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NEW AND ELEGANT EDITIONS
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GREEK AND LATIN CLASSICS:
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A
دار

какди
Fires
TREATISE
ON
DYNAMICS.
CONTAINING
A CONSIDERABLE COLLECTION
OF
Mechanical Problems.
BY
1749-1866
WILLIAM WHEWELL, M.A. F.R.S.
FELLOW OF TRINITY COLLEGE, CAMBRIDGE.
CAMBRIDGE:
Printed by J. Smith, Printer to the University;
FOR J. DEIGHTON & SONS, CAMBRIDGE; AND
G. & W. B. WHITTAKER, AVE-MARIA LANE, LONDON.
1823
:
Proz, Alex. Ziwet
gt.
2-26-1923
PREFACE.
THE following pages are intended to convey the ma-
thematical doctrine of the Motion of Bodies, presented with
the symmetry and generality which the science has now
attained, satisfactorily established, and illustrated by its
application to a variety of instances. Our language could
scarcely be said to possess a scientific treatise on Dynamics,
till Mr. Cresswell published his translation of Venturoli;
and it is still presumed that the appearance of that elegant
compendium of the subject may not have at all superseded
the utility of a work like the present, in which the demon-
strations are fully developed, and elucidated by a consider-
able collection of mechanical problems, selected from the
works of the best mathematicians, and arranged with their
solutions under the different divisions of the science.
From the time when Mechanics began to be cultivated
as a branch of mathematics, but more particularly while the
great Geometers of the last century were employed upon it,
and before they had succeeded in compressing the whole
science into a few short formulæ, the solution of particular
mechanical problems, interesting from their nature or ap-
plication, was a favorite employment of those who gave
their attention to mathematical pursuits. An immense
418545
iv
PREFACE.
number of such investigations is to be found in the different
Academical Collections of Europe, the work of various
hands, but above all of Euler, whose love for the specula-
tions of the science of quantity, and long life of uninter-
rupted application to them, led him in succession to almost.
every possible question of this kind. Even for a great
number of years after his death, the transactions of the
Imperial Academy of Petersburg were enriched by the
materials of this description which he left accumulated
for it. Among more recent mathematicians however, less
attention has naturally been directed to this kind of re-
searches. They became less attractive as the labours of
those who went first gradually exhausted the problems
which were obvious without being too difficult; and as the
improvement of general methods, and the reduction of
difficulties to classes, made the questions less interesting by
making them less particular, and by diminishing the un-
certainty of success. Perhaps it cannot now be considered
as the most profitable or advisable employment of mathe-
matical talent to exercise it upon particular problems,
distinguished only by the simplicity of their conditions
or the elegance of their results; especially while so many
general methods remain to be cultivated and perfected, and
so many practical questions to be subjected to calculation.
Still, at a certain period of the mathematician's progress,
this employment is instructive, and is generally found to be
interesting. The object of the present work is therefore to
present such a selection of problems as may sufficiently shew
the application of the general formulæ, and may at the same
time satisfy the curiosity which the student may feel with
PREFACE.
regard to the labours of those to whom the science owes so
much. It seems no more than a proper respect for the
Maclaurins, Bernoullis, Eulers, and Simpsons of the last
age, to preserve in our treatises some specimens of the
questions which occupied so much of their attention, and
which probably, if they had not cleared the way for us,
might have occupied much of ours at present. If a mathe-
matician should now give up his mind to such subjects,
without looking to what has been done by his predecessors,
he might easily, in following a favourite path of research,
spend too much of his time in inventing, solving and gene-
ralizing particular problems. Instead of this, a proper se-
lection, such as is here attempted, may shew him what
questions have been undertaken with success, and by
sparing him all fruitless endeavours may leave him at
liberty to employ his powers on those parts of mathematics
where their exertion may really be of service. The in-
vestigations relating to the system of the universe, much
as has been done, still offer an ample field for the display
of the talents, however great, of this and future gene-
rations. Even without an extraordinary degree of the
inventive faculty, much may be executed. The student
who feels a proper admiration for the system of the
Principia, ought to look forwards to the complete de-
velopement of it in the Mecanique Celeste, as the
ulterior subject of his labours; and those who shall
simplify the different parts of that work and reduce them
to the level of ordinary readers, as far as they admit of it,
will deserve to be considered as real benefactors to the
commonwealth of science.
vi
PREFACE.
Perhaps it is not too much to hope that the succession
of mathematical ability which the system of this University
calls out, might, if directed to such objects, enable us to
add much of what is most valuable in modern writers to
the present subjects of our studies. It is undoubtedly
desirable to give a sound knowledge of several branches
of mathematical science, and of the Newtonian System, to
a greater number of persons than we can expect to form
into profound analysts; but if we could save the time
which is now often lost in searching and selecting through
a variety of books, and in unprofitable and unsystematic
reading, we might make the extent of our encyclopedia less
inconvenient than it is sometimes found to be at present.
Instead of balancing the simplicity and evidence of the
mathematics of a century ago against the generality and
rapidity of modern analysis, it might be better to attempt.
to combine them; and if our University were provided
with a course of elementary works written with this view,
and if the higher branches of the science were simplified
and made to correspond with these introductory steps,
we might include in the circle of our studies a larger
portion of the modern additions to mathematical know-
ledge than is now in most cases practicable.
The present Work is in some measure an attempt to
facilitate some of the higher problems of the science,
though the mathematician will find in it little except what
he will consider as elementary. Its object is the mathematical
developement of the doctrine of motion, beginning with the
application of the differential calculus; and hence the fun-
PREFACE.
vii
damental laws and principles from which the science is
derived are only briefly stated at the outset. Some proofs
and illustrations of them are however added in Appendix
(A), and it is presumed that the three laws of motion are
there given in the form which is most distinct and simple,
and which makes them a satisfactory foundation for the
principles to be established on them. The common doctrine
of projectiles and of bodies falling by gravity is omitted,
as naturally belonging to a more elementary treatise. In
the third Book, which treats of the motion of a solid
body, a principle equivalent to D'Alembert's principle
in its simplest form, viz. the equilibrium of impressed
and effective forces, is made the foundation of each pro-
position. The examples given, will, it is hoped, suffi-
ciently familiarize the student with the meaning and
application of this important and extensive law. With
respect to the evidence which is in general offered for
the truth of the principle, it is in fact the extension
of a statical law to a system in motion; and an attempt.
is here made to render this proof more clear and satis-
factory.
After mature deliberation, the plan has been adopted
of dividing the work into short Propositions, and enun-
ciating each at the beginning of its Article. Though this
method interrupts the succession of truths as they grow out
of each other in course of our analytical reasoning, and so
far may be considered as a blemish by the practised analyst,
in an elementary work on such a subject its advantages
seem to preponderate, and especially for the purposes of
viii
PREFACE.
academical instruction. The propositions are a sort of
perpetual recapitulation of what has been done; and the
reader generally attends to the successive portions of the
reasoning more clearly and steadily, from having each of
them separated, circumscribed, and its object pointed out
beforehand besides which, if he meet with any difficulties,
it is easily seen, in a work of such a form, exactly where
and what they are. Thus, though it is not to be imagined
that this work has any claim to be called synthetical from
thus imitating the division of treatises of that description,
it may perhaps, while its demonstrations are in general
entirely analytical, possess some of the advantages of the
synthetical form as a means of conveying information.
Among other matter, the work contains propositions
corresponding to nearly all those in the two first books of
the Principia; and references to these are given where
the propositions occur. Partly also with the object of
making our arrangement correspond with that of Newton,
the motion of bodies in fluids is separated from their free
motion, and forms a second book; an order which pos-
sesses some advantages, though it is different from that
of most analytical treatises. The doctrine of central
forces is deduced for all the cases from the same diffe-
rential equation* which is employed by writers who
treat of the perturbations in Physical Astronomy; and
which, besides its claim to notice on that account,
* The equation
do²
2
d² u
+11-
P
= 0.
h² u
2
ix
PREFACE.
gives all the required results with remarkable simplicity
and uniformity; applying also immediately to the cases
of orbits where the line of apsides is in motion, as in
the cases of orbits
Newton's ninth section; and to
which have an asymptotic circle, first considered by
Maclaurin. The problem of three bodies, as treated in
Physical Astronomy, could not be introduced without
leading to a mass of calculations which would of itself
form a treatise; and the reader is therefore referred to
works professedly on that subject. The problems of
the motions of bodies on surfaces of revolution, of the
motions of various systems of connected points, of curves
of equal pressure, tractories, synchronous curves, &c. in
Chapters V, VI, VII, will be found useful as specimens
of the various problems which have been solved, and
of the artifices which have been employed.
The third Book, on the motion of any system of
rigid bodies, is the one where the work of simplification
was at the same time most difficult and most neces-
sary. The treatise of Atwood, unscientific and cumbrous
in the highest degree, and in some respects erroneous,
was till lately the principal English treatise on this
subject. The more analytical modes of investigation.
presented to many readers great difficulties in this depart-
ment in consequence of their conciseness and generality,
and of the want of examples in which the results of the
formulæ might be followed into detail. By breaking
up the reasoning into distinct and short propositions,
as is here done, it is hoped that the subject is
b
Χ
PREFACE.
rendered more easily accessible; and the examples in
Chap. VII and VIII, besides illustrating the formulæ,
are themselves very curious. Among other exemplifica-
tions are introduced those which are suggested by Captain
Kater's method of determining the length of a seconds'
pendulum, (Arts. 91, 92,) and the proof of the pro-
perty established successively by Euler, Laplace, and
Dr. Young, that in this method the time of vibration
is independent of the bluntness of the axes, if their
radii be the same. (Art. 131, Prob. VII.) The demon-
stration of the existence of three principal axes, (Art.
112,) differs from the one generally given, and is borrowed
from Lagrange: it is recommended, like most of the
processes of that great mathematician, by its remarkable
analytical symmetry. The Appendix contains several
investigations not absolutely essential to the treatise,
but worth notice on various accounts, and most of them
new to the English reader.
Besides the direct advantages of introducing into
the following treatise many problems not usually found
in elementary works, this collateral purpose is answered.
Several of these investigations shew very remarkably the
application and utility of some of those particular cases
and branches of analysis, which might otherwise be
considered as merely subjects of mathematical curiosity.
Thus in finding the time of a body's descent to a centre
of force varying inversely as the distance, we have to
obtain the definite integral of ex² dæ. (p. 15). In
considering the motion of a complex pendulum, (p. 136,)
PREFACE.
xi
we employ the simultaneous integration of n differential
equations, and are led to Daniel Bernoulli's important
principle of co-existent vibrations. We are naturally
introduced to the calculus of variations in solving the
different cases of brachystochronous curves, (p. 161,) and
in proving the principle of least action. (p. 401). In
determining the perturbation of elliptical motion, arising
from the resistance of a medium, (p. 201,) we use the
method of the variation of parameters. In investigating
the motion of a body about two centres of force, (p. 353,)
we have occasion to refer to the criterion of the differ-
ence between particular solutions and particular integrals.
In finding the attractions of spheres by Laplace's method,
(p. 365,) we have to differentiate under the sign of in-
tegration; and in solving the problems of the vibrations
of strings, and of elastic rods, (p. 378, 386,) we have to
use and integrate partial differential equations. The
student will be induced to consider the different branches
of analysis more closely, when he thus sees instances of
their use and necessity, and he may learn at the same
time, that no portion of pure science is to be rejected
as barren and useless; it is impossible to say what
value and what results the extension of applied mathe-
matics may sometime give to it.
From the nature of the work, it must be in a great
measure borrowed from many preceding writers; and it
is neither very easy nor very necessary to point out all
who have been of service. Besides Euler, however, who
is a never-failing guide and assistant, I may mention
xii
PREFACE.
my obligations to the excellent Treatise on Mechanics.
of Mr. Poisson. A considerable portion of the Trans-
actions of learned Societies has been examined, but
without any pretensions to the merit of having made
a complete abstract of the problems on this subject
contained in the different Academical Collections.
many
In going through the detail of so many cases,
properties and methods have occurred, as was to be ex-
pected, which so far as the author is aware, have been
hitherto unnoticed by mathematicians. But in such
cases there is always a strong probability that a more
extended examination of what has been previously
written, would overthrow most of the claims to origi-
nality which, in this late period of the history of
science, can be advanced. In the present instance, the
author would willingly abandon all such pretensions.
for the praise of having written
a useful and per-
spicuous treatise on the subject on which he has laboured.
The Syndics of the University Press have, from
the funds at their disposal, contributed liberally to the
expense of the work; and the author gladly takes this
opportunity of acknowledging his obligations to them.
CONTENTS.
xiii
CONTENTS.
DIVISION of the work...
Definitions and Principles
BOOK I.
Page
1
3
Co
CHAP. II. The free curvilinear Motion of a Point .
THE MOTION OF A POINT IN A NON-RESISTING
SPACE..
CHAP. I. The rectilinear Motion of a Point.
CHAP. III. Central Forces . . .
5
ib.
16
24
CHAP. IV. The Motion of several Points.
66
Sect. I.
Problem of two Bodies
68
Sect. II.
Problem of three or more Bodies.
74
CHAP. V. The constrained Motion of a Point on a given
Line or Surface..
79
Sect. I. The Motion of a Point on a plane Curve.. 81
Sect. II. The Motion of a Point on a Surface of
Revolution
95
Sect. III. The Motion of a Point upon any Surface.. 109
CHAP. VI. The constrained Motion of several Points ..
Sect. I. The Motion of a Rod on Planes
Sect. II. Tractories.
Sect. III. Complex Pendulums
CHAP. VII. Inverse Problems respecting the Motion of
Points on Curves.
112
113
127
130
152
xiv
CONTENTS.
1
Sect. I. Curve of equal Pressure .
Sect. II. Synchronous Curves....
Sect. III. Tautochronous Curves..
Sect. IV. Brachystochronous Curves
BOOK II.
Page
ib.
154
157
. 161
THE MOTION OF A POINT IN A RESISTING MEDIUM .. 171
CHAP. I. The Rectilinear Motion of a Point in a resisting
Medium..
Sect. I. No Forces but the Resistance....
Sect. II. The Body acted on by a constant Force besides
Resistance.
Sect. III. The Body acted on by a variable Force . . . .
CHAP. II. The free curvilinear Motion of a Body in a resist-
ing Medium....
172
173
176
. 181
183
Sect. I.
The Force acting in parallel Lines and con-
stant
184
Sect. II. Any Force acting in parallel Lines •
191
Sect. III. Central Forces.
· 195
CHAP. III. The constrained Motion of a Point on a given
Curve in a resisting Medium
204
CHAP. IV. Inverse Problems respecting the Motions of
Points on Curves in resisting Media....
216
BOOK III.
THE MOTION OF A RIGID BODY OR SYSTEM..
221
CHAP. I.
Definitions and Principles
. ib.
CHAP. II. Rotation about a fixed Axis
226
CHAP. III. Moment of Inertia
233
CONTENTS.
XV
Sect. I.
General Properties.
Page
ib.
Sect. II. Moment of Inertia of a Line revolving in
its own Plane.
Sect. III.
Moment of Inertia of a Line revolving per-
pendicularly to its own Plane.
Sect. IV. Moment of Inertia of a Plane revolving in
its own Plane.
Sect. V.
Sect. VI.
•
236
238
239
Moment of Inertia of a Plane revolving
about an Axis in or parallel to the Plane. 244
Moment of Inertia of a symmetrical Solid
about its Axis
Sect. VII. Moment of Inertia of a Solid not symme-
246
trical..
Sect. VIII. Centre of Oscillation
247
248
CHAP. IV. Motion of Machines
257
Sect. I. Motion about a fixed Axis..
.. ib.
Sect. II.
Motion of Bodies unrolling.
Sect. III. Motion of Pullies..
Sect. IV. Maximum effect of Machines
266
269
274
CHAP. V. Pressure on a fixed Axis...
277
Sect. I. A Body revolving, acted on by no Forces.... ib.
The three principal Axes of Rotation.
292
Sect. II. A Body revolving, acted on by any Forces .. 283
CHAP. VI.
CHAP. VII. Motion of any rigid Body about its Centre of
Gravity
CHAP. VIII. Motion of a rigid Body acted upon by any
. 304
Forces.
322
xvi
CONTENTS.
Page
Sect. I. When the Motions of all the Particles are
in Parallel Planes.
Sect. II. When the Body moves in any manner what-
ever
APPENDIX (A). On the Definitions and Principles
(B). On the Motion of a Body about two Centres of
Force.
(C). On the Attractions of Bodies.
ib.
334
343
346
364
(D). On some particular Cases of the Motions of three
Bodies..
372
(E). On the Vibrations of Strings.
377
(F). On the Vibrations of Springs
384
(G). On the Descent of small Bodies in Fluids.—On
the Ascent of an Air-Bubble.
388
·
(H). General Mechanical Principles .
394
I.
Principle of the Conservation of the Mo-
tion of the Centre of Gravity
ib.
II. Principle of the Conservation of Areas.. 395
III. Principle of the Conservation of Vis
Viva...
IV Principle of least Action..
397
400
A
TREATISE
ON
DYNAMICS.
DIVISION OF THE WORK.
1. THE object of the present Volume is to determine, for
any body or system of bodies, the motion which corresponds to
any forces; and conversely, to determine the forces which corres-
pond to any motion. That is, we have to investigate the relation
of the time, space, velocity, and force, when bodies are in motion
under given circumstances.
The subdivision of the Work will be as follows:
2. It will consist of THREE BOOKS; the two first treating
of the motion of a point, and the third, of that of a body of finite
magnitude. Thus, the former books apply to the motion of a
particle of matter which is conceived to be indefinitely small, so
that the path of every part of it may be considered as the same
line: they include also the case of a finite body which moves so
that all the points of it describe lines parallel and equal to each
other. In these cases there is only a progressive motion, or motion
of translation. The third book extends our reasonings to masses
of determinate magnitude and figure, moving into various positions
while their parts still remain at the same distance from each other
in consequence of the rigidity of the bodies. In this case we may
have, besides a motion of progression, a motion of rotation about
an axis either fixed or variable; and we thus comprehend any motion.
which can exist in a body.
A
2
3. The first and second Books consider the motion of a point
as it moves in a non-resisting space, or in a resisting medium. In
the FIRST BOOK we conceive a body to move freely but if,
instead of this, we suppose it to be retarded by moving through a
medium, as air or water, which does not allow it to pass without
affecting its motion; the mathematical difficulties of the question.
will be somewhat modified: and this case will form the subject of
the SECOND Book. In the THIRD BOOK, which treats of the
motion of finite bodies, we shall not consider them as moving in a
resisting medium.
4. Also, a point or body may either move freely in space, or
may be constrained to move upon a given line or surface: and
these different conditions will give rise to various problems relating
to motions both in resisting and in non-resisting media.
Again, instead of considering a single point only, we may sup-
pose that several points, connected either by invariable straight
lines, or by laws of mutual attraction and repulsion, influence each
other's motions: and in the first. Book we shall, after investigating
the motion of one point, proceed to the case of several.
The problems which we shall have to solve may differ also by
the different suppositions which we make with respect to the force.
They will vary, accordingly as we suppose the force to act in paral-
lel lines, or in lines tending to a centre, &c. and likewise according
to the power or other function of the distance from the centre, &c.
which we suppose to express the force. We may also consider the
motions of bodies when attracted to two or more centres of force.
The centres of force which we assume, are points from which
attraction or repulsion emanates in every direction: that is, to or
from which bodies tend to every side. The intensity of the force is
supposed to be the same at the same distance from the centre, in
every direction, and to vary according to some function of the dis-
tance. There are not, strictly speaking, such fixed centres of force
actually existing in nature; because, though attractions and repul-
sions do appear, they take place to and from bodies, and not
mathematical points; and these attractive and repulsive bodies are
themselves attracted and repelled, and generally are themselves
moving while they influence other motions. Still the introduction of
3
such imaginary centres of force, besides being easily and distinctly
conceived, and containing the most natural mathematical simplifi-
cation of the conditions which actually exist, does, in most cases,
offer an approximate solution of the problem in nature, and leads
to others yet more exact.
The law of attraction which appears to prevail in nature, or
rather the universal law of one of the kinds of attraction which
particles of matter exert, is that the force increases in the pro-
portion in which the square of the distance decreases. All the
attempts to prove this general fact to be a necessary truth, seem
to be completely unsatisfactory. At all events innumerable other
laws are mathematically possible, and will be supposed in the
problems which will occur; aud more especially those laws in
which the force varies according to some other power of the
distance, direct or inverse.
DEFINITIONS AND PRINCIPLES.
5. Velocity is the degree in which a body moves quickly or
slowly. When constant, it is measured by the space described in
a unit of time. This is the same as measuring it by the ratio of
the space to the time. When variable, it is measured by the limit
of this ratio.
Force is that which produces or tends to produce motion.
Accelerating Force is force considered with respect only to the
velocity produced, without regard to the magnitude of the body
moved. It is measured, when constant, by the velocity generated
in a unit of time; or by the ratio of the velocity generated, to the
time. When variable, it is measured by the limit of this ratio.
If t be the time, s the space described, v the velocity, f the
accelerating force acting in the direction of the motion,
2 =
ds
dť
dv
f
dt
….(a).
Moving Force is the product of the accelerating force into the
quantity of matter moved.
4
Momentum is the product of the velocity into the quantity of
matter of the moving body. Hence, moving force is, when con-
stant, proportional to the momentum generated in a given time, as
accelerating force is to the velocity generated in a given time.
6. The following are the Laws of Motion.
LAW 1.
A body in motion, not acted upon by any force,
will move on in a straight line and with a uniform velocity.
LAW 2. When any force acts upon a body in motion, the
change of motion which it produces is the same, in magnitude and
direction, as the effect of the force upon a body at rest.
LAW 3. When pressure communicates motion, the moving
force is as the pressure.
Hence, the accelerating force is as the pressure directly and as
the quantity of matter moved inversely.
The Inertia of a body is its resistance to the communication
of motion; and since the velocity communicated by a given pressure
is inversely as the quantity of matter; the inertia is directly as the
quantity of matter.
When bodies in motion press each other, Reaction is equal and
opposite to Action; that is, the pressures on each other are equal
and in opposite directions.
Hence, the moving forces are equal, and the momenta com-
municated in opposite directions also equal.
Impact is a very short and violent pressure. And hence it
appears, that in impact the momenta gained and lost are equal.
When a force acts to turn a body round an axis, its Moment is
the product of the force into the perpendicular upon it from the
axis.
When the moments of forces are equal, their effects are equal.
BOOK I.
THE MOTION OF A POINT IN A NON-RESISTING SPACE.
7. THE laws and equations of motion are here immediately
applicable, putting for the force ƒ, the value of the extraneous ac-
celerating forces which act upon the point in motion. We shall
first consider the motion in a straight line, and afterwards in a curve.
For the first case we shall require no principles, except the nature
of our ideas of force and velocity: for the second, it is necessary
to introduce also the second law of motion.
CHAP. I.
THE RECTILINEAR MOTION OF A POINT.
8. WHEN a point moves in a straight line, this will be the
line in which the force acts, and we can immediately apply the
equations,
V =
ds
dt'
d v
f
dt
•.(a).
Where t, s, v, are the time of motion, space described, and velocity
of a body, which is acted on by an accelerating force ƒ in the direc-
tion of its motion.
* It is here assumed that the forces and attractions in this case are
independent of the mass of the body acted on,
6
These equations would enable us to obtain finite relations
among the quantities in question, in several cases. For instance, if
the velocity were given in terms of the time, we could find the space
described by integrating the first equation, and if the force were
known in terms of the time, we might in the same manner obtain
the velocity from the second. In general, however, the force de-
pends upon the position of a body, and in this case, therefore, is a
function of s.
9. PROPOSITION. When the force is a function of the space,
to find the velocity and time.
Take the equations
ds
d v
v=
f
;
d t
dt
multiply them crossways to eliminate d t, and we have
ds
do
v.
dt
f. or vd v = fds...
dt'
··(b).
If we put for f its value in terms of s we can integrate equation (b),
and shall thus obtain ffds: whence v is known.
2
After this, t is determined by the equation
v=
ds
dt'
ds
or dt = ;
V
where a value of v being obtained in terms of s, we can integrate,
and find t =
jds
V
and thus the relation of the quantities in question
is completely determined.
10. We may now proceed to the calculation for different sup-
positions of force.
One of the most common and useful suppositions is, that the
force tends to a certain point or centre, and varies according to
Let
some direct or inverse power of the distance from this centre.
S, fig. 1, be the centre of force; SP= r; and when a point is at P
m
let it be urged towards S by a force or ma”, m being a con-
xn,
stant quantity. The quantity m depends upon the attractive power
7
residing in S, and is called the absolute force. It is measured by
the accelerating force at distance 1, for inaking r 1, we have
the force = m.
m
xn
It is supposed that or mx expresses the accelerating force
on a particle P, whatever be the magnitude of the particle: the
moving force or pressure produced by the attraction of S is greater
as the mass acted on is greater.
PROBLEM I. A body P falls from rest from a given point
A, fig. 1, towards a centre of force S, varying as some power of the
distance SP: to determine the motion*.
Let SA = a, AP = s;
x = α
→
s; and take the force to
m
be as some inverse power of the distance and ==
xn
1st, To find the velocity
m
v dv = ƒds
=
X
dx;
.. 2vd v =
2 m d x
xn
2m
(n − 1) xn−1
1) arī + C,
C being an arbitrary constant quantity, to be determined.
شوح
2 m
When r = a, v = 0; .. v²
=
N-
an-
(
124
v =
(2 m) t
(n −
1)}
a²-1 x²-1
This gives the velocity when n> 1.
* This is Prop. 39, Book I, of the Principia.
'To determine the motion,'" definire motum," implies the problems of
obtaining the relation of the space, velocity, and time in finite terms, that is,
freed from differentials.
It will be recollected that the bodies spoken of in this PART are always
to be considered as physical points.
8
If n = 1 this integration fails, and recurring to the differential ex-
a
pression, we have v² =
2 m hyp. log. x + C=2m hyp. log. x
2 m
If n < 1, v² =
(a¹—n — x¹—”).
·
1
N
If the force vary as some direct power of the distance, let
ƒ=mr, and we have
2² =
2 m
n+ 1
(an+1 — x²+1).
In cases where the force varies as some inverse power n, greater
than 1, when x=0, v = inf. or the velocity of falling to the cen--
tre is infinite.
In the same case; when a = inf. v remains finite, and
v² =
2 m
1
(n − 1) an~
;
or the velocity of falling from an infinite distance to the distance r
is finite.
If the force vary inversely as the distance, both these velocities
are infinite.
In all other cases, the velocity from an infinite distance to a
finite one, is infinite; and the velocity from a finite distance to the
centre, is finite.
If the body, instead of falling from rest at the distance a, be
projected upwards or downwards with a velocity V, we have, when
x = a, v = V, if the body be projected in the direction of the
force, and v = V, if it be projected in the opposite direction.
2
In this case, v² = √² +
2d, To find the time;
ds
dt =
v
1
2 m
1
n
a"-
(2 m)ì
(n − 1) ½
an
d x
-
1
x² -
xn
$
9
n
( n − 1 ) . a = ¹
(2 m)
2-1
x=
xz dx
(a”
1
x n − 1 ) ½ 3
which can be integrated only in particular cases: see Lacroix,
Elem. Treat. Art. 169.
1st, We can integrate if
ก
1
m
be a whole number where m'
1
n
and n'′ = n − 1; that is, calling r a whole number, if
n
1
+1
2 + 1
2
N
r, or n + 1 = 2rn
2r; .. n =
Qr
1 3 5
this comprehends the cases = ~ 1,
&c.
-
5' 5' 7'
also n =
I' 3
color
5 7
&c.
5'
m'
1
2d, We can integrate if
be a whole number; suppose
n
1+ 1
2 (n − 1)
1
1 ·
r;
n
1
r+1
.. 11 =
n
r
3
4
this comprehends the cases n = 2.
51
&c.
3
also n = 0
Q' 3
01/09
S 4
&c.
4' 5
Hence the only laws of force expressed by integral powers, for
which we can find the time, are
1
1
force oc const., force o dist., force o
force x
(dist.)'
(dist.)³·
The most simple fractional powers are
1
1
1
force ∞
force∞
force ∞
(dist.)½'
(list.'
(dist.)*
B
10
If the force be repulsive, the process of finding the velocity and
time will be the same as above, except that the signs will be
different.
m
In that case if force =
s = x a;
xn
v d v =
m dx
xn
dx
dt =
7
In many of the integrable cases, it is better to employ par-
ticular methods, than the general substitution for making the differ-
ential expressions rational.
Ex. 1. The force varies directly as the distance:
f
= mx;
v² = C -
v = m³ (a²
ds
dt
ง
:. vd v = mxds =
.vdr
mx² = m (a² — xº),
x²);
dx
m² (a² — x²)²,
mxdx,
t =
1
m s
and if t begin when x = a,
1
.arc
(cos. = 2)
+ C;
arc
(cos.
x
COS. =
a
COR. If with radius SA
and PQ be perpendicular to
a, fig. 2, a quadrant be described,
SA; and if SP = x:
AQ arc (cos. = x,
r, rad. = a) = a arc
(co
Cos. =
X
a
rad. = 1);
rad.
.. velocity at P = m³. PQ,
time in AP =
arc AQ
arc AQ
=
mr a
vel. at S
Ex. 2. The force is constant:
Principia, Book I. Prop. 38.
11
vdv=fds; ..v² = 2fs; the motion beginning when 8=0.
ds
dt =
· . t =
√(2ƒs)'
√
2 s
t being O when s is 0.
f
If the constant force be gravity, represented by g,
ance,
v² = 2gs, and t = 1
or s=½gť².
g
Ex. 3. The force varies inversely as the square of the dist-
t=
m
mdx
ƒ ===, odv = = "dz, v² = 2m (--);
f vdv=
dt=
益
​ar¹ dr
(2m) (ax)
2
x²
α
a
2 m
x dx
(a x − x²) * •
(~~) * {(ax—19)³ — — arc (ver. sin. =
{(ax-x®*
2x
a
Ꮖ
)}+C;
==
and, t being supposed to begin when ra, since ver. sin. 2,
a
2x
α
13/14
³
1 = (-) {(ax~ 1²) + [~~ are (ver.
ver. sin. -
a
=2=-=-) ] } • •
t=
2m
COR. On AS-a, fig. S, let a semi-circle be described, with
centre C; and let PQ be drawn perpendicular to AS meeting it,
PQ= √(SP. PA)= √(ax-r²). AQS=
па
10
2
SP
SC
arc SQ=SC× ang. SCQ= SC ang. (ver. sin. =
a
=
arc (ver. sin. =
*When the force is as the distance, we have for the whole time of falling
to the centre, making r = 0; t
t =
π
2 ms
24
When the force is inversely as the square of the distance we have for
the whole time of falling to the centre, making «=0,
πα
2 (2 m)
12
Hence, time in AP=
124
(~~)*(PQ+arc AQS – are SQ)
1/4
(4) * (PQ+ are AQ).
a
Hence, also time =
( 2 ) * (PQ.; SC+arc AQ. ‡ 40
2 m.
2√2
AC
2.44C)
AC
=
area ASQ*.
(am)
If SQ be produced to meet in R a tangent to the semi-circle at A;
ARSA.
PQ
PS
a V (a x − x²)
až
χ
( - ) *
;
a
(2 m)+
•
AR.
α
Ex. 4. The force varies inversely as the cube of the distance;
√(a² — x²)
v = V m
√m.
; t = a.
ax
√ (a²x²)
Vm
COR. If with centre S and radius SA, fig. 4, we describe a
circle, and make PQ, AR perpendicular to SA, and draw SQR;
v=
√ m
a² e
a
AR; t t =
PQ.
Vm
α
Ex. 5. The force varies inversely as the square of the dis-
tance, and is repulsive:
v = √(2m) { /
X
: dt =
Cam
}
α
x dx
2
√(x² — ax)
COR. If with focus S, fig. 5, and vertex A, the point from
which the body begins to move, we describe a parabola, and take
SQ=SP; SY being the perpendicular upon the tangent QY,
we have
SQ
SQ.d. SQ
d. arc AQ
.d. SQ
QY
√(SQ² — SY2)
SQ. d. SQ
x d x
√ (SQ² — SQ . SA)
2
√(x² — ax)'
Principia, Book I. Prop. 32.
13
.. dt =
Ex. 6.
distance:
2m
H
a
d. arc AQ, and t =
AQ.
•
2 m
The force varies inversely as the square root of the
až
v = 2m³ . { a¹ — x¹}};
2
{a}}
t = {x¹ +2a¹}. {a} − x§}}.
3 m² •
PROB. II. A body acted upon by a force varying as any
power of the distance, falls to the centre from a given distance (a):
to find the whole time of falling to the centre.
By Prob. I, we have, if force =
(n −
d t
--1)*. a
(2m)
(n − 1)³ x
(2m) ½
(n − 1) 3
1)½ x
(2 m)
n-1
n
-1
2
Ꮖ
I d x
72.
{an - I
Z dx
I d x
· {1 -
m
xn
an-1)
一
​{1+
1
x² - 1
1.3
an 2
x
+
2
an-
1
2.4
a2n-
Multiplying and integrating, we shall have
1
+
&c.}.
3π-1
t =
(n − 1)
(2m)
{c -
2
2
x
. X
2 + 1
3n
I a² - 1
an
5 n-3
1.3
2
X
2.4
ɔ̃n — 3 a²n -
2
&c.} :
and this, taken from xa, to a = 0, gives for the whole time
t =
(n - 1) ½
(2m)
n+1
1
1
2
2 a
+
in + 1
23n-
-
1
1.3
1
+
+ &c.}.
2.4 5n
3
4
14
COR. 1. From different distances the times of falling to the
same centre are as
2+1
2
a
Hence if force oc dist. time ∞ 1, or is constant,
if force ≈ 1, or is constant, time ∞ √(dist.)
1
if force ∞
time o dist.
(dist.)'
1
if force ∞
time (dist.),
(dist.),
1
if force ∞
(dist.)³ ›
time ∞(dist.)²,
1
if force ∞
time ∞ (dist.).
(dist.)³ ³
COR. 2. In all these cases the time is greater as the distance
is greater; but if the force vary in a higher direct ratio than the
simple power of the distance, the contrary will be the case.
Thus if force (dist.)², time ∞
if force ∞ (dist.)³, time ∞
x
1
(dist.)
1
(dist.)
COR. 3. The integration for finding the time when the force
is inversely as the distance, is not properly included in this case,
and is considered in the following problem. .
PROB. III. When the force varies inversely as the distance,
to find the whole time of the descent to the centre.
Let any distance SP, fig. 1, = r; hence,
J'
و
v² = 2m hyp. log. —
mdr
m
ƒ
v dv =
d r
dr
1
dt =
V
(2 m) s
√hy
hyp. log.
↑
a
15
1
And our object must now be to integrate this expression from
r = a, to r = 0.
Let
a
a
.. hyp. log. x², r
r
>
a
e.z.2;
a
r
ex29
dr=
2axdx
:. dt =
1 2adx
ez2
;
(2m)
ezi
; and time to centre
2 a
(2m)‡ Se
Se-a² dx,
2
from x = 0 to x = x
Now let there be a curve BQ, fig. 6, of which the ordinates
are CO = u
= u, OQ = z: and let its equation be ≈ = e
e-112. Let
this curve revolve round the axis CB, parallel to z, through a quad-
rant, so as to generate the surface BQQ'. We may find the solid
content thus generated by supposing the plane CBQO to revolve
through an angle de NCn. If CN = u, we shall have Nn
= ude, and if we take a portion of the triangle whose breadth
along CN is du, and conceive standing upon it a prism whose
height is NP or z, the solid content of this prism will be .ude.du:
and the solid content of the wedge BCPn will be the integral
of this from C to N=fzudude=fe-² udu.de=d0fe¬"udu.
And the solid content of the figure when the plane has revolved
through a quadrant, will manifestly be fe~*² udu = { C—e="² },
π
-
Π
4
and if this be taken from C, when u=0, integral =0;
π
4
2
.. content
{1-e-«²}. And if we suppose the solid to be extended
to infinity, so as to comprehend the whole space between the planes
BCX, BCY, and the curve surface, we must make u infinite, and
we have the content =
π
4
But we may find this solid content in another manner, by re-
ferring the surface to their rectangular co-ordinates, CM = x,
MN=y, NP =:: and it will then be equal to ffzdxdy (Lacroix,
Elem. Treat. Art. 247.)
Now u² = ('N° = x² + y², and ≈=
=e
е
e-lx² + y²).
1
16
Hence, content =ƒƒe-x³-ÿ² dx dy
=SS
2
=SS'e¯x²dx. e-y²dy
2
fe-x² dx.ƒe-dy; because in integrating with
respect to y, x may be considered as constant.
.2.2
And for the whole content we must take the integrals from
x=0 to x=x, and from y=0 to y=x; and in this case,
Se-2² dx and Se-ydy will manifestly be equal. Hence, whole
content= Se-xdx) from x=0 to x=¤;
П
4
π
= (Se~x²d x)², from x=0 to x=×,
√ = ƒe-*² dx, from x =0 to r=».
2
And time to centre =
απ
√2m
CHAP. II.
THE CURVILINEAR MOTION OF A POINT.
1
11. WHEN a point in motion is acted on by a force which is
not in the direction of its motion, it will be perpetually deflected
from its path, so as to describe a curve line. The quantity of this
deflection will be regulated by the second law of motion, in the
manner which we shall explain. By that law it is asserted that if
a point at P be moving with a velocity which would in a given
time carry it through the space PR, fig. 7; and if, during its
motion it be acted on by a constant force always parallel to itself,
which would in the same time make it move through a space Pp
17
from rest, it will be found, at the end of that time, in a point r,
determined by completing the parallelogram Rp.
If the force which acts upon the body, be variable in magnitude,
or direction, or both, we can no longer in the same manner find
the place of the body at the end of a finite time from P. The second
law of motion is then applicable ultimately only; that is, to the
motion of the body during an indefinitely small time *. This may
be stated also thus. Let PR be the space which would be de-
scribed in any time in consequence of the velocity; PQ the path
which is actually described in the same time in consequence of the
action of a variable deflecting force; Pp the space through which
the force, retaining the magnitude and direction which it has at P,
would cause a body to move from rest in the same time; Rr equal
and parallel to Pp: then will RQ be ultimately equal to Rr, and
coincident with it in direction.
12. PROP. To find the equations of motion of a body, moving
in a plane and acted upon by any forces in that plane. Fig. 7.
Let t be the time from a given epoch, till the body arrives at
P, and th till it arrives at Q, so that h is the time of motion in
PQ. Also let AM, MP be rectangular co-ordinates to the point.
P, and be called x and y: similarly, let AN, NQ and AO, OR,
be co-ordinates, parallel to these: PIand RK parallel to AN. Let
the force at P be called P, and the angle which it makes with x,
be called a. Also let the velocity at P be called V, and the angle
which it makes with r, be called 0.
We shall then have by supposition PR=Vh: and Pp=Ph²;
because Pp is described by a constant force. (Ch. I. Ex. 2.)
Hence, PH=Vh cos. 0; RH= Vh sin. 0.
Also if Rs, sr be parallel to AM, MP,
Rs = Ph² cos. a; sr=Ph² sin. a.
* The second law of motion is proved by experiment in the case of a
constant force; and it is manifest that the effect of a variable finite force for
an indefinitely small time may be considered to be the same as if it were
constaut.
C
18
But by Taylor's Theorem, considering x and y as functions of t,
and dt constant,
dx
d² x
h²
AN=x+
h +
+ &c.
dt
dt² 1.9
dy
d- 4
d² y
h²
NQ = y +
•h+
+ &c.
dt
dt²
1.2
dx
d² x
h²
=
Hence, RK MN- PH=
.h +
+ &c. - Vh cos. 0.
dt
dt² 1.2
dy
day h²
KQ= IQ — RH=·
.h +
+ &c.
Vh sin. 0.
2
dt
dt² 1.2
Now since Rr ultimately coincides with RQ, we have ultimately
Rs, RK equal, and also sr, KQ. Hence, ultimately
d x
V
dt
cos. 0) h +
d²x h²
d t² 1.2
+ &c. = ĘPh*. có
Cos. a.
dy
V sin. 0
d t
0) 1
d² y
h²
h+
+&c. = Ph². sin. a.
dt² 1.2
Whence we must necessarily have, equating coefficients of h,
d x
dy
V cos. 0=0,
V sin. 0=0,
dt
dt
d² x
ď² y
- P cos. a
= P sin, a.
dt
d t
dy
dx
Hence,
= velocity in r,
velocity in y,
d t
d t
d² x
d² y
= force in r,
= force in y.
d t²
d t²
2
If we represent by X and Y, the whole forces which act on the
point in the direction of x and of y, we have
d²
x
ď² y
= X, and
= Y;..........(c)
•(c)
d t²
2
d t²
24
where dt is constant, and X and Y positive, when they tend to
increase x and y.
r
A
19
COR. 1. It is clear that if we had referred the path of the
body to three rectangular co-ordinates, x, y, z, and if we had
made X, Y, Z, represent the whole forces in the directions of these
co-ordinates, we should have had, by reasoning exactly similar,
d² x
dt
=
X,
dy=y, dz
ར.
=Z.........(c')
d t²
t²
2
13.
These equations enable us to solve various problems re-
specting the motions of bodies acted on by any forces. If the
motion be known, we can, from them, find the forces in the direc-
tions of the co-ordinates, and by compounding these, the whole
force which acts upon the body. If on the other hand, the force de-
pends, in a known manner, on the position of the body, we can, by
resolving it in the proper directions, find X, Y, Z, in terms of x, y,
and z; and we shall then, by integrating the equations, have the
motion of the body determined. If we can eliminate t, we obtain
a relation among the co-ordinates which defines the curve described
by the body. We shall have instances of these various applications
in what follows.
Ex. 1. To find the forces which must act upon a point, so
that it may describe the arc of a parabola with a uniform motion.
If x, y, s, represent the abscissa, ordinate, and curve of the
parabola, we shall have, since the velocity is constant,
ds
dt
= c, a constant quantity;
ds2
¿² =
dt²
dx² + d y²
dť²
Now if 4 a be the principal parameter of the parabola, we have
y = 2 √ ax; :. dy = dx
dx²
dx
dr²
Hence, c²=
d ť²
( 1 + c ),
a
and
dt
a
+
a
2
20
Differentiating,
2 dxd2x
d t²
2
d² x
c² a dx
x²
2
(1 + - ) ²
(1+
2
c˜ a
2
dt2
2 (a + x)² 9
which gives the force parallel to the abscissa.
y²
y dy
Again, x=
;
.. dx =
4 a
2 a
dy2
y
2
•*. c.²
dy²
dt²
2 dy d² y
2
d to
d t²
4 a²
4 a²
2 2
c
4a² + y²
2
+1 1):
;
; and differentiating,
8a2 c2 ydy
C
(4 a² + y²)²
2
27
d²
У
4 a* c* y
4 a² c² y
d t²
(4 a² + y²²)²
(4 a² + 4 ax)²
c² y
4 (a + x)²'
which gives the force parallel to the ordinate, the negative sign shew-
ing that it tends towards the axis.
If S, fig. 8, be the focus, and SA the axis of the parabola; AD
AS-a, and DN perpendicular to AD, so that DN is the
directrix; we shall have DM or NP = a + x. Hence, the forces
in NP and PM are respectively as
2 AS
and
PM
PN² › NP²;
or as
MK
PN²⁹ PN²
PM
and
Hence, the
PK being the normal, and therefore MK=2AS.
whole force on P, which is compounded of these two, is in the
direction PK*, and proportional to
PK
PN² *
* It may easily be shewn that if a body move uniformly in any curve,
the force which retains it is perpendicular to the curve.
7
21
Ex. 2. A body is acted upon at every point of its path, by
a force which is proportional to its distance from a given centre
towards which it tends to find its path.
Let the centre of force C, fig. 9. be made the origin of rectangular
co-ordinates CM, MP: the force in the direction PC is every where
proportional to PC. Resolve it in the directions PM, MC; and
these lines will be proportional to the resolved parts. Hence, we
shall have
force in direction of r= mx, force in direction of y =
my:
m being some constant quantity, and the negative signs indicating
the direction of the forces.
ď² x
Hence,
mx,
dtⓇ
d'y
dt2
my;
2 d x d² x
d t²
dy
Q d y d² y
2 m x dx,
= − 2 mydy;
dť
integrating,
dx²
d t
dyⓇ
C ma
mx²;
= D − my² :
d t²
We may
where C, D are arbitrary quantities depending on the velocity and
direction of the body's motion at some given point.
evidently, without restricting their values, put for C and D, mh³
and m k²; and thus we have
dy
d x
√(C-mx²)
= dt
dt =
V (D- my²);
dr
or
√(h² — x²)
dy
√(k² — y³)'
Integrating, we have a=ẞ+y;
where a is the arc whose sine
is
ẞ
is the arc whose sine is
Y
,
and yan arbitrary arc. There-
h'
k'
fore, we have
sin. asin. ẞ cos. y + sin. y cos. ß:
or if n be the cosine of y, and consequently √(1 — n²) its sine,
ľ ny + √(1 − ‚‚²) .
=
h
k
»²). √
(1-
22
transposing and squaring, we get
x²
2
n² y²
2
+
h² k2
2 n xy
h k
(1 − 2²). (1
y
༧°
k²
=1~n²
2
y² n² y²
+
k²
k²
2
x2 y2
Whence +
Q n x y
= 1 − n² :
112
k²
hk
which is the equation to an ellipse referred to rectangular co-ordi-
nates measured froin the centre. (Wood's Alg. Part IV.) Hence,
the curve described is an ellipse, of which the centre is C.
The axes of the ellipse may be thus found: let the tangent at P,
fig. 9. be parallel to Cr, whence CP and CD will be conjugate
diameters, and hence
CP² + CD² = a²+b², PM. CD = a.b,
where a and b are the semi-axes.
Now, since
dy
√
2
dx
'k² — y²
h² = x²;
2
it is manifest that when the tangent is parallel to Cx, we have y=k:
hence to find x = CM, we have
x²
2
2n x
+1-
h
2
= 1 − n² ;
h
X
h
--n=0, x=nh = CM; .. CP2 = CM²+MP² = n² h²+k².
Also, to find CD, put y=0, and we have
x²
12=1—n²; .. x=h √(1 — n²) = CD.
Hence a²+b²=h² + k², ab=kh √(1−n²), whence a, b are
known.
To find the position of the major axis CA, or the angle ACx=0,
we may proceed thus. Differentiating the equation to the curve, we
find
で
​NY
dy
h
hk
dx
Y
N X
/
k²
h k
23
Now at the point A, the curve is perpendicular to CA; and
hence at that point the normal passes through the centre; there-
fore (Lacroix. Art. 65.)
ny
y dy
dx
r²
2
h k
18
Ꮖ
;
x;
༤
N X
hk
y
k2
x y
ny
xy
nx²
;
whence we find
h²
hk
k²
h k
y³, h²
h² - k² y
22
+-
1=0; and hence
nhk x
S
2nhk
r
h² - k²
У
1
X
2 tan. 8
= tan. 20;
L
1
tan.20
hence is known.
To find the time of describing any portion, we have
d x
dt
=
d x
√(C — mx³)
√ m √ (h² — x²) '
I
'. t =
arc
√ m
(sin.
x
sin. =
+ const.
the constant being determined by the place of the body at a given
time.
1
For a whole revolution, we have time =
2π. Hence,
√AZ
it is independent in the size of the orbit.
CHAP. III.
CENTRAL FORCES.
14. THE equations of the preceding Chapter would enable us
to determine the motions of bodies acted on by any forces whatever,
and of course, among the rest, in the case where the force is
supposed always to tend to a centre, and to be represented by some
function of the distance from that point, of which we had an instance
in the last example. But problems respecting the action of central
forces as these are called, are of such importance, and lead to such
simplifications of our general formula, that it is convenient to con-
sider separately this application of our reasonings.
15. PROP. A body acted upon by a central force will
describe a curve lying in one plane.
If we consider a body, moving in any direction, to be acted on
by a force tending to a given centre, it is clear that the body will
be deflected from its rectilinear path, and will describe a curve.
And this curve will be in one plane, namely, the plane passing
through the centre of force and the original direction of its motion.
For the body, by the combination of its original velocity with the
action of the force, will, in a small time, describe a path by the
second law of motion, (see p. 17.) which will be in the plane
in which both the motion and the line drawn to the centre are.
And at the end of this small time it will tend to move on in the
same plane; but being deflected as before by the central force,
which is still in the same plane, its actual motion will still be in
the same plane. And similarly, after the lapse of any number of
such intervals, that is, of any finite time, its motion will be still in
the plane in which it originally was.
Hence, it is only necessary to consider the two equations,
Art. 12, which belong to the motion of a body in a plane.
25
16. PROP. A body being acted on by a central force, the
sectorial areas described are proportional to the times.
Let the centre of force S, fig. 10, be the centre of rect-
angular co-ordinates SM=x, MP=y; let SP=r; and let the
whole force at P, in the direction PS, be called P, P being a func-
tion of r.
Also let the angle ASP =v; then, P being resolved
into its components X and Y parallel to SM and PM respectively,
we have X=- P cos. v, Y= − P sin. v. Hence, equations (c)
become
fx
Px dy
Py
dť
r
dt
gu
multiply by y and r respectively, and subtract;
hence
integrating,
-
x d² y — y d² x
dt²
x d y-y d x
dt
= 0
=h, a constant quantity.
Now sector ASP = ASNP~ SNP, MN, being a parallelogram;
.. diff. of sector ASP = diff¹. of ASNP - diff¹. of SNP
xy
=xdy-d.
2
x dy — y d x
Q
hdt
ht
Hence sector ASP
the time and the area beginning at A.
2
And any portion of this sector is proportional to the time of de-
scribing that portion*.
COR. 1. Also since we have ht=2 area described in t, if we
make t = 1, (the unit of time, for instance 1",) we have h =
2.area described in time 1.
COR. 2. If SP =r, and angle ASP=v, we shall have
diff¹. of sector ASP =rdv; (Lacroix, Elem. Treat. Art. 111.);
hence rdv = hdt.
Principia, Book I, Prop.1
D
26
17. PROP. A body being acted on by a central force, to find
the velocity at any distance from the centre.
Take the equations
ď² x
Px d'y
Py
;
dte
2
dt²
r
Multiply by 2dx, 2dy, and add ;
2 d x d²x+2dy d²y
2
dt2
=
2P (rdr+ydy)
- 2 Pdr,
r
because x² + y² = r², and xdx + ydy = rdr :
or d.
dx² + dy²
dt²
2
2P dr.
If s be the length of the curve
dx² + dy²
dt
2
ds²
= (velocity)².
dt2
Hence the velocity will be known by integrating the expression
ds²
d.
dt²
2 Pdr,
which may be done when P is a function of r; and we shall have
ds2
dr²
= C 2f Pdr.
If C be the (velocity) at any distance a, the integral SP dr
must be taken between the limits a and r.
COR. 1. If the velocity at the distance a be given, the velocity
at the distance r will be the same, whatever be the path which the
body describes. For the integral depends only on the force P,
and not on the path.
=
2
COR. 2. If angle ASP v, as before; ds² = dr² + r² dv²;
(Lacroix, Art. 110.).
27
dr² + r² dv²
.d.
2 Pdr.
dt²
18. PROP. A body being acted on by a central force, to find
the polar equation to the curve described.
r² dv
By Cor. 2. to Art. 16, dt =
h
dr²+redv²
dt²
2
h² dr²
r4 dv²
.. by Cor. 2. to Art. 17.
+
h²
2
الله
h² dra
h
2
d.
+
2 P dr.
r4 dv²
22
1
d r
Now make
=u, and, .'.
du, dr =
du
2
ጥ
น
:. d
h² du²
dv²
2 P du
+ h² u²
u²
2 h² dud² u
2 P du
or
supposing dv constant,
dv²
or
ď² u
+ u
dv²
-
..(d).
h² u²
+ 2h²udu =
P
2
2
दर
= 0:.
=
which is the equation of which we shall make use most commonly
in the consideration of orbits described about a centre. It may be
employed either in determining the law of force, when we know the
curve, and consequently the relation of u and v, which is called the
direct problem of central forces; or if P be known in function of r,
and therefore of u, we may, by integrating, find the relation of u
and v, which gives us the nature of the orbit; this is called the
inverse problem of central forces.
19. PROP. The orbit being given, to find the time of de-
scribing any part of it, and the velocity at any point.
dv
r² d v
The expression for the time is dt =
-
:
h
hu²
and having expressed dv in terms of u, or u in terms of v, we can
find ↑ by integrating.
28
เ
}
With respect to the velocity, we may thus obtain an expression
for it. We have ds = dr² + r²dv³, where ds is the differential
of the curve;
4
r² dv²
and dt² =
;
h²
ds²
... (velocity)²
h2 dr² h²
dr
+
dt²
p4 do
or, since
2
du,
2
= h²
(du² + u²)
COR. 1. Multiply the equation
d2 u
P
+ u
= 0,
dv²
h² u²
by 2h du, and we have.
2 dud² u
h².
2 P du
+h². Qudu
dv²
=
u²
.. h²
dv2
(du² + u²)
+r) -
Pdu
Q
= C
С
2
u³
h² (du
+u"
u²) =
рез
2
0;
= = = C + 2 2f Pdu = C - 2f Pdr,
which agrees with Art. 17.
COR. 2. If we draw SY a perpendicular upon the tangent,
and suppose p
=SY, it will be easy to see that we have
p
rdv
r² dv
or p =
ds
ds
hdt
ds
h
and
;
ds
d t
P
Hence by what precedes, p =
therefore the velocity is inversely as the perpendicular on the tangent*.
Principia, Book I. Prop. 1. Cor. 1.
It has been usual among
English Mathematicians to define a spiral by the equation between the
radius vector r, and the perpendicular on the tangent p. This is virtually
only a differential equation to the curve, but its use is sometimes convenient.
PROP.
29
20. PROP. When bodies revolve in circles, having the centre
of force in the centre, to determine the periodic times †.
In this case r is constant, and x = r cos. v.
rdv
constant, by Art. 17,
>
dt
dt being constant, d²v = 0.
Also, since r is
the velocity, is constant; and therefore,
Hence xr cos. v, dx=
— r sin. v.dv, d²x — — r cos. v
r cos. v.dv²;
which substituted in the equation
ď² x
Px
dt2
-
r
PROP. To obtain the central force in terms of and p.
By Cor. 2, Art. 19, we have p =
h d t
ds
; and differentiating,
h²
d s²
2 h² d p
p3
2
p²
dt-
ds²
= d.
2 Pdr, by Art. 19. ;
dt
.. P
h² dp
p³ dr
COR. The velocity in the curve at any point is equal to that generated
by the force Pat that point, continued constant, and acting ou a body while
it moves through one-fourth the chord of curvature drawn at that point
through the centre of force.
h2
P.pdr
For (velocity)²
= 2 P. chord of curvature,
p²
dp
2pdr
(for chord =
dp
Lacroix, Note H.)
= (velocity)² by force P through chord,
because for constant forces, (velocity)² = 2ƒs.
+ Principia, Book. I. Prop. 6.
30
rdv²
d v²
P
give
P, and
dt2
dť²
p
dv
The fraction is the angular velocity, and if 2π be four
dt
right angles, and T the time of a revolution,
dv
2 π
Απ'
2
P
dt
T
T2
r
2 π V r
T=
VP
rdv
velocity
√(Pr).
dt
COR. J. If Por, Tis constant,
velocity ∞ r,
velocity ∞ Vr,
velocity ∞
Vr
velocity ∞ -
"
P∞1, Tx Vr,
Рос
Po 1, Tor,
Рос
22
γ
P× 1, Tar²,
γ
COR. 2. Similarly if the variation of T were given, that of P
would be known.
21.
We shall now proceed to determine the paths and motions
of a point acted on by any central force whatever, beginning with
the cases in which it is proportional to some power of the distance
from the centre.
PROB. I. Let the force be directly as the distance, or P
m
mr = m being a constant quantity.
,
=
u
In this case the equation (d) becomes
ď² u
+u
dv²
h² u³
m
2
3
=0
= 0 ;
which might be integrated: but we have already solved this case
31
by a different method, and found the orbit to be an ellipse, the
centre of which is the centre of force*. See Chap. II. Ex. 2.
If the force be repulsive, and as the distance, the process
is nearly the same, and a hyperbola is described, the centre of which
is the centre of force.
PROB. II.
Let the force be inversely as the square of the
m
mu².
=m
distance, or P =
r
Hence, by (d);
ď u
dv²
m
h²
= 0.
m
To integrate this, let u-
= w;
h²
d2 w
+ w = 0;
dv
(Lacroix, Art. 280, 281.)
k v
and if €*º represent a particular value of w, we have
k² + 1 = 0, k = ± √ − 1.
Hence, the general value is
w= Ce°√ =¯ + C'e¯°N=T
(C+C)
C
C
(c°1/-
9
C₂
- C'=
(making −
making C + C' = C₁, C − c′
C₁ cos. v + C₂ sin. v.
Hence, u=
and
du
dv
2
-
C₁ cos. + C₂ sin. v +
C₁ sin. v + C₂ cos. v.
Principia, Book I. Prop. 10.
་
m
1
32
du
Now when v = 0,
C2,
dv
du
when vπ,
=
C₂ i
dv
hence, between v=0, and v=π, there must be a value of v which
du
makes
= 0: let this value of v be a;
d v
-
C₁ sin. a + C₂ cos. a = 0; C₂ cos. a =
C₁ sin. a;
1
m
+
h2
r
cos. a
m
h²
C₁ cos. a cos. v+ C₂ cos. a sin. v
C₁
cos. a
(cos. a cos. v+sin. a sin. v) +
C₁ cos. (v - a)
cos. a
m
+
h²
Let r' and r'" be the values of r, for v=a, and v=π+α;
both being supposed positive; hence,
2
h²,
1
C₁ m
+
r
cos. a
C₁
F
ՊՆ
h²
cos. a
+
m
h²
; and, adding and subtracting,
C₁
+
J'
cos. a
2'
Hence, the equation becomes
1
↑
cos. (v − a) + ½
a) + { //
1
+
"
2rr
.. r
"
"
r'tr'' + (r" -r') cos. (v − a)
Now r', rare opposite parts of the same line : let r'"+r'=2a,
r"-r'=2ae; .. r'r" a² - a² e², and
グ
​a (1-e²)
1+e cos. (v - a)
the equation to an ellipse, v-a being measured from the vertex
nearest to S*.·
* Principia, Book I, Prop. 11.
33
If r" be less than r', e will be negative, and the angle v- a
will be measured from the larger portion of the axis.
The curve may assume different forms by the alteration of the
arbitrary quantities C1, C2. If u ever become 0, or negative, the
form of the curve is no longer an ellipse.
Now the two values corresponding to v=a, and v=π+a, being
those for which
du
d v
Hence, if any value of u be negative, one of these will be so.
And hence the curve will no longer be an ellipse, if either
=0, are manifestly the greatest and least values
of u.
C₁
cos. a
M
C₁ m
+ or
h2
+ be negative.
cos. a h2,
If for instance, the latter be negative, we may suppose
1
C₁
M where 7" is positive, whence, as before,
cos. a h2
7'
27"r
(?'"' + r') cos. (v − a)+r"— r'"
and making 7" — r' = 2 a, r'" + r′ =2ae, (supposing 7" >r') we have
r =
a (e² – 1)
ecos.(v-a) + 1
the equation to an hyperbola.
If we have
C₁
cos. a h²
m
M
C₁
+
O, we get, putting for
h2 cos. a
1
m
{cos.
}
h
7=
cos. (v− a) + 1}; and if
Σα
1+cos. (va)
the equation to a parabola.
h²
=2a,
m
And similarly if
C₁ m
+
cos. a h
O, the curve will be a parabola,
but in a different position.
Hence, in all cases the curve will be a conic section: and sup-
posing C₁ positive, it will be
E
34
h²
m
an ellipse if
C₁
m
+
cos. a
C₁
a parabola if
cos. a
C₁
an hyperbola if -
cos. a
h²
be positive, or
+ = 0, or
h²
m
m
C₁
h²
C₁
cos, a
m
:
h²
cos. a
m C₁
h2
cos. a
+ be negative, or <
PROB. III. A body being projected from a given point, with
a given velocity, in a given direction; and acted on by a given
force varying inversely as the square of the distance; to find the
trajectory described *.
This problem might be solved by the preceding formulæ, but
more simply as follows.
In fig. 11, let a body be projected from a distance SPD, in
a direction PY making the angle SPY-8, and let the velocity at
m
P=V; force; Sp=r.
2m
By Art. 17, velocity² = C—2ƒP dr = C - 2 mdr = C + 2m
r
Sm
2
{
-하
​D
and when r=D, velocity = V; ... velocity² = V² + 2 m
But by Art. 19, Cor .2, velocity=
on the tangent.
h
p
, p being the perpendicular
Hence, at P,
2
h²
V²; because perp'. = D sin. ♪
D² sin.² d
h2
at p,
p
= V² + 2m {-}}
;
Ꭰ
D2 sin.28
2 m
.. dividing,
=1+
2
Vs
D
P
* Principia, Book I, Prop. 17.
35
11
1
2 D² sin.² 8
Ρ
now in the ellipse
in the hyperbola
and the expression for
12
2 m
V2 D³ sin.2 8
Ρ
+
Ρ
b²
1
+
2 a
b2
2 a 1
b²
p
भ
+
2 m
V² D² sin.2
1
in the trajectory may manifestly be made.
to agree with one or the other of these, as the part of it independent
of r is positive or negative.
Hence, the trajectory will be an ellipse
1
2 m
if
D2 sin.2 8
V² D³ sin.²
28
be negative;
2 m
2 m
that is, if 1 <
V² Di
or if V² <
D
Similarly, if V² >
2 m
D
If V² =
2 m
D
>
the curve will be an hyperbola.
the curve will be a parabola.
In the case of the ellipse, we must have
1
2 m
1
b2
V² D³ sin.² 8
D² sin.2 S
2 m
V2 D² sin. & D
{-2}
2 m);
2 a
2 m
62
1
2 a
V² D² sin.² d
D
Ve
2 m
* For p²
and p²
b² r
2a-r
b² r
2a+r
in the ellipse,
in the hyperbola, by Conic Sections.
36
Hence, a and b, the semi-axes of the ellipse, are known: and
a² - b⁹
hence, e² =
is known.
a²
To find the position of the major
axis we have
also.
* We may
D =
a (1 − e²)
1+e cos. (v− a)’
find the position of the major axis in the following manner
If PV, fig. 11, be the chord of curvature at P, by Note, p. 29,
√2 =
2 m 1
PV ; .. PV
D² 4
2 D² V2
is known.
ጎ
2 CD2
4 SP.PH
AC
SP+PH'
Also, by the property of the ellipse, PV=
.. PH=
SP.PV
4 SP — VP ;
whence PH is known: aud it must make with PY
an angle equal to SPY, and hence the point H is known. Hence C is known,
and AC (SP+PH): from which data the dimensions and position of the
ellipse are easily found.
=
From the same principles we may solve other problems where the ve-
locity in the conic section is concerned.
PROP. To compare the velocity in the ellipse with the velocity in a
circle about the same centre of force and at the same distance.
Let PV be the chord of curvature at P.
2 m PV
In ellipse, velocity2-
4
2 CD2
PV=
AC
m.PV
2 p²
2 SP. PH
2r (2a-r)
AC
a
.. velocity² =
m(2a-r)
ar
m.
m
And in the circle velocity.r
by Art. 20.
a
24-1
1
Hence, velocity in ellipse: velocity² in circle ::
-
2a-r: a.
ar
r
This agrees with Newton Book I, Prop. 16, Cor. 3, 4, 6, 7.
And
37
where v is the angle which determines the position of P, and is
therefore known. Hence, cos. (v — a) is known, and hence a, which
determines the position of the major axis.
COR. 1. It appears by Art. 10, that the velocity2 from an in-
2 m
D
finite distance = ; hence, the trajectory will be an ellipse, a
And in the same manner may the velocities in the other conic sections
be compared.
PROP. A body is revolving in an ellipse, and the force is suddenly
altered in the ration: 1; to find the alteration which takes place in the
orbit, fig. 12.
If m be the force before the alteration, n m will be the force after. Let a
be the semi-axis major before, and a' after; and let r be the radius vector
where the alteration takes place.
Then since the velocity at this point may be considered as belonging to
both the first and the last orbit, we have
m (2 a− r)
nm (2 a' - r)
velocity2
;
ar
a'r
.. 2 aa-ran (2 a a'r a);
nar
:.a'=
2(n-1)a+r
(2 a − r) r
Hence, 2 a-r=
2 (n−1) a+r
This is the value of PH', S and H' being the foci of the new orbit, and
PH' will be in the same line with PH; hence the new orbit is known.
If PH' be infinite, the new orbit will be a parabola with its axis parallel
to PH. This will be the case if
2 (n−1) a+r=0;
if n=
2a-r
2 a
PH
SP+PH®
If n be less than this, the new orbit will be an hyperbola, and PH' must
be measured in the opposite direction.
In nearly the same manner we may find the alteration produced in the
orbit, if the velocity be suddenly altered in any ratio.
38
parabola, or an hyperbola, as the velocity is less than, equal to, or
greater than that acquired from infinity.
Cor. 2. In the first case
1
1
V2
.. V2
;
= 2m
2 a
D
2 m
{
:
D
2 a
therefore V is the velocity acquired by falling from a distance 2 a
to a distance D, (see p. 11.). Hence 2 a is the distance from the
centre at which a body must begin to fall, so that when it reaches
the curve, it may have the velocity of the body in the curve; and
this distance is the same for every point of the curve.
COR. 3. Let the velocity=n times the velocity from infinity, or
V² = n².
2 m
D
1
(1 − n²)
—
; .. by last Cor.
2 a
D
D
1
1-n²
2
n² D² sin.² 8
2a:
b² =
2
1
n
b2
1 - n²
b2
2
2
1-
α
Hence e² = 1
n² D² sin.² d'
= 1 − 4 n² (1—n³) sin. d.
By means of these formulæ, we may, under given circumstances,
find the magnitude and position of the trajectory.
COR. 4.
It appears from the preceding investigation that the
major axis is independent of the direction of projection. And that,
if n be given, the excentricity is independent of the distance of pro-
jection.
EXAMPLE: A body is projected at an angle of 30° with the
distance, and with a velocity which is to the velocity from infinity
as 4 to 5: to determine the ellipse described.
4
In this case n ==
sin. d
5
22
62
b² =
D
1 - ก
28
n² D² sin.2 8
1
n²
2
9
4 D2
9
25 D
39
b 12
b²
= .48;
e
= 1
N
a
25
a
:.7696
e=.877.
And at the point of projection
D =
a (1 − e²)
1+e.cos. (v − a)
25 D
18
=
× .2304
1+.877 cos. (v — a) '
25
× .2304 - 1
18
680
... cos. (v-a)
.877
877
.7753, &c. =
cos. 39º. 10'.
Hence, va= 140° 50'
=
ASP, fig. 11.
PROB. IV. To find the
elliptical orbit*.
time of describing any portion of the
r² dv
fr³dv
We have the equation dt =
(Art.19.); .. t="
;
h
h
m
1
2 a
and since
+
플​.
2
h²
2
a² — a²e² ;
-ae
.. h = √(am). V (1 − e²).
Instead of substituting for r, its value in terms of v, which
would produce for dt an expression not readily integrable; we
shall express the time in terms of another angle u, as follows.
On the major axis of the ellipse Aa, fig. 13, let a semi-circle
be described, and MPQ drawn perpendicular to the axis, and
SQ, CQ joined: and let ACQ = u.
The expression fr du, beginning from A, is twice the area
ASP. Now it is easily seen that
1
area ASP
MP BC
area ASQ MQ
AC'
Principia, Book I, Prop. 31.
40
b
.. area ASP.
a
b
a
>
area ASQ =
b
(area ACQ-SCQ)
a
(† AC . AQ – { SC . MQ)
b
(a.au
ae. a sin. u);
2 a
.. fr³dv = 2 area ASP = ab (u - e sin. u),
and putting for b, a √(1 − e²), and for h its value,
fridv
αξ
t
h
(u-e sin. u)..........(1),
mž
the time being supposed to begin from A.
We have now to find the relation between u and v;
HP2 - HM² = PM2 SP2 - SM²;
PM² = SP² –
.. HP2 - SP² = HM2 - SM²;
(HP+SP)(HP – SP) = (HM+SM) (HM – SM);
2 AC. (2 AC-2 SP) 2 CM.2 CS;
or, dividing by 2.2
and putting a cos. u for CM,
a (a− r) = a cos. u. ae
.. SP = r = a - ae cos. u,
CM-CS
SP
and cos. v=
SM
SP
Cos.u
-e
.*. cos. v =
1 COS. V
Hence
1 + cos. v
.. tan.
2
and tan.
2012
2107
li
||
a cos. u. ---- ae
a-ae cos. u
1 — e cos. U
1 − e cos. u + e COS. U
1 — e cos. U e + cos. u
(1+e) (1
cos. u)
(1 − e) (1 + cos. u)
1+e
1- e
•
tan.
и
2
101
1+ e
.tan.
· e
;
И
..(2).
41
Hence, we can find u in terms of v by (2), and then t in terms of
u by (1); and conversely. The angle u, or ACQ, is called in
Astronomy the eccentric anomaly, ASP being called the true
anomaly*.
COR.
To find the time of a half revolution from A to a,
it is evident that we must take u from 0 to 7: which will give
t =
at
m 3
π.
Hence, the time of a revolution is
2 atπ
m²
If the trajectory be an hyperbola, the calculations will be
nearly the same as in the case of the ellipse.
If the force be repulsive, an hyperbola will be described
having the centre of force in its exterior focus: and its properties
will be analogous to those in the other cases.
PROB. V. In the case when the orbit is a parabola; to find
the expression for the time†.
As before, dt =
And r =
a = 0.
2 a
r² dv
h
; measuring angles from the vertex, so that
1 + cos. v
Also h√(2 ma); see p. 33.
4 ad v
√(2ma). (1+cos. v)²
Hence, dt =
að
(2 m)³
dv
4
COS.
* See Woodhouse's Astron. p. 190, edit. 1.
For expansions of and r, see Laplace, Mec. Cel. Part. I, Liv. 2.
p. 180, &c.
+ Principia, Book I, Prop. 30.
F
42
*
ať
(2 m)+
*
•
dv
do
cos.
2
+ sin.
2
2
ข
cos.
4
-------
212
(2 m)
dv
1+tan.
2
2
2
COS.
we can integrate; and find
and since d. tan.
V
2
COS.
2
αν
t =
=
tan.
2 mt
216
+tan.³
t being supposed to begin at the vertex, where v= = 0.
PROB. VI. Let the force be inversely as the cube of the
m
distance, or P = = mu³*.
The equation in this case becomes
ď² u
+ u
dv2
M U
h²
= 0.
liv
To integrate this equation, let u= €
ek be a particular solution.
(See Lacroix, Elem. Treat. Art. 280.)
* Newton considered the curves described when the force is inversely as
the cube of the distance, and besides the logarithmic spiral, noticed the
curves, Species I, V, and VI; but omitted the examination of the others,
by supposing the body to move from an apse. Principia, Book. I, Prop. 9,
and Prop. 41, Cor. 3. The complete analysis of this case was given by
Cotes in his Logometria; Phil. Trans. 1715; from which circumstance
these curves are sometimes called Cotes's Spirals. From certain ana-
logies observed by Newton, Species I, and V, are called the Hyperbolic
and Elliptic Spiral, respectively. It may be remarked, however, that the
Reciprocal Spiral is sometimes, by foreign writers, called the Hyperbolic
Spiral.
43
and if y,
Thence, k²+ I
integral will be
m
h2
= 0;
be the two values of k in this equation, the general
Yo
u = Cer⁰ + C'εro;
C, C' being any two arbitrary constants.
The curves described will be different, as the values of y
are possible or impossible, and as the arbitrary constants are
positive or negative. We shall consider the different species
thus produced.
m
SPECIES I. Let > 1, and C, C' both the same sign.
h²
Hence, k± √ (m − 1) = ± 7.
જ
Suppose therefore u Ce+ C'e¯Y°;
du
dv
hence, = y {Ce?" — C'e˜Y"}.
du
dv
Now when = 0, Cer = C'e-y, or €27v =
Ε
و
C'
which
C
can always be fulfilled by a possible value of v: let this value
be a, so that
γα
-γα = c;.. C = ce˜ya, C' = cεya.
Cεya = C'e
·{EY (v—α).
€˜Y (v−a)z.
Hence, u = c {
When o = a, u 2c; and since at that point
=
du
= 0, the
d v
As v
curve is perpendicular to the radius, or there is an apse
increases, u increases, and therefore r diminishes, and when v be-
comes infinite, r becomes O. Hence, the curve is such as is repre-
sented in fig. 14.
* An apse is a point where the curve is perpendicular to the radius vector,
and where, consequently, in general, the radius vector will be either a maxi-
mum or minimum.
44
If C, C' be both negative, the curve will be the same. The
sign can only indicate that the angle v is to be measured in the
opposite direction.
m
2
SPECIES II. Let > 1; and C' = 0.
h²
a
Therefore u=Ce¹º; yv=hyp. log.hyp. log., if a = ·
Hence, the curve is the logarithmic spiral, fig. 15.
r
dr
dr
m
Differentiating, ydv=
;
=~=
"'
rd v
h²
√(-1):
:
hence, √(m − 1) is the co-tangent of the constant angle SPY,
h²
2
which the tangent makes with the radius vector: and therefore
√m
h
h
is the co-secant, and
the sine of SPY.
√·m
h
Let
sin. ß; .. h = sin. ß √m.
V m
If C = 0, the curve will be the same.
m
SPECIES III. Let > 1, and C' negative.
h2
Therefore u = CεY'
Cert C'e
уч
Now when u=0, CeY = C'e-y and e2yv =
C'
for which
C
there is always a possible value of v let this value be a; and
let Cera = C'e <= c,
γα
γα
..C=ce¯¥ª; C'=ce: and hence u = c
γα
= c {e
{EY (v− a)
- E
− y(x − a)}.
When va, since u=0, r is infinite.
As v increases, u in-
creases, and r decreases; and when v is infinite, u is also infinite,
andr vanishes. Hence, the form of the curve is that in fig. 16,
va being the angle ASP.
