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Cornell Hniversity Library

THE EVAN WILHELM EVANS
MATHEMATICAL SEMINARY LIBRARY

THE GIFT OF
LUCIEN AUGUSTUS WAIT

Maas MATHEMATECH og giay vo |

 

7348-1

 
wii
DYNAMICS.
Cambringe:
PRINTED BY C. J. CLAY, M.A. AND SON,

AT THE UNIVERSITY PRESS,
A TREATISE

ON

DYNAMICS.

BY

W. H. BESANT, D.Sc, F.RB.S.

LECTURER OF SsT JOHN’S COLLEGE, CAMBRIDGE.

LONDON:
GEORGE BELL AND SONS.
CAMBRIDGE: DEIGHTON, BELL AND CO,
1885

+r
PREFACE.

THE object of the present Treatise is to introduce the
mathematical student to some of the earlier and easier
branches of Kinematics and Kinetics.

In the chapter allotted to Kinematics, I have deduced the
expressions for velocities and accelerations, as far as possible,
from the definitions and axioms of the subject.

In the applications to Kinetics, or, in other words, in the
combination of these expressions with the Laws of Motion,
for the determination of the motion of a particle or of a
system, I have adopted the same plan of operations.

I have assumed, and made free use of, the methods of
Analysis, for the performance and simplification of the
requisite calculations.

The methods employed, and the order of thought which
is followed, are those which during my experience as a teacher
I have found to be most effective in the elucidation and
development of the ideas of Kinetics.

The majority of students do not easily or rapidly absorb
general ideas, and they are most effectively assisted by the
gradual development of a subject through simple cases, and
illustrative examples.
vi PREFACE.

With this view I have endeavoured to explain the
application of the Laws of Motion to the determination of
the motion of a particle and of systems of particles, com-
mencing with easy cases, and leading up to a few of the
interesting and important cases of the motion of a body in
space.

My especial object has been to illustrate the direct appli-
cation of the Laws of Motion, and thereby to produce a
treatise of an elementary character, but of Educational utility
to the student who is commencing the study of theoretical
Kinetics. .

Iam very much indebted to Professor W. H. H. Hudson, of
King’s College, London, for useful suggestions and criticisms,
and to Mr G. B. Mathews, Professor of Mathematics at the
North Wales University, and Mr J. Brill, B.A. of St John’s
College, Cambridge, for their kind assistance in the correction
of manuscripts and proof sheets.

I venture to hope that the explanations and illustrations
of the text, and the numerous examples appended to the
several chapters, will be of assistance to the student in
mastering the elementary ideas of the subject, and pave
the way for the consideration of the higher branches and
the more difficult problems of the great science of Dynamics.

W. H. BESANT.

September 16, 1884,
ARTICLES
1

3—30

31—53

54—74

75—89

CONTENTS.

CHAPTER I.

PAGE

Introductory . ‘ : i 7 ‘ ‘ 1,2
CHAPTER II.

Differential Equations ‘ : ‘ : j ; 3—5
CHAPTER III.

Kinematics. : q ‘ ; : . : . 6—31

Examples . ‘ s . 7 5 : ‘ ‘ 31—37
CHAPTER IV.

Laws of Motion . 2 4 . 3 ‘ ‘ 38—51
CHAPTER V.

Rectilinear Motion . Fi ‘ ‘ . 5 ‘ 52—68

Examples . 3 7 3 . S 5 ‘i ‘ 69—82

CHAPTER VI.

Accelerations Parallel to fixed Axes . ‘ “ « » 88—95
Examples , . , rr Sr ro

ail
Vili

ARTICLES
90—121

122—140

141—157

158—168

169—176

177—201

202—229

230—251

CONTENTS.

CHAPTER VII.

Radial and Transversal Accelerations
Examples

CHAPTER VIII.

Tangential and Normal Accelerations
Examples

CHAPTER IX.

Motion in Three Dimensions
Examples

CHAPTER X.

Hodograph and Brachistochrone
Examples * Z %

CHAPTER XI.

Two Particles acting on each other .
Examples

CHAPTER XII.

Energy and Momentum
Examples

CHAPTER XIII.

Equations of Motion .
Examples

CHAPTER XIV.

Motion of a system in Three Dimensions.

Examples

PAGE
104—181
131—144

145—165
165—180

181-196
197—204

205—215
215—219

220—225
226—228

229—248
248—257

258—286
286—298

299—324
325—334
DYNAMICS.

CHAPTER I.

1. THE problems usually discussed under this head are
those which relate to the geometrical connections between
given motions, or given kinds of motion, and those which
relate to the action of forces, and the motions and changes of
motion produced by forces.

The former belong to pure science, and deal with the
geometry of motion, a branch of mathematics to which the
name Kinematics was applied by Ampére.

Strictly speaking the word Dynamics includes Statics,
the discussion of the equilibrium or balancing of forces, and
Kinetics, the discussion of the effects of forces on the motion
of bodies. Mechanism belongs to Kinematics, including such
problems as result from considering trains of wheelwork or
any connected machinery.

To Kinetics belong the consideration of the forces setting
such machinery in motion, or keeping it in motion, the
problems of Physical Astronomy, and others of important
practical application.

We shall commence by a development of the formule of
Kinematics, and afterwards proceed to consider the applica-
tion of the formule, and of the Laws of Motion, or Laws of
Force, to the determination of the motion of a particle, and
of a system of particles, produced by the action of given
forces, or, conversely, of the forces required to produce given
motions. The idea of a particle, or of a material point,
capable of being set in motion, or of having its motion

B. D. 1
2 DYNAMICS.

affected, by the action of force, is a mathematical abstraction
leading to the simplest forms of Kinetics. The determina-
tion of the motions of the bodies constituting the Solar
System belongs to this class in virtue of the facts that the
Planetary Bodies are nearly spherical in form, and that their
dimensions are very small in comparison with their distances
from each other and from the Sun.

Moreover the mathematical idea of a solid body is that of
a system of particles, and the discussion of the motion of a
single particle therefore naturally precedes the discussion of
the motion of a body or system of particles.

It will be seen that Newton’s Laws of Motion connect the
action of a force ona particle with the accelerations produced,
and lead to the formation of differential equations, the
integrations of which give the solution of the problem of
determining the motion.

It will appear further that Newton’s Laws are sufficient
for the determination of the motion of a system of particles
or bodies, whether rigidly connected or not, and lead, in a
similar manner, to systems of differential equations containing,
in their solution, the motions of the body, or of the various
bodies of the system.
CHAPTER II.
DIFFERENTIAL EQUATIONS.

2. THERE are certain differential equations which occur
so frequently in the discussion of questions in Kinetics, that
we think it worth while, for convenience of reference, to give
a brief solution of them.

(1) The equation, a +yf' («) =F (a), is at once solved
by the integrating factor

F(z)
es
leading to ye f@ | F(a) oe” dat C.
For example, if ay +—_ =tan2, the factor is tan 5, and
> da ~ sin & 2? 2’
therefore
y tans = 2tan 5 — a +C.
(2) TY + nly =0.
Multiply by 2 dy and integrate, then

dx
ty) eee a x da _,1 1
(Z =n’ (ce —y") or dy *n Jeay
". y =ccos (nx+a),
or y =A cos na + B sin na.
4 DIFFERENTIAL EQUATIONS.

(3) TY _ hy =0,
As before,

 

(y= nv (y? +e jy Oak ;
dz dy ~mn [yt +

-, ne=log ——~7—,
from which we obtain
y= A e+ Be™.

The equation oy + n?y = aa +b is reduced to one of the
two preceding by assuming

 

wr 4(QYeq=Fq), let $=

dy _ dz 7@ = d.2
da? da “dy ?° dy’
and the equation becomes, putting v for 2’,

Ft 2S) e= Fy)

which is of the ag (1).

then

dy dy
(5) ne ae

The solution of this equation is effected by the calculus
of operations, leading to

e”™™ D -1
Utena +5(1+ 5045)

D representing the operation © , and the expression affecting

+ by =e" 4 a".

x” being expanded in ascending powers of D,

The complementary function Ae + Be** must be added,
a and @ being the roots of the equation.

m+am+b=0.
DIFFERENTIAL EQUATIONS. 5

If the roots of this equation are imaginary and of the form
a+B4/1, the complementary function takes the form
e*” (A cos Ba +B sin Bz).
“q2

(6) a — n'y = COs ra.

The calculus of operations, or the variation of parameters,
gives as the integral,

 

=— eat + Ae’ + Be.
(7) ae n'y = cos re.
The solution is
= == + A cos nx + B sin nz.

If n=r, then, writing the solution in the form
es a as vas +Acosnz+Bsin na,
n—r :
and finding the limiting value of the fraction when n=7,
we obtain

y=5-sinne +A cos nv + Bsin nex.

72,
(8) oY +a BY + by = cosa.
This leads to

_ D-aD+b = cos ra
I=(p by ae D cos ra = (D — aD +b) Gary aie

the complementary function of course being added ;

_._ (b=7") cos ra + ar sin rx

19 barca

a and @ being the roots of the equation,
2+ az+b=0.,

+ Ac + Bef,
CHAPTER III.

KINEMATICS,

3. Definition. The velocity of a moving point, when
uniform, is measured by the number of units of length passed
over in the unit of time.

If the velocity be not uniform, it is measured at any
instant by the space which would be passed over in the unit
of time if the velocity were to remain the same as it is at
that instant.

To express this idea mathematically, let s be the space,
that is, the number of units of length passed over by the
moving point in the time ¢, and s+ ds the space passed over
in the time t+ 8, so that 8s is the space passed over in the
time 6¢, and, if d¢ be so small that the velocity is not sensibly

changed in the time &t, the limiting value of the expression 7

that is é is the measure of the velocity.

Or we may argue as follows, If v be the velocity at
the time ¢ and v + év at the time ¢ + é¢, then 8s lies between
vot and (v + dv) dé, and therefore ultimately,

v= di .
4, Ifthe velocity of a moving point be variable, it is said
to have positive acceleration, if the velocity be increasing,

and negative acceleration, or retardation, if the velocity be
decreasing.
KINEMATICS. 7

Tf the rate of increase of the velocity be uniform the
acceleration is measured by the increase of the velocity in
the unit of time, and, if variable, it is measured at any
instant by what would be the increase of velocity in the
unit of time if the rate of increase were to remain what it is
at the instant in question.

Mathematically, if » be the velocity at the time ¢, and
v+6v at the time ¢+6¢, the measure of the acceleration,
: see oee «x» OY
in the direction of motion, is —.

dt

Or, if f be the acceleration at the time ¢, and f+ 6f at the
time ¢ + ét, dv lies between fot and (f+ 5/) 5, and therefore
ultimately,

dv

=e

5. The composition and decomposition of velocities and
accelerations.

Supposing a moving point to possess, or to have
impressed upon it, two velocities in different directions,
there arises the question ; what is the resulting motion ?

To solve this kinematical question, we must invent
machinery to represent coexistent velocities.

BF D

A E

Imagine then a point to move uniformly along a straight
line while the line is carried parallel to itself at a uniform
rate. The point will then have two coexistent velocities.

Let EF be the moving line, and, while the point moves
from £ to F, let the line move from AB to CD. Then PE
is to HA as the velocity of the point along the line is to the
velocity of the line, that is, as CD to AC;
8 KINEMATICS,

Therefore APD is a straight line, and the point P actually
moves from A to D.

AD therefore represents the resultant velocity in
magnitude and direction.

This proposition is called the Parallelogram of velocities.

In the same way, if a point have two coexistent accelera-
tions, taking AB and AC to represent the velocities added
per unit of time, or which would be added per unit of time,
due to the accelerations at the instant in question, it follows
that AD represents the resultant velocity superposed, or
which would be superposed, per unit of time.

This is the Parallelogram of accelerations.

Conversely, any velocity or acceleration, represented by a
line AD, can be decomposed into two velocities or accelera-
tions, AB, AC, in any assigned directions,

6. Change of wnits in the measures of velocities and
accelerations.

If v be the measure of a velocity, the meaning is that v
units of length are passed over in the unit of time.

If a feet and ¢ seconds be the units, and if v’ be the
measure of the same velocity when a’ feet and ¢’ seconds are
units, it follows that

av_av
or?
for each expression represents the velocity in feet per second.

If f be the measure of an acceleration when a feet and ¢
seconds are units, the meaning is that

the velocity per ¢ seconds added in ¢ seconds = fa in feet;

*. velocity per second added in ¢ seconds =~,

d

*. velocity per second added in one second 2 .
If f’ be the measure of the same acceleration when a@

feet and ¢’ seconds are units, it follows that Le is the measure
KINEMATICS, 9

of the same acceleration referred to a foot and a second, and

. fa _fe
ee Ee

= f°

7. Angular velocity and angular acceleration.

If a straight line turn round in a plane it is said to have
angular velocity, and if this angular velocity be variable it is
said to have angular acceleration.

If @ be the inclination, at the time #, of the moving line
to any fixed line in the plane, then, exactly as in Articles (8)
and (4) the angular velocity is de or 6, and the angular

a 2 os
acceleration is ay or 0.

It must be observed that this is quite independent of any
motion of translation which the line may have, and simply
measures the rate of turning round,

When we speak of the angular velocity of a point P,
moving in a plane, about a fixed point O in the plane, we
really mean the angular velocity of the straight line OP.

If the velocity and direction of motion of P be given, a
simple expression can be obtained for its angular velocity
about a fixed point 0.

_ For if OP=,, and if p be the perpendicular from O on
the direction of motion of the point, and v its velocity, the
angular velocity is equal to the resolved part of the velocity
perpendicular to OP divided by OP, and therefore

aye ap al?
r r

8. Expressions for accelerations.

If z, y, 2 be the co-ordinates of a point, referred to a
system of fixed axes at right angles to each other, x, y, and z
are the distances of the point from the planes of yz, 2a,
and xy,
10 KINEMATICS.

The velocity parallel to x is thg rate of increase of the
distance from the plane yz, and, as in Art. (3), is represented

by - or by 4, employing fluxional notation.

And, similarly 4% and =

parallel to y and z.

If u, v, w be these velocities, the accelerations parallel to

the axes are, by the same reasoning as in Art. (4), = i 7 , and

, or y and 4, are the velocities

: ay. and

dae dy az
dt’ dt

dt’?

- ,or w%, 0, and w; that is, are or &, i,

and 2.

We shall in all cases limit the use of the symbols 4, # to
the case in which the time is the independent variable.

9. It must be carefully observed that the velocity of
a point in any direction is the rate of change of the distance
in that direction, and is equal to the limit of the change of
distance divided by the change of time, when that change
is indefinitely small.

And similarly, the acceleration in any direction is the rate
of change of the velocity in that direction, and is equal to
the limit of the change of velocity divided by the change of
time, when that change is indefinitely small.

10. Radial and transversal velocities and accelerations.

 

 
KINEMATICS. 11

Let r, 6 be the polar co-ordinates of a moving point,
and u,v the radial and transversal velocities, that is, the
velocities in direction of OP and perpendicular to OP.

P being the position of the point at the time ¢, and P’ at
the time ¢+ d¢, and if OP’ =r + 6r,

OP'cos80— OP _dr_.
St =a
OP'sin 80 rd0 _ a
3% dE

If u+ du, v + 6v, be the velocities at P’ in direction of

and perpendicular to OP’, acceleration in direction OP

(u + Su) cos dO — (v + dv) sin dd —u

u=limit of

 

 

and v=limit of

 

= limit of SE
_d_ db
dt dt

2, .
=5a—0 (G) at o8

acceleration perpendicular to OP
= limit of 2 + 6v) cos dO + (w+ du) sin dé —v
~ «bt
dv dO dO, drdé
ata ae tae a

4, 9291 2 (.d0
=7 0+ 270 =" a" a

It should be noticed that, ifr =0, that is, if the moving
point is passing through the origin, these expressions are

¥ and 276.

 

11. The expressions for radial and transversal accelera-
tions may otherwise be obtained in the following manner.

The component accelerations parallel to w and y being #
and ¥,"
12 KINEMATICS.

the radial and transversal accelerations are respectively
& cos 6 + ¥ sin @, and ¥ cos 6 — #sin 0.
Putting r cos @ for 2, and rsin @ for y, these expressions

become i . .
r—ré*, and ré + 276.

