of moments, 139
of moments with respect to the
centre of mass, 205
d'Alembert's, 345, 480
Hamilton's, 371
of Least Action, 374
Variational, 370
Product of inertia, 191
Projectile, Motion of a, 93, 424
Quadric, Central, 194
Radius of gyration, 129
Eankine, 10
Rectilinear motion, 49
Relative velocities, 177
Resistance, Graph of the, 84
Resisting medium, Motion in a, 81
Resultant, 2
of parallel forces in a plane, 21, 23
of n couples, 31, 36
of n forces in a plane, 32
of n forces in space, 36, 38
axis, 39
of two velocities, 87
Riemann, 381, 445
Rotation of the axes, 454
about a fixed axis, 127, 136
of a plane lamina, 139
of a rigid body, Chap. VI
Routh, 134, 202, 225, 246, 337
Ruled surfaces, 176
S
<r, 197
Evaluation of, for a rigid system.
208
Transformation of, 202
Sabine, 190
Sand tunnel, 185
Ship's stabilizer, 236, 247
Simple Harmonic Motion, 64, 415
Simple pendulum, 97, 419
Small oscillations, 333
Smooth curve, 99
Solution of trigonometric equation,
12
Hamilton's Equations, 410, 432
Sommerfeld, 236, 246
Space curve, Motion on a, 100
Spherical pendulum, 306
Stabilizer, Ship's, 236, 247
Stationary, 359
Strings, Elastic, 58
Symbolic vector operator, 254
Tautochrone, 98
Tennis ball, 282
Top, 220, 438
Torque, 29
Torsion pendulum, 139
Transformation of <r, 202
Contact, 390, 399, 413
Canonical, 389
of Hamilton's Equations by con-
tact transformations, 400, 413
Translation, 159
Transmissibility of force, 22
Triangle of forces, 4
Trigonometric equation, 12
theorem, 44
Two body problem, 114, 879, 427, 434
TychoBrah6, 115
Tyndall, 166
Units, Absolute, 52, 55, 79
Change of, 76
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