45
If p be the perpendicular from S upon the tangent, we have,
(p. 28,)
1
p
2
=
ds2
r * d v²
do
2
=c² {e? (v− a)
r² dv²+dr²
r² dv²
= u² +
du²
dv²
€
v (v − a)} 2 + c² y² {EY (v− a) + €¯v (v − a) z z
−
2
=c² (1+ y²) {e²r (v—a)
and when = a,
v
=
-€− 2 y (t − a)} + 2 c² (y² — 1),
1
2
c² (1+ y²). 2+2 c² (y² − 1)
p
2 2
4 c²²; and p =
I
207
And, hence there is an asymptote BZ to the curve, parallel to SA,
at a distance SB =
1
207
Similarly, if C′ be positive and C negative.
SPECIES IV. Let
ጎቢ
1.
h²
In this case we must return to the original equation, which
here gives us
12
d2 u
= 0; ..
1
a
= C, u = C (v — a): r=
C (v—a)
î
- a
du
d v²
d v
it being supposed that when u = 0, v=a.
For this position r is infinite: any other
is reciprocally as the angle v-a, or ASP.
a, or ASP.
this case is the Reciprocal Spiral, fig. 17.
value of r, as SP,
Hence, the curve in
If a circular arc PQ be described with centre S, PQ =
r (v− a) = a; and hence, is at every point the same.
It is manifest that the curve will have an asymptote BZ, such,
that SB=a.
SPECIES V. Let < 1.
ጎ
h2
m
In this case the values of k in the equation + 1
= 0,
h²
46
1
are impossible. Let them bey. Therefore for the
general integral of the equation we have
u = Cε rv √ = 1
√ = 1
= { (C + C' ) (ε Ÿ³ √ = ¹ +6¯yo√ = 1)
1
+ } (C− C′) (e¥© √ = ¹ _ 6—YON—1)
= C₁ cos. yv + C₂ sin. yv,
1
2
making C₁ = C + C', and C₂ = √ — 1 (C— C').
du
Hence, =
y C₁ sin. yv + y C₂ cos. yv;
dv
Co
du
and when
<= 0,
tan. yv =
: for which there is always a
d v
C₁
value of v, whether C₁ and C₂ be of the same or of different signs.
Let a be this value
:. Ca
2
=
;
sin. ya
C₁
2
.. u. =
cos.ya
C₁
cos. ya
{cos. yv cos. ya+sin. yv sin. ya}
cos. y (v − a),
C₁
cos. ya
C₁
1
cos. y (v− a)
U=
and r=
,
a
a
or, making
cos. ya α
when v = a, r = a, and there is an apse.
π
2'
cos. y (v-a)
When y (v-a) =
r is infinite, and, therefore, may be parallel to, an asymptote; to
find the position of the asymptote, we have
1
du²
2
= u² +
cos.²y (v—a)
2
y² sin.² y (v - a)
p
dv²
a²
2
+
a²
and when y (v — a) =
2014
-12
The form of the curve is given in fig. 18.
2
2
a
2
2 ?
= SB2.
a
m
1
h
47
PROB. VII. To determine in what cases each of these curves
will be described.
We may observe, that in the case where the body describes a
ď² u
circle, and consequently where =0, we have, in a circle,
mu
น
h
m
dv²
=0, and = 1, or h= √m, the area in time 1.
h²
m
Now the species varies as in the curve is greater than, equal to, or
h²
less than 1: that is, as h, the area in time 1 in the curve, is less than,
equal to, or greater than Vm, its value in the circle. So that if
the area in a given time be less than that in a circle with the same
force, we shall have Species I, II, or III; if the areas be equal,
we have Species IV; if the area in the curve be greater, we have
Species V.
In these two latter cases it is clear, that since the area is
not less than it is in the circle when the radii vectors are the same,
the velocity will be greater than it is in a circle. In the three first
cases we may thus compare those velocities.
In the circle whose radius is r, since r. velocity = h² = m,
we have velocity
m
-
2
= mu²*.
In the curve, velocity² = hº (u² + du³).
v
dv2
But u = CεY + C'e¯Y",
2
(du)²
=
v
= {yC€¹® — yÑ€˜Y ° } ?
= y² u² - 4y² CC'.
Hence, velocity² = h² (1 + y²) u° − 4y°CC'
= mu² — 4y²CC'; since h²(1+y³)=m.
If the velocity be less than that in a circle, we have CC' negative,
*This is also the velocity from an infinite distance.
48
and therefore the curve is Species I.
that in a circle, we have Species II.
be greater, we have Species III.
If the velocity be equal to
If the velocity in the curve
If the force be repulsive, the equation will resemble the one
for Species V, and the curve, which we may call Species VI, will
be as in fig. 19.
PROB. VIII. To find the time of describing any portion of
the curves in Prob. VI.
In Species I, if we suppose the angle v to be measured from
the apse, and consequently a=0, we shall have
dv
d v
1
hdt = r² dv =
=
2
и
c² { ε Y " + ε ~ Y " } ²
200
1
€
d v
11
ال
+1} 2
1
.. ht =
= C
2c² 7
2 γυ
होक
may suppose the time to begin when v=0: on this supposition
We
we have
1
1
a² €
27v
1
1
t =
; if a = SA=
2 c² yh (2
27 v
€
+
27r
yh e +1
2c
Similarly, we should find
a²
in Species II, t
27h
a
2
2 cot. BVm
{ 1 − e−2yt'},
v and t being measured from the point where the radius vector = a.
2 a²
2
I
In Species III, t =
yh e
V h ε 2 r
v
if a = ; V = ASP,
1
2c
being measured from SA, and t being the time from P to the centre.
In Species IV, t =
a² 2
a²
=
hv
; being measured from SA,
√m.v
and t being the time to the centre.
49
In Species V, t =
the apse.
y h
tan. yv; v and t being measured from
T
COR. 1. In order to find the time of describing a given angle
8, we must take the value of t between the values v and v+d; we
shall thus have in Species I,
hr =
11
1
22
20 Y
1
2 む
​€
+1 €
1
2 y (v + d)
+1
જ
€2 r (0 +8) — € 270
y v
૪
Y (ε ² Y³ + 1) (ε ²
1
Ć²
22² 7
(E
1
2
+
2 y (x +ô)
+1)
€
€
Qc²y {e7°+e¯Y°} { €? (©+8) +€˜~ (?+d) }
Υ
8
€
•
7125
2γ
r and r being the radii at the beginning and end of the given angle.
Similarly,
in Species II, hr =
in Species III, hr =
Cerd
€
7172,
γ
- E
rire,
(erò
27
in Species IV, hr = d . 1' 1 12,
in Species V, hr =
sin. yd
COR. 2. In all the cases, the times of successive revolutions
in the same spiral are as the extreme radii.
Let a straight line SRQP, fig. 14, drawn from S, cut the
spiral successively in R, Q, P: thence, since in this case 82 is
constant, we have
time from P to Q: time from Q to R :: SP.SQ : SQ.SR
:: SP : SR.
G
50
Let the force be inversely as the 5th power of the
PROB. IX.
m
distance, or P =
=mù.
5
ď² u
тиз
Therefore, by (d), do
+ u
0;
h2
and multiplying by 2du and integrating,
du²
тия
dv²
2
+u
= C;
2h2
C being an arbitrary constant; hence,
du²
d v²
тия
C − u² +
2/2
This equation cannot be integrated generally by the common
methods.
When the right hand member is a square, it becomes simple;
that is, if 4 times the product of the extreme terms be equal to the
square of the middle term;
2 Cm
h2
if
= 1; if C
=
h2
2m
On this supposition,
2
du
1
h
u² V m
V m
2
2
= ±
= ± +
dv
√2
√ m
h
h V 2 m
u² };
and we shall have two different Species as we take the + or the sign.
2h du
In the first case,
h²
m
h
√ m
2
+ u
u
= dv V2;
= √2.(va);
√ m
... hyp, log.
h
√ m
a being the value of v, when u=0.
51
When v=a, r is infinite; as u increases and r decreases,
V m
h
v increases, and when u =
Vm'
or r =
h
v is infinite.
Hence, the curve, fig. 20, has what may be called an asymptotic
√ m
circle with radius SA =
to which circle it perpetually
h
1
p
du²
d v²
approximates, but which it never actually reaches.
We have =u² +
and when r is infinite, or u = 0, p=
distance of the asymptote BZ from SA.
In the second case, we have
h2
тия
= u²+
u² +
h2 тия
+
2 m
Qh²
છુ
2m
2h2
ૐ
√(2m)
;
which is SB, the
h
chdu
V m
น
u²
ท
h
h2
= dv √2;
.. hyp. log.
И
√ m
h
u +
√m
= √2. (va),
a being the value of v when u is infinite.
α
Hence, hyp. log.
h
u+
√ m
h
= √2.(a-v).
и
√ m
When va, u is infinite, and 7=0; as u decreases, or r increases,
h
√ m
or r =
a-vis
V m
h
See
v also increases; and when u =
infinite. Hence, the curve has, in this case also, an asymptotic
circle, and is situated within it, as it was before without it.
fig. 21. SQ comes to SA when v = 0; and ASP = a
COR. 1. We shall now compare the velocity with that in a
circle.
v.
52
In a circle with radius = r, velocity2
m
= тия.
2.4
= Pr, (see Art. 20.)
==
In the curve, velocity2 = h² (u² +
du²
dv²
ht
4 тия
+
2m
2
h²
тия
Now when r = SA, or u²
,
= velocity in curve
+
m
тич
= mu² =
2
2
velocity in circle; which it manifestly should be,
because as the radius approximates to SA, the motion approxi-
mates to circular motion.
In the first case u is always less than
velocity is always less than that in a circle.
h
and hence the
Vm
>
In the second case u is always greater than this value, and the
velocity is greater than that in a circle at the same distance.
COR. 2. To find the velocity, so that one of these curves may
be described.
2
Let, at any point P, the angle SPY = ß, SY being a per-
pendicular on the tangent. Therefore h² = velocity. SY2-velocity
r² sin.2 B.
?
Now let the velocity be e times that in a circle at the distance
SP: that is, velocity2 = emu: hence,
9
h4
тия
&² mu² =
2 m
+
; .'. (2e² — 1) m² u¹ = h* ;
u4
2
but h² = e²mu¹
sin.2 B
h2 √(2€² − 1)
2
U
;
... sin.² ß
2
€² mu²
2
€
ดู
Hence, if e be given, we can find
sin. ß; and hence, the
direction in which the body must be projected to describe the curve.
It will belong to the first or second Species, as e is less or greater
than 1.
53
2
2
Also e² sin.²ß= √(2e² — 1); e* sin.* ß − 2e²+
1
1
sin.* ß
sin.* ß
1
2
Ε
.*. €² sin.² ß-
+
sin." B
V(1-sin.4 B)
sin.² ß
2
1 ± √(1 − sin.¹ (ß)
€
sin.4 B
and the first or second curve will be described, as
lower or the upper sign.
we take the
COR. 3. By equating the values of ASP in the two species, we
should find for the same angle ASP, fig. 20, SP'. SP = SA².
PROB. X. Let the force vary inversely as any power of the
distance, or P=mu".
Therefore
ď² u
dv²
mu”
+ u
h²
multiplying by 2du, and integrating,
2
0 ;
(n − 1) h²
du²
2 mun
1
+ u²
C ;
dv
du
whence,
dv;
√{c-
u² +
2 mun
(n - 1) h²)
and if the expression on the first side be integrable, we can find the
relation between u and v.
To find the time, we have
d v
dt =
hu²
du
hu²
√ {c-v² +
2 m u"
(n − 1) h²)
The quantity C will depend upon the velocity, and will be
known, if we know the velocity for a given point; which may be
called the velocity of projection, if we consider this point as the
beginning of the motion.
For we have
velocity2 = h²
h² (u²
du2
u² +
2 m un
I
dv²
+h²C,
(n-1)
=
V,
and if, when u=a, we have velocity
2 ma"
1
V
√² =
(n − 1)
+h'C; whence C is kuown.
54
It may be convenient to compare the velocity with that ac-
quired by falling from an infinite distance. Let q be the velocity
acquired by falling through any space towards the centre. Therefore
mdr
q d q
-
2m
qⓇ
(n − 1) p²
I
+ const.;
and if q be the velocity acquired from infinity,
const. = 0, q²
2m
(n − 1) p² - 1
2 mun-
N - 1
Hence, if at the point of projection, when u = a, the velocity
be e times that from infinity, we have
2man-1
€2
2 man
-
- 1
+ h² C.
N
1
2 ma'
N
1
... h² C = (e² - 1).
du
COR. At the apsides we have
= 0;
d v
2 mu"
¿. C − u² +
= 0,
(n − 1) k²
2
or, putting for C its value,
(n − 1) h²
2
(e² − 1) a"-1
• u²+u"-¹ = 0.
I
2m
This may have four roots possible, {for instance, if n = 5, and
(n − 1)² h4
4m²
> 4(e² — 1) a¹,} but only two give apsidal distances;
in fact the other two are always negative.
PROB. XI. In the particular case where the velocity is equal
to that from infinity*, to find the curves.
* If the velocity be at one point that from infinity, it will be so at all
points. For, by Art. 17, Cor. 2, if the velocity at the distance a be that
from infinity, it will, at the distance r, be the same as if the body had con-
tinued to descend in a straight line.
55
V
have
du
dv=
Here in the last Prob. C=0, and we can integrate. For we
du
3
$2 mu"
2 m un
2
и
u
(n − 1) h²
((n − 1).h²
}}
3
2 mu"
Let
(n − 1)h²
= y²; .'.
(n − 3) du
2 dy
;
... cos.
d v =
(v − a) =
n
3
И
2
dy
N
2'
y
3` y√ (y² — 1) '
n − 3
1
(∞ — (2)
У
arc (sec. = y);
hV(n−1)
-
√(2m). u z
-3
Hence, if n>S;
h₂ (n − 1)
√(2m)
n
N· 3
2
r
= COS.
(v − a).
V
h√(n-1) 1
3 - n
Similarly, if n < 3;
= COS.
(a — v').
√(2 m)
3-λ
↑ 2
C1
In the first case, it is manifest that when the first side is = 1,
-3
✓ (2 m)
or † 2
h √(n − 1) '
the figure has an apse. It is symmetrical on
the two sides of this apse, and diminishes as v'
7
increases.
— a= ASP, fig. 22,
When
N 3
2
π
T
(v − a)
or v
α
we have
Q
n — z '
r=0, and the curve passes through the centre, as in fig. 22.
In the second case, r increases as a
v=ASP, fig. 23, increases,
1
du² Q m un
when
3 - n
2
π
(α — v)
or a
3 - 11
curve is parallel to it. To find the nature of the infinite branch we have
1
π
r is infinite, and the
2
u +
p
d v
9
;
(n − 1) h²
infinite; hence, the branch AZ has no asymptote.
and when r is infinite u = 0, and
p is
56
PROB. XII. Let the force vary inversely as any power of the
distance; it is required to find the conditions requisite that the orbit
may have an asymptotic circle. See Prob. IX.
1
P=mu", and as before in Prob. X.
d v =
v=
du
2
√{c-u² +
√
d u
2 mun
(n − 1)h²Š
2 mun
/{c-
u² +
-
(n − 1) h²)
Now if the orbit have an asymptotic circle, of which the radius
is it is manifest that the value of v, taken from any value of u up
C
to u=c, will be infinite. That is, the integral on the right hand side
must be infinite when u=c. Also u―c is necessarily a factor of
du
the denominator, because when u=c,
O, and therefore C – u²
d v
1
2 mun
+
=0. But if the denominator has two factors u-
-C,
(n − 1)h²
For in that case
√ {(u — c)². Q}
>
the integral will be infinite for u=c.
v=S
du
Q involving u"-3, and inferior powers of u.
And if we put
u=c+z, it is manifest that Q will become A+ Bz+ &c. and
ย
d z
=√ √ {z². (A + Bz + &c.) }
=S
the other terms involving
z=0, v becomes infinite.
dz
2
Z√ A
2
{
Bz
1 + &c.}
2 A
Bz
2AVA
+ &c.;
hyp. log. z
VA
direct powers of z.
Hence, when u≈c,
57
We shall therefore have an asymptotic circle if there be the
factor uc twice in the denominator of dv; that is, if the equation.
have two roots c, c.
2 mun - 1
(n − 1) h² — u² + C =0
2
But in this case the equation.
2 mu"
h2
2 mcn
2
-
Ad
1
(n-1)h
2
and
2 mc"
h
2 u=0, has one of these roots; therefore
c²+ C = 0, or C=c² -
2
2 c0, or h² = m c²-³.
Now, in the curve, velocity=h2
2 mcn
1
(n − 1) h²'
= h² (du²
(du² + "')
= h²
il
(putting for C its value,) =
น
2 m u² - 1
c)
1)ke + C);
(n − 1)h²
2 mu"
η - 1
Q mun
1
(putting for hits value,) =
N - 1
1.
+ c² h² -
+mcn-1
2 m c²·
-
(n − 1)
2 mcn
n-
1
-
1
M
N-
1
1
{2 un−¹ + (n−3) c²−¹}.
Let at any point the value of u be b, and the angle SPY, which
the tangent makes with the radius vector, ß; and suppose that at
this point the velocity is e times that in a circle.
velocity2 force × radius (Art. 20.) = mu² .
1
U
Now, in a circle
= mun−1.
mu²-1. And
Hence, when u = c, velocity² = m c²−1
manifestly should be.
H
velocity in a circle; as it
58
when u=b, velocity in circle = mb-1, and velocity in curve
1
2
= ε² mbn-¹.
Hence, e² mb" ~ :
2
m
1
1
{2 b²-¹ + (n − 3) c²−1} ;
n-l
1
and {(n − 1) e² — 2} b^− ¹ — (n − 3) c²−1;
1
1
2
- 2) = =
(n − 1) e².
:. c=b
Also h² = 2.velocity sin.²
h'
b²
But h² = mc2-3;
3
2
3
= €
2
n-3
1
β B = {z. c^mb"- ¹.sin.° ß
3
b2
= ε². m bn - ³ sin.2 ß.
2
2
... c²-³ — e² bn-³.sin.' ß; and c=b.(e. sin. B)s.
2 (n
3
2
Therefore (e sin. (3)² = {("— 1) e² — 9
n—3
77-3
− 2)2n-2
And sin. B = !. {(n = 1)² - 27
€
n-3
1
n-1
which gives the relation between the velocity and the direction of
projection, in order that the curve may have an asymptotic circle.
The radius ( = -),
(= }
1
of the circle, is easily found by the pre-
1
ceding formulæ. If is greater than , the circle is an interior
one as in fig. 20: that is,
C
(n-1) e² - 2
2
if
> 1,
-
22 3
if (n-1) e-2 > n−3,
2
if e> 1, or if € > 1.
If on the contrary e be less than 1, the circle is exterior to the
curve, as in fig. 21.
It is clear that we must have > 3.
N
59
In nearly the same way we may find the conditions requisite for
the description of an orbit, with an asymptotic circle, when the force
is represented by any function whatever of u.
PROB. XIII. Let the force consist of two parts, one of which
varies inversely as the square, and the other inversely as the cube,
P = mu² + m'u³;
of the distance.
ď² u
m
m'u
+ u
= 0;
d v²
hⓇ
h²
ďu
or d'u+
dv²
+(1
To integrate, let (1)
d² w
dv²
И
- m²) u
-
or u = w +
m
h²
m
h² — 1 M
m
h2
= 0.
(1)
2
w,
m'
h²
+(1) w = 0; or if 1 – = y',
m)
dv²
ď w
+ y²w=0:
of which, by nearly the same process as in Prob. II, we shall find
the integral to be
W = C₁ cos. yv + C₂ sin. yo;
m
..u = C₁ cos. yv + C₂ sin. yʊ + h²-m
may be transformed in exactly the same manner as in Prob. II;
This
du
that is, let a be the value of v, which makes
= 0, then the
d v
value which gives yoπ + ya will also make
du
= 0; and if
d v
1
1
goll
//
be the values of u, corresponding to these values of v,
we shall have
60
U
u =
p
17
}
cos. y (v− a) + 1/2 ? + };
which is the equation to the curve described, if r′ and r" be
positive.
This manifestly agrees with the equation to an ellipse, of which
the focus is in the centre of forces, except in having y (v-a)
instead of v α. Hence, the curve may be thus described: if, in
fig. 24, and 25, Ap be an ellipse of which the focus is S; and
Sp being any radius, if we take ASP =
ASP
Y
; so that ASP
being v-
v-a we may have ASpy (va); then SP=Sp may
ber, and the equation just found for r will be satisfied; there-
fore the curve APB thus described will be the path.
APB will be without the ellipse Ap, fig. 24, if y be less than
m'
unity; that is, if 1 - <1, or if m' be positive. If n′ be
h²
negative, or the force be P=mu² - m'u³, the path described will
be within the ellipse, as in fig. 25.
In both cases we shall have an apse B, corresponding to an
apse b in the ellipse; at which point ASb=y. ASB, and SB = Sb.
Hence, since ASb, we have ASB =
ASB is the angle between the apsides.
ASI
π
Y
√(1±
m'
h2
After describing an angle BSA' ASB, the body will come
again to an apse at A', and so go on perpetually revolving about S;
and approaching to it, and receding from it alternately.
The line of apsides SA retains always the same position, when
a body describes an ellipse as in Prob. II. In the case of the
present problem, this line, which is at first in the position SA,
fig. 26, 27, would after one revolution come into the position SA',
after a second, into the position SA", and so on; the angles ASA',
A'SA", &c. being equal. Hence, this line is said to revolve round
S. If it revolve in the direction of the body's motion, as in fig. 26,
61
it is said to move in consequentia, or to progress; if it move in the
opposite direction, as in fig. 27, it is said to move in antecedentia,
or to regress. It appears by what has preceded, that the first or
the second of these cases will occur, as the part of the force m'u³
which varies inversely as the cube of the distance, is additive, or
subtractive*.
3
If P=mu² + m'u³ so that we have ASB =
π
/(1-
m'
*
h²
m'
it is manifest that we must have
1; and therefore h² > m'.
h²
When h² m', the body will fall into the centre without coming
to a second apse, as might be shewn by integrating the equation
de u
m'
dv2
(−1)
m
И
= 0.
h²
d2u
m
du
m
When h²=m',
dv²
h²
છું
= 0;
(v − a);
dv
h²
m
(v — a)²
u -
h²
+c; supposing that u=c when v=a.
In this case the body approaches the centre by an indefinite
number of revolutions.
PROB. XIV. Let the force be represented by any function of
the distance; it is required to find what value the angle between
the apsides approximates to, when the orbit becomes very nearly
a circlet.
It is manifest, that if we project a body perpendicularly to the
radius vector, with a velocity very little greater or less than the velocity
in a circle for the same distance and force, the path of the body
will not differ much from a circle. With many laws of force, the
body will revolve perpetually between its greatest and least apsidal
distances, as in last Prob. fig. 26, 27: and the angle between the
This corresponds with Principia, Book I, Prop. 43, 44.
† Principiu, Book I, Prop. 45.
62
apsides will depend both upon the velocity and the law of force.
As, however, the velocity approaches more nearly to that in a circle,
the angle between the apsides will tend nearer and nearer to a cer-
tain limit. This limit it can never reach, because when the velocity
becomes accurately that in a circle, the apsidal distances are equal,
a circle is the curve described, and there is no longer, properly
speaking, an angle between the apsides, as every point is an apse.
But if we find this limiting angle, it may serve to indicate what is
the angle between the apsides, when the difference of the higher
and lower apsidal distances is small, but finite.
Let P = u²pu; when u is a function of u; so that P may
be any function whatever of u;
.. by (d),
d² u
dv²
фи
+ u
=
= 0.
h²
2
Now at the point where the body is projected perpendicularly
to the radius, let u=c; and for any other point let u=c+z, z
being small.
Then pu=c+ q'c.z + g″c
2
z²
1.2
+ &c.
Also if 1:1+d were the ratio of the velocity to the velocity²
in a circle at the point of projection, we should have
in the circle, velocity force × radius (Art. 20) = c² pc
.. in the curve, velocity
=
2
c & c
1+8
;
velocity
...
•. in curve h²
Фо
2
c² c ( 1 + s ) ;
1
[
=cpc.
C
d2 z
(1 + d) c
+c+z-
dv²
2
фс
therefore, substituting in the
original equation, we have
{pc+p'c.z+p″c.
z²
+&c.} = 0;
1.2
d² z
or
dr²
2
+ (1 -
сф
cp"c
z²
Z
&c.
Фо
фо 1.2
0.
сб
c p' c
S
&c.
фо
And when the orbit becomes indefinitely near a circle, & be-
63
comes indefinitely small, as does z; and hence, z2, zd, &c. may be
omitted in comparison of z:
сф'с
ď z
hence,
dv²
+ (1 - )
z — cd = 0.
фс
c o'c
If we make 1
=y², we shall have, as in Prob. XIII,
фо
сб
for the integral of this equation,
z = C₁ cos. yʊ + C₂ sin. yo+
22
and u =c+z = С₁ cos. yv + C₂ sin. yv +c+
сб
જ
which indicates the same kind of orbit as is described in the last
problem. And here, as there, we shall have
π
A = the angle between the apsides
Y
Π
сф'с
V{1_cpc
фо
Ex. 1. Let the force vary as any power of the distance,
P=mu”=u² .mu”-2; .. pu=mu”—2; p'u= (n − 2) mun−³ ;
фи=
(n − 2) mc²
mcn - 2
2
=
= 3 — N ;
.. the angle between the apsides =
π
√(3 — n) *
=, which agrees with Prob. I, of this
When n = −1, A =
Chapter;
π
when
n = 0,
|
A =
√3
π
when n = 1,
'n
A =
when
n = 2,
Vai
A, which agrees with Prob. II;
when
n = 3,
A is infinite;
64
and when n> 3, the expression is impossible. In fact, in this
case, if the body leave one apse, it will never reach another, but
will go off to infinity, if the velocity be greater than that in a circle,
and fall to the centre if the velocity be less *.
I
If n be a little greater than 2, the apsides progress slowly:
thus the apse will advance about 3° in one revolution, or 10 in a
semi-revolution, if n = 2
4
243
Ex. 2. Let the force consist of two parts, each varying as
any power of the distance;
2
P = mu" + m'un' = u² (mu"-² + m'un² − 2);
..фи=mu" - 2
фи=ти +m'un' - 2
(mu”
3
p'u = (n − 2) mu² –—³ + (n′ — 2) m'u¹¹ −3;
+ (n' − 2) m'cn'
2
(n − 2) mc'
2
= 1
mcn
2
C
+ m'cn'
2
2
N 2
(3-n) mc' + (3 — n') m' c².
mc"
2
+ m'cn'
2
T
whence A
>
is known.
V
2
Ex. 3. Let the force vary as the sine of the distance from the
centre: the distance being considered as an arc.
Let 9 be the distance, which, in this variation is considered as
a quadrant; and m the force at that distance: then,
sin. r
sin. q: sin. :: m : m.
force at distance r: the
r
sin. q
sines being taken to such a radius that q is a quadrant.
But, if the sines of the corresponding angles be taken to radius
* This is also true if the velocity be not nearly equal to that in a circle,
as might be shewn.
The Student will find an investigation of the angle between the apsides,
in some cases, when the orbit is not nearly circular, in the Transactions of
the Cambridge Philosophical Society, Vol. I, Part I, p. 179.
•
1
65
1, they will be in the same ratio: and q : r
;
::
014
29
the
angle corresponding to r
π
πη
.. force = P=m. sin.
=m.
= m. sin.
2q
2 qu
m
П
= u²
sin.
2
ม
Qqu
m
π
...фи
sin.
2
И
2m
p'u=
3
น
Q qu
sin.
2qu
π
m
π
π
COS.
u² * Q qu²
2qu
Hence, y² = 1
сфс
фо
π
π
= 3 +
cotan.
;
2gc
2qc
1.
where
-
C
is the radius of the circle to which the orbit approximates.
1
If we make
C
=a, we have
z² = 3 +
If a = 0,
па
2q
y²
2
-
cotan.
па
29
4, y = 2.
I, Y = 3 +
π
If a = 1/1/19,
7°
If a = q, y²
= 3.
π
π
The angle between the apsides varies from to
according
Q
√3
to the different magnitudes of the circle described.
+
I
R
CHAP. IV.
THE MOTION OF SEVERAL POINTS.
22. IN the last chapter we have supposed a single body to be
acted on by forces tending to fixed mathematical points, and on
that supposition have calculated its motion. But we may suppose
those points, from which the force enianates, to be themselves move-
able bodies, acted upon by their mutual forces, or by any others,
and we may then have to calculate the motions of each of these
bodies. This is in fact the problem which occurs in nature; for
we do not there find forces tending to mathematical points, but re-
siding in physical bodies, and connected with their material pro-
perties; and, with respect to those forces which we have most
frequently to consider, depending entirely on the quantity of matter.
By considering the motions of the planets, and other bodies of
which the universe is composed, it was discovered by Newton, that
each of them exerts upon all the others a force which is at different
distances inversely as the square of the distance; and that at the
same distance from each, the forces with which a point would be
impelled towards them, are directly as their quantities of matter.
It further appeared, that this force or attraction exerted by each
mass, is the result of an attraction exerted by each particle of
which it is composed; so that we may conceive every physical
point in the universe to exert a force varying inversely as the square
of the distance.
We shall consider more particularly bodies exerting forces of
this kind; but it is manifest, that any other hypothesis of the
variation of the force is equally possible, speaking mathematically,
and may occasionally be introduced in our problems.
67
་
23. The forces which are exerted by these bodies, are of the
kind which we have called accelerating forces; that is, they are
measured solely by the velocity produced in a given time, and are
entirely independent of the mass moved. Thus, if a body M exert
upon a particle P, a certain accelerating force, which is represented
by f, it will, under the same circumstances, exert upon 2 P or 3P
the same accelerating force; though it is manifest, that for this
purpose the pressure or moving force exerted, or the weight pro-
duced in the particle 2 P or 3P, must be two or three times as
great, respectively, as it was in P. At a given distance ƒ is pro-
portional to M.
By the third law of motion, the accelerating force ƒ is pro-
portional to the pressure exerted on P directly, and to the mass of
P inversely. Hence, by what has been said, we have
Mxfx
pressure on P
P
and hence, pressure on P∞ MP, and =µ MP, suppose: µ being
the same for all bodies.
Hence, if a body M act upon any particle at a distance r, the
accelerating force which it exerts may be represented by, where
m
m is proportional to the body itself. By properly assuming the
unit, m may be considered equal to the body. And to this force
acting upon the particle, estimated in the proper direction, we
may apply the equations of motion in the same manner as if it
tended to a fixed point. We shall now proceed to the different
cases of the problems to which we are led by these consider-
ations.
We have taken both the bodies M and P as points. If they
be spheres of finite magnitude, the effect will (in the case in
which the force varies inversely as the square of the distance) be
the same as if they were collected at the centre; if they be of any
other form, an irregularity will be introduced into the variation of
the force this will be shewn in treating of the attractions of
bodies.
68
SECT. I. Problem of two Bodies.
24. PROP. When two bodies are acted on by no forces except
their mutual attractions, their centre of gravity will either remain at
rest, or move uniformly in a straight line. The motion will
evidently be in the same plane.
Ꮖ
referred to rectangular
and let their masses be
Also let (r) represent
Let P, Q, fig. 28, be the two bodies,
co-ordinates x', y' for P, and x", y" for Q;
m', m', respectively, and their distance r.
the function of r, according to which their force varies; so that
m'o (r) is P's action on Q, and m" (r) is Q's action on P.
Resolving these forces parallel to x' and y', respectively, we have by
equations (c),
Ꮖ
dex'
x″ – x′
x
ďý
y
m" (r)
=
dt2
r
dt²
2
m”p (r) Y″ — y′..
y'
•(1),
↑
"
d²x"
Ꮖ
-x d³y"
- m'p (r)
m' & (1)
-
· .(2).
dt
ጥ
dt2
Now, multiplying equations (1) by m', and equations (2) by m", and
adding those which stand under each other, we have
m'dx'+m"d²x" m'd²y′+m'd³y'
:
dt
2
= 0,
<= 0;
dt2
integrating these, we have
dx'
dx"
ní
+ m"
A, m'
dy
+ m"
dy"
= B.
dt
dt
dt
dt
"
If x and y be the co-ordinates of the centre of gravity of m',
m", we have by the formula for the centre of gravity
m'x' + m″x
m'y' +m"y"
x =
m'
"
, y
"
m² + m
m² + m²
dx
A
du
B
Hence,
=
dt
'm' + m²
dt
m² + m
dx dy
But
manifestly represent the resolved parts of the velocity
dt
dt
69
of the centre of gravity, in directions parallel to the co-ordinates
x and y, respectively; and it here appears, that these parts are
Hence, the motion of the centre of gravity is uniform
in these directions, and consequently, uniform and rectilinear in its
constant.
own.
If A=0, and B=0, the centre is at rest.
This proposition was first stated by Newton; and, as we shall
see afterwards, is true of any number of bodies*.
25. PROP. The motion of each body about the centre of
gravity, is the same as if that point was a centre of force, the law
of which was the same as the law of the attraction of the bodies t.
Let us suppose
x' =x+x₁ y' =ÿ+Y₁,
x"=x+x2 y″ =y+Y2 ;
so that x1, Y1, x2, y2, indicate the co-ordinates of the points P, Q,
measured from the centre of gravity.
Then since, as appears above,
d²r
Ꮖ
= 0,
dt2
d²y
d t
= 0,
d²x²
Ꮖ
we have
ď² x 1
&c.;
dt
and equations (2) become
dt
λ y z
2
d² x 2
X2 — X 1
тф
·mp (r)
dt²
ተ
dť²
2
·m p (r) Y 2 — Y₁
Y½—Y1;
but it follows from the property of the centre of gravity, that if we
make GQ=r2, we have r =
m'+m"
m'
Te
X2-X1 PO GK
X2
.
-
Also,
P Q
GQ
12
↑
Principia, Cor. 4. to the Laws of Motion.
+ Ibid. Book I, Prop. 61.
70
Hence, our equations become,
for the body Q,
'm' + m"
X2
d²
ď x 2
m' φ
Te
2.)
m'
12
dť
d² y
ď²y 2
m'q (m
'm' + m'
m"
Y 2
Te
m'
r2
dt2
Now these are the equations we shall have, if we suppose the
centre of gravity a fixed point, to which a force tends, represented
m"
m² + m²
by m'p (m² -
m'
T2 2); therefore the motion of the body Q about
the centre of gravity, will be the same as if such a force resided in
that point.
It might in the same manner be shewn, that the body P will
move about G as if there were in G a force = m"p
'm' + m²
my
:).
And it is evident, that if (r) represent any power of r,
'm'+m"
m
Φ
Hence, the law of the force about G will be the same as that of
the attraction to P or Q supposed fixed.
) will vary according to the same power of r¸.
COR. 1. The angular velocities of P and Q round G will
always be equal, and PG, QG will always be in a given ratio.
Hence, the figures described by P, Q are every way similar*.
COR. 2. The velocities of P and Q relatively to G, will always
be parallel, in opposite directions, and proportional to GP and
GQ.
PROB. I. Let the force vary inversely as the square of the
distance, and P, Q hàve no angular motion originally : it is required
to determine their motions+.
It is manifest, that since P has no angular motion round G, it
will descend in a straight line to G. Similarly, Q will descend to
G in a straight line. And QG, PG, will be described in equal
*Principia, Book I, Prop. 57.
+ Ibid. Book. I, Prop. 62.
71
times, so that the bodies will meet in G. For, since the accelerating
forces on P and Q are inversely as P and Q, that is, directly as GP
and GQ, the velocities will be in the same proportion, and cor-
responding portions of GP, GQ, will be described in equal times,
so that the whole will be described in the same time.
Hence,
also, after these bodies meet, they will go on together with the same
velocity and direction with which the centre of gravity moved before
they met.
By last Article, since (7) is here r-2, the body P will move
towards G as if there were in G a force =m"
'm' + 'm'
m
"
"
:)
113
772
(m' + m")"
1
2
Hence, if a be the original distance of P from G, and P be
supposed to have no original velocity towards G, we have, by
Chap. I, Ex. 3, time of P falling to G
a²
2 m
#13
(m' + m″)²
π (m' + m″) a
m""
π
2√2
Similarly, if b be the original distance of Q from G,
time of Q falling to G =
(m² + m") b³
π
m'
2 Vo
a
b
And since =
these are equal, agreeably to what has just
m
ท
been said.
In the same manner, by Chap. I, Ex. 3, we might find the
velocity at any point, and the time of falling through any portion of
the distance.
PROB. II. Let the force vary inversely as the square of the
distance, and let the bodies P, Q have any velocities whatever
originally; it is required to determine their motions*.
*Principia, Book I, Prop. 63.
72
It has already been proved, (Art. 24.) that the centre of gravity
G will move uniformly in a right line; and that (Art. 25.) P and Q
will describe about G similar figures; P moving as if it were acted
on by a force
force
m':
13
#13
m
1
(m' + m")*° r₁
2
1
(m' + m″)² ' r₂
29
and Q as if it were acted on by a
placed in G. Hence, the curves described
about G by P and Q will be similar ellipses, with G in the focus;
and if we knew the original velocity of P and Q about G, we might
determine the ellipse, as in Chap. III, Prob. III.
The velocities of P and Q at any moment, and consequently at
the beginning, will be compounded of two velocities; viz., that
which the whole system has in consequence of the motion of the
centre of gravity, and the velocity of each point P and Q about this
centre. Now these last velocities are, by what has already been
said, (Art. 25, Cor. 2.) parallel to each other, in opposite direc-
tions, and proportional to GP, GQ, or to m" and m'. The whole
original velocities being known, we may thus find the separate parts
of them.
Let P and Q have the original velocities p and q, making with
QP angles
= a, ẞ, respectively. Let the velocity of the centre G
resulting from them be c, and y the angle which this velocity makes
with GP: let also v and v" be the velocities of P and Q about G,
and let be the angle which this motion makes with GP or SQ;
which will be the same for both bodies.
;
Now the component of the velocity of P resolved parallel to PQ
is
p cos. a: but since p is compounded of the motion c of the
centre of gravity, and v' about the centre of gravity, its component
will be c cos. y+v' cos. p. And thus equating the expressions
for the components of the velocities of P and Q, parallel and per-
pendicular to PQ, we have
=c cos. y—v″ cos. &;
p cos. a=c cos. y+v′ cos. P; q cos. ß=c c
p sin. a=c sin. y+v'sim. Q; q sin. ß=c sin. y—v″ sin. 9.
Multiply the two upper equations by m' and m", and add them,
observing that m'v'=m"v";
.. (m'p, cos. a+m'q cos. B) = (m+m") c. cos. y.
F
73
In the same way, the two lower equations give us
m'p sin. a+m"q sin. ẞ=(m'+m") c. sin. y.
By adding the squares of these two equations, we have
2
m² ² p² +m" ² -
q²+2m'm" pq cos. (a − ẞ) = (m' +m")°c².
√ {m²² p² +m"²q²+2m'm"pq cos. (a — ß)}
whence, c =
m'+m"
—
By taking the quotient of the same two equations, we have
m'p sin. a+m'y siu. B
α
Again, subtracting the upper equations, and also the lower, we get
tan. y =
m'p cos. a+m"q cos. B
p cos. a-q cos. B=(v+v") cos. =v'
α
p sin.
ß=(v'
$=v′
m'+m"
m"
m' + m²
cos. ;
sin. P.
a− q sin. ß=(v′ + v″) sin. $=v′
Adding the squares of these equations, we have
2
p² + q² - 2pq cos. (a — ß) = v
· (a− v'² (
whence v =
"
m"
m² + m²
m'
m' + m'
//
m
'm'+m'
m"
)
2
:
√ {p²+q² −2pq cos. (a−ẞ)};
√ {p²+q² - 2pq cos. (a-B)};
and dividing the same two equations
tan.
-
p sin. a q sin. B
p cos. a q cos. B
Hence, we know the velocity and direction of projection of P
round G, and we can therefore, by Chap. III, Prob. 3, find the
conic section described. And, combining the motion in this, with
the motion of the centre of gravity, which we have also found, we
have the motion of P.
COR. 1.
By Art. 25, Cor. 1, it appears that the curve de-
scribed by P relatively to Q, will be similar to the curve which
P describes about G. If a, be the semi-axis of the ellipse which
P describes round G, and a the semi-axis of the ellipse which P
describes relatively to Q, which is also in motion; we shall have
ar: a
: a :: m" : m² + m".
K
74
1
a force =
2.
COR. 2. Since an ellipse with semi-axis = a,, is described by
113
m"3
(m² + m²) 2 ⋅ r₁
we shall have the periodic time T by
113
Chap. III, Prob. 4,
Cor.; putting a, for a, and
m
(m' + m')²
for m;
2 a
π
2 π α, * (m' +m")
2πα
..T=
-
113
m
m"
11 3/2
(m' +m") ½
(m' +m")²
2
by last corollary.
COR. 3. If the body Q were at rest, and P were to revolve
about it, at the same distance from it as in last Cor.; the ellipse
would have its semi-axis majora, and we should have for the
periodic time T',
T' =
Ωπα 3
m'2
Hence, T T' :: m'" : (m' + m″)} *.
COR. 4. If P were to revolve round Q at rest, in an ellipse
of which the semi-axis major was A, we should have for the
periodic time T",
T"
2 π A ‡
ΩΠΑ
m" ½
And we may find A, such that T", about Q at rest, may be
equal to T, about Q in motion. For this purpose,
ΩΠΑ
m" }
A*
Σπα
(m' +m″) ½
:'. a : A :: (m'+m″)* : m″³ †.
SECT. II. Problem of three or more Bodies.
26. If we suppose three bodies to act upon each other, we
shall no longer be able generally to find the paths described as in
the former case. Any two of them would describe regular orbits
as in the preceding Section, but these will be changed by the action
Principia, Book I, Prop. 59.
+ Principia, Book I, Prop. 60.
75
of the third; and, in consequence of this change, an orbit will be
described completely different in kind from the former one. In
particular cases, however, the third body will only slightly alter the
regular orbit described by the two others, and in these instances,
this slight deviation from the regular orbit may be approximated to
by particular methods. As this is the problem which nature actually
presents to us in the case of the Earth, Moon, and Sun; the great
importance, joined to the great difficulties of it, have made the
problem of three bodies very celebrated; and since it was first
suggested, it has employed a large portion of the attention of the
best mathematicians, down to the present day. It does not, how-
ever, fall within the plan of the present Treatise: the student will
find the different steps of the solution in Professor Woodhouse's
Physical Astronomy.
27. PROP. If any number of bodies be acted on only by
their mutual attraction, their centre of gravity will either be at rest,
or will move uniformly in a straight line.
Let P, Q, R, &c. be any number of bodies distributed in
space, of which the masses are m', m", m", &c. And let them be
referred to co-ordinates parallel to three rectangular axes, viz.
x', y', z'; x″, y″, z″; x", y'", z z″, &c.
Also, let 1,2 be the distance of m' and m'
m",
71, 3••
T2, 3°
&c.
...of m' and m
"
..of m" and m
///
And let the law of attraction between P and Q be p (r,, 2), between
P and R, (1, 3); between Q and R, † (r2,3), &c.
Hence, since each body attracts all the rest, by resolving the
forces, and applying equations (c'), we have
d² x'
x′ — x″
x' - x'"
m" ("'1, 2)
+m" (1193)
+ &c.
dt
dex"
1'1, 2
x' - x'
T193
"
m' (1'1, 2)
dt²
dex""
dt²
m' p (r1, 3)
7193
7192
x' - x""
+m" (1293)
m" (r2, 3)
で
​- X
1293
x
7293
+ &c.
+ &c.
x"
&c. = &c.
76
Multiply the first equation by m', the second by m", the third
by m", &c. and add; and we have
m' d²x' + m" d²x" +m" d² x"" + &c.
1
d t²
Similarly, we find
= 0.
m'd²y' +m" d³y" +m"d³y" +&c.
dt2
m' d² z +m" d² z″ + m" d² z" + &c.
Integrating, we have
וי
dt2
dx"
+ m² + m
dt
= 0;
= 0.
dx""
+ &c. = A,
:
dt
dy"
dx'
m'
dt
m'
dy
dy"
+ m
dt
dt
dz
dz"
dz"
m'
+ m
+ m
+ &c. = C.
dt
dt
dt
+m"" + &c. = B,
dt
Now, if x, y, z, be the co-ordinates of the centre of gravity,
we have
m' x' + m"x" +m""x" + &c.
30
m' + m'"+m"" + &c.
Y
m'y' +m"y"+m"y"+&c.
///
m² + m" + m" + &c.
m' z' +m" z″ +m" z"" +&c.
"
m' + m'' +m" +&c.
Hence, if for the sake of abbreviation we make m'+m" +m"
+ &c. = M, we shall have
dx A dy
B dz C
dt M' dt M' dt M'
and therefore the resolved parts of the velocity of the centre of
gravity are uniform; and hence, this centre moves with a uniform
velocity, and in a straight line.
77
28. PROP. Let any number of bodies, whose magnitudes are
m', m", m" &c., act upon each other with forces which are directly
as the distances: it is required to determine their motions*.
The distance of any two bodies m', m', being r, the force of
m" on m' may be represented by m'r, and the part of it parallel to
Ꮖ X
x'
x, will be m'r
7'
= m″ (x′ — x"); similarly, the force of m'
m"
on
m', parallel to x, will be m" (x − x"); and similarly for the other
bodies, and also for the other co-ordinates y, z. Hence, by (c'),
ď² x'
dt
m" (x'′ — x″) + m" (x′ − x″) + &c.
d²x"
m (x' — x″)+m"″ (x" − x″)+ &c.
dt²
d2x"
m' (x' — x″)
dt2
m” (x" — x″") + &c.
&c.
&c.
d²x'
dt²
which may be put in this form,
= (m' +m" +m" + &c.) x' — m'x' — m"x" —m"x", &c.
dx
=
(m'+m"+m”"+&c.) x" — m'x' — m'"'x'
-M X
&c.
dt2
dx"
dt2
(m'+m"+m" +&c.) x″ — m'x' — m'x'
Ꮖ
m x
&c.
>
&c. &c.
or, (observing that if we make m+m'+m" + &c. = M and x, y, z,
the co-ordinates of the centre of gravity, we have
m'x' + m″x" +m"x""+&c. = Mx,)
d²x
dt2
= M (x − x) ;
d²x"
1
dt2
= M (x" − x);
Principia, Book I, Prop. 64.
78
!
d² x"
dt2
= M (x""-x),
&c. &c.
Similarly, we should have
d² y
dť
= M (ý – y), &c.
d² z
= M (z — z), &c.
dt2
Now '-x, &c. are the co-ordinates of m', &c. measured from the
centre of gravity; and it has already been seen that
d² (x' — x)
dt2
d² x'
&c.:
27
dt2
hence it appears by comparing these equations with Chap. II,
Ex. 2, that the motion about the centre of gravity is the same as if
there were no force but one residing in the centre of gravity, and
equal to M× distance. Hence, the bodies will all describe ellipses
about the centre of gravity, as a centre. And the periodic times in
these ellipses will all be the same. Their magnitude, eccentricity,
the positions of the planes of the orbits, and of the major axes,
may differ in any manner.
Also, the motion of any one body relative to any other, will be
governed by the same laws as the motion of a body relative to a
centre of force, varying as the distance; for if we take the equations
d² x'
d² x"
= M' (x' — x),
= M' (x" — x),
dť²
dt2
and subtract them, we have
d² (x' — x")
= M (x′ — x″);
dt²
and similarly for the ys and zs, from which it appears that the
motion of m' about m" is of the kind described.
CHAP. V.
THE CONSTRAINED MOTION OF A POINT ON A GIVEN
LINE OR SURFACE.
29. Ir we suppose a body to slide along a surface, as the in-
side of a bowl which is perfectly smooth, it is evident that we
cannot apply immediately the reasonings of the preceding Chapters;
for by the impenetrability of the surface, the body is perpetually
deflected from the path in which it would move in consequence of
the action of the forces alone. Since the surface is supposed
perfectly smooth, the force which it exerts upon the body must be,
at each point, perpendicular to it. For there is no reason why
the direction of the action of the surface should be inclined on
one side rather than the other, to this perpendicular; if any lateral
force do exist, it is attributed to the defect from absolute smooth-
ness, and called friction.
This perpendicular force exerted by the surface, varies per-
petually during the motion of the body: it is always such as exactly
to keep the body in the given surface, that is, to resist the tendency
to move through the surface. Now, if we were to suppose the
body to move freely, and a force of the same magnitude and direc-
tion as this reaction, but not arising from the contact of the body
with a surface, always to act upon it, its motion would be exactly
the same.
But in this case, we may calculate the motion by the
formulæ of the preceding Chapters, introducing among the forces
this new one, and making it such as always to keep the body in the
surface.
30. PROP. The reaction of the surface does not increase or
diminish the velocity of the body.
This will be seen when we come to deduce the formula for the
motion. But it appears to be nearly evident from this consideration;
80
that the force of reaction is entirely employed in deflecting the body.
If it were not perpendicular to the direction of the motion, we
might resolve it into two, one perpendicular to this direction which
deflects the body from its path; and one in the direction of the
motion, which alone would affect the velocity.
We may, as the simplest case, consider the body to move on a
curve line, lying in one plane. This will happen when the original
motion and the forces are all in one plane, and that plane also every
where perpendicular to the given surface. For then the reaction
will be in the same plane, and there will be nothing to deflect the
body from it. It will therefore be sufficient to consider the motion
ason a plane curve. The next simplest case will be when the
surface is one of revolution, and all the forces act in planes passing
through the axis. We shall afterwards consider any surface what-
ever, with any forces.
Instead of supposing the body retained by the reaction of an
impenetrable surface, we may suppose any other means. For in-
stance, if a body be fastened by an inextensible string to a given
point, it can move in the surface of a sphere: and the conditions
of its motion will be the same as if it moved on a smooth spherical
surface: the tension of the string supplying the place of the reaction
of the surface. And by supposing the string during its motion to
wrap round other curves and surfaces, the surface to which the
body is confined may become any whatever.
We may also, instead of supposing the body to move on a curve,
conceive it to move in a curvilinear tube, indefinitely narrow, the
body being considered as a point. The difference between this case
and the former one, will be, that in this, the reaction can operate in
any direction, whereas before it could only act on one side of the
body. So that in the tube, if the reaction were to become first
nothing and then negative, (or opposite to its former side) the body
would still be retained; while, in the other case, it would, on this
supposition, fly off from the curve and describe a path in free space.
In this manner also we may suppose a body to be compelled to
move in a curve of double curvature.
81
་
SECT. I.
The Motion of a Point on a plane Curve.
31. PROP. When a body moves on a curve, acted on by given
forces, to determine its velocity.
Let a body move on the curve PA, fig. 29, referred to the co-
ordinates AM, MP, which are represented by x, y. And let the
forces which act upon it be resolved into X, Y, parallel, respectively,
to these co-ordinates. Besides these forces we have to consider
the reaction of the curve, which is in the direction PK, perpendi-
cular to the curve, and which being represented by PK, may be re-
solved into PL, PH. If we call the reaction R, we shall have the
resolved parts in PL, and in PH parallel to AM,
PL
Ꭱ . and R.
PK
PH
PK' respectively;
or, (since the triangles PLK and TMP are manifestly similar, PT
being a tangent),
Ꭱ .
MT
PT'
and R.
MP
PT
dx
; that is, R
ds
and R dy
d s
supposing ds = √(dx²+dy³) = the differential of the curve AP.
Hence, collecting the whole forces in the directions PH and
MP, we have, by equations (c),
ď x
dy
= X + R
d t²
ds
d²
ਨੰ: ਪ
dr
=Y Y - R
d t
d s
Now, to eliminate R, multiply by 2 dr and 2 dy, respectively,
and add; and we have
d t°
2 d x d² x + 2 dy d² y
= 9Xdr+.Ydu,
and
2f(Xdx+Ydy).
dx² + dyⓇ
d t²
This expression is the same as when the body moves freely.
Hence, it appears that when a body, acted on by given forces,
moves from one given point to another, as from B to P, the velocity
L
1
ๆ
82
is the same, whatever be the path it takes, and whether it moves freely
or be constrained to move along a given curve.
If we suppose the body to be acted on by a force in parallel
lines, we may suppose the axis of r to be parallel to these lines;
and we have then Y=0.
If the force be also supposed to be constant, as for instance,
gravity, and to be measured upwards, we have X-g; and
2fXdx=C-2gx. Or, if we put C=2gh, h being an arbitrary
quantity, we have
velocity2
=
dx² + dy²
dt²
= 2g (h-x).
Here, when x=h, the velocity is =0; therefore h is the height
from which the body begins to fall. Also since the velocity depends
on x alone, it appears that it is the same, whether the body fall down
the perpendicular DM, or down any curve BP of the same vertical
height.
If we suppose the force to tend to a centre, and to vary as some
function of the distance from it; the centre of force may be made the
origin of co-ordinates A, fig. 30. Let AP=r, and the force in
PA=P, a function of r. Hence, we have
X = − P ², Y= - P²/,
-
7'
S(Xdx+ Ydy) = -fP
x d x + y d y
-fPdr;
r
dx² + dy²
dt²
C-2fP dr.
Or if _PAM = v,
dr² + r² dv²
2
velocity2
dt2
=C-2/Pdr.
Since the velocity depends on r alone, it is the same whether
the body fall down the curve BP, or down the line BQ (making
AQ = AP) acted on by the same force. And if the velocity in
:
;
!
83
the curve and in the straight line AB, be equal at any correspond-
ing equal distances from the centre, they will be equal at any other
equal distances*.
32. Having found the velocity in terms of the co-ordinates,
or of the radius vector: and knowing moreover the nature of the
curve, we can find the time of the motion, as will be seen in the
following examples.
ds
Since dt=
when the force acts parallel to x,
we have
v
ds
dt =
√ {2g (h− x)}
ds
And when the force acts to a centre, dt =
=
√(C – 2ƒPdr)
PROB. I. Let AP, fig. 29, be a cycloid with its axis vertical:
to determine the motion upon it when the body is acted on by
gravity.
We have here dy=dx/2a-; (Lacroix, Art. 102.), where
T
a is the radius of the generating circle;
2 a
.. ds=dx
x
ds
dt=
Va
dx
√2g √(h− x)
g
√ (hx — x²) '
a
..t=C
C-
√². are (ver. sin. = 25)
g
g
{
П
•
h
arc (ver. sin. = 2)};
for when x=h, t=0, and arc≈π.
h
And for the whole time of descent to A, t=π
Principia, Book I, Prop. 40.
zion
84
COR. 1. Hence, the time of descent is the same whatever be
the arc BA.
COR. 2. If the cycloid be completed, fig. 31, the body, after
descending from B to A, will ascend by the velocity acquired, to b,
in the same horizontal line with B: for the height up which the
body must ascend to lose the velocity, will be the same as that down
which it descended to gain it. And the time up A b, will equal
the time down BA.
When the body comes to b, it will have lost all its velocity: it
will then descend to A, and rise again to B, and so go on oscillating
for ever on the supposition that the surface is perfectly smooth.
COR. 3. By Lacroix, Elem. Treat. Art. 103, the evolute to
the cycloid EPA e, consists of two semi-cycloids, CE, Ce. Hence,
if a body be suspended by a string of proper length, which wraps
round the curves CE, Ce, and oscillates, the conditions of the
motion of the point P will be the same as those of a body upon a
cycloidal surface, just investigated.
Ifl be the length of the pendulum curve EOC;
1=4a, and time in BA
π
مة
.. time of an oscillation from B to b = π
g
COR. 4. The time which a body would employ in falling down
the vertical length l is
1.
.. time of oscillation: time down pendulum
The semi-cubical parabola, with its axis vertical, is another case
in which we can integrate the expression for the time.
PROB. II. Let AQ, fig. 31, be a circle whose radius is c: the
body being acted on by gravity.
y= √(2cx − x²); ds=
ds
c d x
dx
√(2cx − x²)
dt
2g. V (hV −x)
C
−
√2 g´´√ {(h—x) (2cx — x²) } '
Veg
A
85
C
dr
√2g' √ {(hx − x²) (2c—x)}'
e might integrate by expanding (2 c - x). But we may do
We
it better thus.
Let arc
dr
X
(sin.
in. =
; ...de=
2 √ (h x − x²)°
Also since
√
18
sin. 0, x = h sin.20, 2c-x=2c-h sin.* 0
x=
=2c (1-d² sin.² ): putting d² =
dt
C
√(4gc)
h
Hence,
2 de
20
√ (4gc)˚ √(1 − ♪² sin.²
d Ꮎ
g√(1-d² sin.20)
Ꮎ)
The integral of this is an elliptic transcendent, and may be
found by the methods of approximation, or the tables, given by
Legendre for such functions. The integral must be taken from
x=h, that is, from 0
π
- and for the whole arc 4P, it must be
2
taken to x=0, and therefore ◊ =0.
If the oscillations be small, h is small; and therefore d.
this case, we may approximate by expanding.
In
We have thus
d Ꮎ
S
1.3
S√(1—8° sin.²0)
=ƒdo {1+
sin.20 +
S4 sin.40+&c.}.
2
2.4
Now, to find sin." Ode, where m is an even number, we have
(Lacroix, Elem. Treat. 205.)
1
ƒ de sin."0=
m
1
cos. sin.m- Ө +
1
fde sin.m-20;
m
m
and taking the integral from 0 to 0 =
and
π
Ela
the first term vanishes;
86
· 1
=== fde sin.m-20
də
m
S de sin.™ 0 =
m
Similarly, do sin.m²- 20 0 =
=0
π
m
3
(e=0)
-40
π
m
e=
2do sin.m-4
São
and so on to ƒ'de sin.² 0 =¦ƒ ao = 1. T
and
Hence, do sin." 0 =
√(1
Ꮷ Ꮎ
d Ꮎ
S² sin.
d0 2 2
(m − 1) (m − 3)…………..1 π
m. (m 2).
·
from 0=0 to 0
Ø)'
from 0
Svo
S-√(1-d'
√ : {¹ + () +
t =
√ (1 − d² sin.³0)'
1
1
2014
;
2
П
is the same as
2
to 0=0. Hence, we have
2
2
1.3.5.83.
П
(1.3.d²)² + (1.5.5.0 ) + &c.}. T.
2.4
2.4.6
If we neglect all the terms after the first, we have t =π
2
4g
which is the time of descent down a cycloidal arc of which the radius
C
is Hence, if AF AC be the axis of a cycloid AP, the
4
times in PA, and QA are equal. In fact, C is the centre of
curvature of both, and they may, for small arcs, be supposed to
coincide.
If we take two terms, we have, since d' =
h
2c'
π
t
2
{1+
π
4
{1+
But if AB, the chord of the whole descent=k, we have h=
k-
2 c
π
k2
:.
.. t =
+
16 c
2
87
Hence, if a pendulum oscillate on each side of the vertical
1
through an arc, the chord of which is of the length,
10
kc
1
C
10
1
and the oscillation is longer by
of the whole.
1600
PROB. III. Let the curve be the hypocycloid, and the force
tend to the centre of the globe, and be as the distance.
A hypocycloid is a curve, APD, fig. 32, generated by a point P,
in the circumference of a circle GPe, which rolls along the inside
of the circumference EFD of another circle. The circle EFD is
called a circle of the globe, and the rolling circle FPG the wheel
or generating circle.
Suppose, that when the describing point P was at A, the
diameter Pe of the generating circle was in the position EA, co-
inciding with CE. And suppose that when A comes to P, E comes
to e, so that FE = Fe. Now when the point is at P, the circle
is turning on the point F; hence, the motion of P at that point
will be perpendicular to FP; and hence, a tangent at P will be at
right angles to FP, and will therefore pass through the point G.
Let CY be a perpendicular on PG. And let CF-a, OF=b;
CPr, CY = p.
PY
GY2
Now by similar triangles,
CF
CG
or
-
— —
r² = p² - (a — 2b)² = p² — e² = p²;
a
putting ea- 2b = CA.
(a ~ 2b)²
e² (a² — 7.²)
Hence, p²
a² - e²
(See Mr. Peacock's Collection of Examples, p. 195.)
Also, ² - p² =
a² (1.² - e²)
a²
88
Now, we have in spirals, ds =
√(a² — e²) r dr
a V (r² — e²)
-
rdr
2
√(n² — p²)
-
in this case.
mr2
And, Pmr; f Pdr =
; C− 2ƒ Pdr=m (h² — r²),
2
making C=mh; where h is the height CB, when the body
begins to move.
√(a² - e²)
rdr
Hence, dt =
a V m
√ {(n² – e²) (h² — p²)}
rdr
2
to integrate this, let (r²-e²) = u; :.
√ (r² — e²)
h² — r² = h² — e² — u²,
du;
and
V (a²-e²)
dt =
a V m
du
√ (h² — e² — u³) ³
√ (a² — e²)
t:
t =
arc
a v m
(co
น
√(r² — e²)
COS. =
=
√ (h² — e²)
√(h²-e²))
which is = 0, when r = h, as it should be.
Hence, the time of falling to A, found by making r=e, is
t =
√ (a² — e²). π
a v m
COR. 1. Since h does not appear in this result, the time of
descending to A is the same, whatever be the arc; that is, the
descents are isochronous*.
Since e-a-2b, a²-e²=4ab-4b2;
. . t = π
b
b2
2
✓ Com
am
m
*Principia, Book I, Prop. 51.
89
COR. 2. If am force at E,
be put = g, and if we suppose
b
a large compared with b, so that
a
may be rejected, we have
t = π
g
which coincides with the time in a common cycloid (see Prob. I.),
as it manifestly should; for if a be very large, ED will be nearly
a straight line; and the force in DA constant and parallel to EC.
COR. 3. Hence, also the oscillations in such a curve are iso-
chronous, and the time is found as above. The time of an oscillation
π V (a² — e²)
a v m
If we suppose the Earth spherical and homogeneous, the force,
in proceeding from the surface to the centre, varies as the distance
from the centre. Hence, if we were to suppose a body to move
on a hypocycloid, generated by the rolling of a circle on the interior
of the circumference of a circle within the Earth, and concentric.
with it, the descents and oscillations in such a curve would be
isochronous.
COR. 4. Instead of supposing the body to move on a curve,
we may suppose it to be suspended by a string, which, during its
oscillations, wraps round the curves SD, Sd, the evolutes to the
portions AD, Ad of the hypocycloid. The motions will then be
the same as before, and the oscillations of such a pendulum will
therefore be isochronous.
The evolutes SD, Sd are also hypocycloids*.
Let PO be the radius of curvature, CZ a perpendicular upon
it, CO=r', CZ = p',
(Lacroix, Elem. Treatise, Note H, p. 668.)
rdr
PO=
dp
e √ (a² — r²)
erdr
Ρ
:.dp
=
√ (a² — e²)
√ {(a* — e²) (a* — r²) } '
*Principia, Book I, Prop. 51.
M
90
PO=
√ {(a° — e³) (a² — 2.²)}
e
Also, 02=OP+CY=
CZ = √(»º − p²) =
V
—
a² V (a² — r²)
e V (a² — e²)'
2
a V (r² — e²)
√ (a² – e²) •
at (a² — r²)
Hence, squaring,
e² (a² — e²)
p¹² — p¹²,
12
a² (r² — e²)
p'².
(a² — e²)
Multiply the first by e², and the second by a², and add, and we have
a² = e² r²² — e² p¹² + a²p'²;
Let
p
a²
e
12
a² - e² 2'2
er
a² - e²
a'
a/2
that =====
= a so that
a
12
12
a
a
2
a
e²
2.12
γ
2 12
a² (a
a
1
12
a²
a
212)
2
which is the same as the equation which we had to the hypocycloid,
only putting a for e,
Hence, take CS
and a' for a.
a²
e
>
a third proportional to CA and CE;
and describing a circle with radius CE, suppose a wheel, whose
diameter is ES, to roll on this circle, so as to describe the hypo-
cycloids SD, S'd; these will be the evolutes of AD, Ad. And
a pendulum oscillating between them will always have its extremity
in the hypocycloid DAd.
The length of the hypocycloid is thus found*,
ds=
√(a² — e²).rdr
it V (r² — e²)
* Principia, Book I, Props. 48, 49.
91
.. s
√ {(a^—e*) (»² — e²)}, measuring from A.
a
And for the whole AD, s =
Similarly, SD=Sd=
a
a² — e²
for r=a.
12
24
a
a
a'
a4 - a² e²
a²e
a² - e²
COR. 5. If the length of the pendulum = 1,
whence, a² - e² = le =
la²
a
a² - e²
e
= 1,
e
And time of oscillation = π
√(a² - e³)
a Vm
=
πνι
√(ma')'
COR. 6. By Chap. I, Ex. 1, the time of falling to the centre
П
C will be
π Vď
•
2 Vm
Hence, time of an oscillation:
2 V (ma')
time to centre: Vl: Va: VSA: VSC*.
If the curve be an Epicycloid, and the body be acted upon by
a repulsive force, varying as the distance from the centre of the
globe, we shall obtain similar results: and the oscillations will be
isochronoust.
* Principia, Book I, Prop. 52.
+ If we resolve the force in these isochronous cases, so as to obtain the
portion which acts along the curve, we shall find, that this portion is as the
length of the curve from the lowest point; consequently, the force which
accelerates the body, is as the space to be described; and, therefore, the
time to the lowest point is independent of the original distance by Chap. I,
p. 14.
Conversely, in any curve the descents to a given point will be isochronous,
if the force be such, that the resolved part of it along the curve is as the arc
dx
from that point. If the force be P, and act parallel to r, P
will be the
ds
ལ
resolved part.
Ex.
92
33. PROP. When a body moves upon a curve, to find the
magnitude of the reaction R, which is also equal to the pressure
upon the curve, we resume the equations
ď²x
= X + R
dy
;
dt
ds
dy
dx
= Y — R
d t
ds
dť
2
Hence, R =
Multiplying the first by dy, and the second by dx, and subtracting,
we have
dy d²x-dx dy
=
Xdy-Ydx+Rds.
Xdy - Ydx dyd²x - dxd y
2
+
ds
2
d t² ds
ds³
3
If
P
be the radius of curvature, we have
P
-
dy d² x — dx ď³y
hence, R =
+
ds
pdť
Ydr - Xdy ds²
Of the two parts of which this expression consists, the first is
what we obtain by supposing X and Y resolved perpendicular to a
tangent to the curve. The second,
force which would retain the
a circle whose radius is
p.
d s²
pdť
or
129
velocity2
P
و
is the
body in the curve if it were moving in
This latter is called the centrifugal
force, and arises entirely from the tendency of the body to go in a
Ex. To find the force which must act in parallel lines, that the descents
in a circle may be isochronous.
If c be the radius, and 0 the angle which the radius CQ, fig. 31, makes
with the vertical, x will be = c ver. sin. 0, and s = c0; ..
dx
ds
sin. 0.
0
Hence, P sin. 0 must vary as ce, and P must vary as
the circle may be isochronous.
in order that
sin.
'
This problem is in the Principia, Book I, Prop. 53.
93
straight line, instead of the curvilinear path in which it is com-
pelled to move it is greater as the curvature is greater.
If the body be acted on by no force, R =
ds2
pdt2
velocity*
p.
In the case where the body is acted on by gravity only,
R =
ds
gdy ds²
+ 2
pdť²
In the case where the force P tends to a centre at the origin of
co-ordinates,
xdy-y d x
Ydx
Xdy
= P
= Prdv;
Т
for xdy - ydx = r² dv, (see Art. 16.).
2
R =
Prdv ds²
+
ds
pdť
PROB. IV. A body oscillates in a cycloid acted on by gravity:
to find the tension of the string. Fig. 31.
dy
g.
R = g⋅ ds
ds
+ 29
pdt2
x
we have dy=dr√/2a-7, ds=dra, dy = 2a-x
x
Ꮖ
dx
p = 2 √ {2a (2 a− x)}, (Lacroix, Art. 103),
Hence, R = g√2a
2 a
x
+
2a+h-2x
√(4a² - 2 ax)
x ds
ds²
dt
g (h-x)
√{2a. (2a-x)}
2 a
= 2g (h — x).
=g.
2 a
h
√(2a − h)
When x = h, R = g
go
√(4a² — 2ah)
√(2a)
2a+h
h
When x = 0, R= g
θα
= g (1 +
;
94
the part g arises from gravity, the other part
force.
g h
from centrifugal
2 a
If h=2a, or the body fall from the highest point; pressure at
A=2g.
PROB. V. A body oscillates in a circular arc acted on by
gravity: to find the tension of the string.
(c-x) dx
√(2cx − x²)'
Ꮖ
dy C-X
ds=
cdx
√(2cx − x²)³ ds
= 2g (h− x);
=
C
dy =
ds²
2
p = c, d t²
dt²
c-x
. . R = g ·
2g (h-x)
c+2h-3x
+
g.
•
C
C
C
c+2h
When x = 0, this gives R = g
If hc, or the body
C
fall through the whole quadrant, R = 3g; and the tension at the
lowest point is three times the weight.
If the body fall through the whole semi-circle from the highest
point of the circle, h=2c, R=5g, and the tension at the lowest
point is five times the weight.
This tension, by the increase of x, may become 0, and after-
wards negative, or opposite to its former direction: the body will
then tend to approach the centre of suspension.
It will=0, when c + 2h - 3 x = 0, or x
=
c + 2 h
3
If a body move on the upper convex surface of a circle, it will
only remain upon it while its pressure is towards the centre. It
will fly off and describe a parabola, when the pressure becomes = 0;
c + q h
that is, when x=
; x and h being measured from the lowest
3
point of the circle.
95
The parabola which the body describes, will be one which has a
common tangent with the circle at the point where the body leaves it,
and the velocity at that point = √ {2g. (h− x)} = √ {2g.
h-
3
Sincer is less than 2 c, we have c + 2h < 6c, or h< C
If h be greater than this, the body will not move along the circle
at all, but will leave it at the highest point.
By a similar application of our formulæ we might easily find
the tension, when a body oscillates in a hypocycloid.
SECT. II. The Motion of a Body on a Surface of Revo-
lution.
34. PROP. To find the motion of a body upon a surface of revo-
lution, and acted on by forces in a plane passing through the axis.
Let CP, fig. 33, be the curve by the revolution of which the
surface is described, AC its axis, and let AMN, be a fixed plane
perpendicular to this axis, AM, MN, NP three rectangular co-
ordinates to the point P; represented by x, y, z, respectively. Also
let OLP be a plane parallel to AMN; and let OP = r. Then
since AO = NP = z, and OP ≈r, knowing the nature of the curve
CP, we know the relation of r and z.
The forces which act upon the body when at P are the reaction
and the extraneous force: let this latter, which is by supposition in
the plane POA, be resolved into two, P in the direction PO, and
Z in the direction PN. Also the reaction is manifestly perpen-
dicular to the curve CP, and in the plane AOP; and if we resolve
it, as in the case of a plane curve, we shall have its component
dz
in direction PO = R
ds
dr
in direction NP = R.
supposing ds = √(dz² + dr²).
;
ds
dz
Hence, the whole force in PO = P + R
and if we resolve
ds
96
this parallel to x and y, we shall have, since OL, LP, OP are
x, y, r, respectively,
component in x =
(P + R
d z
X
R ds
r
in y =
(P + R
dz
Y
ds
r
dr
2+ R
ds
Also force in z =
Hence, by the equations (c') we have
d² x
dt2
ระบ
dt2
2
ď² z
dt²
Hence, we find
(P + R
(P + R
dz
dr
ds
)
dz)
ds
Z + R as '
ds
Ꮖ
y
.(1).
r
x ď² y − y d² x
= 0; or d
dt
(rdy-9dr)
= 0....(2),
dt
dx d² x + dy d² y + dz dz
dt2
Zdz
r
´dx² + dy² + dz
dy²+
dt2
x dx + y dy
which since
d.
(dx²
2
Now
d t²
2
dr²'dť²
dz
is known in terms of r.
dr
2
dz² dz² dr²
²
x d x + y dy
P
R.
ds
2'
(dz xdx + y dy drdz)
r
dr, becomes, multiplying by 2,
2 Pdr 2 Zdz....(3).
-
ds
and from the nature of the curve CP,
dz
Let =p, p representing the tangent
dr
of the angle which the surface at P makes with the horizon.
dr²
Therefore
d t²
Ρ
2
·
dr.2
dt²
2
?
97
Also let the angle LOP=v, and we shall have, (as in
Art. 16, 17,) xdy
ydx = r²dv;
dx² + dy² = dr² + r² dv².
Hence, the equations (2) and (3) become
2
r² dv
d.
= 0,
dt
(dre
γ
r² dv²
dr²
d.
+
di"
dt²
+ p²
- 2 Pdr – 2 Z dz.
d t²
If we integrate the first of these, we obtain
r² dv = hdt; h being a constant quantity,
or dt =
2
r² do
h
If
The second of the two equations just found, might be integrated,
if we could integrate the right-hand side - 2 Pdr 2 Zdz.
we put for P, Z, and z, their values in terms of r, this expression
will become a function of r, and its integral will be a function of r.
Let f (Pdr + Zdz) = Q: hence
dr² 2
r² dv²
+ +p².
dr²
dtⓇ
C — 2Q.
dt2 dt
2
To find the path of the body, put for dt its value
1.2 dv
h
and
we have
(1 + p²).
h² dr² h²
+
C − 2 Q ;
p4 dv
2
.. dv=
√(1 + p²). hdr
r V {(C − 2Q) 12
(e).
h²
Hence, dt =
(ƒ).
√(1 + p²).rdr
2
√ {(C − 2 Q) r² — k² }
If equation (e) be integrable, we have v in terms of r, whence
the locus of N is known. And r being known, we know z; for
dz = pdr.
If the force be always parallel to the axis, we have P=0;
and if also Z be a constant force, as gravity, and = g, Q = f Zdz
=g, and is to be expressed in terms of r.
N
98
If the force tend to a fixed point in the axis, we may make this
the origin A. Let AP, and the force in PA
fore PP'
2
2
ช
= P'; and Q =ƒ(Pdr + Zdz)
=ƒP'
γ
r dr + zdz
r
=SP'dr';
P'; there-
because r² + z² = r². And if P' be a function of r' we find Q
by integrating.
dr
There will be apsides when
= 0, and therefore when
dv
(C - 2Q) 1.2
h² = 0.
35. PROP. To find under what circumstances a body will de-
scribe a circle on a surface of revolution.
For this purpose it must always move in a plane perpendicular
to the axis of revolution; r and z will be constant; and as in
Art. 20, r cos. v=r; therefore
d² x
dť
r cos. v.dv²
d t2
;
rdv
d2 x
and if V be the velocity, V =
dt
dt
7'
172 cos. v
Hence, the first and third of equations (1) in last Article become
V2
dz
P+R
ds
r
dr
0 =
Z+ R
ds'
V2
dz
P + Z
..(g).
dr
If the force be gravity acting in the direction of the axis OA,
so that P=0, 2=g;
V2
dz
g
·
r
dr
Since ǹ = 2 area in time 1 = Vr, h² = gr³ dz
h
dr
gr³p.