12. Case of uniform motion in a circle.

If x and @ are both constant, and if @ = a, the transversal
acceleration vanishes, and the radial acceleration =—o’r;
that is, the resultant acceleration is directed to the centre
of the circle and is equal to the radius multiplied by the
square of the angular velocity.

13. Velocities and accelerations referred to two axes, at
right angles to each other, moving in a plane about the origin.

 

If Ox, Oy be the positions of the axes at the time ¢,
Oz’, Oy' at the time t+ dt, 2Oz' = 60, and uw, », velocities
parallel to x and y at the time ¢,

NL (a + da) cos80—(y+dy)sind0—a« dx dd
ae ag St ~ at 9 Gt’
oe + 8x) sin 86 + (y + 8y) cos 80— y _ =¥ do
8t ft” ag

 

 
KINEMATICS, 13

If u+6u, »+8v be the velocities of P’ parallel to 2’
and y/,
acceleration parallel to Ox
L (u + du) cos 86 — (v + Sv) sin 86 —u
= Lt di ,
ai ,d8_ ae, (d_ yO gd
“ab de ae * (ai ve ae dt’
and acceleration parallel to Oy
=r, et™ sin 86 + (v + dv) cos 86 —v

 

 

ot z
as dv do d’y dey? ao dx dé
= ae 9 () bea ee

In fluxional notation, the velocities are
&— 6, and y¥ +26;
and the accelerations are
u—v6 and 6 +u6,
or a — yO — 26 —2y6, and 7 + 26 — y@ + 2006.
14. Case of a point moving on the surface of a right
circular cone.

 

 
14 KINEMATICS,

PN being perpendicular to the axis, let OP=r, and the
angle between the moving plane VPO and a fixed plane
“AON =6, and let Q be the projection of P on a plane
through O perpendicular to the axis of the cone.

d’.ON  d’r

 

Acceleration parallel to ON = Geo ype Oe (1),
acceleration in direction NP = that of @ in direction OQ,
d’. OQ dQ\*
=“ *—09 a) , by Art. (10)..eeseees (2),
dr. ; ;
= Gpsin a—rsina (J) ;
and perpendicular to the plane VPO
=, dt Of es | , a0
Sane (r sina) =—— a (« ay

Multiplying (1) by cos a, (2) by sin a, and adding, we find
that the acceleration in the direction OP

2 2
=r sinta (3) =f—rsin’ a6’.

Multiplying (1) by sina, (2) by cos a, and subtracting, the
acceleration in the direction of the normal PG to the surface

dé\? .
=rsinacosa () =frsin a cos a6”,

15. Case of a point moving on the surface of a sphere.

 
KINEMATICS. 15

OAC being a fixed plane, and OC a fixed radius, and PN
perpendicular to OC, take ¢ as the angle between the planes
OAC and OPC, and 6 as the angle COP.

The accelerations in the directions, ON, NP, and per-
pendicular to the plane OPC, are respectively

a a, . J dd\?
ap (4 08 8), yp (asin 8) — asin 0 (2) :

Tk Py ag Sl

and zane a (2 sin OF ‘

Multiplying the second of these by cos 0, and the first by
sin @, the difference of the products is the acceleration in
direction of the tangent at P to the meridian curve CP, and
is equal to e :

. af ~ a sin 6 cos 0. ¢’.

Similarly the acceleration in the direction PO

=a? + asin’. ¢’.
16. In the general case in which the position of a

moving point at any time is defined by the polar co-ordinates
r, 0, p, the accelerations of P in the directions perpendicular

 

 
16 KINEMATICS.

to Oz in the plane CPA, and perpendicular to that plane are
the same as the accelerations of Q, and are therefore
2 \ 2 :
= (r sin 0) —r sin 6 (2) and ao Gi sin? 0 a ‘
the former in the direction NP and the latter perpendicular
to the plane CPA.

Also the acceleration parallel to Oz
@.ON @
= = A cos 6).

If p, t, « be the component accelerations in the directions
OP, PT perpendicular to OP in the plane CPA, and
perpendicular.to that plane, it follows, by resolving’ the above
accelerations, that

p= — re? = r¢? sin? 0,
T=70 + 276 —rd* sin 0 cos 6,
o=rd¢ sin 6 + 27¢ sin 0 + 2rG6 cos 0.

_ 17. Tangential and normal accelerations of a point moving
in a plane curve.

E

 

 

 
KINEMATICS. 17

If v be the velocity at P of a point moving in a curve,
and if s be the arc OP of the curve measured from some
fixed point O,

y=
~ dt"
The tangential acceleration at P
= (v + dv) cos 86—v_dv_ d's
=n 8t ~ de at
and the normal acceleration :

= (v+6v)sindp¢ _wdd_vdd ds_v°
ae ae aE pe

if p be the radius of curvature'‘at P.

 

If the motion of the point be uniform motion in a circle,
v and p are constant, and, if w be the angular velocity, v= ap.
The tangential acceleration is then zero, and the normal
acceleration, measured inwards, is equal to op, as in

Art. (11).

18. Accelerations of a point moving in a tortuous curve
in directions of the tangent, the principal normal, and, the
binormal.

If x, y, 2 be the coordinates of the point referred to fixed
rectangular axes, we have,

dr _de ds
dt ds dt’
Vx dsd«, /ds\ dx
and therefore a See t (5) a
ay _ ds dy , (ie) ay
at dt? ds (a ds*’

de de de, (ib)
dé dé ds" \dt/ ds**
18 KINEMATICS.

Now oe ‘ ay fe are the direction-cosines of the tangent,
ds’ ds’ ds
and if p be the radius of absolute curvature,
da dy az
Pp ds’ » P ds? » p ds? >
are the direction-cosines of the principal normal.
The above equations therefore prove that the resultant

: a  @
acceleration of the point is compounded of the acceleration on

2
in direction of the tangent and of the acceleration : (=) in

direction of the principal normal.

It follows at once that there is no acceleration in direction
of the binormal.

Or, the direction-cosines of the binormal being pro-
portional to

dy@z dz@y dzd«a dx’d@z dad’y dy d«

ds ds* ds ds*’ ds ds* ds ds*’ ds ds* ds ds*’
if we multiply the above equations by these three quantities
respectively, and add them together, the right-hand member

vanishes identically.

19. These results can also be obtained by the consider-
ations that the osculating plane is the plane containing two
consecutive tangents, and that the consecutive osculating
plane is obtained by an infinitesimal twist round the tangent.

The circles PA, PA’, in the accompanying figure, are
consecutive circles of curvature, the angle between their
planes, 67, being the angle of torsion, and the circles being,
in general, small circles on the sphere of curvature. The
circles may be in certain cases coincident, or either of them
may be a great circle.

If P' be a consecutive point on the circle PA’, de be the
angle of contingence, that is, the angle between the tangents
at Pand P’, v the velocity at P, and v + du at P’,
the changes of velocity in directions of the tangent PT,
the principal normal P#, and the binormal are
KINEMATICS, 19

(v + 8v) cos Se — v, (v + dv) sin Se cos dy,
and (uv + 8v) sin de. sin 87;

 

 

dividing by S¢, we obtain in the limit, the expressions
dv vw
= and —
at p
from the first two, and the ratio of the last expression to
o¢ vanishes in the limit.

The Principles of Relative velocities and Relative accelera~
trons.

20. If P be a moving point, A, B, C,...other moving
points and O a fixed point, the velocity.of P in any direction
is the sum of the velocities, in the same direction, of P
relative to A, of A relative to B, of B relative to C, and so
on, and of the last moving point relative to O.

For if u be the velocity of P in the direction considered,
U,) Uy)-..U, of the moving points in the same direction,

Us (U —U,) + (U, — Uy) Fereeee + (thy, — Up) + Uns
which establishes the statement.

The same principle is equally true of accelerations,

du_d d :
for Ti = age ~ tq aa Mt aise vy |
that is the acceleration of P is the sum of the relative
accelerations.

21. By aid of the foregoing principle, the expressions of

Art. (18) are at once obtained.
2—2
20 KINEMATICS,

For u=velocity of P relative to NV + velocity of NV

=—y¥6 + &,
and so v=y+20.
Again, the acceleration of N in the direction Oz
=#— 26",
and that of P relative to V in the same direction
=~ (y8 + 2y8);

therefore the acceleration of P in the direction Ox
= & —«f — 6 — 246;

and similarly the acceleration of P in the direction Oy
= ij — 6 + 26 + 200.

22, We can also obtain the expressions for the accelera-
tions in the following manner.

 

The velocities of a moving point P, at any time ¢,
parallel to two moving directions Ow, Oy can be represented
by u= ON, v= QN,
and the accelerations parallel to 2 and y will be the rates of
change of « and v in those directions and will therefore be

u%—v0, and 6+.

Observing that ; ;

u=a@—y6, andv=y+26,
we obtain the expressions given at the end of the last
article.
KINEMATICS. 21

23. Component velocities and component accelerations
of a@ moving point referred to two axes turning round the
origin nm any given manner.

 

Let 6 and ¢ be the inclinations, at any instant, of Ox and
Oy to some fixed line in the plane, and represent by w the
angle «Oy or ¢ — 0. :

The velocity of P relative to V is yp perpendicular to
PN, and this, by the triangle of velocities HPN, decomposes
into — yd cosec w, and yd cot a, ;
parallel to # and y. '

The velocity of V in direction Ox is #, and perpendicular
to Ox is «6.

The latter, by the triangle of velocities ONF, decomposes
into —26 cot a, and 26 cosec a,
parallel to # and y.

Hence, the velocity of P being compounded of its
velocity with regard to N, and of the velocity of NV, the
components parallel to # and y are respectively

& — «0 cot w — yd cosec a,
y + 26 cosec w + yd cot w.

Tn exactly the same manner the component accelerations

are at once seen to be
22 KINEMATICS.

a — 06" — (a6 + 2a) cot w — (yp + 29d) cosec w,
4 — yO + (wb + 246) cosec w + (yp + 2y¢h) cot

24. The foregoing expressions can be obtained in a
different manner.
Thus if u and v be the component velocities,

weos 0+ 000s = 5 (aw cos O-+y cos $),
usin d-+vsin = 4 (sin 8 + ysing),

and if f, f’ be the component accelerations,

foosd+f’cosp= 5 (a cos @ + y cos ®),

2
f sin @ +f’ sin = Llesin 6 +ysin 4).

The solution of these equations will give the expressions
for u,v, f and f’.

25. Particular illustrations of the use of the principle of
relative velocities and relative accelerations.

(1) A point P describes an equiangular spiral with
uniform angular velocity round O, and a point Q describes an
equal spiral with the same angular velocity round P; tt is
required to find the path of Q.

Since r = ae®ets, we have

$=seca.r=r6 cosec a,
so that, if ~OP be the velocity of P, in the direction PT,
pPQ is the velocity of Q relative to P in the direction QF,
the angles OPT, PQF being equal and constant.

Hence OQ is the actual velocity of Q.

Further if QE, QF represent the component velocities in
direction and magnitude, the resulting velocity of Q is
represented by QG@ the diagonal of the parallelogram, and the
angle OQG = PQG — PQO=PQOF+ FQG — EGQ= PQF;
hence it follows that the path of Q is an equal spiral.
23

KINEMATICS.

s
ane ona R
ty

a

oO
A circle rolis uniformly inside a circle of double its

(2)
radius; it is required to find the acceleration of any carried

pownt.

 

ed

r
If P be the carried point on the radius CB, and if OB
produced meet the circle in A, it follows, since the angle
QCB = 2 COB, that the arcs QA, QB are equal.
24 KINEMATICS,

Therefore A is a fixed point and B moves along the line
AOA’.

Since OC, CB make equal angles with OA it follows
. that the angular velocities of the lines OC, CP are equal.

Hence, if w be this angular velocity, the acceleration of C
is in the direction CO and =o’. CO, and the acceleration of
P relative to C is in the direction PC and = w*. PC;

therefore, by the triangle of accelerations, the resulting
acceleration of P is in the direction PO and = w’. PO.

(3) A circle rolls on a straight line ; it is required to find
the acceleration of the point of the circle in contact with the
line.

If 6 be the angle through which the circle has rolled
from any assigned position, and a the radius of the circle, a9
is the linear space traversed by the centre C and therefore a6
is the acceleration of the centre.

The accelerations of the point of contact P, relative to the

centre are a” in direction PC, and a parallel to the line and
in the direction opposite to that of the motion.

Compounding these with the acceleration of C, it results
that the acceleration of P is in direction PC and is equal to

a6’, or to aw’, if w be the angular velocity.
This result, may otherwise be obtained as follows:

Let # and y be the coordinates of the point initially in
contact ; then
2=a0—asin 6, and y=a—acos 0,
from which #=a0—a(cos0.6—sin@. @),
ij =a(sin 0.6 + cos 0. &),
and, putting @=0, #=0, and 7 = a6’,

we hence infer that if p be the radius of curvature, at the
point of contact, of any curve rolling on a straight line, the
acceleration of the point of the curve in contact with the line
is w’p in the direction perpendicular to the line.
KINEMATICS. 25

Or we can give a proof directly from the definition of
acceleration.

wy

 

 

 

77 a

_ Let the curve P@ roll on a straight line, the point P
rising from the point A.

When at A, P has no velocity, and when Q is the point,
of contact, and is therefore the instantaneous centre, the
velocity of P is w. PQ perpendicular to PQ.

Taking PQ as an infinitesimal arc, and £# as the centre of
curvature, the velocities of P parallel and perpendicular

to the line are w. PQsin oO and w. PQ cos 66, if PHQ = 80.

Hence if 5¢ be the time the accelerations are the limits of
wo PQ se and w PQ ay , and, as PQ = pd8, the first of
these ultimately vanishes, and the second = op.

(4) A circle of radius b rolls on a circle of radius a; tt
is required to determine the acceleration of the point of
contact.

Taking the are BP equal to the arc AP, and o as the
angular velocity,
a+b,

o=(¢+d)=——9,

 
26 KINEMATICS.

 

 

and the accelerations of P relative to C, in the direction PC
and perpendicular to it, are

bw’ and bo.

The accelerations of C, in the direction CO and perpen-
dicular to it, are

; (a +b) @ and (a+ d) 6,
2.2
or DB aa be
at+6
Compounding these with the relative accelerations of P,
it results that the acceleration of P is in the direction PC
and is equal to

ab ,
arb”
Replacing a and b by radii of curvature this expression

gives the acceleration of the point of contact of any curve
rolling on a fixed curve.

26. We have considered in Art. (7) the angular velocity
of a line moving in a plane; we shall now find it necessary to
consider the angular velocity of a rigid system of lines
and points in space.
KINEMATICS, 27

The angular velocity of a rigid system about an axis is
the rate of increase of the inclination of a plane fixed in the
system, passing through the axis, to a plane passing through
the axis, fixed in space.

Parallelogram of angular velocities.

Imagine that a rigid system has two coexistent angular
velocities , @ about two axes OA, OB. Construct a

P,

iN

 

O MI A.

parallelogram AOBC such that OA and OB are in the
ratio of the angular velocities, and take a quantity Q such

that
OA: OB: OC :: 0: 0': 0.

If P be any point in the plane AOB, the velocity of P,

due to the two angular velocities, is equal to
oPM+o'PN,

perpendicular to the plane.

Now, OACB being a parallelogram, we know that

PL.OC=PM.0A+PN. OB;

oe 0O.PL=a.PM +0’ PN,
and therefore the velocity of P=.PL, which is the
velocity due to an angular velocity about OC.

The line OC therefore represents the resultant angular
velocity.
28 KINEMATICS,

Hence it follows that angular velocities are subject to the
parallelogrammic law, and can be compounded and decom-
posed in the same way as linear velocities.

In other words, an angular velocity is a vector.

27. If a rigid system be in motion about a fixed point,
there is always one line in the system which has no motion
and about which the system is turning.

It is clear that the motion of the system is completely
determined by the motions of any two given lines OP, OQ of
the system. Now, at any instant, OP must be moving in
some plane and therefore must be turning round some
straight line in the plane through OP perpendicular to the
plane of motion of OP.

Similarly OQ must be turning round some line in the
plane through OQ perpendicular to the plane of motion
of O@.

If then OC be the line of intersection of the two planes
through OP and OQ, perpendicular respectively to their
planes of motion, the motion of the system is completely
represented by a state of rotation about OC.

Any state of motion of a rigid system about a fixed point
can therefore be represented by a single angular velocity, or
by three coexistent angular velocities about three lines
through the fixed point.

28. Velocities and accelerations of a point referred to
three moving axes at right angles to each other.

Let 6, 0, 6, represent, at any instant, the angular
velocities of the system of axes about the axes themselves, or
rather, about the lines fixed in space with which the axes
are, at the instant, coincident.

If u, v, w be the component velocities,
u=velocity of P relative to K+that of K relative to V
+ that of V relative to O

=a—y9, + 20,,
KINEMATICS. 29

and similarly
v=y—20,+20,,
w=s— 20,4 0,

 

 

 

 

 

 

xz
K; | PL
/O “TL xv
‘M
y
For the accelerations,
let OL=u, OM=», ON=w

represent the component velocities ;

then the accelerations parallel to the axes are, on this
scale, the velocities of P, and are therefore

u— v0, + w8,
19 — wi, + U8,
w—ud, + v0.

Or, the acceleration of P relative to WV, in the direction
Ox, being w#—v6,, and the acceleration of N in the same
direction being w@,, the acceleration of P parallel to Ow is
the sum of these two, and the accelerations parallel to Oy
and Oz are obtained in the same manner.

29. If the position of a moving point P be defined
by the length r of the tangent PQ to a given curve, and the
deflection ¢ of the tangent,
30 KINEMATICS,

the acceleration parallel to PQ relative to Q = #— r¢’,
and that of Q in the same direction = 5, if the are OQ =s,
.. acceleration of P in the direction QP=7—r¢? +8

 

 

Ce

and, similarly, acceleration perpendicular to PQ
=rgét+ anb+ oe

if p be the radius of curvature at Q.

30. A point moves on a given curve, while the curve
turns round a fixed point in its plane: it is required to
find expressions for the accelerations of the point.
KINEMATICS. 31

If OP=r, the accelerations of the point P of the curve
are wr and r@ in the direction PO and perpendicular to

PO.

If the moving point be passing over the point P its
accelerations are those of P compounded with its accele-
rations relative to P.

These relative accelerations are due to the angular
motion of PY and to the motion on the curve, and, in
the respective directions of the tangent PZ’ and the normal
at P, are, if s be the are AP measured from a given point A
of the curve,

3?
§ and 2os + p ,
in accordance with the observation of Art. (10).

It is an instructive exercise to obtain these expressions by
taking a consecutive position of the moving point on the
curve, when twisted through a small angle wdt, and actually
resolving its velocities, relative to P, in the directions PZ’
and perpendicular to it.

EXAMPLES.

1. A railway train is moving at the rate of 30 miles an
hour, when it is struck by a stone moving horizontally and at
right angles to the train with the velocity of 33 feet per
second. Find the magnitude and direction of the velocity
with which the stone appears to meet the train.

2. Assuming that the earth describes a circle uniformly
about the sun in a year, that the distance of their centres is
240 radii of the sun, and that the radius of the sun is 100
times that of the earth, find the measure of the velocity of the
vertex of the earth’s shadow, taking the sun’s radius as the
unit of length and a year as the unit of time.
32 EXAMPLES,

3. If one point move uniformly in a circle, and another
move with equal velocity in a tangent to the circle, what are
their relative paths ?

4. The radius of the earth being 4000 miles, the latitude,
r, of a place at which a train travelling westward at the rate
of 1 mile per minute is at rest in space is given by

cos A = 507

5. Acircle revolves with uniform velocity about its centre.
The centre moves with varying velocity along a straight line.
Find the velocity parallel to this line at any instant of a point
on the circumference, and deduce the acceleration of the centre

necessary for this point to be always moving at right angles to
the line.

6. A point moves in a curve in such a way that its direc-
tion of motion changes at a rate varying as the velocity directly
and the whole space described inversely. Prove that the cur-
vature varies inversely as the are.

7. A wheel revolves uniformly about its centre C, which
is fixed, and a particle A moves uniformly in a straight line
through the centre; describe the path of a point # in the
wheel relative to A, (1) when CA is in the plane of the wheel,
(2) when CA is perpendicular to that plane.

8. Ifthe resolved parts of the velocity of a moving par-
ticle perpendicular to its distances from two fixed points are
constant, and equal to one another, its velocity varies as the
square root of the product of its distances from these points.

9. Aparticle A moves ina straight line, a second particle
B always moves towards A and keeps at a constant distance
from it. Find the path of B, and shew that its velocity is a
mean proportional between the velocity of its projection on
the path of A and the velocity of A.

10. df a point be situated at the intersection of the per-
pendiculars, drawn from the angular points of a triangle to
the sides respectively opposite to them, and have three com-
EXAMPLES. 33

ponent velocities, represented, in magnitude and direction, by
its distances from the angular points of the triangle, prove
that its resultant velocity will tend to the centre of the circle
circumscribing the triangle, and will be represented by twice
the distance of the point from the centre.

11. The tangent at a point P of a parabola meets the
tangent at the vertex in Y and the axis in JT. If Y move
with uniform velocity, shew that 7’ moves with uniform ac-
celeration: if 7’ move with uniform velocity, the velocity of ¥
varies inversely as AY.

12. If the velocity of a point be resolved into any number
of components in a plane, its angular velocity about any fixed
point in the plane is the sum of the angular velocities due to
the several components.

13. A point moves in a plane curve and sounds as it
moves. At a fixed point 0 in the plane the whole sound
produced is heard simultaneously. Shew (i) that if the point
moves uniformly, the curve is an equiangular spiral—(ii) if
the velocity of the point vary inversely as the distance of C
from its line of motion, the curve is a reciprocal spiral.

14, A point moves in the are of a cycloid so that the
tangent turns uniformly; prove that the acceleration of the
point is constant.

15. If the axes Ow, Oy revolve with constant angular
velocity w, and the component velocities of the point (xy)
parallel to the axes are

; a—b 2p?
a+b oy, e+e wx,
prove that the point describes relatively to the axes an ellipse
in the periodic time
aw a+b
@° ab

 

Prove that the locus of the points about which the angular
velocity of a point moving in any manner is, at the same
instant, the same, is a circle,

RT 3
34 EXAMPLES.

16. If the acceleration of a falling body be the unit of
acceleration and a velocity of 60 miles an hour the unit of
velocity, find the units of length and time.

17. Ifthe angular velocity of a particle about a given
point in its plane of motion be constant, prove that the trans-
versal component of its acceleration is proportional to the
radial component of its velocity.

18. If the acceleration of a falling body be the unit of
acceleration, and if a velocity of a yard per minute be the
unit of velocity, find the units of space and time.

19. Ifa lamina move in its plane so that two fixed points
in it describe straight lines with accelerations f, /’, shew that
the acceleration of the centre of instantaneous rotation is

JP FP HF 05 0

sin 0

3

-@ being the angle between the lines.

20. A lamina moves in its own plane so that two points
fixed in the lamina describe straight lines with equal accelera-
tions; prove that the acceleration of the centre of instan-
taneous rotation is constant in direction.

21. A point P moves with uniform velocity in a circle;
@ is a point in the same radius at double the distance from
the centre, PR is a tangent at P equal to the are described
by P from the beginning of the motion; shew that the
acceleration of the point # is represented in magnitude and
direction by RQ.

22. In two different systems of units an acceleration is
represented by the same number, while the velocity is
represented by numbers in the ratio 1 : 3. Compare the
units of time and space.

23. If the time is a quadratic function of the space
described, prove that the tangential acceleration is propor-
tional to the cube of the velocity.
_EXAMPLES. 35

24. A point moves in an ellipse so that the velocity
varies as the square of the diameter parallel to the direction
of motion: prove that the resultant acceleration at any
instant will be in direction of the line joining the point with
the middle point of the perpendicular from the centre on the
tangent at the point.

25. A point moves in a plane in such a manner that
its tangential and normal accelerations are always equal, and

its velocity varies as ec, s being the length of the are
of the curve measured from a fixed point; find the path.

26. Ifacurve roll in contact with a straight line with
uniform velocity, shew that the acceleration of the point in
contact with the straight line varies inversely as p, but if
with uniform angular velocity directly as p; p being the
radius of curvature of the curve at the point in contact,

27. A curve rolls along a straight line, the point of
contact moving uniformly along the line. Shew that the

acceleration of the centre of curvature of the rolling curve at
2
the point of contact is proportional to Be in the curve at the
8
point.

28. If the motion be referred to two axes one of which
is fixed, and the other revolves about the origin in such
a way that the line joining the origin to the particle is
equally inclined at an angle $@ to the axes, shew that the
component acceleration parallel to the fixed axis (€) is

& — (26 + £0) cosec 0,
What is the other component ?

29. Ifthe radial and transversal accelerations of a particle
be each proportional to the velocity in the direction of the
other, the path of the particle is given by the equation of
the form

2) -2 ir 2 C
(8) Gleave anares
3—2
36 EXAMPLES,

30. If the perpendiculars from a point Pon axes Ox, Oy
of which Ox is fixed, and Oy revolves uniformly, are & 7
respectively, prove that the accelerations of P parallel to
the axes at the instant when they are perpendicular
are j—2Ew +7’, and £; the angular velocity of Oy
being o.

31. The centre C of an elliptic wire is moved with
uniform velocity along a fixed line CY in its own plane
whilst the wire is in contact at P with a fixed line PY
perpendicular to CY; shew that the acceleration perpen-
dicular to PY of the point of the curve in contact at
Po x’ yp’, where CY =p, and a, y are the co-ordinates
of P referred to the axes.

32. A point moves in a plane with an angular velocity
w, and the plane is turning round the radius vector with an
angular velocity w’; prove that the accelerations in the plane
are #— o'r, and ro + 2rw, and that the acceleration perpen-
dicular to the plane is raw’.

33. A point P moves on a straight line OP which is
made to describe uniformly a right circular cone about an
axis OA, while OA sweeps out uniformly a right circular
cylinder ; find an expression for the acceleration of the point
P in the direction OP. ;

34, A point P moves so that its velocity is compounded
of two constant velocities, one of which is in a fixed direction
and the other is perpendicular to the line joining P to a fixed
point. Find the orbit described by P.

35. A plane is moving about an axis perpendicular to it,
and a point is moving in a given curve traced on the plane;
in any position w is the angular velocity of the plane, v the
velocity of the particle relative to the plane, r its distance
from the axis, p the perpendicular on the tangent, s the arc
described along the plane, prove that the acceleration along
the tangent to the curve is

dv dw 2 ar
0 (Greg) oe a
EXAMPLES. 37

86. The velocity of a point moving in a plane is
the resultant of two velocities v and v along two radii
vectores 7 and r’ measured from two fixed points at a
distance a apart. Prove that the corresponding accelera-
tions are

du ve’ 2 2 2
dit apy + oF),

ba ate (r?—r? +0’).

and dé * Qrr®

37. <A circle of radius @ rolls on a second fixed circle of
the same radius and a point P on it traces out a cardioid.
If O and C be the centres respectively of the moving and
fixed circles, prove that the accelerations of P along and
perpendicular to PO are

2a (26? — 6 cos 6 + Osin 6),
_ OF. Oy se o
4a sin 5 (sin5 4 -—6@ £08 5) ,
@ being the angle OC makes with a fixed line.
38. Two circles are taken, and the motion of a point is
given by the component velocities (u, u’) in the directions of

two tangents drawn one to each circle. Shew that the com-
ponent accelerations in the same directions are respectively,

du , {1 cos |

ut (7-4 ;

du fl ome)

at (G-F) |

where (J, I’) are the lengths of the tangents, and (¢) their
mutual inclination.

and
CHAPTER IV.

31. THE preceding discussions belong to the domain of
pure reason; we have now to introduce the facts of nature,
and to employ the results we have obtained in the solution of
actual cases of motion.

For this purpose the laws enunciated by Newton are
sufficient, and, once enunciated, the solution of any problem
concerning the motion of a body or a system reduces itself to
the integration of differential equations of the second order,
and the interpretation of the solutions.

The introduction of the principles of momentum and of
energy will in many cases enable us to determine the motion
of a body or a system in a simple manner and without the
intervention of differential equations of the second order.

32. <A particle of matter is supposed to be a very small
body, but possessing a sensible mass and capable of being
acted upon by forces, and, for theoretical purposes, two
particles are supposed to differ from each other only in the
case of their having different masses,

Force is any cause which tends to change the state of rest
or motion of a particle or a body. The weight of a body for
instance is the force of the action of gravity upon it, and is
found experimentally to be proportional to the mass or to the
quantity of matter in the body.

Expeiiments cannot be made with particles, such as we
have imagined, but experiments made with bodies of various
shapes and sizes lead to the enunciation of, and belief in
THE LAWS OF MOTION. 39

the laws of motion as applied to a particle, or to a body of
any kind ; and the results of theoretical calculations, tested
by experiment, and applied, on a large scale, to the motions
of Planetary Bodies, have led to a profound conviction, in
the minds of students of mechanical science, of the truth of
these laws.

THE LAWS OF MOTION.

33. First Law of Motion.

_ Ifa particle be not acted upon by any external forces, it
is either at rest or it moves uniformly in a straight line.

Second Law of Motion.

Change of motion is proportional to the acting force, and
is in the direction in which the force is acting.

Third Law of Motion.
Action and Reaction are equal and opposite.

34, With regard to the first of these laws, it is only
necessary to remark that it is confirmed by the perpetual
experiences of all ordinary phenomena.

Any change of motion of a body is seen to be due to the
action of some force, and the more we can eliminate the
action of external force the more nearly we find that the
motion of a body approaches to that of uniform motion in a
straight line.

35. The Second Law contains really two distinct state-
ments.

The first is the enunciation of the principle of the physical
independence of forces, namely, that each force produces its
full effect in its own direction.

To illustrate this consider the case of a ball receiving,
simultaneously, two impulses in different directions. Sup-
posing the velocity due to each individual impulse, when
applied alone, to be known, then the second law tells us that
40 THE LAWS OF MOTION.

the ball is at once imbued with two coexistent and given
velocities in given directions, and the motion of the ball, due
to the two simultaneous impulses, is immediately determined
by the parallelogram of velocities.

The second is the quantitative relation between the
magnitude of the acting force and the change of motion
produced.

Definitions. If m be the mass of a particle and v its
velocity, the product mv is called its momentum or quantity of
motion, and the rate of change of momentum is mv or m/f, if
J is the acceleration of the particle.

The assertion of the second law is that the momentum
produced in some given time is proportional to the mag-
nitude of the acting force. If a constant force P acting
on a body of mass m, produce in the unit of time the velocity
J, mf is the momentum acquired and therefore

P «mf.
and, if we choose the units so that the unit of mass is that in
which the unit of force produces the unit of acceleration, we
obtain P=mf.
This equation really contains all the kinetics of a single
particle.

In any case whether the force be constant or variable,
it is proportional to the momentum which it is capable of
producing in a given time, or, which is the same thing, to
the rate of change of momentum, and therefore, in general

P=mf.

Even if the mass which is acted upon be variable, the
effect. of force upon it is the production of momentum,
and the measure of the force is the rate of that production,

d
th = +
so that P ait (mv),
or, in other words, force is the time-flux of the momentum.

36. Impulses. We have spoken of an impulse as an
action producing velocity instantaneously in a given mass,
THE LAWS OF MOTION. 41

and such an action is proportional to the momentum which
is apparently at once produced by it, so that if Q be the
measure of an impulse producing a velocity v in a mass m,
Q = mv.

_ There is no real difference between a momentum produced
gradually and a momentum produced in a very short interval
of time.

The application of a powerful time-microscope to the

latter case would present the appearance of a force gradually
accumulating momentum, and the final result is that which

is spoken of as being instantaneously produced.

If P be the measure of a very large and variable force
acting for a short time 7, and producing the momentum mu,

then m= | : Pdt.