PROB. VI. To determine the time of a pendulum performing
a conical revolution.
99
A body suspended by a string SP from a point S, fig. 34, will
be retained in a spherical surface whose centre is S, and may, by
properly adjusting the velocity, revolve in a circle, SP describing
a conical surface, which is the motion here spoken of.
In a spherical surface of which the radius is c,
z = AS -- √(c² — ‚¹²);
dz
=
dr
√(c² — r²)'
172
g r
p²
√ (c² — p²) ·
If we draw PT a tangent at P, OT=
እ?
√ (c² — p²)
V² = 2g. 1OT';
hence, the velocity of P is that acquired by falling down half OT.
The circle described by P has its circumference
we have the time of a revolution =
Ωπη
V
2π (c² - r²)
2πr; hence,
2TVSO
Vg
√ g
COR. 1. By Prob. I, Cor. 3, the time of a double oscillation
of a pendulum whose length is 1, would be 2 π
g
Hence, in
order that this oscillation may employ the same time as the revolution,
we must have = SO.
COR. 2. The last corollary is true for any surface, PS being
a normal.
COR. 3. If a body were to revolve in a parabolical surface,
the times of all circular revolutions would be the same. For in
this case, SO the subnormal is constant.
PROB. VII. A body moves in a spherical surface acted on by
gravity, and so as not to describe a circle, to determine its motion.
Let c be the radius of the sphere. We have Q=gz, and
C − 2 Q = 2 g (k − z), k being an arbitrary quantity.
au
Also r²=2c-22, z being measured from the surface:
2
2
rdr=(c−z) dz; 1+p²=1+
(c — =
(c — :) * °
Upp
100
Hence, by equation (f) dt =
dz
In order that
d t
√(1+p²)rdr
√ {(C − 2 Q) r² — h² }
cdz
V{2g(k-2)(2cz— z²) — h² } *
may= 0, the denominator of the right-hane
member of this equation must=0; that is
2g (k—z) (2 cz - 2º) — h²=0;
h²
3
or z³-(k+2c) z² + 2k cz -
0;
2 g
which equation will necessarily have two possible roots; because, as
the body moves, it will necessarily reach one highest and one lowes.
point, and, therefore two places when
has also a third possible root.
dz
dt
0. Hence, the equation
Suppose it to be identical with
(z − a) (z – B) (≈ − y) = 0:
where a is the greatest value of z, and ẞ the least, which occur
during the body's motion.
c d z
√(2 g). √ {(a — z) (≈ — B) (y − z)} ´
Hence, dt =
To integrate this, let 0=arc
(sin. = √ B);
G
α
d z
.. do
d Ꮎ -
ช Թ
a B.
Also sin² =
2 √ {(z — B) (a− ß)}. \/\/ {1
dz
2✓ {(a−z)(x − ẞ)}'
z - B
β
a - B; · · ≈ = ß + ( a − ß) sin² 0.
And y-z=y- {ß+(a−ß) sin. 0} = (y − ß) { 1 − d²sin.20};
if d = √ a - B
γ-β
MU
101
2 cd0

.. dt =
√ {2g(y-ß)} • √ { 1—8° sin.² 0}
to be integrated from zß to za; that is, from 0=0, to 0
this expanded in the same manner as in Prob. II, gives
2
1.3.82
2.4
2 c
t =
√28 (7-B)
{
1+
+
(1.8
1.3.583
2
π
+&c.
2
π
:
2
+(
2.4.6
which is the time of a whole oscillation from the least to the greatest
distance.
hdt
Also dv
2
hd t
2cz-z
and is hence known in terms of z.
36. PROP. A body acted on by gravity moves on a surface of
revolution, whose axis is vertical: when its path is nearly circular,
it is required to find the angle between the apsides of N's path,
fig. 33.
In this case, fd z = gz=Q.
And if at an apse r=a, z=k, we have
h²
(C — 2g k) a³ — h²=0; .. C=2+2gk.
Hence, equation (e) becomes
dv =
}}
Let = 1 + 1;
p
√(1+p²). hdr
a
z) r²
__
h². (a² — r²)
-
a²
"√ {2g(k − 2);
√(1+p²).
hdr
√ { 2 g (k − c ) → h² (= − 4 ) }
· 2) —
dr
// = dw,
10%
102
d v =
1
√(1+p²). hd w
(k
(k
z)
z) h²
( - )}*
4
√ {eg
(-2) -
h² G
(
— z) h²
dr² _ 2 g (k − 2) — 4"
dw2
d v²
h² (1 + p²)
1
2
1
- )
a
a
2
;
It is requisite to express the right-hand side of this equation in
terms of w.
Now since at an apse we have w≈0, z=k, r=a,
d z
we have generally z = k +
w +
dw
d² z
we
dw² 1 1.2
2
+ &c.
the values of the differential coefficients being taken for w=0.
And dz=pdr-pr²dw,
d² z=-2 pr dr dw-rdwdp; or, making dp = qdr
=-(2p+qr) r d r d w = (2p+qr) r³ dw².
And if P1 and 91 be the values which p and q assume when
w=0, and ra, we have for that case,
ď² z
dw²
z = k—p₁ a² w + (2 p₁+q₁ a) a³
Also = (+10)
a
a
a
1
Ωω
+
+w².
2
a
2 = (2 p₁ + q₁ a) a³,
•
w2
&c.
19
qi
Hence, 2g (k − 2) — h²
2
(3)
becomes
a
2
w²
2 g (p, a² w−(2p, +q, a) a ³ . 20 + &c.) - h² (22
- +
3
το
Q
a
w²
2012).
But when a body moves in a circle of radius=a, we have hª
=gr³p=ga³p, in this case, (Art. 35.). And when the body moves
nearly in a circle, h will have nearly this value. If we put h²=(1+d)
gap, we shall finally have to put =0, in order to get the
3
103
ultimate angle when the orbit becomes indefinitely near a circle.
Hence, we may put hg a p₁, and
2
3
2 g (k − z ) − h² ( ; — — — ) becomes {3ga³p₁+ga* q₁} w² + &c.
2
a
2
Ρι
in which the higher powers of w may be neglected in comparison
of w²;
d w²
3
ga³ {Sp₁+q₁a} w²
(3 p₁ + q₁a) w³
dv²
h² (1+p²)
P₁ (1+p³)
1
(3 p₁ + g₁ a) w²
P₁ (1 +- p₂²)
again omitting powers above w²: for p=p₁+Aw+ &c.
Differentiate and divide by 2d w, and we have
d² w
d v²
3 p₁ + q₁ a
P1
W=
ΤΟ
Nw, suppose.
១
P₁ (1 + p₁²)
Of which the integral, taken so that v=0 when w=0, is
w= C sin. v√✅ N.
で
​And w passes from 0 to its greatest value, and consequently r
passes from the value a, to another maximum or minimum, while
the arc v ✅N passes from 0 to π. Hence, for the angle A between
the apsides we have
AVN=T; A =
π
.. where N
3 p₁ + q₁ a
I
P₁ (1 + p₁²)
PROB. VIII. Let the surface be a sphere; and let the path
described be nearly a circle: to find the horizontal angle between
the apsides.
Supposing the origin to be at the lowest point of the surface,
we have
ď z
2'
dp
2
z = c − √(c² — r²); p
—
=
q
2
dr
dr
104
a
c²
2
2
.. Pi
√ (c² — a³)
91=
(c² – a²)
1+Pi
2
C
2 - a
C
2
.. N=
4 c² - 3a²
2
C
c²
Hence, the angle between the apsides =
пс
√(4 c² - 3a²)
The motion of a point on a spherical surface, is manifestly the
same as the motion of a simple pendulum or heavy body,,sus-
pended by an inextensible string from a fixed point; the body being
considered as a point, and the string without weight. If the pendulum
begin to move in a vertical plane, it will go on oscillating in the
same plane, in the manner already considered in Sect. I, of this
Chapter. But, if the pendulum have any lateral motion, it will
go on revolving about the lowest point, and generally alternately
approaching to it, and receding from it. By a proper adjustment
of the velocity and direction, it may describe, by Art. 35, a
circle; and if the velocity, when it is moving parallel to the horizon,
be nearly equal to the velocity in a circle, it will describe a curve
little differing from a circle. In this case, we can find the angle be-
tween the greatest and least distances, by the formula just deduced.
Пс
Since A
√ (4c — Sa³) ;
If a=0, A =
П
Է
—
; the apsides are 90° from each other, which
also appears by observing, that when the amplitude of the pen-
dulum's revolutions is very small, the force is nearly as the distance;
and the body describes ellipses nearly; of which the lowest point
is the centre.
If a = c, A = π = 180°; this is when the pendulum string is
horizontal; and requires an infinite velocity.
C
If a = 1/1;
so that the string is inclined 30° to the vertical;
Ωπ
A =
= 99º 50'.
√13
105
If a² =
ર
; so that the string is inclined 45° to the vertical;
A = π
==113º. 56′.
3 c²
If a = -
4
so that the string is inclined 60° to the vertical;
Ωπ
A
=136°, nearly.
PROB. IX. Let the surface be an inverted cone, with its axis
vertical: to find the horizontal angle between the apsides when the
orbit is nearly a circle.
Let a be the radius of the circle, and y the angle which the
slant side makes with the horizon.
Hence, z=r tan. Y, p=tan.y, q=0;
3 tan. Y
= cos.2
2
tan. y. sec.y
3 cos. y:
.. N=
π
and the angle between the apsides
=
cos. Y√√√3
2 T
If y=60°, angle =
= 120°.
حت
PROB. X.
parameter is c.
Let the surface be an inverted paraboloid whose
r² = c z ;
Р
d z
2 r
p =
q=
dr
C
2710
C
6 a
+
C
.. N=
If a = 2,
ordinate,
2
C
sa (1+
C
2 a
C
4 a
c²
4 c²
c²+4a²
or the body revolve at the extremity of the focal
N=2; angle between apsides =
π
106
PROB. XI. Let in fig. 35, OP vary inversely as CO§.
Let AC = h, h − z
C4
dz
3c4
12 c
z =
73; p =
=
dr
; 9
до в
2015
9c¹
12c4
4
4
:. N =
a
a
3c4
(1+
+ 9c²)
α
.8
a8
8
28
a³ + 9c8 ·
This is negative, and hence the angle is impossible, and the
body will never come to a second apse. If the velocity be at all less
than that in a circle, the body will go on descending continually
down the funnel PA.
37. PROP. When a body moves on a conical surface, acted on
by a force tending to the vertex; its motion in the surface will be
the same, as if the surface were unwrapped, and made plane, the
force remaining at the vertex.
Let A, fig. 33, be at the vertex, AP r', force P', and the
angle which the slant side makes with the base = y; •'. z = r
tan. Y, p = tan. y, 1 + p² = sec. y.
2
Also Q = f(Pdr + Zdz) = f P'dr'.
Hence, by equation (e), Art. 34,
sec. Y
.hdr
d v = rv{(C- of P'dr') r² — h²} '
or putting h' cos. y for h, dv'. sec. y for dv, and for r its equal
r' cos Y?
dv' =
h'dr'
12
r' V {(C — 2ƒ'P'dr') r'² — h²}
Now do is the differential of the angle described along the
eonical surface, and it appears that the relation between v' and '
will be the same as in a plane, where a body is acted upon by a
central force P'. For we have in that case, (see page 27,)
d
and integrating,
12
12
Sh² dr² h²)
!* dv
12
\ r 4 d v¹²
+
2 P'dr';
12
107
h² dr²
r'+dv
h12
+ = C - 2fP'dr',
12
which agrees with the equation just found.
38. PROP. When a body moves on a surface of revolution, to
find the reaction R.
Take the three original equations, (1) p. 96, and multiply them
respectively by rdz, ydz, and rdr; and the two first become
xd xdz
dt²
Pr²dz
dz² x²
R
r
ds
y d² y d z
Pydz
dz² y
R
dt
2
Ꭲ
ds ľ
Add these, observing that x² + y² = r², and we have
(xd²x + yd²y) dz
+yd'y)
dz²
dt²
Prdz Rr ;
ds
and the third is
rdrdz
dt²
dr²
Z r d r + R r
ds
Subtract this, observing that dz² + dr² = ds², and we have
(x d²x + yd³y) dz — rdrdz
dt2
=
r (Zdr Pdz) - Rrds.
But x² + y² = r², xdx + ydy = rår,
x ď² x + y d² y + dx²+ dy² = rd²r + dr³.
Hence the equation becomes
(dr² — dx² — dy³) dz¸rd zdr— rdr d² z
=r (Zdr- Pdz) - Rrds,
+
dt2
dt2
and dr² = ds² - dz².
Zdr - Pdz
drd²z - dzd²r
Hence, R=
+
ds
dt² ds
(dx² + dy² + dz°
ds²) dz
+
rdt²ds
Now if p be the radius of curvature of CP at P,
108
ds³
P
drd2z dzd²r
(Lacroix, Traité du Cal. Diff. Art. 351.)
and dr² + dy² + dz² = do²,
do², σ being the arc described,
Pdz ds² 2 (do² - ds²) dz
+ +
pdt2
Zdr
therefore R=
ds
ds²
Here it is manifest that
dt2
rdt
ds
is the square of the velocity resolved
in the curve PC, fig. 33, and that
do² 2
J
di2
is the square of the
velocity resolved perpendicular to OP in the plane OLP. The
two last terms, which involve these quantities, together form that
part of the resistance which is due to the centrifugal force; the
first term is that which arises from the resolved part of the forces.
From this expression we know the value of R; for we have, as
before,
do²
dt2
= C − 2ƒ (Pdr + Zdz);
do² - ds²
r² dv²
h2
also
vel.2 in PU =
dt²
2
dt²
ds²
h²
Hence,
= C
dt2
C − 2f (Pdr + Zdz) -
2
PROB. XI. To find the tension of a pendulum moving in a
spherical surface.
Retaining the denominations of the Prop. p = 0;
C − 2 (fP dr + Zdz) = 2g (k − z),
dr
C
ช
ds
C
r = √(2cz – z²);
dz
V(2cz
-
2) dr
C
2
ds
C
C
dz
√(2cz
z²)
Τ
h2
2 g (k
Z)
Hence, R =
g (c
2)
2.2
+
+
с
C
C
g. (c + 2 k
32)
C
109
and hence it is the same as that of the pendulm oscillating in a
vertical plane with the same velocity at the same distances: see
Prob. V.
SECT. III. The Motion of a Point upon any Surface.
39. PROP. To find the velocity, reaction, and motion of a
body upon any surface whatever.
Let R be the reaction of the surface, which is of course in the
direction of a normal to it at each point. And let e, e', e", be the
angles which this normal makes with the axes of x, y, and z re-
spectively; we shall then have, considering the resolved parts of R
among the forces which act on the point, by the formulæ (é),
ď² x
= X + R cos. €,
dt²
ď y
· · (1)
= Y + R cos. €
é',
dt²
d²
2 = 2 + R cos. e €".
dt
Now the nature of the surface is expressed by an equation
between x, y, z and if we suppose that we have, deduced from this
equation,
d z = pdx + qdy, so that p =
d z
dx' q =
dz
dy'
P, q, being the partial differential coefficients of z; (Lacroix, 125);
we have for the equations to the normal of the point whose co-ordi-
nates are x, y, z, (Lacroix, Elem. Treat. No. 143); x', y', z', being
co-ordinates to any point in the normal;
x²
x + p (z′
z) = 0,
y
= y + q (z′ — z) = 0.
Hence, it appears that if PK, fig. 36, be the normal, PG, PH
its projections on planes parallel to rz and yz respectively; the
equation of PG is x' − x +p (z' — z) = 0; and hence GN+p.PN=0,
and GN= p.PN, similarly the equation of PH is y-y
+q (2-2)=0, whence HN+q. PN=0, and HN=-q. PN.
110
Ph
And hence, cos. ecos. KPh=
PK
Ρ
GN
✔(PN²+NG+HN²)
√(1+p² + q²)
HN
√(PN²+NG² + HN²)
P g
cos. e'=cos. KP g
PK
9
√(1+p²+q²)
1
Whence, cos.e" √(1 − cos.² e — cos.²
cos.² e')=
4
√(1 + p² + q²)
Substituting these values; multiplying by dx, dy, and d z*,
respectively, in the three equations; and observing that dz-pdx
-qdy=0, we have
dx d²x + dy d'y + d z d² z
dt2
=Xdx+Ydy+3d..
Hence,
dx²+dy²+dz²
2
d t²
=f(Xdr+Ydy+~dz),
and if this can be integrated, we have the velocity.
If we take the three original equations (1), and multiply them
respectively by-p, -q, and 1, and then add, we obtain
d2 x
2
d² z
dť
d t² d t2
= −pX−qY+Z+ R √ (1 + p² + g³).
- p.
•
d t²
-9·
2
day
+
But d z
pdx+qdy;
d² z
ď² x
ď²y
d p d x + d q dy
hence,
2
d t² d t.2
P
+q
d t²
+
;
d t²
substituting this on the first side of the above equation, and taking
the value of R, we find
P
X+qY-Z
R=
√(1+p² + q²)
+
dp d x + dqdy
2
dť² √(1+p²+q²)
* Here dzindicates the differential of z considered as a function of both
the quantities x and y. It is what Lacroix, Elem. Treat. 126, indicates by d(z).
111
and p, q,
If in the three original equations we eliminate R, we find two
second differential equations, involving the known forces X, Y, Z,
which are also known when the surface is known, com-
bining with these the equation to the surface, by which z is known
in x and y, we have equations from which we can find the relation
between the time and the three co-ordinates.
Ꮖ
40. PROP. To find the path which a body will describe upon a
given surface, when acted on by no force.
In this case, we must make X, Y, Z each≈0. Then, if we
multiply the three equations (1) of last Art. respectively by
−(qdz+dy), pdz + dx, qdx − pdy, and add them, we find,
<<
− (qd z+dy) d²x + (pdz+dx) d² y+(qdx − pdy) ď² Z
= Rdt²
(q dz+dy) cos. e
+ (pdz+dx) cos. e'
+ (qdx − pdy) cos. e
"/
or, putting for cos. e, cos. é', cos. e", their values,
Rdt2
√(1+p²+q²)
{p(qdz+dy)-q(pdz+dx)+qdx-pdy}=0.
Hence, for the curve described in this case, we have
(pdz+dx) dy = (pdy− qd x) d² s + (q dz+dy) d² x*.
*If we make ds constant, in this we have dr dx+dy ď²y+dz d² z=0,
and eliminating d²r we obtain
{p d z+dx) d x +(qd z+dy) dy} d² y={(p d y − q d x) d x
−(qdz+dy) dz } d² z.
But the first side=d² y { (pdx+q dy) dz+dx²+dy²}
= d²y { dz²+dx²+dy² }=d s² d²y.
And the second side=d² ≈ { (p d x-dz) dy — q (dx² + d z²) }
= d² z { — qd y² − q d x² — q dz² } = − q d s² d² z.
Hence, the equation becomes, omitting d s²,
‹l² y+q d² c=0.
112
This equation expresses a relation
x, y, z, without any regard to the time.
among the differentials of
Hence, we may suppose r
the independent variable, and dr=0; whence we have
(pd z+dx) d² y=(pdy-qdx) d² z.
This equation, combined with dz=pdx+qdy, gives the curve
described, where the body is left to itself, and moves along the
surface.
The curve thus described is the shortest line which can be drawn
from one of its points to another, upon the surface.
The velocity is constant, as appears from the equation
velocity²=2f(X dx + Ydy + Z dz).
By methods somewhat similar we might determine the motion of
a point upon a given curve of double curvature when acted upon
by given forces.
CHAP. VI.
THE CONSTRAINED MOTION OF SEVERAL POINTS.
41. IN the present Chapter it is intended to consider some
problems in which several material points are connected with each
other by means of cords or rods, not possessing inertia or weight, and
generally inflexible and inextensible, while one or more of the points
are compelled to move on given lines or surfaces.
Some of these problems will approach in their nature to those
in which we consider the motion of a finite body. For two material
points connected by a rigid rod may be considered as a finite body,
and accordingly, the formula will be the same as those for a rigid
body in some cases. But by considering the tension of the rod
113
or string, which connects two points, as one of the forces which
acts upon them, we can reduce these problems under the formulæ
for the motion of points. And as this tension will always be the
same for both the bodies which the rod or cord connects, we shall
be able to eliminate it, and to obtain the motions of the points.
The tensions here spoken of are of the nature of those which we
have called moving forces, or pressures. Consequently, the acceler-
ating forces which they produce upon the bodies are, by the third law
of motion, as the pressures or tensions directly, and as the mass of
the bodies inversely. If the tension of a string which acts upon
a body m be equal to the weight of a mass p, and if g be the
accelerating force of gravity, we have
P g
m
= accelerating force produced by tension.
SECT. I. The Motion of
The Motion of a Rod on Planes*.
PROB. I. Fig. 37. P, Q, are two material points, connected
by an inflexible rod PQ; PQ falls by gravity in a vertical plane,
while P moves along a horizontal plane: to define the motion
of Q.
N.B. If PQ have at first no motion, except in a vertical
plane, it will continue to move in a vertical plane, because all
the forces are in the same plane.
Let APN be the horizontal line, G the centre of gravity of
P, Q'; GH, QN vertical; let A be a fixed point, AP=x,
AN=x', NQ=y'. And let p = the tension of the rod PQ;
(that is, the force with which it presses P in the direction QP,
and Q in the direction PQ;) also, let O be the angle NPQ; then
the resolved parts of the tension will be p. cos. O parallel to AN,
and p. sin. parallel to NQ. And, if m, m', be the masses of
O
P, Q, respectively, the accelerating forces will be
pg cos. O
on P
;
m
pg cos. O Ꮎ
'm
and
Pg sin. O
on Q.
m'
* The motion of a rod without weight connecting two heavy bodies, will
be the same as the motion of a heavy rod: this will appear hereafter.
P
114
Hence, we have, by equations (c),
d² x
pg cos. O
dt2
d² x'
dte
m
pg cos. O d²y
mi
dt2
pg sin. O
m'
= 0.
md² x + m'd²x
dt²
Let x = AH =
mx+m'x'
; ... d² x = 0.
m+m'
an
……….(1);
Hence, the point H will either remain at rest, or move uni-
formly along AN.
If we make PG=b, QG=c,
x=APAH-PH-x-b cos. 0; .. dx bd cos. 0,
x' = AN = AH+HN=x+c cos. 0;
QP
PG
Ꮎ
Also, y = QN= .GH =
-
Ꮎ
. dx= cd² cos. 0.
m + m²
b sin. 0.
278
Hence, the first and last of equations (1) give, substituting
these values, and multiplying by m, m',
mb
d² cos. O
dt²
=pg
pg cos. ;
d² sin. 0
Ꮎ
(m+m')b.
dt²
=pg
pg sin. 0-gm'.
Multiply the first of these equations by 2 d. cos. 0=-2 sin. 0d0,
and the second by 2 d. sin. 02 cos. O de, and add them, and we
have
m b
2d cos. 0.d² cos. 0
dt²
2 d sin. O d² sin. O
+ (m+m) b
dt2
2 gm' cos. Ode.
115
This integrated gives
d.cos.0\2
mb. (
dt
+ (m+m²) b (
(d sin.s)
sin. 0) *
= C − 2 gm' sin. 9 :
=C-2 gm' sin. 0:
dt
or, {m sin.20+(m+m) cos.° 0} b.
which reduced, gives
d02
b. ==
dt2
ᏧᎾ
dt²
C-2 gm' sin. O
m + m² cos.* 0
To determine C, suppose that when PQ reaches the line AN,
C
do
or when =0, the angular velocity
is a: hence,
dt
ba² =
dee
dt
b (m+m' cos.² ()
m + m
ba² (m+m') — 2 gm' sin. 0
….(2).
There may be a position when the angular velocity is 0; sup-
pose that there is 0₁; then, we shall have
ba² (m+m') — 2 gm sin. 0₁ = 0...
0,0........(3);
d02
dt²
2 gm sin. 0₁
b
01 sin. 0
m+m' cos.20
1
:
If the rod fall from rest in a given position, ₁ is known and
hence, the angular velocity at any point. To determine the position
at any time, we must obtain t in terms of 8, by integrating again;
and hence in terms of t.
π
Q
If the line fall from a vertical position ₁
; hence, to find
the angular velocity a, acquired when it becomes horizontal, we
have, by (3),
a² =
2 g m
b (m +m');
::. b² a² =
Q bgm
m+m'
Now ba is the linear velocity of the point G at H, and. it appears
116
from this, that the space through which a body must fall freely to
acquire this velocity, is
GP.
bm'
m+ m²
7, a third proportional to QP and
If it were supposed that P were constrained to move in a ho-
rizontal groove AP, while PQ were so connected with it, that PQ
could descend below the horizontal position, so as to revolve en-
tirely round P; the mechanical conditions would be the same, and
the expression obtained above for the angular velocity is true in this
case also.
Since
Ꮷ Ꮎ
dt
√ba²
'ba² (m+m') — 2 m'g sin. 0
b(m +m' cos.2 0)
the motion is possible, so long as the numerator of the fraction is
positive. The negative part is greatest, when
numerator is always positive, if
π
◊
hence, the
2
Q m g
ba² (m+m') > 2 mg, or if a² >
α
(m+m') b
In this case, if H be at rest at first, the line PQ will go on
revolving about the point H; the centre of gravity G ascending
and descending in a vertical line.
If H be at rest, the path described by Q will be an ellipse,
whose semi-axes are GQ and PQ,
C
x²
y2
For
(GO)
2
+-
(RN)²°
c²
= 1, ΟΙ +
C (b+c)²
=1:
Q
the equation to an ellipse, of which the semi-axes are b+c and c.
The velocity of P =
bd.cos. O
dt
d Ꮎ
- b sin. 0.
dt
2 m'g sin. 0
··(4).
'ba² (m + m')
√//ba²
= - b sin. 01/ ba
b (m + m' cos." )
If H be in motion, and move uniformly so as to carry the whole
line PQ in the direction AX, the relative motion of the points P, Q
will not be altered. The system will have the motion of rotation
117
already investigated, combined with the uniform motion of translation
in the direction AX; and the point Q by the mixture of its elliptical
motion with this rectilinear one, will describe a kind of trochoidal
curve.
Suppose Q to hang vertically from P, and a given velocity V
to be communicated to P: to determine the motion of P and Q.
d²
d2x
The velocity of G parallel to AX is constant, because.
0,
d t²
and = constant. Hence, G will retain the velocity parallel to AX,
d x
d t
which it has at first.
But at first, when Q is at rest, and P moves
V.QG
QP
with the velocity V, the velocity of G parallel to AX will be
And at first the angular velocity of Q round P will be
V c
b+c
V
π
b+c
; hence, by (2), putting for 0,
Q
Va
(b+c)'
m+m'
2gm
m
mb
whence, a is known; and hence the motion of the system is
completely determined.
The quantity g which represents the vertical force may be of
any value. If we make it = 0, we have the motion of a rod moving
freely, for instance, on a horizontal plane, while one end moves in
a rectilinear groove.
In this case
ᏧᎾ
d t²
a² (m + m')
m+m'cos."0"
This velocity is a, when the rod coincides with the groove, and
increases to a
V
'm+m'
m
when it is perpendicular to it.
If we make m=0, we have, by (2),
d 02
ba² - 2 g sin.* 0
d t
b cos. O
118
Here H coincides with N; we might find 8 by integrating. It
is easily seen that the curve described is a parabola.
PROB. II. The same things being supposed, except that the
plane AX is inclined to the horizon; to determine the motion.
The rod is still supposed to move in a vertical plane.
If we resolve the forces parallel and perpendicular to the plane AX,
we shall find that the point I will descend down the inclined plane
in the same manner as a heavy point would do: that is, acted upon
by a constant force g sin. , being the inclination. And that the
angular motion will be obtained in the same manner as in the last
problem, except that instead of g we shall have g cos. 1, the force
perpendicular to the plane.
If the line AX be vertical, the motion of PQ, with respect to
AX, will be the same as in the last problem when we made g=0.
PROB. III. A rod PQ, connecting two material points P, Q,
acted on by gravity, moves so that one of them slides along a horizon-
tal plane: to define the motion.
This differs from Prob. I, in not supposing the motion to be in
a vertical plane.
Let AL, LP, fig. 38, be rectangular co-ordinates of P in the
horizontal plane, AM, MN, NQ co-ordinates of Q,
AL=x, LP=y, AM=x', MP=y', NQ=z', PQ=b.
Also let NOM=0, and QPN=0; hence,
z′ = b sin. 0, x′-x=b cos. 0 cos. P, y' —y=b cos. O sin. P.
Ρ be the tension of PQ, m, m' the masses of P, Q.
Let
And resolving the forces, we have
pg
cos. O sin. O,
os. O sin. Φ,
. . . .(1).
ď² y
ď² x
pg
cos. ◊ cos. ¿,
d ť²
m
d t²
d'x'
2
d t²
pg
m
os. O cos. O,
COS
d² z
d t
=
d'y' p g
2
d t²
pg
m'
m
sin. 0-g.
m
119
md²x+m'd²x'
Hence, we have
md²y+m'd²y'
=0;
=0.
d t²
d t2
From which it appears that the projection of the centre of gravity
of P, Q, on the plane of xy moves in a straight line with a uniform
velocity.
Also
d² x' - d² x
d t²
m
+
m
Pg cos. Ө
cos. P,
d²y' - d²y
d t²
..(2).
( + - )
m
pg cos. O sin. Ø
Ꮎ
And y'-y=b cos. O sin. p, x-x=b cos. O cos. ;
Ꮖ
(x' — x) (d² y' — d²y) — (y′ − y) (d² x′
Ꮖ – d²x)
dt²
=0.
Whence it appears that N describes about P areas proportional
to the times; (see Art. 16,) and therefore
b² cos." 0 do hdt: h being a constant quantity. . . .(3).
аф
Again, equations (2) may be put in this form
+
d². cos. cos. O
dt²
=
G
m
m
1) PE
pg
cos. cos.
O
cos. &;
b
ď.
pg cos. O sin. p.
dt
d. cos. @ sin. 4 = ( — + - ) P5
And multiplying by cos. Ø,
cos. & ď³. cos. O cos.
•
m
+
m
m
b
sin. 4, and adding
+ sin. Ø ď³ cos. Ø sin. O
pg
cos. O dt.
m
b
But d. cos. cos. &=cos. pd . cos. § – cos. O sin. pdo,
d². cos. O cos. P
d. cos. O sin.
d². cos. ◊ sin.
:
=cos. pď² cos. 0 – 2 sin. pdø d. cos. 0
- cos. O cos. odp³ — cos. 0 sin. pďp,
= sin. pd. cos. + cos. O cos. odo,
=sin. pď² cos. 0+2 cos. pdp.d. cos. ◊
cos. O sin. pd¤²+cos. O cos. Pď³p.
120
Hence, cos.pd². cos. O cos. + siu. Od² cos. O sin. O
2
becomes d² cos. 0 — cos. Odp²;
h²d t²
or since, by (3), dp³
2
it becomes
4
6ª cos.* 0
2
1
had t
2
6$ cos.³ Ø
And the above equation is equivalent to
2
ď² cos. Ô
"
d² cos. 0 -
hⓇ dt2
b4 cos.³ Ø
G
+
๓.
m
1) PE
cos. Odť²....(4).
b
Also the equation in z' may be written thus,
d² sin. 0 = (.sin. 0 –
:)
·5) at²
đt....
.(5).
Eliminate p by multiplying these by m sin. 0, and (m+m') cos. 9.
and subtracting
mh² sin. Odt²
(m +m') cos. O ď² sin. 0
cos. ◊ ď² sin. ◊ — m sin. Od² cos. Ø -
64 cos.³ 0
3
(m + m² ) &
cos. Odť².
Multiply by 2de, and we may put the equation in this form,
2 (m+m') d sin. O d² sin. 0 + 2md cos. Od² cos. O
+
2
2mh²d cos. Odť²
4
3
b₁ cos.³ 0
2 (m + m) d. sin. Od t².
b
Integrating, (m+m') (d sin. 0)²+m (d cos. 0)² --
mh² d t²
2
b4 cos. Ꮎ
= Cdt²
2 (m+m')
m') g
b
sin. 0.dt;
d02
dt2
= {c⋅
-
{(m+m') cos.² 0 + m sin.º 0}
2 (m+m') g sin. Ø
b
'm h²
2
+
b cos.
2
121
ᏧᎾ
dt
2
{m+m' cos.² 0} = C-
2(m+m'). g sin. ✪
m h²
2
+
(6).
b
b4 cos.20
d02
dt²
When = 0, (m+m') = C +
mh2
64°
π
ᏧᎾ .
We can never have 0 =
;
for then
is infinite.
dt
d Ꮎ
To find when
=0, we have a cubic equation.
But it is
dt
manifest´that if ✪ be
will also satisfy it.
Having found
a value which satisfies this equation, T Ꮎ
in terms of t by (6), we find & by equation (3).
If in equation (3) we eliminate dt by (6), we have do=d0 × a
function of 0. And if we put b cos. 0 = PN=r, d0 =
=
whence do dr x a function of r,
orbit of N about P.
If m = 0, we have
dr
2
√(6² —²)
which is the equation to the
doe
dte.
2
m' . cos.² 0 = C.
2 m'g
sin. 0.
b
And when = 0,
d02
C
dt²
m'
In this case we might integrate. The body Q will describe a
parabola.
PROB. IV. Two points P, Q, fig. 39, are connected by a rod,
and one of them P, slides along a vertical line AZ: to define the
motion.
Let AP≈; and let AM = x', MN = y', NQ = ', three
rectangular co-ordinates of the point Q from a fixed point A.
Let QPZ=0, MAN=4, PQ=b ;
... x′ = b sin. 0 . cos. O, y=b sin. 0 sin. 4, z'—z=b cos. 0.
Q
122
And if p be the tension of PQ, the equations of motion will be
d²x'
pg
= sin. O sin.
dť² m
pg
d²y'
sin. 0 cos. 0,
2
dt2
m
d² z
pg
dt2
=8+
Po, cos. 0,
ď² z
pg
m
dt²
g
cos. A
m
•(1).
Hence, as in last problem, we find
-
x'd³y' — y'd²x'
d t²
2
= 0, and b² sin.² Od¶ = hdt
d²x'
dt2
Also, cos. $ + sin.
•
(2).
dt2
d'y pg
PE sin. 0,
which, transformed as in last problem, gives
m
2
d² sin. 0 — sin. Odp² = sin. Odť²;
2 pg
m'b
•
h2dt2
.`. by (2), d² sin. ◊
pg
64 sin.3 0
m'b
sin. Ꮎ dt ...
2
· (3).
And bď² cos. 0 = d²z' — d²z; hence, by equations (1),
pg
d² cos. O • = ( = + = 1 ) PR
m
m
cos. 0...
Eliminating p, as in last problem, we obtain
·(4).
Ꮎd
Ꮎ
2. (m+m') d. sin. Od sin. 0+2.md. cos. Od cos. O
9
2h² (m+m') dť² . d sin. 0
64 sin.3 0
Integrating, (m+m) (d sin. 0)²+(d. cos. 0)²
= 0.
d02
dt
•
2
h² (m + m') dť²
+
= Cdt²,
64 sin.² 0
h² (m+m')
4
b¹ sin.² 0
(m+m' cos.² 0) = C
(5).
Hence, the angular velocity of PQ in a vertical direction is known.
And hence, by (2), we may, as in last problem, find the differential
equation to the orbit described by N round A. AN describes areas
proportional to the times by (2).
The relation between t and
may be expressed by means of an
elliptical arc. Equation (5) may be put in this form,
m+m' cos.² 0
dt = sin.20 də².
h² (m+m')
C
64
C cos.² 0
123
m + m' cos.²
= sin.² Od 0²
D-C cos.20
Making D=C -
h² (m+m')
64
Now, if 1 and a be the semi-axis minor and major of an ellipse,
☛ the arc measured from the extremity of the semi-axis, and its
abscissa from the centre along the axis-minor; we shall have
σ
2
do² = d§² .
1 + (a² − 1) §³
If, in the value of dť² we make C
cos. 0 =
2
d}
Ꭰ
dť² = de · C
1 - 2
cos.² = D², we have
§; — sin. Odo
· Ꮎ =
-
Dm'
m + हु
C
Ꭰ-ᎠᏑ
=
with ~.do³, if we make
√2.đ§; and hence,
m
dr².
Dm'
1+ हु
Cm
m
which agrees with
Dm'
a² - 1 =
whence a is known.
Cm
t
0
;
1+C=√m × elliptical arc
.*. cos.
V
When 0 =
π
2'
C
D
(abscissa=cos.
=abscissa of elliptical arc (t+C). √
m
=0, and the arc=0, and if at that time t is t₁,
C=-t₁.
Also, when =
3 п
2
=0, and the arc = a semi-ellipse; hence,
T
V c
- semi-ellipse, gives the interval between two horizontal
m
positions of the rod.
When 0=0, the angular velocity is infinite; hence, the rod will
never coincide with the vertical line.
124
This is true only if the body have some angular motion hori-
zontally, for if it move at first in a vertical plane, it will continue to
do so, and its motion may be calculated by Prob. III.
PROB. V. Two points P, Q, connected by a rod, slide along
two given inclined planes, acted on by gravity; to determine the
motion.
The motion is supposed to take place in a vertical plane.
ทุ
Let AX, AY, fig. 40, be the two planes, making with the
horizon angles ẞ, y. Let PQ make an angle ✪ with AX, and ʼn
with AY. AP=x, AQ=y, PQ=b; tension of PQ=p, masses
of P, Q=m, m'.
Resolving the forces in the lines AX, AY, we have
d2x
2
d t²
pg
m
cos. — g sin. ß
d²y
. (1).
pg
dt2
n
m'
cos. ŋ — g sin. y
Now x
Hence,
m' cos. nd³x — m' cos. Oď³y
d t²
g (m' sin. y cos. O — m sin. ß cos. n). . . .(2).
У
b sin. 0
sin. (B+y)
b sin. n
, y
sin. (B+y)
Substituting these values in (2), multiplying some of the terms
by 2d0, and others by - 2dn, (since de = — dŋ) and integrating,
we have
b
-
2
sin. (B+y)
{m (°
d. sin.
dt
")² + m²
m' (
(d-sin. 6)"}
=
dt
C-2g (m' sin. y sin. 0+m sin. ß sin. n);
b
γ
ᏧᎾ
sin. (ẞ+y)˚ dť²
·
(m cos.² n+m' cos.20)
= C-2g (m' sin. y sin 0+m sin. ẞ siu. ŋ). . . .(3).
Hence, the angular velocity is known.
This may be reduced as follows. Let the angle which PQ
125
makes with the horizon be y. Then, 0 = ß+4, n = y−4.
Also, let C2gn; then the right-hand side of equation (3)
becomes
2g {n-m' sin. y. sin. (B+)—m sin. ß. sin. (y)} :
expanding, the part in brackets becomes
n
(m' + m) sin. ẞ sin. y cos. V
(m' sin. y cos. ẞ + m sin. ß cos. y) sin. ¥.
Suppose (m+m) sin. ẞ sin. y=r cos. &r
m' sin. y cos. ẞ – m sin. ß cos. y=r sin. d)
.(4)
which always give possible values of r and d; and our expression
becomes
n—r cos. d cos. 4-r sin. § sin. y, or n − r cos. († –d).
Hence, equation (3) gives
d02
dt2
2g sin. (B+y)
b
By equations (4), tan. d
n—r cos. (↓ -8)
2
m cos.² (y-4)+m' cos. (B+y)'
m' cot. B-m cot. y
m+m'
;
and hence, by Statics, (Chap. VII, Prob. 13.) & is the value of ↓
in the position of equilibrium of PQ. When y=d, the angular
velocity is greatest. About this position PQ oscillates both ways;
(except its velocity be too great). We shall find the limit of the
oscillations by making
d Ꮎ
dt
N
=0; :. n—r cos. († —♪)=0; cos. († — d) = 2.
ጥ
And is the cosine of a positive arc, and of a negative arc of
ጥ
equal magnitude. Hence, the limiting positions PQ, P'Q', make
equal angles with p the position of equilibrium.
If we know the velocity at any point, we know C, and con-
sequently n, by equation (3); r is known by equation (4).
If the velocity be so great that n exceeds r, there will no longer
be a limit to the angular motion of the rod. If we suppose XA,
YA to be grooves in which P, Q, slide, instead of planes; and to be
126
produced beyond A, which will not alter the mathematical con-
ditions; the rod PQ will perform complete revolutions, and the
limits to its positions will be curves XYX'Y', fig. 41. The end P
will oscillate backwards and forwards through XX', and Q through
YY'.
If we suppose P, Q to move in two grooves, not affected by
gravity, we must in equation (3), make g=0. Hence,
b
ᏧᎾ?
sin. (B+y) dt²
2
(m cos.²n+m' cos.² 0) = C.
If the grooves cross each other at right angles, and m=m', PQ
will revolve with a uniform angular velocity.
PROP. VI. The rod PQ descends, one end sliding along a
horizontal and another along a vertical plane: to determine its
motion.
The rod is supposed to move in a vertical plane.
This is a particular case of last problem.
fied by beginning with the equations (1) properly modified; but it
will be sufficient to apply our equation (3).
It might be simpli-
Making, in that,
B = 0, y = right angle, n = comp. 0; we have
d02
b. (m sin.2+m' cos.2 0) = C-2gm' sin. 0.
dt2
If we suppose a to be the angular velocity when the rod is
horizontal, or when 0=0;
ba²m' = C.
If we suppose that when the angular velocity is 0, 0 is 0,,
0=C—2gm' sin. 0₁; .. ba² = 2g sin. 0¸.
1
Hence, the angular velocity a acquired by falling from rest through
an angle 0, to a horizontal position, is independent of m, m'.
When Q comes to the horizontal plane
velocity² of Q = b²a² = 2gb sin. 0₁;
hence, the velocity is the same as if Q had fallen freely through the
space b sin. 0, which is its actual descent.
If ba² > 2g, there is no position where the velocity=0, and the
rod will revolve in the manner described in the last problem.
127
SECT. II.
Tractories.
42. When one end of a string or rod is drawn with a given velocity
along a given straight line or curve, which is called the Directrix,
so that a body fastened to the other end is dragged along by it,
and made to describe a certain path, this path is called a Tractory.
The motion of the body at any instant will depend upon its pre-
ceding motion, and upon the tension of the string. The body is
considered as a material point.
PROB. VII. A point P, fig. 42, moves uniformly along a straight
line AX, drawing along with it, in the same plane, a body Q, by
means of a string PQ: to find the tractory described by Q*.
*This problem is sometimes solved without taking account of the ten-
dency which the angular motion, generated in PQ, has to continue itself.
This mode of considering it would be true, if we were to consider Q to move
upon a plane, and the friction to be such, as instantly to destroy any motion.
communicated to the body. In that case, Q, fig. 43, would always move
in the direction QP, in which the string draws it; QP would be a tangent;
and the curve would be determined by the condition, that the tangent QP,
intercepted by the abscissa, is a constant quantity. Hence, if AN=x,
NQ=y, rectangular co-ordinates; QP = a; we have
NP =-
y d x
dx
dy
dx2
•·. QP = a = y
√(1+1+);
d y²
√(a² — y²)
.. dx
dy
Y
dx =
and by integrating we get the relation of x and y. (See Mr. Peacock's Ex-
amples, p. 174, for the properties of this curve.)
This curve has an asymptote, as represented in fig. 43. We may reserve
for it the name of Tractrix, by which it is frequently designated; giving to
the curves, which really solve the problem in the text, the name of Tractory,
which is analogous to the names of other curves which occur in Mechanics.
If the directrix be a circle, and the friction be such as immediately to
destroy the motion of the body, it will move in a curve, the tangent of which,
intercepted by a circle, is a constant quantity. This curve is sometimes
called the Complicated Tractrix. See Cotes, Harm. Mens.
For
·
128
A
་
We might solve this by the equations of motion: but we may
find the curve more simply by the following reasoning.
If, when any system is in motion, we suppose the space in
which it is, to move uniformly, so as to carry all the parts of the
system in parallel directions, and with equal velocities; the relative
motion of the parts, and their action upon one another will remain
the same as before, by the second law of motion.
While P is moving with a uniform velocity along the line AX,
let this line and the space containing PQ move in the opposite
direction XA, with an equal uniform velocity. Therefore, by what
has been said, the angular motion of PQ will remain the same as
before. And P, having two equal and opposite velocities, will be
at rest. Now, if a body fastened to a string, revolve round a fixed
point P, it will revolve uniformly. Hence, the angular motion of
PQ round P when fixed, will be uniform; and therefore it will be
uniform when P moves uniformly along AX.
Hence, the motion of Q arises from a uniform angular motion
round P while P moves uniformly in a straight line: and hence
its path will be a cycloid, or a trochoid, within or without the
cycloid, (see Mr. Peacock's Examples, p. 186.). Take a point R
in the radius PQ, produced if necessary, such that the velocity of
R round P at rest, may be the same as the velocity of P along
AX. Let a circle be described with radius PR, and let NO,
parallel to AX, be a tangent to it. Then, if the circle RO roll
uniformly along the line NO, the point Q will trace out a trochoid,
which will be the tractory in question. For, in this case it is ma-
nifest that P the centre of the circle moves uniformly along the
line AX, and that PQ revolves uniformly round P; which are
the conditions requisite.
For the tractory, when the directrix is any given curve, see Euler, in
the Nova Acta Acad. Petrop. 1784. He has there also considered the
problem, taking a finite friction into their account: and likewise, what he
calls Compound Tractories, where there are more points than one attached
to the string.
129
!
The curve will be a cycloid if R coincide with Q; a trochoid
or prolate cycloid if R be without Q; and a curtate cycloid if R
be within Q.
If BC be the original position of PQ, when Q is at rest, and
if the point P begin to move along AX, BC will be a tangent to
the tractory. The curve will be a trochoid of the first kind, and C
will be its point of contrary flexure. If BC be perpendicular to
AB, the curve will be a cycloid.
PROB. VII. Supposing the same things, and that PQ does
not move in the plane APQ; to find the tractory.
By reasoning similar to that of last problem, it will appear that
the motion of Q will be determined by supposing P to move uni-
formly along the directrix, and PQ to revolve uniformly round P:
its motion relatively to P, being always parallel to a certain fixed
plane. Hence, the path of Q will be an oblique helix, which may
be supposed to be described on the surface of an elliptical cylinder,
of which the axis is AX.
PROB. VIII. The point P, fig. 44, moves uniformly in the
circumference of a circle BP, drawing the point Q: to find the
tractory described by Q.
The motion is supposed to be in the plane of the circle BP.
Let AX be any fixed line; AM, MPx, y, and AN, NQ=x',
y', rectangular co-ordinates to P and Q. The tension of PQ=p,
and the mass of Q=m. Also, lett be the time; and BP, pro-
portional to the time, nt, QPO= 4, PO being parallel to AX,
AP=a, PQ=b. Hence, by resolving the forces,
d²x'
pg
d²y'
pg
d t²
cos. P,
sin.,
m
d t²
m
x = a cos. nt, y = a sin. nt;
d² x
d²
Q
an' cos. nt,
=
an sin. nt;
d t
d t²
R
130
bd² cos. O
d t2
d²x' - d²x
pg
cos.+an² cos. nt,
d t²
m
bd² sin.
d t²
2
d²y' - d²y
d t²
pg
2
sin. +an² sin. nt;
m
b sin.pd³cos.¿—b cos.pdªsin. — an² (sin. cos.nt-cos.& sin.nt);

dt2
bd2
d t²
=
=
2
an² sin. (§— nt).
Let (p-nt)=&; then d²=ď³; and substituting and mul-
tiplying by 2d, and integrating, we have
The angle
=
dx² 2an2
(C + cos. Y).
-
b
d ť
-nt is QPO-RPO=QPR. The angular
velocity of PQ will be greatest when -nt, or when PQ
coincides with PA in direction, as at P' Q'. If C be less than
unity, PQ will oscillate about PA, while P moves uniformly in
the circle. If C be greater than unity, Q will revolve about P
with a variable velocity, while P revolves about A uniformly. The
Least velocity will be when -nt=0, or when AP, PQ are in a
straight line, as at Q".
PROB. IX. Let P move uniformly along a given straight
line, while Q is drawn along, and also acted on by gravity: to find
the motion of Q.
By reasoning, as in Prob. VI, it will appear that the motion of
Q with respect to P, will be the same as if P were fixed. Con-
sequently, if PQ move in the same plane, it will be the motion of
a circular pendulum, and if not, it will be the turbinatory motion
of a point in a sphere. This, combined with the rectilinear motion
of the point P, will give the actual motion of Q.
SECT. III. Complex Pendulums.
PROB. X. P, Q, fig. 45, are two bodies, of which the first
hangs from a fixed point, and the second from the first, by means
of inextensible strings AP, PQ: it is required to determine the
small oscillations.
131
Let AM=x, MP=y, AN=x₁, NQ=y₁, AP=a, PQ=aj⋅
Mass of Pm, of Q = m₁; tension of AP=p, of PQ=P₁.
two
Hence, resolving the forces p, P₁, we have
ď y
d t²
d²
y
dt
=
Pig Yi-y
m
1
pg y
а1
m
a
..(1).
Pig Yi-y
m1
а 1
By combining these with the equations in x, x₁, and with the
x²+y²=a², (x,- x)² + (yı—y)² = a₁²;
1
we should, by eliminating p, p, find the motion. But when the
oscillations are small, we may approximate in a more simple
manner.
Let ß, B₁ be the initial values of y, y₁. Then manifestly,
P, P₁ will depend on the initial position of the bodies, and on their
position at the time t: and hence we may suppose
p=M+Pß+Qß₁ + Ry+Sy₁+&c.: and similarly for P1.
Now, in the equations of motion above, p, P₁, are multiplied by
y, y₁-y, which, since the oscillations are very small, are also very
small quantities (viz. of the order ẞ). Hence, their products
with B will be of the order 62, and may be neglected, and we
may suppose p reduced to its first term M.
1
M is the tension of AP, when ß, ß₁, &c. are all = 0.
Hence,
it is the tension when P, Q hang at rest from A, and consequently,
M=m+m₁; similarly, the first term of P1, which may be put for
it, is m₁. Substituting these values, and dividing by g, equations
(1) become
d²y
m1
=
gdt2
+
.mai
m a
m + m₁
y +
mi
Yı 1
ma
та1
(2).
d°y,
Y
Yı
&dta
a1
a1
Multiply the second of these equations by X, and add it to the
first, and we have
132
d²y +λd'y,
g dt²
1
༢
ma1
(m² +
m+m, _ =)
y-
· ma
a₁ 1
a 1
m₁
½) y₁:
ma 1
and manifestly this can be solved, if the second side can be put in
the form―k (y+λy₁); that is, if
m1
k = +
m + m₁
λ
;
mai
m a
α1
λ
ՊՈՆ,
kλ =
;
α,
та1
m1
α
m₁a₁
1 1
or a,k
+
+
1
m
a
m₁ = (a₁
M
Eliminating X, we have
m₁
1
m a
(a¸ k − 1) λ
k − 1) (
(a,k−1)a, k- = (a,
(a, k
—
m
1
m₁
+
M
1
“
a
Hence, (a, k)º — (1 + 2) (1+ª) a¸k= −
1 2
•. (3).
1
+ ma₁).
Mi a 1
(4).
a
m a
From this equation we obtain two values of k. Let these be
denoted by ¹k, k; and let the corresponding values of λ be 'X, X.
Hence, we have these equations
d² y +¹λ d³y,
gd t2
d² y + ²λ d³ y,
gd t
¹k (y +¹λy),
· ²k (y +²λy,).
And it is easily seen, as in Prob. II, Chap. III, that the integrals
of these equations are
1
y+¹λy₁ =¹C cos. t V(kg)+¹D sin. t √('kg),
1
y+²λy₁ =²C cos. t √(kg)+2D sin. t √(kg).
1
1 2 ˚C
2
€,
¹C, ¹D, ²C, 2D being arbitrary constants. But we may suppose
'C' =¹E cos. 'e, 'D⇒NE sin. ¹e, 'C=*E cos. ²e, 'DE sin. 2
where ¹E, 'E, ¹e, "e, are other arbitrary constants. By introducing
these values, we find
133
y+¹λy₁ =¹E cos. {t √('kg)+'e}{
་
2
y+λy, = 'E cos. {t V(kg)+'e})
From these we easily find
1
2x ¹E
•.(5).
.cos. {t V(kg)+'e} +
1E
¹²E
λ -
2 E
cos. {t V(kg)+e}
(6).
Y₁ = 1 — 2 • cos. {t V (kg)+'e} + 1—2 cos. {t V(®kg)+e} }
1
1
•
1
The arbitrary quantities ¹E, 'e, &c. depend on the initial position
and velocity of the points. If the velocities of P, Q=0, when
t=0, we shall have ¹e, 'e each=0, as appears by taking the differ-
entials of y, Y₁•
2
If either of the two ¹E, E, be=0, we shall have, (supposing
the latter case, and omitting ¹e)
1
¹E
Y
2x - x
cos. t vkg,
1
¹E
yi
cos. t¹kg.
Hence, it appears that the oscillations in this case are symmetri-
cal: that is, the bodies P, Q come to the vertical line at the same
time, have similar and equal motions on the two sides of it, and
reach their greatest distances from it at the same time. It is easy
to see that in this case, the motion has the same law of time and
velocity as in a cycloidal pendulum; and the time of an oscillation,
in this case, extends from when t=0 to when t √('k g) = π,
t =
π
√(kg)
or
Also if ẞ, B, be the greatest horizontal deviation of
P, Q, we shall have
y=ẞ cos. t V(kg), y₁=ẞ, cos. t √('kg).
In order to find the original relation of ß, ß, *, that the oscil-
*The oscillations will be symmetrical if the forces which urge P and Q
to the vertical, be as PM, QN, as is easily seen. Hence, the conditions
for symmetrical oscillation might be determined by finding the position of P,
Q, that this might originally be the relation of the forces.
134
lations may be of this kind (the original velocities being 0), we must
have, by equation (5), since 2E=0,
B+λß₁ =0.
Similarly, if we had ß+¹λß₁=0, we should have ¹E=0, and
the oscillations would be symmetrical, and would employ a time
π
√(kg) *
When neither of these relations obtains, the oscillations may be
considered as compounded of two, in the following manner.
pose that we put
y=H cos.tv (kg)+K cos. t V(kg)....(7),
1
=
Sup-
omitting ¹e, 2e, and altering the constants in equations (6); and sup-
pose that in fig. 45, we take Mp H cos. t V(kg). Then p
will oscillate about M, according to the law of a cycloidal pendulum
(neglecting the vertical motion). Also p P will=K cos. t √(²k g).
Hence, P oscillates about p according to a similar law, while p
oscillates about M. And in the same way, we may have a point q
so moved, that Q shall oscillate about q in a time
q
oscillates about N in a time
π
V(kg)
π
√(kg), while
And hence, the motion of
the pendulum APQ is compounded of the motion of Apq oscillat-
ing symmetrically about the vertical line, and of APQ oscillating
symmetrically about Apq, as if that were a fixed vertical line.
When a pendulum oscillates in this manner, it will never return
exactly to its original position, if √¹k and Vk are incommen-
surable. If 'k and V2k are commensurable, so that we have
m√¹k=n√/²k, m and n being whole numbers, the pendulum will
at certain intervals, return to its original position. For let
t√(kg)=2n; then t (kg) will = 2 mm; and by (7),
y=H cos. 2nπ + K cos. 2mπ=H+K,
which is the same as when t=0. And similarly, after an interval
135
such that t√('kg)=4nπ, 6nπ, &c. the pendulum will return to
its original position, having described in the intermediate times,
similar cycles of oscillations.
Ex. Let m₁ =m, and a₁=a, to determine the oscillations.
Here equation (4) becomes,
a² k² — 4a k=-2, ak=2± √2.
Also, by equation (3),
a k=3-λ; .. 'λ=1+ √2; λ=1— √2.
Hence, in order that the oscillations may be symmetrical, we
must either have
B+(1+ √2) B₁=0; whence ß₁ = −(√2—1) B:
or ẞ-(2-1) B₁ = 0; whence B₁ = ( √2+1) B.
The two arrangements indicated by these equations are represent-
ed, fig. 46, and fig. 47. The first corresponds to ẞ₁ =( √ 2 + 1) ß,
or QN=( √2+1) PM. In this case, the pendulum will oscillate
into the position AP'Q', similarly situated on the other side of the
line; and the time of this complete oscillation will be
π
√ {{(2-12)}
a
sion
π
√(2- √2)
g
In the other case, corresponding to B₁ = −(√2—1) ẞ, Q is
on the other side of the vertical line, and QN=( √2—1) PM.
The pendulum oscillates into the position AP'Q', the point O re-
maining always in the vertical line; and the time of an oscillation is
π
√(2+ √2)
✓
g
The lengths of simple pendulums which would oscillate re-
spectively in these times, would be
a
2- √2'
a
and
2+ √2
>
or 1.707 a and .293 a.
136
19
If neither of these arrangements exist originally, let ß, ß₁, be
the original values of y, y₁, when t is 0. Then making t=0 in
equations (5), we have
2
¹E = B + (√2 + 1) ẞ₁, ²E=ẞ−( √2 — 1) ẞ₁.
19
And these being known, we have the motion by equations (6).
1
3.
PROB. XI. Any number of material points P, P2 P .Q,
fig. 48, hang, by means of a string without weight, from a point A:
it is required to determine their small oscillations in a vertical plane.
Let AN be a vertical abscissa, and P¸M₁, P₂M₂, &c. horizontal
ordinates; so that
1
2
AM₁=x₁, AM₂=x2, AM₂=x3, &c.
1
AM½
P₁M₁=y₁, P₂M₂ =Y2, P3M =Y3, &c.
2 2
AP₁=σ₁, P¸Ð½=ɑë, P₂P¸=a3, &c.
1
2
tension of AP₁=P₁, of P₁P½=ν‚ of P2P3=P3,
=P2, &c.
mass of P₁m,, of P₂m, of P3=m3, &c.
Hence, we have these equations, by resolving the forces parallel
to the horizon,
Pig Y₁ 1 P2g Y2 - Yı
J² Y 1
+
d t²
1
m₁ ai
m₁
A2
ď² y 2
Pag Y2 - Y1
P3g Y3 - Y2
+
d t2
M2
a2
m2
аз
.(1).
ď² y 3
P3g Y3-Y2
P4g Y4-Y3
+
d t²
2
m3
ძვ
M3
a4
ď² y n
Png Yn Yn-1
d t²
mn
an
And, as in the last problem, it will appear that P1, P2, &c. may,
for these small oscillations, be considered as constant, and the same
as in the state of rest. Hence, if m₁ + m₂+ m3...+m₁₂ = M,
1
P₁ = M, P₂ = M - m₁, P3M-m¸-m₂, &c.
1
Also dividing by g, and arranging the above equations, may be
put in this form
137
d2
ď² yı
g d t
P1
P2
(P₁₂+ Pa)

P 2Y 2
1
Y₂+
m₁ az
ď² y z
Y 2
P2Y 1
P2
P3
+
Y2+
P3Y3
gd ť²
mą aq
mz az
m2a3
m2 az
..(2).
ď Y 3
P3 Y 2
Pa
P4
+
gd t²
mz az
mz az
M3
m3 a4
:).
P4 Y 4
Y3 +
mz as
ď² y n
Pn Yn - 1
Pn Yn
mn an
mn Un
gdť
The first and last equations become symmetrical with the rest if we
observe that
Yo
= 0, and P₂+1
n = 0.
Now, if we multiply these equations respectively by 1, λ X', X”,
&c. and add them, we have d'yı +λdy + X'dys + &c.
1
3
2
g dť²
={-
P1
m₁ a₁
Pe
λpg
+
Y₁
m₁ az
+ { P
+
m₁ ac
Pa
λ
λP3
P3
λ
то аз
(P² + P) + XP² } y
a
Mz Az
+
P4
m3 a4
-) +
+
Y3
m4a4
P3
Yo
mz az
+
(n-3)
Pn
λ'
mn - 1 an
/ (n-2)
Pn
2) Pal
Mn an
Syn⋅
And this will be integrable, if the right hand side of the equation
be reducible to this form
k (y₁+λ y₂+λ y3+&c.).
S
138
That is, if
h=
Pi P2
XP2
+
m₁ai m₁a2
m2 a 2
P2
Κλ
kλ =
m1a2
+>(
P2
+ P3)
XP3
m2 a2
m2 a 3
mz az
⚫(3).
kλ=
Xp3
m2 a 3
+x'(
P3
P4
+
+
mz az
Mza4
та
m4a4
(n-3) Pn
λ'(n-2)
Pn
+
mn-1 an
mn an
kλ(n-2)=
3
If we now eliminate X, X, X", &c. from these n equations, it is
easily seen that we shall have an equation of n dimensions in k.
Let¹k, 2k, ³k. ."k be the n values of k; then, for each of these there
is a value of X', X", X"" easily deduced from equations (3), which
we may represent by ¹λ, ¹x', ¹x", ¹\"", &c., ²X′, X", "X""', &c.
Hence, we have these equations, by taking corresponding values of
λ and k,
ď³y₂+¹λ ď² y₂+¹X'ď³y%+&c.
2
2
gdt
2
3
d³y₁+²λď³y½
'2+ ²X"d² y z + &c.
1
gdt
and so on, making n equations.
=
¹k. (Y₁+¹λy₂+¹X'y₂+&c.)
3
- ²k (y₂+²λY₂+²X′y3+&c.)
Integrating each of these equations we get, as in last problem
Y₁ +¹λY₂+'\'Yç +&c. =¹E cos. {t√('kg)+'
1
3
2
€
Y₁+λy₂+X'¥3+&c. = E cos. {t V(kg)+²e} { • • • •(5).
2
&c.
1 2
&c.
¹E, 'E, &c. ¹e, e, &c. being arbitrary constants.
From these n simple equations, we can, without difficulty,
obtain the n quantities y₁, y2, &c. And it is manifest that the
results will be of this form
y₁=¹H₁ cos. {t√('kg)+'e} +°H¸ cos. {t √(kg)+³e} +&c.
1
y₂='H₂ cos. {t √('kg)+'e} +²H₂ cos. {t √(kg)+'e}+&c.
2
2
&c. &c.
2
(6),
139
1
where 'H₁, ¹H₂, &c. must be deduced from B1, B2, &c. the original
values of y₁, y2, &c.
If the points have no initial velocities, (i. e. when t=0) we shall
have 'e=0,e=0, &c.
1
2
2
3
We may have symmetrical oscillations in the following manner.
If, of the quantities ¹E, E, ³E, &c. all vanish except one, for
instance, "E; we have
Y₁ + ¹λY ½ + ¹λ'Yç +&c. =0,
1
Y ₁ + ² λ Y ½ + ² X Y 3 +&c. =0,
1
Y₁ +³ λ Y½ + ³ X' Yç
2
+&c. =0,
(8).
Y ₁ + "λ Y½ +” X Y ç+&c.="E cos. t V(kg), omitting
2
1
2
n
3
From the first n-1 of these equations, it appears that y₂, ya'
Y2, Y3'
&c. are in a given ratio to y; and hence, y₁+"y₂ + "Xys + &c.
is a given multiple of y, and=my, suppose. Hence, we have
my₁ = "E cos. V("kg);
or, omitting the index n, which is now unnecessary,
my₁ = E cos. t V(kg). Also, if y2 = €2Y1,
my₂ = Ee, cos. t √(kg), and similarly for y, &c.
Hence, it appears that in this case the oscillations are symmetrical.
All the points come into the vertical line at the same time, and
move similarly and contemporaneously on the two sides of it. The
relation among the original ordinates ẞ₁, ß, ß, &c. which must
subsist in order that the oscillations may be of this kind, is given
by the n-1 equations (8),
ẞ₁ + ¹λß₂ +¹λ'ßg + &c. =0,
ẞ₁ + ²λ ß₂+ ²X² ß;+&c.=0,
B₁ + ³λ ß₂ + ³ X² ß₂+&c. =0,
19
&c. &c.
These give the proportion of ẞ₁, B, &c.; the arbitrary constant
"E, in the remaining equation, gives the actual quantity of the
original displacement.
140
2
3
Also, we may take any one of the quantities 'E, E, E, &c.
for that which does not vanish; and hence obtain, in a different
way, such a system of n-1 equations as has just been described.
Hence, there are n different relations among B₁, ẞ₂, &c., or n
different modes of arrangement, in which the points may be placed,
so as to oscillate symmetrically*.
19
The time of oscillation in each of these arrangements is easily
known; the equation
my₁ = "E cos. t √("kg),
shews that an oscillation employs a time t =
π
√("kg)
And
hence, if all the roots ¹k, 2k, 3k, &c. be different, the time is
different for each different arrangement.
If the initial arrangement of the points be different from all those
thus obtained, the oscillations of the pendulum may always be con-
sidered as compounded of n symmetrical oscillations. That is, if
an imaginary pendulum oscillate symmetrically about the vertical
π
line in a time (kg); and a second imaginary pendulum oscillate
√(kg)
about the place of the first, considered as a fixed line, in the time
π and a third about the second, in the same manner, in the
√ ´(kg) '
time
П
√ (³kg)'
; and so on; the nth pendulum may always be made
to coincide perpetually with the real pendulum, by properly adjust-
ing the amplitudes of the imaginary oscillations. This appears by
considering the equations (6),
*We might here also find these positions, which give symmetrical oscil-
lations, by requiring the force in each of the ordinates P,M,, P,M,, to be as
the distance; in which case the points P₁, P₂, &c. would all come to the
vertical at the same time.
If the quantities ✔'k, √²k, &c. have one common measure, there will
be a time after which the pendulum will come into its original position. And
it will describe similar successive cycles of vibrations. If these quantities
be not commensurable, no portion of its motion will be similar to any pre-
ceding portion.
141
y₁ = 'H₁ cos. t √(kg) + ºH, cos. t√(kg) + &c.
&c. = &c.
1
This principle of the co-existence of vibrations is applicable in
all cases where the vibrations are indefinitely small. In all such
cases each set of symmetrical vibrations takes place, and affects the
system as if that were the only motion which it experienced.
A familiar instance of this principle is seen in the manner in
which the circular vibrations, produced by dropping stones into still
water, spread from their respective centres, and cross without
disfiguring each other.
If the oscillations be not all made in one vertical plane, we
may take a horizontal ordinate z perpendicular to y. The oscilla-
tions in the direction of y will be the same as before, and there will
be similar results obtained with respect to the oscillations in the
direction of z.
We have supposed that the motion in the direction of x, the
vertical axis, may be neglected, which is true when the oscillations
are very small.
Ex. Let there be three bodies all equal; (each=m,) and also
their distances a₁, a2, aз, all equal; (each = a).
Here p₁ = 3m, p2 = 2m, p3=m, and equations (3) become
5 — 2λ,
ak
αλλ
akλ =
2 + 3λ
λ',
αλλ' = -
- λ.
λ + .
Eliminating k, we have
50 2)፡
5λ — 2λ² =
2 + 3λ − x',
50' - 2λλ' = -
λ
+ λ,
or,
λ = 2x² - 2λ
-
2,
λ
-
-
-
ร
20 - 4
4λ' αλλ' = - λ, λ' =
…. (እº Ωλ — 2) (2λ-4) = λ,
or, λ³-3x²
3)።
+ λ + 2 = 0,
which may be solved by Trigonometrical Tables. We shall find
three values of λ.
142
Hence, we have a value of X' corresponding to each value of X;
and then by equations (8),
ΟΙ
B₁ + ¹λß₂ + ¹λ' ß3 = or
B₁ + ²λß₂ + ²X² ß₁ = 0)
Whence we find ẞ₂ and ẞ, in terms of ß₁.
We shall thus find*
3
B₂ = 2.295 B₁,
2
or B₂ = 1.348 B₁,
or B₂
.643 B1,
according as we take the different values of X.
•.(8′).
And the times of oscillation in each case will be found by taking
the value of ak 5 2λ; that value of A being taken which is
not used in equation (8′). For the time of oscillation will be
given by making t √(kg) = π.
19
If the values of B1, B2, B3 have not this initial relation, the
oscillations will be compounded in a manner similar to that de-
scribed, p. 135, in the Example for two bodies only.
PROB. XII. A flexible chain, of uniform thickness, hangs
from a fixed point: to find its initial form, that its small oscilla-
tions may be symmetrical.
Let, in fig. 49, AM, the vertical abscissa = x; MP, the hori-
zontal ordinate = y; APs, and the whole length AC = a;
.. AP—a— s.
s. And, in the same way as in Prob. XI, the tension
at P will, when the oscillations are small, be the weight of PC,
and may be represented by a-s. This tension will act in the
direction of a tangent at P, and hence, the part of it in the direction
PM will be tension x
dy
ds
>
or (a − s)
dy
ds
Now, if we take any portion PQ=h, we shall find the hori-
zontal force at Q in the same manner. For the point Q, supposing
d's constant,
2
3
dy
dy
d² y
h
ď³ y
h²
becomes
+
+
+ &c.
d s
ds
d s²
1
d s³
1.2
*Euler, Com. Petrop. tom. VIII, p. 37.
143
Also, the tension will be (a-s-h). Hence, the horizontal force
in direction NQ, is
(a-s-h) (dy
d² y
h
d³ y h²
+
+
d s²
2
ds 1
ds³ 1.2
Mia, &c.)
Subtracting from this the force in PM, we have the force on
PQ horizontally
3
dy
h
= (a− s)
+
d³ y
h2
ds²
1
d s3
1.2
+&c.)
h
- h
d s
(dy + d²y 4+ &c.)
ds
And the mass of PQ being represented by h, the accelerating force
(=
pressure.g
mass
is found. But since the different points of PQ
is found.
move with different velocities, this expression is only applicable
when his indefinitely small. Hence, supposing Q to approach to,
and coincide with P, we have, when h vanishes,
d² y
dy
accelerating force on P = (a− s)
ds2
ds
But, since the oscillations are indefinitely small, a coincides
with s, and we have
ď² Y
dy
accelerating force on P (a- x)
dx2
dx
Now, in order that the oscillations may be symmetrical, this
force must be in the direction PM, and proportional to PM, in
which case all the points of AC will come to the vertical AB at
once. Hence, we must have
ď y
dy
(a-x)
dx²
dx
ky........(1),
k being some constant quantity to be determined.
This equation cannot be integrated in finite terms.
a series, let
To obtain
144
y = A + B (a − x) + C (a − x)² + D (a− x)³ + &c.
dy
d x
B − 2 C (a − x) — 3 D (a − x)² + &c.
1.2C+2.3 D (a-x) + &c.
ď y
dx²
2
ď y
dy
Hence, O=(a− x)
+ky; gives
dx2
dx
0 = 1.2 C (a−x) + 2.3 D (a− x)² + &c.
+B+2C (ax) + 3D (ax)² + &c.
+kA+kB (a-x)+kC (a− x)² + &c.
Equating coefficients, we have
B=-kA, 22. C kB, 3D = kC, &c.
=
-
2
C=*A
-
3
k² A
k³ A
.. B = − k A,
D =
&c.
24.32,
k²
and y = A {1−k (a−x) + ₂ē (a-x)2
3
kc³
2°32
(a−x)³ +&c.}..(2).
Here A is BC, the value of y when xa.
k² a²
ka
k³ a³
3 3
When x=0, y=0;.'. 1-ka +
+&c.=0..(3).
22
22 32
From this equation k is to be found. The equation has an infinite
number of dimensions, and hence k will have an infinite number of
values, which we may call 'k, 2k, 3k... "k,... ; and these give an
infinite number of initial forms, for which the chain may perform
symmetrical oscillations.
The time of oscillation for each of these forms will be found
thus. At the distance y, the force is kgy; hence, by Chap. I,
Ex. 1.
π
time to the vertical =
2 √ (kg)
; and time of oscillation
7
√(kg)
* The greatest value of ka is about 1.44, (Euler, Com. Acad. Petrop.
tom. VIII, p. 43.). And the time of oscillation for this value, is the same
as that of a simple pendulum, whose length is a, nearly.
145
The points where the curve cuts the axis will be found by
putting y=0. Hence, taking the value "k of k, we have
n k² (a - x)²
nk³ (a-x)²
0 = 1-"k (a− x) +
+ &c.
22
22.32
which will manifestly be verified, if
1
nk =
"k (a− x) =¹ka, or "k (a− x) = 2ka, or "k (ax) — ³ka, &c.
2
because ¹ka, ka, &c. are roots of equation (3).
That is, if
x = u
(1 - 1 1/2),
1 k
or = a
(1)
or = a
(1 - 2/42)
3 k
"k.
&c.
>
1
n k
าย
Suppose ¹k, 2k, 3k, &c. to be the roots in the order of their
magnitude, ¹k being the least.
1
1
Then, if for "k we take 'k, all these values of r will be ne-
gative, and the curve will never cut the vertical axis below A.
If
If for "k we take 2k, all the values of x will be negative
except the first; therefore, the curve will cut AB in one point.
we take ³k, all the values will be negative except the two first,
and the curve cuts AB in two points; and so on.
3
Hence, the forms for which the oscillations will be symmetrical,
are of the kind represented in fig. 50. And there are an infinite
number of them, each cutting the axis in a different number of
points.
If we represent equation (2) in this manner, y=Ap (k, x);
it is evident that y=¹Ap (¹k, x), y=² A¢ (² k, x), &c. will each
satisfy equation (1). Hence, as in Prob. XI, if we put
1
a
y = ¹Ap ('k, x) + ² A $ (² k, x) + &c.
and if ¹A, A, &c. can be so assumed, that this shall represent
a given initial form of the chain, its oscillations will be compounded
of as many co-existing symmetrical ones, as there are terms ¹A,
A, &c.
T
1
146
SECT. IV. The Motion of Bodies connected by Strings.
PROB. XIII. Two bodies, connected by a string passing over
a fixed pully, move on two given curves: to determine their motions.
The motions are supposed to take place in a vertical plane.
Let A, fig. 51, be the pully, AM a vertical line, BP, CQ,
the given curves, and MP, NQ horizontal ordinates to the places
of the bodies.
AM=x, MP=y, AN = x', NQ=y'; AP=r, AQ=r',
curve BP=s, CQ=s': mass of P=m, of Q = m'; re-action of
surface BP=n, of CQ = n'; tension of the string PAQ = p.
Then, by resolving the forces which act upon each of the
bodies, we shall easily find the equations
d² x
d to = g
pg x
ng du đ y
pg y ng dx
;
+
m
r
m
ds' d t²
m
m
ds
·(1).
d² x'
dt2
p'g x' n'g dy' d'y'
p'g y'
n'g dx'
g
+
m r
m
ds'' dt²
2
m
r m' `ds'
Multiply by 2 mdx, 2mdy, 2 m'dx', 2 m'dy', and add the
two upper ones, and the two lower ones; and we have
2 m (dx d²x + dy d³y)
d t²
x dx + y dy
= 2mgdx
2pg.
T
-
d t²
2 m' (dx'd²x' + dyd³y')
2
2mg dx' - 2pg
2
2
Add these equations, observing that ²= x²+y², whence
x'dx' + y'dy'
x
x d x + y dy
x'd x' + y'dy'
dr =
; similarly, dr' =
gol
γ
and rr a constant length APC; .. dr + dr'=0. Thus we
have
=
dx d² x + d y d³ y
2m.
+2m'.
dt2
dx'd²x' + dy' d³y'
dt²
=2g (mdx+m'dx').
147
Integrating, we have
m
dx² + dy²
d t²
dx² + dy"
12
+ m'
= 2g (mx + m'x') + C.
dt2
If we suppose a, a', to be the values of x, x', when the velo-
cities are 0, we shall have
ds²
m dt²
+ m'
ds'2
d t²
. − .
= m.2 g (x − a) — m' . 2 g (a′ — x'). . . . (2).
The quantities x-a, a'x' are the spaces through which P
has descended and Q ascended: 2 g (x − a), 2 g (a′ — x′) are the
squares of the velocities which would have been generated in
falling freely through these spaces.
The product of the mass and square of the velocity is some-
times called the vis viva of a body: hence the equation just found,
shews that in our problem.
The sum of the vis viva of the bodies in constrained motion, is
the sum or difference of the vis viva which they would have had, if
they had descended freely through the same vertical spaces *.
The sum if both descend the difference if one ascend.
By introducing into equation (2), the relations among x, x', s, s',
given by the nature of the curve, and by the condition r + r' = a
constant quantity, we have the equations which determine the
motion.
Ex. Let Q, hanging freely, draw P along a horizontal plane,
fig. 52.
Let the original position of Q, when the bodies begin
to move, be D; AD=a, AB= c; length of string PAQ=1:
AP=r, AQ=1-r.
dr
Velocity of Q -
; and BP = √(r² — c²) ;
dt
dr
.. velocity of P=
ተ
–
√(r² - c²) ・ dt
* This is the principle of the conservation of vis viva, which applies not
only to this problem but to all cases whatever of mechanical action.
148
P neither ascends nor descends: hence, the equation (2) gives
22
dr²
dr²
m.
p²
+ m'.
d t²
d t²
2
= 2 m' g (l — r — a),
dr²
d t²
2 m' g (l — a − r) (p² — c²)
(m + m') r² – m'c²
which gives the relation between r and t.
PROB. XIV. A body P, fig. 53, is fastened to two equal
weights Q, Q', by strings passing over pullies A, A', equidistant
from it, and in the same horizontal line: to determine its motion.
The vertical line PE will bisect AA', and the body being acted
upon by equal forces on the two sides of the vertical line, will not
be drawn from it, and we shall only have the vertical motion to
consider.
Let AE= EA'=a, EP=x; mass of P=m, of Q = Q′ =m':
tension of PA, or PA'=p; AP=AP'=r: hence, the accele-
rating force which each string exerts vertically upon P will be
pg x
m
d² x
dr² =g-
dt
2
;
therefore
2pg Ꮖ
from the motion of P
m
. (1);
d²r
d t2
pg
=g
from the motion of Q
m
2 mdx d²x+4 m' dr d²r
x²=a²,
(for since r² — x² = a², rdr-xdx=0.)
Integrating,
d t²
2
= 2 mg dx — 4 m'g dr:
1
dx
dr²
dr2
m +2m'
d t²
dt2
=2mg (x-b) - 4 mg (r−c)....(2).
Supposing that when the velocities are 0, x=b=EB, r=c=AB.
Hence, c² a² + b².
149
But, since r√(a²+x²), dr=
{m
m +
dx²
=
2 m²x²) dx²
a² + x²) dť²
2
2 dt 2
x²
a² + x
d t² ma² + (m + 2 m') x
x d x
Va²+x); this becomes
√(a²
2
= 2 mg (x − b) — 4 m'g (r−c),
2
·
2 g {m (x − b) — 2 m′ (r — c)} .
-
To find when the velocity again becomes 0, we must have
m(x-b) — 2 m' (r−c) = 0;
or, putting for r its value √(a²+x²), transposing and squaring,
(4 m′² — m³) x³ — 2 m (2 m' c — mb) x + 4 m² a² - (2 m'c — mb)² = 0,
—
the two values of x in this equation give the points where the ve-
locity is 0; hence, b is one of them; and if b' be the other, we
shall have
b+b' =
2m (2 m'c - mb)
4 m² m²
4 mm'c — (4 m²² +m²) b
b =
4 m²²
m²
2
If 2 m'>m, make EB'b', and P will go on oscillating
between B and B'. If 2 m' <m, b' will be negative, and P will
never come to a second point where its velocity is 0;
The velocity of P will be greatest when it is in the position in
which there would be an equilibrium: for then the force is 0,
and therefore
ď² x
= 0.
d t°
m a
That is, when x =
√ (4m²² — m²)
>
see Statics, Chap. IV,
Prob. II.
PROB. XV. A weight Q draws a weight P over a fixed pully
A, fig. 54, P in the mean time making small oscillations: to de-
termine the motion.
150
When the oscillations are very small, we may here, as in
Prob. XI, suppose the tension to be the same as if P did not
oscillate and it will thus be found.
Let m, m' be the masses of P, Q; then m―m is the mass em-
ployed in producing motion; and m+m the mass moved: hence,
the accelerating force on Q is
m'
m
g
m' + m
But the force on Q downwards is manifestly the excess of its
weight downwards, above the tension which acts upwards. Hence,
if p be the tension,
pg
g
M
m' — m
m' + m
2m' m
g; p=
m² + m
Let AM vertical=x, MP horizontal=y, AP=r, AQ=l—r;
ď² y
pg y
2 m'
m g y
.(1).
m² + m r
d t
m
CASE 1. When m'=m, accelerating force on Q=0;
is constant.
dr
Let
dt
-
b, P being supposed to ascend; .. dt =
and equation (1) becomes
b² d² y
dr²
y
ky
or
"
k being constant.
r
dr²
r
..
dr
dt
d r
b
d² y
Let the integral of this be y = Ar+Br²+Cr³+&c.
= 1.2.B+2.3 Cr+ &c.
dr²
= 0;
ky
+
kA+kBr+&c.
Τ
k A
k² A
whence B:
D=
2
22.3
k Ꭺ
22.32.4
&c.
kr²
k² 3
2
r³
3
k³ 24
..y = A { r·
y=A {r•
+
2
223
22 32 4
+ &c.}.
J
151
By making y=0, we have an equation of an infinite number of
dimensions; shewing that the curve described by P cuts the axis in
an infinite number of points.
In order to determine the points where y is a maximum, we
must have
k2
2.2
3
k³ 4.3
dy
0; or 1-kr +
+&c.=0,
22
22 32
d r
(agreeing with equation (3), Prob. XII,) which gives the points
of extreme deviation of the pendulum from the vertical. The times
of successive oscillations will be as the differences of the successive
values of r, because r diminishes uniformly *.
CASE 2. When m, m' are unequal.
d² r
m
m
dr
m'
m
Here
g;
gt; velocity being
d t²
m² + m
d t
m' + m
O when t is 0,
m' — m
m gt2
m'
-
r=a
-
m' +m
=a-ngt²; making
m² + m
m
= n;
2 m²'
whence
=1+1.
m' + m
And equation (1) becomes
ď y
(1 + n) gy
;
d t²
a ng
whence y may be found by series in terms of t.
* This problem has been differently and erroneously solved by some
authors. No solution but an approximate one is attainable. Euler, Com.
Petrop. tom. VIII, p. 137, &c. obtains an equation not integrable, and then
observes, “Ita ut determinationem hujus motus oscillatorii, quo corpora A
et B cientur, dum filum super trochleam uniformiter promovetur, pro
casu desparato declarare simus coacti."
The equation in Case 2, is integrable for some particular values of P and
Q: for instance, if Q=3 P.
1
CHAP. VII.
INVERSE PROBLEMS RESPECTING THE MOTION OF POINTS
ON CURVES.
IN the fifth Chapter we supposed a body to move on a curve,
the curve being given, and the motion being the thing to be deter-
mined. In the present one we shall collect several questions which
have occupied the attention of mathematicians, in which some
property or consequence of the body's motion is given, and the
curve is required.
SECT. I. Curve of equal Pressure.
43. PROP. To find the curve on which a body, descending
by the force of gravity, presses equally at all points.
Let AM, fig. 55, be the vertical abscissa =x, MP the hori-
zontal ordinatey; the arc of the curve s, the time t, and the
radius of curvature at P = p, p being positive when the curve is
concave to the axis; then, R being the re-action at P, we have by
Art. 33, page 93,
R =
gdy
+
ds pdt
ds²
2
.(1).
But if HM be the height due to the velocity at P, AH = h, we
have, by Art. 31,
ds2
2g (h-x).
dť²
dsdx
Also, if we suppose ds constant, we have p
; and if
day
the constant value of R be k, equation (1) becomes
153
ค
k
dx
gdy _ 2g (h− x) dy
k
dsdx
ds
7
ď y
;
dx
ds
2
(a) = √(h-x). -dy. -1)
g 2 √(h− x)
ds
-
x)
The right hand side is obviously the differential of √(h − x) .
~.√(h− x) = √(h− x).
hence, integrating,
k
g
dy
k
C
√(h-x)
dy
+ C,
ds
.(2).
dy
;
ds
==
ds g
If C = 0, the curve becomes a straight line inclined to the
horizon, which obviously answers the condition. The sine of
inclination is
k
g
In other cases the curve is found by equation (2), putting
√(dr²+dy²) for ds, and integrating.
If we differentiate equation (2), ds being constant, we have
d²y
ds
Cdx
2 (h− x)
; p
dsdr 2 (h-x)
ď²y
C
.(3).
And if C be positive, p is positive, and the curve is concave to the
axis.
dy
ds
We have the curve parallel to the axis, as at C, when = 0,
k
that is, when
=
g
C
; when x = h
√(h− x)
C² g²
k
dy
When x increases beyond this, the curve approaches the axis, and
is negative; it can never become < −1; hence, the limit of
ds
k
C
as B, is found by making
- 1;
g
√(h-x)
C² g²
or, x = h
(k + g)
U
154
If kg, as the curve descends towards 2, it approximates per-
petually to the inclination, the sine of which is
k
If k >g,
g
fig. 56, there will be a point when the curve becomes horizontal as
at D, after which it will ascend in a form similar to the descending
branch.
C is known from equation (2) or (3), if we know the pressure
or the radius of curvature at a given point.
If C be negative, the curve is convex to the axis. In this
case the part of the pressure arising from centrifugal force diminishes
the part arising from gravity, and k must be less than g, fig, 57.
SECT. II. Synchronous Curves.
44. In fig. 58, let AP, AP', AP", &c. be curves of the same
kind, referred to a common base AD, and differing only in their
parameters*: a curve P, P', P', &c. cutting them, so that the arcs
AP, AP', AP", &c. may all be described in the same time, by a
body descending from A by gravity, is said to make them synchronous.
PROP. To find the curve which cuts a given assemblage of
curves, so as to make them synchronous.
Let AM vertical = x, MP horizontal=y; y and x being con-
nected by an equation involving a. The time down AP is
S
dx
√(2gx)'
the integral being taken from x=0 to x=AM; and
this must be the same for all curves whatever be a. Hence, we
may put
ds
=
k......(1),
√(2gx)
k being a constant quantity, and in differentiating, we must suppose
* Any constant line is called a parameter, which occurs in the equation
to a curve, and by its different values gives different magnitudes to corre-
sponding portions of the curve. Thus the radius of a circle, and the semi-
axis of a cycloid are parameters.
155
a variable as well as r and s. Let ds = pdx, p being a function
of x and a, which will be of O dimensions, because dr and ds are
pdx
Hence, S√(2g)
quantities of the same dimensions. Hence,
k, and
differentiating
pdx
V(2gx)
+qda = 0...
(2),
q being the differential coefficient of f√(2x)'
pdx
>
with respect to a.
S
pdx
√(2gx)
Now, since p is of O dimensions in x and a, it is easily seen that
is a function whose dimensions in r and a are,
because the dimensions of an expression are increased by 1 in
integrating. Hence, by a known property of homogeneous functions,
(see Lacroix, Elem. Treat. Art. 266,) we have
Ρ
√(2gx)
q
=
x + q. a = 1½ k;
k
p V x
a V(2g)'
2 a
substituting this in equation (2), it becomes
pdx
kda
pd a V x
+
V(2gx)
2 a
a √(2g)
= 0.... (3),
in which, if we put for a its value in x and y, we have an equation
to the curve PP'P".
If the given time k be the time of falling down a vertical height
h, we have k =
p (adx
V
g
>
and hence, equation (3) becomes
xda) + da V(h x) = 0........(4).
Ex. Let the curves AP, AP', AP" be all cycloids of which
the bases coincide with AD.
Let CD be the axis of any one of these cycloids 2a, a being
the radius of the generating circle. If CN=x', we shall have as
before
156
-
2a
ds = dx' √²
ds dx
=
✓
Hence, p =
✓
2 a
;
2 a
and since x′ = 2a
2a. X
2a-x
√(2a). (adx — xda)
√(2 a−x)
I
Let
a
stituting,
Ꮖ
X,
; and equation (4) becomes
+ da V(hx) = 0........(5).
=u, so that adx - xda a²du, x = au; and sub-
=
a² du V2
√(2—u)
+ da V(hau) = 0;
du V 2
da Vh
+
√(2u — u²)
a*
= 0;
2 Vh
= C.
Va
12. arc (ver. sin. u)
When a is infinite, the portion AP of the cycloid becomes a ver-
tical line, and x = h; .'. u = 0; .'. C=0.
Hence, 2
2h
=
a
ver. sin. V
a
..(7).
From this equation a should be eliminated by the equation to the
cycloid, which is
y = a
a. arc.
(ver
ver. sin. = -
-) - √(2ax − x³)..
√(2ax − x³). . . . . ...(8),
a
and we should have the equation to the curve required.
Substituting in (8) from (7), we have
y = √(2ah) - √(2ax — x²),
da Vh
dy =
√(2a)
xda+adx-xdx
√(2ax-x²)
;
157
!
and eliminating da by (5),
dy
dx
2α-x
√(2ax − x²)
√(2a - x)
√x
But differentiating (8) supposing a constant, we have in the cycloid
dy
dx
√x
V(2a-x)
And hence the curve P, P', P", cuts all the cycloids at right
angles *.
The curve PP'P" will meet AD in the point B, such that the
given time is that of describing the whole cycloid AB. It will meet
the vertical line in E, so that the body falls through AE in the
given time.
COR. If instead of supposing all the cycloids to meet in the
point A, we suppose them all to pass through any point C, fig. 65,
their bases still being in the same line AD; a curve PP' drawn so
that the times down PC, P'C, &c. are all equal, will cut all the
cycloids at right angles. This may easily be collected from the
preceding reasoning.
SECT. III.
Tautochronous Curves.
point be the same,
We shall consider
45. If a body move upon a curve, the curve is said to be
tautochronous, if the time of descent to a given
from whatever point the body begin to descend.
the body as descending to the lowest point.
PROP. When a body is acted upon by a constant force in
parallel lines, to find the tautochronous curve.
Let A, fig. 59, be the lowest point, D the point from which
the body falls, AB vertical, BD, MP horizontal. AM=x, AP =s,
AB=h, the constant force =
g.
Hence, the velocity at P= √{2g (h—x)},
ds
dt =
√ {2g (h — x)} '
* For the subnormal of the former coincides with the subtangent of the
latter, each being
y (2a-x)
√x
158
and the whole time of descent will be found by integrating this from
x=h, to x = 0.
Now, since the time is to be the same, from whatever point D
the body falls, that is, whatever be h, the integral just mentioned,
taken between the limits, must be independent of h. That is, if
we take the integral so as to vanish when x=0, and then put h for x,
h will disappear altogether from the result. This must manifestly
arise from its being possible to put the result in a form involving
Ꮖ
X
2
only, and functions of as &c.; that is, from its being
h
of O dimensions in r and h.
h2
Let ds=pdx, where p depends only upon the curve, and does
not involve h. Then, we have
t =
t=
=-√
1
pdx
✓ {2g (h− x)}
Spdx
1 pxdx
dx
+
+
2 ht
2.4 h&
1.3 pa'da + &c. };
√(2g). ht
and from what has been said, it is evident, that each of the quanti-
pdx p x x
ties fd, fd, and generally, must be of the
2n+1
h 2
h 2
2n+1
2
2n+1
сх
form
;
that is, Spa" dx must = cx 2
; hence, pxdx
2n+1 c
2n+1
cx
ds=dx
Va
Ꮖ
or, if
c = a*, p =
Q
ક
2 n + 1
2
2n-1
2
x dx; p=
which is the property of a cycloid*.
* Without expanding, we may reason thus. If p be a function of m
Ρ
dimensions in æ, √(h—x) is of m−1 dimensions; and as the dimensions of
pdx
an expression are increased by 1 in integrating,h-x)
sions in a, and when his put for x, of m+
to be independent of h, or of O dimensions.
Therefore p = a² X as before.
is of m+dimen-
dimensions in h. But it ought
Hence, m + 4 = 0, m =
159
46. PROP. When the body is acted upon by a force tending to
a centre, and varying as any function of the distance, to find the
tautochronous curve.
Let
Let S, fig. 60, be the centre of force, A the point to which the
body must descend; D the point from which it descends.
SA=e, SD=ƒ, SP=r, P being any point, AP=s.
=
Now we have velocity2 C 2f Pdr, (Art. 31. p. 82.),
or, if 2f Pdr = $(r), velocity² = $ (ƒ)
velocity² =0, when r=f.
t =
=S ·
ds
• (ƒ) − $ (r); because the
Hence, the time of describing DA is
"
taken from r = ƒ, to r = e. And
whatever is D, the integral so
Let p (r) —$ (e) =z, $ (ƒ)~
√ {p (ƒ)~$ (r)}
since the time must be the same
taken must be independent of f.
(e)=h, ds=pdz, p depending only on the nature of the curve,
and not involving f. Then
t =
-S
pdz
√(ḥ − z)'
pdz
taken from z = h, to z = 0
S√(h-2)'
'(h — z) '
from z=0, to z=h.
And this must be independent of f, and therefore of
(f), and of h.
Hence, after taking the integral, the result must be 0, when z=0,
and when his put for z, must be independent of h. Therefore it
must be of O dimensions in z and h. But if p be of n dimensions
p
√(hz)
will be of n-dimensions, and
in z, or if p = c z",
pdz
Sh
V(h
p = √
Z
2)
whence the curve is known.
If v be the angle ASP, we have
of n + Hence, n = 0, n =
+ ½
- ½, and
Therefore ds=dz
√ ÷ =ørar√
C
Z
$(r) − $ (e)
ds² - dr²
ds² = dr²+r²dv², dv² =
2
whence we may find a polar equation to the curve.
160
Ex. 1. Let the force vary as the distance, and be attractive.
Here P = mr, (r) = mr² ;
&
= $(r) — $ (e) = m (r² — e²); dz = 2mrdr,
2 =
ds=dz
אן.
=2 mr dr
✓
C
=rdr
2
ช
m (p² — e²)
√
4cm
p² - e²
ds
When re,
is infinite, or the curve is perpendicular to SA at A.
dr
If SY, perpendicular upon the tangent PY, be called p, we
have
2
ds² - dr²
dr²
p²
e
: 1
1
ds2
ds²
27
4 cm r
p²
e² − (1 − 4cm) p²
4cm
If e = 0, or the body descend to the centre, this gives the
logarithmic spiral.
In other cases let 14cm =
e² (a
2
2
(a² — r²)
e2
;
a
a² — e²
..4cm =
2
a
a²
2
the equation to a hypocycloid, see p. 87.
If 4 cm = 1, the curve becomes a straight line, to which SA is
perpendicular at A.
If 4 cm> 1, the curve will be concave to the centre, and
will go off to infinity.
Ex. 2. Let the force vary inversely as the square of the dis-
tance.
27
m
P =
p² = 1,2
and as before, we shall find
2.5 (7 — e)
(r
2 m ce
161
SECT. IV. Brachystochronous Curves.
47. The brachystochron is the curve, down which a body must
descend from one point to another, so that the time of descent
may be less than that down any other curve, under the same cir-
cumstances.
PROP. A body being acted upon by a force in parallel
lines, in its descent from one point to another; to find the brachys-
tochron.
Let A, B, fig. 61, be the given points, and AOPQB the
required curve. Since the time down AOPQB is less than down
any other curve, if we take another curve AOpQB, which coincides
with the former, except for the arc OPQ, we shall have
time down 40+time down OPQ+time down QB, less than
time down AO+ time down Op Q+ time down QB:
and if the times down QB be the same on the two suppositions,
we shall have
time down OPQ less than time down any other arc Op Q.
The times down QB will be the same in the two cases, if the
velocity at Q be the same. But it has been seen, (Art. 31,) that
the velocity acquired at Q is the same, whether the body descend
down AOPQ, or AOpQ*. Hence, it appears that if the time down
AOPQB be a minimum, the time down any portion OPQ is also
a minimum.
Let a vertical line of abscissas be taken in the direction of the
force, and perpendicular ordinates OL, PM, QN be drawn, it
being supposed that LM = MN. Then, if LM, MN be taken
indefinitely small, we may consider them as representing the dif-
ferential of r: on this supposition, OP, PQ, will represent the
differentials of the curve, and the velocity may be supposed constant
in OP, and in PQ. Let AL=x, LO=y, AO=s; and let dx, dy,
ds represent the differentials of the abscissa, ordinate and curve at
Q, and v the velocity there; and dx', dy', ds', v', be the corres-
* This is true, whenever the body descends in a non-resisting space, or
when the forces are necessarily the same in the same points.
X
162
ponding quantities at P. Hence, the time of describing OPQ
will be
d s
ds'
+ ;
O'
บ
which is a minimum; and consequently, its differential is equal 0.
This differential is that which arises from supposing P to assume
any position, as p, out of the curve OPQ; and, as the differentials
indicated by d arise from supposing P to vary its position along the
curve OPQ, we shall use & to indicate the differentiation, on hy-
pothesis of passing from one curve to another, or the variations of
the quantities to which it is prefixed. We shall also suppose p to
be in the line MP, so that dr is not supposed to vary.
considerations being introduced, we may proceed thus,
S
jds
ช
S ds'
+
v
=0........(1).
These
And v, are the same whether we take OPQ or OpQ; for the
velocity at p = velocity at P. Hence, dv=0, do=0: and
dv′=0:
dds dds
+
= 0.
ช
Now, ds2dx²+dy²; .. ds. dds=dy.ddy, because ddx=0.
Similarly, ds'.Sds' =dy'. Sdy'.
Substituting the values of 8ds, & ds' which these equations give,
we have
dy.ddy dy ddy'
+
ds.v
ds.v
= 0.
And since the points O, Q, remain fixed during the variation of
P's position, we have
dy+dy' = const. ddy'
Substituting, and omitting &dy,
dy dy
ds.v ds.v
= 0.
Sdy.
Or, since the two terms belong to the successive points O, P,
their difference will be the differential indicated by d; hence,
163
dy
dy
= constant..
d.
=
= 0;
ds.v
ds.v
..(2),
which is the property of the curve; and v being known in terms of
x, we may determine its nature.
COR. 1. Let the force be gravity: then = √(2 g x);
dy
d s V (2 g x)
dy
1
= constant,
d s V x
Vai
Va being a constant quantity;
dy
ds
a
which is a property of the cycloid, of which the axis is parallel to x,
and of which the base passes through the point from which the body
falls.
COR. 2. If the body fall from a given point to another given
point, setting off with the velocity acquired down a given height:
the curve of quickest descent is a cycloid, of which the base co-
incides with the horizontal line, from which the body acquires its
velocity.
48. PROP. If a body be acted on by gravity, the curve of its
quickest descent from a given point to a given curve, cuts the latter
at right angles.
Let A, fig. 63, be the given point, and BM the given curve;
AB the curve of quickest descent cuts BM at right angles.
It is manifest the curve AB must be a cycloid, for otherwise
a cycloid might be drawn from A to B, in which the descent would
be shorter. If possible, let AQ be the cycloid of quickest descent,
the angle AQB being acute. Draw another cycloid AP, and let
PP' be the curve which cuts AP, AQ, so as to make the arcs
AP, AP' synchronous. Then, by Art. 44, PP' is perpendicular
to AQ, and therefore manifestly P' is between A and Q, and the
time down AP is less than the time down AQ; therefore, this latter
is not the curve of quickest descent. Hence, if AQ be not per-
pendicular to BM, it is not the curve of quickest descent*.
* The cycloid which is perpendicular to BM may be the cycloid of
longest descent from A to BM.
164
7
49. PROP. If a body be acted on by gravity, and if AB, fig. 64,
be the curve of quickest descent from the curve AL to the point
B; AT, the tangent of AL at A, is parallel to BV, a perpen-
dicular to the curve AB at B.
If BV be not parallel to AT, draw BX parallel to AT, and
falling between BV and A. In the curve AL take a point a near
to A. Let a B be the cycloid of quickest descent from the point
a to the point B; and Bb being taken equal and parallel to a A,
let Ab be a cycloid equal and similar to a B. Since ABV is a
right angle, the curve BP, which cuts off AP synchronous to AB,
has BV for a tangent, (Art. 44.). Also, ultimately a coincides
with AT, and therefore Bb with BX. Hence, b is between A and
P. Hence, the time down Ab is less than the time down AP,
and therefore, than that down AB. And hence the time down a B
(which is the same as that down Ab,) is less than that down AB.
Hence, if BV be not parallel to AT, AB is not the line of quickest
descent from AL to B.
50. PROP. If a body acted on by gravity descend to a given
point C, fig. 65, setting off from a curve BM, with a velocity ac-
quired in falling from a given horizontal line AD, the curve of
quickest descent cuts the curve BM at right angles.
As before, BC the curve of quickest descent, will be a cycloid,
by Cor. 2 to Art. 47.
If possible, let QC be the cycloid of quickest descent, making
CQB an acute angle. By Cor. 2 to Art. 47, the base of this
cycloid will be in the horizontal line AD. Let OPC be another
cycloid, of which the base is in AD. And by Cor. to Art. 44, if
PP' cut off synchronous arcs PC, P'C, PP' will be perpendicular
to the curves PC, QC. Hence, P' will fall between Q and C, and
the time down PC, being equal to that down P'C, will be less
than that down QC. Hence, if QC be not perpendicular to QB,
it cannot be the curve of shortest descent.
From this it appears in what manner a cycloid must be drawn,
so that it may be the curve of quickest descent from one given
curve to another.
165
If the body descend from rest, from the curve BM, fig. 66, to
the curve CN, by the action of gravity, the curve of quickest de-
scent will be a cycloid, of which the base is the horizontal line BE,
which cuts CN at right angles, and which is so situated, that the
tangents to BM at B, and CN at C, are parallel.
If the body descend from the curve BM to the curve CN, the
velocity being that acquired in falling from the horizontal line AD,
the curve of quickest descent will be a cycloid, of which the base is
the horizontal line AD, and which cuts both the curves BM, CN,
at right angles.
For the brachystochron, when the length of the curve is given,
see Mr. Woodhouse's Isoperimetrical Problems, p. 122.
51. PROP. Supposing a body to be acted on by any forces
whatever, to determine the brachystochron.
Making the same notations and suppositions as before, AL,
LO, fig. 61, being any rectangular co-ordinates; since as before,
the time down OPQ is a minimum; we have, by equation (1) of
last Article,
б
{d
S
ds'
+
=0.
= 0....
..(1),
dds
Sds'
d s & v
ds' d v'
+
= 0.
V
V
Now we have as before dds =
21/2
で
​dy. Sdy
>
supposing & dr=0,
ds
dy'. Sdy
dy'. Sdy
Sds' =
ds
ds'
dv=0, for v is the velocity at O, and does not vary by altering
the curve.
v = v + d v ; . . S v = d v + & dv=ddv.
dv;
v′
dy.ddy
Hence,
ds. v
dy.Sdy
ds.v
ds'. d d v
= 0.
が
​12
1
1
d v
Also
v + d v
บ
V
22
; for dv, &c. must be omitted.
166
Substituting this in the second term, we have
dy. Sdy dy ddy dy.dv.ddy ds'.Sdv
+
0,
ds.v
ds.v
ds.v²
12
ข
or
(dy - dy)
dy'.dv
ds & dv
0.
+
ds.v²
12
V
Sdy
dý
dy
dy
Now as before
is d.
dy
And in the other terms we
ds' d s
d s
may, since O, P, are indefinitely near, put ds, dy, v, for ds', dy,
and multiply by -v, we have
dy.dv ds & dv
v': if we do this,
dy
d.
+
d s
ds ช
ช
•
v Sdy
= 0........(2),
which will give the nature of the curve.
If the forces which act on the body at O, be equivalent to X in
the direction of x, and Y in the direction of y, we have
vdo = Xd +Ydy, (Art. 31);
:. dv =
ddv =
Ꮖ
Xdr+Ydy
v
Yddy
V
because Sv=0, 8dx=0; also X and Y are functions of AL and
LO, and therefore not affected by d.
Substituting these values in the equation to the curve, we have
dy
dy Xdx + Ydy
ds Y
+
0;
d.
ds ds
2
ข
v
dy
dx Xdy-Ydx
= 0,
or d.
ds
ds
v2
which will give the nature of the curve.
COR. 1. Ifp be the radius of curvature, and ds constant, we have
ds dx
dεy
:p being positive when the curve is convex to AM;
167
The quantity
d.
dy
ds
22
P
22
P
11
dx
P
; and hence,
Xdu– Ydr
dy
ds
is the centrifugal force, and therefore that part
of the pressure which arises from it. And
Xdy – Ydr
d s
is the
pressure which arises from resolving the forces perpendicular to the
axis. Hence, it appears then in the brachystochron for any given
forces, the parts of the pressure which arise from the given forces,
and from the centrifugal force, must be equal.
COR. 2. If we suppose the force to tend to a centre S, fig. 62,
which may be assumed to be in the line AM, and P to be the whole
force; also SA = a, SP=r, SY, perpendicular on the tangent
PY=p; we have
Xdy – Yd x
ds
= force in PS, resolved parallel to YS = P
v² = C− 2ƒP dr, (Art. 31);
C-2fPdr _ Pp
P
=
P
r
rdr
Also p=
;
dp
Ppdr
.. C-2f Pdr =
dp
2dp
-2 Pdr
and integrating,
Ρ
C-2/Pdr'
C'
p²=C′ { C− 2 ƒ Pdr}; whence the relation of p and r is known.
If the body begin to descend from A, C-2ƒPdr must = 0
when ra.
Ex. 1.
Let the force vary directly as the distance.
168
P=mr, C−2f Pdr = velocity² = m (a² — r²), p² = C′ m (a² — p²),
which agrees with the equation to the hypocycloid, p. 87.
Ex. 2. Let the force vary inversely as the square of the dis-
tance,
m
C-2f Pdr =
2. ՊՆ
2 m
P =
2 J
ጥ
a
2m C a- p
a
r
2 m C
p² =
= c².
2
putting c² =
a
r
r
a
r² — p²
3
+
c² r - c²
c² a
ጥ
Ꮎ
de =
pdr
r √ (2.² — p²)
c dr
3
c √ (a− r). dr
2
√ (r³ + c³r− c² a)
"'
2
When ra, de=0: when r³ +c²r− c²a=0, de is infinite,
and the curve is perpendicular to the radius as at B. This equation
has only one root.
a
If we have c = -
2, SB =
a
B being an apse,
if c =
10:
α
a
SB=
a
a
if c=
SB= >
>
30
10
a
a
if c =
SB
3
2
n° + n
n² + 1°
52. PROP. When a body moves on a given surface, to deter-
mine the brachystochron.
Let x, y, z be rectangular co-ordinates, x being vertical; and
as before, let ds, ds' be two successive elements of the curve: and
let dx, dy, dz; dx', dy', dz' be the corresponding elements of
x, y, z; then, since the minimum property will be true of the
indefinitely small portion of the curve, we have, as before, sup-
posing v, v the velocities,
X,
169
ds
V
+
d s'
V
min.
Jd s
d s
+
= 0 0..
·(1).
V
The variations indicated by & are those which arise, supposing
dx, dr' to be equal and constant, and dy, dz, dy' and dź to vary.
Now,
ds² = dx² + dy²+dz²; .. ds dds =dy ddy +dz 8dz,
similarly,
Also, the extremities of the
dy+ dy' = const.
:. Sdy + Sdy' = 0;
ds' Eds' = dy ddy'+dz'ddz.
arc ds+ds being fixed, we have
dz + dz'const.
Sdz + 8d z = 0.
Hence, Sds =
ds
dy ddy +
dz
8dz
ds
Sds'
dyddy - 18d=
dz'
Sd z
ds
(2).
And the surface is defined by an equation between x, y, z, which
we may call L=0. Let this, differentiated, give
dz = pdx + qdy
Hence, since dx, p, q are not affected by d,
S d z = q Sd y
q8dy.
··(3).
..(4).
For the sake of simplicity, we will suppose the body to be
acted on only by a force in the direction of x, so that v, v' will
depend on a alone, and will not be affected by the variation of
dy, dz. Hence, we have by (1),
dds
V
+
I dy
v'ds'
dds
70
= 0; which, by substituting from (2), becomes,
dyl
vds.
dz'
Sdy +
v'ds'
dz
vds
S d z = 0.
Therefore we shall have, as before,
vds
dy
d. Sdy + d.
d z
S d z = 0 ;
vd s
Y
170
and by equation (4), this becomes
dy
dz
d. +q.d = 0...
vds
vds
whence the equation to the curve is known.
·
(5),
If we suppose the body not to be acted on by any force, v will
be constant, (Art. 39;) and the path described will manifestly be
the shortest line which can be drawn on the given surface, and
will be determined by the equation
dy
dz
ds
d. +qd. =0...
ds
•
.(6).
If we suppose, as in Note, p. 111, ds to be constant, we have
d² y + qd* z
z = 0;
which agrees with the equation there deduced, for the path when
the body is acted on by no forces. Hence, it appears, that when
a body moves along a surface undisturbed, it will describe the
shortest line which can be drawn on that surface, between any
points of its path.
For a more general investigation of the nature of the Brachy-
stochron, see Poisson, Traite de Mec. No. 288, &c.
1
BOOK II.
THE MOTION OF A POINT IN A RESISTING MEDIUM.
53. THE preceding reasonings go upon the supposition, that a
body in motion, and left to itself, would move on for ever with a uni-
form velocity. This would be true, if the body moved in a perfect
vacuum; but when the motion takes place in any fluid or medium,
whose density is finite; then will be a retardation, arising from the
resistance which the medium offers to the motion of the body: and
if the body be acted on by any extraneous forces, the effects of the
resistance will be combined with those of the forces, and the curves,
&c., described by the body, as calculated in the preceding Book,
will be modified in a manner which we shall consider in this.
The resistance arises, in part, from the friction and tenacity of
the fluid, but principally from its inertia, that is, from the force
which the body, moving through the medium, necessarily exerts in
putting the fluid particles in motion. Hence, it will be in a di-
rection opposite to the body's motion: and if, cæteris paribus, the
velocity be increased, the resistance will also be increased; for the
body will strike more particles, and with greater violence. The
law, according to which the resistance varies with respect to the
velocity, is to be deduced from experiment: in most fluids, and for
a moderate velocity, it appears to be nearly, but not accurately,
as the square of the velocity. In the following problems we shall
suppose various laws of resistance.
The resistance is of the nature of a pressure, or moving force,
and may be represented by a weight. Hence, its effect on the
body, that is, the accelerating force (or, as it may here be called,
the retarding force,) is to be found by dividing this resisting force
by the mass of the body.
172
The quantity of the resistance, (considered as accelerating or
retarding force,) will of course depend upon the law, and upon
a constant coefficient. Thus, if the resistance vary as the square
of the velocity (v), it may be supposed equal to kv². The quantity
k will vary with the density, and other circumstances of the fluid,
and also with the mass, magnitude, and form of the body.
We may also represent the absolute quantity of the resistance
thus. Since it assumes different values as the velocity varies, we
may suppose a velocity such, that with it, the resistance shall be
equal to gravity (g). Suppose to be this velocity, and let the
resistance vary as v", v being any velocity. We have then
Vn: v :: g:
gvn
V n
= resistance to velocity v.
V is such, that if a body were moving downwards with that
velocity, it would move on for ever uniformly, for the action of
gravity downwards, and of the resistance upwards, would exactly
counteract each other; and the motion would be the same as if the
body were acted on by no force at all.
If we suppose the resistance = ko", we have k =
g
76
CHAP. I.
RECTILINEAR MOTION OF A POINT IN A RESISTING
MEDIUM.
54. THE formulæ dv=fdt, ds=vdt, vdv=fds, are applicable
to all cases of this kind, (v being the velocity, f the force in the
direction of the motion, and s the space described,) provided we
put for f the whole force arising from the attractions, &c., and
diminished by that arising from the resistance. The forces in this
Chapter are supposed to act in the line of the bodies' motion.
173
SECT. I. No Forces but the Resistance.
As an example of this case, we may suppose a body to move
on a horizontal plane perfectly smooth.
PROP. The resistance varying as any power of the velocity,
to determine the motion.
Let the resistance=kv": hence, in this case f= -kv", where
the negative sign is used, because the force tends to diminish the
velocity.
First, for the space,
vd v =
konds,
d v
kds
VN − 1 ›
1
ks
(n − 2) vn — 2
+ C.
And if v, be the velocity of the body when s=0,
1
(n-2) ks=
2
(2n) ks = v₁²,
1
V1
»−2, if n > 2,
v²-n, if n < 2.
If n=2, this integral fails, and we must return to the original
equation.
Second, for the time,
dv=fdt = kv"dt,
d v
kdt =
ht=
and if t = 0, when v=v1,
(n-1)lt=
1
(n − 1) vn−1 + C;
v² )
1
1
1
vn -
1
"
if n > 1;
-
(1 − n) kt = v,
1
N
vl.
v¹-", if n < 1.
174
If n=1, the integral fails, and must be obtained differently.
Ex. 1. Let n=1; our expression will be
k s = v₁ — v ;
or, the velocity lost is as the space".
And in this case, kdt =
dv
.. kt hyp. log.
=
จ
1
;
ข
therefore, if times increase uniformly, velocities decrease in geome-
trical progression †.
When v=0, s =
V₁
k
t = ∞,
that is, in losing its whole velocity, a body will employ an infinite
term, but will describe only a finite space.
Ex. 2. Let n= =2.
In this case,
dv
ข
kds =
.. ks hyp. log.
Again, kt=
VI
ย
1
;
ข
v
201
... 1 + k v₁ t =
or, if T
ข
kvi
k (T+t)=
ข
Hence, if the times (T+t) increase in geometrical progression, the
reciprocals of the velocities increase, or the velocities decrease, in
the same progression +.
*Principia, Book II, Prop. 1.
Ibid. Prop. 5.
† Ibid. Prop. 2.
1
If tx
S is constant, and v₁-v∞
Ibid. Prop. 6.
175
21
Also
हा
V
= e*' ;
1
ks
:. kt =
€
{e** — 1},
V1
ks
€
1 + kv₁t = e**, or kv₁ (T+t) = €**.
Hence, if the times (T+t) increase in geometrical progression, the
spaces increase in arithmetical progression, and the spaces in the
successive intervals are all equal.
When = 0, we have s and t both infinite, or the body will
describe an infinite space, and move for an infinite time before it
loses all its velocity.
Ex. 3. Let n = 4. In this case,
1
1
2ks=
12
21
V1
1
3kt=
اس
3
3
V
SCHOLIUM. It is manifest, that if we suppose the body to
move till all its velocity be destroyed,
if n<1, both space and time are finite,
if n = 1, or > 1 and 2, space is finite, and time infinite,
if n=2, or > 2, both space and time are infinite.
55. PROP. The resistance being = hv + kv², to determine
the motion*.
v dv
d
v
;
Here ds =
f
h + k v
1
h + k vi
.. S=
hyp. log.
h+k v
d v
dv
dt =
f
h v + k v²
*Principia, Book II, Prop. 11, 12.
176
11
t =
||
1 sd v
h
k d v
h + kv)
v
ย
{hyp.log.
21
byp. log.
+hyp. log.
v (h + k v₂)
When v=0, s=
h+kvil
h + k v S
vi (h+kv)
hyp. log. (1+1),
t = inf.
k
SECT. II. The Body acted on by a constant Force
besides Resistance.
56. The constant force may be supposed to be gravity (=g), and
the body to move in a vertical line, upwards or downwards. In the
former case, the force which acts upon it is the sum of gravity and
the resistance; in the latter, it is the difference. Hence, the
motions in ascent and descent are not similar, and cannot be obtain-
ed from the same equation, as in the case of a vacuum. They may
however easily be obtained separately, as in the following examples.
PROP. A body is acted on by gravity, and also by a resistance
varying as the velocity; to determine the motion*.
For the descent, f=g-kv;
v d v
ds=
cd v
f
g - k v
g d v
k g
k v
- dv}
*Principia, Book II, Prop. 3.
It may be observed, that the constant force of gravity, urging a body to
descend in a resisting medium, is not the same as the whole force of gravity
g in a vacuum. It is only the relative gravity of the body, with respect to
the fluid; that is, its weight, diminished by the weight of an equal bulk of
the fluid.
177
... s =
S=
hyp. log.
g - kv₂ v
k²
g-kv
k
supposing g ko positive,
and v₁ being the velocity when s = 0.
It is manifest that the greatest value which v can assume, is that
which makes g— kv=0, or v =
k
g
= V suppose; so that k
This velocity V is called the terminal velocity; it is the limit
to which the velocity perpetually approximates as the body descends,
but which it never actually attains in a finite time.
g
If we put
for k, we have
v2
V ~ V 1
V
S=
hyp. log.
(v — v1).
g
V-v
go
d v
d v
Again, dt =
=
ƒ
g-kv
.. t =
hyp. log.
g-kv₁
g-kv
V
V - v
hyp. log.
g
From this we should easily obtain the relation between s and t.
Supposing kv-g positive, we have
κυ
kv₁-g
V2
S=
hyp. log.
21 Ú
+
v₁
V V
VV
hyp.log.
+- (v₂-v).
k
kv - g
k
g
บ
V g
Hence, if v be at one point greater than V, it will always con-
tinue so.
This belongs to the case when the body is projected
downwards with a velocity greater than the terminal velocity. The
velocity will, in this case, decrease and approach to the terminal
velocity as its limit.
The ascent may be obtained in nearly the same manner by
making f= ·(g + k v).
57. PROP. A body is acted upon by gravity, and by a resis-
tance varying as the square of the velocity: to determine the ascent
and descent *.
* Principia, Book II. Prop. 8, 9.
2
178
For the ascent: ƒ = −(g + k v²),
ds=
v d v
v d v
g
8 + kv 2 3
1
S =
hyp. log.
2k
When v=0, s =
whole height ascended,
hyp. log. (1 +
When is small, or the density small,
1
8 + k » ₁ 2
g + k v
k v 1
2
which gives the
2k
g
whole height
1 Skv₁
2k
2
1 k²v, 4
g
2
gé
1
2
K v, 4
2 g
4 g
neglecting terms beyond k
The first term is the ascent in a vacuum, (Book I. Chap. I.
Ex. 3.) and therefore the second is the defect of height produced
by a medium of small density. It appears from the expression that
this defect is as the square of the height ascended.
Again, dt =
1
t:
-
d v
j
d v
g + k v
√eg) · {are (tun. = r;
№2130
When v = 0, time of whole ascent =
If v be infinite, t =
height ascended is infinite.
T
2V (kg)
- arc (tan.=v√)}
√ (k g)
are
(tan.=v,√!:).
and is therefore finite, but the
When k is small, expanding the arc by the formula,
tan.3
arc tan.
+&c.
3
1
we have time=
√(kg)
2100
¤ tp
neglecting higher powers
gg
3
3
k v, ³
3.g.2
179
f
v dv
8- kv²
1
g- kv,2
of which expression the first term is the whole time of ascent in a
vacuum, and the second the diminution of the time by the resist-
ance.
For the descent: ƒ = g
ds=
v dv
-
kv2,
; and if g-kv² be positive,
g
S
hyp. log.
;
2 k
g- k v²
or if k = 1, V = √
V2
S =
2g
hyp. log.
V² 2
V₁
V² -²
But if g-ko² be negative,
1
s= hyp. log.
2 k
V72
kv, ² -
kv-g
k v²
1
2
2
g
hyp. log.
2g
v² - 17 2
V is the terminal velocity as before; and the former expression is
for the case when the body is projected with a less, and the latter,
when with a greater velocity than V.
તેજ
d v
dt =
t =
11
1
{
dv v k
ƒ¯¯g — k v´ ¯¯ 2,√(kg)
1
-
2√(kg)
{hyp.
hyp. log.
+
dv Vk
V g + v Vk Vg - v V k
V g + v v k
V g + v₂ V k
V g − v₁ Vk
+hyp. log. / } (0, < V)
√KS (U₁<V)
V g − v V k
+ hyp. log. V-1
V(
V+で
​hyp. log.
2g
V + V₁
V
Ꮴ - i
- v,
V+c)
1
hyp. log.
2g
Ꮴ
V
V + 21
When o₁> V, the expression is similarly integrated.
01
If the body fall from rest, v₁ = (),
V2
S=
Qg
172
hyp. log. 7:
V+v
V
t =
2g
hyp. log. 7-v
で
​180
The last equation gives
}
V-0
V+o
2 g t
V
€
When t becomes considerable, the right hand side of this equation
is very small, and very nearly V. Hence, though the body
v =
never acquires the terminal velocity, after the lapse of a certain finite
time, it comes very near it, and the velocity afterwards may be
considered constant.
58. PROP. The resistance varying partly as the velocity, and
partly as the square of the velocity: to determine the ascent of a
body by gravity*.
-
Resistance = hv+kv²; f= −g− hv − k v²;
ds=
v dv
1
f
v dv
g+ hv + kv² ·
h
hv
h²
Let v +
= u, v² + +
= w
2k
k
4k²
2
2
h²
g+hv + kv² = g
=
+ ku².
4k
and we shall have two different cases,
In the former case,
h²
as is less or greater than g.
4k
h
udu
du
2k
ds=
h2
g
an
+ ku²
4k
+
S=
+
4 kudu
4 kg — h²+4k² u²
1
2 k
h
2hdu
4 kg → h² + 4 k² u² ›
hyp. log, (4kg-h² +4k² u²)
arc.
k
(tan.
2 ku
tan. =
+c.
V(4kg-h).
* Principia, Book II. Prop. 13, 14.
1
181
+
(tan
tan. =
1
h
k
1
2 k
arc.
hyp. log. 4k (g+hv+kv²)
2kv + h
-
√(4kg - h³)
g + h v₁ + k v₁²
2
+ C
hyp. log.
2 k
g + h v + k v
2
h
+ 1/2 · {
{arc.
2kv+h
arc. tan. =
arc. tan. =
√ (4 kg − h²)
-
2 k v₁+ h
√(4kg - h²)S,
h²
In the latter case, when
is greater than g, we have
4 k
4 kudu
2 h du
ds=
2
+
4k² u² - (h² - 4 kg)
4 k² u² — (h² — 4 kg)
and both the terms may be integrated by logarithms.
do
dv
Also dt =
f
g + hv + k v²'
which may be integrated in the same manner.
In this case the terminal velocity is found by solving the
equation g-h V-k V² 0. One of the roots of this will be
negative, and the other gives the velocity.
=
SECT. III. The Body being acted on by a variable Force.
59. In this case, to make the problem more general, we may
suppose that the density of the medium, or its power of producing
resistance, is different in different places. On this supposition k
will be variable.
PROP. A body is acted on by a force P in the line of its
motion, and by a resistance which is as the square of the velocity:
to determine its motion.
Lets be the body's distance from a given point towards which
the force tends.
In the descent towards the given point ƒ = − P + kv²,
vdv=-Pds+kr'ds,
182
1
or, vdv-kv³ds Pds......(1).
-
ds
s
Now d.v³e-f2kds = Qvdve-12k dı – Q kv²ds e-ƒ½kdı;
hence, the first side of equation (1), multiplied by 2€
comes integrable.
Multiplying and integrating, we have
k
v² € - S 2 k d s
-
2/ Pdse-2ds+C,
S s
s
v² = 2 e² kd ¹ƒ Pdse - 25 kds + Ce² kds
-ƒ 2 kd s be-
..(2).
If the integrals on the right hand side of equation (2) can be taken,
the velocity is found. And this being known, the time may be
found.
as s.
2
Ex. Let the force be inversely as s², and the density inversely
h
Here k h being constant· ƒkds=hfds = h hyp. log. s
S
S
€2ƒkds = s-2h; and the force being =
2 / +2
•*. v² = Cs²h
SPdse -²/d¹ = S
mds
-2fkds
And if v=0, when s=
=α, C =
2m
2
2 m
m
(2 h + 1) s2 h +1
2 m
(h + 1) s
2 m
(2h+1) a² + 1;
(2h+1) a
D
2
E-3
ds
For the time, dt =
; and putting for v its value, the possi-
V
bility of the integration will depend on the value of h.
In the same manner, if the force vary as any power of the
distance, the velocity may be found.
If the density vary inversely as the square of the distance, and
the force as any inverse power of it, the integration is possible.
183
If the density vary inversely as the cube of the distance, and
the force as any inverse odd power, the integration is possible.
And generally, if the density vary as any inverse power r, of
the distance, the integration is possible, if the force vary as any
inverse power of the distance, whose index is contained in the
series, r, 2r-1, 3r-2, &c.
CHAP. II.
THE FREE CURVILINEAR MOTION OF A BODY IN
A RESISTING MEDIUM.
60. THE second law of motion, and the results deduced from
it, are true when the body moves in a resisting medium, provided
we comprehend the resistance among the forces which operate to
change the motion. Hence, we may apply to this case also, the
equations (c), Art. 12, (the motion being in the same plane,) viz.
ď² x
ďy
= X,
dť
Y,
d t²
taking care to include in X, Y the resolved parts of the resistance
in the directions of x, y.
Let R be the resistance in the direction of the curve, and
opposite to the body's motion. Then it is clear, that the resolved
parts of this in the direction of x and y are
dx
R
ds
dy
R ;
ds
where the negative sign is used, because when x and y are in-
creasing, the resistance tends to diminish them. Adding these to
184
the other forces which act on the body, we have the whole values
of X and Y, and may proceed to determine the motion nearly in
the same manner as in the former book.
If the motion of the body be not all in the same plane, we shall
have, besides the two equations already mentioned, a third
and 2 will involve a term
dz
ds
d² z
d t
Ꭱ . arising from the resistance.
و
SECT. I. The Force acting in parallel Lines and
constant.
2;
61. PROP. Let a body, acted on by a constant force, as gravity,
(=g), be projected in a uniform medium, of which the resistance is
as the velocity: it is required to find the curve*, fig. 67.
Let x, measured from the point of projection A, be horizontal,
y vertical, s the curve, t the time.
our equations become
ds
In this case R=k. ; hence,
dt
d² x
dt
- k.
2
dx d²y
dt' dť²
2
Integrating,
1
dy
g-
– k
..(1).
dt
dx
hyp. log.
dt
-
= C − kt; hyp. log. ( g + k) = C' — kt,
and when t = 0, if a be the velocity of the body, and a the angle
which its direction makes with the horizon, we have at that time.
dx
= a cos. a,
dt
dy
= a sin. a.
dt
Hence, obtaining the values of C, C', and reducing,
dx
dy
a cos, a ɛ
αε € g+k.
=(g+ka sin. a) e
•.(2).
dt
dt
Now integrate the first of these, and substitute in the second ;
Principia, Book II, Prop. 4.
185
kt
x
a cos. а ε
k
a cos. a (1—€˜kt)
‚'.
•. 1 — €
kt
k
— kt=hyp. log. (1
k x
a cos. a
kx
,
+ constant
for x=0, when t=0;
k x
ki
;
€
= 1 -
a cos. a
a) ; :. dt
a cos. a
.. =
dx
- kr
kx
a cos. a
Putting the values of dt and of e¯** in the second of equations (2),
€
dy
g+k (a cos. a—kx).
dx
= (g+ka sin. a) (
a cos. a—k x
kx);
a cos. a
dy
dx
213
10%
-
g+ka sin. a
ka cos. a
1
k´a cos. a- kx
(3).
Multiplying by dr and integrating, y being=0, when x=0,
g
g
y =
(tan. a+
ka cos. a) * - .hyp. log.
a cos. a
a cos. a―k x
.(4),
Ꮖ
k²
the equation to the path of the projectile.
COR. 1. Expanding the logarithm, this becomes
y=tan.a.x +
gx
gx
kacos.a kacos.a
gx²
2
1
g k x³
3
2
2'a cos.*a 3 a³cos.³a
+ &c.
3
where, omitting the terms which destroy each other, the two first
terms represent a parabola, the curve described in a vacuum ; and
the remaining ones the alteration introduced by resistance.
COR. 2. It is manifest, that r cannot become greater than
a cos. ɑ for if it were so, a cos. a-kx would be negative, and
k
its logarithm impossible. Hence, if we take AB=
a cos. a
k
a vertical line BD will be an asymptote to the descending branch.
COR. 3. If we suppose the curve to be continued backwards
beyond A, it will not have an asymptote to the branch AO, but
Ал
186
it will perpetually approximate to a certain angle of inclination with
To find this angle, make, in equation (3), x infinite
the horizon.
and negative: we have thus,
g
dy
tan. a +
dx
ka cos. a
for the ultimate position of the tangent at 0.
COR. 4. If we draw AE parallel to this tangent, we have,
since AB =
a cos. a
;
k
dy
BE = AB.
a sin. a g
+
dx
k
k2
a sin. a
Also
BC AB. tan. a
a=
Hence, CE =
g
k
a
Also
AC AB sec. a =
k
Hence, AC CE ka g: initial resistance: gravity.
:
COR. 5. To find when the body is highest, we must have
dy
= 0; hence, by (S),
dx
X =
a² sin. a cos. a
g+ka sin. a
62. PROP. Let a body, acted on by gravity, be projected in a
uniform medium, in which the resistance is as the square of the
velocity it is required to find the curve, fig. 68.
The co-ordinates are measured from the point of projection A,
as before.
ds2
We have here Rk.
29
and our equations become
dt
* We shall have PQ CE hyp. log.
AB
MB'
187
dsdx
d2x
dt
=
k.
2
d t
…….(1).
ď² y
d s d y
dy
- k
g
d t²
d-t²
The first gives
d2 x
dx
11
kds.
d t
d t
d x
= C.eks
.. (2).
dt
C being a constant quantity, which is to be so taken that when
J x
s = 0,
may be the velocity of projection in a horizontal di-
d t
rection. Hence, if a be the velocity, and a the angle of projection
BAC, we have
a cos, a = C;
ks
€** dx
dx
a cos, aɛ -ks, dt
dt=
a cos.a
d t
Let at any point dy = pdx, whence p will be the trigonometrical
tangent of the angle TPN, which the curve at P, or its tangent
PT, makes with the horizontal line PN. Hence, we have
d² y
= dpdx + pd²x,
which values, substituted in the second of our equations (1), give
d²x
dp d x
pds dr
dt2
+ P
d t²
2
g-k.
d t²
and by the first,
d2x
pds dx
P
k
d t²
d t2
.. dpdx = - gdť²..
gd ť........(S),
and putting for d t its value already found, and omitting da,
g eks d x
dp
a“. cos.“ a
...(4).
188
This equation expresses the nature of the curve. It may be
made integrable, by multiplying it by the equation
dx √(1+p²) = ds, which gives
dp √(1+p³)
geeks d s
2
a
2
cos. a
.(5).
Integrating this equation, we have
p√(1+p²)+hyp. log. {p + √(1+p²)} = C —
g€
2ks
ka² cos.² a
sin. a
When C is to be such, that when s=0, p=tan. a =
cos. a
...(6).
Hence,
sin. a
sin. a + 1
2
cos. a
+ hyp. log.
= C
C·
cos. a
g
ka² cos." a
2ks
€
We may eliminate between equations (4) and (6), and thus
obtain dr in terms of p and dp; and, by integrating, x in terms of
p: we have also dy=pdx, and hence, y in terms of p.
we have d t² =
2
dp d x
g
Finally,
and for dx we may substitute its
value, and thus by integrating, obtain t in p. If these integrations
could be performed, and p eliminated, we should have a complete
solution of the problem.
COR. 1. If we make k0 in the value of dp V(1+p²), we
have an equation which belongs to a projectile in a non-resisting
medium; the original velocity and direction of projection being the
same as in this problem.
Let AP be the curve in a resisting, Ap in a non-resisting
medium; and let these be taken, so that the tangents PT, pt are
always parallel; therefore the values of p are the same in the two
cases; APs, Ap=S. Then by (5),
gds
aks ds
2
a² cos.2
ɑ
And
a² cos.*
in both,
a
t
dp √(1+p²), making k=0;
= dp V(1+p²). Hence, p being the same
=
189
2ks
2 kd Se²**.2 kds; 2kS =
supposing S and s to begin together.
And hence,
2ks= hyp. log. (1+ 2 k S).
2ks
€
– 1,
COR. 2. The quantity p will be positive, and a will increase,
to a point which is found by making p=0. After that, p will be
negative, and will become numerically greater and greater without
limit, as the curve becomes more nearly vertical.
COR. 3. The curve has a vertical asymptote BD, to the de-
scending branch Pz.
To prove this, eliminate es in (6) by means of (4), and we
have
pV(1+p²) + hyp. log. {p+ V(1+p²)} = C +
Dividing by på,
+ 1) +
hyp, log. {p+ (1+p²)} C
p
2
dp
k d x
dp
-
+
P
kp¨dx
Now, when p becomes very large and negative, all the terms on
the first side become very small, except 1*. Hence, this must
ultimately be equal to the term on the other side; and we have,
taking the value of 1, because p decreases as a increases,
- 1
1 =
dp
kp dx:
and dx =
dp
k p²
This, which is nearly true when p is very large, may be used in
finding the value of x, in the interval between p≈ a large quantity,
and pinfinity. And, as the value of r corresponding to the
former quantity is not infinite, if the value in this interval be finite,
the whole value of the abscissa will be finite, and the curve will
have a vertical asymptote. Now by integrating, we have
* Hyp. log. ✓(1+p²)-p is infinitely small compared with r
when p is infinite; and a fortiori compared with p².
190
1
x = const. + ;
kP
the constant being known by giving p and x, known finite values.
And when p becomes infinite and negative, we have x = constant,
which is finite, and there is therefore a vertical asymptote.
COR. 4. The curve has an oblique asymptote EH, to AO
the ascending branch continued backwards.
1
Put, in the equation 2 ks = hyp. log. (1+2 k S), S = — 2 k
and we shall have s infinite. Hence, if we take, in the parabola
described in a non-resisting medium, an arc Ao=
1
2 k
and draw
a tangent oe, the curve AO becomes ultimately parallel to oe.
To shew that there is an asymptote parallel to this line, we
must prove that AH is always finite, H being the intersection of
the tangent with the axis.
y
AH =
-Ydx
dx
ydp
- x; d. AH = yd.
dy
dy
p²
Also, if the ultimate value of p be called n, we shall have ulti-
mately, the abscissa being -x,
y = nx, s = − x √(1+n²); and by (4),
2
geeks dr
a²
€
x
a cos. a
dp
2
will be finite, if the integral be finite for
P
=-Sydp
H=-
.. AH=
the portion, when p becomes n, and when x and y become in-
finite.
That is, if
S
g x dx
- 2 k x √ (1+n²)
€
2
a² cos.² a.n
be finite when x is infinite; which it will be seen to be by inte-
grating.
COR. 5.
Hence, it appears that the ascending and descend-
191
ing branches are not similar.
The body rises more obliquely, and
descends more vertically than it would do in a vacuum*.
SECT. II. Any Force acting in parallel Lines.
63. PROP. Let the force act parallel to the ordinate y, and be
equal to P; to find the equations of motion †.
We shall here have
2 d x d² x + 2 dy
d2x
dx
d² y
= R
-
d t²
2
ds' d t²
2
ď²
Y
#
Hence,
d t² 2
ds2
or, d.
d t²
Also, by (1),
dy
P - R ...... (1).
ds
2 Pdy - 2 Rds;
2 Pdy-2 Rds..........(2).
dx d²y - dy d² x
d t²
- Pdx;
or,
2
dx² dy
d.
d t² d x
- Pdx.
We may in this suppose dr constant; and differentiating on
this supposition, we have
ď² y
dy
- P...
d t²
(3).
* For the determination of other circumstances in the curve, and for me-
thods of constructing the curve approximately, see Journal de l'Ec. Polyt.
tom. IV, p. 204. This part of Mechanics has been called Ballistics.
When the angle of elevation is small, we may modify our formulæ and
obtain an approximation for that case.
For the path of the projectile when the resistance varies as (velocity)",
see John Bernoulli's works, tom. II, p. 293. This is the problem which Keil
proposed to foreign Mathematicians as a challenge, and of which he was
unable to produce a solution when Bernoulli called upon him to compare
it with his own.
+ Principia, Book II. Prop. 10. This problem, which was erroneously
solved in the first Edition of the Principia, gave rise to some of the most
angry disputes in the controversy between the English and Foreign Mathe-
maticians of that period.
192
d.
Putting in (2) the value of dt from (3), we have
Pds2
2 Pdy + 2 Rds.. .. .. .. . . · ·(4).
ď y
This is an equation to the curve, if the resistance be known, and
conversely.
COR. 1. If P be constant and=g, since d.ds² = 2 dyd²y,
we have
2 gdy
gd s² dy
(d*y)²
2 gdy + 2 Rds;
gd s ď³ y
:. R = -
..(A).
2(d² y)²
COR. 2. In this case, if R be as the square of the velocity,
ds
let R = Q
Q.
gd s ²
2
d t²
2
ď² y
by (3);
d³ y
•. Q
·(B),
2 ds d² y
which gives the density.
Ex. 1. A body moves in a semi-circle, acted on by a constant
force in parallel lines, the resistance varying as the density and
square of the velocity; it is required to find the variation of the
resistance and density, fig. 69.
If we measure r from C the centre, and call the radius a, we̱.
have
y = √(a² — x²),
x d x
dy =
d² y
2
2
Ꮖ
√(a²
2
a² d x²
x²)
(a² - 2²),
3 a²x dx³
a d x
ds=
√(a² — x²)
>
3
d³
y
5
(a² — x²) ž
Hence, by (A), R =
Q (d² y)²
gd s d³ y
3g a³r
3
3x
=
,
2 at
. g.
o a
193
1
gd s²
Also velocity
ď y
R
Hence, density
velocity
g (a² — x²) *
a
2
Зах
2 (a² — x²)³
Hence, the density =0 at B, when x=0; and is infinite at A.
Between B and a the expression will be negative; and in order
that the body may describe the arc a B, it must be propelled by the
medium, which physically speaking, is absurd.
Ex. 2. Let the body, acted on by a constant and parallel force,
move in a parabola of any order; it is required to find the resistance
and density.
Let BN, fig. 69, be a tangent at the highest point; and let NP
vary as any power of BN. Then if CM=x, MP=y, CB=b,
we shall have y=b-cx"; and differentiating
1
dy=— nc x²-¹ dx, ds=dx √(1+n² c² x² n −²),
d² y = — n(n − 1) c x²-² dx², d³y= − n (n − 1) (n − 2) c an-³ dr³.
Then by (4), R =
3
n(n − 1) (n − 2) cx” − ³ √(1+n³c²x²n−²)g
2n² (n − 1)² c²x²n=2
(n − 2) √(1 + n² c²x²n-2) g
2n (n - 1) cx² + 1
If n be greater than 1, this value of R will manifestly become
infinite, when x=0, and negative, when x is negative. Also if NP
do not generally vary as some power of BN, yet if the curve be
symmetrical with respect to a line BC, it is manifest that ultimately,
when BN is small, NP may be considered as proportional to some
power of it. Hence, any symmetrical curve will require an infinite
resistance at the vertex B; and hence, conversely, the curve described
in a medium, the density of which is every where finite, cannot be
symmetrical with respect to its ascending and descending branches.
Let us now take an example of a curve not symmetrical.
Ex.
3. Let the curve OPQ, fig. 70, be an hyperbola of any
order, of which one of the asymptotes CB is vertical; it is re-
quired to find the resistance and density at any point *.
Principia, Book II, Prop. 10. Ex. 3.
B B
f
194
[
Let AB = a, BC=b, AM=x, MP=y. And suppose NP
1
to be as
CNn
Also let MB=z, so that a
z=X.
b x
bz
MN =
= b
a
α
1
1
NP which varies as
varies as
CNn
MB"
с
Let NP
zn
- y = b-bz - C
α
Z"
c
dy = (-) dz,
− i)
a zn+
ds. = - dz√ {1 +
d z
= = = √ {~
Y
ď y
ช
+
( −
b nc
a zn+
b z
1
)}
( = + 2 ) } .
a
n. (n + 1) c d z²
N + 2
zn
a(n+1)(n+9) cả
Zn + 3
ช
78
Making these substitutions in the expression for the density, we
have by (B)
d³y
Q =
2 ds d² y
n(n + 1) (n + 2) z² + 2
2n (n + 1) zn+3
n + 2
b Z
2
b
2
√ { ²² + ( 01/2
NC
a
nc
"V+C-21
√ { ²² + ( ²/z
a
zn
Now this does not vanish for any value of z, except z=0.
195
Hence, if the density be every where finite, the curve will approach
more nearly to this form than to the symmetrical curves before-
mentioned.
SECT. III. Central Forces.
upon
64. PROP. A body moves in any resisting medium acted
by any force tending to a centre; it is required to find the curve.
described.
Let, in fig. 71, SM=x, MP = y, be rectangular co-ordinates
to the place P of the body. SP=r; and P the force, any function
of r, and tending to the point S. Also let R be the resistance
which acts in the direction of the curve, and depends upon the
velocity, and on the density of the medium. And let s be the
curve described.
The resolved parts of the force P are P and P in directions
force R be resolved,
x
T
Pm parallel to MS, and PM. Also if the
d x
dy
R
and R
ds
d s
the components in the same directions are
hence, we have (the body moving in the direction AP),
d2x
dt
2
Rdx dy
P x
ds
d t²
- 2 P.
Py
2
x d x + y dy
Rdy
ds
(1);
Ꭱ
− 2 R. dx² + dy²
ds
r
2 d x d x + 2 d y dy
dt2
2
or since dx²+dy² = ds², 2dxdx+2dyd² y=d.ds²,
ở tỷ ở, các+yd!=rdr,
=r
ds2
d. di
2 P dr - 2 Rds.
.(2).
dt2
Again, the original equations give us
y d² x - xd³ y
y d x - x d y
R.
dť
ds
196
4
But if we call ASP=v, we have
y dx − x dy = r²dv, yd² x − xd²y=d. r²d v;
d.r² dv
d t²
r2 dv
- Ꭱ .
.......(3).
ds
ds
R is a function of
the velocity, and of r, if the density vary
d t
ds
Hence, putting for R its value, we may eliminate
be-
dt
with r.
tween the equations (2), (3), and obtain an equation in r and v,
which by integration will give the curve.
65. PROP. Under the same circumstances, it is required to find
the relation between the radius vector (r), and the perpendicular
upon the tangent (p),
r'dv=pds, and equation (3) gives
d.pds=
Rpdť²,
or pd²s+dp ds =
—
Rpdť².
And equation (2) gives
dsds Pdrdt2 - Rdsdt.
Eliminating R, we have
2
2
ds² Ppdr
ds²dp= Ppdrdt;
dt³
dp
(C).
Let q be one-fourth the chord of curvature through S: then
pdr
2q,
dp
d s²
2
= 2Pq*.
dte
* Hence, the velocity in the curve is equal to that acquired by falling
down one-fourth the chord of curvature by the action of the constant force P;
the same property which obtains in a vacuum.
197
And substituting in equation (2) of last Article,
d. Pq = -
Pdr Rds;
d. Pq+Pdr
.. R =
ds
·(D).
When the relation between p and r is given, this formula enables
us to find the resistance, and conversely.
COR. 1. If R = Q
dse
dt²
2
=Q.2Pq; we shall have
Q =
d. Pq+Pdr,
·(E).
2 Pq ds
Ex. 1. To find the resistance in the circle, the centre of
force being in the circumference, and the force as any power of
the distance. The body moving towards the centre.
dp
Ρ
=
Here
m
p =
a
P =
доть
2 dr
; q
=
12
m
Pq=
pdr
2 dp
=
dr
ds
S =
4
4p-1; dPq+Pdr =
.. by (D), R =
m (5—n)
4 pr
O being the angle which
√6-2
m (n-1) dr
4 pn
m (n − 5)
4pn
V(-5)
a
2
;
a
mdr
+
n
m (5—n) sin. ◊
=
4pn
makes with a line through the centre.
ds²
m
ds2
Also
= 2Pq =
; .'. if R=Q
Q
dt2
n
2 r
dt²
(5-n) sin. O
2 r
Principia, Book II, Prop. 17, 18.
198
Ex. 2. To find the resistance in the logarithmic spiral; the
force as any power.
Here, p = ar;
ds²
事
​pdr
2 dp
=
= 2 Pq = Pr;
dt
d. Pr+Pdr
dr
.. R=-
:.
; or since ds =
ds
√(1 - a²)'
2
α
the negative sign indicating that the body is moving towards the
centre;
d. Pr
R =
2 dr
+ P) . V (1 − a²³).
m
m
And if P-
Pr=
-->>
доде
R={-
n — 1
m
+
R = { - " == ¹ . == + =} √(1−a®)=
m
(3 - n) m
√(1 − a²);
-
2
2
zin
R =
(3 — n)
2
m
√(1 − a²),
=
and, if R = Q. velocity Q. Pr =
Q =
(3 — n) √(1 — a²)
2 r
Qm
Jin - 19
Hence, the density is inversely as the distance from the centre *.
If n>3, R and Q will become negative, and it will no longer
be possible for a body to move in this spiral towards the centre;
but if the motion be from the centre, we shall find in the same
manner
Q =
(n − 3) √(1 − a²)
2 r
* Principia, Book II, Prop. 15, 16.
199
From the expression for R, it appears that (fig. 72,)
3 - n
R: force to centre ::
Py: Sy.
2
If n=2, this gives
R: force to centre :: Py: 2 Sy.
Hence, when the force varies inversely as the square of the dis-
tance, and the density inversely as the distance, the resistance
being small, a body may, as in fig. 72, approach the centre in a
logarithmic spiral not much differing from a circle.
This motion requires a particular relation of the velocity and
direction. If the velocity be not that which the direction requires,
the path described will no longer be a logarithmic spiral. But if
the resistance be small, the curve will still be a spiral, approaching
the centre in successive oblong revolutions, as in fig. 73. And if
the density, instead of being inversely as the distance, follow any other
law, the orbit will still be of a similar form. We shall in a future
Article consider the effect of resistance in altering an elliptical orbit.
66. PROP. The resistance varying as the square of the ve-
locity and the density and the force being given; it is required to
find the polar equation to the trajectory.
Taking the equations of Art. 64.
ds²
d.
2 Pdr-2 Rds··
(2),
d t
d.
r² dv
d t²
r² dv
R
•
(3).
ds
ds²
Let R=Q; dt being constant, (3) becomes
2
d t
..
d.r² dv =
Qds. rdv;
integrating, dv=hdte /Qds
(F), h being a constant quantity.
ds
hdse jods
ds²
;
dt
r² dv
dt
h²ds²-21Qds
p4 dv²
う
​200
!
ds2
d.
2 Qds.
dt²
2
r² dv²
h² ds²e-esQ as
d
And by (2),
ds²
+ h² €¬?! Qds
sods.d.
r*do²
ds
2
h² ds² e-
2
- 2 S Q d s
d.
2 Pdr - 2 Qds.
dt2
4
r² dv²
ds²
ds2
putting for R, Q
and for
its value.
dt2
2 J
dt²
2
Hence, equating,
ds²
2
h² €-es Qds d
d.
=-2 Pdr,
r4 dv²
ds2
2 PdreQds
d.
+
= 0.
+
r* dv²
h²
dr
But if we make
=
u; whence
2
— du,
T
d s2
dr² + r² dv²
du²
r¹ dv²
p4 d v²
2
dv2
+ u³ ;
and substituting, differentiating, and reducing,
ď² u
+ u
dv²
Pe²SQds
h² u²
2
= 0........ (G).
This differs from the equation (d), Art. 18. for motion in a
vacuum, only in having the last term multiplied by €²Qds
If Q be small, the equation (G) will enable us to approximate
to the effect of resistance in altering the orbit, as will appear
the following Article.
in
67. PROP. A body acted upon by a central force, varying
inversely as the square of the distance, moves in a medium of small
density (Q); to find what alteration will be produced in the eccen-
tricity, and in the place of the major axis.
2
-ƒ
By (F), r² dv=hdte Qds;
or, if h' be the area in time 1 at the end of time t,
"dv = h'dt, h′ = he¯
h' so as
(1).
201
Also by (G), since P=mu²,
ME
2 f Q d s
ďu
do 2
+ u
h²
ďu
m
or
+ u
= 0.
dv²
h'2
=0...
.(2);
If h' were constant, we should find the integral as in Book I,
Chap. III, Prob. 2, and it would be
u = C cos. v + C' sin. v +
m
h'²
(3);
but since h' varies a little, u will differ a little from this form. We
may suppose u to be still expressed by the preceding formula, C
and C' being now taken variable, and determined so as to satisfy
equation (2).
Differentiating (3), we have
du
C sin. v+C' cos. v + cos. v
d v
d C
dv
d C'
m
+ sin. v
+ d.
d v
h's
dv
But since we have yet only one condition to determine C, C',
may assume another: let therefore
we
d C
dc'
m
cos. v
+ sin. v
+d.
dv
dv
h'
= 0... .(4).
dv
Hence, will be reduced to its two first terms; and differ-
du
d v
entiating again,
ď u
dv²
- Сс
C cos. v- C' sin. v
substituting in (2), it becomes
dC
sin. v
dv
dC'
+ cos. v
dvi
dC
d C'
sin. v
+ cos. v
=0..
•
(5).
dv
d C
C c
:
202
Hence, by (4) and (5), we have
m
d C
- cos. v. d.
v.d.