0
If in any given case we could find this short time 7, then
the mean value of the measure of the force would be mv/r.

The expression | " Pat, or the time integral of the force,

is called the impulse of the force, or, briefly, the impulse. If
the time + be infinitely small, and the force P infinitely large,
a finite momentum mv may be instantaneously acquired.

37. In the particular case of the action of gravity, if W
be the weight of a body, and g the acceleration of a
falling body, it follows that

W=mg.

From this it appears if we take a pound as the unit of mass,
the weight of a pound is g units of force, so that, g being
32-2 when a foot and a second are units, the unit of force is
approximately equal to the weight of half an ounce. This is
' the British absolute unit of force,

38. Parallelogram of forces. If a particle be acted
upon by two known forces in given directions, we know from
the second law that the particle has coexistent, in these direc-

tions, two known accelerations.
A2 THE LAWS OF MOTION,

The parallelogram of accelerations gives the resultant of
these two, and therefore it follows that the resultant of the
forces follows the same law, and the parallelogram of. forces
is at once established.

39. The angular momentum of a particle about a fixed
axis is the moment of its momentum about the axis; and
therefore, if v be the component, perpendicular to the axis, of
the velocity of the particle, p the distance between the axis
and the line of this velocity, and A the angular momentum,

h=mwp.
Now consider the motion of a particle in a plane, and let

T and N be the acting forces in directions of the tangent and
normal to its path,

ae
Y

aE a

 

 

 

 

a 0

The time-flux of the angular momentum about the axis
perpendicular to the plane through the fixed point O
=h=mip + mop.
The first term of this expression, mip, is equal to Tp

and represents the moment about the axis of the tangential
force.
THE LAWS OF MOTION. 43

During an infinitesimal period of time 8¢, the velocity
has changed to v+6v, and the direction of motion has
turned through an angle d¢, and therefore the change of
momentum in direction of the normal PZ is m (v + 8v) sin 8¢,
or mudd.

Now, if OZ be the perpendicular on the normal
vip =v.0Z. 84,
and therefore modp = mvdo . OZ.

But, N being the normal force Nét is the change of
momentum in direction of PZ, and therefore

Nét=mvd$, or N=mvd
and consequently mup= md. OZ = N. OZ,
which is the moment of the normal force.

*

Hence we obtain
h=T.OY+N. 02,
that is, the time-flux of the angular momentum is equal to
the moment of the acting forces. ;

In the case of motion in a tortuous curve, or in any
manner in three dimensions, the same result is ‘true if the
figure represent the projection, on a plane perpendicular to
the axis, of the path of the particle.

Other methods may be adopted to obtain this result.

For instance, we can transform to polar coordinates, by
observing that pds = 7°80, each being the double of the area
. of an infinitesimal triangle, so that

h=mps = mr’,
and h= ma (r°6).

But we know that ae (r°6) is the expression for the

transversal acceleration, and therefore, if Q be the transversal
force in action,

a
Qam.- 50°),
44 THE LAWS OF MOTION.

so that h=Q.r,
and Qr is the moment of the acting forces.

If A be the vectorial area, or the area swept over by the
radius vector, ; .
A=}r'd,
and therefore h=2mA,

and, A being the rate at which area is being swept over per
unit of time, / is the product of a mass and an area, and also,
being a vector, or directed quantity, it can be represented by
a straight line, and is subject to the parallelogrammic law.

40. The third law expresses the fact that if two bodies
act on each other in any way, either by contact, or through a
connection by means of strings or rods, or by mutual attrac-
tion or repulsion, the force which one body exerts on another
is exactly the same in amount but opposite in direction
to that which the other body exerts on the one.

Some important theorems are at once deducible from the
third law, which are of the greatest utility in the discussion
of the motion of systems of particles or bodies,

Take the case of a system of bodies, attracting or repelling
each other, acting on each other by contact for a finite time,
or by mutual impulse. In this case any momentum which is
produced or destroyed in any assigned direction is accom-
panied by the production or destruction of an equal momen-
tum in the contrary direction.

Hence it follows that, if no extraneous forces act on a
system of bodies, the total momentum of the system in any
assigned direction remains constant.

This is the principle of the conservation of linear mo-
mentum.

Again the moments of two equal and opposite momenta
about any straight line fixed in space being equal and
opposite in direction, i¢ follows that the angular momentum
about any gwen axis, which is. defined to be the sum of the
moments of momenta of all the particles of the system, remains
constant provided that no eaternal forces act on the system.
THE LAWS OF MOTION. 45

This is the principle of the conservation of angular
momentum.

_ 41, One immediate consequence of the preceding results
is that if no extraneous forces act on a system, its centre
of inertia is either at rest or moves uniformly in a straight
line.

For if & be the distance from any fixed plane of the

centre of inertia
= (m).&= 2 (mz),
« being the distance from the plane of a particle m.
This gives ;
= (m) § = (mé),
or &(m).£ is equal to the total momentum perpendicular to

the plane, and as this is constant, £ is constant; and the
same thing is true of any other direction.

42. Again, as any acting force produces momentum in
its own direction, it follows that the sum of the forces acting
on a system in any assigned direction is equal to the rate of
the change, that is to the time-flux, of the total momentum
of the system in that direction; and that, for impulsive
actions, the sum of the impulses in any direction is equal to
the immediate change, in that direction, of the momentum of
the system.

Further, since the aggregate of the forces which are at
any instant acting on the particles of a system have for their
resultant the system of extraneous acting forces, the moments
of these two systems about any fixed axis are equal.

But the former are producing changes of angular mo-
menta, and their moments about the fixed axes are equal to
the rates of change of the angular momenta of the particles.

Therefore the sum of the moments, about any fixed axis,
of the extraneous acting forces is equal to the rate of change,
that is, to the time-flux, of the angular momentum of the
system about that axis.
46 THE LAWS OF MOTION.

For impulsive actions, the sum of the moments of the
extraneous impulses about any fixed axis is equal to the
immediate change in the angular momentum of the system
about that axis.

43. We now see that the principle of the conservation of
angular momentum, as stated in Art. (40), should be, more
generally, given as follows.

If the extraneous forces, acting on a system, have no
moment about a given aais, the angular momentum of the
system about that axis remains unchanged.

44, These principles of motion, which are derived im-
mediately from Newton's laws, constitute the whole of the
Kinetics of a system.

lf Q be the linear momentum of a system in any
direction, and P the sum of the acting forces in that
direction, the connecting equation is

Q=P.

If H be the angular momentum about any assigned axis,
fixed in space, and ZL the sum of the moments of the acting
forces about that axis, the connecting equation is

Hal,

For impulsive actions, the corresponding equations will

‘be, if Q’, H’ be the new values of @ and H,

Q-Q=K, and H’- H=G,
K and G being the sums, in the directions considered, of the
impulses and impulsive couples.

It will be seen that linear momentum and angular mo-
mentum are quantities of the nature of vectors; that is, they

can be represented by straight lines, and are subject to the
parallelogrammic law.

45. A rigid body is considered to be an aggregation of
particles, or molecules, bound together by the forces of
internal mutual attractions which are in all cases equal and
opposite.

It follows therefore from the preceding articles that the
linear and angular momenta imparted to a rigid body by any
THE LAWS OF MOTION. 47

extraneous forces are independent of the shape, size, or
~ nature of the body, but depend only on the acting forces ;
and hence that the motion of the centre of gravity of a body
is the same as if it were a single particle into which is
concentrated the mass of the body, and the rotation of
’ the body about the centre of gravity is independent of the
motion of that point, and depends on the moments, about
axes through it, of the acting forces.

46. We can now state, in general terms, the principle
embodying the preceding discussions.

The forces affecting the particles of a body or system of
any kind, are the extraneous acting forces, and the internal
forces, due to mutual pressures, or to mutual repulsions or
attractions, and these systems together are the exact equi-
valents of the system of time-fluxes of momenta.

But the system of internal forces, which is made up of
sets of equal and opposite forces, has no resultant, and there-
fore it follows that

The system of time-fluzes of momenta is the exact
equivalent of the system of acting forces.

47. Energy is capacity for doing work, anda system may
possess energy of motion, or energy of position, or both, the
former being due to the relative motions of the bodies which
constitute the system, and the latter to their relative posi-
tions.

‘The energy of motion is called Kinetic energy and is
measured by the expression 4 2 (mv’), v being the velocity of
a body, m, of the system.

As applied to a single particle in a field of force, if P be

d
the acting force, P = mi =m 7 Z = mv cr =4 a (mv’),

so that force is measured by the space-flux of the kinetic
energy.

The energy of position, or the Potential energy, of a
system is the work which it is capable of doing in virtue of its
- configuration, that is, the relative position of the bodies of the

system, or it is the work which has been expended upon it
‘in moving it from a certain defined configuration to its pre-
sent configuration. it beino understand the work done by
48 THE LAWS OF MOTION.

a force is the product of the force by the space through
which it is exerted.

Suppose for instance that two equal spherical balls, each
of mass m and radius a, attract each other with a constant
force P, and that they are placed at rest with their centres at
a distance 2c.

If the zero configuration be when they are in contact, the
work P (2c—2a) must have been done to separate them.
This is the initial potential energy of the system of the two
balls, their kinetic energy being initially zero.

The system being left to itself, the balls will approach
each other, and when their centres are at a distance 2a, the

kinetic energy of each is $m. om. (c— ), and therefore the

kinetic energy of the system is 2P (¢—). But in this con-
figuration the potential energy of the system is P (2a — 2a),
The sum of the kinetic and potential energies is therefore
P (2c — 2a), and is constant during the motion.

For another instance take the case of two balls connected
by an elastic string and pulled apart; potential energy is
thus stored up, and if the balls be let go, kinetic energy is
acquired which is the exact equivalent of the loss of potential
energy due to the contraction of the string.

48. The preceding is a very simple case of a great
general principle, which we now proceed to enunciate.

The Principle of energy. In any conservative system the
sum of the kinetic and potential energies is a constant quantity.

As applied to a mechanical system, the meaning is that,
in a conservative system, there is no loss of energy by
conversion of kinetic energy into heat, or by internal friction
producing loss of kinetic energy without gain of potential
energy.

_There is no doubt that in any system the principle is
universally true and that, in all cases, any loss of ordinary

mechanical energy is accounted for by its conversion into
heat or some other form of energy.

In other words we may say that the gain of kinetic

energy of a system is equal to the work done by the forces
of the system, which is in effect the loss of potential energv.
THE LAWS OF MOTION. 49
49. For a single particle, since P = md, we obtain
acy pe
Po= mii =} 5 (me ).

Since v=, the left-hand member of the equation re-
presents the rate at which work is being done, and the right-
hand member is the rate of increase per unit of time, or
the time-flux, of the energy. Hence it follows that the work
done in any time gives the change of kinetic energy during
that time.

Further it follows that, in a conservative system, that is,
in a system in which there is no transformation into heat or
other forms of energy, the change in the kinetic energy
of the system is entirely due to, and is measured by the work
done by the acting forces.

Internal friction, collisions and explosions may produce or
destroy energy, but, in what we have called a conservative
system, such modes of developing or losing energy are not
supposed to exist. In fact, from a mechanical point of view,
forces of the character referred to are of the nature of forces
extraneous to the system, by means of which the total energy
of the system may be increased or diminished.

50. In the chapter of the Principia on “ Axiomata
sive Leges Motus,” the concluding paragraph, which is often
quoted in support of the Law of Energy as a fundamental
law is the following :

Nam si estimetur agentis actio ex ejus vi et velocitate
conjunctim; et similiter resistentis reactio estimetur con-
junctim ex ejus partium singularum velocitatibus et viribus
resistendi ab earum attritione, cohzesione, pondere, et ac-
celeratione oriundis; erunt actio et reactio, im omni instru-
mentorum usu, sibi invicem semper zquales, Et quatenus
actio propagatur per instrumentum, et ultimo imprimitur
in corpus omne resistens, ejus ultima determinatio determi-
nationi reactionis semper erit contraria.

This passage really shadows forth the principle of energy
in its modern form, and indeed states the principle as far as
it could be stated at the time when. the Principia was
written,

B.D. 4
50 THE LAWS OF MOTION.

It was not until Count Rumford began to make obser-
vations, and to draw inferences from his observations, followed
by a host of other investigators, that the principle of energy
presented itself in the form which now renders it the one
principle of the greatest utility in the discussion of natural
phenomena.

The magnificent intuition, which forms the opening
Chapter of the Mécanique Analytique, and is the basis of
operations of that great work, is, in effect, only a particular
case of the general principle of energy.

51. The preceding discussions of this chapter are
sufficient for the purposes of the present treatise; but for
elaborate accounts, historical and critical, of the laws of
motion and the science of energy, the student will consult
the Natural Philosophy of Sir W. Thomson and Professor
Tait. ;

The student will find valuable expositions of the same
ideas, from elementary points of view, in Matter and
Motion by the late Professor Clerk Maxwell.

Formation of the equations of motion of a particle.
52. We hence see that the solution of any problem on
the motion of a particle depends upon the equation
= mf
If a, y, 2 be the coordinates of a particle referred to
three fixed axes, and mX, mY, mZ be the component forces
parallel to those axes, then, since the acceleration in any
direction is entirely due to the resultant force in that
direction, the equations of motion are
méi=mX, mj=mY, mZ=mZ.
If the axes are in motion about the origin, the equations are,
dividing by m,
& —v0,+wd,=X
d —w6,+u0,=Y
w— ub, +00, = Z.
If we refer to the tangent, the principal normal and the
binormal, the equations are
a2

s=S,-=N, o=T,
p
THE LAWS OF MOTION. 51

me forces in the respective directions being, mS, mN, and
mT.

If we use cylindrical coordinates, the equations are
#-1r0=R, 764+276=T, 2=Z,
mR, mT, and mZ being the forces.

And, if we use any other system of representing the ac-
celerations the equations are formed in the same manner.

The integration of these equations, and the determination
of the constants of integration by means of the initial
circumstances of motion, constitute the solution of the
question under discussion.

53. Equations of motion of a system of particles.

The inferences which have been drawn from the laws of
motion, when expressed in mathematical forms, give the
equations of motion of a system.

Thus if m be the mass of. a particle of the system, whose
coordinates are 2, y, z, the rates of change of momenta
parallel to the axes, sometimes called effective forces, are
mé, mij, and mz.

We have shewn that the system of these quantities is
exactly equivalent to the system of acting forces.

If then X, Y, Z, be the sums of the acting forces resolved
parallel to the axis, and LZ, M, N, the sums of the moments
of these forces about the axes, we at once obtain

Ym = X, Imi = Y, Sm7 = Z,
2m (yz — 2i/) = L, Sm (zt — a2) = M, Ym (aij — y#) = NV.

As in the case of a particle these equations can be
presented in various forms, and in a subsequent chapter
some of their applications and developments will be con-
sidered.

At present they are placed on record, as being immediate
consequences of Newton’s Laws of Motion.
CHAPTER V.

RECTILINEAR MOTION.

54. THE simplest case of motion is that of a particle in
a straight line under the action of forces in that line, and the
equation of motion in that case is

méi=mX.
If the force be constant and equal to mf,
&=fandé¢=/ft+u,
wu being the initial velocity.

Integrating again, «=43/+ut+a,if a be the initial
value of «.

The equation may also be written in the form

dv
Vv We = Ss
leading to v=u? + Ifa.

55. Motion of two weights connected by a fine string
passing over u smooth fixed pulley.

If m, m’ be the masses, 7 the tension of the string and
the distance of m from the pulley,

mé= mg — T,
and similarly a —# being the distance of the other weight,
—mié=m 9g - T.
RECTILINEAR MOTION. 53

Solving these equations we find that
2mm'g

m-+m'”

 

Mm
oS eee m9, and T=

56. Motion of a particle, initially at rest, acted upon
by a force toa fixed point varying as the distance from that
pornt.

If the force be mua, the equation of motion is
= — px
or &+ pr =0,
the solution of which is
a= A cos/ut+B sin /ut,
this gives
é=—A Jusin Jutt+ BJycos Jpt.
If initially «=a, and #¢=0; then B=0,
and & = a COS apt.

The interpretation of this. equation is that the particle
oscillates through the centre of force between the positions

x=aand «=—a, the time of a complete oscillation being
2a
Was
If we multiply the equation of motion by 2é and integrate,
we obtain a = p (a? — 2°),

shewing as before that the velocity vanishes when «= + a.