Since by (1)
and if Q and
m
12
equal to unity; d
2
Mc²S Q d s
Ε
d
h2
m
h'
, dC - sin. v.d.
=
M
Q Qd se² Qd s
h
m
Qds be small, so that e2ds may be considered
m
'm
•
2 Qds.
h'
2
Hence, we have
d C
m
h²
2 Q cos. v.ds, d C'=-
Now, if in equation (3), C, C',
to the beginning of a revolution,
C'e sin. a, and we shall have
じ
​m
h²
.2 Q sin. v. ds.... (6).
and h be supposed to belong
we may make C = c cos. a,
m
m
ch²
U
u =
h²
+c cos. (v - α) = {1+
h²
cos. (v~a)},
m
which agrees with the equation to an ellipse.
a
2 = {1 + e cos (v− a)}, if
b2
m
α
ch²
and
hⓇ 62
29
e.
m
And we shall find the variation of a and of e in the course of a
revolution in the following manner.
Make C=cos. a', C'c' sin. a', supposing c', a', variable;
and we have
cos. a'd C' sin. ad C
c'dá',
sin. ad C+ cos. a'dC = dc.
Hence, putting for dC, dC′ the values from (6),
M
¿'da'
2Q sin. (va) ds
h2
m
(7).
dc'
2Q cos. (v – a').d's
h
203
For a first approximation we may suppose the expression for r
and for ds to be the same as in an ellipse; and Q being any
function of the distance, it will easily be seen, that, supposing e
small, and neglecting powers of it, we have
Qds = {A+eB cos. (v-a')} dv,
A, B being constant quantities. Hence, by (7),
c'da':
dc
m
h²
m
h2
{2A sin. (v -- a')+eB sin. 2 (v – a')} dv
{2A cos. (v-a')+eB [1+cos. 2(v-a')]} dv
•(8).
In order to find the effect in a whole revolution we must inte-
grate these expressions through 2π. And the values when v =
B
let them when v=2+ß be a+Da, c+ac,
m
being a, c, e,
29
h²
M
m
Δ
e+ Ae, 72 + A2, then we shall have from the first of equations
h2
h2
(8) making cand a constant in the integration on the right hand
side,
Hence, Aα=0, and the
after a whole revolution.
c' ▲ a = 0.
A
position of the major axis is not altered
From the second of equations (8), making a constant in the
integration,
m
Δε
2пе В.
h
me
m▲
e
M
But since c =
we have Ac =
h²,
h²
+e, neglecting
}
powers of Ae, &c.;
m
ከ
m
also,
h'2
h
h
2 f Q d s
2 m
h
ƒ {A÷cВ cos. (v—a')} dv.
204
Hence, the integral being taken through 27, we have
Δ
'm
h²
2 m
•
2 пA;
h2
m
2 m
m
Aete
2TA=
h²
2
h2
h²
2
2 пе B,
2Tе
Ae=
2πе (2A+B).
Hence, it appears that by the effect of the resistance of a medium
of small density the eccentricity diminishes perpetually, while the
major axis contains the same position, see fig. 73.
Also, if we consider the remaining terms in Qds, which will
be of the form, E cos. 2 (va), F cos. 3 (va), &c. the ex-
pressions in (8) arising from these, will, when integrated through
2π, become=0; hence, the truth of our result does not depend
on supposing e small.
CHAP. III.
THE CONSTRAINED MOTION OF A POINT ON A GIVEN
CURVE IN A RESISTING MEDIUM.
68. THIS case might be treated in the same manner as those
in the last Chapter, by considering in addition to other forces, the
re-action of the curve in the direction of its normal, and resolving
this in the direction of the co-ordinates. It will, however, generally
be more simple to consider only the forces along the curve. It has
already been seen, that the alteration of velocity in the curve is
entirely due to the resolved parts of the forces in the direction of
the curve; and the same is true in a resisting medium, if we con-
205
sider the resistance, which is wholly in the same direction. Hence,
we may apply the formulæ for rectilinear motion to the motion along
the curve, as will be seen in the following examples.
PROB. I. A body moves on an inverted cycloid, with a vertical
axis, in a medium, the resistance of which is as the velocity; to
determine the motion, fig. 74.
Let s be the arc AP from the lowest point A.
of gravity in the direction of the curve is as AP;
force at P =
gs
1
, g being gravity; call this fs.
Then the force
and if AC=l,
Also, v being
the velocity, let the resistance be 2kv, when 2k is used rather than
k, to avoid fractions, as will be seen. Then, by the formulæ,
dv force.dt, we shall have, when the body descends,
d v =(fs — 2k v) dt;
and the same expression is true for the ascent for s becomes
negative, and the term which arises from gravity changes its sign as
the force changes its direction. Now we have v =
decreases as t increases.
ds
dt
; for s
Hence, making dt constant, we have
d's
dt
(fs + 2k15)
ds
dt;
dt
d² s
ds
+ 2k.
+fs = 0.
dt"
dt
We shall obtain a particular integral of this by making se
being constant ;
m=
... m² + 2km + f = 0,
k± √(− 1). √(ƒ− k³); or if ƒ — k² = hˆ,
m = − k ±h V (− 1).
Hence, particular integrals are
— [k − h √( − 1 )] t
~[h+h √(−1)] †
S=E
$ = €
S
;
m t
', m
206
and the complete integral
s=C¸€¯[k—h√/(−1)]† +C₂€¯[k+h√(−1)] t
= €
kt
E
- k t ht√(-1)
{C₁e
ht√(− 1)}
+ C₂€˜ht √(− 1) z
{C₁[cos.ht+(-1).sin. ht]+C₂[cos. ht — √(−1)sin. ht]}.
པ་
And if C₁+C₂ = C, (C₁ — C₂) √( − 1) = C',
- kt
S = €
(C cos. ht+C' sin. ht).
C and C' must be determined by considering, that at the beginning
of the motion, when we shall suppose t=0, we have s = a
ds
dt
the arc AD; and the velocity = 0.
k t
Now =Є
ds
dt
Hence, when t=0,
{(C'h- Ck) cos. ht - (Ch+C'k) sin. ht}.
a = C,
·
J
0 = C'h-Ck; .. C' =
= £ '.'
kt
kC
h
ka
=
h
(h cos. ht+k sin. ht),
ає
h
ds
dt
a (h² + k²)
h
kt
af
kt
€ sin. ht =
sin. ht.
h
If the body descend, and then ascend till its whole velocity is
destroyed, it will have performed one oscillation. This will be
when
ds
dt
= 0, or sin. ht = 0, which will first be the case when
Hence, if T be the time of an oscillation
ht=π.
π
π
π
T=
h
√(ƒ − k²)
g
k³)
207
This is independent of a, and hence the oscillations occupy the
same time whatever be the length of the arc. The cycloid, which
is the tautochronous curve in vacuo, also possesses that property in
a medium whose resistance is proportional to the velocity*.
COR. 1. To find where the velocity is the greatest. This will
be the case when the accelerating force is 0; that is, when
fs+2k.
ds
dt
= 0.
Or, f(h cos. ht+k sin. ht) - 2 kf sin. ht=0;
h
.. tan. ht =
k
COR. 2. To find the decrement of the arc, that is, the difference
of the arcs of descent and ascent. At the extremity of the arc of
ds
ascent we shall have
dt
•.ht.
= 0, and .. ht=π. If we suppose the
arc of ascent to be numerically represented by b, we shall have,
putting
П
b for s, aud for t,
h
b = a ε
κπ
h
;
К п
h
},
.. (a−b) = a {1- €
and if k be small, neglecting its powers
μπ
a − b = a
h
Tha
ggo k2
√ (§ - k²)
COR. 3. The body, after reaching the extremity of the arc of
ascent, will again descend and ascend on the first side. Let the
arc through which it ascends be a₁; we have then
μπ
2κπ
h
h
a₁ = bε
be
= α €
Principia, Book II. Prop. 26.
1
208
If the body make 2 (n + 1) oscillations, and return to distance an
on the first side, we shall have
2 k π
4k π
T
2 nk w
h
h
h
anan - 1
= An-2 €
=αЄ
ає
k
Hence, if a and an be observed, we may find
h
་ག་
and thence k.
PROB. II. A body moves in a cycloid as before, the resistance
being as the square of the velocity; to determine the motion*.
g's
Let the force of gravity in the curve = =fs; resistance
kv²; and in descent, by the formulæ vdv
v dv
-
-f's kv²) ds,
which is also true in the ascent as in last Problem.
2
·
Let v² = 2%; . vdv = dz;
V
-
dz = − (fs — 2kz) ds;
force. ds,
2ks
dz-z.2kds = fsds. Multiply by e-2**;
.. dze
2ks
Integrating both sides
ΖΕ
2 ks
-
k s
-ze 2ks.2k ds = - fe²*³ sds.
€
2 k s s d s
= C-ƒƒ
ƒs €¯îks
-
2 k
fse
2 k s
+
£
aff
2k
€
2 ks
=C+
=C+
2k
z=Cc2ks +
4k2
$ (Qk s + 1).
4k²
- 2 ks d s
'ds
$
Now at the beginning of the motion, let sa, and v=0, and
z=0;
..0 = Ce²ka +
2
f
(2ka + 1);
4k2
Principia, Book II. Prop. 29.
209
f
4/
.. C =
fe
a ka
4/2
{2ka+1},
—
-
{2ks+1 − (2 ka+ 1) €− 2 k (e − ³) } .
Let the body perform a whole
numerical value of the arc of ascent.
and O for v, and therefore for z;
0 = 2 kb + 1
G
kb
e
oscillation, and let b be the
Therefore putting -b for s,
(ka + 1) e
— Ik (a + b)
Ωκα
whence (1−2kb) e² *¹ = (1+2 ka) e˜ºka ;
from which may be found from a.
COR. 1. To find the decrement of the arc of ascent.
The equation last found may be put in this form,
1-2 kb
1+2 ka
=-2k (a+b); and expanding,
€
1 - 2 ka + 4k² a² — 8 k³ a³ + &c.
-2kb+4k²ab-8k³ a²b+ &c.)
8 k³ a b + &c.}
1- 2k (a + b) +
=
4k² (a+b)² 8k³ (a+b)3
1.2
1.2.3
+ &c.
which gives, omitting the identical terms, and dividing by 4k,
a (a + b) — 2k a² (a+b)+&c. =
(a + b) 2 2k (a+b)s
1.2
1.2.3
Dividing by (a+b), &c. we have, omitting k², &c.
+ &c.
2 k
b=a
(5 a²-2ab-b²).
3
Hence, the decrement of the arc =
(5a2ab - b³); or,
3
4 kaⓇ
(since ba nearly,)
>
3
nearly.
COR. 2. To find where the velocity is the greatest.
dv
d z
We must have
O, and therefore
= 0;
ds
ds
I k − k (ka + 1 ) € ~ ? k (a− s)
2k
.. = 0;
2 k (a~s)
€
=1+2ka, 2ka - 2ks=hyp, log. (1+2ka).
D D
210
If we expand, and neglect powers of k, we shall find
s=ka².
PROB. III. A body oscillates in a circle, the resistance being
as the square of the velocity; to find the velocity at any point.
If, in fig. 75, the radius CP=1, arc AP=s, ACP=0, we shall
have the force of gravity along the curveg sin. 0, and the resistance
= k.
ds²
dt2
= force in direction of the curve, we have
Hence, since
d² s
d tº
d's
d t²
d s²
2
− g sin. 0 + k
d to
Or because s=10,
d Ꮎ
d02
i
g sin. 0+kl.
dt²
d t²
Which is true both for ascent and descent.
This may be thus integrated. Multiply by
2d0
and integrate,
and we have
d02
2 g cos. 0
+ 2 kl
d t²
2
if
od 02
de
do.
dt²
S
d 02
Let fa do=x;
d t²
dᎾ
dz
d ť
d Ꮎ -
d z
2 g cos. 0
Ꮎ
+ 2klz.
d Ꮎ
l
Which is a linear equation of the first order, and may be in-
tegrated. Transpose, and multiply by d0e-2k10, and we have
dze-2k10-ze — 2 klo 2 kl do Ꮎ =
2g
7
-2k 10 cos. 0.de.
Where both sides are integrable. Therefore
2g
≈ 6 − 2k10 = 25 Se2k 10 cos. 0d0+C.
211
..
Se
Now feme cos.0d0 = -
€
тө
cos. A
m
1
m
(1+
(1 + - ) Se− mo
M
kl
€
{-
mu Ꮎ
€
COS.
m
m0 sin. 0
m
Od =
Se-me cos. O do
:. Sε-2k10 cos. O d0 =
And hence our equation becomes
ર
€
2k 10
€
1+47
2g sin. 0-2kicos. O
L
1+ 4k² (=
d z 2g cos. 0 + 2k l sin. O
1+4kc
Hence,
d Ꮎ
m
And hence, the angular velocity is known.
os.✔
+
тв
1
m.
m
Se-mo.sin.Ode
ƒ€¬mo cos.0d0}
{sin. 0 — m cos. 0 };
{sin. 0-2 kl cos. }.
+ Cε 2 klo
+ Qkl C € 2 kl 0 __
dee
dt2
If a be the value of 0 at the beginning of the motion, we have
de
d t
11
2g
cos. a +2 k l sin. a
1+ 4k 2222
+ 2kl C € 2 k la =0;
2g (cos.0+2 klsin.0-(cos. a+2 klsin. a) -2 kl (a—0)
l
2
1+ 4k² ²
COR. 1. If we make 2k1=tan. ß, we shall have
do² 2 g cos. B
d t²
-
•
{cos. (0-ẞ) - cos. (a− ß) €—(a—0) tau. ß}.
k. 2gl
Here tan. B
g
resistance to velocity acquired down /
gravity
П
COR. 2. If we make a
kla
+ ẞ, this becomes
d 0°
2 g cos. B
d t²
cos. (0 — ß).
Hence, we have this curious property. If in fig. 76, we take
AT horizontal, such that AC: AT:: gravity: resistance to velocity
acquired down AC, and make DE a quadrant, the body, setting
off from D, will move as if it were acted on only by a uniform force
g cos. ẞ in a direction parallel to AT.
212
In this case, the angular velocity is greatest at E, when 0=6,
and vanishes again when
Gog
β
π
™, or 0 = − ( − ß).
The whole arc described DA d is a semi-circle, and the decrement
of the arc of ascent is 2ß.
COR. 3. Generally, to find when the angular velocity is great-
de
dr²
est; the expression for must be a maximum.
Therefore
— sin. (0 — ß) — cos. (a — ß) tan. ẞe~(a−0) tan. ß=0.
-
And if the resistance, and
therefore tan. ß, be very small,
sin. (0 — ß) =
cos. (a – ẞ) tan. ß.
Or, putting 0 - ẞ, and ß, for sin. (0 – ẞ), and tan. ß,
0 =ẞ {1 - cos. (a — ß)} = 2ß. sin.'
a
= 2ß. sin.2 nearly.
α
2
-ß
PROB. IV. To find the time of oscillation in a cycloidal arc,
in a medium, the resistance of which is as the square of the velocity.
We have, (see Prob. II,)
2k2
{2ks+1−(2k a+ 1) e − 2 k (a − s) } .
v²=2x=
v² = 2 z
Put n for 2k,
and
y
for a―s, whence, s
s = a -y,
y, and
z?
ef
{1 + n s − (1 + na) ɛ−ny}.
Expanding the exponential, we have
of
22
n"
心
​2
(Qay — y²)
(3uy-y³)
n
1.2.3
+
2
124
1.2.3.4
n
(4ay' —y') - &c.}
N
= ƒ {2ay — y³. — "/ (3 ay³ — y³) + (4 a y³ — y¹) - &c.}
3
-
2
3.4
= ƒ {A - Bn+C &c.} suppose;
213
*
1
ย
B
{ A − Bn + Cu² − &c.} −}
VC + " + "² ( . . - 1) + &c.}
+n.
ds
And dt =
v
2A4
=
1.3. B2
C
2.4.A
2
dy
;
whence, t will be found by integrating
V
each term of the above series multiplied by dy.
To find the time of descending to the lowest point, (down DA,
fig. 74), we must take the integrals from y=0, to y=a.
Now Say =S
A
dy
√(2ay — y²)
π
And to the lowest point = 7.
(3
2
=arc
(v
ver. sin.
sin. = 24).
Also ƒ Bdy = f (Say³-y³)dy
2 A3
6(2ay-y3)
a
and to the lowest point ==.
Again, ƒ (1.3 B²
S
5
12.4.1 / /
=S
A
4
6
C
QA
a² y* dy
5
24(2 ay — y²) —
+
and, to the lowest point:
a²
24
=
arc
a
24
dy
2
y²
a
6 (2 ay — y³)½ ›
3 B²-4AC
=√s B²
3
8AŽ
a² Q y³ — 3 a y²
18 (2ay-y)
sin. = 2),
(ver
er. sin.
π
a²
18
Hence, omitting terms involving, &c., we have
time down DA
げに
​na n² a²
•
dy
J
+ + ( - )}·
6
24
The expression for the time up AD would be the same, except
that the resistance acts in the opposite direction, and consequently
it will be had by making negative. Hence, if AE be b, we
shall have
214
π
n b
n² b²
time up AE
AE =
π
+
fl2
6
24
2
1)}
And the time of an oscillation through DAE
1
n (a - b)
n² (a³ + b²)
π
Vf
π +
+
6
24
And it has already been seen that
4ka2
2na
b = a
= a
3
3
hence, the time of an oscillation
=
1
Vf
π
Vf
{
π +
{1+
2
n² a²
na
+
9
2 a
n²
2 2
n² a
12
(-3)}
√f {1 + Ka² } .
π
=
24
Vf
6
The quantity ƒ is, I being the length of the pendulum; and
π
is the time of an oscillation in a vacuum.
Hence it appears,
that when the arc described (a) is small, the time of oscillation is
the same as in a vacuum very nearly.
COR. 1. The arcs described become smaller and smaller, in
consequence of the resistance, and the times of the smaller oscil-
lations are somewhat shorter.
The excess of the time in the medium above that in the va-
cuum, is as the square of the arc described *.
COR. 2. The same things are true for small circular oscil-
lations +.
* Hence, Cor. 2, to Prop. 27, Book II, of the Principia, is erroneous.
It is there asserted, that the excesses of the times above those in a vacuum,
are as the arcs themselves.
For the direct proof of this Proposition in the case of circular arcs,
see Poisson, Traité de Mec. Art. 273.
215
PROB. V. A body oscillates in a cycloidal arc: the resistance
being small, and varying as any power of the velocity; it is re-
quired to find the decrement of the arc of ascent.
It will be shewn, that if the resistance vary as the th power of
the velocity, the decremental arc will be as the nth power of the
arc of descent *
Let resistance=kv"; the other denominations as before;
…. vdo
02
2
07
(fs-kv") ds,
· = ƒ (a² — s²) + k ƒ v² ds.
And v² = 2ƒ (a² — s²) nearly, since k is small;
· · ƒ v²ds = (2ƒ)ễ ƒ (a² — s²)¶ds, nearly.
n
2.2
= (eƒ}ƒ {a" — "a"-'s² +
(2f
= (2ƒ)* {a*s=
2
an
na"-283
+
2.3
n(n-2)
2.4
an-484-
s¹ - &c. }ds
&c.}ds
•
− &c. } + C;
n (n-2) a"-455
2.4.5
nan-283
.. v° = 2ƒ (a²—s") + 2k (2ƒ')* {a"s — """" +&c.}
+&c.} + C.
2.3
Now v=0, when s=a. And when v becomes O again, suppose
· s = −a+d, and neglecting powers of 8, and products kd, we have
d
0 = 2ƒ. 2ad + 4k (2 ƒ)*
.. 8 = 2ka" (2ƒ)*~' { 1. —
And hence, is as a”.
{-
a²+1
n
na" +1
+ - &c.};
2.3
23+ (-3)-&c.}.
2.4.5
COR. If & be known, k may be found †.
Principia, Book II, Prop. 31. Cor.
† Ibid. Prop. 30. Cor.
CHAP. IV.
INVERSE PROBLEMS RESPECTING THE MOTION OF POINTS
ON CURVES IN RESISTING MEDIA.
69. HERE, as in the corresponding Chapter in the first Book, the
curve is to be determined from some property of it which is given.
Most of the problems, however, which occur, present considerable
difficulties, and lead to very complicated calculations. We shall
therefore take only one of the most remarkable and celebrated of
the questions of this kind. The solution is nearly that given by
Laplace, Mecanique Celeste, Prem. Par. Liv. 1, No. 12.
PROP. To find the tautochronous curve in a medium, of
which the resistance varies partly as the velocity, and partly as the
square of the velocity; the body being acted on by gravity.
In fig. 74, let the vertical abscissa from the lowest point.
AM=x, AP=s; the velocity=v, and the resistance = 2 hv + kvª.
The resolved part of gravity in the curve will be g.
making dt constant, for the descent
dr
ds
and hence,
d² s
d t²
dx
=
g.as
+ (2 h v + k v²);
ds
or since v
dt
ď's
di
d x
+ S⋅ d s
+ch.
ds
ds?
k
0........ (1).
dt
d t²
217
Similarly, in ascent,
d² s
dt
dx
+ g
ds
+2h + k
dt
ds
ds
...(2),
dt2
s being positive.
1
To integrate equation (1). Let s = -
hyp. log. (1-z) ;
k
ds
=
1
dz
dt
k
1 - 2
dt
d² s
1
dz2
1
1
d² z
=
dť² k (1 − z)² ' dť
Substituting these values, we have
+
T
dt²
1
1
d² z
π ī — z dť²
+ gk (1-2)
dx 2 h
+
1
dz
= 0;
dz
k l-z
dt
d² z
..
dt2
dz
dt
dx
+2h + k²g (1 − z)².
·
= 0....
.... (3).
dz
Now, suppose the last term of this equation to be expanded in
a series of powers of z. It will not involve any power of z below
for at A, where s = 0, and therefore z = 0, we must have
x∞ ·´`·
2 × s², and .. x œ z²;
dx
xz.
dz
·
Let therefore
dx
k² g (1 − z).
=
Az+Bzẞ+&c. where ß> 1…..
·(4);
dz
d² z
dz
..
d t²
+2h
+ A z + Bzß +&c.=0.
dt
Let T be the time of descending to the lowest point, and put
Then our equation becomes
T-t=t'.
d² z
dť½
2 h
dz
dť
+ A% + B÷B + &c. = 0..
·(5).
E E
218
If we omit the terms after Az, we can integrate this equation,
and find a factor which will make the first part of the expression
immediately integrable. This factor is
€
-t' (h— y √ − 1)
for if we differentiate,
we find
€˜t' (h—y √ − 1) (dz
where y= √(A — h²),
dt
(h+y V−1)z},
€
6-t' (h− y √ − 1) { d°z
dz
2 h
2
dť
dť
{+ (1² + y²) z} dt.
t'
e
Hence, multiplying equation (5), by "(h~~~−1), and
integrating, we have
-
€ − t (h− y √ − 1) (dz
{πť ~ (h+y v-1)
Idť
+ Bƒ z³ dť' €˜ť'(h−y√ − 1)
€
zB t'
= }
+&c. + C = 0.
Putting for "-1 its value cos. t'y + V(−1). sin. t'y,
and taking the impossible parts, we find
e-th
dz
{(37) — hz) sin. t'y — yz cos.
dt'
-
t'y}
+ C = - Bƒz³dt sin. t'ye¯th+&c.
But when t=0, t=T', s=0; .'. z=0; and .. C'=0, the
integral on the right-hand side being taken from t=0. Also
when t'=T,
d s
= 0,
0, and
d z
= 0.
dť
d t
Let at that point z=2; and we have
€--Th{h
{h sin. yT+y cos. yT} 2 = Bfz³dt' sin. yte
the integral being taken from t'=0, to t' = T.
- ht'
+&c.
Now, when the oscillation is indefinitely small, and therefore
219
z indefinitely small, the second side vanishes (for ẞ> 1). Hence,
we have in this case,
h sin. y T÷y cos. y T=0, tan. y T=
Y
· · (6).
h
But the time T is, by the supposition of tautochronism, inde-
pendent of the arc, and we must therefore have the same equation
for it, when z is not indefinitely small. Hence, we must have
β
0 = Bƒ½³ dt' sin. y t'eht' + &c.
between the limits t'=0, and t = T. But this cannot be the case
except B=0.
For the factor sin. yt'e-ht' is always positive
between the limits; and when z is small, the first term may repre-
sent the whole expression. Hence, the integral cannot be = 0;
and therefore we must have B=0.
Hence, we have from (4),
k² g (1 − z)²
dx
dz
=
= A z
A z ;
or, since
COR. 1.
we have
d z
→ks
€
and
(1 − z)²
= ke**.ds,
dx
k g €¯k s
-ks
e
= A (1
€
(−ks),
ds
kg d x = Ads (e*s — 1).
If we expand e**, divide by k, and then make k=0,
If we expand e
gdx
A sds,
which is the expression for the tautochron in a non-resisting me-
dium.
COR. 2. The expression for T in (6),
coefficient of the square of the velocity.
does not involve k, the
Also, if the resistance
involved terms mv³+nvª +&c. this expression would be the
same.
220
COR. 3. For the arc of ascent we may proceed in the same
manner, making
1
S=
hyp. log. (1 + z); and we shall have
kgdx = Ads (1 — €¯*³).
For the brachystochron in a resisting medium, see Woodhouse's
Isoperimetrical Problems, p. 141.
BOOK III.
THE MOTION OF A RIGID BODY OR SYSTEM.
70. THE Conclusions of the preceding part of the Work are true
of the motion of a point: some of them are also true of a body of
finite magnitude, if we suppose it, during its motion, always to
retain the same position, so that any line drawn in the body may
continue parallel to itself. But they are no longer necessarily true
if the body have any rotatory or angular motion; that is, a motion
such that a line drawn in the body, and retaining its position in it,
revolves successively into different directions.
principles and new formulæ are to be applied.
In these cases new
CHAP. I.
DEFINITIONS AND PRINCIPLES.
71. WHEN a solid body has any rotatory motion, its parts act
upon one another, and thus modify the effects of the other forces
by which they are acted on. Every body, under such circumstances,
may be considered as a machine; and by means of its rigidity or
other properties, the forces which are applied to one part pro-
pagate their effect to another. So that each particle both presses
with its own force, and serves to form levers and rods by which
229
the pressures of other particles are communicated. The laws
according to which this connexion of different particles modifies
the effect of the forces which move them, are to be the subject of
our consideration in the present division of the Work.
When a body revolves continually about the same fixed line,
this is called a permanent axis. This axis may be merely a
mathematical or imaginary line: the points where it meets the
surface of the body are called poles.
A body, while revolving, may change its axis of rotation, so as
to revolve first about one line, and then about another. And we
may suppose, that this change of the axis to be perpetually and
gradually taking place; so that the body revolves about the same.
axis only for an indivisible instant, and the next moment proceeds
to revolve about an axis immediately contiguous to the former, and
so on continually; the poles of the axis moving perpetually upon
the surface of the body. In this case, the axis at any instant is
called the momentary axis.
A motion of translation is distinguished from a motion of rota-
tion by the latter, the body merely changes its position about
some point; in it (as for instance, its centre of gravity;) by the
former, this point changes its place, and is transferred to some new
point of space.
By combining a motion of translation with a rotation round a
variable axis, we may produce any motion whatever; conversely,
the motion of a body in any manner whatever, may, at any instant,
be resolved into a motion of translation of the centre of gravity, and
a motion of rotation about an axis passing through that point*.
* This may be thus proved. If the centre of gravity be in motion, let
an equal velocity be communicated to the body in the opposite direction, so
that that centre may be at rest. Then let the centre of gravity of the body
be made the centre of a sphere, fig. 77, and let the points of the body be
referred to the sphere's surface by lines drawn through them from the centre.
Let P be one of these referred points, and in an indefinitely small time let P
move to p. Draw a great circle PQ perpendicular to Pp, and in this small
time
223
72. The principles upon which our reasonings must depend
are the laws of motion, which have been already applied to points.
The most important principles which will be requisite in applying
them to the cases in question, are the two following.
Principle I. Instead of the forces which act upon any body in
motion, we may substitute those which are equivalent to them
according to the principles of Statics.
Thus a force P (Pp, fig. 78.) acting at a certain distance (CP)
to turn a body round an axis (C), exerts the same effort as a force
2P (P'p') acting similarly at half the radius (CP').
Thus any force which acts to turn a body round an axis, acts
effectively upon all the particles; the body itself transmitting the
action after the manner, and according to the laws, of a lever.
Principle II. If the particles of a system when unconnected,
would move so as always to have the same relative situation, we
may suppose them to be connected, and their motions will remain
unaltered.
Thus if the particles P and Q, fig. 78, which are moveable
separately in the same plane about the centre C, be acted upon
by such forces that their angular velocities round C would always be
equal, we may suppose P, Q connected by a rigid rod PQ, and
they will move in the same manner as before. For the bodies have
no tendency to increase or diminish their distance from each other,
and therefore exert no force in PQ, and cannot disturb each other's
motions by means of the rod PQ.
This principle is an extension of the one in Statics, that when
a system is in equilibrium any two of its particles may be supposed
to be connected.
time suppose an arc PQ to come into the position pq; therefore
pq=PQ,
and Qq must be perpendicular to PQ. Let the arcs QP, qp produced
meet in O; then O will have been at rest during this small time. And if
we draw a line from the centre of gravity to O, all the points of the body in
this line will have been at rest, and therefore the body during this instant has
been revolving round this line as an axis, (Euler, Theor. Corp. Sol. Cap. II.
Th. 9.)
224
These principles may be applied immediately to the motions of
any bodies. We may also deduce from them a general theorem,
equivalent to what is called D'Alembert's Principle, and reduce
our mechanical conditions into equations by means of it.
The forces which really act upon a system are called the im-
pressed forces. The forces which must act upon each of the points
of the system, (supposing them unconnected,) in order to produce
the effect which really takes place, are called the effective forces.
73. PROP. If any forces whatever act upon any points of a body
or system, the impressed forces, and the effective forces on all the
points of the system, would produce an equilibrium by acting on the
same system at rest; (according to its statical properties;) the
latter forces being supposed to act in a direction opposite to that in
which the forces are impressed.
Let accelerating forces P, Q, R, &c. act upon a system; and
let p, q, r, &c. be the masses of the particles on which they
respectively act; also let m, m', m", &c. be the other particles of
the system.
Let the effect produced in an indefinitely small time be that
which arises from compounding velocities a, ß, y, w, w', w', &c.
with the original velocities of the points p, q, r, m, m', m", &c. And
let P', Q', R', M, M', M", &c. be the accelerating forces which
would produce this effect.
Then Pp, Qq, Rr, &c. are the moving forces impressed; and
P′p, Q'q, R'r, Mm, M'm, &c. the effective moving forces; and it
is to be shewn, that the latter are statically equivalent to the former.
If the forces impressed are not equivalent to P'p, Q'q, R'r, Mm,
M'm', &c. let them be equivalent to kP'p, kQq, kR'r, kMm,
k M'm', &c.*. And since P'p, &c. would produce in an indefinitely
* This is always possible. For, from the nature of the system, the
virtual velocities of the parts are given. Hence, in the equation of equi-
librium given by the principle of virtual velocities, k P', k 2', &c. will alone
be unknown. Also, since P', Q', &c. are known, k will be the only un-
known quantity in the equation, and may be assumed so as to satisfy the
equation.
225
small time velocities a, &c.; k P'p, &c. acting on the particles,
supposing them unconnected, will produce velocities ka, &c. And
since the system can have the velocities a, ß, &c. communicated to
its parts without destroying their connexion, the velocities ka, kß,
&c. can be communicated without affecting that connexion. Hence,
by Principle II, the forces k P'p, &c. acting upon the system, sup-
posing it rigid, will produce the same velocities ka, &c. And
therefore by Principle I, the impressed forces Pp, &c. which are
supposed equivalent to k P'p, &c. will produce in the system the
velocities ka, &c. But they produce the velocities a, &c. There-
fore k=1, and Pp, Qq, Rr, are equivalent to P'p, Q'q, R'r,
Mm, M'm, &c. Therefore Pp, Qq, Rr, along with P'p, Q'q,
R'r, Mm, M'm', &c. acting oppositely, will produce an equilibrium
Q. E. D.
COR. Each force may be
effective force and the force lost.
valent to P'p and Ap; Qq to
Cr, &c.; the forces Ap, Bq, Cr, &c. are the forces lost.
Mm, M'm', &c. are the forces gained".
considered as equivalent to the
Thus, if the force Pp be equi-
Q'q and Bq; Rr to Rr and
This being premised, the following proposition is true.
Also
When any forces act upon a system, and produce motion, the
forces lost and gained would balance each other, acting upon the
system in opposite directions.
It has already been shewn, that
–
—
·P'p, Q'q, R'r, -Mm,
M'm', &c.
would balance Pp, Qq, Rr. That is, they would balance
P'p, Ap, Q'q, Bq, Rr, Cr.
Therefore Mm, M'm, &c. would balance Ap, Bq, Cr.
Or Ap, Bq, Cr would balance Mm, M'm', &c. acting in the
opposite direction.
It is in this form that the principle was enunciated by
D'Alembert.
* It is evident that Mm, M'm', &c. may be classed with the forces
Ap, Bq, &c. for the force which acts on m is 0.m;
Mm, the effective force, and — Mm, the force lost.
considered as the force gained.
F F
which is equivalent to
Hence, Mm may be
226
1
CHAP. II.
ROTATION ABOUT A FIXED AXIS.
74. WE proceed to determine the angular motion produced when
forces act upon a body moveable about a fixed axis. We consider
the effect in producing motion only. The other effects of pro-
ducing pressure upon the axis, and affecting its motion when it is
moveable, will be investigated afterwards.
PROP. In a system consisting of any number of points m, 11,
p, q, &c. fig. 79, in the same plane, moveable about an axis C,
perpendicular to that plane, a force Facts to turn the system; to
find the effective accelerating force on any point.
Let F be a moving force which acts perpendicularly at an arm
Cf. And let M, N, P, &c. be the effective accelerating forces
on m, n, p, &c. Therefore Mm, Nn, Pp, &c. are the effective
moving forces; and they are perpendicular to CM, CN, CP, &c.
because the motion is so.
Hence, we have
Impressed force.... Facting perpendicularly at an arm CF,
Effective forces ....Mm, Nn, Pp, &c. acting perpendicularly at
arms Cm, Cn, Cp, &c.
Hence, by Art. 73, and by the general proposition of the lever,
F.Cf-M.m.Cm - N. n. Cn - P. p. Cp - &c.=0.
But since m, n, p, &c. must all move with the same angular
velocity round C, their linear velocities must always be as Cm, Cu,
Cp, &c. and therefore all alterations of the velocities must be in
this ratio, and the accelerating forces which produce them in the
same ratio. Hence, we have M N :: Cm Cn; therefore
227
11
M.Cha
M.Cp
N =
; similarly P =
&c. And substituting
}
Cm
Cm
Cn
Cp²
M.p.
&c.=0.
Cm
Cm
these values in the equation
F.Cf-M.m.Cm-M.n.
Whence M =
Similarly N =
F.Cf.Cm
m. Cm² +n. Cn²+p. Cp²+&c.
F. Cf. En
m. Cm² + n . Cn²+p. Cp² + &c.
F. Cf. Cp
m. Cm²+n. Cn²+p. Cp²+ &c.
P =
and so on for any other point.
F.Cƒ³
And the effective force on ƒ=-
m. Cm² +n. Cn+p. Cp+ &c.
1
The effective force at a distance 1 from the axis
F.Cf
m. Cm² + n. Cu+p. Cp²+&c.
* As this Proposition is the foundation of the whole doctrine of rotatory
motion, we shall shew how it may be deduced from elementary laws, inde-
pendently of D'Alembert's Principle.
Let F be statically equivalent to a number of accelerating forces M at m,
N at n, &c. which act perpendicularly to Cm, Cn, &c., and are as Cm, Cn,
&c. Therefore
Cn
F.Cf-M.m. Cm-N.n.Cn, &c.=0; and N=M &c.
&c.=0; and M=
C'm'
F. Cf.Cm
m.Cm²+n. Cn
Cn²
.. F.Cƒ—M .m.Cm— M.n
Cm
&c.
Now, by Principle I, Art. 72. M, N, &c. will produce the same effect as F.
Also, if Cm, C'n be supposed unconnected and moveable independently about
C, and if M, N, &c. act on m, n, the accelerations of m, n, will be propor-
tional to Cm, Cn, and therefore the angular velocities of Cm, Cn will be
equally increased, and they will retain the same position with respect to each
other. Hence, by Principle II, Art. 72. we may suppose Cm, Cn, &c.
to be connected, and the system to become rigid, and the effect will still be
the same. Hence, when Facts, the effective force in m is M as before.
A
228
If instead of the force F we have any forces acting in any
manner, we must substitute instead of F. Cf the moment of these
forces about the axis C; that is, the sum of each into the perpen-
dicular upon it from C; those being taken negative which tend to
turn it in the opposite direction.
COR. 1. Since the effective accelerating force on f
m
·
F
2
Cm
Cn²
Cp2
+n.
Cf2 Cf²
+p.
It appears, that the resistance which m opposes to the com-
munication of motion, is the same as that of a mass m.
Cm²
Cf²
placed
at f, and acted upon immediately; and similarly of the other par-
ticles.
COR. 2. It appears by the demonstration, that the effective
forces on different points are as their distances from the axis C.
COR. 3. If the force F, and the radius Cf be constant, the
effective force on each point will be constant; the motions will be
uniformly accelerated, and the formula for such motions may be
applied. If F be variable, the formulæ for variable motions may
be applied.
COR. 4. If the force which acts be the weight of any body,
this body must be included among the bodies m, n, p, &c. in the
denominator.
Thus if a system of material points in horizontal planes, m, n,
p, fig. 80, be moved about a vertical axis AC, by a weight W acting
perpendicularly at the radius Cf, by means of a string passing over
a pully B; W moves with the same velocity as a body at the
extremity of the area Cf; and therefore the same effective force is
employed in moving W, as if it were at f. Hence, we have
effective force on ƒ
1

weight of W. Cf²
W.Cf+m. Cm²+n. Cn²+ &c.
and the effective force on W is the same.
COR. 5. The quantities W, m, n, &c. in the denominator in the
last Corollary, are the masses of the bodies; the weight of W in the
numerator is a moving force. If g represent the accelerating force
of gravity, the weight of W is Wg.
229
COR. 6. If the lines m, n, p, &c. be not in the same plane
perpendicular to the axis, if Cm, C'n, C'p, &c. be their perpen-
dicular distances from the axis, the same formula will be true,
putting these lines for Cm, Cn, Cp, &c.
Or, if we take a plane Cmn, perpendicular to the axis, and
refer the points of the system to this plane, by lines parallel to the
axis; if m, n, p, &c. be the points thus referred, the same formulæ
will be true.
The denominator of the fractions which express the effective
forces in the preceding formulæ, is the sum of each particle mul-
tiplied into its distance from the axis. This sum is called the
Moment of Inertia with respect to this axis*. It occurs perpe-
tually in considering the subject of rotation.
eng
If the system, instead of consisting of distinct material points,
be a continuous body of finite magnitude, the momentum of inertia
will be the sum of each point into the square of its distance from
the axis, and will consist of an indefinite number of terms. The
sum of these terms may be found by the integral calculus, as will
be shewn in the following Chapter.
If the points be m, m1, M2, M3, &c. and their distances from
the axis, Cm, Cm1, Cm2, Cm3, &c. the moment of inertia may
be represented by (m. Cm²). And if F be a moving force which
Σ
:
* The inertia of a body is its effect in resisting the communication of
motion in a single point, it is as the mass simply; but in a body revolving
about an axis, the effect of a particle in resisting motion depends on the
distance from the axis, like the effect of the force acting on a lever. The
effect on a lever is as the product of the force and distance, and this product
is called the moment; the effect of the inertia of the mass in resisting rota-
tory motion, appears from the above investigation to be as the product of the
mass and square of the distance, and hence, this product is called the moment
of inertia. And the sum of these products is called the moment of inertia of
the system.
230
acts perpendicularly at a distance Cf, we shall have the accele-
rating force at the point where the force acts =
F.Cƒ²
Σ (m. Cm²) *
*
If forces act upon every point of the system, the effect may be
calculated by the same principle as before, as will be seen in the
next Problem.
75. PROP. A system of material points, moveable about a
horizontal axis, has all its parts acted on by gravity; it is required
to determine the accelerating force.
Let C, fig. 81, be the axis, aud m, n, p, the points. Draw
a horizontal line through C, meeting vertical lines through m, n, p,
in d, e, h. Then the moving forces impressed are the weights of
m, n, p.
Let M be the effective accelerating force upon m;
therefore in the same way as befere, the effective accelerating
forces on n, p, are
M. Cn
M.Cp
Cm
Cm
And the effective moving forces are
M.n.Cn
M.m,
Cm
M.p.Cp
Cm
Now Cd, Ce, Ch are perpendicular to the directions of the former,
and Cm, Cn, Cp to those of the latter; also, if m be the mass of
one of the bodies, its weight or moving force is mg, and so for
the rest. Hence, by the equilibrium between the impressed and
the effective forces, we have, by Art. 73.
m.g.Cd+n.g.Ce+p.g. Ch
= M.m.Cm +
:: M =
M. n. Cn²
C m
M.p.Cp
+
C m
(m.Cd+n.Ce+p. Ch) Cm.g
m. Cm²+n. Cn² +p. Cp²
COR. 1. If we had supposed more bodies, we should have
had a corresponding number of terms, both in the numerator and
denominator.
231
COR. 2. There will be negative terms in the numerator, when
any of the bodies are on the other side of the vertical through C;
the terms in the denominator will always be positive, because the
bodies all move in the same direction round C; and therefore the
effective accelerating forces are always in the same direction.
COR. 3. The effective accelerating force ou any other point of
the system, asn, will be
N =
(m. Cd+n. Ce+p. Ch) Cn.g
m. Cm²+n. Cn² +p. Cp²
COR. 4. If G be the centre of gravity of the system, and if
a perpendicular from G meet Chin H, we have
(m.Cd+n. Ce+p. Ch) = (m+n+p). CH.
And if 0 be the angle which CG makes with the vertical,
CH = CG. sin. 0.
Hence,
M=
(m+n+p) CG. sin. 8. Cm.g.
m.Cm²+n. Cn²+p.Cp²
;
,
or, denoting m+n+p by Σm, and the denominator by Em. Cm²,
whatever be the number of bodies,
M=
Cm.CG.g sin. 0. Em
Σ (m. Cm³)
76. PROP. To find a point of the system which shall be
accelerated exactly as much as a single point in the same position.
If O be any point in CG, we shall have
accelerating force on 0 =
O
CO.CG.g.sin. 0.2m
Σ (m. Cm²)
Now, if a single particle were placed in O, and all the rest
removed, we should have
accelerating force on particle in Og sin. 0.
1
1
232
And we have to find O, so that these accelerating forces may
be equal. For this purpose, we must have
CO.CG.Σm = Σ (m. Cm²);
Σ. (m. Cm²)
.. CO=
.(a).
CG.Em
The point O is called the Centre of Oscillation; a single point
placed in O, would, in any position of CG, be acted on by the
same accelerating force as when O is a point in the system; and
therefore, the oscillations of CO and of the system would be ex-
actly the same as if we had but one particle O.
COR. 1. The time of oscillation of the system, is the same as
that of a simple pendulum, whose length is CO. Hence, if we
make CO=1, we shall have the time of one of the small oscilla-
tions = π
COR. 2. When we know the moment of inertia, and the place
of the centre of gravity, the centre of oscillation with respect to the
axis C is found by the formula
CO =
Σ (m. Cm²)
CG.Em
And this is applicable, whether the system consist of distinct
points, or of finite bodies.
!
CHAP. III.
MOMENT OF INERTIA.
77. In the present Chapter we shall find the moment of inertia
of a variety of different bodies, and with respect to any axis. From
this it will be easy to deduce the position of the centre of oscillation.
We shall in the first place, prove a property of the moment of
inertia, by means of which, knowing this moment with respect to
any axis passing through the centre of gravity, we can find it with
respect to any other axis parallel to the former.
SECT. I. General Properties.
PROP. The moment of inertia of any system, with respect to
any given axis, is equal to the moment about an axis parallel to
this, passing through the centre of gravity, plus the moment of
the whole body, (collected in its centre of gravity,) about the given
axis.
Let fig. 82 represent any system, moveable about an axis C;
and let m, n, p, q, be the particles of it, referred to a plane per-
pendicular to the axis. Let G be the centre of gravity of m, n, p, q.
Draw md perpendicular on CG.
Now, Cm² = CG² + Gm² + 2 CG. Gd.
Similarly, if ne, ph, qk, be perpendicular on CG,
Cn² = CG² + Gn-2 CG. Ge,
&c. &c.;
G G
234
... m . Cm² + n. Сn²+p. Cp²+&c.
= m.CG²+n.CG²+p. CG² + &c.
+ m. Gm² +n. Gn²+p. Cp²+&c.
+ 2 CG (m. Gd-n. Ge+p. Gh- &c.)
And, since by the property of the centre of gravity,
m. Gd-n. Ge+p. Gh-&c. =0;
we have
m. Cm² + n. Cn²+p. Cp² + &c. = (m+n+p+&c.) CG²
+m. Gm²+n. Gn+p. Gp²+&c.
Or, moment of inertia round C-moment of (m+n+p+&c.) at
Ground C+ moment round G.
COR. 1. We may represent this theorem thus, whatever be
the number of bodies;
Σ (m. Cm²) = CG²Σm +Σ (m. Gm²).
COR. 2. Knowing the moment of inertia round G, we may,
from this expression, find the moment round C.
COR. 3. The moment round G, the centre of gravity, is less
than that round any other axis C, parallel to G.
78. Since the expression for the moment of inertia of a system,
consisting of a finite number of points, is (m. Cm²), m being
the portion of the mass which is at the distance Cm; when the
number of points becomes indefinite, the expression will evidently
become fd M. Cm², where d M is the differential of the mass at
the distance Cm. For we approach the true value by dividing the
mass into portions smaller and smaller indefinitely, and taking the
sum of each into the square of its distance; but by the nature of
the differential calculus, we thus approach the differential of the
mass at each point, multiplied into the square of its distance, and
the integral of this. Hence, if r be the distance of any point from
the axis, and d M the differential of the mass corresponding to dr,
moment of inertia = fr³d M*.
2
* This may be proved more distinctly thus.
Let AB, fig. 83, be any body, and Cm=r, C'm' = r', be two distances
from
235
This moment will be the same as that of the whole mass, col-
lected at a certain distance. If we call this distance k, we have
k² M = fr²dM*.
We may con-
We proceed to find kM in different bodies.
sider bodies as being either physical lines, surfaces, or solids, and
apply our formulæ to each. The moment itself will depend upon
the thickness of the lines and surfaces, and the density of the sub-
stance; but if the mass be homogeneous, the line k will depend
only upon the geometrical form of the system. The other quan-
tities will enter as multipliers on both sides of the equation
2
k² M=fr²d M.
Hence, in finding k we may suppose the thickness of lines and
surfaces, and the density of the mass, each to be unity.
from the axis M the mass included in the distance r from the axis, M'
that included in the distance r'. And let the moment of inertia of the former
portion = R, and of the latter = R'. Therefore, the moment of the portion
m n m', included between the two distances, will be R'- R. And the portion
itself will be M'—- M.
Now R'-R is evidently greater than (M'— M) r²,
and less than (M' — M) r'²;
R'
Ꭱ .
is > r², and < r²².
M
M'.
But, as M' and M become ultimately equal, r² and ² become ultimately
equal, and therefore
ultimately
R' - R
is ultimately equal to either of them. Also,
M'- M
R' R
-
M'— M
is
d R
dM
therefore, we have
d R
d M
= r², dR= r²d M, R= r²d M.
M‚_R=/
C
* The centre of gyration is defined to be the point at which the whole
mass must be collected, that the rotatory motion communicated by a given
force may be the same as before.' It appears by the last Chapter, that any
point, whose distance from the axis is k, possesses this property. We may
call k the radius of gyration.
236
SECT. II. Moment of Inertia of a Line, revolving in its
own Plane.
79. If any distance from the axis be r, and ds the differential of
the length of the line, corresponding to ds, it appears by what has
just been said, that we may suppose the thickness and density of the
line each 1, and dMds. Then k² M = r²ds.
2
=ƒ
We may take, for the centre of revolution, any point with respect
to which the relation of ds and r is most simple; and then by
Art. 77 find the moment of inertia about the centre of gravity,
and from that, about any other point.
Ex. 1. To find k for a straight line revolving about an axis
perpendicular to it in its middle point, fig. 84.
Call any distance Cm=r, CA
2ak² = ƒ r³dr=
=
CB=a, M=2a,
and, the integral being taken from r =
2.3
+ constant.
3
a to r = a,
2ak²
=
3
2 a³
3
a²
k² =
3
a² M
when M is the
3
Hence, momentum of inertia = k² M
mass of the line, and its thickness and density may be any whatever.
(The thickness must necessarily be small, that it may be considered
as a line).
Ex. 2. A straight line about an axis perpendicular to it through
any point of it; (D, fig. 84.).
Let mass M, AB = 2a, CD=b.
By Art. 77, mom. ab. D=mom. ab. C+ mom. of M at dist. CD,
a²M
KM =
3
k²
N
a
2
+b2M,
+ b².
3
237
Ex. 3. A straight line about any axis perpendicular to the
plane in which it is.
AB fig. 85, about E,
AB = 2a, CD = b, DE = c; EC² = b² + c².
As before, k2M = mom. round C+ mom. of M at dist. EC.
2
=
... k² =
a² M
+ (b²+c²) M;
3
2
a
+ b² + c².
3
Ex. 4. A circular arc about the centre of the circle.
If a be the radius, r always = a, and mom. of inertia =
.. k² = a².
2
a² M;
Ex. 5. A circular arc about an axis perpendicular to its plane
through its centre of gravity, fig. 86.
Let C be the centre of
the arc PAP, G its centre of gravity. Then by the rules for the
centre of gravity, (Statics, Chap. VI, Ex.
29.)
CG
CA. chord Pp
arc Pp
aq
= ;
P
a being the radius, p the arc, q its chord.
Then, by Art. 77, mom. round C = mom. round G+CG². M ;
.. mom. round G = mom. round C – CG². M,
a
k² Ma°M -
M,
9
a² (
2
p
If PAP be a semi-circle, q=2a, p=πα,
k² = a² (1-3)
Ex. 6. A circular arc about an axis perpendicular to its plane
through its vertex A, fig. 86.
238
By the two last examples we shall here easily find
k²
2
= 2a² (1 -
· (1 − 2).
If the arc be a whole circumference, q=0, k²=2a².
SECT. III. Moment of Inertia of a Line, revolving
perpendicularly to its own Plane.
80. In such cases the figure revolves about a line in its own plane,
or parallel to it. If this line divide the figure symmetrically, as
shall have d M = 2 differential of arc AQ, of
which NQ = r is the ordinate and if AQ = s, k² M = 2 ƒ r²ds.
AC, fig. 87, we
Ex. 7. A circular arc about a radius through its vertex A,
fig. 37.
Let CA = a, ds
adr
2
√(a² — p²) 3
2ar dr
—
k². arc PAp = √ √ (a² ~ 7")
S
= C—ar V (a² — r²) + a³ arc
(sin = :-)
:).
a
And this, being taken to begin when r≈0, gives, putting p for the
arc PAP, and c for PM,
a²
2
k² p
2
2
1² =
Q
19
—
p.
» — a c √ (a² — c²),
ac V (a² — c²)
Ρ
If the arc be a semi-circle, c=a, k² =
The same is true if the arc be a circle, and .'. c = 0.
239
SECT. IV. Moment of Inertia of a Surface, revolving
in its own Plane.
81. If the mass be a surface revolving about an axis perpendicular
to it, we shall have dM=fdr.rde, taken between proper limits
of 0; k³M = ƒƒ r³drdě.
The integrations may be performed in any order. If we in-
tegrate first for r we shall have
k² M=
orde
4
**
the value of being taken which belongs to the boundary of the
figure. The same expression might also be obtained by conceiving
the figure divided into triangles of indefinitely small width by lines
drawn in it from the axis.
Ex. 8. To find the moment of inertia in any triangle, about
an axis perpendicular to its plane, and through one of its angles.
Let ACB, fig. 88, be the triangle, and C its axis; CD per-
pendicular on AB=h. CB= a, CA = b, AB = c; DCM=0,
CM=r,
0 = angle (cos.
до
COS. =
4);
; do =
Ꮎ
hdr
r V (r² — h²) ³
hr² + 2h³
3
12
√(r² - h')+constant,
hr³dr
k² M = √ 4 √ ( ² — h²)
and for the triangle ADC, we must take the integral from r=h to
b²+2h²
rb, which gives
h√ (b² — h³).
12
a² + ch²
have
12
h V (a² — h²).
Hence, for ACB we have
Similarly for BDC we
a² + 2h²
b² + 2 h²
k²M=
hv (a²
hV (a - h) +
h √ (b² — h²).
12
12
a²+ch
COR. 1. If the triangle be isosceles, k²M =
M.
6
240
COR. 2. Let AB be bisected in E, and DE=q ;
.. BD = √(a² — h²)
√(a²—h³) == + q, AD= √ (b² — h²) = — — q⋅
And BC BD² = A‚C² — AD²,
2
or a² - ( + q)² = b² − ( − 9)˚ ;
Also a²+2A′ =a' + 2
... a² — b² = 2cq.
+ 9) '} = 3a² − 2
α
-
(
+ 9) ',
+ 4) .
− q)'} = sb°
− 9 ) } = 36° — 2 (
{a"
2
—
(
b² + 2 h² = b² +
2
{
b²
−
—
(
(
(
− q)
Hence,
h
k²M=
12
+ 9
h
· 1/2 {3a² († + 9 ) − 2 (
= {( −
(a²+b²) c + 3 (a³ — b²) q
12
3
-
)² + 3 6° ( − q ) − 2 ( − 9 ) ' }
C³
— -6cg"},
2
or, since 3 (a² – b²) q = 6cq²,
cq
h
2
k²M=
(a² + b²) c
12
ch
11
{3 (a²+b²) — c²}
24
M
{ 3 (a² + b²) − c² } ;
12
... k²
3 (a² + b²) - c²
12
Ex. 9. A triangle about an axis through its centre of gravity.
By the property of the centre of gravity, (Statics, Chap. VI,
Ex. 3.)
2 (a² + b²) — c²
CG2=
9
241
}
M
{3 (a² + b²) — c² } = M.
12
c²·
2 (a² + b²)—c
9
2
+ kⓇM.
[ { a² + b² + c } = M (k + k + f
K²M M
36
12
h, k, l being the distances from G to the angles A, B, C.
If the triangle be equilateral,
a²
12
KM = M.
=
M
12
4
Ex. 10. A parallelogram about an axis perpendicular to it
through its centre of gravity.
ABDE, fig. 89, the parallelogram, AB = 2a, AD = 2b,
BD=2c.
Let G be the centre of gravity of ABD; .. by Statics,
GC² =
2 (4a²+4b²)—4 c² 2 (x² + b²) — c²
36
9
And mom. of ABD round C = mom. round G+ABD.GC²
2 (a² + b²) — c²
j4n² +4 b² + 4c²
= ABD
+
36
(a²+b²)
= ABD
S
and doubling both sides, since 2 ABD = M,
a² + b²
a² + b °
KM = M.
and k²
3
3
9
Hence, it is independent of the angles of the parallelogram.
Ex. 11. Any regular polygon about an axis perpendicular to
it through its centre.
The polygon may be divided into isosceles triangles, and for
each of these, mom.
a²+ec²
6
A, a and c being the radius of
the circumscribed and inscribed circle, and A one of the triangles ;
.. RM =
a² + 2 c²
6
Q
a²
n A =
M
6
6
(1+2 cos.²),
H 11
242
M
6
(2 + cos. 2=) M,
π
1º
2
1 + 2 cos.
or, if I be a side, k²M =
N
M.
24
75
sin.2
n
Ex. 12. A circle revolving about an axis perpendicular to it
through its centre.
Here r is constant, and
12 M =
a, the radius;
=√
a* d Ꮎ
do
2пал
4
,
integrating for ;
4
4
πα
a² M
a
k2
2
2
Ex. 13. An annulus whose external and internal radii are a, b,
4
k² M=
π (a* — b¹)
a² + b²
2
(π a² — π b²)
a² + f ²
M,
10
2
k² =
a² + b²
2
Ex. 14. An ellipse revolving about the centre.
If O be measured from the extremity of the major axis,
2
1 - e² cos. "0
-
2,2
Ꮎ
.. k² M
= b²S
+
= v+S
d Ꮎ
(2-2 e cos."0)3
d Ꮎ
2
(2 — e² — é² cos. 20)*
do
putting n for
び
​(2-e²)²
n cos. 20)²'
e
2-e
I
243
To integrate, let P =
2 cos. 20 de
sin. 20
n cos. 20'
and the integral = S,
2 n sin.220d0
d P =
1
n cos. 20
(1 − n cos. 20)²
==
2 cos. 20 d0 – 2 nd 0
(1 − n cos. 20)²
d0
2 (1-n²) do-(1-n cos. 20) de
(1-n cos. 20)°
N
2(1-n²)
de
1
Q d Ꮎ
N
(1 − n cos. 20)²
n 1 − n cos. 20'
2 (1-n²)
ᏧᎾ
:. P =
. S
n
-n cos. 20
And this latter integral is (Lacroix, Elem. Treat. Note K),
1
arc
√(1-n²)
(co
cos. 20 - n
1
COS. =
1
n cos. 20
√(1 − n²)
- nº4: suppose,
1
1
.. S
P+
A.
2(1-n²)
2 ( 1 − n²) ⇓
-
=
If we take the integrals from 0=0, to 0π, and then double
them, in order to obtain the whole value of S, we have
パ
​S=
1
arc (cos. = 1)
=
... k² M =
12)
(1 — 1
Ωπ
n
(1 — n
2 π b+
(2 — e²)² (1 − m²)³ ³
e
and putting a √(1 − e³) for b, and √(2¿¹) for 11,
k² M
πα
a¹
4
4
— e²)
(2 — e²) √(1 − e²).
But the area M=πa² √(1—e²);
:. k²
a² (2—e³) a² + aˆ (1 − e')
a² + b²
4
4
4
244
a²
When e=0, k*
as in a circle.
a
When e=1, b=0, だ
​4
82. If the surface be bounded by rectangular co-ordinates, we
may thus find its moment when revolving in its own plane.
Let the centre of motion C, fig. 90, be the origin of co-ordinates
CN=x, NQ=y. And we have
moment of Qq about Cmoment of Qg about N+ CN². Qq,
(by Ex. 1, Art. 79.),
y²
2y+x². 2y.
3
And multiplying by da, and integrating,
2
3
k° M = 2ƒ ( + x²y) dx.
S &
2
Ex. 15. A parabola revolving about its vertex. Here y¹²,
X
2aª
1º M = 2ƒ° (“²²² + @²x² ) d x = 2 (24²x² + 20¹³),
K²
3
7
Q a²x²z
a³ xî
15
7
3x²
and M
a x
x²; • . k²
y
+
3
7
x and y being the extreme ordinates.
SECT. V. Moment of Inertia of a Plane revolving about
an Axis in or parallel to the Plane.
83. When any plane revolves about a line in it, we may call r
the distance of any point from the axis; and if dz be the differential
of the axis, fdzdr integrated with respect to z, will be the portion
at distance r; and ffr²dzdr the moment of inertia.
2
3
r³ dz
; and is given in terms
The integral JJ'r² dzdr is = f²;
3
of z.
245
Ex. 16. A circle revolving about a diameter.
Let radius=a, z measured from the vertex;
•. a—z= √(a² — p²),
rdr
d z =
√(a² — r²)
;
k² M==
r4 dr
√(a² — r³)
which, integrated from z=0, to z=a, gives for a semi-circle,
пая
Co
k² M =
TA
a²
a²
And since M =
k² =
2
4
The same expression is true for a whole circle.
Also for an ellipse revolving about either principal axis, 2a
being the other.
Ex. 17. A circle revolving about a line parallel to its plane,
at a distance c from its centre; radius = a.
A diameter being drawn parallel to the axis of rotation, we
have, by last Example,
moment round diameter =
2
M.
4
And therefore by Art. 77,
2
ll
KM =
M+ & M,
4
a
k² =
+ c².
4
Ex. 18. An isosceles triangle about its perpendicular.
Perpendiculara, base 26; z measured from vertex,
3
bc
"'
a
k² M =
3 a³
263
=Sz³dz =
6324
b³ a
=
6a³
3
6
bc
But M=ba; .. k² =
6
246
SECT. VI. Moment of Inertia of a symmetrical Solid
about its Axis.
84. When we have a solid of which all the sections perpen-
dicular to the axis are similar, it may, by planes perpendicular to
this axis, be divided into indefinitely thin slices, and the moment of
inertia of the whole, will be the sum of the moments of these parts;
and ultimately, it will be the integral of a portion of the moment
which corresponds to dz the differential of the axis. This portion
will be found by taking the moment of the plane, which is the
section of the solid, (found by last Article), and multiplying it by dz.
Ex. 19. A cone revolving about its axis.
The momentum of a circle, radius
= 1,
is
Tr
2
4
And in the
The momentum of a differential slice
4
4.4
N²
cone r is as z: let r=nz.
4
π. r¹ dz
π n¹ z dz
"
2
2
πη 25
k² M =
and taken to z=α,
10
πηνας
π b4 a
10
10
if nab, the radius of the base.
Π
π b² a
362
And M
;
•. k²
3
10
Ex. 20. A sphere about a diameter,
k²
2 a²
5
Ex. 21.
A hollow sphere, of which the external and internal
radii are a, b,
a5 — b5
k² =
5
3
a
b³
Ex. 22. An ellipsoid about its axis: the semi-axes of the
largest section perpendicular to the axis of rotation being a, b.
だ
​K² =
2
a² + b²
5
247
Ex. 23. A parallelepiped, k=
a²
Ex. 24. A cylinder, k
a² + b²
3
Ex. 25. A hollow cylinder, k2=
a² + b²
2
SECT. VII. Moment of Inertia of a Solid not symmetrical.
85. When the solid comes under this description, different
methods may be used, as in the following Examples.
Ex. 26. A cylinder about an axis perpendicular to its own,
through its middle.
Let its radius=b, its length =2a.
A circle perpendicular to the axis of the cylinder, and at a
distance from the axis of rotation, has its moment, (Ex. 17.)
X
== b² (x²+²),
3
+
•., & M = S =
:. k² b²
V
{
x
+ } d z = = b²
(
3
= = b²
{ 2
{ a²
3
3
a² + b² a } = 2 = b²a
x
" - }
4
2° b2
+ 1}.
and since M=2π b²a, k² =
be
+
3
4
Ex. 27. A cone about an axis perpendicular to its axis at
its vertex.
Fig. 91. AB the base, CD the axis of rotation, MP any section
parallel to the base,
CM=x, MP=nx, CA=a, AB=b=na.
By Ex. 17, moment of circle MP round CD
=
circle . (2²
n² x
x² +
4
11
TIE (1+
TN
a¹;
248
k²
4
n° x²
TN X
5
2. }^ M=ww² (1 + ") ƒ x^dx = (1 + ")
=
2
N
2
πn² a5
2
S
4
'α² (1 +
5
and M =
! = = n² ¹² · a ;
3
2),
3
2
N
2
5
b
. . k² = ³ a² ( 1 + 1 ) = (~² + ²).
5
3
5
4.
Ex. 28. A cone about an axis perpendicular to its axis through
its centre of gravity.
The distance of the centre of gravity from the vertex is
hence, by Art. 77,
moment round CD= moment round GH+CG2. M,
3
160
b2.
(a² + =) M = k² M +
9 a²
M,
16
За
;
4
4
3 (a²+4b²)
k2
80
And similarly for other figures of revolution*.
SECT. VIII. Centre of Oscillation.
86. PROP. To find the momentum of inertia of a given body,
by a practical method.
Let the body AB, fig. 92, be suspended from an horizontal axis
C; and let a vertical line be drawn from the axis, which will give
* The expressions in this Chapter may also be found by using the
following formulæ in rectangular co-ordinates. The body revolves about the
axis of z.
For planes,ƒƒ(x²+y²) dx dy.
Where, after the first integration, in a for instance, limits are to be put
in terms of y.
X
For solids, f(x²+y²) d x d y d z =ƒ ƒ (x²+y²) z dx dy.
Where must be put in terms of x and y,
formed, and the limits introduced as before.
No. 348, 349.
and then the integrations per-
See Poisson, Traitè de Mec.
249
the plane Cb, in which the centre of gravity is. Let the body be
moved, so that Cb may be exactly horizontal (CB), and let the
weight Q be ascertained, which, acting vertically at B, will sup-
port in this position. If W be the weight of the mass, we shall
have
Q. CB
W =
CG
Now let the body hang from C, and make small oscillations,
and let it make n oscillations in a time t.
oscillation is, by Cor. 1, Art. 76.
t
n
π
◇ being the centre of oscillation ;
CO
g
Then the time of one
g t²
.. CO =
And by Art. 76.
2
π² n²
Σ (m. Cm²) = CG. CO.Σm =
M being the mass of the system ;
Q
W
CB.g.f. M,
2
π N
Q
tⓇ
.
.. k²=
W
2
2
π² n²
CB .g.
37. PROP. The centres of suspension and oscillation are re-
ciprocal.
That is, if the centre of oscillation be made the point of sus-
pension, the former point of suspension will become the centre of
oscillation.
In fig. 81, CO =
But by Art. 77.
Σ (m. Cm²)
CG.Σm
Σ (m.Cm²) = Σ (m.Gm²) + CG². Em ;
or, putting M for Σm the mass, and k² M for ZΣ (m. Gm²),
k²
CO = + CG, and GO =
CG
k²
CG'
2
and CG =
GOʻ
I 1
250
If therefore, we suspend the body from O, C will be its centre
of oscillation.
COR. 1. CO depends on CG alone, and will be the same, so
long as CG is the same. Hence, if with centre G, (fig. 93,) and
radius GC, we describe a circle; (in the plane of oscillation, that
is, a plane perpendicular to the axis of suspension ;) CO is the same
from whatever point of this circumference we suspend the body.
And therefore, the time of oscillation is the same.
COR. 2. Also, if we describe a circle with radius GO, the
time of oscillation will be the same, whether the body is suspended
from any point in the circumference 00', or CC'.
88. PROP. To find from what axis a body must be suspended,
that it may oscillate in the least time possible.
That is, amongst all the axes which are parallel to one anothér.
The time of oscillation is that of a simple pendulum, whose
length is CO, fig. 93, C being the axis of suspension, O the centre
of oscillation. Therefore, the time will be least, when CO is least.
.. by last Art. CG +
and taking the differential coefficient,
k²
= min.
CG
k²
] -
=0
= 0; .. k=CG.
CG2
Hence, the time will be the least when CG=k.
2
k²
CG
In that case, GO= = k also.
Hence, the least time of oscillation will take place when the
body is suspended from K, so that L being the centre of oscilla-
tion, KG = GL.
COR. 1. The least time of oscillation is equal to that of a
simple pendulum, whose length is 2k.
251
COR. 2. If we describe a circle with radius GK, the time.
will be the same when the body is suspended from any point in
this circumference KK'.
COR. 3. It has been seen in last Art., that the times of oscil-
lation (t), are the same when the point of suspension is in any
point of the circumferences CC', 00'; for a point between these
circumferences, the time is less than t; and least for a point in
the circumference KK. For points within OO', or without CC',
the time is greater than t, and becomes infinite as the point of
suspension approaches the centre, or goes off to an infinite dis-
tance.
89. PROP. To find the centre of oscillation in given figures.
We have, (Art. 87.)
moment of inertia about axis
CO =
mass. CG
k2
Also, GO=
CG
..(a).
.(a');
k² M being the moment of inertia about an axis through the centre
of gravity, and parallel to the axis of suspension. And either of
these formulæ, by the assistance of the preceding part of this
Chapter, will enable us to find O.
Ex. 1. To find the centre of oscillation of a straight line, sus-
pended from its extremity.
1
By Ex. 1, Art. 79, k² =
α
a
3
2 a being the length of the line;
.. by (a'), GO = for CG = a. And CO = a +
3
Ex. 2. A line AB, fig. 94, oscillating in its plane.
G the middle point, CG=1, AB=2a;
a
8 100
k2
a²
GO =
31
.. it is independent of the angle COA.
252
Ex. 3. A circle AB, fig. 95, oscillating in its own plane.
CG=1, radius = a; and by Ex. 12, Art, 81, k² =
2
a²
2
k2
a²
.. GO=
21
If C be in the circumference,
GO = 1/2
a
За
CO =
2
axis.
The same is true for a cylinder, about a line parallel to its
Ex. 4. A circle oscillating about an axis in its own plane;
(about CD, fig. 95.)
By Ex. 16, Art. 83,
2
a
2
a
k² =
2
; GO=
4
47°
If the axis be a tangent to the circumference,
a
l=a, GO =
CO=
>
4
5 a
4
Ex. 5. A globe about any axis.
Distance to the centre of the globe=l, radius = a.
By Ex. 20, Art. 84,
k²
2
2
2 a²
5
GO: =
2a2
51
Σα
2 a²
CO=1+
5 1
Ex. 6. A cone about its vertex.
Axisa, radius of base = b.
By Ex. 27, Art. 85,
3
moment of inertia =
5
(a² + b²)
3 a
M. Also CG =
4
.. CO =
(a+
b2
4 a
4 a
+
5
b2
5 a
253
That the centre of oscillation may be in the centre of the base,
we must have CO=a; whence b=a, and the cone is a right
angled one.
Ex. 7. Let a solid, composed of two equal cones set base to
base, oscillate in the direction of its axis.
Let G, fig. 96, be the common centre of the bases, GB a
horizontal radius, and therefore parallel to CD the axis of rotation.
We must find the moment of inertia of the figure about GB.
Now for the cone GA, we have, if F be its centre of gravity, and
FK parallel to GB,
moment about GB = moment about FK + GF². cone,
3 (a² + 4b²)
a²
by Ex. 28,
cone +
• cone
16
80
2 a²+3b2
cone.
20
And hence the whole moment, or
2a²+3b²
2
k² M =
.2.cone;
20
2a²+362
:. k² =
And if CG=1,
20
k²
2a+3b³
CO = 1 +
ī
1+
20/
90. PROP. Having a system composed of several separate
bodies, whose centres of gravity and oscillation are known; to find
the centre of oscillation of the whole.
In fig. 97, let g, o, be the centres of gravity and oscillation of
m; g', o', of m'; g", o", of m"; and so on. And let G be the
centre of gravity, and O that of oscillation for the whole system.
Also let gh, g'h', g"h", &c. be perpendiculars on CG.
moment of inertia of m about C = m. Cg. Co,
of m'...... = m'. Cg'. Co',
&c.
= &c.
254
.'. whole moment = m. Cg. Co+m. Cg'. Co'+&c. ;
.. CO =
m. Cg. Co+m' . Cg'. Co'+&c.
(m+m² + &c.) CG
m.Cg.Co+m' .Cg'. Co+&c.
m. Ch+m'. Ch' +&c.
91. PROP. To find practically the length of a pendulum, which
oscillates seconds.
If we know the exact length of a simple pendulum which makes
a given number of small oscillations in 24 hours, we can find the
length of a pendulum which shall oscillate in any given time as
1 second. But it is impossible to form a pendulum which may,
with sufficient regard for accuracy, be considered as a simple pen-
dulum; that is, as a single point suspended by a string without
weight. It is necessary, therefore, in our experiment, to find the
distance between the centre of suspension, and of oscillation, of the
oscillating body. And the difficulty of the case is to determine
accurately this latter point; for the unavoidable irregularities of
figure and density make its geometrical determination include errors
which the delicacy of the inquiry renders important.
To avoid these sources of inaccuracy, Captain Kater has in-
geniously employed the property of a compound pendulum, proved
in Art. 87; viz., that the centres of oscillation and suspension are
reciprocal. It follows from that property, that if a pendulum have
two centres of suspension, and oscillate on them, first with one end
uppermost, and then with the other; so that the times of oscilla-
tion in the two cases may be exactly equal; the distance of these two
centres will be the length of the equivalent simple pendulum, what-
ever be the irregularities of form or composition in the instrument.
The manner in which this effect was produced, was as follows:
A brass pendulum CD, fig. 98, was furnished with two axes,
from which it could be suspended; one passing through C, and
the other through O. Besides the principal weight D, it was
provided with a smaller sliding weight F, which could be moved
along the stem CD; and this weight was to be moved till the
number of oscillations in a given time, (as 24 hours,) was the same,
whether the pendulum was suspended from C or from 0.
255
}
F was placed in such a position, that by moving it from O, as
to f, the number of oscillations about C in twenty-four hours was
increased; and by the same change, the number of oscillations about
O in the same time was still more increased. We shall afterwards
consider this position mathematically. The adjustment was thus made.
Let the weight be at F, and let the number of oscillations in
10 about C be 606, and about O be 601. Now let F be moved
to f; and let the oscillations in 10m be 607 about C, and 609
about O, (because the latter are more affected than the former).
Then, the proper position of the slider is somewhere between Fand f.
Let it be placed at f', bisecting Ff; and let the oscillations in this
case be 606, and 606; then, the proper position is between ƒ
and f'; and so on. Observing always, that if the number of
vibrations about C be the greater, the slider must move toward C;
and if the contrary, it must move towards O. By this means,
continually halving the distance last moved, we may make the
oscillations about C and O approach within any required degree of
exactness. The distance of C and O being then measured, will
give the length of a pendulum which makes a known number of
oscillations in 10 minutes *.
* There were, in Captain Kater's experiments, a number of contrivances
which it would detain us too long to describe. Besides the slider F, he had
another moveable weight E; and he made the numbers of oscillations nearly
equal by means of this, before he attempted a more accurate adjustment by
the slider.
The axes through C and O, were made with knife edges, which resting
on planes of agate, turned as nearly as possible on a mathematical line. It
is however true, as has been proved by Laplace, and will be shewn here-
after, that, if they had been cylinders, their distance would still have given
exactly the length of the simple pendulum. The method of determining the
number of vibrations in 24 hours was elegant; it was done by placing the
pendulum in front of a clock pendulum, oscillating nearly in the same time;
and observing the intervals at which the two pendulums coincided. Cor-
rections were also to be made, for the magnitude of the arc vibrated; for the
bouyancy of the atmosphere; for the temperature, &c. The reader will find
Captain Kater's account of his method, and its results, in the Phil. Trans.
for 1818, p. 33.
256
92. PROP. To determine the effect produced by the change
of position of the moveable weight in last Article.
We shall consider any pendulum, and a small weight moveable
along the line passing through the centres of suspension and
gravity.
Let m be the mass of the pendulum, independent of the move-
able weight; h the distance of its centre of gravity from the centre
of suspension C, 7 the distance of its centre of oscillation. And let
u be the moveable weight, λ the distance from C of its centre of
gravity, or of oscillation, supposing them to coincide because it is
small. And let L be the length of the corresponding simple pen-
dulum; then, by Art. 90,
2
mhl + u x²
L =
mh + uλ
2
dL =
; and L and X being variable,
µ ² λ² + 2 m µ hλ - muhl
(mh+µλ)²
μ² (^—a) (A+B)
(mh + µ λ)²
dλ;
where ua = √(m² h² + mµh!) — mh ;
μβ √(m² h² + mµhl)+mh.
dλ
Let Cxa; then, if F be below x, dL will be negative, when
dλ is negative; and hence, if F be moved towards C, L will be
shortened, and the number of oscillations about C in a given time
will be increased. And this increase will be the slower, as F comes
nearer to x. At x a small change in F's position would produce no
effect, and, when F is between C and x, the effects would be
contrary to the former ones.
Let the distance CO of the two points of suspension be 2a; and
M being the middle point of CO, let F be the position of the moveable
weight, when the oscillations about C and about O are performed
in the same time; therefore in this case L = CO2a.
d.
MF; therefore, when L = 2a, λ = a + 8. Hence,
Let
mhl + u (a +8)²
2a =
mh + u (a + d)
:.1 = 2a +
u (aº — d²)
mh
257
8'?) - 1}·
Hence, μa mh
{V(
(1+
2
Σμα
μ² (a² - 8²
+
m² h²
mh
Expanding, and neglecting higher powers of
бле
mh '
a = a
2mh
Hence, x is between M and C, and near to M.
Similarly, if Y be the point where the weight F does not
affect the oscillations about O, y is between M and O, and near
to M.
If F be between x and y, we shall, by sliding F towards O,
lengthen the oscillations about C, and shorten those about 0, and
vice versa. But, if F be in Cx, or in Oy, we shall, by sliding it
towards the middle, shorten both sets of oscillations; and by sliding
it from the middle lengthen both, though in different degrees.
CHAP. IV.
MOTION OF MACHINES.
93. WE shall in the present Chapter apply the preceding
principles to determine the motion of the mechanical powers, and
other simple combinations.
SECT. I. Motion about a fixed Axis.
PROP. To determine the motion of weights on a wheel and
axle, fig. 99.
Let P draw up Q by means of strings wrapping round two
cylinders A, B, which have a common horizontal axis.
Let a, b,
be the radii of the cylinders respectively; and Mk² the moment
KK
258
of inertia of the machine AB, about its axis. We shall then have
impressed forces, Pg at distance a,
of which the moment is Pga-Qgb.
-
Hence, by Cor. 4. to Art. 74, we have
Qg at distance b;
(Pa-Qb) ga
2
છે
2
Pa²+Qb²+Mk²
accelerating force on P =
downwards;
accelerating force on Q
(Pa – Qb) g b
Pa²+Qb²+Mk²
upwards.
And these being constant, the motion may be found by the
formulæ for constant forces.
COR. 1. If Qb> Pa, the force will be in the opposite direc-
tion, and Q will descend.
COR. 2. If Tg be the tension of the string by which P hangs,
P is impelled downwards by its weight, and upwards by the tension.
Hence, the moving force on P is Pg- Tg, and the accelerating
(P − T) g ;
force
P
;
(P-T)
g
P
T=
(Pa - Qb) ga
Pa²+Qb²+M k² ’
P (Qb² + Qab + Mk²)
Pa²+Qb²+Mk2
Similarly, if T' be the tension of the string by which Q hangs,
T' =
T=
Q (Pa²+Pab + Mk²)
2
2
Pa²+Qb²+Mk²
COR. 3. The pressure on the centre of motion arising from
P, Q, will be the sum of these tensions, (see next Section)
.*. pressure on the centre =
PQ (a+b)²+(P + Q) Mk² g
Pa²+Qb²+Mk
94. PROP. To determine the motion of weights acting on a
combination of wheels and axles, fig. 100.
259
The wheels and axles may act on each other, either by means
of teeth as at D, or by strings passing round both, and turning them
by friction as at D'; or in other ways: the mechanical conditions
of the problem are the same in all these cases.
Let a, b be the radii of the first wheel, and of its axle; a', b'
of the second, a", b" of the third, and so on. Then the impressed
moving forces are Pg acting at A, and Qg in the opposite direction
at B. By Statics, the latter would be counterbalanced by a force at P
equal to Qg.
valent to
•
b b'b"
aa'a".
Pg – Qg
Hence, the moving force impressed is equi-
bbb"
a a'a"
112
at A, at distance a from C.
Let Mk², M'k, Mk2 be the moments of inertia of the
respective wheels M, M', M" about their centres, (including the
axles).
And let be the effective accelerating force on P or on A.
Then, since the accelerating forces are as the velocities, the acce-
lerating force at D or E will be ; that at D'or E' will be
b x
a
b b'x
a á
; and that at B or Q,
b b' b″x
aaa"
Now, since the effective accelerating force at P is x, that at
any distance r from C, in the wheel M, is ; and if m be a
rx
a
particle at that distance, the effective moving force is
this is equivalent in its moment round C, to a force
mr x
a
mr²x
aⓇ
And
acting
at A. And hence, the whole effective moving forces in M are
equivalent to a force
valent to
x Σm r²
a²
acting at A; that is, they are equi-
x. Mk²
at A.
a²
Similarly, the effective moving forces in M' are equivalent to
260
bx M'K²
12
a
a
at E; which is, by the property of the machine, equi-
valent in its effect to turn the system round C, to a force
b²x
2
a
M'k'2
12
at A.
a
And the effective moving forces of M" are equivalent to
M'k"2
112
at E'; which is, with respect to C, equivalent to
b b'x
a a'
b² blz x
12
a²a'
a
M'k'2
at A.
112
a
The effective accelerating force on Q is
b b' b" x
;
which gives
a
aa'a"
moving force
bb'b"x
a a' a"
b² b'² b'ex
Q, equivalent to
at A.
a² a² a"
Hence, we have the moment of the impressed forces about C
=ga (P - Q
bb'b").
a a'a"
And the moment of all the effective forces about C
k²
2
12 112
b² b2b″ 2
= Pax + Qax 12 112
+ Max. + M'ax.
a
2
a² a¹2
a
b² k'2
b² b¹² k112
+ M"ax.
12
a²
a
a² a² a
2
Equating these (by Art. 73,) and putting n, n', n" for
we have
Ꮖ
x =
(P − Qn n' n″) g
k²
P+Qn²n²n" + M + M'n
a
2
b
>
a
k/2
12
12
+ M"nn'
11:2
a
a
b' b"
a"a",
The accelerating force on Q=nn'n'x. These forces are constant.
95. A machine was constructed by Atwood, to measure the
spaces and velocities of bodies descending by gravity, in order to
261
compare them with theory.
It is represented in fig. 101. Two
equal weights P, Q, hang by a fine string over a fixed pully M. One
of them is made to descend, by placing upon it a small weight D,
and the times and spaces of the motion are observed. The weights
at P and Q are inclosed in equal and cylindrical boxes; so that
the effect of the resistance of the air will be the same upon both,
and will not affect the motion. And the effect of friction is nearly
removed by making the axis of M very slender, and causing each
end of it as C to rest upon Friction Wheels, as M, M'*. The
times are observed by means of a pendulum, and the spaces by a
scale of inches BF, down which P descends. To determine the
velocity, P is made to pass through an opening MN, in a stage
fastened to the scale BF; and the weight D, which is too large to
pass, is left resting on M, N. Therefore, after passing the point
E, P will move uniformly with the velocity acquired. When it
has passed through a given space EF, is stopped by striking the
stage F, which is there fixed to the scale.
The body P being allowed to descend from rest at a given
point B, descends till D is heard to strike the stage M, N, and the
time is noted; it then descends till P is heard to strike the stage F,
and the time is noted: the space EF, divided by the interval of the
*To shew that these wheels will diminish the effect of friction, we may
consider friction as a force acting in a tangent to the axle. If the axle C
rested on immoveable surfaces, and the friction were F, its effect at A
Fb
a
would be But if the axle C rest upon friction wheels, their circum-
ferences will turn with the circumference of the axle, and between them there
will be no friction. The friction will take place at the axles C, C'; and if we
b'
suppose it to be F at the angle C', this will be equivalent to at the
b
b
circumference of the axle C, and to F
at A. And as there are four ends
a a'
bb'
of the axles C' for one C, the friction with friction wheels is 4 F.
a a
Hence, by means of such a contrivance, it is diminished in the ratio of
a: 4b′, supposing F to be the same in both.
262
latter times gives the velocity; and the space BF, and the time of
describing it, being known, we can compare our results with theory.
The velocities are small, both because D is small, and because the
wheels F, M', M" are to be moved, and their moment of inertia
enters the denominator of the accelerating force.
Observing, that besides the friction wheels M', M", there are
two others at the other end of the axis A; calling the moment of
each of these M'k', and of M, Mk2, and the radii of the wheels
and axles a, b, a', b', we have
accelerating force on P =
Dg
2
k2
2P+D+M+ 4 M'
a
b 12
k
12
a a
The effect of the inertia of the wheels is the same as if a mass
k2
M+4M'
a²
+ 4M'.
bk2
a a'²
12
were collected at the circumference of M.
The reader will find in Atwood's Treatise on Rectilinear and
Rotatory Motion, Sect. 7, an account of experiments made with
this machine. They all agreed accurately with the formulæ for
constant forces.
96. PROP. To determine the motion of weights on a lever,
fig. 102.
Let P, Q, be attached to the extremities of a lever whose arms
are a, b; and let M be the mass of the lever, and h the distance of
its centre of gravity. Let PQ be any position in which it makes an
angle with the vertical. Then a cos.
Then a cos. 0, b cos. 0, h cos. ◊ are
the perpendiculars from the
of the forces of P, Q, M.
(Pa+Mh−Qb) g cos. 0, to
be the moment of inertia of the lever itself, we have
accelerating force on P=
acting perpendicular to CP.
centre upon the vertical directions.
And the moment of the forces is
make P descend. Hence, if Mk²
(Pa+Mh-Qb) ga cos. O
Pa²+Qb²+Mk²
The space described by P in dt is-ad0; hence, by the
formula v dv=fds, we have, v representing P's velocity,
263
1
vdv=
Ꮎ dᎾ
;
and integrating,
(Pa+Mh-Qb) ga cos. Ode
Pa²+Qb²+Mk²
2
2 (Pa+Mh− Qb) ga sin. O
v² = C- Pa²+Qb²+Mk²
If the lever descend from rest from a position AB, let AB make an
angle 0, with the vertical, and we have
1
2 (Pa+Mh-Qb) ga
Pa²+Qb²+Mk²
રે
ᏧᎾ
-
1
(sin. 0₁ sin. 0)=v²=a²
dt
And hence, by integrating, we should find the relation of 0 and t.
If P descend through a quadrant, we have, at the lowest point 0,
2 (Pa+Mh- Qb) ga
v2
=
Pa² + Qb² + Mk²
97. PROP. A body moveable about an axis C is struck at a
given point by a given mass with a given velocity; to determine
its motion, fig. 103.
Impact is, properly speaking, a violent pressure continued for
a short time. Now if any force act at a distance a from the axis
of a body whose moment of inertia is Mk2, the effect produced at
Mk2
a²
any instant will be the same as if a mass were collected at the
distance a. Hence, the whole effect produced will be the same as
if such a mass were substituted for the body, whatever be the time
which the charge employs. And hence, the effect of perpendicular
impact at a distance a will be the same as if it took place upon a
Mk2
placed there.
mass
a²
In fig. 103, let a mass P impinge directly on a system CA,
with a velocity ; and let CA be a perpendicular on P's direction.
If CA=a, the effect will be the same as if P impinged on
Mk2
2
a²
Let the substances be supposed inelastic; and the bodies will both
move with the same velocity after the impact; and since, by the
third law of motion, the mass multiplied into the velocity will be
the same before and after the blow, we shall have, if x be the
velocity of A after the stroke,
264
x (P + MK²) = PV,
x =
a
L
PV a2
Pa+Mk
If the body be acted upon by no force after the impact, it will
revolve uniformly. If it move about a horizontal axis, and be
acted on by gravity, it will ascend till all the velocity be destroyed,
and then descend, and so oscillate.
If the bodies be elastic, we must apply the rules for impact in
that case.
On this supposition, P and M will separate after the
impact. And if the impact be not direct, we must, supposing the
bodies perfectly smooth, take that part of it which is perpendicular
to the surfaces at the point of contact.
98. An instrument depending upon these principles was con-
structed by Robins for the purpose of measuring the velocities of
musquet and cannon bullets, and called the Ballistic Pendulum. It
consisted in an iron plate CA, fig. 104, suspended from a hori-
zontal axis at C, and faced with a thick board DE. When this
was at rest, a bullet was fired into it as at P, which caused it to
move through an arc MN. The chord of this arc was known by
means of a ribbon fastened to the pendulum, as at N, and sliding
through a slit at M, so that when drawn to the length MN it did
not return. The ball stuck in the wood, and was prevented from
going through by the iron.
Let O be the centre of oscillation of the pendulum, including
the bullet. Then the motion of the pendulum will be the same
when left to itself, as if all the matter were collected in O. And
hence, the arc through which O will move will be that down which
it would acquire the velocity which it has at the lowest point. If
✪ be this angle, the velocity acquired in describing it would be that
acquired down the versed sine of ; or down a perpendicular height
CO. ver. sin. 0. Let CO=1; .. velocity of O at lowest point
= 2 sin.
=V(2gl ver. sin. Ø)
Ꮎ
√(gl).
2
265
But since the velocity of P at the lowest point is by last Article
PVa
2
2
Pa²+Mk²'
the velocity of O, which is to this as CO to CP, will be
PV al
Pa²+Mk²
Ꮎ
= 2 sin.
√(gl) by what has been said.
If h be the distance from C of the centre of gravity of the
pendulum, including the ball,
Pa² + M k²
1 =
Pa²+Mk² = (P + M) h l ;
(P+M) h
.. PV al = 2 sin.
2
IQI
(P + M) h l √ (g l),
0 P+Mh
V = 2 sin.
√(gl).
P a
If the pendulum after being struck by the ball, makes n oscil-
lations in a minute, we take
time of oscillation =
60
N
;; ·. √(gl) =
g
0 P+M 60g h
And, 2 sin.
=
2° P
πNa
60g
=
ПП
This agrees with Dr. Hutton's formula.
dividing the chord MN by the radius CN.
We have 2 sin. by
Dr. Hutton himself however, in his own experiments, found the
velocity of balls, by suspending the cannon which he used, and
observing the arc through which it was driven by the recoil. The
same formula is still applicable, M now representing the weight of
the cannon and its appendages without the ball. For the effect will
be the same, whether a velocity be communicated to the pendulous
body by the impact of the ball, or its reaction. And the momentum
communicated at the axis of the cannon will be PV, because the
momentum communicated to the ball in one direction, and to the
pendulum in the other, must be equal.
LI
266
It is found by experiments of this kind, that the velocity of
musquet and cannon bullets varies from 1600 to 2000 feet per
second.
SECT. II. Motion of Bodies unrolling.
99. PROP. A cylindrical body unrolls itself from a vertical
string, the other end of which passes over a fixed pully, and
supports a weight; to determine the motion, fig. 105.
One end of the string is supposed to be fastened to the surface.
of the body M, so that it cannot slide, but can only descend by
unrolling.
Let the tension of the string ABP be Tg; (if we neglect the
inertia of B, it will be the same throughout). Let the moment of
inertia of M be Mk2, and CA=a: and the effective accelerating
force on C downwards=x.
The effective moving force on P downwards is (P-T) g, and
the accelerating force is (
T
(1 - 27) 8
(1 - g. And this will be the effec-
tive accelerating force upwards on any point of the string BA, as A.
Now since C descends by an accelerating force x, and A ascends
T
by an accelerating force (1 − 1)
g, the relative accelerating
force of A round C is x + (
T
(
P
g=y suppose. And hence,
the effective accelerating force round C, of a point at a distance r
from Cis"
a
round C is
:
and the moment of all the effective accelerating forces
yΣmr²
Mk² y
or
The impressed forces on M
a
a
are the weight Mg acting downwards at C, and Tg acting upwards
at A. Hence, to establish an equilibrium between the impressed
and effective forces, according to Art. 73, we must have the forces
equal,
=
or Mx Mg-Tg;
and their moments with respect to C equal,
T
Mk²y
Mk (
or
{x + (1 - 4) 8} = Tga.
a
a
267
Eliminating T, we have
M
Mk² {x + g
P
x=
(g − x)} = Ma³(g — x),
Pa²+(MP)k²
Pa² + (M+P) k² • g⋅
COR. 1. Since Tg=M(g~x), we have
2 MP L2
2
T = Pa² + (M+P) k²
The accelerating force on P, is
COR. 2.
(1 - 4) 8
T
Pa'-(M-P) k²
Pa²+(M+P) k² · 8 ·
g==
g.
COR. 3. It is not necessary that the whole body should be
cylindrical, but only that the part of it from which the string unrolls
should be a cylinder, of which the axis passes through the centre
of gravity. The vertical plane of the string must be perpendicular
to the axis of the cylinder, and pass through the centre of gravity.
COR. 4. If the figure be a cylindrical shell of small thickness,
k=a,
accelerating force on C =
accelerating force on P =
M g
•
2 P+ M³
2P - M
2 P+M
g,
tension =
2 MP
2P+M'
COR. 5. If the figure be a solid homogeneous cylinder, k²
accelerating force on C =
accelerating force on P =
tension =
P+M
3P+M 8,
3P-M
3P+M•8,
2. MP
3P+M
a²
268
100. PROP. A cylindrical body unrolls itself from a vertical
string, the other end of which is fixed; to determine the motion,
fig. 106.
If we assume P, in last Article, to be such that it shall neither
ascend nor descend, we may suppose the string AB to be fixed at
the point B, and the motion will be the same as before. We must
therefore in this case, have the accelerating force on P = 0;
or, Pa² – (M – P) k² = 0, whence, P =
Mk2
a² + k²·
Hence also, T=
Mk2
a² + h² •
And accelerating force on C =
(M − T) g
M
a² g
=
a² + k²·
COR. 1. If the figure be a cylindrical shell, k=a;
accelerating force on C = 5, T =
2'
M
COR. 2. If the figure be a solid cylinder, k² =
a
2
2 g
M
accelerating force on C =
T
3
3
2 a²
COR. 3. If the figure be a globe, k² =
;
5
g
accelerating force on C =
55, T =
2 M
7
7
COR. 4. If the string, instead of being vertical, be laid along
an inclined plane as BA, fig. 107, the same conclusions are mani-
festly true; putting for g the force of gravity down the plane, which
is
g
•
sin. inclination. The tension will also be T' g. sin. inclination.
COR. 5. If M, instead of rolling by means of a string, roll
down the plane in consequence of the friction entirely preventing
its sliding, the results will be the same. The tension of the string
is now replaced by the effort which the friction exercises to prevent
269
sliding. Hence, when a body rolls down an inclined plane, the
accelerating force is if it be a hollow cylinder, if it be a solid
5
1
2
cylinder, and if it be a globe, of the force with which a body
would slide down the plane, if friction were removed.
SECT. III. Motion of Pullies.
101. PROP. One body draws another over a single fixed pully;
to determine the motion, fig. 108.
Let Mk2 be the moment of inertia of the pully, a its radius.
And let r be the effective accelerating force on P downwards;
which is therefore the accelerating force on the circumference of
the pully M, and on Q upwards. Let Tg be the tension of the
string AP, and T'g of BQ. Hence, the force impressed at the
circumference of the pully is Tg- Tg', and therefore,
x=
(T – T') ga²
Mk2
.(1).
But the accelerating force on P = x =
(P − T) g
Р
and the accelerating force on Q
(T' — Q) g .
=
Q
.. Px (PT) g, Qx= (T' — Q) g,
(P + Q) x = (T' − T) g + (P-Q) g………. (2).
Multiply (1) by Mk², and (2) by a², and add;
.. Mk²x + (P+Q) a²x = (PQ) ga*,
ľ
(P − Q) g a
Mk² + (P + Q) a²
COR. 1. The tensions of AP, and BQ, are respectively
(Mk² + 2 Qa²) Pg
(Mk² + 2 Pa²) Q g
2
Mk² + (P+Q) a²'
Mk+(P+Q) a² *
COR. 2. Hence, when strings are in motion about pullies, the
tension of each string is no longer the same throughout its length.
270
1
A part of the tension of PA is employed in turning M; and it is only
the remainder which is continued along the cord, so as to act in BQ.
The same results might have been obtained from Art. 93, by
making the radii of the wheel and axle equal.
102. PROP. In the single moveable pully with the strings
parallel; to determine the motion, fig. 109.
f
Let P, Q, be the weights; Mk2, M'k', the moments of the
pullies; a, a' their diameters. And let the tension of AP-Tg,
of BD=T'g, of EF=t'g. Then, if x be the accelerating force.
x
on P, will be the accelerating force on Q, because it moves with
2
half the velocity. Also, the accelerating force at the circumference
of M will be x: and since, while E remains fixed, the centre of
M' rises with half P's velocity, the relative motion of E round the
centre, is half P's velocity, and therefore, the effective accelerating
force at the circumference of M' round C' is
Ꮖ
And if we consider the forces which act upon M', we have
Impressed forces, T'g, t'g upwards, Qg downwards;
Q including the weight of M'.
Effective forces, upwards for Q, and
X
2
cumference, turning M' round C'.
Hence, by Art. 73, we must have
Ꮖ
(T' + t') g − Qg = Q ₂;
x M'k'
2 a
12
at the cir-
and, considering the moments with respect to C',
}
(T' — t'′) a'g =
M'k'½
Ꮖ
a
X
M'k'2
.. 2 Tg = Qg + Q
+
12
a
271
Also, we have, as before,
P-T
P¯¯¯g, or (P− T) g = Pr,
х
P
x =
(T-T') ga²
Mk2
or (T- T') g =
Mk2
a²
Q2 x;
add these, and the former one
Q g
Q x
M'k'2 x
T'g=
+
+
Q
22
a'2
and we have
Q g
Pg=
2
+ (P +
10:10
Mk
M'k'2
+
a²
2
+
22 a¹2
12
(P - 2)
g
.. x =
Q
Mk2 M'K2
P+
+
a²
+ 12
22a/2
x;
from this also the tensions might be found.
103. PROP. In the system of moveable pullies, where each
hangs by a separate string; to determine the motion, fig. 110.
The strings are supposed parallel.
M, M', M", M" the pullies, Mk², M'k2, M"k", &c. their
moments; a, a', a", &c. their radii. Let x be the effective accele-
rating force on P; then will be the accelerating force on M';
x
Ꮖ
on M":
22
X
on M""; and these will also be the effective ac-
23
celerating forces producing rotation at the circumferences of
M, M', M", &c. Then, by reasoning with respect to each
pully, as we have done for M' in last Article, we have
Mk2
(P− T) g=Px, (T-T') g=
X.
a²
272
X
81%
(T′ + t') g − M′ g − T″g = M' 2, (T′ − t') g =
Ꮖ
(T″ +t") g− M″ g−T" g = M" 2, (T″ −t") g =
and so on.
25
M'k'
a
72
M"k"2
all2
Eliminating t', t", &c. from each successive pair, we have
Pg = Px + Tg,
Mk2
Tg= x + T'g,
a
2
M'k'2
M'x
30
T' g =
12
a
T" g
112
a
X
23
M" "2
22
22
+
+
M'g + Tg
T"
2
2
+
M" x
+
M" g
+
T"" g
23
2
2
and so on. Therefore,
Mk2
Pg=Px+ x +
+
a
2
M'k'2
112
M" "2
a
1124
Ꮖ
+
12
22
M'x
22
M" x
+
+ &c. +
M'g
+
24
2
The law of continuation is manifest.
the figure) is that which immediately
the effective accelerating force on Q,
X
24
M"g
22
+ &c.
Tig
+ &c. + 1,35
23
And the last tension (T" in
raises Q. Hence, we have
Q) g
x (TIV
Q =
Q
Ꮖ
.. Tg = @g+
Qg +
23
Substituting this, and obtaining the value of x, we have
(P -
M'
M"
&c.
2
22
3)
x=
Mk2
M'
M'2 M"
M" "2
P+
a²
+ +
22
+
22 a'r
2/2
+
24
24 a
44
+ &c.
and similarly for any number of pullies.
う
​273
By similar reasoning, we shall have the accelerating force in
the system of pullies, when each is attached to the weight. But
more immediately in all such cases by the following Proposition.
104. PROP. To find the accelerating force on any machine
whatever.
Let P be one of the bodies of the machine, and let P' be the
mass, which, placed at P, would preserve the equilibrium. Then
the weight (P – P') g is the impressed force, which produces the
motion.
Let u be the velocity of P, and v, v', &c. the velocities of any
'other bodies m, m', &c. in the system. Then, if x be the effective
V X
U
will be that on m, and
M V X
u
the ef-
accelerating force on P,
fective moving force. Therefore, the forces which must balance
each other by Årt. 73, are (P— P′) g in one direction, and
m'v'x
тих
26
, &c. in the opposite.
и
Now u, v, v', &c. may be considered as the virtual velocities of
the points where these forces are applied.
ciple of virtual velocities,
Hence, by the prin-
MVX
m'v' x
V
(P-P')g.u- Px.u-
. V
U
U
(P — P¹).
.v— &c. = 0;
(P − P′) g
:: x =
m v²
ጎ
m'v'2
Σmv*
P+
+
+ &c.
P+
u²
นี้
Let the motion of any mass, as M, be
a moment taking place about a fixed axis.
2
UZ
considered as for
This is always pos-
sible, (see Note, p. 222). Let a be the distance of the centre of
gravity from this axis, and a its velocity; then, if m be'a particle at
the distance r, m's velocity
mr²
a²
α
2
να
==
And for the whole of M,
a
2
α
a²
Σmv² =Σ
mv² = 2 "r" a² = = = £ mr² = 2 M (a² + kº),
a
M M
274
using the same notation as before,
= Ma² + M.
k² a²
2
Q š
Hence, in the denominator of the accelerating force x, we shall
have, for each mass M, two terms in the denominator, such as we
α
have just found. It may be observed, that is the angular velocity
of M about its centre.
a
It will be seen by comparison, that this includes all the pre-
ceding propositions of this Chapter.
SECT. IV. Maximum effect of Machines.
105. PROB. I. A weight P, acting at a wheel, produces rotation
in a mass which moves about an axis passing through the centre of
gravity; it is required to determine the distance at which P must
act, that the angular velocity, generated in a given time, may be
the greatest possible.
Here the accelerating force on P is
29
f=
Pa² g
Pa² + Mk²
P acting at a radius a.
And the velocity generated in time t in
the circumference at which P acts, is ft. And hence,
angular velocity =£t
a
; ..
2
f
a
= max.,
Pa² + MkⓇ
a
= max.
છે
Pa²+Mk2
min.,
a
Mk2
Mk2
Pa+
= min. whence P -
= 0,
a
a²
M
a = k
p
PROB. II. P raises Q by means of a wheel and axle, as in
Art. 93; the axle being given, to find the wheel, that the time of
Q's ascending through a given space, may be the least possible.
275
The accelerating force on Q is
f=
(Pa-Qb) gb
Pa² + QU² + M k²
And, as this is constant, we have t =
which will mani-
festly be least when ƒ is greatest. Therefore, we must have
Pa-Qb
Pa“ + Qb° + M k²
= max.
If we suppose a to vary, k will also vary in a manner depending
on the form of the wheel; but if we suppose M to be small,
we have, neglecting it,
Pa-Qb
=max.
2
Pa²+Qb²
and differentiating, supposing a variable,
P (Pa² + Qb²) — 2 Pa (Pa- Qb)
Pa²
-
2 Qab — Qb² = 0;
= 0;
Q b
P
{¹ + √/ (1 + })}·
P
If P be small compared with Q, this will give nearly
a=
2 Q b
P
+
2
The weight P must act at a little more than twice the distance
at which it would balance Q.
PROB. III. The wheel and axle, and the weight P, being given;
to find Q, so that the momentum communicated to it in a given time,
may be the greatest possible.
The accelerating force ƒ on Q, is the same as before, and the
velocity acquired by it in a given time, is v=ft. Aud hence, we
must have Qft a maximum, or Qf a maximum;
276
(Pa- Qb) Q
Pa²+Qb² + Mk²
max.;
.. (Pa-2 Qb) (P a² + Qb² + Mk²) -- b² (PQa- Q° b) = 0.
And hence, Q is found by the solution of a quadratic equation.
If we neglect M, we have
Pa²
Q:
b2
{√ (1 + - ) = 1} :
a
and if we suppose b small, compared with a,
Pa
P
Q
26
8
Q is nearly half the weight which P would support.
PROB. IV. A body revolving round an axis, strikes another
given body P in a direction perpendicular to the radius; to find the
distance at which the impact must take place that the velocity com-
municated may be the greatest possible.
Let be the distance of the striking point from the axis, and
M² the moment of inertia with respect to the axis. Any pressure
acting at the distance, will produce the same effect as if there
were collected at that distance a mass
Mk2
2.2
Hence, the impact,
which is only a short and violent pressure, will produce the same
effect, as if such a mass were to impinge on the given body P.
Let a be the angular velocity, then ra is the velocity at the point
of impact. And when a body A impinges with a velocity a on B at
Λα
rest, the velocity after impact is
A ÷ B
elastic, (Elem. Treat. on Mechanics, Art. 166.). Hence, the ve-
locity communicated to P is
; supposing the bodies in-
Mk2 a
T
MK
Mk² ar
Mk² + Pr² 2
+ P
277
which is to be a maximum. Whence we find
M
r = k
✓
P
If the body be in any degree elastic, the result will be the
same, for the velocity of B in that case, is
(1+e) Aa
A+B'
(Elem.
Treat. on Mechanics, Art. 167.).
CHAP. V.
PRESSURE ON A FIXED AXIS.
106. WHEN a body revolves about a fixed axis, the axis in
general suffers some pressure, depending upon the form and motion
of the body, and on the forces which act upon it. The different
parts of the system influence each other's motions by the intervention
of this axis; and it supplies, as it were, the difference of the forces
impressed, and the effective forces, so that they may balance each
other. D'Alembert's Principle, Art. 73, being general for all the
forces which act upon a system, will enable us to find the pressures
in question.
If a body, not acted upon by any forces, have a rotatory motion,
it will retain it for ever, (neglecting friction, &c.) and go on with
a uniform velocity. If it be acted upon by external forces, its velocity
will be variable. We shall consider successively these two cases.
SECT. I. A Body revolving, acted on by no Forces.
PROP. A system acted on by no forces, revolves about a fixed
axis with uniform velocity: to find the pressure on the axis.
1
278
Let z C, fig. 111, be the axis of rotation, a Cy a plane perpendi-
cular to it, in which Cr and Cy are at right angles. The system may
be referred to three rectangular co-ordinates, parallel to Cx, Cy,
Cz, which we shall call x, y, z. And let M be any particle of the
body, and MO perpendicular to the axis. M describes a circle
about O, and for this purpose, it must be retained in a circle by
a force in the direction MO, the magnitude of which force is known
from Art. 20. If w be the angular velocity, and OM=r, the
effective accelerating force in MO=rw³. Hence, if m be the mass
of the particle at M, mrw² will be the effective moving force in
мо.
2
The force in MO may be resolved in the directions MN, NO,
parallel to Cx and Cy. And we shall have for the particle m,
MN
moving force parallel to rmrw²
= mx w²;
MO
NO
MO
to y = mrw² = myw².
And effective forces, analogous to these, act on each of the points
of the system.
Also, the moments of these forces will be
about the axis Cx,
about the axis Cy,
0, and my zw²,
mxzw, and 0.
And if represent the sum of all the products corresponding
to different points analogous to m, we shall have for the whole ef-
fective forces, wΣmx, w² Emy, and for their moments about Cx,
and Cy, w² Emyz, wmxz: these forces acting to diminish r
and y.
ω
2
ω
2
The impressed forces are none except the reaction of the axis,
which is the pressures upon it inverted. We may reduce these
pressures to two, acting at given points.
* They cannot always be reduced to one, as will be seen.
When forces are reducible to two equal ones, acting on a line in opposite
directions, at different points, the effect produced is of a peculiar kind, and
may be called Torsion. There is no tendency to a change of place, but only
of position. The axis about which this torsion takes place, passes through
the centre of gravity, as will be shewn.
279
If the axis be supported at two points A and B, we may sup-
pose the forces to act at these points. Let U and V be the forces
which act at those points, both being in plaues perpendicular to
the axis; for it is evident that there will be no pressure in the di-
rection of the axis. Let U make an angle with the plane xz,
and Van angle with the same plane. And let CA = a, -CB=b.
Then the pressures on the axis will be
parallel to x,
U
cos. O at A,
V cos. Y at B,
to y,
U sin. at A,
V sin. Y at B.
Hence, U cos.
+ V cos. Y = w²Σmx,
U sin. & + V sin. y = w²Σmy;
because the forces must be equal.
And Ua cos. O
Vb cos. Y
4 = w² Σmxz,
Ua sin.
Vb sin. ↓ = w²Σmyz;
because the moments of the forces, with respect to Cr and Cy
must be equal.
From these four equations, a and b being known, we may find
the four quantities U, V, P, Y, (or rather, U cos. p, V cos. Y,
U sin. p, V sin. †.)
And we shall
107. We may suppose C to be taken so that the plane r Cy,
fig. 112, passes through the centre of gravity G.
be able to reduce all the forces to
(1) A force CR in the plane x Cy.
(2) Two equal forces FS, F'S' at equal distances CF, CF',
and acting in opposite directions.
These latter forces will produce no effect, except a torsion
round C. The tendency to a motion of the centre of gravity will
entirely result from the force CR.
PROP. It is required to find the force which acts at C, and the
forces which tend to turn the system round C.
280
Let the former force be R, and make with the plane rz an
angle p; let the forces at F and F' be each S, and make with
xz an angle σ; and let CF = CF' = ƒ.
impressed forces parallel to x,
We shall then have
S
S
R cos. p at C,
cos. σ at F,
cos. σ at F',
2
S
S
to y, R sin. p at C,
sin. σ at F,
sin. σ at F'.
And considering the moments of the forces with respect to the
areas Cr and Cy, we have, as before, by the conditions of equi-
librium with the effective forces,
R cos. p = w²Σmx, R sin. p = w²Σmy,
Sf cos. σ = w²Σmxz, Sf sin. σ = w°Σmy z.
The two former equations give
tan. p
Σmy
Σmx
•
,
R = w² V {(Σmx)² + (Σmy)²}.
If x and y be the co-ordinates x and y for the centre of gravity,
h its distance from the axis, and M the whole mass, we have
y
tan. p = R = w²M√(x² + y²) = w²Mh.
Hence, the force R passes through the axis and the centre of
gravity; and its magnitude is the same as if the whole mass were
collected in the centre of gravity.
If the axis pass through the centre of gravity, the force R is 0.
The two latter equations give
tan. σ =
Σmyz
Σmxz
; $f = w² V {(Σ m x z )² + (Σ my z )²}.
It appears by this, that we cannot determine the force S, or the
distance ƒ, but only their product, or the moment of the forces in
SF and S'F' to turn the axis round C.
+
281
S any value, and find the
As we suppose ƒ larger
COR. 1. Hence, we may assign to
corresponding distance f, and vice versâ.
and larger, S becomes smaller and smaller, and when ƒ is indefinitely
large, S' is indefinitely small. Hence, in this case the force in SF
would not affect the value of R, even if it were not neutralized by
that in S'F'; and we may suppose the two portions S and S to
act at the same point F.
Thus it appears, that forces which tend at the same time to
give a rotatory motion, and a motion of translation, to a system,
may be resolved into two; a finite force, producing the motion of
translation, and an indefinitely small force, acting at an indefinitely
great distance, to produce the rotation. This latter force, being
indefinitely small, would not affect the motion of translation, if it
were transferred to the centre of rotation. This resolution is fre-
quently used by Euler.
COR. 2. The directions of the forces R, S, change as the
position of the system changes, and in the case we are considering,
these directions revolve uniformly round the axis. So that to keep
the axis at rest during a whole revolution, it must oppose the
tendency to motion in every direction with sufficient force.
COR. 3. If the system, instead of consisting of separate particles,
be a continuous body, we must put the differential of the mass
instead of m, and the integral, instead of the sum : we shall then
have
tan. σ =
Sy z d M
fxzd M
Sf = w² V {(fxz d M)²+(fy zd M)²}.
COR. 4. S cannot =0, except fxzd M = 0, fyzd M = 0.
PROB. I. A plane revolves about an axis perpendicular to it;
to find the pressure on this axis, fig. 113.
We may suppose C to be placed at the intersection of the plane
and the axis; we shall then have =0, and consequently S=0.
z= O,
If G be the centre of gravity,
plane, the force R acts in CG,
and M represent the mass of the
and is = w² Mh; CG being h.
2
The same is true for any body which is symmetrical with respect
to the plane a Cy.
NN
282
PROB. II. A uniform straight line revolves about an axis
meeting it at any angle; to find the pressure on the axis ; fig. 114.
Let AB be the line, G its centre of gravity; M any point,
Dz the axis, GC, MO perpendiculars upon the axis.
CO=%,
OM=x, CG=h, and if the angle ODM = a, GMs,
have x = h + s . sin. a, z = s. cos. a, y=0, d M = ds.
GA=GB=a. Then
fxzdM = f(h+s. sin. a). s. cos. a.ds
= COS. a
3
hs² s³ sin. a
+
17. 9).
3
a to sa, gives
we shall
And let
and the integral taken from s =
2 a³
3
Ma²
2
Cos. a. sin. a
α.
sin. a. cos. a.
3
3
Ma²wⓇ
Ματω
2
Also, fyzdM=0; .'. Sƒ =
sin. a. cos. ɑ.
3
And the effort to turn AB round G is the same as if we had a mass
M
3
placed at A, and revolving round an axis GH parallel to CZ.
2
Besides this, we have R = Mhw² acting at G as before.
PROB. III. A straight line revolves in any position not meeting
the axis; to find the pressures; fig. 115.
G the centre of gravity, GC a perpendicular on the axis. Let
the axis of x be parallel to CG. And let the position of the line
be such, that it makes an angle with the plane of xy, and that
its projection GK in that plane makes an angle ʼn with x.
η
GM=s, CG = h, GA
h, GA = GB = a.
s sin. 0 =
s cos. O . sin. ท
MK = z;
Also let
We shall then have
KL; s cos. O cos. n = GL;
.'. s cos. ◊ sin ʼn = y, h+s cos. 0 cos. n = x,
SxzdM = f(h+s cos. O cos. n) s sin. 0. ds
hs'
$3
sin. +
cos. O sin. cos. + const.
2
3
283
which, taken from s=—a, to s=a, gives
2 a³
3
cos. O sin. ◊ cos. n,
Syzd M=f's cos. O sin. n. s. sin. 0. ds.
2a3
=
3
cos. . sin. . sin. n,
tan. σ tan. n;
2 a³w²
3
2
Μάω
MaⓇwⓇ
Hence, Sf=
sin. . cos. 0 =
sin. O cós. 0.
3
3
If from C a perpendicular CT be drawn on KG; the axis
will be drawn in the direction CG by the force R, and at the same
time will have an effort to turn round the axis CT by the force S.
SECT. II. A Body revolving acted on by any Forces.
108. PROP. Let any forces act upon a system moving round
a fixed axis; to find the pressure on the axis.
Let the forces at each point be resolved in the directions of the
co-ordinates; 1, Y1, 21, X2, Y, Z, &c. the co-ordinates of their
points of application; and suppose
X, X2, &c. the parts parallel to r,
Y₁, Y, &c.
►
Z1, Z2, &c.
to y,
to z.
The pressures, and the forces which support the axis, may still
be reduced to two, acting at different points of the axis, but not in
planes perpendicular to it. For the sake of simplicity, let one of
these act at the origin C. We may resolve each of these forces in
the directions of the axis, and perpendicular to it. Let U, V be the
forces perpendicular to the axis, and T the sum of the forces in the
axis. Let U pass through the origin, and make an angle with x;
let V act at a distance b from the origin along the axis, and let it
make an angle with the line of r; and let the symbol Σ be used
to represent the sum of all the products with respect to X1, X2, &c.
284
We shall then have for the whole impressed forces, U, V, T, acting
to diminish x, y, z,
ΣX-U cos. -V cos. Y, parallel to x,
EY− U sin. ¢ - V sin. Ý,
ZZ-T.....
And their moments
to y,
. to z.
Vb sin. + (ZyYz) with respect to Cx,
Vb cos. +
(2x-Xz) with respect to Cy,
Σ (Yr-Xy) with respect to Cz:
these forces acting to turn the system in directions yz, xz, xy.
The effective forces are the same as in the former section,
adding the effective forces which accelerate the motion round the
axis. These act upon each particle, and are as the distance from
the axis; if F be the force at distance 1, in direction xy, Fr will
be the force at a distance r; Fmr will be the moving force, and
its resolved parts parallel to Cr and Cy will be
Fmy and
Fmx. And the moments about Cx, Cy, and Cz arising from
this force will be - Fmxz, Fmyz, Fmr².
Hence we have the whole effective forces
- w²Σmx - FΣmy parallel to x,
2
- w²Σmy + FΣmx
to y,
to z;
and their moments in directions ry, xz, yz,
w² Σmy z - FΣmxz about Cr,
w² Σm x z + FΣmy z...... Cy,
FΣmr²
z.C
These effective forces inverted (that is, with their signs changed)
must produce an equilibrium with the effective forces: (Art. 73.)
hence, we have these six equations,
285
ΣΧ
U cos.
ΣΥ - U sin.
- cos. ++w²Σmx +FΣmy
V sin. + wmy - FΣmx
- Y+w³Σmy
= 0,
= 0,
Vb sin. +
(Zy
-
Yz) — w²Σmy z+FΣmxz = 0,
Vb cos. † + Σ (Zx
-
Xz) - w²Σmxz - FΣmyz = 0.
ΣΤ
Σ7 - Τ = 0,
==
Σ (Yx-Xy) - FΣmr² = 0.
From the last, F is determined; and then the four first equations
serve to determine U, 4, 4, and Vb.
PROB. IV. Let a body which is symmetrical with respect to a
plane passing through the centre of gravity, revolve about a hori-
zontal axis, so that this plane may be vertical; the body being acted
on by gravity only, to find the pressure on the axis.
Let C be taken at the point where the plane of symmetry
meets the axis. In this case Zmxz, Σmyz each=0, because
every positive value of zm will have a corresponding equal negative
value. Also if x and y be the co-ordinates x and y of the centre
of gravity, and M the whole mass; Σmr= Mx, Σmy:
Σmx=Mx, = My.
Let r be vertical, and y horizontal; then ΣY=0, ΣZ=0, and
ΣX= Mg. Also ZXy=Mgy. Hence, our equations give
Mgy
FΣmr² = 0,
or, putting Mk for Emr², and h sin. O for y, h being the distance of
the centre of gravity from the axis, and the angle which h makes
with the vertical,
=
gh sin. 0 Fk2, F=-
And the four equations become
gh sin. O
k2
Mg– U cos.
–V cos.
- U sin.
− V sin.
Vb sin.
+w²Mx+FMy=0,
+ My – FM =0,
W
= 0,
Vb cos. Y = 0.
286
+
Hence, Vb = 0, and we may make V = 0, b remaining indeter-
minate. This is also seen by considering that the whole pressure
will obviously pass through the origin. Hence, putting for F its
value, and for x, y, h cos. 0, h sin. 0, we have
Mg− U cos. = − M (w³h cos. 0
h²
k2
2 g sin. 0),
2
h2
k² g sin.²0+g),
h²
and U sin. = M (wh sin. 0 +
k²
g sin. ◊ cos. Ø).
or U cos. = M (w²h cos. 0
Let the force U at the axis be resolved into two; R in the
direction of the line passing through the centre of gravity, and S
perpendicular to this line.
Then taking the forces in the direction
of x and y, we have
h2
R cos. + S sin. 0 = M (w²h cos. O
k2
g sin." 0+g),
h²
g sin. cos. ).
k²
R sin. - S cos. 0 = M (w²h sin. 0 +
Multiply by cos. 0, sin. Ø, and add, and we have
R = Mw²h+Mg cos. 0.
Multiply by sin. 0, cos. 0, and subtract, and we have
S = Mg
Mg sin. 9 (1-1).
2
In the value of R, the first term, Moh, arises from the cen-
trifugal force, and is the same as if the mass were collected at the
centre of gravity. The second, Mg cos. 0, is the resolved portion
of the weight in the direction of a line passing through the centre
of gravity. The forces U and V, which support the axis, are sup-
posed to act opposite to the pressures on it. R and S in this case
will act upwards, if O be less than a right angle.
If the body were a line, revolving in a vertical plane, about
one of its extremities, its length being a, we should have h=-,
a
287
1
a
2
k =ª². Hence, in this
3
I
Hence, in this case, S = Mg sin. 0. And when the
line is horizontal, S=
Mg, or the pressure downwards is one
fourth the weight of the body.
PROB. V. A uniform straight line revolves about a horizontal
axis meeting it in any position; to determine the pressure upon the
axis.
In fig. 114, let CG be considered as the direction of gravity;
notation as in Prob. II.
The forces will be more distinctly conceived by separating them
in the following manner.
We have the forces arising from the rotation, calculated in
Prob. II, of this Chapter.
2
They are
(1) A force Mhw acting in CG.
(2) A force whose moment is Ma² w², sin. a. cos. a
3
to turn z C about C in the direction z Á.
We have also the forces arising from gravity. The force of
gravity on each point may be resolved parallel to CG, and per-
pendicular to the plane zCG; and if be the angle which this
plane makes with a vertical plane, we shall have, arising from the
forces parallel to CG, '
(3) A force Mg cos. O acting in CG; and we have to find
(4) The pressure arising from the forces perpendicular to
the plane zCG.
These pressures will be perpendicular to the plane zCG, and may
be represented by Q and S acting in the axis at distances e and f
from C. Hence, the impressed forces are Q, S, and Mg sin. 0,
and the moment is Qe+ Sf about CG. ·
—
—
Now if F be the effective force at distance 1, acting to diminish.
gh sin. O
0, we have F =
k2
; and at the point M, effective force
gh sin. 0
gh sin. O
Ꮎ
ka
xds, and its moment round CG =
xzds.
kⓇ
288
Observing that x=h+s sin. a, z=s cos. a, we have
Lxds=hs +
201
sin. a = 2 ha = Mh,
2
3
fxzds=h cos. a
S
+ sin. a. cos. a
3
203
2
a
sin. · a cos. ɑ.
3
3
sin. a. cos. a= M.—
Hence, the whole effective force is Mg.
and its moment about CG is Mg.
therefore,
-Q-S+Mg sin. 0 Mg.
h² sin. O
k²
ha² sin. O sin. a cos. a
3k2
h² sin. O
K
ha² sin. Ø. sin. a . cos. a
Qe+ Sf= Mg
•
3 k²
1
Whence the forces must be determined.
If we suppose Q to be the force acting at the centre C, and
Sf to be the moment of the force of torsion, represented by an
infinitely small force at an infinite distance, we have S= 0, e = 0 ;
whence
Q = Mg sin. 0 (
1
h²
2
k²
Sƒ= Mg sin. 0.
2
ha sin. a cos. a
3k2
These two forces, with the three before-mentioned, give the whole.
pressure on the axis.
The last tends to turn z C about CG as an
axis. Hence, on the whole, besides the pressure at C, there will
be a tendency to turn Cz about a line inclined to the plane z CG.
We may find k by the Chapter III.
For Sf we may substitute two equal forces acting at equal
finite distances from C in opposite directions. The forces will be
SS
and the distances f, Sf being of the proper value as above.
2
289
109. PROP. When a body revolving about an axis is acted on
by a single force in a plane perpendicular to the axis, to find the
pressure on the axis.
This is a particular case of the general proposition, and might
be so considered; but as it leads immediately to the centre of
percussion, we shall consider it separately.
Let P be the force, and let it act parallel to the axis of r, at a
point of which the co-ordinates are y', z'. Let the body be sup-
posed to be at rest. And let the reaction of the axis be resolved
into a force S parallel to x at an ordinate z'", and a force T parallel
to y at ordinate z". Let as before F be the effective force which
produces rotation about the axis z, at distance 1. Then Frm is
the effective moving force of m at distance r, -Fym, and Fxm,
the parts parallel to x and y. And - Fxzm and Fyzm are the
moments of these forces about the axes x and y respectively. Also,
Fr2m is the moment about z. Hence, we have
the whole effective forces
- FΣym, FΣxm, 0; and their moments
-FZxzm, FΣyzm, FΣr²m.
The impressed forces are P, S and T; and therefore we have
the whole impressed forces
parallel to x, P—S, to y, -T; to z, O; and their moments
about x, Tz""; about y, - Pz+Sz"; about z, -Py.
Hence, P-S= − FΣym,
T= - FΣxm,
- Pz+Sz" = FΣyzm; Tz" - FΣx zm,
=
- Py' = FΣr³m,
Hence, P-S=
PΣym
·y';
Er² m
S=P (1 − y Zym),
-
S2" = P (1 - 12yzm)
Σrm
If the body, instead of being at rest, have any angular velocity,
this will produce a centrifugal force in the parts, the result of
which will, by the first section of this Chapter, be a pressure on
O o
290
the axis in the line passing through the centre of gravity. Hence,
if we do not consider this force, S will be the same as before.
COR. 1. If the centre of gravity be in the plane yz, we shall
have Σrm=0, and T=0. And if the figure be symmetrical with
respect to the plane y z, we shall have Σxzm=0, and Tz””=0.
Zxzm will also = 0 in many other cases.
COR. 2. To find at what point P must be applied, in order
that the effect of S' upon the axis may be 0. We must have S = 0,
and Sz" =0*;
.. 0 =P (1 - 1 Zym)
Στη
;
0 =P (2
y' Zyzm
Σr²m
Zyzm).
Σr² m
Hence, y' =
; =
Lym
y Σ y z m
Σr² m
Zyzm
Zym
110. PROP. To find the Centre of Percussion of any system.
If a body be struck, in a direction perpendicular to the plane
passing through the axis and the centre of gravity, the centre of
percussion is the point at which the impact must take place, that
it may produce no pressure upon the axis.
P, in the last Article, represents any pressure, and therefore, the
conclusions will be true of a sudden and violent pressure, which is
an impact.
The impact is perpendicular on the plane y z, in which the
centre of gravity is.
And we have, by Cor. 2, to last Article,
y' =
Σ m² m
Lym
Σr²m
Mh
>
if h be the distance of the centre of gravity from the axis. Hence,
* It does not follow, that if S=0, Sz"=0; for Sz" expresses the force
of torsion exerted by S, which may be represented by an indefinitely small
force acting at an indefinitely great distance. The same is true of T
and Tz"". See p. 281.
291
the distance of the centre of percussion from the axis is the same
as that of the centre of oscillation
Σ
Ź
Z y z m
Lym
Σy zm
Mh
If the system be symmetrical with respect to a plane perpen-
dicular to the axis, and through the centre of gravity, and if c be
the value of z corresponding to this plane, Zxzm=Mhc, and
z=h. Therefore, the centre of percussion will be in the same
plane; and of course, in the line passing through the centre of
gravity.
In other cases, we must find Erzm by integrating.
If Exzm do not vanish, the whole force on the axis will not
vanish. There will be a tendency to twist the system round the
axis x.
Ex. 1. A semi-parabola ABC, fig. 116, is moveable about its
axis AB to find the point O, at which it must be struck perpen-
dicularly to its plane, that there may be no stress on the axis.
:
If z be measured along the axis, and y perpendicular to it, we
have
and Y
Σr°m=ff y°dzdy = ƒ y°dz,
3
is to be taken from 0 to V(az), a being the parameter ;
· · Σ r² m = { ƒ a * 2 * d z =
..
3
And if b be the whole length AB, this =
Σyzm=ffy z dydz=f2_zd z =
2 a
15
2 a b ÷
15
= 2dr = = far" d= =
zd z dz
Sazd
aze
6
6
3
a
= = = ab³
b²
4
£ym=ffydyd:=f&d==ż fazd; = = =
z
4
ab
292
5
Σr² m
2 a b
4
8
y
Lym
15
af²
at bì,
15
Σy zm
a b³
4
Ź =
11
Lym
6
ab²
213
b.
3
Hence, we find the point O by making
8
AL= AB, LO= BC.
3
15
Ex. 2. Let ABC be a right-angled triangle,
AL = 2 AB, LO = ½ BC.
CHAP. VI.
1
THE THREE PRINCIPAL AXES OF ROTATION.
111. THERE are in every body or system, three lines so situated,
that if the body revolve about any one of them, the pressure upon the
axis is 0. The same lines also have this property; that the moment
of inertia about one of them is greater, and about another less, than
it is about any other axis. These three lines are called the principal
ares: they are all three at right angles to each other. The pro-
perties, &c., of these axes will be the subject of the present Chap-
ter.
PROP. To find the moment of inertia of a system about an
axis, passing through the origin of co-ordinates, and making with
the axes of x, y, z, any angles a, ß, y*.
* The equation cos.²a + cos.³ß + cos.3y = 1, connects a, ß, y.
293
Let Cx, Cy, Cz, fig. 117, be the axes of co-ordinates, and CI
the axis of rotation. Let M be any point, and MO=r a perpen-
dicular on CI. Also, let the co-ordinates of the point M be
x, y, z, and CM= D.
If the angles which CM makes with Cr, Cy, Cz respectively,
be a', B', y': and & the angle which CM makes with CI, we shall
have
cos. d=cos. a. cos. a'+cos. ß. cos. ß' + cos. y . cos. y' *.
Ꮖ
But, cos. a
&c.;
D
Ꮖ
... cos. d
cos.a +
a+
D
Y
D
Z
cos. B +
cos. Y.
D
Also, r² = D² sin.² d= D² - D² cos.² 8; and D²=x²+y²+z²;
•• r² = x² + y²+zz
-
x² cos.“ a — y² cos.² ß - z² cos.² y
2yz cos. ß cos. y − 2 x z cos. a
2 x z cos. a . cos. y — 2xy cos. a cos. ß.
Putting 2 sin. a for x2-x2 cos.2 a, &c., multiplying by m, the
particle at M, and taking the sum indicated by Σ;
Σr²m = sin.² a Σx²m+sin.² ßΣy'm+sin.❜g Σz² m
-2 cos.ẞcos. yZyzm-2 cos.a cos. yZxzm-2cos.acos. ẞZxym.
* Let a sphere, fig. 118, described with centre C, meet Cr, Cy, Cz,
CM, CI, in x, y, z, M, I. Then, arcs Ir, Iy, Iz, will be a, ß, y;
Mx, My, Mz, will be a', ß', y'; MD will be d. And we shall have, since
xzy is a right angle, and az a quadrant
cos. ß
COS. a
cos. Izx=
sin. Izx=cos. Izy =
;
sin.
sin. Y
ช
૪
Cos. a'
cos. B'
cos. Mzx=
sin. y
sin. y'
.. cos. Iz M =
sin. Mzx=cos. Mzy=
cos. a. cos. a′+cos. ß.cos. B
α
sin. a. sin. a'
And, cos. cos. Iz M. sin. a . sin. a'+cos. a. cos. a'
d
= cos. a. cos. a+cos. ß. cos. B'+cos. y. cos. y'.
294
1
1
If we put
Σ x²m = f, Σy³m = g, Σz²m = h,
Σyzm = F, Σxzm = G, Exym = H,
2
Σr²m=f sin.² a + g sin.² ß+h sin.² y
2 F cos. ß cos. y −2 G cos, a cos. y~2 H cos, a cos. ß.
is
If the three quantities F, G, H, be each equal to 0, the ex-
pression will be simplified. We shall shew hereafter, that
always possible to place the axes of x, y, z, in such a position,
that F, G, H, shall=0. In that case,
Σr²m=f sin.* a +g sin.² ß+h sin.2
in. y.
112. PROP. To find the position of three axes of rectangular
co-ordinates x, y, z, such that
Σy'z'm=0, Σx'z'm=0, Ex'y' m = 0.
Let x, y, z, be rectangular co-ordinates, having the same origin
as x', y' z', and a known position with respect to the body.
Let 'make with x, y, z, angles, whose cosines are a, b, c,
y'.
a', b', c',
a", b", c".
Hence, by note, p. 293, the cosine of the angle which x and y
make, is aa+bb'+cc'; but this angle is a right angle;
.. aa' + bb' + cc' =0
α
Similarly, aa" + bb″ + cc″ =0
Also,
a'a'' + b'b" + c'c" = 0.
a² + b² +
· .
· (ß) *,
c² = 1
=1
c²² = 1
112
a
+
b²² +
c²
112
1
12
12
a²² + b²² +
12
by Note, p. 292.
* We have also ab+a'b'+a″b″=0
ac+a'c'+a"c"=0
bc + b'c' + b"c"=0
a² ta²² ta'z
112
1
..(7).
82 + 612 +6112
1
c² + c²² + c²² =1
295
These are six equations of condition.
Now let the moment of inertia be found for any axis, making
angles a, ẞ, y, with x, y, z.
B,
It will be
2
f sin.²a+g sin.2 ẞ+h sin.2 જ
- 2 F cos. ß cos. y − 2 G cos. a. cos. y −2 H cos. a. cos. ß.
12
But, if this axis make angles a', B', y', with x', y', z', and if
Σx²m = X, Σy" m = Y, Σz" m = 2; since by supposition,
Σy'z'm=0, &c., the moment of inertia will be
2
X sin.2a + Y sin. ß' + % sin.² y'
and these two expressions must be equal.
The latter is equal to
X+Y+Z-X cos.² a Y cos. B' - Z cos.²y',
-
and X+Y+2=Σ (x²²+y'² + z′²) m=Σ D°m, if D be the distance
of m from the origin; and this is
1
= Σ (x² + y² + z²) m=f+g+h.
Also, ƒ sin.² a=ƒ−ƒ cos.* a, &c.: hence, substituting, the ex-
pressions to be made equal, are
ƒ cos.2 a +&c. +2 F cos. ẞ cos. y +&c.
and X cos. a+&c.
By the Note, p. 293, we have
cos. a′ = a cos. a+b cos.ẞ+c.co
cos. ß′ =a' cos. a+b' cos. ß+c' cos.
ولا
,لا
cos. y' = a″ cos. a+b" cos. ß+c" cos. y.
2
Hence, X cos.² a' + Y cos.² ß' + 2 cos. y' becomes
X (a² cos.² a+b² cos.* ß+c² cos.y
cos. ẞ cos. y +2 ac cos, a cos. y +2 ab cos. a cos. ẞ)
+2bc cos.
19
+ Y (a²² cos.° a+b²² cos.² ß + c²² cos.² y
α
+2b'c' cos. ß cos. y+2 a'c' cos. a cos. y +2 a'b' cos. a cos. ß)
2
2
+ % (a"² cos." a+b" cos. B+c" cos. y
+2″bc" cos. ß cos, y +2 a″c" cos, a cos. y+2a"b" cos. a cos. ß);
296
which must be identical with
f
ƒ cos.² a+g cos.2 B+h cos.² y
+2 F cos. ẞ cos. y + 2 G cos. a. cos. y+2 H cos. a. cos. B.
Equating coefficients of cos.2 a, &c., we have
112
112
=ƒ••.
Xa²+Ya²+Za"² = ƒ. . . . . . . .(1),
Xb² + Yb'² + Z'b″² = g........ (2),
Xc² + Yc²+Zc2h........(3),
Xbc + Yb'c' + Zb"c" F... ... (4),
=
Xac + Ya'c' + Za"c" = G......(5),
Xab+Ya'b'+Za"b" = H... ... (6),
(1) a +(6) b + (5)c, gives
Xa (a²+b²+c²)+Ya' (a a' + bb'+ cc') + Za″ (a.a"″ +bb" +cc″)
=fa+Hb+Gc;
or, by the equations of condition (ß),
Xa=fa+Hb+ G c
Similarly, (2)+(4) c + (6) a gives Xb=gb+Fc+ Ha....(7).
(3)c+(5)a+(4)b.... Xc hc+Ga+Fb
=
From these three equations, and a+b²+c² = 1, we must find
X, a, b, c.
By the two first, we get
Eliminate c,
(X−ƒ) a − Hb - Gc=0,
(X — g) b — Fc - Ha=0.
.. {(X−ƒ)F+GH} a — {(X− g) G+ FH } b=0 ;
Eliminate b,
. b =
(X−ƒ) F+GH
(X − g) G+ FH • a.
.. {(X−ƒ)(X−g) — H²} a — {(X − g) G+ FH} c=0;
.'. c =
(X −ƒ) (X − g) — H²
α.
(X-g) G+FH
297
Substitute these values in the third of equations (7), viz.,
(X-h) c-Ga - Fb=0, and we have
(X-h).
(X ~·ƒ) (X − g) — H²
G-F.
(X-g) G+FH
(X−ƒ) F÷GH
=0;
(X − g) G+ FH
-
or, (X-f)(X-g) (X − h)
— {(X−ƒ) F² +(X−g) G² + (X − h)H2} - 2 FGH=0;
a cubic equation, from which X may be determined.
X being known, we can find a, b, c, by the equations
་
(X−ƒ) F÷GH
f=
(X-g) G+FHa,
C =
(X −ƒ) (X − g) – H²
(X-g) G+FH
a² + b²+c² = 1.
And hence the position of the axis x' is determined.
а,
If we make the same combinations as before, of equations
(1), (2), (3), (4), (5), (6); only using a', b', c', instead of a, b, c,
we shall have the same final result, with the difference of Y instead
of X. If we use a", b', c", instead of a, b, c, we shall have Z
instead of X in the result. Hence, the same cubic equation will
give X, Y, Z; and therefore, these must be its three roots.
hence, there is only one system of three axes, possessing the pro-
perty required; for the first root of the cubic giving one axis, the
second and third roots give the two other axes of the same system.
And
This cubic has necessarily one root possible; and it may be
shewn that the other roots are also possible. If they be impossible,
suppose them to be of the form m +n √(− 1), and m n √(— 1).
The quantities a, b, c, are possible when X is so; and for one of
the impossible roots, the corresponding quantities a', b', c', will be
of the form p+q V( − 1), p′ + q √(−1), p″ +q″ V(−1); and
for the other root, a", b", c', they will be of the form p+q√(−1),
p' — q' √( — 1), p" - q" V(-1). Now, a'a'"+bb" + c'c" =0;
•· p²+p²²+p + q² + q'² + q″² = 0,
112
12
which cannot be, if p, q, &c., are possible. Therefore, the roots
are not of the form supposed.
P P
298
113. PROP. Given one principal axis, to find the other two.
Let the given principal axis coincide with the axis of x. There-
fore, a=1, b=0, c=0. And putting these values in the four
equations in X, a, b, c, we find
X=f, 0=H, 0=G;
the first equation is equivalent to X-f=0: substituting the two
latter values in the cubic, dividing out the factor X-f, and putting
Y for X, we have
(Y— g) (Y− h) — F² = 0.
and the value of Y, obtained from this, must be substituted in
the equations
Yb' = gb' + Fc' +Ha',
12
a²² + b²² + c²² = 1.
But H=0, and a' = 0, since y' must make a right angle with x;
.'. (Y— g) b′ = Fc′ ; b'²+c'² = 1.
=
If O be the angle which y' makes with y, b' cos. 0, c' = sin. 0;
c
Y-g
... tan. 0=
b
F
2 F(Y-g)
.´. tan. 20 =
F² — (Y — g)² '
But
Y² − (g+ h) Y+gh — F² = 0.
Add (h-g) Y-(h-g) g = (Y—g) (h — g);
.'. (Y— g)² — F² = ( Y − g) (h—g) ;
... tan. 20
2 F
g-h
(8);
this gives two values of 0, differing by a right angle, which deter-
mines the required positions of y' and z'.
114. PROP. To find the moment of inertia about any axis, in
terms of the moments about the principal axes.
299
Let A be the moment about the axis of a: then, if r be the
distance of a particle m from that axis,
r² = x²+y², A =Σr² m = Σ (x² + y²) m = Σx²m+Σy³m=g+h.
Similarly, if B be the moment about the axis of y, and C about the
axis of z, we shall have
B =ƒ+h, C=ƒ + g•
Now, it is shewn at the end of Art. 111, that if f, g, h, belong
to axes for which Σyzm = 0, &c., we shall have for the moment of
inertia (µ), about any other axis,
But, since
Q
2
μ= ƒ sin. a + g sin. B+ h sin.2
µ
7.
cos.² a + cos.² ß+cos.² y = 1, sin.² a = cos.² ß+cos.²
similarly,
sin.² ß = cos.²a+cos.²y, and sin.' y = cos.² a + cos.³ß;
and substituting these values,
2
µ = (g+h) cos.³ a +(f+h) cos. B+(f+g) cos. y
2
= A cos.² a+ B cos. ß+C cos.2
7.
115. PROP. Of the moments A, B, C about the principal arcs,
one is greater and another less than the moments about any other
axes.
Let A be the greatest of the three, and C the least, and μ the
moment about any axis, and, putting 1 - cos. ß-cos. y for cos.*a,
cos.*
2
9
µ = A (1 − cos. ẞ- cos. y) + B cos. B+C cos.² y
= A-(AB) cos.2 B-(AC) cos. y,
and A-B, A-C are positive; ..μ is less than A.
Similarly, μ= A cos.' a +B cos. B+C (1—cos.
=C+(AC) cos. a+(B-C) cos. ß;
..μ is greater than C.
2
If two of the moments as B, C, be equal
2
µ = A cos.² a + B (cos.² ß+ cos.² y)
= A cos.² a+B sin. a.
a
cos. B)
If all the three A, B, C arc equal, µ = A for every axis.
300
116. PROP. To find all the axes for which the moments are equal.
If
μ be the moment, all these axes are connected by the
equation
2
A cos.² a + B cos. B+C cos.
(A-C) cos.² a +(B-C) cos.
2
2
y=µ, a constant quantity,
و
ẞ= μ- C.
radius is 1, to be described
If we suppose a sphere whose
with its centre at the origin; and suppose the axis to meet it in a
point, which we may call the pole of rotation; the co-ordinates of
this pole parallel to Cx and Cy are cos. a and cos. ß, and these
determine the projection of the pole on the plane xy. Hence,
the above equation is the equation to the projection, on the plane of
xy, of the locus of all the poles for which μ is the same, or, as we
may call them, the equi-momental poles. And it appears by that
equation, that this projection is an ellipse with its centre C, and its
semi-axes in the direction of x and y, equal to
C
A-C
and
μ C
B-C'
=
and xy is the plane perpendicular to the axis of least moment.
Similarly, (AB) cos.2 B+(AC) cos. y A- µ,
and hence, the projections of the locus of the equi-momental poles
on a plane yz perpendicular to the axis of the greatest moment are
ellipses.
Again, (A — B) cos.² a − (B− C) cos.³ y = µ— B.
Hence, the projection on the plane perpendicular to the plane rz
of the mean moment is a hyperbola.
Thus, in fig. 119, PQ is the locus of equi-momental poles on
the surface of a sphere. And the projection of PQ on the plane
xy is the ellipse MN; on the plane yz it is the ellipse RQ, and
on the plane az it is the hyperbola PO.
In fig. 120, are represented the loci of equi-momental poles,
on the surface of a sphere concentric with the body. If we make
μ= B, we have the locus a quadrant of a great circle y B, and its
projection BC on xz a straight line. When μ> B, we have
curves PQ, P'Q, &c. approaching nearer to r as μ is larger.
When «< B, we have curves pq, p'q', &c., approaching to C as µ
is smaller.
301
117. We shall now find the principal axes in given figures.
PROB. To find the principal axes of a given parallelogram.
Let, in fig. 121, AB=2a, AD=26, angle A=(. And let
the axis of x be perpendicular to the plane in its centre C;
y parallel to BA, z perpendicular to BA. CP=y, PM=z.
And let CHu, HM = v.
X
It is manifest, that the axis of x is a principal axis, and we have
to find the other two by Art. 113. The figure being supposed to
be divided into particles by lines parallel to AB and AD; one of
these at M will be dudv sin. (. Also, we shall have
y=v cos. (+u, z=v sin. (,
g=fy'd M=ff (v cos. (+u)² dudv sin. (.
And, integrating for u, =
which taken from u =
(u+v cos. ()³. dv sin. (,
a to u = a, gives
g=
реа
2a³+6av cos.2
९
dv sin. (
3
2a³v sin. ( _ 2av³ sin. ( cos.³ (
3
+.
This taken from v= b to v = b gives,
3
g=4ab sin. (
a²+b² cos.²
cos.² (
3
3
Also⋅ h=fz'd M=ffv sin.³ (du dv=2a sin.³ (vdv
3
4 a b³ sin.³ (
2
ૐ
b² sin. C
= 4 ab sin.
b sin.
3
3
And F=fyzd M=ff (u+v cos () v sin. (dude,
and integrating for u, =ƒ½ (v cos. (+u)². vdv sin.² (;
which taken from u = a to u = a, gives
F=f2av dv cos.
sin.
4ab³ cos. (.sin.*
* ફ્
=2a cos. (sin.² (v²dv=
CO
= 4ab sin. C.
b³ sin. ( cos. (
>
1
302
2 F
26² sin.
Hence, tan 20=
cos.
5
g
h
a² + b² cos.² (
cos.
- b
b2 sin.2
b² sin. 2
2
a² + b² cos. 2)²
which gives two values of 0, corresponding to the two principal
axes in the plane of the parallelogram.
To find the momentum with respect to one of these principal
axes, we have by the formula, Art. 111, observing that fred M=0,
a = 1, B = 0, y = " 0,
2
2
frad Mg sin.20+h cos.2 0-2 F sin. 0 cos. 0
2
a²+b² cos.²
= 4ab sin (
५
sin.20+
b² sin.* (
cos.20
3
3
cos. 0}
26² sin. (. cos. (
3
4 ab sin. C
3
sin. cos.
{ a² sin.²0+b² (cos. (sin. cos. O sin. ()² }
S
4ab sin. Ï {a² sin.² 0+b² sin.² ((—0)}
3
2 a b sin. ( { a² (1 − cos. 20)+b² [1
3
2
a² (1 − cos. 20)+b² [1 − cos. 2 (( − 0)]}.
Now, a² cos. 20+b² cos. 2 (0)
= a² cos. 20+b² cos. 2 (cos. 20+b² sin. 20. sin. 2
=cos. 20 {a²+b² cos. 2(+b²
2
= cos. 20 {a² + b² cos. 2(+
· {a² +
=cos. 20
sin. 2( tan. 20}
64 sin.2 27
(
a²
2
+ b² cos. 2(
a*+2a²b² cos. 2(+64)
a²+b² cos. 2(
1
2
(a² + b² cos. 2)²
Also, cos.² 20
2
1+tan.² 20
a¹+2a²b² cos. 2(+64"
· · fr²d M =
2 a b sin. C
3
{a² +b² ± √(a¹+2a²b² cos. 2}+b*)},
and the two values give B and C.
303
COR. The two moments B and C together are equal to
a² + b²
4ab sin. C,
3
the moment A about the axis r.
This proposition is general for plane figures.
In any symmetrical plane figure, the principal axes are, the
axes of symmetry, and the axis perpendicular to the plane.
In a sphere, all the axes are principal axes.
In a cube, the same is true.
In a parallelepiped, the lines perpendicular to the surfaces are
principal axes, and the moments are as b² + c², a² + c², a² + b².
In any figure of revolution, the axis of revolution is a principal
axis, and any other in a plane perpendicular to this through the
centre of gravity, is a principal axis.
118. PROP. The principal axes are axes about which the body
can revolve permanently.
By Art. 107, Cor. 4, if the body revolve about an axis z, such that
Exzm=0, Σyzm=0; the effort to turn the axis round the
centre of gravity of the body will be 0; and therefore if the point
of the axis, where the perpendicular from the centre of gravity
meets it, be fixed, the axis will be fixed. And if, besides this con-
dition, the axis of rotation pass through the centre of gravity, the
pressure on this axis will=0.
but the body left to itself, it will
axis z for which Exzm = 0,
manent axis of free rotation. Similarly, if faydm=0, y and x are
also permanent axes of free rotation.
Hence, if the axis be not fixed,
still revolve about this axis; the
Zyzm = 0, is therefore a per-
CHAP. VII. ·
MOTION OF ANY RIGID BODY ABOUT ITS CENTRE OF
GRAVITY.
119. OUR object at present is to obtain the equations for the
motion of rotation of any body about its centre of gravity, supposing
that point to be fixed. It has already been shewn, that when forces
act upon any body, the centre of gravity is the centre about which the
rotation, when separated from the motion of translation, does take
place, and it will be seen hereafter, that the motion of the centre of
gravity will be the same as if the same forces acted immediately on it.
Let three rectangular axes, fixed in space, pass through the
centre of gravity C, fig. 122, and let x, y, z, be co-ordinates
parallel to these axes. Let three rectangular axes, fixed in the
body, and moveable with it, likewise pass through the centre, and
let x1, y1, 21, be co-ordinates parallel to these. We shall also for
the sake of simplicity suppose these latter axes to be the principal
axes of the body.
Let, as in Art. 112,
r make with x1, Y1, Z1, angles whose cosines a, b, C,
Y
•
Z
a', b', c',
a", b″, c",
we shall then have the same equations of condition (B), (y), as
in p. 294. Also, if D be the distance from the origin C, of any
point, by p. 293,
or,
Ꮖ
D
Z1
a
+63 + c
D
D'
c21)
• • (€) ;
x = a x₁ + by₁ + c z1
α
similarly, yax₁ + b'y₁ + c'z₁
1
1
z = a″x₁ + b″ y₁ + c″zı
Yı
305
Ꮖ
Also, 1/3 =
a
+
tá
D
+a" D'
2
1
or x₁ = ax+ay+a'z
Y₁ = bx+by+b″z
(8).
Z1
Z₁ = cx + cy+c″z.
120. PROP.
Whatever be the motions of the different parts
of the body, it may at any instant be considered as moving round
some axis *.
This axis is called the axis of instantaneous rotation.