If the force be repulsive the equation of motion is
& = px, or &— pa =0,
leading to
a = AcVit + Be Ve,
this gives,

z =Ju ar — Jp Be,
54 RECTILINEAR MOTION.

and introducing the initial conditions we find that
a=A+B,and O=A-B,

so that 2=5 (Vit + e Vet) = a cosh - /p.t.

As before we can obtain the velocity at once in terms of
the distance from the equation, 4” = ys (2"— a’).

57. Motion of a particle, initially at rest, under the
action of a force varying inversely as the square of the
distance from a fixed point.

In this case, the force being supposed attractive, @ = — par*.

Multiplying by 2#, and integrating, we obtain

# = (a*—a"), or 2 + hf ai
da J2pV a-a’
where a is the initial distance of the particle from the centre
of force.

Assuming that the motion is towards the centre of force,
we must take the negative sign, and we then have

 

 

pace a
OS oa a eS cee
adz Jaz—x Jac—x? Jaa—a’
d Ge ata ce ae
an 7 lte Jaa i 5 208 a?
when t=0, z=a, and.’. C=0,

and t= = \aa—# 4 cos 20 —a .
2y 2 @

Putting «=0, we obtain the time from the initial
position to the centre of force, which is

a?

2 Qu"

At this point the velocity is infinite, and, as the particle
passes through the centre of force, the direction of the
force changes, and the motion of the particle is retarded.

Imagine now that the particle is projected away from the
RECTILINEAR MOTION. 55

origin at the initial distance c (less than a) with the velocity

2 (67 — a").
The equation of motion is the same, and we obtain
# = Qua + C.
Initially x=cand £=,/2u(c*—a);
a C=— 2yua",

@ = 2p (a —a"),
and &= /2u (e@?— a),
taking the positive sign as the motion is outwards. This
shews that the motion is exactly reversed, and that the
particle will traverse the distance from #=c to =a and

then come to rest, the time being exactly the same as in the
inward motion from «=a to #=c.

This is an instance of Mechanical Reversion, and hence it
appears that the particle, in the first case, on arriving at the
centre of force, will pass through and repeat its previous
motion in exactly the reverse order, and thus perform
complete oscillations in the time

“a
Qa J

58. We have found that in the case of Art. (56), the
time to the centre of force is independent of the initial

distance, and, in the case of Art. (57), is proportional to a*.
The consideration of dimensions enables us to predict
each of these results.

In each case, if a be the distance and yw the acceleration
at the unit of distance, the time must depend upon a and pm.

In the first case, if we assume that t o au’, and observe
that ¢ is of no dimensions in line, and that, wx being an
acceleration, » is of no dimensions in line, and of —2
dimensions in time, we see that

p=0and 1=— 2g,

so that ¢ o x and is independent of the distance.
be
56 -RECTILINEAR MOTION,

In the second case yu is of three dimensions in line, since =

is an acceleration, and therefore
p+3q=0, and 1 =— 2¢,
at

so that : t o—=

N/m
Conversely, if the time to the centre be independent of the
initial distance, and depend only upon y, and if we assume that
the force is proportional to some power of the distance, say the
n‘® power, then, #a” representing an acceleration, which is of
one dimension in line, and » being a function of the time only,
it follows that n=1, so that, if the force vary as some power
of the distance, the only possible law is that of the direct

distance.

59. Motion of a particle initially at rest under the
attraction of a solid sphere, the particles of which. attract
according to the law of nature.

If « be the mass of the sphere and m the mass of the
particle, the force on the particle when outside at a distance
ris mur”, and the case is therefore solved in the preceding
article.

If however the particle, on arriving at the sphere, be
supposed to enter into a fine straight tube, in the line of its
motion, passing through the centre, the force at the distance
T, less than the radius of the sphere, is

oe ae eee
BTPr =, OF 5 TPT,
p being the density.

This being proportional to the distance the motion inside
the sphere is determined as in the case of Art. (56).

If the tube extend through the sphere, it will be found
that the time of a complete oscillation is

sf/si00r 4/5 + (eee z tf S sin WS sya
RECTILINEAR MOTION, 57

where a is the radius of the sphere, and b the initial distance
of the particle from the centre of the sphere.

60. Motion in a straight line, in which a centre of force,
the attraction to which varies as the distance, is moving with
a given constant acceleration f.

If the original position of the centre of force be taken as
the origin, the equation of motion is

#=—p(e—$fe),
or, ifa—tfPe=r,
F+f=—pr, or #+u(r +£)=0,
the solution of which is

rttna cos Vat + B sin J yt.

Introducing the initial conditions, we find A and B, and
thus determine a in terms of the time.

61. Motion of a heavy particle, suspended from a fixed
point by an elastic string.
If # be the length at the time ¢, and 7 the tension,

the equation of motion is

mé = mg — T=mg-r.-—, by Hooke’s Law, .

.
or #+™ (w-a—-™¥) =o,
ma,

whence a-a— 9 A cos, / 4 Bin: a

: x ma ma
Suppose that initially the particle is held at the distance

a, the natural length of the string, and then let go; that is,

when ¢ = 0, let r=a, and «= 0, then

ag 4 8 (4 > 4)
al x ¢ COS 4, wa ‘
58 RECTILINEAR MOTION.

This shews that the particle descends through the space
2a <8, and then rises again to its initial position and

continues to oscillate, the time of a complete oscillation being

n/m,

The range of oscillation can be obtained at once by
the principle of energy, for the particle will fall until the
gain of potential energy developed by extension is equal to
the loss of potential energy due to the fall.

Now the potential energy of a stretched elastic string
=} (Tension) (Extension),
and therefore, if z be the total fall,

eee = 94 oe
$(a2)e mgz, or z 2a >

62. Fall of a heavy particle in a resisting medium
when the force of resistance 1s proportional to the velocity.

Measuring w downwards the equation of motion is

c= —ke,
which gives é+ ka = gt,
if the particle fall from rest,
and.’ : ae” = [gte™ dt, [Chap. 11.]
t kt
or L= ¢ - q + fe ee

taking «= 0, when ¢=0.
Hence & = - t er"

and, if ¢ increase indefinitely, ¢ = f ‘

This is called the terminal velocity.

63. Fall of a heavy particle in a medium, the resistance
of which varies as the square of the velocity.
RECTILINEAR MOTION. 59

Measuring x downwards the equation of motion is

dv _ ty OP aa
oe 0 ke? or + 2h = 29,
from which ve = . e+ 0, [Chap. 11]

and, choosing the origin so that v=0 when 2 = 0,

v= (l-e«™).

In this case the terminal velocity is a ‘.
Ms ice ang
dz gNe*—1’

“ OTs Py Sea 1,

from which we obtain,

kx = log (

Further,

ele =)
5 :
Or we may use the equation,
0 =9g- ke’,
which leads to
ie ; gvtt. = Net a

gg) Nw 41 Nae NER

and gives the same value of x as before.

64. Motion of a heavy particle projected vertically up-
wards in the same medium.

In this case, measuring # upwards,
dv
v——=—g—kv’.

dc
Integrating and taking w as the initial velocity,

g + ky? hens et

gtk

 

2
6U RECTILINEAR MOTION.
shewing that the ak rises to the height
ap loe (1 + ew a
g

Further il = o A ku’) * — 9g,

bo o
wodkg .t = sin
Ju 1+ — pe 14+—
which gives _
Zp k

a a/ a .usin (hg. t) + cos (J/kg)

Or, starting from the equation 9 =— g — kv’, we obtain

- Jgkt= tan*(v J = tan”? (u J)

and therefore

 

— sin™

 

 

w/t
on/s= tan./kg.t uk cos Jlig.t—s/g sin Jig.t. t.
Tt ua/* tan vig t ~ J/g cos Jkg- ttuJksin Jig.t”

giving the same value of « as before.

65. A particle moves from rest, in a medium the resistance
of which varies as the square of the velocity, under the action of
a force to a fimed point varying as the distance.

In this case

dv
= = ky’ — px,
and therefore _
ees oe = oe aie—a) ai (1 — ea), [Chap. 11]
observing that v= 0 when =a.

66. Motion of a piece of uniform chain in a straight line,
under the action of forces in that line.

Taking a fixed point O in the line, let x be the distance
from O of one end A of the chain and take r as the distance
from A of a point Pof the chain, —
RECTILINEAR MOTION, 61.

The motion of the element PQ (ér) depends upon the
tensions at P and Q and the acting force.

Oo A PQ B

If m be the mass of unit length, the mass of the element
is mér, and if mérX be the force acting upon it, the equation
of motion is

mor .@=8T' + mor. X,

taking 7 as the tension at P, and observing that the accele-
ration of every point of the string is the same as that of the
point A.

Integrating this equation over the length of the string we
shall obtain an equation for determining « in terms of the
time.

Suppose for instance that the force is repulsive and varies
as the distance from the point O, or that
A=p (a+r).
Integrating,
r

2
mrié= T+ mp (re + 5) + C,

and, observing that 7’'= 0, when r=0 and when r=a,

so a
cL=p (2 - 5) 5
the solution of which is the same as in previous cases,
Substituting for # we find that
T=impr (a—-71).

67. Direct impact of elastic balls on each other.

If two elastic balls impinge directly on each other, that is,
if the line joining their centres be the line of motion of each
ball, the effect of the impact is an immediate change in the
momentum of each ball.

But, since action and reaction are equal and opposite the
momentum added to one ball is equal to that which is lost
62 RECTILINEAR MOTION.

by the other so that the total momentum remains un-
changed.

This gives one equation of motion.

For another we appeal to experiment; and assume the ex-
perimental law that, if e be the coefficient of elasticity, the
relative velocity of the two balls after impact is reversed in
direction and is to the relative velocity before impact in the
ratio of e to unity.

Hence if. the ball m impinge with velocity u on the ball
m’ moving in the same direction with velocity w', and if
v and v’ be the velocities after impact, both measured in the
same direction as before, we have the equations,

mvt+ my = mu+m'w,
vw —v=e(u—v),
from which we obtain
(m+m’) v' = u(m+em) + wu (m' — em),
(m+) v = u(m— em’) +u' (m' + em’).
It is worth mentioning that, in all cases of the impact of

elastic bodies, energy is lost by impact; if the elasticity be
perfect, that is, if e = 1, no energy is lost.

If two elastic balls impinge obliquely on each other, all
that is necessary is to resolve the velocities parallel and
perpendicular to the line of centres; the motions perpendi-
cular to the line of centres are unchanged, and the preceding
equations determine the changes of motion along the line
of centres,

68. In the case of Art. (55), it is required to examine
the effect of suddenly attaching a weight, mass p, to any point
of the ascending string.

The mass », having no momentum before it is attached,
acquires momentum instantaneously, and if m be the
descending body its motion is suddenly checked, while the
portion of string between mw and m’ is slackened and m’ rises
freely.
RECTILINEAR MOTION, 63

_ Aft be the velocity with which the two are moving at the

instant before mw is attached, and uw’ immediately afterwards,
(m + p) u’ = mu,

since the momentum in the direction of motion is unchanged.

The impulsive tension @ of the string is given by the
equations
m (u'—u)=~ Q, pu'= Q,
the effect of the impulse on each body being change of
momentum.

Subsequently, if ¢ be the time which elapses before the
lower string becomes tightened,
m — pb

t—ho? =wt+4——" gt’;
Bm tg = ait h ae}

this determines ¢, and therefore determines the velocities of
m’ and of « and m at that time.

A jerk then takes place, and the momentum of the
system in direction of motion remaining unchanged the new
velocity is at. once determined and the subsequent accele-
ration is

m-h—m
m+p+m I

69. A straight piece of uniform chain lying on a smooth
horizontal table recewes at one end a given impulse in direction
of its length ; it is required to determine the motion and the
umpulsive tension at any point.

B p 2172

 

Let m be the mass of the chain, and a its length; then if
v be the velocity produced by the impulse,

Q=m,
and, if 7’ be the impulsive tension at a point P,

eae
a

for the mass of BP is set in motion by the impulse 7.
64 RECTILINEAR MOTION.

70. A heavy uniform chain is suspended by one end above
a horizontal table, its lower end being just above the table; of
it be allowed to fall, it is required to find the pressure on the
table.

We have seen that force is measured by the rate of pro-
duction, or destruction, of momentum.

As the chain falls, the table receives an infinite number
of infinitely small impulses, and the result is that a finite
varying pressure is produced, which, added to the weight of the
portion coiled up at the instant considered, gives the pressure
on the table at that instant.

When a length z has been coiled up, the velocity is J 29a,
and therefore the portion coiled up in a small time 6¢ is
5¢ /2gx, and the momentum of this portion, which is de-
stroyed in the time 6¢,

MEM Nigam uy
M being the mass of the chain and a its length.

Hence momentum is being destroyed at the rate of 2Mg -

per unit of time; and therefore, adding the weight of the coil,
the pressure on the table is three times the weight of the coil.

71. One end, B, of a heavy uniform chain hangs over a
small pulley A, and the other 1s coiled up on a table at C; of
B preponderate it 1s required to determine the motion and the
tension at C.

It is easily seen in this case that all internal tensions
neutralise each other, and that the momentum of the system
in the direction of motion is due to the external forces acting
on the system in that direction, that is to gravity, and the
reaction of the table.

This reaction is equal to the weight of the coil on the
table, and the resulting force in direction of the motion is the
difference of the weights of the two straight portions of
chain.
oO

RECTILINEAR MOTION, 6

If AC=a, and AB=za, and if v be the velocity, the
momentum = 4 (w+a)v, w being the mass of an unit of
length ; therefore

d
5 (eta) 1} =9 (ea),
or, since dao dy P= (
i den tare v=g(x—a),

the integral of this equation is
3
(a + a)? t= 29 (Gata) + C,
the constant being determined by initial conditions.

If for instance # = a, initially, that is, if # be just greater

than a, C= ga’,

(w+ aot = 2 (@— a) (e+ 20).

Also, (the tension at C) x t= momentum generated in the
time é¢ by the action of the tension
= (uvdt) v;
therefore tension = pv,
Or, we might have arranged the process thus :
calling the tension 7,

di
p(e+a) 05" = pg (w—a)—T

is the equation of motion of the portion of chain CAB,

and, as above, 7'= pv’.
\

72. A spherical raindrop, as tt falls, receives continually
by precipitation of vapour, an accession of mass proportional
to ws surface; neglecting the resistance of the air, it is re-
quired to determine its motion.

If a be the initial radius, and e the thickness of the shell
deposited in the unit of time, the radius (r) at the time ¢
=at+e.
B. D. 5
66 RECTILINEAR MOTION,

Hence the momentum at the time ¢
= mp (a+ et)? 2,

and © (4mp ar ell 0) =9.4mpate ,
dv 3ev
and therefore i + a 9:
am pee
and = ate asa:

This article and the two preceding were published in 1873,
in a paper in the Mathematical Journal for that year.

73. Fall of snow down a sloping roof.

Imagine the snow to be just supported by friction or
adhesion, and that a very slight downward impulse is given
to the top line of the snow just below the ridge of the roof.

In that case the snow will slide down from the top and
gradually set the whole in motion.

Take 6 as the breadth in motion, and mas the mass of
unit area ;

then, neglecting the friction on the mass in motion which
is practically very slight, the equation of motion is

. (mbxé) = mbag sin a,

or xt + a = gx sina,
which gives &* = 2 gxsina,

and shews that the acceleration is one-third of that of a mass
sliding freely.

74. The equilibrium and motion of a heavy ball, supported
by a vertical jet of sand.

We shall consider the case of a cylindrical homogeneous
jet, supporting the ball symmetrically, and assume that the
reflected particles of sand do not interfere with the ascending
particles,
RECTILINEAR MOTION. 67

The weight of the ball will be equal to the rate at
which momentum is being destroyed, or created in a reversed
direction, when resolved vertically.

In other words the weight will be equal to the resultant,
which is clearly vertical, of the negative time-fluxes of the
momenta.

We shall assume that the velocity of the jet is consider-
able, so that we may neglect the changes in the velocities of
its particles due to the action of gravity.