By differentiating equations (e) with respect to t, observing that
for a given point, x₁, y₁, Z₁, are constant, we have
dr
da
d b
dc
= x1
+ Yı
+ %1
dt
d t
d t
d t
dy
d a
d b'
dc'
= x1
dt
dt
+ y r d t
+%₁
• •(n).
dt
d z
da"
db"
dc"
x1
+ y
+%1
dt
dt
dt
dt
Which give the motions of any point in consequence of the change
of the angles whose cosines are a, b, c, a', &c.
system for which the velocity = 0; that is,
dx
dy
dz
If we put
d t
0, = 0,
dt
:0
0, we find the points of the
dt
..(1),
X1
yı 1
..(2),
0……….
..(3).
x₁ da + y₁ db + z₁ dc=0..
x₁ da' + y₁db' + z₁ dc' = 0....
x₁da" + y₁db" — z₁ dc" =0.
And (1) c + (2) c' + (3) c" gives
C
x₁ (cda + c'da' + c"da") + y₁ (c db + c'db' + c'db″)
+ z₁ (cde + c'de' + c"de")=0.
12
Also c² + c'²+c"² = 1; .. cdc + c'de + c"de" = 0.
C
And we shall have similar equations by taking
(1) b + (2) b′ + (3) b″, and (1) a + (2) a' + (3) a″.
* See proof also in Chap. I. of this Book, Note, p. 222.
Q Q
306
The differentials da, &c. are taken with respect to t: we will
suppose*
cdb + c'db' + c"db" = pdt,
and since cb + c b′ + c"b" =0,
bdc + b'dc' + b'dc"-cdbc'db'-c'db"--pdt.
Similarly, we will suppose a dc + a' dc′ + a″dc"=qdt;
whence, cda + c'da + c"da" = -q dt,
and bda + b'da′ + b″da" =rdt;
whence, adb + a'db' + a'db" - rdt.
And our equations become
PY₁ = q x₁ = 0, r₁x pz₁ =0, qz1
1
ry₁ = 0.
And hence, the points for which the velocity is O, lie in a straight
line, passing through the origin. Therefore, this line is, for an
instant, immoveable, and the body revolves round it; and these are
the equations to the axis of instantaneous rotation.
121. PROP. To find the angles, which the axis of instanta-
neous rotation makes with x1, Y₁ Z₁. If IC, fig. 122, be the axis
NM, MI, be x1, y1, 21, IN
of instantaneous rotation, and if CN,
will be perpendicular to Ca,, and
CN
X1
cos. ICx,
CI
√(x₁² + y₁² + 21²)
2
2
1
px1
2
√(p² x₁² + p²y₁² + p² z₁²)
2
px1
2
√(p² x₁² + q²x₁² +r²x₁²)'
x1
by the equations just found.
p
cos. IC x₁:
2
√(p² + q²+ r³)
q
Similarly, cos. ICy₁:
√(p² + q² + r
cos. IC z1
√(p² + q² + 2.²) •
2
* The quantities p, q, r, are the angular velocity resolved respectively
parallel to the planes yz, xz, and x y.
307
122.
PROP. The angular velocity about the axis IC is
2
√(p² + q² + r²).
Take a point P in the axis Cz₁ at a distance 1 from C; so that
for it ₁ =0, y₁ =0, %1=1.
Therefore, by (e), x = c, y = c', z=c'.
y=c',
dy2 dz² dc² + d c² + dc"²
dr²
x
And velocity of P =
2
+
+
d t2 d t² d t²
d t²
Now- pdt=bdc + b'de' +b"de", qdt = adc + adc +a"de';
0=cdc + c'de + c'de".
qdt=adc+a'dc'+a"dc';
Adding together the squares of these three equations, and taking
account of the equations (B), we have
(p² + g²) dť² = dc² + dc² + dc"2;
.. velocity of P = √(p² + q³).
Also if we draw PQ perpendicular on CI,
2
PQ= sin. IC z₁ = (1-cos. IC =₁)
√(1
IC₁)
√(p² + q²)
√(p² + q² + r²)
velocity of P
... angular velocity of P =
PQ
√(p² + q² + r²).
And the angular velocity is necessarily the same for all the points,
in consequence of the rigidity of the system.
123. PROP. To find the velocities parallel to ₁, Y₁, 21, in
terms of p, q, r.
The position of the axes parallel to ₁, y₁, 1, is perpetually
varying, but taking their position at any moment, we may resolve
the velocities and the forces in their directions. And we may trans-
form the expressions for such quantities in the directions x, y, z,
into corresponding expressions for the same quantities (velocities
or forces), in the directions 1, y1, 21, by the formulæ for the trans-
formation of co-ordinates; for a co-ordinate a will have the same
relation to x, a co-ordinate to the same point, as a velocity or force
in the direction of a, has to its resolved part in the direction of r.
dx dy dz
Now
X, Y, Z.
are the velocities in the directions of
>
dt
d t dt
308
?
Hence, we have by (C),
dx
velocity parallel to x₁ =a.
+ a
dy
d z
"}
+
a
dt
d t
d t
dx
to y₁ = =b.·
+ b'
dy
ď z
+b"
d t
d t
d t
d x
dy
dz
to 21 = c.
+ c
+c"
dt
dt
dt
But from equations (n) by the same reductions as in Art. 120.
d x
dy
d z
a + a.
+ a"
q Z1
ry 1,
dt
d t
dt
d x
dy
dz
b
+b².
+ b"
= rxı — pz1,
dt
d t
dt
d x
C
+ c'.
dy
d z
+ c"
=
= Pyi qx1.
dt
d t
d t
Hence, it appears that the quantities qry1, rx₁ - pz1,
py1 q x1,
which are O where the velocity is 0, do at other
points represent the velocities in the directions parallel to ₁ to yı
and to Z1.
124. PROP. To find the effective forces parallel to x1, yı, Z1,
in terms of p, q, r.
For the sake of abbreviation, let
qz1
And by the last
ryiπ, rx 1
pz1X, PY1 qx1
p.
dx
a
+ a
dy
+ a"
d z
d t
d t
dt
dx
dy
d z
b
+ ő
+ b"
X,
d t
d t
d t
d x
dy
d z
C
+ c
+ c
p.
d t
d t
d t
Hence, eliminating, and taking account of equations (ß),
d x
dt
dy
dt
= aπ + bx + cp,
= a'π + bx + c'p,
309
And
dz
dt
d² x
d t
d² y
d t²
d2z
d t
= a″π + b″ x + c'p;
απ
= adπ + bdx + cdp + πda + xdb + pdc,
= á' dπ +b'dx + c'dp + n'da' + xdb' + pdc',
a'dπ+b"dx + c"dp+rda" + xdb" + pdc".
d X d² y
29
d² z
2
d t²' d t² d t²
are the components of the effective ac-
celerating forces in directions x, y, z; hence, if π', x', p', be the
effective accelerating forces in directions x1,
1, 21, we shall have
π = a + ά.
d² x
d t²
d²y
d² z
+a".
d t²
d to,
and by the values just found, this becomes, observing the equations
of condition,
π'dt = dπ - rx dt + q pdt;
similarly, dt = dx + rædt - ppdt,
p'dt = dp
Or restoring the values of π, X, P,
1
qπ dt + pxdt.
'dt = z₁ dqyı dr + q (p yı — qx₁) dt — r (rx₁ — pz1) dt,
x dt = x₁ d r —
p' d t = y₁dp -
124. PROP.
z₁ dp + r (qz1 - ry₁) d t − p (py₁-qx₁) dt,
1
x₁ dq + p (rx₁ − p z₁) d t − q(ry₁ — q z₁) d t.
-
A body being acted upon by given forces; to
find the equations of its motion in p, q, r.
By Dalembert's principle, Art. 73, the forces impressed must
be equivalent to the effective forces. In expressing this equivalence
by the equations of equilibrium, we may refer them to any axes;
we shall refer them to the axes x1, y1, Z1, at any moment; for
though these axes are not fixed, the statical properties of the system
being true for any axes, are true for the position which the axes
X1, Y1, Z1, have at any moment. The effective forces π', x', p', on
a particle m, whose co-ordinates are 1, 1, Z1, have been found
in the last Article: their moment with respect to the axis Oz1
is (xix-yi)m, and the whole moment with respect to this axis is
Σ (x₁X' — y₁ π')m,
Σχ'
1
310
and putting for π', X', their values from the last, this becomes
d t
+ x₁r (q %1 − ry₁) − x₁p (pyı — 9x1)
dr
dp
2
X1
X1 Z1
d t
Σ
d q
dr
2
-y1 21
+ Yı
d t
d t
x1
Yıq (pyı − q x1)+yır (rxı − p z1)
1 1
. m.
Now, Ex₁y₁m = 0, Σx₁ z₁m = 0, Zy,zim = 0; and p, q, r,
are constant in the integrals expressed by 2; whence this becomes
2
2
dr
Σ(x₁² + yı²) m
+ Σ (x₁² — y₁²). m. p q.
d t
2
2
But (x²+y12) m = C ;
2
Σ (x₁² — y₁²) m = Σ {(x1² + z₁²) − (y₁² + z₁²)} m = B− A ;
·
whence the moment of the effective forces with respect to the axis
of ≈1, becomes
dr
C + (B − A) p q·
dt
dq
B + (A - C) pr;
dt
Similarly,
dp
and
A
+ (C − B) qr,
dt
are the moments of the effective forces with respect to the axes of
y₁ and z₁ respectively.
Y1
Now, let the moments of the forces impressed with respect to
the axes T1, Y1, 1, be respectively N, N', N": hence, we shall
have, by Dalembert's principle,
dr
N=C + (B− A) p q
d t
dq
N' = B
+ (A − C) pr
· .(0);
dt
dp
N" = A. ¹²² + (C − B) qr
dt
which are the equations of motion.
If the forces which act on any particle m, be resolved into
X₁, Y₁, Z₁, in the directions of x, y₁, %1, respectively, we shall
Y1,
have their moment with respect to the axis Cz₁ = (x₁ Y₁ − y₁ X₁) m ;
and the whole moment for this axis will be
1
311
similarly,
Σ (x₁ Y₁ − y₁ X₁) m = N;
1
Σ (%1 X1 − x1 Z₁) m = N',
Σ (y₁ Z₁ - z₁ Y₁) m = N".
Z1
COR. If the motion take place parallel to a fixed plane, the
expressions may be simplified. (The plane will be one of the
principal planes.)
Let the plane be that of xy, and of x1, y1; then, z and z1 are
always in the same direction. Therefore, c" =1, c'=0, c=0,
a"=0, b″=0,
a² + a²² = 1, ab + a'b' = 0,
a"
b² + b²² = 1,
p=0, q=0,
rdt=bda+b'da'.
If be the angle which r₁ makes with x,
a=cos., a'=sin. &; b
=
sin. O, b'cos. & ;
аф
dt
bda + b'da' = sin.2 dp+cos. &dp; r= ;
and the first of the equations (0) becomes
N=C.
аф
dt²
which may be deduced independently.
125. PROP. The quantities p, q, r, being known, to determine
the position of the body.
Let the plane x Cy, fig. 123, cut x Cy, in the line NC, and
let the angle CN=4, NCx=; and let the angle which the
planes x Cy, x, Cy₁ make, be ; this angle will be the same as
Z1 Cz.
Let a sphere be described about the point C with radius 1, and
let the points x, y, ~; x1, Y1, 1; N; be upon its surface; we
shall then have
x₁N=4, xN=Y, ₁=0; also angle N=0.
And, N≈₁ Nz₁, quadrants.
And joining every two of the points x, y, z, x1, Y1, 21, by
arcs of great circles, we shall have the following equations by the
properties of spherical triangles,
a=cos.x1x=cos. O sin. Ø sin. +cos. O cos. Y by triangle x1 Nx,
=cos. y1 x = cos. O cos. Ø sin. - sin. & cos. Y.
Y
y₁ Nx,
312
c=cos. %₁x=sin. . sin. ; for in triangle z₁ Nx, the angle z₁ Nx
Z1 X
;
is the complement of
a=cos.x₁y=cos. O sin.
b': =cos. y₁y=cos. O cos.
c'cos. z₁ysin. O cos.
is the complement of 0;
a"
=cos. x12=
b"
cos.
- cos. sin. Y by triangle ri Ny,
cos.
+ sin. Ø sin. Y
.y₁ Ny,
; for in triangle z₁ Ny, the angle z₁ Ny
sin.0.sin. by triangle r₁ Nz where x₁Nzis+0,
=cos. y₁zsin. cos. O by triangle y₁ Nz,
c" = cos. 21 z = cos. 0,
Hence,
z=cos.
pdt=bdc+b'de +b"de"
{cos. cos. sin. - sin. & cos. }
π
2
. {cos. O sin.
de + sin. O cos. d¥}
+{cos. O cos.
cos. + sin. o sin. }
• {cos. O cos.
<
do sin. O sin. d}
+ sin. cos. . sin. 0 .dė
=cos.²0.cos.p.de – sin. 0. sin. ¿dy + sin.²0 cos. odł
=cos. desin. Ø sin. dy;
.. pdtsin. O sin. pdf-cos. pd0.
Similarly,
qdt=ade+a' de'+a"de" = sin. O cos. dy+sin. p.do.
And,
rdt=bda+b'da'+b"da".
0
O Y
Now, da=-sin. 0 sin. p sin. do
.. bda=
sin. - sin. cos. } do
Y
cos. Y cos. O sin. y} dy;
sin. - sin. & cos. }.
+{cos. O cos.
+{cos. O sin.
{cos. O cos.
sin. 0. sin.
. sin. Y do
+{cos. cos.
sin. Y
sin. - sin. cos. ¥}˚do
+ {cos. O sin. &
cos. Y
cos. O sin. †}
{cos. O cos.
and we shall have b'da' by putting
sin. - sin.
π
+ for
2
cos. } dy,
in this expression.
313
Hence, bda+b'da'
sin. cos. sin. cos. o d0 + {cos.² 0 cos.² +sin.*p}dø
0 d¥.
- {cos. O cos.²+cos. O sin.² p} dy.
Also, b"da"
=sin. O cos. . {cos. 0.sin. p.de+sin. 0 cos. pdp}.
Hence, rdt=bda + b'da'+b"da" = do— cos. Ody.
Hence, having found p, q, r, we must determine , Y, 0, by
means of the equations
pdt=sin. & sin. Ody- cos. odł
qdt=cos. sin. Ody+sin. ode
rdt=do-cos. Ody
......
• •(c).
And ,, being known, the position of the body is completely
determined.
N, N', N", may be functions of p, . 9. Hence, the six
P, †. 0.
equations (0) and (1), will determine the quantities P, †, 0, p, q, r.
126. PROP. A body revolves about its centre of gravity acted
upon by no forces; it is required to integrate the equations already
found.
Take the equations (0),
Cdr + (B− 4) pqdt=0
Bdq + (A − C) prdt=0
Adp+(C−B) qrdt=0
(K).
Multiply by r, q, p respectively, and add, and we have
Crdr+Bqdq+ Apdp=0;
.. Cr²+ Bq²+Ap² = h².
h being a constant quantity.
..(^),
Again, multiply equations (), by Cr, Bq, and Ap re-
spectively, and we have
C²rdr + B²qdq+A²pdp=0;
·· C²r²+B°q°+A²p=k, a constant quantity......(µ).
RR
314
Again, multiply by c, b, a, and add
C {cdr + (qa-pb)rdt}+
B {bdq+(pc−ra) qdt}+A {adp+(rb−qc)pdt} =0.
But it may be proved *, that
(qa-pb) dt=dc, (pc-ra) dt = db, (rb-qc) dt = da;
.. C.d.rc + Bd.qb+A.d.pa=0,
Crc +Bqb +Apa
ι.
Similarly, Cre' +Bqb' +Apa' = l'
Crc"+Bqb" + Apa" =l″.
If we add then three, we get
· · . (v).
C²r² + B²q²+A°p² = l² + l'² + l″ ² = k².
127. PROP. When a body revolves, acted on by no forces, there
exists a plane, to which it may be referred, which plane is invariable
in position.
If we draw a line m C, making with x, y, z, angles of which
the cosines are
Apa✈ Bqb+Crc
√(A²p² + B²q² + C²¾‚‚²)
&c.:
since these quantities are constant, this line will have the same
position during the whole motion of the body, and a plane per-
pendicular to it will be fixed.
**Take the three equations
a dc + adc' + a"de"=qdt,
bdc+b'dc' + b'dc"--pdt,
ede + c'de' + c"de"=0.
Multiplying the first by a, the second by b, the third by c, and add them ;
and we find, taking account of the equations of condition,
de=(ag-bp) dt.
And in the same manner we shall find
db = (cp — a r) dt, da=(br-cp) dt.
And so for the other similar quantities.
315
This plane is called the Principal Plane of Moments *.
COR. 1. We may thus find the angles which the line mC makes
X1 Y1 Z1.
with x
cos mСx₁ = a.cos. m Cx + a' cos. m Cy+a". cos. m Cz
Bq
Ар
cos. m Cx₁ =
similarly, cos. m Cy,
k
に
​Cr
cos. m Cz₁
た
​COR. 2. If we take the plane perpendicular to mC for the
plane of xy, we have
a"
Ар
k
AP, b" = -
Bq
k
Cr
C"
k
Ap
Bq
Cr
or sin. O sin.
sin. cos.
cos. 0 =
θα
k
に
​に
​PROB. I. All the three moments A, B, C being equal, and
the body not being acted upon by any forces; to determine ils
motion.
In this case the equations (4) become
dr
dq
=0,
dt
dt
dp
0, =0.
d t
Hence, p, q, r are constant quantities.
Hence, the position of the axis of instantaneous rotation is fixed
with respect to the body, and the motion of the poles of instanta-
neous rotation is therefore nothing, and hence this axis is fixed in
space.
All the axes of the body possess the properties of principal
axes: therefore let the axis of instantaneous rotation coincide with
* If we consider at any instant the momentum of each particle of the
system; (that is, the product of its velocity and mass;) and if we resolve
this momentum parallel to a given plane passing through C, aud multiply
the resolved part by the perpendicular from C, so as to get its moment, we
may obtain the sum of the moments of the particles referred to this plane.
And the plane for which this sum is the greatest, is the plane found above,
and hence denominated the Principal Plane of Moments.
316
Cz₁; hence, p=0, q=0, r≈constant; and equations (1) become
sin. sin. Od-cos. pd0=0,
cos. O sin. Od+sin.
the two first give de=0,
.. O and are constant.
d0=0,
do-cos. O dy =rdt.
аф
and sin. Od 0, whence dy =0;
Hence, do=rdt, and
аф
d t
=r, a constant quantity.
Therefore the body will revolve about a fixed axis and with a
constant velocity.
If the motion have been produced by a single force, acting at
a point of the system, (impact or pressure,) the axis of rotation will
be perpendicular to the plane which passes through the direction of
the force and the fixed centre of gravity.
These conclusions are applicable to a sphere, a cube, &c.
PROB. II. If two of the moments, A and B, be equal, and the
body be not acted upon by any forces; to determine the motion.
(This will be the case for all figures of revolution, the axis of
revolution coinciding with the axis Cz₁).
The equations («) here become
Cdr =0,
d q
A
+ (AC) pr=0,
d t
dp
A
— (A–C) qr=0.
d t
From the first of these it appears that is a constant quantity,
and the third being differentiated and divided by dt, gives
ď p
A
(A — C) r
dg
•
dť
dt
dy
and substituting from this the value of
in the second, we have
d t
d'p, (A-C)
+
• p = 0),
A
317
whence we easily obtain
p=a.sin. (nt +y); putting n for
and supposing A > C; and we have from this
q=a cos. (nt+y);
(A-C) r
A
where a and y are two arbitrary quantities, to be determined from
the circumstances of the motion at a given time.
2
Sincer is a constant quantity, and also p²+q a constant
quantity, viz. a², it appears that if IC be the axis of instantaneous
rotation, ICz₁ is a constant angle. Hence, the axis of instan-
taneous rotation describes, with respect to the body, a conical
surface about Cz₁.
By substituting these values in equations (¿), we obtain
adt.sin. (nt+y)=sin. . sin. Od-cos. pd0,
$.sin.
adt.cos. (nt+y)=cos. . sin. Ody+sin. de,
rdt=dp-cos. Ody.
Take the plane of ry to be the principal plane of moments;
we then have, by Cor. 2 to Art. 127,
Ap A a
sin. sin.
sin. (nt+y),
に
​k
Bq
A a
sin. cos.
cos. (nt + y),
k
k
Cr
cos. 0 =
>
k
.'. tan. &=tan. (nt+y);
$=nt+y;
аф
dt
=N.
A a
Cr
A a
sin. 0 =
; cos. 0 =
; tan. 0 =
k
に
​Cr
and is constant.
And by the third of equations (4)
k (r-n) k
dy
1
аф
dt
cos. 6
d t
-:)
Cr
Α
و
318
Hence, the body revolves uniformly about C21, while CN, the
line of nodes, moves uniformly round C in the plane xy.
1
1
*
Suppose, that at the first instant, when t=0, the instantaneous
axis CI is in the plane z,Ca, and makes an angle & with Cz1;
and that the angular velocity about it is e. Then, by the formulæ
of Art. 121, since IC is perpendicular to Cy, we have
cos. IC z₁ =cos.
cos. d=
√(p²+q²+r²)
√(p²+q³+r²)
cos. ICr₁ = sin. d
Ρ
cos. IC y₁=0=
2
9
√(p² + q² + r²)
2
Also, e = √(p²+q² + r²)
Hence we have, since t=0,
..q=0.
COS.
б
=2, sin. d=
a sin. Y
a cos. Y
π
0:
;
y=
€
€
€
A
... r = e cos. d, a≈e sin. d, tan. 0 =
=
tan. d.
C
And n=
A-C
A
A - C
· e cos. d.
A
2
2
Hence, (see Art. 126,) k² = A²a²+ C²r² = Æ°e² sin.²d + C²e² cos.² §
2
= €³ { A² — (A² — C²) cos.* d}.
6
αψ
A2 - C2
Hence,
E
1 –
cos.
A
2
28).
dt
аф
If CA, nearly, n is small; and since = ", and
d t
dy
dt
nearly, the motion of the body round the axis Cz, will be slow,
and the motion of Cz₁ about Cz, comparatively quick.
When e and d are positive, I is within the quadrantal space
x₁ Y₁ z₁ fig. 123; and it appears that in this case is negative,
1
* Since any line in the plane r₁C, is a principal axis, we may take
for C'x, the intersection of æ¡Cy, and ~¡CI.
319
•
and Nx, y, which was in the preceding Articles supposed below
Nry, is here above it.
1
π
We have, when t=0, &=y=. Hence, N is the pole of
z₁₁, and z, which is a quadrant distant from N, is in 1x1.
Therefore in this case I is in the arc z₁z. And any moment may
be considered as that when t=0. Therefore I, the pole of
instantaneous rotation is always in 13. And since and p²+q*
are constant, CI always makes the same angle with Cz.
Since zz₁ =0, z₁I=d,
1
r
we have z I=0+d, sin. zI= sin. O cos. d+cos. O sin. 8,
sin. z I
sin. z₁I
= cos. 0 (tan. 8
+1)=
Cr
A
k
(-4 +1)
(A-C) r
k
do nA
аф
(A − C) r
Also,
;
dy k
k
.'. sin. ≈₁I.dq=sin. zI.dy.
If with centres z, z₁, we describe circles in fig. 124, on the
surface of the sphere, since do is the angle described in dt by the
body about ≈₁ and sin. z₁I the radius; sin. z₁I.do will be Im, the
arc of the circle which a point would describe in dt. Similarly,
dy
Z₁₁₁,
d is the angle which the surface 11 y1, describes about z;
sin. z I is the radius: therefore sin. I.dy is the arc In de-
scribed by a point about z in dt. And Im In.
Hence, if the circle Qn be fixed, and Pm carried by the pole
I of the body, roll on Qn; the angular motion will be exactly the
same as when the body is left to itself; and this supposition re-
presents the motion of the body in the Problem. The reasonings
C- A
will be the same if AC. In this case, we shall have n=
The point
A
7.
will be between, and I, and the circle described
about z will roll with its interior circumference on the outside of
the circle described about z.
320
PROB. III. When all the moments are unequal, and the body
is acted upon by no forces; to determine the motion.
Equations (k) become in this case
dr
са
+(B−A) pq=0,
dt
dq
B
+ (AC) pr=0,
dt
dp
A
dt
+ (C− B) qr=0.
Let pqr. dt = do, and we have
Crdr + (B− A) do=0,
Bqdq + (A−C) dq=0,
Apdp + (CB) do=0;
2 (A-B)
.'. 7.² =
C
$ + c²,
2 (C-A)
Q
$+b²,
B
2 (B — C)
$ + a²,
A
a, b, c being the values of p, q, r, when is 0;
:.dt
аф
pqr

√{(2¢
B-C
A
+ a²) (2¢
аф
C- A
B
+b²) (2¢
A- B
+c²)}
And integrating, we have t in terms of , and hence
And hence, we have p, q, r; and by Art. 125, the
body.
C
in terms of t.
position of the
COR. By Art. 114, we have the moment of inertia at any time
Σrm= B+C 7
A cos.² a + B cos.² ß + C cos². Y
321
=
A p² + Bq² + Cr²
p² + q² + r²
2
hº
2
p² + q² + r
2)
by Art. 126.
· · (p² +q²+r²) Σr'²m=h².
Hence, the sum of each particle into the square of its velocity, con-
tinues constant during the motion.
PROB. IV. A solid body revolves about its centre of gravity,
so that its axis of rotation coincides nearly with one of its principal
axes; to find the motion.
If IC be the instantaneous axis, always nearly coincide with Cz,,
sin. IC z₁ =
√(p²+q²)
√(p² + q² + r³)
.. p and q must both be small, Cdr=0, nearly, and is nearly
constant, and the velocity of rotation
= n;
•.Bảq+(4-C)npdt=0, Adp+(C–B)nqdt=0;
... as before, p=a sin. (n't+y), q=ß cos. (n't+y),
where n'=n√(A — C). (B– C)
B=a\
✓
AB
(A - C). A
(B-C). B'
and knowing p and q, we might find 4, 4, 0.
Now if IC and zС coincide, a = 0, and ß = 0.
Hence,
a and B are small when ICz, is small, and if n be real, p and q will
always be small, but if n' be imaginary, p and q become exponentials,
and increase beyond small values; and the solution is not applicable.
In the first case, the axis will oscillate about Cz₁. In the second
case, the axis IC will leave C₁, and oscillate about another of
the principal axes.
T
The first case will happen, if (A — C). (B — C) be positive;
.. if C be the greatest or least of the moments A, B, C.
The second case, if C be the mean moment.
Hence, if in the second case the body at first-oseillate accurately
about Cz₁, and if that axis be disturbed ever so little, the axis will
entirely leave its position.
S s
322
COR. 1. We may prove that p and q cannot increase beyond a
certain limit.
2
2
-
A p²+ Bq² + Cr²=h²……..(\), A²p² -+- h²q² + Cr² = k² ....(u),
(u) - (X) C gives A (A — C) p² + B (B - C) q² = k² — Ch².
Hence, if p and q are small at first, and A-C and B-C of the
same sign, p and q will always remain small.
2
COR. 2. The limits of p2 and q² are
p* <
k² - Ch²
2
A (A − C)
k²- Ch²
q² <
B (B-C)
CHAP. VIII.
MOTION OF A RIGID BODY ACTED ON BY ANY FORCES.
E
128. We shall consider, in the present Chapter, the motion
of a body, whether it be in free space, or constrained to move upon
a given plane. The same principles are applicable in both cases,
if we include, among the forces which act upon the body, the
reaction of the plane which it touches, and then eliminate this
reaction.
In a large class of problems of this kind, one of the principal
axes moves parallel to itself, and consequently, all the particles
move in planes perpendicular to it; for instance, if a body bounded
by a cylindrical surface of any form roll upon a plane. We shall
take this more simple case separately.
SECT. 1. When the Motions of all the Particles are
in Parallel Planes.
323
PROP. The body being acted upon by any forces, the motion
of the centre of gravity will be the same as if all those forces acted
at the centre.
By Dalembert's principle, Art. 73, the impressed and effective
forces must be equivalent, and their moments about any point also
equivalent; the former consideration will determine the motion of
translation of the centre of gravity, and the latter the motion of
rotation about it.
If x', y' be the co-ordinates of any particle, the velocity in the
dx'
direction of x' is
dt
and therefore the effective accelerating force.
is
d'x'
d t²
dex'
2,
and the effective moving force m
Thus we have
dt²
Emd²x'
whole effective force parallel to x'
; to y
dt
อยู่
Σmdy'
d t²
Also, if X and Y be the impressed accelerating forces on any
particle m,
whole impressed forces are Σm X, Em Y.
Hence we have
Σ
md°x'
dt2
-Ση Χ, Σ
m d'y'
dt²
= Σm Y.
Let x, y be the co-ordinates of the
co-ordinates of m from the centre
y' = y + y;
.2m2=0, Σmy=0; .. Emdr=0, Emdy=0;
centre of gravity; and x, y the
of gravity, so that x=x+x,
m d²x'
.. Σ
Σ
dt
m d'x
dt2
d2x
= M. d'
dt
2
(M being Σm the whole mass ;) because x is the same for all par-
ticles. Similarly, Σ
d²x
dt2
md²y'
= M
d² y
dt
dts
;
છે
Em X dy ZmY
M' dť²
M
....
… . (§),
324
Which are the equations that would result if all the forces were
applied to the centre of gravity.
COR. The forces m X, m Y, are here understood to be pres-
sures; and as impact is only a short pressure, the results are true
of impact. The forces m X, mY, are measured by the momenta
generated in a time 1"; the force of impact may be measured by
the whole momentum which it would generate.
129. PROP. The body being acted upon by any forces, the
motion of rotation will be affected as if the centre of gravity were
fixed, and the same forces were applied.
The moment of the effective force with respect to C is
Ση
´x' d² y' — y' d²x'
d t2
And putting for x', x + x, and for y', y
+ y, it is
x ď²
Y
x d² y
+
+
dt2
d t2
diz
2
x d² y
x d² y
+
d t²
Σ.m
y d² x
y d² x
d t²
d t²
,
y d² x
dt2
y d² x
d t
And observing that Em
M, Em x=0, Emy=0, Σmd'x=0,
Σm d² y = 0; and that x and y
are not affected by ; it becomes
x
2
M.
x d² y - y d² x
d t²
+ Σ.m
mzdy = y dr
Also the moment of the effective forces is
2
d t²
Σ, m (Yx' — Xy')=Σ.m (Yx - Xy+ Y x − Xy)
= M (Y x − Xy) + Σ . m (Y x − X y).
Equating, we have, observing that the terms multiplied by M
are equal by last Article,
x d³y - y d³ x
Σ.m
Σ. m (Y x
Xy)....
d t
·(π).
Which is the equation that would result if the centre of gravity
were fixed.
325
1
COR. If in fig. 125, GA be a line always parallel to a line
fixed in space, GM a line in the plane of ry, fixed with respect
to the body, M any particle, of which the co-ordinates from G are
x, y: it will be seen as in Art. 16, that (x d² y − y d²x) is the
-
second differential of the sector AGM; and hence, since GM is
constant, if GM=r, AGM=0, we have x d'y -y dx = r² d² 0.
Therefore
Σ.mr².
d² 0
d t
= Σ.m (Yx - Xy).
Or if as before, Σ. mr²= M k²,
d Ꮎ Σ.m (Yx-Xy)
d t
Mk2
(p) *.
130. PROP. To find the centre of spontaneous rotation.
The centre of spontaneous rotation is the point which remains
at rest for an instant when the body is put in motion by any force.
Thus, if a body GC, fig. 126, be acted upon by a force PC,
it will, for the first instant, revolve round some point 0, which may
be thus determined.
Let the force in PC be perpendicular to GC, and = P; and
the mass of the body being M, the space Gg described by G in
Pt2
a small time t, will be
the force being supposed constant
M' 2'
for the time t. And by this motion of translation, any point, as O
will be transferred to o, O o being equal to Gg. Let GO=1,
CG=h; and by Art. 75, the accelerating force causing O to re-
volve round C is
t is
Phl t²
Mk² 2
Phl
Mk²
and the space generated by it in the time
And if this be equal to Oo, the point O will be
* These formulæ might be used in solving many of the problems in
Chap. VI. So long as Mk2 is the same, the motion will be the same
whatever be the form of the body. Thus if, in Prob. V, fig. 40, instead of
a straight line PQ, we had a body of any form, of which one point slides
along AX, and the other along AI, the motion would be the same as is
investigated, p. 124.
326
1
carried backwards by the rotation, just as much as it is carried
forwards by the translation, and will be absolutely at rest: that is, if
P t2
M 2
Phl t2
2
Mk²' 2'
k²
h
or
Hence, by (a) C and O, are so situated that if one be the
centre of suspension, the other is the centre of oscillation, or
percussion.
COR. If the force which acts at C be a force of impact, the
point O will be the same, Gg and Oo being in this case described
by the uniform velocities which the impact generates.
PROB. I. To find at what distance from the Earth's centre a
force must have acted, to generate at the same time its progressive
and rotatory motion.
This problem stated generally, is, having given the velocity of
the centre of gravity and the angular velocity in a body revolving
about a permanent axis; to find at what distance from the centre of
gravity a single force would produce them.
Ph
P
Let P be the force acting at a distance h; then and
M M k*
are the forces which accelerate the centre of gravity, and a point
at distance 1 about the centre of gravity; and hence, the whole
velocities generated in any times will be as these forces. If v be
the velocity of the centre, and a the angular velocity, measured by
the space described by a particle at distance 1, we have
P Ph
M' Mk2
k² a
:: V:
v: a; :. h =
V
In the case of the Earth, let its radius be taken = 1, and let the
unit of time be 1 day: therefore, a=2π. And the radius of the
Earth's orbit being r, the space described in 366.24 sidereal days is
2πr, and therefore v =
And since the Earth is a sphere,
Ωπη
366.24
327
12
11
Hence, h =
215
2 366.24
and r =
24266, nearly;
ጥ
1
:. h =
nearly.
165.6
Hence, we may find the centre of spontaneous rotation by the
k2
formula /=
h
66.25, which is a little greater than the Moon's
distance from the Earth.
In the same way for Mars we should have h =
1
of his
464
radius, nearly. And for Jupiter, h = of his radius, nearly. In
6
the latter planet, the centre of spontaneous rotation is only of the
radius from the centre *.
5
131. The following problems refer to such motions as the rock-
ing of a cradle, the oscillation, or as it has been called, titubation of
a body on a curved base, and the rolling of a body not spherical.
We shall solve some of them, first neglecting, and afterwards con-
sidering, friction.
PROB. II. An ellipse with its plane vertical, rolls upon an
horizontal plane which is perfectly smooth; to determine the motion,
fig. 127.
Let P be the point of contact with the horizontal plane,
CM=x, MP=y, CA=a, CB=b, the semi-axes; PN vertical,
CN horizontal; and let the angle ACH be 0,
b
dy
b
x d x
Y
√(a² — r²);
but at P,
a
d x
a
√(a² — x³)
d x
sin. 0
a V (a³ — x²)
= tan. 0; ..
dy
cos.
hence, r=
a² cos.0
√(a³cos.³0+b²sin.20)
;
and
У
=
b x
b² sin. 0
2
√(a*cos.¹0 + b²sin20)
* John Bernoulli's Works, Vol. IV. p. 284.
328
'.
Hence, if y be CH or NP,
y = x cos. 0 + y sin. 0 = √(a cos.20+ b² sin.20).
Also CN=x sin. 0 — y cos. ◊ =
(a² b²) sin. ◊ cos.
√(a²cos.²
√(a cos.20+b² sin.20)
2
Now let R be the reaction upwards at the point P, and R. CN
its moment; M the mass of the body, Mg its weight, Mk its
moment of inertia; the centre of gravity being supposed to be at C.
Then by Art. 128, and 129,
ď² y
R
g,
d t2
M
d² 0
d t²
R (ab²) sin. O cos. O
Mk² V (a² cos.² 0 + b² sin.º 0)
Ө
But since
Y
√(a² cos.² 0 + b² sin.² 0);
(a²
b²) sin. O cos. O d 0
dy
V(a cos. + b² sin.20)
*
And the second equation becomes, multiplying by dě,
R
dy; and eliminating R by the first,
Ꮷ Ꮎ d Ꮎ
dt2
2
•
Mk2
d Ꮎ d Ꮎ
dy
d y d² y
+
+
d t²
k² dtv
g dy
k²
0.
Multiply by 2k² and integrate,
dt
d t2
k² d02 dy
dt
2
+ +2gy C.
Put for y and for dy their values, and this becomes
{k
k² +
Q
4
2
Ꮎ
a*e* sin. cos.20) d02
2
a cos.*0 + b sin. 0 dt
+ 2g V (a² cos.²0 + b² sin.²0) = C.
d Ꮎ
dt
Hence the angular velocity is known. Cis to be deter-
mined by knowing the value of this velocity for a given value of 0.
329
If when CA is
C = k² a² + 2ga.
vertical, the angular velocity be a, we have
And when CA becomes horizontal,
d02
4.2
d t²
+ 2 gb = k² a²+2ga;
d Ꮎ?
2 g (a - b)
a² +
;
d t²
k²
therefore, a point at the distance k from C acquires as much
velocity as if it had fallen vertically through the difference of the
axes CA, CB.
Since there is no lateral force, the centre of gravity C will as-
cend and descend in a vertical line. The surface will slide along
the horizontal line PQ. If there be so much friction as to prevent
this sliding, it will roll; the centre of gravity will have a lateral
motion, and the problem will no longer be the same.
2
The figure may be an elliptical cylinder with a horizontal axis.
In that case, k² (a²+b²). Or it may be an elliptical spheroid,
APB being one of its principal sections. In this case, k²
(a+b²). It may be any figure in which the vertical ellipse is the
part which touches the horizontal line, and the centre of gravity is
at C.
If the plane were inclined at an angle e, the rotatory motion
would be the same, putting g cos. e for g. And the body would
move along the plane, so that the motion parallel to the plane
should be that of a point sliding down an inclined plane. For the
part of gravity parallel to the plane could not affect the rotatory
motion, since its result would pass through the centre of gravity.
PROB. III. When the major axis of the ellipse, in last Problem,
is nearly horizontal; to determine the small oscillations.
Ф
Let be the angle which CB makes with the vertical; ..
П
-0. And when is very small, sin.*0=cos.*p = 1
cos.³0 = sin.²=4³. And, by last Problem, neglecting 4*,
(k² +
a*e¹ س \ dز
2
b2 d t
C − 2 g V (b² + a³é³¤²)
еф
= C-28 (b + ²²²).
26
- p³,
−
TT
330
(And if =ß when
And neglecting ß³
³
аф
=
dt
= 0)
a² e²
g
=
=
b
(B² - $³).
2
do
a² e² g
(B² — p²);
d t²
k2 b
keb
.. if
= 1,
a² e²
-
dt; t =
аф
&g° V (B³ — p³)
and for a whole oscillation, this must be taken from =ß to
Ф =-ẞ. Hence, if time of an oscillation = T,
. arc
дв
(sin. =
;
g
T = T
It appears from this, that the length of the isochronous pen-
dulum is 1=
ι
k2b
22
ae
a² e²
PROB. IV. Let the body roll on a horizontal plane, the part
which comes in contact with the plane being a portion of a common
cylinder with its axis horizontal; to determine the motion, fig 128.
Let a line be drawn through the centre of the cylinder, and
through the centre of gravity; let the latter point be at a distance
c from the former; let O be the angle which this line makes with
the vertical; then it will be found by proceeding as in Prob. II,
that
2
do2
d t²
2 cg (cos. 0-cos. B)
k²+c² sin.² 0
where Mk is the moment of inertia about the centre of gravity.
If the body perform small oscillations, we shall have for the
k
length of the isochronous pendulum, 7 =
с
331
In these cases, the centre of gravity ascends and descends ver-
tically, and goes twice up and down while the body goes once back.
and forwards.
PROB. V. A cylindrical body is supported on a horizontal
plane and oscillates; the friction being such as to prevent all sliding;
to determine its motion, fig. 128.
Let C be the centre of the circle AP, G the centre of gravity;
P the point of contact with the horizontal plane; PC will be
vertical; let GN be horizontal. Let a be the point in which A
comes in contact with the plane: then, AP=aP.
Let CA=a, CG=c, aH=x, HG=y; ACP=0,
x=AP-PH-a0-c sin. 0,
y=CP
CN=a c cos. 0.
Now, the forces which act upon the body are the force of gravity,
and the force which arises from the pressure and the friction at P.
Let these forces at P be composed of a horizontal force Q, and
a vertical force R. And, by Art. 128, the effect on the motiou
of the centre of gravity will be the same as if all the forces were
applied immediately at that point.
Hence,
d²x
Q
ď y
R
g.
dt"
M'
dt t²
2
M
And, by putting for x and
У
their values,
d (ade- ccos. O dė)
d t2
Q
M'
d (c sin. O de)
R
dt²
2
+g
M
Also, by Art. 129, the effect of the forces in producing rotation
about the centre of gravity, is the same as if it were fixed, and the
forces Q, R, acted on the body; hence,
d20
dt2
Q.GH+R.GN
MK2
Q (ac cos. 0)+ Rc sin. 0
Mk²
332
Substituting for Q and R, &c. we find
k².
+
2d0d20 2 (a de c cos. Od☺) d (ad0 — c cos. O d☺)
di
d t²
2
2 c. sin. Oded (c sin. d☺)
+
d t²
+ 2 gc sin. Od0=0.
Integrating,
ᏧᎾ
do
k²
+(a-c cos. 0)².
d ᎾᎿ
+c² sin.20
d ť
-
dť²
2
2 gc cos. 0C;
dt
or, (k²+a²+c² 2 ac cos. 0)
B being the value of ✪ when the velocity is 0.
If we make k² + (a − c)² = b²,
c)² = b², we have
do²
dt²
=2gc (cos. O cos. ß),
have
b² + 4 ac. sin.
a)
d02
dt²
= 4 gc (sin.²
В
sin.2
2).
If we suppose ẞ and small, and neglect 04, ß²0², &c. we
d02
gc
(B² — 0²),
2
dt² b2
whence it appears, as in Prob. III, that the length of the isochro-
nous pendulum is
b²
(a — c)² + k²
a² + k²
=c−2a+
C
C
C
If the curve be not a circle, the results will still be true for
small oscillations, if we take C the centre of curvature.
G may be at any distance from C, and hence it may be beyond
A, as when a body hangs by means of an axis passing through it,
and supported on a plane PQ, fig. 129.
PROB. VI. To find the correction due to the length of a pen-
dulum, for the thickness of its axis.
When it is requisite that a pendulum should oscillate very ac-
curately about a horizontal line, it has an axis as B in the pendulum
333
M, fig. 129, of which the section is triangular; and this, with the
edge O downwards, being placed with each end upon a hard plane,
the pendulum will turn round the axis O.
But, if the axis have a curvilinear section as CP, in the pen-
dulum N, fig. 129; or, if the edge be blunt, the pendulum no
longer oscillates about a mathematical line. In this case it is
required to find from what point the pendulum must be suspended,
that it may oscillate in the same time.
Let the body oscillate about a fixed point S, C being the centre
of curvature, and CS being 8, CG = c, and CP = a. Then, the
length of the pendulum
= SG +
k²
SG
k2
c-d+
C
+8=3
12
k² 8
+
2
C
C
omitting powers of 8, because CP, and therefore CS is small.
Hence, that the time of this may be the same as the time of
the pendulum with the axis CP, we must have, by last Problem,
c-2a+
Hence, omitting
C
1.2 k² S
a² + k²
= C
c - d + +
C
C
which is small, we have
c²
2 ac²
c² - k²
If c >k, the point S will lie towards G, if c< k, it will be be-
yond C.
CS is always greater than CP.
PROB. VII. The pendulum having two axes which are isochro-
nous to each other, (as in Art. 91.), supposing them cylindrical, to
find the corresponding length of the simple pendulum.
In Captain Kater's experiments, mentioned Art. 91, the pen-
dulum was supposed to turn about a mathematical line; if this
supposition be not true, we shall find the consequence of the
alteration.
Let c, c', be the distances of the centres of curvature of the two
blunt axes from the centre of gravity; a, a' their radii. Then, since
334
the pendulum oscillates in the same time about both, the length of
the isochronous pendulum must be the same for both.
by Prob. V,
(c' — a')² + k²
(c− a)²+k²
; ..
k²
C
C
(c — c') (c − a)² + (c— c′) k²
=c+c·
Hence,
c' (c — a)² — c (c' — a′)²
c-c'
(c — a)² — (c' — a')²
(c-c') c
(c-c')
2
a² - a¹²
12
+
C - c
2 (ca-c'a')
c-c'
If we suppose a and a to be equal, we have
l=c+c-2a,
the exact distance between the surfaces of the two axes.
;
Hence, the method of finding the length of the pendulum is
equally correct, whether the edges be sharp or not.
If S, S', be the points from which the body would oscillate in
the same time as about the surfaces P, P', we have SS' = PP'.
SECT. II. When the Body moves in any manner whatever.
132. PROP. The body being acted upon by any forces, the
motion of the centre of gravity will be the same as if all those forces
acted at the centre.
If x', y, z, be the co-ordinates of a particle m; X, Y, Z, the
forces on it; we shall have, as in last Section,
md²x' md²y'
effective forces, Σ
Σ
Σ
d t²
2
d t2
,
md z
dt2
Q
•
impressed forces, Σm X, ΣmY, ΣmZ.
And as before, if x, y, z, be the co-ordinates of the centre of
gravity, x'=x+x, &c. we have
m d² x
Σ
= Σ
dt2
m d² x
dt2
d²x
= M.
&c.;
dt²
29
d² x
Zm X
dy
Σm Y
d² z
Σm Z
;
d t2
M
dt2
M
d t²
2
M
whence the proposition is true.
335
133. PROP. The body being acted on by any forces, the motion
of rotation will be affected as if the centre of gravity were fixed,
and the same forces were applied.
This might be proved as in the corresponding proposition of
last Section; but it appears also thus.
Let the centre of gravity have a velocity V, and let the effect
produced on it by the forces, be the same as if a single force P
acted there. Let there be communicated to each point of the
system a velocity equal and opposite to V. Then, the centre of
gravity will be at rest; and the forces which communicated these
velocities will not affect the rotation about that centre, because
their resultant will pass through that point*. Let there also be
communicated to each point of the system a force equal and op-
posite to P. Then, the centre of gravity will have no tendency to
move, and for the same reason as before, the rotation about that
centre will not be affected.
Hence, if we suppose the centre of gravity to be fixed, and the
same forces as before to act upon the body; the effect on the
motion of rotation will not be altered.
COR. 1.
If the system be at rest when a force acts upon it,
the instantaneous axis, about which it begins to revolve, must be
perpendicular to the plane in which the line of force and the centre
of gravity are. For otherwise the effective forces could not be
equivalent to the force impressed, since the former would all be
in planes parallel to each other, and oblique to the latter.
COR. 2. In this case, the velocity communicated in the first
instant is the same as if this instantaneous axis were fixed, and may
be found by Art. 74.
COR. 3. To find the circumstances of the motion of a body of
any form, acted on by any forces, we must find the motion of the
* If equal and parallel forces be communicated to each point of a system,
their resultant will pass through the centre of gravity; for the force of
gravity acts in this manner, and the centre of gravity is the point through
which the resultant passes in that case.
336
centre of gravity by the formulæ of Book I, and determine the
motion of rotation by equations (0), Art. 124.
PROB. I. A solid of revolution terminated by a point, (PAB,
fig. 130.) moves so that the point (P) is always upon a given hori-
zontal plane; to determine its motion.
Let PO be the axis of revolution; G the centre of gravity;
Gz₁ in the direction of PO, and Gx₁, Gy₁, perpendicular to it,
moveable rectangular co-ordinates. And let Cx, Cy in the ho-
rizontal plane, and Cz perpendicular to it, be fixed rectangular
co-ordinates.
The forces which act on the body, are gravity (g), downwards
at G, and the reaction of the plane (R), upwards at the point P.
Let a", and ", as in Art. 119, be the cosines of the angles which
the vertical line PK, or Cz, makes with Gr₁ and Gy₁; then, the
components of R at P, parallel respectively to Gr₁ and Gy₁, will
be Ra" and Rb"; and if GP-1, the moments of this force, with
respect to the axes Gy, and Ga₁ will be - Rld" and R1b" re-
spectively. The moment, with respect to Gz₁, will manifestly
be 0.
If c" be the angle which PK or Cz makes with Pz,, as in
Art. 119, we have GH=lc"; and for the motion of the centre of
gravity, M being the mass,
d².lc" R
d t² 2
M
g;
d²c"
or R= Mg + MI
d t²
. (1).
And by equations (0) for the motion of rotation, observing that
B=A,
Cdr 0.....
Adq+(AC) rpdt - Rla"dt.
A dp — (A – C) qrdt=Rlb″dt.
From (2) we have r=0, a constant quantity.
..(2),
.(3),
..(4).
337
Also, taking (3) b" + (4) a", we have
A {a"dp+(b″r−c″g) pdt+b"dq+(c″p− a″r) qdt}
+C (a"q-b"p) rdt=0.
Whence, reducing as in Note, p. 134, and integrating, 4
A (a″p+b″q)+Crc"=k .......(5).
Again, (3) q +(4) p gives
A (pdp+qdq)=Rl (pb″ — qa″) dt.
Or, putting - dc" for (pb″ — qa″) dt, and for R its value from (1),
A (pd p+qdq)+Ml²·
dc"de"
dt²
+Mlgdc" = 0.
Multiply by 2, and integrate;
2
2
A (p²+q²) + M i (dc" )² + 2 Mlge"=h. . . . . (6).
l
dt
Ꮎ Ꮎ
Now let be the angle GPK; then c" cos. 0, dc" = − sin. Ode.
And by Art. 125, we find
a"
b″ = —
sin. O sin. p, b'" sin. O cos. ;
and hence, by equations (2),
pa" +qb" = — sin.20
dy
d t
2
; p²+q² = sin.² 0
t
(d)² + (²)².
2
dt
Hence, equations (5) and (6) become
d&
A sin.2 0
+Cr cos. 0 = k
d t
ᏧᎾ
(A+ M² sin.20)
+ A sin.² 0
dx²
..(7).
+ 2 M gl cos. 0=h
d t
d t²
dy
Eliminating
we find
dt
(A sin.2 0+ AM12 sin.4 0)
de²
dt2
= A sin.² 0 (h—2 Mgl cos. 0)
− (k − Cr cos. 0)³………….
O)²..
...(8).
U U
5
4
4
338
If we deduce from this the value of dt, we shall find dt
=F0.d0; F0 being a function of 0; and by integrating this,
we have t in terms of 0, and in terms of t. And hence, by (7),
we have in terms of t; and hence, by equations () we have .
These integrations cannot be performed in finite terms.
The quantities h and k are to be determined from the given
initial circumstances of the motion, by equations (7).
PROB. II. The body having at first no motion except a rotation
about its axis PO; to determine its motion afterwards.
Let the original velocity about PO=e, and the original incli-
nation0₁. And since at first.
dy
ᏧᎾ
0,
0, we have
dt
dt
аф
d t
r= =e; and by equations (7),
Ce cos. 0₁=k, 2 Mgl cos. 0,=h.
1
Hence, equation (8) becomes.
2
(A² sin.² + AM 1² sin.^0)
1
d 02
dt
2
=2 AM g l sin.² 0 (cos. 01- cos. 0) — C²² (cos. 01~ cos. 0)².
The expression on the right hand side consists of two factors;
cos. 01 cos. Ø, and (putting 1 cos.20 for sin.²
2 AM gl-C²e² cos. 01+ C²e² cos. 0-2 AM gl cos.² 0.
If we put
C²
4 AM g l
2
=m, and
2
Ꮎ
0),
a=me² - √(1-2 me² cos. 0, +m³e¹),
2
2
M E
B=me² + √(1-2 m e² cos. O₁ + m²e³).
the above expression becomes
1
2 AM gl (cos. ◊ — a) (ẞ — cos. Ø).
Hence, we shall have
d 02
d ť²
2 M g l (cos. §¸ ~ cos. 0) (cos. ◊ — a) (ẞ – cos. A)
sin.20 (A+Ml sin. 0)
It is easily seen that B is greater than cos. ₁, and that a is
339
less than 1; let a=cos. 02, and 02 and 01, will be the greatest and
least values of 0. The axis will oscillate perpetually between these
inclinations, and the rotation will continue for ever.
If PO be inclined beyond a certain limit, the sides of the
body POB will touch the horizontal plane, and the rotation can no
longer continue in the same manner. If we suppose that PO is
susceptible of all positions above a horizontal one, the rotation will
not be stopped if 0, be less than a right angle. That is, if
2
2
cos. 0, 0, or if me²- V(1-2 me² cos. 01+m²e¹) > 0,
>
1
2
if 2 m e² cos. 0¸ > 1; if
Ε >
or €² >
S
2
2 AM gl
2m cos. 0,
C² cos. 01
1
If e, the original velocity, be less than this, the body will fall.
=
If 0102, the axis will always retain the same inclination. This
supposes
cos. 0,=me - V(1-2 me² cos. 0₁+m²ε*),
1
1
and cannot be except e, the velocity of rotation, be infinite.
PROB. III. The body and the axis having velocities commu-
nicated to them; to determine the conditions that the axis PO may
always retain the same inclination.
In this case 0, and therefore c", is constant; and by equation
will be constant ;
(1) of Prob. I, R = Mg. Also, by (7),
dy
dt
let it = d.
Then, p= sin. O sin.
= - a″d;
dt
q
;
p
.. a" = −}; similarly, b″ — —
and equations (3), (4) become
Adq+(Ar−Cr+Mg¹) pdt=0,
Adp- (Ar-Cr+Mg¹) qdt=0.
340
Hence, we find as in p. 316, making
(A−C)rd+Mgl
A d
= €,
pa sin. (et + y), q = a cos. (et+y).
Substituting in the first and second of equations (1), and dividing,
we have
tan. (et+y)= tan. ; et + y = 8,
And by the third of equations (1), r = ε —
But r=
аф
dt
= €.
cos. 0.8.
Ade - Mgl
(AC) 8
б
Hence, Cde+(AC) cos. 0.82 = Mgl.
Hence, & and O being known, e is known.
If &
=0, we have e infinite, agreeing with last Problem.
The results in the preceding Problems are applicable to the
motion of a spinning top, considering it as upon a perfectly smooth
plane, and supported on a mathematical point. It has appeared,
that under these circumstances the top will, if a sufficient velocity
be communicated to it, go on revolving for ever; but, in consequence
of the absence of friction, the motion will be a good deal different
from that which we observe actually to take place in such cases.
The centre of gravity in our Problem either remains at rest, or
moves up and down in a vertical line, and cannot have any curvilinear
motion. The axis can never become more vertical than it was at
the beginning of the motion, though it will at intervals return to the
incl ination which it then had. But in the experiment, the top, if
inclined at first, will approach to a vertical position, which it will,
as near as the senses can judge, attain and preserve for some time;
and the centre of gravity will frequently describe a curve approach-
ing to a circle, while the foot of the instrument remains stationary.
These differences of theory and practice appear to be attributable
to the effects of friction*.
* Euler thus explains the effect of friction in causing a top to raise itself
into a vertical position. "The friction will perpetually retard the motion of
the
341
134. A Problem of great consequence, depending on the prin-
ciples of this Chapter, is that of the Precession of the Equinoxes,
or motion of the nodes of the Earth's equator on the ecliptic, and
similar motions in the heavenly bodies. This motion in the Earth
arises from the attraction of the Sun on the Earth, which, in con-
sequence of the spheroidal form of the latter, produces a force
tending to turn its poles, one towards, and one from, the Sun. And
this combined with the Earth's rotation, produces a motion like
that described in p. 319. If, in fig. 124, xy be the fixed plane
of the ecliptic, x, y, the Earth's equator, and z, its north pole,
N, the intersection of xy and x₁₁, moves along xz in a direction
opposite to the diurnal rotation. The investigation of this subject.
belongs to Physical Astronomy; but, in order to shew the nature of
the action exercised, we shall take the inverse Problem, the motion
being given to find the forces.
1
PROB. IV. A figure of revolution, turning uniformly on its
aris, retains the same inclination, while the nodes of its equator
move uniformly on a fixed plane; to find the forces by which it is
acted on.
The notation remaining as in Art. 133, fig. 124, let
аф
d
dt
dt
=e, the velocity of rotation;
8, the velocity of the node; constant.
¿
If this happen
the point P of the instrument, and at last reduce it to rest.
before the top fall, it must then be spinning in such a position, that the
point can remain stationary. But this cannot be if it be inclined. Hence, it
must have a tendency to erect itself into a vertical position." Theor. Mot.
Corp. Solid et Rigid, Suppl. Cap. 4.
This property has been applied to obtain an artificial horizon. Since the
axis tends to become vertical, a plane perpendicular to the axis tends to
become horizontal; and for a considerable length of time may be considered
as accurately so. See Phil Trans. 1752.
Mr. Landen's solution of this Problem appears to contain several
mistakes. See bis Math. Lucubr.
342
Hence, by equations (1),
p=8 sin. sin. 0, q=8 cos: sin. 0, re- cos. 0,
and B = A.
Hence, since d, e are constant, we have, by equations (0),
N =0,
N'Ade sin. o sin. 0
+(A–C) de sin. & sin. 0-(A — C) S² sin. & sin. ◊ cos.
{Ce+(A–C) & cos. } 8 sin. O sin. p,
N" Ade cos. o sin. 0
=
+(C− A) de cos. Ø sin. 0 − (C – A) d² cos. O sin. O cos. ◊
= {Ce+(A − C) & cos. } & sin. O cos. p.
Hence, N'-F sin. 4, N" F cos. ; F being constant.
=
These forces N', N" will be supplied, if we suppose a force F
to act any where in the circle z Z1, fig. 124, urging z₁ towards z.
APPENDIX.
APPENDIX (A) to the INTRODUCTION, p. 4.
On the Definitions and Principles.
ds
dt
and f=
d v
dt
may be considered as the
THE equations v =
mathematical definitions of velocity and force. They express that
the velocity is the limit of the ratio of the increment of the space,
to the time in which it is described; and the force, the limit of
the ratio of the increment of the velocity, to the time in which it is
described. And though these definitions are perhaps not the
simplest descriptions of the vague and popular meanings of the
words velocity and force; they may be shewn to agree with those
significations as far as they go; and to be the limitations to which
we are naturally led in making those notions exact and measurable.
The quantities are greater or less according to our definitions, when
they are so according to the common ideas; and the definitions
are capable of being applied to any portions of time however small,
which is requisite for the purpose of considering velocities and
forces perpetually variable.
In fact, we may look upon space and time as the two variables,
whose relations we have to investigate, and consider the general
Problem of Dynamics to be this, "to find the place of a body at
the end of a given time." The space being thus a function of the
time t, it becomes convenient to give a name to the first differential
ds
d
coefficient ;
ds
dt
dt
and to the second,
; we call the first,
dt
velocity, and the second, force.
344
We shall here briefly state the proofs of the laws of motion.
LAW 1. A body in motion, not acted on by any force, will
move on in a straight line, and with a uniform velocity.
First, it will move in a straight line. For if it do not, it must
move in some curve, and it must depend upon external circum-
stances, towards which side the convexity must lie, and how great
the curvature must be. But, when a body's motion is influenced
by external bodies, those bodies are said to exert force upon it,
which is contrary to the supposition. Hence, a body influenced
by no force, cannot describe any path but a straight line.
Next, it will move with a uniform velocity. As we remove the
known causes which retard a body's motion, we find that we
remove the retardation, and this without limit; so that it is evident,
that if we could entirely remove the external causes of retardation,
the body would not be retarded at all; there is no internal principle
which tends to diminish the velocity.
The common causes by which motions are retarded, and finally
stopped, are friction and the resistance of the air. If a wheel turn
on a very smooth axle, it will revolve for a long time; and the
longer, as we remove more of the friction by making the axle
smoother; and if we also diminish the resistance of the air by
making the wheel revolve in an exhausted receiver, the motion will
continue still longer. We can never quite remove the friction or
the resistance; and it is on that account, that the rotation cannot
be made to continue for ever without diminution.
LAW 2. When any force acts upon a body in motion, the
change of motion which it produces is the same, in magnitude and
direction, as the effect of the force upon a body at rest.
Both the original motion, and the change of motion com-
municated, are retained in their own directions. Thus, in fig. 7, if
the body be in motion with a velocity which would carry it through
PR, and be acted on by a force which would carry it through Pp
in the same time, it will at the end of that time be found at the
point r, PRrp being a parallelograin.
The proofs of this law are of the following nature.
345
A body let fall from the top of the mast of a vessel in motion,
will fall down the mast, (if vertical ;) thus retaining the horizontal
motion of the ship, as well as the motion communicated by gravity.
A body thrown across the deck by a person on board, will in the
same manner proceed in the direction in which it is thrown relatively
to the vessel; thus both retaining the motion of the vessel, and obey-
ing the force by which it is projected.
The motions impressed on bodies by the same agent, are the
same, whatever be their direction with respect to the direction of
the Earth's motion. Thus a pendulum oscillates in the same
time east and west, or north and south.
The motions impressed on bodies in different parts of the
Earth, are the same, relatively to the Earth, if the forces be the
same; thus shewing, that besides the motions impressed, they
retain the motions of the parts of the Earth where they are, which
vary infinitely in velocity.
LAW 3. When pressure communicates motion, the moving
force is as the pressure.
This is proved from experiment. The pressure is the weight
which produces motion, and the moving force is measured by the
momentum generated in a given time. Thus, in fig. 107, when
two bodies P, Q are suspended over a pully, if P be the heavier,
P-Q is the mass whose weight is the pressure producing motion;
and if we neglect the pully, P+Q is the mass moved; and this,
multiplied into the velocity generated in a given time, is proportional
to P-Q. This is found to agree with experiments. Atwood's
machine, fig. 101, is the one with which the greatest number of
experiments were made. In this, the mass of the wheels over
which the string passes must be allowed for. See Art. 95, and
Atwood on Rect. and Rotatory Motion, Sect. 7. Also Mr. Smeaton's
Experiments, Phil. Trans. Vol. LXVI.
Action and Reaction signify the mutual pressures of two bodies
which influence each other's motions. Action and Reaction are
sometimes defined to be the momenta gained and lost; and in that
case, in order to prove the equality of action and reaction, it is
necessary to shew, that these momenta are as the pressures which
produce them.
This third law of motion is also proved by shewing, that in the
Xx
346
case of impact, the momenta gained and lost by the mutual collision
are equal. Newton, Scholium to the Laws of Motion, Principia,
Book I. Impact is in some respects the simplest case of pressure,
because in it the consideration of time does not enter.
But pro-
perly speaking, we cannot consider the proof of the third law of
motion in this case, as sufficient to establish it in all others.
The Inertia or resistance of different bodies to motion is, at
the same place, proportional to the weight, but does not vary, as
the weight does, according to the different action of gravity, &c.
The Inertia of Rotation, or resistance to the communication of
rotatory motion, does not depend on the mass only, but also on its
distance from the axis, as is seen in Book III.
The third Book supposes' the general property of the lever to
be established; namely, that the effects of forces to turn a body
round an axis, will be the same, when the sum of their moments, or
products by their perpendiculars from the axis, is the same. See
Statics, Art. 26.
It supposes also the composition of forces to
be demonstrated; which is, however, included in the general pro-
perty of the lever.
APPENDIX (B) to CHAP. III. Book I.
On the Motion of a Body about two Centres of Force.
THIS Problem is remarkable both for the elegance of the re-
sults to which our investigations lead us, and for being almost the
only addition which has been made since Newton's time to the
exact solution of inverse problems of central forces.
This acqui-
sition we owe in the first place to Euler, who in the Transactions
of the Academy of Petersburg, (Nov. Com. Petrop. 1764,) pub-
lished about 1766, examined the question of the motion of a body
acted on by two centres of force, when it is supposed to move always
in one plane. His analysis of this case was complete, but he informs
us, that in his first attempts he was led into a mistake by the method
which he employed. This he detected, by the absurdities which
resulted from supposing one of the forces to vanish, which of course
347
reduced it to a known problem; and in investigating the origin of
the inconsistencies in his solution, he was led to the complete so-
lution. So that he attributes his success to this most fortunate
error," as he calls it.
Nearly at the same time he published in the
Berlin Memoirs, (Hist. de l'Acad. Royale des Sciences, Berlin 1760,
published 1767;) an examination of the cases in which the curve
described is an algebraical curve.
At the end of his first paper, Euler had promised a solution of
the problem, when the motion is not in the same plane. This he
performs in the Nov. Com. Petrop. for 1765, (published 1767,):
where he gives a new method of obtaining the differential equations:
and it is this which, with some modifications, is adopted in the next.
Before the publication of this last paper of Euler, Lagrange /
had taken up the problem, and had written a Memoir, in which,
with very great elegance and simplicity, he solves it for the motion
in a curve of double curvature. This Memoir appears in the Me-
langes de la Soc. de Turin for 1766—1769, (published some years
afterwards). In the same Volume there is also another Memoir of
Lagrange, containing a discussion of the laws of force for which
the integration is possible; and an observation founded upon this,
that it cannot be applied to the motion of the Moon, which the
Student will find explained at the end of the problem as here given.
Legendre has more recently examined the cases of this Problem, in
his Exercices du Calc. Integ. (tom. II, p. 366.).
The body will not necessarily perform its motion always in the
same plane, but will in general describe a curve of double curvature.
It is manifest, however, that if we suppose the direction to be origi-
nally in the plane passing through the two centres, there will be
nothing to deflect the body from this plane, and hence it will con-
tinue in it during its whole motion. We shall first, therefore, con-
sider this, the simple problem.
PROB. I. A body, acted upon by forces tending to two fixed
centres, and varying inversely as the square of the distances from
their respective centres; moving so as to be always in the same plane ;
to determine its motion, fig. 131.
Let the motion be in the plane, passing through A, B, the
centres of force. Let P be any point, AP=1, BP=r'; the
348
and exert upon P,
forces which A and
AB=2c; PM perpendicular to AB; and
MP=y; hence, x+x=2 c. Also,
r² = x² + y²; r¹² = x²² + y².
m
m
m'
29
1일
​respectively.
r
AM= x, BM=x';
Now, if we resolve the force, in direction PA, into two
forces parallel respectively to y, and to x, as in Art. 12, we shall
have the resolved parts
m x m
r ༡༢
m' x' m'
of B's force will be
12
r
T
r
y
ཏྭཱ '༤
: similarly, the resolved parts
Hence, observing the di-
rections of the forces, we have
mx
force in r =
3
J'
+
m'x'
;
13
force in y
my
my
13
3
And hence, the equations of motion (c), Art. 12, are
d² x
dt².
Sm
X
m' x
'x'
3
3
ጥ
ď² y
(my
my
+
d to
13
r
..(1),
. . (2).
We shall indicate these equations by the figures which stand
opposite to them, and operations performed upon these equations
by the usual algebraical symbols; a mode of notation which will
be easily understood.
Thus, (1) dr + (2) × dy, gives
dx d² x + dy d² y
d t²
(mrdr
m'r'dr
+
3
73
the right hand side being reduced by observing, that
xdx + y dy = r dr; -x'dx+ydy = x'd x' + y dy = r'dr'.
349
Now this equation integrated and multiplied by 2, gives
dx²+dy²
2
d t²
m
m'
=2 +
+....
p
C
(3).
Here C is a constant quantity which depends upon the velocity at
a given point; and the right hand side of the equation gives the
velocity at any point.
We shall now make another integrable combination of equations
(1) and (2); observing that dr = d²x',
-
(2) × x' + (1) × y gives
Y-
x'd²y — y d'x'′
d t²
2mcy
since +2c;
3
(2) x x -(1)xy gives
xd²y - y d²x
2 m'cy
13
dts
Multiplying by rdy-ydx, and by x'dy-ydx', respectively, we
have
{x'd²y - yd²x'} {xdy−ydx}
d t²
2 m cy (xdy − y dx)
-
3
r
{xd'y — yd°z} {x'dy — ydz' } _ _ Q m'cy (x'dy - ydz')
d t2
13
r
Add these equations together, and the sum becomes integrable;
for the numerator of the first side will be the differential of
{x'dy—ydx'} {xdy-ydx}. Also
r dx − x dr
d.
T
2
y' d x - x y dy
3
r2dx-xrdr
3
(x²
dx
(r° +y) dr - r (rdr+ydy)
3
; whence one of the terms on the second side.
X
Similarly, d. — = 9'da'z'ydy
13
350
Hence, the integral of the sum just mentioned is
{x'dy-ydx'} {xdy—ydx}
X
Ꮖ
=2mc.
: 2 mc. - +2 m² c
dt²
, + 2 Dc....(4),
ጥ
where D is a constant quantity, which depends upon the direction
and velocity of the motion at a given point. If we eliminate t, we
have the equation to the curve.
We shall transform equations (3), and (4), by the following as-
sumptions.
Let r=c(u+v), r' = c (u — v).
Now, by the triangle APM,
2
4 c² + r²-r²²
4 c² + 4 c² uv
x=
;
4 c
.4 c
x'
x = c(1+uv); hence also, r = c (1 − uv) ;
and y² = r² — x² = c² (u² + v² — 1 — u² v²) = c² (u² — 1) ( 1 − v');
therefore, dx = c(udv+vdu), dx' =
2
c(udv+vdu);
y = c√ (u² — 1) V (1 − v°), dy = cudu√)=0²
Hence, we obtain
2
2
2
น
u² - 1
c v d v √ / u² – 1
1-72
dx² = c² {v² du³+2 uvdu dv+u°dv²},
dy²
2
.. dx² + dy² = c²
{u'aw.
{du²
Also, xdy-ydx
z2
U
2
2
2
-Luvdudv+v²d v²
u²
V
+ dv²
- 1
1 — 72
૧
ገ
2
1
บ
=
c² {udu (1 + uv) √
-
U
S
2 1
U
vdv (1+uv) √ √ ¹² — !
2
1-
V
— c² { vd u V (u² — 1 ) √′ ( 1 − v²) + u dv √ (u² − 1) √ (1 − v²)}
{du
- v²
2
du (u + v)
dv (u + v)
2
2
351
Similarly, x'dy—y dx'
*
น
= c° {du (u−v) √ 13—2)² + dv (x − v) √² = }} ;
hence, (xdy — yd x) (x'dy — y dx')
4 2
1
1
= c* (x² – v®) {du° != ** – dv² ²²-}}.
u — 1
1-c
2
2
Substituting these values in equations (3) and (4) they become,
(dividing by c² and c4),
2
(u² — v²) { du²
2
=
m
{ "
1
dvⓇ 1
+
2
·
dť² u² - 1 dt 1-v
u + v
+
U
m
"
(u² — v²) (du² 1 — v²
{"
·
dtu
d ť u² - 1
1 + uv
+uv
u + v
V
+ c)}
+ c}.
dv² u²
dt
(5),
1
1 - u v
+
- m²'
+ D}.. (6
IL ー​で
​(6).
Now to eliminate dɩ, we take (5) × (u³ — 1)+(6); which gives
(u² - v²)² du² ი
u² 1 dť
~~ {(m + m') u + C (u² — 1) + D}. . . . .(7).
୯୯
Similarly, to eliminate du, (5) × (1 − v²) — (6) gives
(u² — v²)° dv²
น
212.dt²
2
{ − (m − m') v + C (1 − v²) — D }. . (8).
If we now divide (7) by (8), we obtain
1- v² duⓇ
---
(m+m') u + C (u² − 1) + D
u²-1'd v² (m − m') v + C (1 − v°) — D'
If we make D− C = E, we have
du √ (u² − 1) { E + (m + m') u+Cu²}
dv
(v² − 1) { E + (m − m') v+Cv²}
.(9),
which gives us immediately an equation in which the variables are
separated; viz.
}
352
du
√ {(u² — 1) [E+(m+m') u+Cu²]}
dv
√ {(v² − 1) [E ·+ (m − m')v + Cv³]}
. (10).
If we could integrate equation (10), we should have the equation
to the curve described. The variable quantities are separated in
this equation, but the integrals are what are called Elliptic Tran-
scendants, and cannot be obtained in finite terms.
There are,
however, an infinite number of cases, in which (10) belongs to a
remarkable class of equations, to which the researches of Euler and
others have shewn how to find algebraical integrals; namely, when
the quantities C, E, m, m' have such relations, that, making u=ap,
v=by, it can be reduced to the form
μαφ
2
3
4
√(a+ßp+y&²+ dp³ +€¢¹)
v d¥
2
√(a+BY+j¥²+d¥³+€¥³)
µ, v, being any whole numbers whatever. See Lacroix, Traité du
Calc. Diff. et du Calc. Integ. tom. II, No. 702.
But the expressions thus obtained are of so complicated a
form, that we shall not attempt to examine the solutions which
they offer.
There are two remarkable cases, in which we can more simply
find the curve described. If we have u = a, a constant quantity,
such that
(a² − 1) { E+ (m +m') a + Ca² } = 0,
it is manifest, that both sides of equation (9) vanish, and hence,
a, satisfies the equation. Similarly, if we invert the equation,
and suppose
U
(B² − 1) { E + (m − m') B + CB² } = 0,
−
it is clear, that v=ß will satisfy it.
v=
But though u= a, or vẞ satisfy equation (9), it does not
necessarily follow, that they give the curve described in a particular
They will not do this, except they be integrals derived from
case.
353
the complete integral by giving particular values to the constants.
If, instead of being such particular integrals, they be particular
solutions, (Lacroix, Art. 293.) of the differential equation, they no
longer solve the problem.
is a
Now we have this rule, by which we can determine whether
an equation which satisfies a given differential equation in u, v,
particular solution or a particular integral. (Lacroix, Art. 297,
298). Put
du
dv
dp
P, and take the value of
; every particular
du
dp
solution will make
infinite.
du
In this case, if we call VU and VV the numerator and deno-
minator of the fraction in (10), we have
du
p =
VU
ᏤᏤ
dv
and if U', V' represent the differential coefficients of U and V;
dp
U'
V' U dv
du
2 VU. V V
2 V VV' du
U'
V'
dv
2 V U V V
2 putting for
its value;
du
U' V V — V' VU
2 V VU
which will be infinite, when U=0, except U'0 at the same time,
and when √ = 0, except V'=0 at the same time.
Hence, if U=0, and U'≈0, we have not a particular solution,
but a particular integral, and consequently a solution of the
problem. Similarly, if V=0, and V'=0.
Let us consider the equations u=a, U=0, U'=0, or
u = a,
=0]
(u² — 1) { E+ (m+m) u+ Cu³} =
Qu {E+(m+m')u+Cu²}+(u² − 1) { (m+m') + 2 Cu} = 0}
Y Y
(11).
354
Now the first equation u = a, whence r+r'2 cu=2ca, shews
that the curve is an ellipse: the other equations determine the
values of C and E.