 

If m be the mass of the unit of volume of the sand, and u
its velocity, the quantity which impinges on an elementary
zone in the time 6¢ is

m. 2aasin 8. acos 056. ust,

and the normal component of the momentum of this quantity
of sand is
2arma’u’ sin @ cos’ 6606E.

Multiplying this by 1+e, where e is the coefficient of
elasticity between the ball and the sand, we shall obtain the
quantity of motion created in the normal direction outwards
round the zone, and, dividing by &, we then obtain the
negative time-flux of the momenta, which is the pressure on
the ball.

5—2
68 RECTILINEAR MOTION.

The last result, when multiplied by cos@, will be the
resultant vertical pressure on the zone, and, if the breadth of
the jet subtend an angle 2a at the centre of the sphere, and
M be the mass of the sphere, it follows that

Mg =} wma’? (1 +e) (1 — cos*a).

If the ball be in motion, suppose that, at any instant its
vertical velocity is v; then the equation of motion of the
ball is

Mo = } wma? (u— v) (1+) (1 — cos* a) — Mg,

or 07 =a as (Qu — 0’).
Integrating we obtain

ge
Qu —v = Ce”;
and, if we suppose the ball to start upwards with a velocity
v, it will have its velocity destroyed after ascending through
the space
uw 2u

om ee Qu—v'’

 

and will then be under conditions consistent with equi-
librium.

The time in which this takes place is obtained by inte-
grating the equation

b= — 7, Quo —'),

and the theoretical result is that an infinite time must elapse
before the ball absolutely loses its velocity.

The simplest method of illustrating the idea of this article
is to employ a jet of water. The ball rises and falls inter-
mittently, and is occasionally at rest for a sensible time. As
in the case imagined of a jet of sand, the pressure is due to
the rate of destruction, and of creation in the contrary direc-
tion, of the momenta, in directions perpendicular to the
surface of the sphere, of all the elementary cylindrical shells
which constitute the impinging stream.
RECTILINEAR MOTION. 69

EXAMPLES.

1. A smooth wedge on a horizontal plane is moved from
rest with an uniform acceleration; find the direction and
amount of the acceleration that a heavy particle placed on
its inclined plane surface may be in equilibrium relative
to it.

If the acceleration be given, find the motion of the
particle, supposed initially at rest, upon the inclined surface.

2. Particles slide from a fixed point down rough planes
to points in the surface of a cone, whose axis, passing in
direction through the point, is vertical, and vertex upwards.

Shew that, if the vertical angle of the cone= 2 tan ,

the particles will all have the same velocity on arriving at
the cone.

3. Give a geometrical construction for determining the
straight line of quickest descent of a heavy particle from a
given point to a given curve.

If the curve be a conic, with its vertex upwards, the
length of the line of quickest descent from the focus to the
curve is equal to the latus rectum.

4, Determine the motion of a particle, initially at rest,
under the action of a force to a fixed point varying inversely
as the cube of the distance.

5. A train goes from one station to another a mile off,
being uniformly accelerated from rest in the first two-thirds
of the distance, and being brought to rest by uniform
retardation in the remaining one-third of the distance, and
taking 3 minutes to perform the journey. Find the ac-
celeration and retardation and the maximum velocity ac-
quired.

6. A hyperbola is placed in a vertical plane with its
transverse axis horizontal; prove that when the time of
descent down a diameter is least, the conjugate diameter is
equal to the distance between the foci.
70 RECTILINEAR MOTION.

7. Find the locus of points from which inelastic particles
may be let fall on a smooth inclined plane, so as always to
have the same velocity on arriving at the same horizontal
line in the plane.

8. Two equal weights are fastened to the extremities of
a string and are then hung over two small smooth pullies
A, B which are in the same horizontal line. If a third
equal weight be fastened at the middle of the horizontal
portion AB of the string, shew that it will descend a distance
equal to two-thirds of AB, and find the velocity in any
position.

9. Two bodies, 2P and P, are connected by an in-
extensible string, which passes over a fixed smooth pulley;
determine the motion and the tension of the string.

If, after the motion has gone on for one second, another
body P, having no velocity, be suddenly attached to the
descending body 2P, determine completely the subsequent
motion of the system.

10. Two elastic balls impinge obliquely, so that their
directions of motion are interchanged: prove that the
product of their velocities is unaltered, and that if their
masses be equal their directions make equal angles with the
line of impact and that their velocities are unaltered.

11. In an Atwood’s machine, if the string can only bear
a strain of one-fifth of the sum of the weights at its two ends,
shew that the larger weight is not less than 13 times the
smaller weight, and that the least acceleration possible is

V3
57

12. If two elastic balls impinge on each other with
equal and opposite momenta, their kinetic energy will be
reduced in the ratio e”: 1.

13. <A particle is projected in a resisting medium
towards a centre of attractive force which varies as the
inverse cube of the distance: the resistance of the medium
varies as its density and as the square of the velocity, the
density varying as the inverse cube of the distance from the
RECTILINEAR MOTION. 71

centre: prove that the particle’s velocity on reaching the
centre is independent both of its initial distance and of its
initial velocity.

14, A particle starts from rest at a distance b from a
fixed point, under the action of a force through the fixed

point, the law of which at a distance wz is w (1 - <) towards

2
the fixed point when x is greater than a, but »(a-3)
from the same point when # is less than a: prove that the
b? — a?

b

15. In a single moveable pulley when there is equi-
librium the power and the weight hang by vertical strings;
the weight being doubled and the power being halved, prove
that the tension of the string will be unchanged.

 

particle will oscillate through a space

16. If in the second system of pullies there are n strings
at the lower block, prove that the upward acceleration of W
due to a power P will be

nP —W
mP+ ws
If when W has an upward velocity v, the weight P reach
the ground, prove that there will presently be upon the
string an impulsive strain
nP Wy
vP+W

17. A series of n elastic spheres whose masses are
1, ¢, e, &. are at rest, separated by intervals, with their
centres on a straight line. The first is made to impinge
directly on the second with velocity u. Prove that the final
kinetic energy of the system is 4 (1—e+e") w.

18. Two weights P and Q, connected by a string passing
over a smooth pulley, are held at a distance c above a hard
inelastic horizontal plane and let go. After a series of
72 RECTILINEAR MOTION.

impacts by the heavier weight Pon the plane, the system at
length comes to rest. Shew that the whole time of motion is

ies P + Q Cc

2a/ mae + te

3V24/ 5 oe
19. Two balls, of elasticity e, moving in parallel direc-
tions with equal momenta, impinge; prove that, if their
directions of motion be opposite, they will move after impact
in parallel directions .with equal momenta; and that these
directions will be perpendicular to the original direction
if their common normal is inclined at an angle sec" (1+ e)

to that direction.

20. Prove that, in order to produce the greatest de-
viation in the direction of a smooth billiard ball of diameter
a by impact on another equal ball at rest, the former must
be projected in a direction making an angle

sin ¢, /1=8
cV 3-e

with the line (of length c) joining the two centres; e¢ being
the coefficient of elasticity.

 

21. A ball A strikes directly a ball B at rest, B strikes
directly C at rest and is then again struck by 4; if A
is reduced to rest and B and C move with equal velocities,
prove that

A:B:0:3:1:2,
all the balls being perfectly elastic.

22. Three balls of masses A, B, C are at rest with
their centres in a straight line; if A be projected towards

B, shew that there will be four impacts if aa

2 2 2
pene , and only three if it be> ee gs ee

2e 22 * gS
2
and only two if it be > —
RECTILINEAR MOTION. 73

23. A weight P hanging vertically just supports a
weight W in that system of pullies in which there is only
one string. Shew that, neglecting the masses of the pullies,
if P and W be interchanged their centre of gravity will
descend with an acceleration

CW = Py
We WP+ Ps

24. If the weight (P), on a wheel and axle, suspended
from the wheel preponderate over the weight (W) suspended
from the axle, prove that the acceleration of P is

aP—abW
IVW+@P ’
where a and 6 are the radii of the wheel and axle, the inertia
of the wheel and axle being neglected.

If an additional weight (w) be suddenly attached to W,
find the impulsive tensions of the two strings.

25. Two trains of equal weight are being drawn along
smooth level rails by engines, one of which exerts a constant
tractive force, while the other’s rate of working is uniform.
Prove that if their velocities at two instants are equal, the
second train moves through the greater distance during the
interval between the two instants, and that they are working
at the same rate, at the end of half this interval.

26. Two given points are in the same vertical line;
shew that the locus of the points in a vertical plane through
them, from which the times of descent to the two points are
the same, is a rectangular hyperbola.

Shew also that, if two equal circles be in the same
vertical plane, the locus of the points from which the times
of shortest descent to the circles are the same is a rect-
angular hyperbola.

27. A particle moves in a straight line under a centre of
attractive force w’r in that straight line ; if it be initially at a
distance c from the centre of force and be projected in the
(4 RECTILINEAR MOTION,

straight line with velocity V, shew that it will arrive at
the origin in a time equal to

pe

- I +__/.__. ,
B 10. af 7p? + pc
28. A cycloid has its base horizontal and vertex up-

wards; prove that the time of falling down any radius of
curvature is constant.

29. The mean time of descent down a given inclined
plane of unknown roughness is equal to twice that down
an equal smooth plane, all coefficients of friction, for which
motion is possible, being considered equally probable.

30. A heavy particle is attached by an elastic string to
a fixed point on a smooth horizontal table; the particle is
drawn out along the table till the string is double its natural
length (a) and it is then let go; find the velocity of the
particle in any position and shew that it will return to

the starting point after a time = 2y/ (7 +2): the modulus

of elasticity of the string being equal to the weight of the
particle.

31. Two equal particles which mutually repel one an-
other with a force varying as the distance between them are
connected by a light elastic string; find the condition that
the motion may be oscillatory; and assuming that the par-
ticles would rest in equilibrium with the string stretched to
twice its natural length find the amplitude of the oscillation
if the particles just meet.

32. Two particles start simultaneously from the same
point and move along two straight lines, the one with
uniform velocity, and the other from rest with uniform
acceleration. Prove that the line joining the particles at any
time is always a tangent to a fixed parabola.

33. An infinite number of particles are arranged along a
curve; they move normally to the curve with velocities
which are always proportional to the perpendicular from the
RECTILINEAR MOTION, 75

origin on the tangent to the locus at any instant. Prove
that they will always lie in a similar curve with the origin
for a centre of similitude; and that if they move so as to ap-
proach the origin, they will reach it together after an infinite
time.

34, ABCD is a smooth tube in a vertical plane bent up-
wards at Band C, BC being horizontal, and the angles ABC,
BCD, being equal and obtuse. An inelastic particle slides
from a given point in AB. Find the time it will take to
reach its highest position in CD; and if this time be equal to
that of the next similar stage of motion, and angle ABC = 135°,
shew that the original vertical height of the particle above
BOC=2 BC.

35. A smooth circular hoop, radius a, rests on a smooth
horizontal table; a small spherical mass is projected in any
direction from the centre of the circle with velocity v: prove
that the whole time that elapses until the nth impact is

a 2 — et sas e”
. v grt gh
where ¢ is the coefficient of elasticity.

Also find the ultimate velocity of the ring and sphere
after an infinite time.

36. An elastic string is extended between two fixed
points to double its natural length, and a particle of mass m
is fastened to the middle point of the string. If the particle
be drawn towards one of the fixed points through half its
distance from that point, and then let go, find the greatest
velocity which it subsequently acquires.

If a be the natural length of the string, prove that the
time of a complete oscillation is m/ma “af

37. A particle is placed initially at a distance a from a
centre of force the attraction to which varies inversely as the
distance; prove that the time of arriving at the centre of

. Tv
force is a a,"
be
76 RECTILINEAR MOTION.

38. The upper extremity of a piece of chain, hanging
vertically is made to move upwards, with a given acceleration ;
find the tension at any point of the chain.

39. An endless elastic string, modulus A and natural
length 27, is placed in the form of a circle on a smooth
horizontal plane, and is acted upon by a force from its centre
equal to wr per unit mass of the string. Shew that its
radius will vary harmonically about a mean length

2adc + 2arr— mpc,
m being the mass of the string, if 273A > mpc.
Examine the case when 27 = myc.

40. <A particle is placed within a straight tube of length
a and small section and the ends of the tube closed. If the
tube is placed on a smooth horizontal plane and projected in
the direction of its length, prove that the distance advanced
by the tube between the 1st and (n+1)" impacts between
the particle and the ends of the tube is 5 (e"—1); the
masses of the tube and particle being equal and the coefficient
of restitution between the particle and the ends of the tube
being e.

41. Two equally elastic balls, notin the same vertical
line, are dropped upon a hard horizontal plane: if the balls
ever come again simultaneously to positions of zero velocity,
prove that their initial heights above the plane are as

(e" — 1) to (e"— 1)’,
where m and n are positive integers, and e is the modulus of
elasticity.

42. P isa point to which the time ¢ of sliding from A
and B in straight lines isthe same. Find another point @ for
which the times from A and B are also ¢; and shew that if
t’ be the time from Pto Gor @ to P

#~t* _AP.BP  AQ.BQ
f+t? AQ.BQ” AP. BP!

 
RECTILINEAR MOTION. 77

43. A particle is projected with a given velocity in a
medium in which the resistance varies as the cube of the
velocity; find the time in which it will traverse a given
distance, and the velocity which it will have at the end of a
given time.

44, A string hangs over a fixed pulley; a weight of two
pounds hangs at one end, and a pulley at the other ; over the
pulley hangs a string, carrying a weight of one pound at each
end; when the whole is in equilibrium, any force is applied
to one of the smaller weights: shew that, when it has pulled
it down three inches, the other one pound weight, and the
two pound weight has each risen one inch; shew also that, if
the motion of the weight to which the force was applied
be stopped in any gradual manner, the whole will be brought
to rest, and the distances travelled by the weights will be as
oe koto.

45. Two inclined planes of the same altitude and in-
clinations a, 8, are placed back to back with an interstice be-
tween. Two weights P, @ are placed one on each plane at
. the bottom and connected by a string which passes over
two small smooth pullies at the top and under a moveable
pulley, weight W, which hangs between the two planes, the
free portions of the string being parallel. Find the least
value of W in order that both weights may be drawn up: and
if they arrive at the top at the same time, prove that

4 (sin? a — sin’ 8)

 

W
_ 2sina+sinasin8+sin?8 2sin8+sinasin§+sin’ a
= P 0 ie
46. <A particle, of mass m, initially at rest at a distance a

4

) to the

origin, r being the distance; find the time in which it arrives
at the origin.

from the origin, is acted upon by the force mu ¢ + "4

47. Examine the motion of a particle, initially at rest in
a medium the resistance of which varies as the velocity,
under the action of a force to a fixed point varying as the
distance.
78 RECTILINEAR MOTION.

48. A train travels at the rate of 45 miles an hour.
Rain is falling vertically, but owing to the motion of the
train, the drops appear to fall past the window at an angle
tan” 1:5 with the vertical. Find the velocity of the rain-
drops.

tf the raindrop were divided into 1000* equal spherical
portions, prove that the cloud so formed would have a velo-
city of 044 ft. per second.

The velocities are supposed to be full speed velocities, and
the resistance of the air to vary as the square of the product
of the velocity into the diameter of the drop.

49. Two imperfectly elastic balls, equal in size, but
unequal in mass, are placed between two perfectly hard
parallel vertical planes, to which the line joining the centres
of the balls is perpendicular, each ball being initially at a
distance from the plane nearest to it, inversely proportional
to its mass. The balls approach each other, with velocities
inversely proportional to their masses; prove that every
impact will take place at the same point as the first does,

50. In an Atwood’s machine the heavier body P is
perfectly elastic, and Q is perfectly inelastic, and they start
from rest at the same distance a above a fixed horizontal
plane; and when P impinges on the plane and rebounds
with unchanged velocity, Q strikes against a fixed obstacle
and is reduced to instantaneous rest: determine the subse-
quent motion, and shew that the two bodies are again at
instantaneous rest when P is at a height Pa+(P+Q) above
the horizontal plane.

51. A piece of uniform chain hangs vertically from its
upper end, with its lower end just above a smooth inclined
plane; if it be let go, find the pressure on the plane during
the fall.

52. A heavy chain is suspended from one end above a
rigid horizontal plane; on the other side of the plane in the
vertical line of the chain, there is a centre of force the
attraction to which varies inversely as the square of the
distance ; find the motion of the chain, and the pressure on
the table during the fall of the chain upon it.
RECTILINEAR MOTION. 79

53. The potential at the distance r from a fixed centre
being yr", find the motion of a particle originally at rest at a
distance a from the centre, and prove that the time of

oscillation is
4 et r(*)
=a T n

nat” —- oo.
2h T (5 + *)
2°01
54, An elastic flexible ring of natural radius a is stretched

upon a circular cylinder whose radius is (1 + es =) a, and is
e

then pushed off so that all the points of the ring quit the
cylinder at once. Assume the law of compression to be that
the compressed length is to the natural length as) : P+X
where A is the pressure which would halve the length of the
string, and is equal to the tension which would double it
according to Hooke’s law. Prove that in the subsequent

motion the least radius of the ring will be , where e is the

number 2°718281828.

55. A rope passes over a smooth pulley, having a weight
attached to one end, and a monkey hanging at the other
end, just balancing the weight. The monkey suddenly starts
off, runs up a certain length of the rope at a uniform rate, and
then holds on; determine the whole motion, and prove that,
if the monkey exert his whole strength in climbing, and be at
all fatigued by the effort, he will be certainly jerked off.

56. An uniform string, whose length is / and weight per
unit of length w, hangs over a small smooth pulley with its
ends just in contact with a horizontal plane; it the string be
slightly displaced, shew that when one end has risen through
a height h the pressure of the string on the plane is

U
w (20log p= -4),
and its resultant pressure on the pulley is

i— 2h

wh Tok
80 RECTILINEAR MOTION.

57. Two particles of equal mass m are placed in two
smooth straight tubes, between which the shortest distance is
c and the angle 22.

The accelerating effect of the attraction of either upon
the other is mf(p) at distance p. Each particle is initially at
rest, one at the foot of the perpendicular, the other at a very
small distance 8 from it, shew that their respective distances
from it at the time ¢ will be

3 B {cos (t / win a) F cos (t/ pcos a)},
where 2uc = mf (c).

58. A ball is supported by a uniform jet of sand, in the
form of a thin conical surface, impinging upon it symmetri-
cally as regards its vertical diameter: prove that the weight
of the ball is Qu (1+e)cosacos(a+); where Q is the
quantity of sand discharged from the jet per unit of time, v the
velocity of discharge, ¢ the coefficient of elasticity between
the ball and the sand, a the angular radius of the circle
in which the jet impinges, @ the semi-vertical angle of the
jet.

59. Two equal buckets are connected by a string without
weight passing over a smooth pulley, and over one of the
buckets a heavy chain is held by its upper end, with its lower
end just above the base of the bucket; if the upper end
be let go, prove that the equilibrium may be maintained
by pouring water gently and uniformly into the other bucket,
provided the weight of water which can be poured in is
three times the weight of the chain. After the chain has
entirely fallen in, find its pressure on the bucket in which it
lies, supposing the flow of water then to cease.

60. One end of a heavy chain length 3a is fastened
to a small smooth ring through which the chain is passed so
as to be in equilibrium with a length a hanging freely. Prove
that if the free end be slightly displaced downwards its
velocity V when the length of the free portion is is given by
the equation
(a — a) (a + 5a)

=
OE ae BaF
EXAMPLES. 81

61. A heavy chain, of length 4a, is coiled up on a
horizontal table, at the distance a from one edge of the
table, and one end of the chain is then drawn out at right
angles to the edge and just over it; the height of the table

above the floor being a, investigate completely the motion of
the chain.

62. A chain of given length is at rest on a smooth hori-
zontal plane, with one end fastened to a point on the plane,
under the action of a repulsive force from that point varying
as the distance. If the chain be set free, find the initial
change of tension at any point, and the subsequent motion of
the chain.

If the chain impinge upon a vertical wall perpendicular to
its own direction, find the pressure upon the wall at any sub-
sequent period.

63. On a certain day between one and four o’clock in the
afternoon $ an inch of rain fell. Assuming that the drops
were indefinitely small and that their terminal velocity was 10
feet per second, find the impulsive pressure in tons per square
mile consequent on the rain-drops being reduced to rest,
assuming that a cubic foot of water contains 1000 oz. and
that the rain fell uniformly and continuously.

64, <A fine string without weight passing over a smooth
pulley, supports two equal scale-pans; a heavy chain is held
by its upper end above one of the scale-pans, its lower end
being just above the scale-pan; if the upper end be let go,
determine the motion completely, and find, at any time, the
pressure on the scale-pan.

65. A meteor is seen to fall vertically to the Earth,
leaving a bright trail behind it. If the resistance of the air
be constant and equal to £, the rate of loss of matter burnt

off be 7

entering the atmosphere, shew that the velocity after falling
through a distance z is given by
2 iv eg h}

y= —__—_**

(2A)

Rv, and if MW, Vbe its mass and velocity just after

 

+$(2-A gtk,

where 7X = R-
B.D. 6
82 EXAMPLES,

66. An indefinite quantity of a uniform string is coiled
in a heap on the floor of a room and escapes into the room
below through a hole in the floor; shew that the velocity of

escape can never exceed ,/ga, aise ais the height of the
hole above the floor of the room below.

67. A chain whose density varies as the distance from
the end A is coiled up close to the edge of a smooth table
and the end A allowed to hang over. Shew that the motion
is uniformly accelerated and the tension at the edge of the
table varies as the fourth power of the time elapsed since the
commencement of motion.

68. A pulley is fixed above a horizontal plane. Over the
pulley passes a fine string which has two equal chains fasten-
ed to its two ends. In the position of equilibrium a length e
of each chain is vertical, the remainder of the chains being
coiled up on the table.

If now one chain be drawn through a distance ne and
then let go, prove that the system will next come to rest,
when the upper end of the other string is at a distance mc be-
low its mean position, m being given by the equation

(L—m) e* = (1+n)e".

69. <A flexible chain hangs in equilibrium over a smooth
vertical circle with one end fixed to the extremity of a
horizontal diameter and a portion hanging vertically at both
sides of the circle; if the fixed end be set free, shew that the
equation for determining (y) the distance of the lowest point
of the chain from the horizontal diameter during the first
part of the motion is

(1-9) Ge (i-t) =9 (v9)

where / is the length of the whole string and 2c is the circum-
ference of the circle.
CHAPTER VI.

ACCELERATIONS PARALLEL TO CQ-ORDINATE AXES.

75. WE now proceed to illustrate the- use of the

equations

méi=mX, my =m,
by the consideration of some cases to which these equations
are the most easily applied.

If the forces are given, and the initial circumstances of
motion, the equations determine the path of the particle;
and, if the path be given, with some other condition besides,
the forces are determined.

76. Motion of a heavy particle in a vertical plane.

Measuring y vertically upwards the equations of motion

are
&=0, ¥=-g.

If w be the initial velocity and a the inclination to the
horizontal plane of its direction, we obtain
é=ucosa, y=usina—g,
and if the point. of projection be the origin,

a=utcosa, y=utsin a —dgt’.

Eliminating ¢ we obtain the equation to the path of the
particle,
ge
Y = GAD OB costa

6—2
84 ACCELERATIONS PARALLEL TO

which is a parabola, having its axis vertical and vertex
upwards, the co-ordinates of the vertex being
u* sin 22/29 and u’sin® a/2g.

The latus rectum is 2u*cos’a/g, and the height of the
directrix above the point of projection is u*/2g.

The direction of motion at the time ¢ is given by the
equation 5

pe ee
U COS a

and, if v be the velocity at the point (a, y),

V=H +P =u + ot — 2ugt sina =u? — 2gy,
shewing that the velocity at any point is equal to the velocity
due to a fall to that point from the directrix.

The range of a projectile on the horizontal plane through
the point of projection is at once seen to be
u’ sin 22/9.
To obtain the range on an inclined plane, perpendicular
to the plane of motion, through the point of projection,
measure a and y parallel and perpendicular to the plane.

Then, if @ be the inclination of the plane,
#=-gsmP, ¥ =—gcosP, °
. x=ucos(a—f).¢t— 49 sin Br’,

y= usin (a — 8).t—49 cos Bt.
‘Putting y = 0, and eliminating ¢, we obtain

 

_ 2u* cos asina—B

g cos’ B
which is the range on the inclined plane, and is a maximum

2

when a=} (5 + p) , that is, when the direction of projection

bisects the angle between the inclined plane and the ver-
tical.

To find the direction of projection in order that the
particle may pass through a given point is the same thing
as finding the direction of projection in order that the
range on a given inclined plane may be a given quantity.
CO-ORDINATE AXES. 85

Hence from above cosasin a— is given,

and therefore sin 22 — 8 is given.
Tf then a’ and «” be the two values of a, the expressions
2(@’—8)+8 and 2(a”’—£)+ are supplementary, and
therefore

(a8) +(e") =2 45-5},

shewing that there are two directions of projection and
that they are equally inclined to the direction of projection
which gives the greatest range.

77. .Conversely, if the given path be a parabola, and it
be given that the force is parallel to its axis, we have
oy? = daa,
with the equations miéi=mX, y=0,

so that y =c, a constant,
2

and therefore t= ond sal,
2a 2a

shewing that the force is constant.

78. To find the force perpendicular to the axis under
the action of which a conic section can be described, we have
the equations, yf? = 2le — ne’,

&£=0, mi=mY,

From these we obtain

é=c, and yy=(l—-nz)c,

Pe?

hence yy +4? =—ne’, and therefore 7=— ra :

shewing that the force varies inversely as the cube of the

ordinate. .
Conversely, if it be given that the force is perpendicular

to the axis of # and inversely proportional to y’, the equations

of motion are #=0, j=-h

ee Le
and therefore a=, y=p & + 5 :
c and 6 being constants.
86 ACCELERATIONS PARALLEL’ TO

lda by

Hence, aay = eis ;
and therefore = +a= be Ve +y,

which is the equation of a conic section.

79. Itis required to find the force perpendicular, to the
asymptote, under the action of which the curve, 2° +y°=a’,
can be described.

In this case, if P be the force,
ie de P
ti "759 BRS WY = 75
Hence &—Yy=c,
and from the equation of the curve 2° +y’y=0.
These equations give

ae cy? je ou?
Bay? UA ay
actatny :
and i= "Grey whence the force;
a/ 4 ry
and the velocity at any point = a ‘

80. In all cases in which force acts in parallel lines, the
velocity in the direction perpendicular to the force remains
constant, and therefore the time of traversing any arc of the
orbit is obtained by observing the space, passed over in the
direction perpendicular to the force, and dividing this space
by the constant velocity.

%
81. Motion of a particle under the action of a force to
a fixed point proportional to the distance from that pownt.

If r be the distance, and yr the force on a particle of unit
mass, the equations of motion are

&=—prosd=—pa, j=—prsind=— py.
The integrals of these equations are, Chapter II.
w=Acos/ut+Bsin Jut, y=COcos/ut+ Dsin J ut,
CO-ORDINATE AXES. 87

the constants being determined by the initial circumstances.
of the motion.
Eliminating ¢ we obtain the path
(Cx — Ay)’ + (By — Dx) = (BC — AD)’,
which is an ellipse.

82. In this case we may usefully employ oblique axes.

O being the centre of force, and A the point of projection,
take OA as the axis of x, and the line through O parallel to
the direction of projection as the axis of y.

If f, f be the component accelerations parallel to the

axes “
2 2

fsina= z sina), and eee sina),
di de

so that f=4& and f'=j7.
The component forces are — px and — my, and resolving
perpendicular to the axes, we obtain
Gsna=—pesina, ysina=— pysina,
or g=—pat, Y=—py.
Initially z=a, &£=0, y=0, y=,

and therefore aw=acosJ/mt, y= s sin /t,
fo

2 2
and the path is 7 + i =I,

an ellipse of which 2a and ae are conjugate diameters.
be
If the force be repulsive the equations of motion, employ-
ing oblique axes, are :
=e, Y= py,
and therefore
w= AcVit + Be- Vue, y= Cet 4. Dem Vet,

and, with the same initial eonditions,

a, Nv aye u a
e=5 (eVet 4. Vat), v-5 Ge" Vuty,
88 ACCELERATIONS PARALLEL TO

and the equation of the path is

a
a vw ?

a hyperbola, of which 2a and 2” T are conjugate diameters.

In each case if 2a and 2b be the conjugate diameters
v=J/p.b,

and, as any point P of the path may be regarded as the

initial point, the velocity at P is equal to Ju. CD, where
CD is conjugate to CP.

83. General deductions from the equations of motion.

Taking =X, and y= Y, |
we obtain aij — ya = 2 Y —yX,
@ p24
or iv) =™.

if mN be the moment of the forces about the origin.
The angular momentum of the particle is the moment of
its momentum, which is
m (ay — yt).
If then we denote by mh the angular momentum, we
obtain
mh = mN,
that is, the time flux of the angular momentum is equal to
the moment of the forces.
Again we know that, if A be the area swept over by the
radius vector,
2A = xy — yd,
and .’. 2A=N.

If N be zero, that is, if the force be central, A is constant,
so that the area swept over is in that case proportional to
the time.

This is Newton’s Proposition I.
CO-ORDINATE AXES, 89

_ , Further we obtain, multiplying by 2#, 27, adding, and
integrating,

at ytae { (Xde + Yay),

or Lm (a2 + 9’) = { (Xda + Yay).

The left-hand side is the kinetic energy of the particle,
and the right-hand side is the work done by the force, so
that we here have the principle of energy, for a single particle
in a field of force, deduced from the equations of motion.

84. Motion of a heavy particle in medium the resistance
of which varies as the velocity.

Measuring x and y horizontally and vertically from the
point of projection, and taking mkv as the force of resistance
the equations of motion are
da
ds :
and therefore, if w and v be the initial components of the
velocity,

i im EL pe
=— ke, and j = — hs =" —g= ky — 9,

#=— ks

&tka=u, yt+tky=v- gt,

leading to a= ; d=e*,
at gto _ (v 2) -H
Y=E ETB GtE a

85. Motion of a particle in the same medium under the
action of a force to a fixed point varying as the distance from
that pownt.

The equations of motion are
Bt+ki+pe=0, ytky+py=0,
and these are integrated as in Chapter II.
If, initially, «=a, y=0, #=0, and y=», it can be
shewn that

2 2
Co oe
a av v
90 RESISTING MEDIA.

 

__ 2any aia
and Me alae where n=A/ p 4

86. Motion of a heavy particle in a medium the resistance
of which varies as the square of the velocity.

The equations of motion, if mkv’ be the force of resist-
ance, are

#=- bt 2 ahs Yoig— ha.

From the first, &=ue™,
and .. y =p& = upe™, putting p for oe :

Hence 7=wv ap e™ — kupse™ = we e™ — ksy,
and the second equation becomes

ap GF om _
Ne ae =0,
ds
dz’

integrating, we obtain -

Multiplying by 2 which is equal to 2,/1+p*, and

p Ji + pi+log(pt+Ji+p) +o "=C.

If then « be the initial inclination of the direction of
motion, that is when s = 0, and if p=tan 9,

tan 8 sec 0 + log (tan @ + sec @) + % (e* —1)

= tan a sec a + log (tana + sec a).

We observe that as s increases indefinitely, i decreases
and ultimately vanishes, shewing that «# is ultimately
constant, and therefore that the curve, on the positive
side, tends to a vertical direction.

Again, if t be the intercept of the tangent on the axis of y,

ray —ap nde =—a Pao Ke :
MOVING AXES. 91

_Take « and s both negative, say — a’ and —s’; then as
x’ is certainly less than s’,

ae s
gw is less than =
which vanishes when s’ increases indefinitely.
Hence on the left-hand side, the value of approaches
continually to zero, and 7 is ultimately constant.
Again, ifs=— «,
tan 6 sec @ + log (tan 6 + sec 0) -C =0,
shewing that 6 has then a finite value; for, if we put @=0,
the left-hand number of the equation is negative, and if we put
T
d= 3?
We infer then that the direction of the curve, on the
negative side, is ultimately inclined to the vertical.