The second equation gives either E+(m+m) u+Cu²=0, or
u²-1=0; but it is clear, that the latter cannot be an answer to
the problem, since it gives y=0 for every point. Hence, the
problem requires that
E+(m+m') a + Ca²=0,
(m+m)+2Ca=0.
From the latter,
m+m'
C =
2 a
(m +m')
.:. E =
a, by the former.
2
m+ m² a² + 1
2
α
And therefore DE+ C =
If we call 2a the major axis of the ellipse, we have
2a=r+r'=2ca; .. a =
a
1
e being the eccentricity.
C
e
Hence, C=
m+m
C
;
2
a
and substituting in equation (3), we have
Im
m'
m + m²
velocity = 2
+
"'
2a
.(12).
at the extremity D of the axis minor, rra; and here
velocity²==
m+m'
a
By Prob. II. of Chap. III. it appears, that this is the velocity
which the body would have in the ellipse at D, if there were at A
or at Ba single force equal to the sum of the forces at A and at B.
Hence, conversely, if the body at the point D, equidistant from A
and B, be projected parallel to AB with this velocity, it will
describe an ellipse about the centres of force A, B, as foci.
355
If the body were projected in any other manner, if we know
the velocity, we find a by equation (12); and hence, successively,
e, a, C, D, and E; and knowing C, there we can find the direction
of projection by equations (6), (7), or (8).
Similarly, if we make v=ß, ß being subject to equations
analogous to a in (11), we shall have
r — r = 2CB = 2a suppose;
therefore the curve is a hyperbola, B being its interior focus.
C 1
a β
The eccentricity e ==
Also, E+(m-m') ß + Cß² = 0,
(m − m') + 2CB=0;
m
m'
.. C =
23
(m
(m − m')
2
C
; E=
a
(m-m') a
2
m'
m— m'
+
ア
​2 a
velocity² = 2
If we had originally put
x = c (u —v) r' = c (u + v),
the expressions would have been every where the same, excepting
that we should have -v, for v at each step. Hence, we should
have in this case
V =B,
E − (m − m') ẞ+cß² = 0,
− (m − m²) + 2 Cß = 0;
therefore r'—r=uß=2a,
and the curve is a hyperbola, of which A is the interior focus.
m · m'
m - m'
C
C=
23
2
α
E=
(m— m') ß
m
2
velocity² = 2 +
m'
; D=E+C=
+
m
m
M
m' B²+1
2
β
2 a
356
In order to complete the determination of the motion, we
ought to have the time. Now this is easily obtained from equations
(7) and (8). Retaining the notation of p. 353, we have by the
equations just mentioned,
(u² — v²) dv
ᏤᏤ
V2dt
u² dv
v² dv
u² du
v² dv
√2dt
VV
ᏤᏤ
VU
VV
which reduces it to the integration of expressions of one variable.
In the particular case which we have considered, (when u=a)
equation (8) gives
dt V2
C
(a² — v²) dv
v´ {(v² − 1) [E + (m− m′) v + Cv²] }
(a² — v²) dv V(2 a)
√ {(1 − v²) [(m+m′) a² + 2 (m −m') av + (m+m') v²]} '
PROB. II. When the body does not move in the same plane; to
determine the motion, fig. 123.
Let the path be referred to three rectangular co-ordinates x, y, z;
the plane of xy passing through AB, so that AM=x, MN=y,
NP=z; the plane PMN will be perpendicular to AM. Let also
AB=2c, BM=x', AP=r, BP=r'. Hence,
x+x′ = 2c, r² = x²+y²+z², r²² = x²² + y²+z²,
m
m'
and we shall then have the forces and in the directions PA
22
and PB; and the resolved parts of the
NM, and to PN respectively, will be
T
2/2
12
first parallel to MA, to
m Ꮖ m y m ช
2
2
r
2 ༡༧༧
2 ;
2'
similarly of the other force; and hence, the equations of motion
become
d2r
dt2
Ꮖ
mx
m'x'
3
13
……..(1),
357
dy={my + my}.
dt2
ď z
dt²
Sm
3
m'z'
..... (2),
mz + }....... .. (3).
3
Т
13
Then (1) 2dx+(2) . 2 dy + (3) 2dz gives, (observing that
rdr=rdx+ydy+zd,
r'dr' = x'dx' + ydy + zdz=-x'dx+ydy+zdz),
2dx d²x + 2 dy d²x+2dz d²z
Ꮖ
and integrating,
2
dt²
m
m'
dr²+dy' + d² = 2 { " + = +
dt2
Also, (3) y —(2) z gives
(mdrm'dr
+
2
2
12
; Ca constant quantity.. (4).
y d°z - zd³y
dt2
=0;
y dz―zdy
dt
=h, a constant quantity..(5).
It is necessary to get another integrable combination of our
equations; which we may thus perform; (2) x (1) y gives
·
d (xdy — y dx)
dt²
m'y (x+x')
13
"
2m'cy
13
ጥ
(2) x'+(1) y gives, since — d'x'= d°x,
-
d²
d (x'dy — y dx')
dt²
my (x+x')
203
2mcy
2,3
Multiplying these by (x'dy-y'dx), and (x dy - ydx) respectively,
and adding, we have
d. (xdy — y d x) (x'dy—y dx')
dt²
—
2m'cy (x'dy-ydx') 2mcy (rdy — yda)
13
I'
3
358
Similarly, by using (3) for (2), and z for y, we shall get
d. (cdĩ –zdr) (dĩ − d )
(xdz
d t²
C
2/3
Ꮖ
2 mcz (dz – zdr)
3
d
and
d
Now add these, observing that
x
ጥ
Ꮖ
2
rdx-x dr ( tỷ tổ do (d+dy+z)
and we have
2
y (dyyda) z (rdz – zdr)
23
3
y (x'dy — y dx') — z (x'dz — 2 dx′)
13
グ
​d. (ady − y dc) (cdy gyd)+d.(xdz – zdx) (xdz–zd)
X
d t²
=2cmd+2cm'd
Hence, integrating,
(xảy − y dư) (dyyd)+(rdz− z d) ( dĩ - zac)
d t²
X
x'
=2cm
-
+2cm²
+ 2c D. . . . . .(6).
ጥ
r
Equations (4), (5), and (6), contain the complete determination
of the motion. We shall transform them by the following sub-
stitutions.
Let MP=p, and the angle NMP=0: hence, as in Arts. 16,
17, we have
y²+z² = p²; dy²+dz² = dp²+p³dø², ydz-zdy = p²dp.
So that the equations (4) and (5) become, removing the deno-
minators,
m m'
dx²+dp²+p²dp² = 2 d t²
{ + m² + ...
. . (7),
359
p❜dq=hdt..
To transform equation (6), we may observe that
(xdy—ydx)(x'dy—ydx')
..(8).
xx'dy² — xdx'.ydy-x'dx.ydy + y² dxdx',
(xd
Z
zx)(c dĩ – zd)
=xx'dz²-x dx'.zdz-xdx. zdz+z'dxdx'.
And the sum
xx′ (dp²+p³dµ³) — (xdx' + x'dx) pdp + p³dx dx'
=(xdp-pdx) (x'dp-pdx') + xx'p² do².
2
Substituting in equation (6), and multiplying by dť²,
(x dp – pd x) (x'd p— pdx')+xx'p'dp²
Sm x
m'x'
=2cdt²
+
r
+ D}.
.(9).
We shall now make the same substitution as at p. 350, in the
former case, viz.,
r = c(u+v),
r'=c (u-v).
And we shall have the same values of x, r', &c., as we then had,
p corresponding to y. Hence,
du²
dvⓇ
dx²+dp²=c² (u'—v²) { _du" + due},
2
1
-
2
v²
u².
2
2
— du²
(xd p = pd.x) (x'dp — pdx') = c² (w² — v°) {du? != "" — dv° ""— }}
1
Substituting these in (7), and (9), observing also that
2
x x′ = c² (1 — u² v²);
and, dividing the latter by c²,
1-v
we shall have results analogous to
(5) and (6), of the former case, viz.,
c² (u² —v²) {
2 dt²
C
m
duⓇ
dv²
2
+
2 1
V
-
u + v
+
U
1
m'
2
} +p²dø²
+c}....
+c}……….(10),
360
c* (u° — v²) { dr
1-ve
du².
u²
dv².
+ (1 − u³v²) p³dq²
2
2
u² — ]
m' (1 — uv)
+
U-v
+D}...
· D} … … … … … … (11).
2dt² (m(1+uv)
C
u + v
Now, (10) (u² -1)+(11) gives
c² (u² — v²). du².
2 dt
C
Put for dť² its
that (u' — 1) (1 —
v²)
2
༡z? — j?
2
u²
2
+u² (1—v²) p²dp²
{(m+m') u+C (u² − 1)+D}.
value from (8); multiply by (u²-1), observing
2
p²
C
27
; and transpose; and we have
c² (u² — v²)² du² =
2 p¹d þ² (u² — 1) {(m+m') u + C (u² − 1) + D}
ch²
2
Similarly, (10) (1 —v²)
2.4
И
p¹dp². (12).
афг
— (11), gives
u² - v²
• ༧
c² (u² — v²) dv²
2
+ v² (u² − 1) p² dø²
2
ρ
C
2dt
C
{ — (m ~ m') v + C (1 ~ v³) − D}.
And, reducing this in the same way as the other,
c² (u² — v²)² dv²
2 p¹dp2
4
ch²
(1 — v²) { — (m — m'′) v + C ( 1 − v²) —– D}
v² p*dq²
c²
2
· (13).
If we now divide (12) by (13), we have
h²
2 c
du²
2
d v²
2 c
2222
h²
(u² — 1) {(m+m') u + C (u² − 1) + D } − u²
(1 − v²) { — (m — m'′) v.+ C.( 1 — v²). — D } — v²
361
If we call =n, and D-C=E, we have
h²
du²
d v
2
2 n (u² − 1) { E + (m+m') u + Cu² } — u°
2
2 n (v² − 1) { E + (m − m' ) v + C v² } - v³• • • . (14)
suppose;
-
U
V
dv
by integrating which equation, we have the
and
du
VU
curve described: the variables separated, but the expressions tran-
scendant, as before.
Knowing the relation between u and v, we have, from equation
(12) or (13), the value of do;
for c² (u² — v²)² du²:
4
P Udo²
2
=c° .( − 1) (1 −3)° Ud¢°;
.. do=
аф
(
2
2
−
(u² - v²) du
-
1) (1 − v³) VU
du
(u² - 1) VU
2
d u
+
d u
(1 − v³) VŪ
dv
(1 − v³) √ √ ³
+
√
(u² - 1) VU
and for the time we have, since h=
2
− v²) v
N
—
= p² d = c + √ n { ( 1 − v } d® + (u² = 1) du}
dt =
VV
N
h
3 1914
= c² √ n
12 du
v² d v
NU
√ V
value of u, as to make
But it
It is manifest, as in the former case, that the equation (14) is
satisfied by supposing u = a, where a is such a
the numerator on the right hand side vanish.
as before, that a solution to the problem is
except also U' = 0; that is, we have the path
Z z
But it may be shewn
not given by U=0,
in a particular case, if
362
นะC
2 n (a² − 1 ) { E+(m+m') a + Ca² } − a²=0
2na {E+(m+m') a+Ca³}
+ n (a² − 1) { (m+m')+2 Ca}
α =0
.(15).
Hence, r+r2 ca=2a suppose, and P is always situated
in an ellipse which has for its foci A, B, and which revolves
about its major axis AB with an angular velocity
eccentricity = e
=e=
=
a
a² - 1 =
a² — c²
c²
b2
2
C
1
α
аф
dt
2
To find C, multiply the second of equations (15) by a, and
the third by a²-1, and subtract; we have thus,
n (a² − 1)² {m+m'+2 Ca}+a=0,
1
2
2n (a² - 1)²
m+ m²
C=
2 a
m+ m²
c
2
(
2n6±
4
ся
and substituting this, we find
E=
(m+m') a
+
2 n (a² — 1)²
m+m' a² + 1
1
a² + 1
and D=C+E=
+
a
2n
α
a²
2
1
Hence, by (4),
m
'm'
velocity² = 2
L
で
​30
m + m²
1
2 a
Q n
And we have do and dt, by equations (13) and (7).
If we make vß, we may find a solution in the same manner,
which will give for the place of P a hyperboloid, of which A, B,
are the foci.
If we had supposed the force to vary according to any other
law, the process of determining the motion would, to a certain ex-
363
tent, have been the same.
besides that which we have
and eliminate, and in these, the methods are of considerable com-
plexity. The conditions of integrability are very complet ly dis-
cussed by Lagrange in the Turin Memoirs for 1766-1769:
p. 216, &c., and it there appears, that if P and Q represent the
forces acting towards A and B respectively, we shall be able to
reduce the problem to differential equations of the first order, in
the following cases,
There are, however, only a few cases,
considered, in which we can integrate
1º if Par+
В
2º if Par+ Br³,
3
B'
Q=ar' +
Q=a'r' + B'r'³;
3° if Par+B,³ +yr³, Q=ar' + Br's tyr"
The parts
The first of these three cases is somewhat remarkable.
ar, and ar, or, a. PA, and a. PB of the forces are equivalent to
a force 2 a. PC, C being the bisection of A B, (by Statics).
AB, Hence,
P is acted on by three forces; one to C varying as the distance,
and one to each of the points A, B varying inversely as the square.
Now, by each of these three forces, the body might describe an
ellipse, of which the foci are A and B, and it is a curious cir-
cumstance, that it may be made to describe the same ellipse when
acted on by all the three forces *. The proof of this is contained
in the Memoir just referred to.
It may be worth while to notice how far this problem bears
upon those which the solar system offers to us. The Moon is acted
upon by two forces, tending to the Earth and to the Sun, and each
varying inversely as the square of the distance. If therefore we
could suppose the Sun and the Earth to be fixed points, the path
of the Moon would be that determined in the preceding problems.
But this is not the case, for the Sun also acts upon the Earth, and
though the Earth's distance from it remains nearly constant, and
consequently the Sun's force nearly constant, the centrifugal force
* It may also be observed, that ar+expresses the law of attraction,
which particles must exert, that the attraction to a sphere may follow the
same law as it does to one of the particles. Appendix (C).
364
by which the Sun's attraction is counteracted, operates also upon
the Moon. We may, however, reduce this case to our problem by
this consideration. If we suppose equal and parallel forces to act
upon the Earth and the Moon, the motion of the Moon relatively
to the Earth, will remain unaltered. Let, therefore such forces act,
and let them be equal and opposite to that which the Sun exerts
upon the Earth. Then, the Earth will be acted upon by forces
which destroy each other, and will therefore be at rest: and the
Moon (P, fig. 133,) will be acted on by the forces which tend to the
Earth and the Sun, in the directions PS, PT, (suppose
B B'
and also by a force in PO parallel to ST, which may be considered
as constant. Let ST represent this force; ST is equivalent to
SP, PT; that is, if the force=a. ST, ar, ar', are the parts in
SP, PT. Hence, P is acted on by a force - ar+
a force ar' +
B'
„z to T.
142
B
to S, and
to T. This does not come under the integrable
cases of the problem; and hence, we cannot apply the method to
determine the motions of the Moon. The proper mode of con-
sidering the question of the motions of these bodies S, P, T, acted
on by their mutual attractions, belongs to the subject of the suc-
ceeding Chapter.
APPENDIX (C) to CHAP. IV. p. 66.
On the Attractions of Bodies.
Ir is observed in Chapter IV, that the attractions of bodies
are the collective effect of the attractions of all their particles.
We shall here shew how, from the law of the attractive power of
the elementary parts, we may find the attractions of finite bodies *.
PROP. I. A spherical shell of indefinitely small thickness,
being composed of particles attracting according to a given law;
to find the attraction on any point.
* Newton, Book I, Sect. 12. Laplace Mec. Cel. Liv. II. Art. 11.
365
Let S, fig. 134, be the centre of the spherical shell, SE, its
radius = a; EF, its thicknessa; P the point attracted; PS=r,
PF = ƒ, F being any particle. And let PSE = 0.
Suppose two planes FSP, GSP, passing through SP, to make
with each other an indefinitely small angle FDG = 8, FDG being
a plane perpendicular to PD. Then, FG = DF. 8 a sin. 0. 8.
And if we suppose ESe de, Ee ade, and the solid content
of the small portion of the shell EFGe will be dado a sin. 0.
2
Now since this portion is indefinitely small, its attraction on P
may be considered as that of a single particle at the distance f.
Let (f) be the function of ƒ expressing the law of attraction; then
the attraction of the elementary solid Ge on P will be dade.
a² sin. 0p (ƒ). To reduce this to the direction PS, we must
multiply it by
PD
PF
•
or
slice AEB, towards S, is
a cos.
7-a
2
; hence, the attraction by the
f
Ꮎ
fdade.a sin. $(f).
the integral being taken from A to B.
↑ a cos. O
f
;
The attraction, by varying the angle &, manifestly varies in the
same ratio; hence, for the whole shell we must put 2 for d, and
we have for the whole attraction
2π а²a ƒ də sin, 0 † (ƒ)
α
1-acOS.
f
· a cos. O
A, suppose.
Since ƒ²² = r² – 2ra cos. O + a³,
(1/4)
a cos. O
=
(1/2)
dr
df
f
dr
indicating the differential coefficient where r alone is supposed to
vary.
Hence, A2πa afde sin. 0p (ƒ)
dr
(1/4).
Now let fdf (f) = 1 (f), and we have, since (f) is the
differential coefficient of P1 (ƒ),
(do, (f))
£) = + 1). (²²),
dr
366
Hence, if we take В = 2πa³aƒ do sin. 0₁ (ƒ), we shall have
α
d1
ره
dB
dr
= 2 Ta'a fde sin. Ꮎ (dp₁ (f))
dr
= 2 wa'a ƒ d0 sin. 04 (ƒ). (*)
dr
=A,
for since the variations of and of r are independent, it makes no
difference, whether we perform the differentiations before or after
integration*.
Now, since f² = r² — 2ra cos., 0+a², we have
fdf = ra sin. Od0;
fdf
and sin. Ode =ƒdƒ¸
ra
;
Ωπαα
hence, B =
ƒƒdƒ• $1 (ƒ).
r
The integral is to be taken from = 0, to 0; that is, from
f=r-a, tof=r+a. lfffdf. $1 (ƒ) = ¥ (ƒ), we have for
the whole figure,
Ωπα
B =
{y (r+a) - &(r − a)}.
r
d B.
And the attraction A =
dr
is thus known.
For a point within the shell the process will be the same, except
that the integral must be taken between the limits a +r, and a — r.
Ex. 1. Let the force of each particle vary inversely as the
square of the distance.
* If F be a function of r and 8, and B=ƒ Fd8,
d B df Fde
dr
dr
d B
d F
d2B
But
do
dr
d & dr
Hence, (Lacroix, Elem. Treat. 122.)
ď²B d F
dr
drde
d2 B d F
d B
dF
and
•
de; ..
de
dr
dr
dr
dr
367
$1
1
Here ¢ (ƒ) = }; $\(ƒ)=ƒdƒ-4 (J) = −'};
دشهر
¥ƒ=ƒƒdƒ¢、 (ƒ)= −ƒ,
2 ñ а а
{(r− a)—(r+a)} ·
B-
r
dB
4π a²α
παρα
A =
dr
2
4πα α
T
αα
The surface of the shell is 4πa, and hence, its mass is 4πa'a, and
the attraction is the same as if it were collected at the centre of the
sphere.
Ex. 2. Let the force of each particle vary as any power of the
distance.
Let ¢ (ƒ)=ƒ”, $1 (ƒ)=ƒ”+
fn 1
>
n+1
&(f):
f" + 3
(n + 1) (n + 3)'
B =
2παα
(n + 1)(n+3) r
3
{(r + a)² + ³ — (r − a)" +³ }
2
4πα α 2?? +1 +
(n+2) (n + 1)
p² - 1
.1 2
n+1
2.3
+
(n+2) (n + 1)n. (n-1)
r²-3a²
3_4
2.3.4.5
»-³ a² + &c.} .
d B
And A =
Απαρα
dr
4π a³ar" +
{~ +
(n + 2) (n − 1)
27-202
2.3
+
2,22-4a4+
α
2.3.4.5
+ &c.}
(n + 2) n (n − 1) (n-3)
This series terminates, if n be a whole positive number.
=
If n = 1, or n=2, that is, if the attraction varies directly as
the distance, or inversely as the square of the distance, the terms
after the first vanish, and the attraction is the same, as if the mass
were collected at the centre of the sphere.
m'
Hence, if the particles exert a force which is as mr + the
29
whole force will be the same as if the mass were so collected; for
we may suppose the shell to consist of particles which attract with
forces mr, and of an equal number of others which attract with
m'
forces
368
√
If n=-1, or n=3, the integrations for (f) fail, and we
must employ other methods.
Ex. 3. Let the force vary inversely as the cube of the distance.
1
1
$ (f):
фо
$1 (ƒ)=
& (f) hyp. log. f.
"
παα
B =
hyp. log.
r
r a
r + a
ΑΒ πα
Qar
p
A =
2
2
2
- hyp. log.
dr
a
r+al
Ex. 4. Let the force vary inversely as the distance.
2a
Α=παα
r
a
+(1-2) hyp. log. "+a}.
C
r
a
Ex. 5. The force varying as any power of the distance; to
find the attraction on a point within the shell.
As in Ex. 2, (f)
B =
2παα
(n + 1)(n+3) r
n+ 3
+
fn +
+3
(n + 1) (n + 3) '
(n + 2) (n + 1)
2.3
(n+2) (n + 1) n (n-1)
3
{(a + r) n + ³ — (a − r)n+3}
a² + 1 pz
an
q²−1,^ + &c. },
an−1
¹ pº+&c.} .
•
2.3.4.5
Απας
n + 1 l
{2"
+
d.B
A
= 4πα
Su + 2
a²+1
p
dr
3
+
N
-
2 (n+2) n (n − 1)
3.4.5
If n = 2, or the force be inversely as the square of the
distance, we have A=0; the attractions in different directions
counterbalance each other.
PROP. II. To find the attraction of a sphere composed of
particles attracting according to a given law.
If in the last proposition we put u for a, and du for a, the
thickness of the shell, and integrate from u=0, to u = a, we shall
have the attraction of a solid sphere of radius a.
369
By this means, from the expression for A in Ex. 2, we find for
the attraction of a sphere
Απα
3
3
(n+2) (n - 1) 3pm-2a²
2.2
+
+ &c.
2.3
suc.}.
5
In the cases where the attraction of a shell is the same as if the
matter were collected at the centre, the attraction of a sphere will
also follow the same law. For the sphere may be supposed to
be composed of concentric shells, each of which attracts as if it
were collected at the centre, and therefore the whole will attract as
if all its parts were there collected.
PROP. III. To find the attraction of a circle on a point in a
line perpendicular to its plane, and passing through its centre.
Let BC, fig. 135, be the circle, P the attracted point, SP=r,
SE=ƒ, SE any radius=u, and SF a radius indefinitely nearly
equal to this, so that EF du. Let a small angle FSG=8, then the
quadrilateral EG=u.8. du. And,
.8.du. And, if the law of the attraction be
represented by (ƒ), the attraction of EG is 8.udu.& (f);
which resolved in the direction PS becomes 8.udu.¢ (ƒ)
"'
f
And for the whole annulus, whose breadth is EF, we must put 2π
for ♪; whence it becomes 2πudu. (f)... Hence, the attraction
of the whole circle
r
√(r²+u²),
=2xfudup (D)}; where ƒ= V{°+u),
ད4
the integral being taken from u=0, to u = a, the radius of the
circle.
If $ (ƒ)=ƒ”,
attraction = 2πrf u du (»² + u²)
儿​+1
Απ'
n+1
(x² + u²)
2
22 + 1
{(x² + a³)
24-1
72 1
S
+ constant
— +1}.
20?? + } } .
3 A
370
Ex. 1. Let n= -2, or the force vary inversely as the square
of the distance.
Here, attraction=4π {1
2
2
√(x² + a²
Ex. 2. Let the circle be infinite, and n < − 1.
-
n+1
In this case (2 +a²)
2
becomes 0, and we have, putting-m
for n,
attraction =
Απ
(m − 1) pm—2
If m=2, attraction = 4π, and is the same at all distances.
PROP. IV. To find the attraction of a solid of revolution on
a point in the axis.
We must here multiply the attraction of the circle, found in
the last Proposition, by the thickness dr, for the attraction of a
differential slice; and if we then put for a its value in terms of r,
and integrate, we have the attraction of the whole solid.
Ex. 1. The attraction of a cylinder on a point in its axis; fig. 136.
Απ
attraction
n+1
АП
n+ 1
2
pn + 2 dr}
S {rdr (r² + a²) &
4π ((2²+a²)
((n² + a²)
n+1 n+3
n+ 3
Q
2N+3
n + 3
+ constant
B'S'C' be the two ends of the cylinder, and if
PC=c, PC'=c', we have
If BSC and
PS=b, PS' = b',
4π
3
attraction =
{ c'n + ³ — c²+3
(b'n + 3 — b' n + ³)}.
(n+1)(n+3)
21,
If the force vary inversely as the square of the distance, n=-
attraction 4 { b′ — b— (c' — c)}.
Ex. 2. The attraction of an infinite solid bounded by planes.
In last Prop. Ex. 2, multiply by dr, and we have
attraction
4π
S
dr
m-1 pm—2 ~
Απ
1
1
(m − 1) (m— 3) (bm
3
pm -
3
371
where b is the distance of the attracted point from the surface of
the solid.
If m = 2, attraction = 4π (r — b).
ጥ
If m = 3, attraction = 2π hyp.log..
If m> 3, the attraction is finite, when r is infinite, and we have
attraction =
4π
(m − 1) (m − 3) fm−3
Ex. 3. The attraction of a cone on a point at the vertex.
In fig. 137, let PS=r, ST kr, and putting kr for a in last
Prop.
n + 1
2
-1}
n+1Sr²+²dr {(1+k²)”ï— 1}
4π
attraction =
4 π pN + 3
(n + 1)(n+3)
PA the axis.
Where r is to be made
{ √(1+k³)−1}.
PROP. V. To find the attraction of a straight line upon a point.
at any distance from it.
Let BC, fig. 138, be the attracting line, P the point attracted,
PS perpendicular on BC=r, SE=u, PE=ƒ, and let the force
of each particle be as (f). We may suppose EF an indefinitely
small portion, to be du; and its attraction on P will be $ (f)du,
2
r
Ф
and part resolved perpendicular to BC will be (f) du, where
ƒ= √(x²+u²). This is to be integrated from u=0, to u=a=SB,
for the attraction of SB; and the attraction of SC is to be found
in the same manner, and added to the former.
The attraction of du parallel to SB, will be
U
f
(ƒ)du; which
is to be integrated in the same manner as before; and the difference
taken of the parts belonging to SB and to SC.
Ex. Let the force vary inversely as the square of the distance.
4
1
Here (ƒ) = Fi
372
attraction in PS=
Г
rdu
И
(x²+u²)#
r √ (2.² + u²) '
'
attraction perpendicular to PS=
S
2
udu
1
1
2
(x² + u²)²
√(x² + u
- u³)
2
And this is to be taken for SB and for SC, and combined.
APPENDIX (D) to CHAP. IV. p. 78.
On some particular Cases of the Motions of three Bodies.
THERE are one or two other particular cases in which the
motions of bodies acting upon each other can be accurately
obtained. These cases may be deduced from the following prin-
ciples.
PROP. If any number of bodies be acted upon every where by
forces which are to each other as their distances from a given
point, to which they tend, the bodies may be made to describe
similar curves round this point in equal times.
For if they be projected at equal angles to their distances, and
with velocities proportional to them, they will manifestly in the first
instant describe similar curves. And at the end of this first instant,
the forces will still be proportional to their distances, their direc-
tions will make equal angles with the distances, and their velocities
will be proportional. Hence, they will in the second instant
describe similar curves. And so on for any number of instants.
And hence, the curves will be similar, when they are described con-
tinuously.
Hence, if any number of bodies acting upon each other, be so
placed, that the resultant of all the actions on each tends to the
centre of gravity of the whole, and is proportional to its distance
from that point, they will, being projected with proportional velo-
cities in similar directions, describe similar figures round this centre
in the same times. We shall take two examples".
Laplace, Mcc. Cel. Liv. X. Chap. VI. No. 17.
Y
373
PROB. I. Three bodies in a straight line, m, m', m", fig. 139,
attract each other with forces varying inversely as the nth power of
the distance; it is required to find their positions, that they may
describe similar figures round A the centre of gravity.
We must have the whole force on m: whole force on m' ::
Am: Am', and whole force on m: whole force on m" :: Am; Am".
Hence, if Am=r, Am'=r', Am" =r", we must have
: kr";
force on m=kr, force on m' = kr', force on m”
k being any quantity. Hence,
=
m'
m"
+
= kr
(~+r)"
(r+r)n
m
m"
kr'
.(1).
(r+r)²
(2'' — r'jn
m
m'
+
= kr"
(x + plyn
(r" - r'jn
Eliminating k in the two first, we have
(r+r")"
r + r' \ n
m'r'
m"r
+
(~ + 7'')"
or m'r'+m'r'
C
Let
(+ 2)²
r+r"
r
"
r + r
(x² + r²)n
m r
+
m'r
(p'' — p')"
= 0,
G
´r + r
ก
=0....(2).
mr + m″r
; .. 1+z =
rtr
Also, mr-mr-m"r" =0, because A is the centre of gravity.
"" — x'
Hence, z=
rtr
m"r" m"r' mr-mr' - m'r'
m" (r+r')
m" (r + r')
•. m″rz+m″rz= mr − (m'′+m") r'",
(m— m'z) r
and '
m' + m² (1 + z)
Substituting these values in equation (2), it becomes
(m-m'z)
m'+m" (1 + z)
(
m"
{M(x + 1) + }
m"
m+
=0......(3);
zn
4
374
whence z may be determined, and thence r'; and thence
r"=
mr — m'r'
m
"/
If we multiply the first and second of equations (1) by m, m',
respectively, and subtract, observing that mr — m'r'
r =m"r",
n"r", v we
have the third. Hence, the values of r', r" so found, will answer
the conditions. And if, the bodies being at these distances, they
be projected in similar directions with proportional velocities, they
will describe similar figures.
If the force vary inversely as the square of the distance, equation
(3) becomes
z²(m—m″z) { m′(1 + z)² + m" } - (mz² —m")(1 + 2)² { m'+m"(1+z)}=0,
•. {mm' (1+x)² +mm″ — m'm″z (1 + z)° — m″²z} z
2
2
-- {m m'z² + mm" (1 + z) z² — m'm" —m”² (1 + z)} (1+z)² S
or, dividing by m", and changing the signs,
= 0;
mz² {(1+z)³ — 1 } − m′ (1 + z)² (1 − z³) — m″ {(1 + z)³ — z³} = 0.
−
If m be the Sun, m' the Earth, and m" the Moon, m is large
compared with m' and m". Hence, it appears that z will be small,
and that m³ is of the order of m' and m". Neglecting therefore
higher powers of z, we have
3 mz³ — m' — m" =0; 2=
1
3
V m
m' + m²
m'+m"
3 m
;
which gives z= nearly. Hence, in fig. 139, m'm"
100
1
100
mm'.
If therefore, the Earth and Moon had been placed in the same straight
line, at distances respectively from the Sun, proportional to 1 and
1
1+ and if they had had velocities parallel to each other, and
,
100
proportional to those distances, they would have moved about the
Sun, the Moon being perpetually in opposition. Also at this dis-
tance, which is about four times the Moon's actual distance, she
would have been beyond the Earth's shadow. But then, instead of
occupying an angle of half a degree, she would only have subtended
375
1
an angle of about 8', and, presenting only
of the present full
16
Moon's surface, would have answered, in a very imperfect degree,
the purposes of replacing the light of the Sun.
PROB. II. Three bodies, m, m', m", fig. 140, attract each other
according to any function, $ (s), of the distance s; to find their po-
sitions, not in a straight line, that they may describe similar
figures.
Let the bodies be referred to rectangular co-ordinates in the
plane of the three, and measured from the centre of gravity A.
Let these co-ordinates be x, y, for m; x'y' for m'; x", y', for m".
Also, let the distance mm' be s, mm" be s′, m'm" be s' Hence,
we shall have, for the forces on m,
parallel to x, m'p (s).
x − x'
S
x
+m″p (1') ==—=—="";
S
to y, m'p (s)
Y
-y'
+m″p (5') Y — y'"';
y-y
;
$
S
and similarly for the other bodies m', m".
Now, in order that the three bodies may describe similar figures,
the forces upon them must be proportional to their distances Am,
&c. from the centre of gravity. Hence, the resolved part of the
force on m, in direction of AM, will be proportional to the co-
ordinate AM, or x; and similarly for the others; so that K
being a constant quantity, we shall have
m'
m²
Φ
m
$ (s) (x − x') + m″ $ (s')
Φ
S
− -m"
S
$ (s) (2' — 2) + m" Þ (s')
S
$(s)
m
— x)
(x" — x) + m'
(x − x") = K x
(x' — x') = K x'
. (1);
$ (s″)
(x″ — x') = Kx"
S
"
where, as before, the third equation is deducible from the other
two, observing that
mx+m'x'+m"x"=0.
376
find
If we combine this equation with the first of equations (1), we
$(s)
х
{အာ
$(s)
+(m+m”) P(") } +m'x′ {P() _DO
(s)
= Kx.
S
S
$(
$ (s)
If we suppose s'=s, this gives
K = (m+m' +m")
S
If we also suppose s"=s, the two last of equations (1) will give
the same value of K, and all the three equations will be satisfied.
In the same manner the three equations in y, y' y", will be
satisfied by the same supposition. In this case, (that is, when
s=s′=s″,) it is obvious that the forces tending to the centre of
gravity are K √(x²+y³), &c.: or, if we call r, r', r", the distances
from the centre, the forces towards that point are Kr, Kr', Kr".
Hence, if the bodies be projected in similar directions, with ve-
locities proportional to r, r', r'", they will describe similar curves.
We shall now find the force tending to the centre of gravity.
On this supposition, we have
Similarly,
- m'x' — m'x'
MX =
x;
.. (m + m' + m'") x = m′ (x − x') +m" (x − x″).
(m+m'+m") y=m' (y — y') + m" (y—y″).
Squaring and adding,
(m+m'+m″)² (x² +y³)
= m²² (x − x')² + 2 m'm" (x — x') (x — x")+m"² (x − x″)² }…….
+m³ (y—y')²+2m'm" (y — y')( y − y″)+m" ( y −y″)°
-
But, x²+y²=r²; also (x-x')²+(y — y'²)² = s²
Again, since s′=s'″,
S=
(x − x″)² + (y — y″)° = s².
(x − x′′)? — (x′ — x'')² + (y—y″)² — (y' — y″)² = 0.
And (x − x′)²+(y—y')². .. . . .
Adding
= $2.
2x² — 2xx′ − 2 xx″ + 2 x'x″ + 2y² — 2yy′ —2yy″ +2y'y" = s° ;
or, 2(x-x′) (x − x″)‡2 (y—y') (y—y″) = s².
Ꮖ
(2).
377
Hence, equation (2) becomes
(m+m' +m")² r² = (m"² +m'm" +m"²) s³,
S=
(m+m+m") r
√ (m²² + m'm'' + m²
And putting this for s in Kr, where
112
K = (m+m'+m")
$(s)
S
we have the force to A.
If the force vary inversely as the square of the distance, we
1
have os=. Hence, K is
(m+m'+m").
and the force =
112
(m" + m'm" + ni"²)}
(m²+m' +m″)³r
3
(m'²+m'm" +m″²)}
3 2
(m + m² +m")³rs
If the bodies be properly projected, they will move about
the centre, so as to describe similar paths, (namely, conic sections,)
and always forming an equilateral triangle by the lines that join
them.
If we do not suppose s=s', we shall find that the only way in
which equations (1) can be satisfied, requires that
x : y :: x' : y' ;
whence the bodies are in a straight line.
And therefore this case
is reduced to the last problem.
APPENDIX (E) to CHAP. VI. p. 142.
On the Vibrations of Strings.
THE vibrations of strings stretched between two points, such,
for instance, as musical strings, were the subjects of long and
3 B
378
T
various investigations and discussions among the mathematicians of
the last century. We shall here as briefly as possible, give the
statement and solution of the Problem, in its simplest form.
PROB. I. A uniform string being stretched between two given
points; to find the time of its small vibrations.
Let a be the length of the string between the fixed points; Wg
its weight, and Fg the force by which it is stretched. It is sup-
posed, that the vibrations are exceedingly small, so that the length,
and consequently the tension, may remain the same in all the forms
which the curve assumes. Let APB, fig. 141, be the form at a
time t, and let AM, MP, an abscissa and ordinate, be x and y,
and AP=s. Let AN, NQ be another abscissa and ordinate, and
PQ=h. The tension will be the same throughout, and = Fg;
dy
hence, at P the part of it resolved in the direction PM is Fg. ds
dy
dy d'y
ds d s²
h +
2
At Q, becomes +
ds
d³y h²
3
ds³ 1.2
+&c., supposing ds
constant. Hence, the difference of the forces at P, Q, or the
moving force by which PQ is drawn in the direction MP, is
Jd²y
Fg. h +
d³y h
3
ds³ 1.2
+ &c.}.
ds
2
And since the weight of a length a is Wg, the weight of a length h
Wgh
is
a
; hence, the mass of PQ is
force on PQ is
Fag (dy d³y h
Wh
>
and the accelerating
a
+
W
ld s²
ds 2
+ &c.} .
And when PQ is taken very small, every part of it moves with the
same velocity, and this is the true expression for the accelerating
force. That is, accelerating force on P in MP =
Fag d'y
W ds
Since the vibrations are very small, we may suppose P to move
always perpendicular to AM in MP; and hence, the accelerating
379
d²y
force on P =
Also, for the same reason, we may suppose s
d t
to be equal to x.
Hence, we have
d²
Fagdy
…. . . . (1).
122
2
W
dx
The quantities are put in brackets to indicate that they are
partial differential coefficients; in the one y is differentiated, sup-
posing t only to vary, and in the other, supposing a only to vary.
The former differentiation refers to a change in the position of a
given point of the curve; the latter, to a passage from one point of
the curve to another at a given moment of time.
The ordinate Y will be a function of r and t, which is to be
determined by integrating the partial differential equation (1).
Its integration is giveu, Lacroix, Elem. Treat. Art. 319; and if
= b², it is
Fag
W
y=$(x+bt)+↓ (x − bt). .. . . .. . (2).
It is evident from trial, that this satisfies equation (1), for if we
differentiate twice with respect to x, we find
(day
4″ (x+bt)+4″ (x — bt),
dx²
where "z indicates
d² o z
dz2
Φ
Also, differentiating for t, we have
=
dt²
b°q" (x+bt)+b²f” (x — bt).
And it is manifest, that these values verify equation (1). And the
form of the functions and is to be determined from the initial
circumstances of the string.
Differentiating equation (2), we have
dy
dt
= b {p' (x+bt) — \' (x — bt)}·
And if we suppose that the curve was at rest in any position when
t=0, we must have p′(x) — y′(x)=0. Hence, y(x)=p′ (x),
380
and \ (x) =¢ (x). Let p (x)==ƒ (x), and equation (2) becomes
y =
1IQ
{ƒ (x+bt)+ f (x − b t)} • • . . (3),
dy
dt
b
{f (c+bt)—f ( x − bt)}....(4).
2
Equation (3) gives the position of the curve at any time, and
equation (4) the velocity of any point; 6 is equal to Fag
b
√
If we make t=0, we have y =ƒ (x), which is the equation to
the original form of the string, and hence, the form of ƒ is known.
The original form may be any whatever, and ƒ any function what-
ever, subject to the conditions that it must = 0, when x = 0,
and when x = a. It appears from the theory of partial differential
equations, that the function f, introduced in integration, may be
discontinuous; that is, the initial form of the string APB, may be
composed of different lines, not expressed by the same equation all
the way from A to B.
But, from the nature of the reasoning, we obtain other pro-
perties of the function f. The points A, B, are fixed; hence, we
must have y = 0, when x = 0, whatever be t. Therefore by
equation (3),
0=ƒ(bt)+f(− bt).
Or, if we make bt=u, ƒ ( − u) = −ƒ (u)
(5).
Also, if xa, y = 0, whatever be t;
·.0=ƒ (a+u)+ƒ (a − u), and ƒ (a +u) = − ƒ (a−u) ....(6).
From equation (5) it appears, that the curve represented by
y=f (x), is continued with similar forms on each side of A, the
curve being on one side above, and on the other below the axis.
Also, from equation (6) it appears, that the curve is continued
with similar forms on each side of B, above the axis on one side,
and below it on the other. Hence, the curve is continued inde-
finitely on both directions, in the manner represented in fig. 142,
}
381
and it is the ordinate of this curve which is to be considered as
f(x) in finding y from equation (3). To find the position of a
point P at the end of any time t, we must take MN=MN'=bt,
and MQ=½ (NO+N'O′) will give Q, the position of P after a
time t. And as the curve is continued indefinitely, the time may be
supposed to increase indefinitely, and the same construction will
always determine the position.
The string will perform oscillations, assume a position similar
to its original one on the opposite side of the string, return to its
original position, and so on perpetually; as may thus be shewn :
By (6), ƒ (a+u)=-ƒ (a-u). Let u-a=u'; .. u=a+u',
ƒ (2a+u') = −ƒ(—u')=ƒ (u') by (5), or ƒ (2a+u)=ƒ (u):
sim¹y. ƒ (4a+u)=ƒ (u), and generally,
ƒ (2ma+u)=ƒ (u), m being any whole number.
Hence, the ordinate for an abscissa 2 ma+x is the same as for x,
as also appears from fig. 142.
If, therefore, we assume bt=2a, bt=4a,....bt=2ma, at
the times corresponding to each of these values, we shall have
y= { {ƒ(x) +ƒ (x)} =ƒ(x), and every point of the curve is in its
original position. The intervals between these times are the lengths
of complete vibrations of the string, and we have for these intervals,
bt = 2a, t
=
2a
b
Wa
2
Fg
If the thickness and material of the string be given, W is as a.
Hence, for a given string, the time of vibration is as the length
directly, and as the square root of the tension inversely.
b
If we make bt=a, or t = -
1
,
we have
α
y = ¦ {ƒ(x+a) +ƒ (x− a)}.
And ƒ (x − a) = −ƒ (a− x) by (5), =ƒ (a+x) by (6);
f
•'• y=ƒ (x+a).
And hence, after this time, the form of the curve is ARB, in fig. 141,
382
identical with BDB' in fig. 142, and exactly similar to the original
curve inverted; the greatest ordinate being now at the same distance
from B, as it was from A at the beginning of the motion. Similarly,
3b 5b
at the end of times
ARB.
a
α
&c., the figure will be the same as
a
If we make bt =
we have the figure in the middle of the
2
time between the positions APB and ARB.
Hence y = ± {ƒ ( x + 2 ) + f (x − )}·
f
a
) = − ƒ (¦ − x) by (5);
− -
But ƒ (x −
.y
· · 3y = ž { ƒ
(
+ x ) − ƒ ( − x ) },
−
a
+ x, and
a
J
Q
- correspond to points at equal distances from
the middle point C of AB. If in the original form of the curve,
the ordinates for the portion CB be greater than for AB, the posi-
tion of the portion AEC at the middle of the time of oscillation will
be above the axis AC. In the same manner it may be shewn, that
the position of the portion CFB at that time will be below the axis
CB, and similar to the portion AEC inverted. The string never
becomes a straight line.
If the curve, instead of being at first in the position APB, have
an original position APBDB', fig. 142, A and B' being the extreme
points, and the curve consisting of two equal and similar portions
APB, B'DB, it will oscillate so that the point B will remain fixed,
and each half APB, BDB' will oscillate as if the string were fixed,
at A, B, and at B, B', and the vibrations will employ half the
time. Similarly, if the original form be APBDB'B', the points B
and B' will remain fixed, and the oscillations will employ one-third
of the time, and so on. The points B, B' which remain fixed, are
sometimes called nodes.
The musical tone, or note, produced by a vibrating string,
depends upon the rapidity of the vibrations. If a string a vibrate
383
twice as fast as another A, the note produced by a is said to be an
octave above that produced by A. If a string e vibrate three times,
while a vibrates twice, the note of e is a fifth above that of a; and so
on. And notes where numbers of vibrations are to each other in
ratios so simple as these, are found when combined, to be agreeable
to the ear.
It appears from what has been said, that a string may vibrate
so that no point of it is at rest, except its two extremities, and thus
give the fundamental note; or it may vibrate so that its middle
point is at rest, in which case it will produce the octave to the
fundamental note. Or it may vibrate so as to have two points at
rest, dividing it into three equal parts, and it will then give the fifth
above the octave; and so on.
It appears by experiment, that a wire which performs 240 vi-
brations in a second, sounds the note called C in the middle of
the scale of a harpsichord. It has been proposed however to con-
sider 256, or 28 as the number of vibrations determining this note,
in which case every other note C would also have a power of 2 for
the number of its vibrations. Hence, we may find the note pro-
duced by any string; by finding the number of its vibrations, and
comparing them with the number just mentioned.
PROB. II. To find the form of a string, that it may oscillate
symmetrically, that is, that all its points may come to the axis at
the same time.
In this case, the force which accelerates each particle, must be
as its distance from the axis; that is, as y. Hence, taking the
expression for the force obtained in p. 378, we have
Fag dy
W dx²
2
or
d²
my, m being a constant quantity.
Of this the integral is, if
Wm
y
+
y=0.
dx²
Fag
W m
k²,
Fag
C sin. kx + C' cos. kx;
y
dy
= kC cos. kx
k C' sin. kx.
dx
384
་
dy
And when x = 0, = 0; .'. C′ = 0,
dx
y = C sin. kx.
The quantity C is arbitrary, and determines the magnitude of
the original ordinate. It must necessarily be small, because the
vibrations are supposed to be small.
The time of a point coming to the axis will be
hence, the time of an oscillation to the opposite side, is
π
and
2 V m
>
π
and
V m
the time of a complete vibration, the string returning into the original
position, is
2π
V m
The curve must meet the axis again when x = a, hence,
0 = C. sin. ka ; .'. ka=π, and k=7;
a
..
√
W m π
Fag
Fg
and V m = π
П
2
Иа
a
2 T
Wa
Hence, the time of a vibration =
before.
= 2
√ m
the same as
2
Fg
APPENDIX (F) to CHAP. VI. p. 142.
On the Vibrations of Springs.
As another case of small oscillations, we shall obtain the equa-
tions for the motion of an elastic rod.
PROP. I. A uniform elastic rod BC, fig. 143, firmly fixed at
B, and naturally straight, vibrates by its elasticity; to find the
equations of its motion in small vibrations.
We shall find first the forces which must act at every point to
385
}
keep the rod at rest in the position BPC. Let AB be its position
when left to itself, AN, NQ, an abscissa and ordinate; CQ=s.
And suppose that a force F, acting at every point Q, would keep
the rod BPQC in equilibrium. Since the whole is in equilibrium
the part PQC must be so, and hence, the forces which act on PC
must balance the tendency which the rod has to straight itself at P
by its elasticity. Now the moment of the elasticity is directly as
the curvature, and therefore inversely as the radius of the curvature
at P. Let this radius be p; and let the effect of the elasticity to
turn PC round P be that of a force K acting at any arm c. Hence,
Ke∞. Let Ke=
Kc
сох
р
E
P
And, if AM=x', MP=ý, AN=x, NQ=y, the moment of
any force F about P is F (x' — x); and, as the forces F act at
every point of PC, the moment of the forces on ds is F(x-x) ds,
or F(xx)dx, when the flexure is very small, so that dr and
ds may be conceived to coincide. Hence, F(x-x) dx is the
ƒ
whole moment of the forces on PC, the integral being taken from
x=0 to x = x'. Therefore,
E
: ƒ F(x′ − x) dx.
ds's
dx'
Now, p
d'x' ď² y'
dy'
And ƒ F(x' - x) dx=x'f Fdx-f Fxdx.
If we suppose Fdx taken from r=0, to be = F₁, we shall
have
f Fxdx = fxd F₁ = F, x' -f F, dx.
F₁dx.
Hence, ƒ F(xx) dr becomes fFdx. And we have
Ed²y'
=fF₁dx
dx
12
Edy
= F₁.
dx's
=
SC
386
1
for it is clear that fF, dr, from r=0 to x=x', will be a function
of x', and that its differential with respect to r' will be F
on the same account.
Ed¹y'
dx"
14
d F
dx
=
F,
Hence, F is known at every point.
Now, if when the spring is kept at rest by these forces F acting
at every point, we suppose at each point a force equal to F in the
opposite direction, these forces will counterbalance each other, and
the rod will be left to its own elasticity. But, in this case the points
are manifestly each urged from rest by a force F. Hence, we shall
have = F; or, using the same notation as before,
ď² y
d t²
2
- E
dy
dx¹
d²y
4
2
• • . . (1),
d
which is the equation to the motion. This cannot be integrated in
finite terms, and though we might obtain integrals in series, we shall
confine ourselves to the consideration of symmetrical oscillations.
PROB. II. To find the form of an elastic rod, that it may
vibrate symmetrically.
For this purpose the force which urges each point towards the
axis must be as the distance from the axis. Hence, we must have
d4
E
* Y
d¹y
my
ky suppose.
=my, or
dx4
dx¹
E
..(2).
This equation may be integrated, and gives
y= A cos. kx + B sin. kx + Ce** + D€˜¯**......(3).
A, B, C, D, being arbitrary quantities. To determine them, we
may observe, that when r=0, we must have
d³y
dr
d x
3
d³
y
dx
= F in last Prop. For the same reason,
x=0. Hence,
0 =
-B+C-D, 0=−A+C+D.
.. A=C+D,
B=C-D.
3
= 0, because
ď² y
=0, when
dx²
387
Again, since the extremity B is fixed, we have, when x=a, the
whole length, y=0. And, since that extremity is also fixed in
dy
direction, we have
O, when x = a.
Hence, we find
dx
0=(C+D) cos. ka+ (C-D) sin. ka+Ce+De
-ka
ka
0 = −(C + D) sin. ka + (C–D) cos. ka+Ce*ª — De¯*ª ;
whence
C
sin. ka― cos. ka — ɛ˜k²
€
D
sin. ka+cos.ka+e**
cos, ka + sin. ka+e
cos.ka - sin. kate
-ka
ka
ka
Hence, we find 2+ cos. ka (e*ª +ɛ*ª) =0. . . . . .(4).
€
From this equation k is to be determined: k will have an in-
finite number of different values; and to each of these will correspond
a different form of the curve, determined by equation (3), in which
the rod will oscillate symmetrically.
The least positive value of ka is 1.8751, nearly. The other
3 п
5 π
values of ka are very nearly
&c.
Q
2
2 π
m
The time of a vibration will be
√ m
Now k*, whence
E
m = Ek'. And hence, the time of vibration is
2 π
k² VE
Ωπα
2 2
Q
k² a VE
La
For the same value of ka, that is, for the same form, in rods of
different lengths, it appears that the time of the vibration is as the
square of the length directly, and as the square root of the elasticity
inversely.
Hence, if a rod A' were twice as long as A, and of the same
elasticity, the note produced by A' would be-four octaves below
that produced by A.
wo
The different values of ka in equation (4), taken for the same
rod, give a series of different forms, resembling those in fig. 50,
and differing like them in the number of their nodes. The times of
388
vibration will be successively as the values of
}
k² a²
and the num-
2
bers of vibrations in a given time, and consequently, the note, will
be as ka². Hence, the note produced by the successive figures
will be as
1, 6.2673, 17.549, 34.386, &c.
Hence, the notes which an elastic rod can produce, besides the
fundamental note, are, the fifth above the double octave, somewhat
too sharp the half note above the fourth octave nearly the half
note above the fifth octave nearly and so on.
:
If the rod be not fixed into a support at B, but be subject to
other conditions, for instance, if both ends were fastened, or both
free, we shall have to make some other supposition instead of
dy
dx
putting y=0, and 0, when xa.
will be nearly the same as before.
The rest of the process
We may find E by the equation E=Kcp, where a force K,
acting at an arm c, produces a radius of curvature p. See Statics.
For an examination of the different cases of this Problem, see
Com. Acad. Petrop. for 1741-1743, p. 105, and Novi Com. Acad.
Petrop. for 1772, p. 449. Dr. Young, Elem. of Nat. Phil.
Art. 398. For experiments, see Chladni, Traité d'Aconstique,
p. 92, Biot, Traité de Physique, tom. II, p. 74.
APPENDIX (G) to Book II. CHAP. I. p. 180.
On the Descent of small Bodies in Fluids.—On the
Ascent of an Air-Bubble.
THESE problems do not belong immediately to our subject,
but they depend upon little in addition to the principles employed in
the text and as the latter problem has been imperfectly solved by
Atwood in treating of this subject, and the solution copied into
other works, we shall introduce them here.
389
The resistance on a plane moving perpendicularly to itself in a
fluid, is the weight of a column of fluid, whose altitude is the height
due to the velocity.
The base of the column is understood to be the surface resisted.
It appears probable in the first place, that the resistance should
be as the square of the velocity: for the momentum lost by a body
moving in a fluid is, by the equality of action and reaction, the same
as the momentum communicated to the fluid.
ommunicated to the fluid. Now, the momentum
communicated will be as the product of the quantity of fluid-matter
moved, and the velocity communicated to it. But the quantity
moved will be nearly as the space through which the velocity moves
in a given time; and the velocity communicated will be nearly
that of the body itself. Hence, the momentum, which is in the
compound ratio of these, will be as the square of the velocity
nearly. And this is confirmed by experiment.
The height fallen through to acquire the velocity, is as the square
of the velocity; hence, the resistance is as that height. Also, it is
cæteris paribus as the surface resisted, and as the density of the
fluid. Hence, it is as the surface, the height due to the velocity,
and the density; and therefore it is as the column of fluid, whose
base is the plane resisted, and its height that due to the velocity.
Though this reasoning is far from being exact and conclusive,
experiment confirms the result of it; and shews that the resistance
or moving force retarding the body, is equal to the weight of a
column of the fluid, whose base is the surface resisted, and its height
the height due to the velocity.
Let A be the area of the plane which moves perpendicularly to
itself, V its velocity, D the density of the fluid; then the mass of
the column spoken of is A.
v2
2g
.D; and the moving force of re-
sistance, which is equal to the weight of this column, is AV D.
If the moving body be a cylinder, in which bis the radius of
the base, a the length, and ▲ the density; we shall have A Th²,
and D=the
☛ b² V¹ D≈ the moving force of resistance. Also, the mass
390
of the body is baA: and hence, the accelerating force, (or pro-
perly the retarding force) of resistance,
which corresponds to k in the text, is
DV2
is
; and the quantity
2 Δα
D
2 Δα
The moving force of the resistance upon a globe is the resistance
upon one of its great circles, moving perpendicularly to itself. This
appears by resolving the force on each particle of the spherical sur-
face, in a direction perpendicular to the surface, then resolving this
part again in the direction of the motion, and taking the sum of all
the forces so resolved. It appears by experiment, that the resistance
on the globe is a little greater, in proportion to the resistance on
a circle, than this theory gives it.
I
b²
Taking for granted the truth of the theory, the moving force
of resistance on a globe of radius b, is π62 V2 D. And the
mass of the globe is b³ A. Hence, the accelerating force is
3 DV2
16 A b
: and
3 D
16 Ab
3
corresponds to k in the text.
The force by which a body descends in a fluid, is the excess of its
weight above the weight of an equal bulk of fluid. This appears
by the laws of Hydrostatics. If the body be a sphere, ‡ π b³▲g is
its weight, and Dg the weight of an equal bulk of the fluid,
and hence, the body descends with a moving force π b³ (▲ – D) g,
(A
or an accelerating force
g. And if D be greater than
A-D
A
▲, the accelerating force with which the body ascends will be
D-A
A
g.
PROP. I. To find the descent of very small bodies in fluids.
In p. 179, we have the equation
where V² =
go
k
V2
s=
2g
V2
hyp. log. vv²
29
, V being the terminal velocity.
ཡ ཨགོ
i
391
.. 2ks= — hyp. log. (1
– –
-
V²
→ęks
_³ — [ =
3 Ds
8 Ab
=1-€
Now, if b be small, and s considerable, D not being very small
compared with A, the quantity ẹ
2,2
V2
3 Ds
10 Δ6
will be very small, and
= 1, very nearly. Hence, after the body has descended through
a certain space, it will have acquired very nearly the terminal ve-
locity, and may after that be considered as going on with that
uniform velocity.
Thus, let A=2 D, s=32b, and we have
v²
V2
2
1
Now e-6
403
1
nearly. Hence, is within of V³, and v
1
400
,
within of V. Therefore while a sphere of twice the density of
800
the fluid, descends through 16 diameters, it acquires a velocity within
1
800
of the terminal velocity, and may after that be considered as
moving uniformly.
The terminal velocity
V=
√
k
√16A
16 A b g
3 D
8 Ab
And the height due to it is
3 D
Hence, if small spherical particles be dispersed in a fluid, they
will, as to sense, descend with uniform velocities, for the space at
first during which a sensible acceleration takes place, may be neg-
lected. And if the particles be very small, the velocities will be
1
F
392
very small. Thus, if they be
1
1000
of an inch in diameter, and
1
A=2D, we shall have b =
second.
of a foot: V=.24 feet per
6000
PROB. II. To determine the ascent of an air-bubble.
#
D-A
Δ
go where
In this case the body will ascend by a force
A is exceedingly small, being the density of atmospheric air. Also,
in consequence of the elasticity of the air, the spherule will vary its
dimensions as it ascends, and as the pressure upon it diminishes.
The density of air is directly as the pressure. Let AC, fig. 144, be
the depth of water which is equivalent to the pressure of the at-
mosphere on the surface of the fluid, CO the depth of the spherule
below the surface of the water; then, the pressure on the spherule
is equivalent to a column of fluid of the height AO. The density
of the spherule is as AO, and therefore, as the quantity of air in it
is given, its magnitude is inversely as AO. Let B be the original
place of the spherule, and its radius at B=b; its radius at 0=Y,
AB=a. Then, since the magnitude is as the cube of the radius,
bat
3
y³ : b³ :: a : x, y
=
7 3
Also, the density is as the pressure, and therefore, if c be the
density at B, A =
сх
a
Now, the force upwards is
D-A
Δ
cra
(Da-1) 8-
s Dv²
16 Ду
3 Da
16bc x
2
.v2.
ཀ॰
3
C
૩.
D
Let =m, a very small quantity; and
=n; then, we
166
shall have
393
vd v = -
dv=
2 vdo
a
· (−1) gdx+
2 n v² dx
a
nv²
.dr;
mx
my = -2(-1) gdr:
2
x 3
of which the first side becomes integrable, when multiplied by
6ns
€
Integrating, we have
v² €
6 nr
m
1
бns
773
a
= −2gƒ€ (-1) dr.
If we make x=z, the integral on the right hand side be-
comes
6 nx
3 a dz.e
бn s
fdz
m
·3fdz.z'e
m
M
The latter integral will evidently have each term multiplied by m,
and may be rejected, because m is very small.
gives
6ns
M
m
dze
The former term
6 ns
772
or
3 a
m
m
6 n
a €
2n
N
бns
m
€
+
Z
a
2n
S
6 n°
dze
x²
6 ns
m
2
And all the following terms in the integral will be multiplied by m,
and may be rejected.
v² €
Hence
B
6nx
m
=c+
6ns
ga e
m
n
x+
v²
ga
6
nxt
+ Cε
仇
​16 bga
3 xf
I
+ Се вот
3 D
394
When x= =a, v=0. Hence, O
Се
ga
8 bm
16b g + Cem,
0 =
3
v²
16bg (a
ga
(a
8 bm
€
3
x +
When x is very nearly equal to a, that is, when the bubble is near
to the point B, the latter term is considerable. But when a — x*
is not very small, this term becomes inconsiderable, because m is
very small, and therefore the index a large negative quantity. Hence,
after the early part of the motion, we have
16 b g
3
v²
16 bg a¹
a
で
​3
เ
√16bg
XT
APPENDIX (H).
General Mechanical Principles.
THERE are several properties of the motion of a system of
bodies, some of which Authors have attempted to establish by
peculiar reasonings, and to found upon them different branches of
Mechanics. Properly speaking, however, they are consequences of
the elementary laws of motion; and as they are remarkable either
in themselves, or for the facilities they offer in the solution of
problems, we shall here give the most important of such properties.
I. Principle of the conservation of the motion of the centre of
gravity.
The centre of gravity of a system of bodies moving freely,
moves in the same manner as if all the masses were collected in
the centre of gravity, and all the forces which act upon the system
were transferred to that point, retaining their magnitude and direction.
This is proved in the text, Art. 172, for the reasoning there
will be the same, whether the particles m, &c. be connected or not.
395
Hence, if the particles of a system be acted upon only by their
mutual forces, the centre of gravity will either remain at rest, or
move uniformly in a straight line.
For if mact on m', m' re-acts upon m with an equal force;
and these two forces transferred to the centre of gravity, would
destroy each other. Therefore the centre of gravity, acted upon
only by such pairs of forces, would be affected as if it were not
acted upon by any forces. Hence, it will either be at rest, or move
uniformly in a straight line.
It follows from this, that the momentum estimated in a given
direction, is always the same in a system acted on only by the
mutual forces of the parts. For if x, x, x", &c. be the distances of
the particles m, m', m", &c. measured from a fixed plane in the
given direction, x the distance of the centre of gravity, we have
x =
m'x'
mx + m²x² + m″x" + &c.
m+m+m" +&c.
dx
dx
dx'
dx"
(m + m² +m" +&c.)
=m
+ m²
+m"
+ &c.
dt
dt
dt
dt
dx
And
&c. are the velocities.
dt'
Hence, the sum of all the
momenta estimated in the given direction is equal to the momentum
of the whole system collected in the centre of gravity. And the
latter momentum is constant; therefore the former sum is so.
This may be called the Principle of the conservation of
momentum. What is said of mutual actions, refers either to pres-
sures, attractions, or impacts; and the latter, either supposing the
bodies elastic or inelastic.
II. Principle of the conservation of areas.
If a system of bodies moving freely is acted upon only by their
mutual forces, the sum of each particle multiplied into the pro-
jections of the areas described about a fixed point is proportional
to the time.
Let x, y be the co-ordinates of a particle m, from the fixed
point; and since the mutual action of the bodies will, in all cases,
396
produce equal and opposite moving forces, the moment of the
impressed forces will be 0. And taking the moment of the effective
forces as in Art. 129, we shall have
x d'y - y d²x
dt2
z d'x - x d²z
Σ.m
= 0; and similarly,
Σ.m.
= 0,
dt²
Σ.m.yd² z
yďz – zdy
dt2
= 0.
If a be the projection of the area described by m on the plane
of xy, we have
da
dt
xdy-ydx
2dt
d'a
dt2
xd³y-y d²x
2dt2
And if a', a", &c. be the same quantity for m', m", &c., we have
d2a
dt2
da
dt
Ση = 0, Σm = c, a constant quantity;
.. Σma=ct; or ma+m'a'+m"a" + &c. = ct.
Similarly, if ß, ß', ß", &c. be the projections of the areas described
by m, m', m", &c., on the planes of x z; Y, Y', y', &c. the projec-
tions of the same areas on the plane of yz,
mß+m′ß′+m"ß" + &c.=c't,
my+my+m"y" + &c. = c″t.
If, besides the mutual actions of the bodies, they be acted on by
forces tending to the origin of co-ordinates, the same proposition
will manifestly be true.
In the same manner it would appear, that if instead of the fixed
point, we take the centre of gravity, the same properties will be
true.
Let a line be drawn, making with the axis of z, of y, and of x,
angles of which the cosines are
C
2
c
n
C
√(c²+c²²+c²²)³ √(c°²+c²+c″²)' √(c®
√(c²+c²+c″²)3
397
and
It may
the position of this line will be the same at all points of time*;
a plane perpendicular to it is called the Invariable Plane.
always be found, when the motions of the particles are given; for
we have
da
m + mi
da
+ &c.=l,
dt
dt
d ß
dB'
m
+ m²
+ &c. =c',
dt
dt
dy
dý
m
+ m'
+ &c. =c".
dt
dt
In the same manner we may find a plane of invariable position
passing through the centre of gravity of the system. Thus a plane
passing through the centre of gravity of the solar system, and
determined by these formulæ, will, during all the motions which
the different bodies undergo, retain the same position.
III. Principle of the conservation of vis viva.
The vis viva of a body is the product of its mass into the
square
of its velocity t. The sum of the vires vivæ in any system is
* This plane will be that on which the sum of the projections of the areas
multiplied respectively into the particles which describe them, is the greatest
possible. Upon planes perpendicular to it, this sum is 0. Also, if A, A',
A", &c. be these projections for m, m', m", &c., and m A+m'A' +m"A" +&c,
=Σ.m A, we have
(Σ.mA)²=(Σ.ma)² + (Σ . m ß)² +(E. my)².
See Poisson, Traité de Mec. No. 84. Laplace, Mec. Cel. Liv. I. No. 21.
It is the same plane which is called the Principal Plane of Moments in
Art. 127.
+ The force or quantity of motion of a body is generally understood to
mean the product of the mass into the velocity, and is the same as the
momentum. The conservation of the quantity of force thus measured has
been proved in proving the conservation of the motion of the centre of gravity.
But if the force of a body in motion be measured by the whole effect which
it will produce before the velocity is destroyed, or by the whole effort which
has
398
the same as if its particles, being separate had been acted upon by
the same forces through the same spaces.
Let P be the force which acts on the particle m, p its distance
dp
from a fixed point in the direction of the force; therefore is m's
dt
velocity in the direction of the force. Also, if q be the distance of
dq
the body from a fixed point in the direction of its motion, is its
dt
ďa
velocity, and the effective accelerating force. And if we make
dt
similar suppositions with respect to m', m", &c. we may consider
the velocities of m, m', m", &c. as virtual velocities, since they are
consistent with the connexion of the system; and we have
impressed forces, Pm, P'm', P"m", &c.
with virtual velocities
dp dp
dt' dt d t
dp"
&c.
و
effective forces
md¾q._m'd°q′_m"d²q″
2
&c.
dt
dť²
dt²
dq dq
dq"
with virtual velocities
&c.
dt dt d t
Hence, since these forces must balance each other, we have by the
principle of virtual velocities, (see Statics, Art. 105.),
dp
dp'
mP.
+ m'P'.
d q d³g
2
+ &c. = m.
+ m²
dt
d t
dt³
3
dq'd°q'
dt's
+ &c.
3
{
has been exercised in generating it, without regard to the time, it must
be measured by the mass into the square of the velocity. Thus balls
of the same size projected into a resisting substance, as a bed of clay, will
go to the same depth so long as their weights into the squares of their
velocities are the same. Force thus measured is called vis viva, in oppo-
sition to force measured by momentum, which is proportional to the pressure,
or dead pull, producing it. And it appears from the text, that forces will
always produce a certain quantity of vis riva by acting through a given space,
whatever be the manner in which the bodies are constrained to move.
399
or, Σm.
d q d'q
dt2
-
Σm Pdp...
And multiplying by 2 and integrating, and putting v for
Σ.mv² = C+ 22. mf Pdp
….(1).
dq
dt
(2).
Now 2fPdp is the square of the velocity which the force P
would have generated in a point separated from the rest; hence,
the integral being taken between the same limits, the vis viva is the
same as it would have been in that case.
Let x, y, z, be the co-ordinates of m, and we shall have the
square of the velocity, or
212
dx² + dy² + d=2
dt
Also, if X, Y, Z, be the resolved parts of P, parallel respectively
to x, y, ≈, and if a, ß, y, be the co-ordinates of a fixed point O,
in the direction in which this force acts, so that Omp, we shall
have
X α
X = P.
Y = P.
p
7 - B
p
z = P.
2
p
Hence,
And p² = (x − a)² + ( y − ẞ)² + (= − y)²;
.. pdp = (x − a) dx + (y-ẞ) dy + ( − y) dx.
d?+Ydy+3d:=P.
(x − a) d x + ( y − ẞ) dy + (: − y) d =
= Pdp.
Р
By substituting these values, the equation before obtained becomes
Σ.m.
dx² + dy² + dz²
dt²
=C+92.mf(Xd +Ydy+^d≈).
If the system be acted upon by no forces, we shall have
Σ. mv² = C.
the sum of each particle multiplied into the square of its velocity,
will always be equal to a constant quantity.
400
If the system be acted on by gravity only, let x be vertical, and
we shall have
Σmv² = C — 2Σ.mgx=2Σ.mg (h—x),
h being the height from which m must have fallen to acquire its
velocity at any point.
This Proposition may be employed for the solution of a variety
of Problems respecting bodies acted on by gravity. For instance,
the greater part of the Examples in Chap. VI. Book I, might have
been readily solved by means of it.
The mutual action of the parts of a system may increase or
diminish its vis viva in any degree. Their attraction or repulsion
may augment the velocities, and consequently, the vis viva. Their
collision will generally diminish the vis viva, except they be per-
fectly elastic, in which case, after the impact, the vis viva will be
the same as before.
As the vis viva varies, it may become a maximum or minimium.
This happens when the system passes through a position of equi-
librium.
When the vis viva is a maximum, the body passes a position of
stable equilibrium, when it is a minimum, it passes a position of
unstable equilibrium. See Poisson, Traité de Mec. No. 472, &c.
IV. Principle of least action.
The action of a particle is here measured by the product of the
If the
momentum, and of the space through which the body moves.
velocity be perpetually varying, the action will be the integral of
this product taken for a differential of the space. If m be the mass,
the velocity, and s the space, the whole action of the body is
Smods, where the integral is to be taken between the beginning
and end of the motion. And the principle to be proved is this:
When a system moves in any manner, the actual motion is such,
that the sum of the actions of each particle for the whole motion is
less than if the particles had taken any other paths between the same
points.
401
That is,
mods is a minimum.
To prove this, we must, according to the Calculus of Variations,
prove that the variation of this integral, between the proper limits,
is 0. See Lacroix, Elem. Treat. Art. 351.
In the preceding proof of the principle of vis viva, we have
dq
dp
taken the actual velocities
&c.,
dt
dt
>
&c. for the virtual velo-
cities. But if Sp, &c., dq, &c., be any possible corresponding
variations of p and q, they will be as the virtual velocities; and we
shall have, instead of equation (1), this equation,
Ση
dt²
&q=Σm Pdp..........(3).
8q=2mΡδρ...
Also, as in p. 399, it will appear, that
Pop
=
Xôn+Yông+2.
2
And since dq²= d x² + dy³ + dz², differentiating, &c.
ď² q
d2x
dy
dz
.dq
dx
dy
dt2
dt²
d
+ May + =
dz.
dt²
Hence, considering the first side as the differential of a function of q,
and the second as the differential of a function of x, y, z, we shall
have the variations in the same manner, or
d2q
Sq
=
dt
dt
d²x
d²y
8x +
dt
d²z
Sy + zdi.
dť²
2
Hence, equation (3) becomes
dx
d2z
Ση
Sx
бу
dt
dt
dt2
{d8r+by+d=} = 2m Pôp.
Now &fmvds=f2m dvds;
and Σm dvds=Σmvdds+Emdsdv.
dt
And ds=vdt; ..ds&v=dt.v&v=
d. v²
;
d t
. . E mds & v ==
Σmd.v².
2
3 E
402
2
But Σm v² =
.. Emd. v² = 2Σ.m P&p
C+2 Σ. mf Pdp;
[d²x
d²y
= 2Σ.m
8x +
бу
d t²
d ť²
(dy + d z dz},
бъ
dto
2
Also d s² = d x² + d y² + d z² ;
d s
x
dds =
dråd
8 d x
+
dy
t
dt
x
or vdds =
Hence, transposing &d, and adding;
d x
d 8x +
dy
d t
d t
d Σ mv ds = Σ .m
d²x
Ꮖ
8x +
dy sy
d t
dz 8 d x + dy ddy + d28 dr.
d t
d t
Emds&v+Emvdds, or
d d y + √ z d d z
+
14 8dy + 178dz,
dt
dz
d t
бу
d t
d t
d x
= Σ.md
dt
x
dy
Sx + бу
бх
= dz.m (dz 8x +
dt
&
dt
+
dt
z
S
2
бъ
dy 8y + d=8= }}
d t
. . dƒ & mv ds = ƒd Σm vd s
t
dz
dt
d2z
E
d t
d z
= Σ.m
S d dx
dy
d z
бх
x + dy+ δε
d t
d t
d t
Now the right hand side is to be taken for the limits of the
motion of the points m, m', &c.; and these limits being fixed, dx,
Sy, dz, are each 0.
Hence, dfmvds=0, and fΣmvd s is a minimum.
v²
It follows from this, that the sum of the vis viva of a system in
passing from one position to another, is a minimum. For fmvd s
=f2mvdt. Hence, this sum of the vis viva with respect to the
time is less than if the system had moved in any other manner to
the same position.
When the particles are not acted upon by any accelerating force,
the vis viva is constant. Hence, the sum of the vis viva for any
403
time is proportional to the time. And hence, in passing from one
position to another, the time actually employed is the least pos-
sible.
The principle of least action was announced by Maupertuis as
a fundamental law of Mechanics, and attempted to be proved
a priori. It appears from what has been said, that it is a conse-
quence of the elementary laws of motion.
ERRATA.
Page 48. I. 18 in denominator, for cot. ß read 2 cos. ß.
125. 1. 6 from bottom, for pg read.pq.
130. 1. 7 for
read +.
1. 9 for greatest read least.
1. 14 for least read greatest.
132. 1.
227. 1.
3 from bottom, for ¹E sin. ¹e read ¹E sin.'e, and for
2 E sin.² €.
2 E sin.² e read
1 in numerator, for M. Cm read M.Cn.
1. 9 from bottom, the + should be in the denominator.
229. 1. 11 for into its read into the square of its.
279. 1. 10 for U cos. ↓ read U cos. p.
303. 1. last but one, for fxy dm read Exym.
314. 1. 5 of note, for Multiplying read Multiply.
321. 1. 8 dele be.
1. 3 from bottom, for oscillate read revolve.
330. 11. 5 and 8 for
√
read
347. 1. 12 for in the next read in the following pages.
377. 1. 9 in denominator, for m² read m.
387. 1. 5 from bottom, for four read two.
393. 1. 12 at the beginning, insert or.
!
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22
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28
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29
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30
31
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PLATE N

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36
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35
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PLATE IV
99
103
100
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106
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UNIVERSITY OF MICHIGAN
3 9015 06708 0831
Form 176(8-22-21)30M
B 448346
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