87. Illustrations of the use of the equations of Arts. (18)
and (21).

(1) Case of a weightless rod, of length 2a, carrying a
particle at a distance c from tts middle point, and having rts
ends moveable on two straight wires, at right angles to each
other, and made to revolve uniformly in a horizontal plane.

it is positive.

y

 
52 MOVING AXES.

Taking the wires as axes of # and y, the equations of
motion are
m (# — o's — 2wy) = RB,

m (ij — wy + 20%) = PR’,

with the geometrical conditions,
x=(a+c) cos 6, y= (a—c)sin 0.
Now, since the system of acting forces is the equivalent

of the system of effective forces, the resultant of R and R’
must be in the line P#, and therefore

R(a+c)sin 6=R’' (a—c) cos 0.
From these equations we obtain
(a? + — 2ac cos 26) 6 + 2acé sin 20 + 2ace* sin 26 = 0,
the integral of which is
(a? + c? — 2ac cos 20) 6? = C + 2acw’ cos 26.

If the particle is at the centre of the rod, 6 is constant, so
that, if initially the rod have no motion relative to the wires,
its relative position will remain unchanged.

(2) A smooth plane, carrying at-a point C a centre of
force the attraction to which varies as the distance, revolves’
umformly about a fixed axis, perpendicular to itself; it is
required to determine the motion of a particle on the plane.

Taking the foot of the axis, O,.for origin, and OC for the
axis of x, the equations of motion are, if OC =<,
& — wa — 2oy = — (a —c),
9 — wy + 20% = — py.
The elimination of y leads to the symbolic equation
{D* + 2 (co? + w) D® + (w* — w)"} @=— po (w* —p),
the solution of which is

 

 

{fo + Vut + a} + B cos fo — Vt + A};

       

and y is obtained in a similar form.
If it be required to determine the absolute path of the
particle the equations of motion are

&=—p(x—ccos wt), y=—p (y—Csinat),
MOVING AXES. 93°

the integrals of which are
pec

tf ed 7
Bow

 

cos wt + A cos (Vut +a),

 

y= sin wt + B sin (Vut + f).

Bo

88. Equations of motion of a free string in a plane
under the action of given forces.

¥

 

 

 

If w and v be the velocities parallel to w and y, of a point

P of the string, and 7 the tension at P, the component
tensions are 3 p
pa oy

T ds’ and T ae |

which are functions of the arc AP measured from a given

point A of the string, and the equations of motion of the

element &s are
a
ds

més t=

(7 2) 8s -+mds. X,

. @ dy
| mis i= 5-(T ) ds + més. Y,

t
94 MOTION OF A STRING,

m being the mass of unit length of the chain, or

. ad/,dx
mis = 5,(7 3) +mX,
a er ae

The geometrical equation is given by the condition that
the velocity of Q, in direction of the tangent at P, is ulti-
mately the same as that of P, leading to

dx dy da dy
(u+ bu) 7+ (v+ bv) a i ae

dudx | dv dy

ds ds * ds ds
These equations, combined with the boundary conditions,
such for instance that the tension vanishes at a free end of
the string, are sufficient theoretically for the determination
of the motion of a string in a plane under the action of given
forces.

The same equations apply to the problem of determining
the initial tension at any point of a string, originally in
equilibrium, when some of its constraints are removed.

It will be found however that this problem, and that
of the next article, are more easily treated by making use of
tangential and normal components.

or =0.

89. Motion of a string, or a fine chain, lying on a smooth
horizontal plane, which has impulses applied at one or both
ends.

If Tbe the measure of the resulting impulsive tension at
the point P, and wu, v the velocities imparted to the point P,

mos. U = £(2 “| és, mss.v=4 (7%) 88,

ds\" ds ds ds,
_m@e , dx dT _ may , dy dT
or mu=T oat 7: ie? m=Toeta a
EXAMPLES. 95

with the same geometrical condition as before, i.e.
duda | du dy
ds ds * ds ds

EXAMPLES.

1. Find the law of force parallel to an asymptote under
which a rectangular hyperbola can be described.

2. Prove that the vertical velocity of the centre of
curvature of the path of a projectile is proportional to the
time which has elapsed since the projectile was at the highest
point of its orbit.

3. A particle moves in a plane under the action of forces
to two fixed points in the plane, one attractive and the other
repulsive, and each varying as the distance; if the absolute
intensities be the same, find the path.

4, Find the path of a particle which is in motion under
the action of a force perpendicular to a fixed line and
inversely proportional to the square of the distance from
the line.

5. A particle moves under the action of an attractive
force perpendicular to a straight line and proportional to the
distance from that line; prove that the path is the curve

of sines.

6. Shew that the least inclination to the horizon at
which a particle can be projected so as to strike at right
angles any plane through the point of projection is cos™4.

If the direction of projection be inclined at an angle @ to
the plane, and if the projection on the plane of this direction
be inclined at an angle ¢ to the line of greatest slope,
shew that the range on the plane is

 

es

= = (cos? @ cos’ g + sin” @ sin”? a—4 sin 20 sin 2a cos ¢)},
g cosa *,

where a is the inclination of the plane to the horizon.
96 EXAMPLES.

7. Particles are projected from the same point in a

vertical plane with velocities which vary as ‘(sin @)” 4 @ being
the angle of projection ; find the locus of the vertices of the
parabolas described.

8. Particles fall down chords of a circle to the lowest
point. Prove that the tangents to the circle at the upper
extremities of the chords pass through the foci of the para-
bolas described after leaving the lowest point.

9. Particles slide down chords of a vertical circle to
the lowest point; shew that the locus of the foci of the paths
of the particles after leaving the chords is a cardioid.

10. An ellipse is held with its major axis vertical ;
find a point on the curve such that, if a perfectly elastic
heavy particle slide down an inclined plane to it from
the upper focus and be reflected by the curve, it will fall
to the lower vertex; and shew that in an ellipse, whose
eccentricity is 5, this point will be the extremity of the
minor axis.

11. A bullet is fired in the direction towards a second
equal bullet which is let fall at the same instant. Prove
that the two bullets will meet, and that if they coalesce
the latus-rectum of their joint path will be one quarter
of the latus-rectum of the original path of the first bullet.

12. Chords are drawn, joining any point of a vertical
circle with its highest and lowest points; prove that, if a
heavy particle slide down the latter chord, the parabola,
which it will describe after leaving the chord, will be touched
by the former chord, and that the locus of the points of
contact will be a circle.

13. Through a given point an inclined plane is drawn,
perpendicular to a given vertical plane, and from that point’
a particle is projected in the vertical plane, with a given
velocity, so as to strike the inclined plane at right
angles; prove that the locus of the point on which it
falls, for different positions of the inclined plane, is an
ellipse.
EXAMPLES. 97

14. Prove that the range of a projectile on an inclined
plane is greatest for a given velocity of projection when the
focus of the path is in the plane.

If ¢ and ¢’ be the two times of flight corresponding to any
range short of the greatest, and a be the inclination of the
plane, prove that

C4 t? + 2te’ sin a
is independent of a.

15. Two smooth equal balls are placed in contact on a
smooth table; a third equal ball strikes them simultaneously
and remains at rest after the impact ;. shew that the coefficient
of restitution is 2/3.

16. A person at a distance ¢ from the vertical wall of a
fives court discharges a perfectly, elastic ball from. his hand so
that it shall strike first the floor, then the wall and finally
return to his hand. If Z' be the whole time of motion
and nT the time between the two rebounds, shew that the
velocity and angle of projection measured downwards are
given by

V cos a = 2¢/T, Vsin a= gnT.

17. On the moon there seems to be no atmosphere, and
gravity is about one-sixth of that here on earth. What space
of country would be commanded by the guns of a lunar fort,
able to project shot at 1600 feet per second ?

18. A particle is projected from a point in a smooth
plane, inclined at an angle a to the horizon, in a vertical
plane which cuts the inclined plane in a horizontal line, and
at an angle @ to the horizon. Prove that after m rebounds
the space travelled in the direction of the line of greatest
slope on the inclined plane is
e(1 —e””*)

l-e ’
where a is the horizontal space described, and e the coefficient
of restitution. ‘

B,D. 7

asinatan@.
98 EXAMPLES.

19. A perfectly elastic particle is dropped from a point
on a fixed vertical circular hoop; shew that after two rebounds
it will rise vertically if

2 sin 46 = tan 6,

where 6 is the angular distance of the point from the highest
point of the hoop.

20. The intersection of two planes, inclined to the hori-
zon at the angles a, 8, respectively, is horizontal: a perfectly
elastic particle falls upon the former plane, then rebounds to
the latter, and then rebounds vertically: prove that the ratio
of the distance of the first point of impact from the inter-
section of the planes to the height through which the particle
falls before striking the first plane is equal to

cos (a+ : :
ots, . {2 sin 2a — sin (2a + 28)}.
21. A smooth inelastic ball slides from rest down a length
(2) of a plane inclined 30° to the vertical, and impinges on a
horizontal rail, parallel to the plane, and at a distance from it
equal to half the radius of the ball. Neglecting the thickness
of the rail, prove that the ball will afterwards strike the plane
at a distance 3/ from its point of contact when striking the
rail.

22. A particle is projected from a point at the foot of one
of two parallel vertical smooth walls so as after three reflections
at the walls to return to the point of projection; prove that

e+é+e=l,

and that the vertical heights of the three points of impact
above the point of projection are as

e’:1-é@:1.

23, A sphere is fixed upon a horizontal plane; find from
what point in the plane a particle must be projected, with a
velocity due to falling down a vertical space equal to the dia-
meter of the sphere, so that the focus of its path may be in
the centre; after reflection at the sphere shew that it will
EXAMPLES. 99

strike the horizontal plane at a distance from the point of
projection equal to the diameter of the sphere, if the elasticity
be perfect.

24. Two equal perfectly elastic balls A and B are pro-
jected from the same point, in the same vertical plane and
with equal horizontal velocities; prove that if when they first
impinge A has not yet struck the ground, and B is moving
horizontally, then the ratio of the cotangents of the angles
which the directions of projection make with the vertical
Qn? + 2n+1
: 2n+1
also that when first after the impact one of them strikes the
ground, the other is at a height above the ground

_ 2n(n+1)
— Qn+IyP g’
where »v is the vertical velocity of projection of B.

must = where » is some positive integer, and

25. A ball is projected from a point in one of two smooth
parallel vertical walls against the other in a plane perpendi-
cular to both, and after being reflected at each wall impinges
again on the second at a point in the same horizontal plane
as it started from: shew that

b? =a(1+et+e),
where e is the coefficient of restitution, b the free range, and
a the distance between the planes.

26. A man stands on the upper end of a long rough
plank, of length a and mass M, which lies along a smooth
straight groove on an inclined plane, having its upper end
supported by a cord. The cord is cut, and, at the same
instant, the man starts off, and runs with very short steps
down the plank, at such a rate that the plank does not move;
prove that the velocity of the man at the lower end of th
plank is ,

m+M
a/ 2900008 at

where m is the mass of the man, and 2 the inclination of the
groove to the vertical.

 

 

7—2
100 EXAMPLES,

If the man then jump horizontally so as not to set the
plank in motion, he will alight on the groove at a distance

4a cot? a. wat
m

below the position of the lower end of the plank at the instant
he alights,

Determine also how he must jump so that he may alight
on the lower end of the plank.

27. A stone is projected upwards with velocity /2gc
from a point on the margin of a circular pond, radius c. IEf
all directions of projection be equally probable, shew that the
chance that the stone falls into the pond is

28. A solid smooth cylinder, of radius 1, lies on a smooth
horizontal plane, to which it is fastened, and an inelastic.
sphere, of radius 27, moves along the plane in a direction at
right angles to the axis of the cylinder; find the condition
that it may pass over the cylinder.

If the sphere be elastic, and the coefficient of the elasticity
be greater than 1/8, prove that it cannot in any case pass over
the cylinder; and if e be less than 1/8, find the condition that
the sphere may, after its first ascent, fall upon the top of the’
cylinder.

29. From a point A in one of two vertical lines a par-
ticle is projected with a velocity w at a given inclination
to the horizon, and meets the other vertical line in B: it is
then projected from B with a velocity v at the same inclina-
tion to the horizon and returns to A. Prove that the har-
monic mean between w and v” is constant.

30. A particle is projected horizontally along a plane,
whose inclination (a) to the horizon is equal to the angle of
friction (A) between the particle and the plane; prove that it
will ultimately move down the plane with a uniform velocity
‘EXAMPLES. 101

equal to half the velocity of projection (V) and along a line,
whose horizontal distance from the point of projection

Vv?

gsin a’

 

= 2
~3

If a be less than » and tan’ =n tana, and if the particle
receive a small horizontal impulse and when reduced to rest
again, another, and so on continually, its path down the plane
will not sensibly differ from a straight line inclined to the
: das _ ya 4-1
horizontal line in the plane at an angle =.tan net ay’

31. Find the path of a particle acted on by a repulsive
force always perpendicular to.a given straight line and pro-
portional to the distance from it, the velocity at any point
being that which would be acquired by moving from rest on
the given line to that point.

32. Ifa particle be acted upon by a force always parallel
to the axis of y and proportional to the square of the radius
of curvature at the point, prove that it will describe the curve

ae
- log sec -,

the particle moving parallel to the axis of x at the point (9, 6).

33. The trochoid 2 =a(@—esin @), y=a(1—e cos 9), is
described under the action of a force parallel to the axis of a;
e—cos 0

sin’@ ~

 

shew that the force varies as

34. Ifa particle be moving in a medium whose resistance
varies as the velocity of the particle, shew that the equation
of the trajectory referred to the vertical asymptote and a line
parallel to the direction when the velocity is infinite as co-
ordinate axes, is of the form

y= blog =.
102 EXAMPLES.

35. A body describes the curve whose equation is

a ye
+4 =I,

under the action of a force to the centre. Shew that the
central force is Ar. (ay)"*, and that when n is even, the
periodic time

2A n—Il

(ab)eV ®

where A is equal to the area of the curve.

36. Two particles A and B, of masses 8m and m respec-
tively, lie together at a point on a smooth horizontal plane,
connected by a string which lies loose on the plane; B is
projected at an elevation of 30° with velocity equal tog. If
the string becomes tight the instant before B meets the
plane again, and breaks when it has produced half the im-
pulse it would have produced if it had not broken, and if the
particle rebounds at an elevation of 30°, shew that the elasti-
city of B is equal to 5/9.

37. A parabola, having its vertex at A and its axis co-
incident with AB the diameter of a semicircle, is described so
as to cut the semicircle in P; prove that, if a body move in
the semicircle under the action of a force perpendicular to
AB, the time of moving from A to P varies as the difference
between AB and the latus rectum. Prove also, that if a
second body move from A to P in the parabola in the same
time under the action of a force perpendicular to its axis, and
the velocities in the two curves at P be equal, the latus
rectum of the parabola is 4 AB,

38. A parabola is described under an acceleration f to
the focus, and another a parallel to the axis; shew that

2 (f+a+ tao

‘ °

39. Two particles are projected in parallel directions
from two points in a straight line passing through a centre of
EXAMPLES, 103

force, the attraction to which varies as the distance, with
velocities proportional to their distances from the centre.
Prove that all tangents, to the path of the inner, cut off,
from that of the outer, arcs described in equal times.

40. OA is a smooth tube; OB a light rod perpendicular
to it; B, a fixed point in OB, a centre of force attracting with
force wr a particle P in the tube OA. The system being
made to revolve with uniform angular velocity w on a hori-
zontal plane about O, determine the motion of P; and shew
that, if ~ > o, P will oscillate with period 27/,/u — o”.

41. A rod revolves about its middle point with uniform
angular velocity a and has at its extremities two centres of
force varying as the distance one attractive and one repulsive
of the same absolute intensity; supposing a particle placed
in the plane of rotation in a line perpendicular to the rod
through its centre, shew that its path will be cycloidal, the
time from one cusp to another being 27/a.

42. A smooth horizontal disc rotates with angular velo-
city J about a vertical axis at which is placed a particle
attracted to a certain point of the disc by a force whose
acceleration is sx distance; prove that the path of the
particle on the disc will be a cycloid.

43. A particle moves under the action of two constant
forces in the ratio of nine to one, whose directions rotate in
opposite directions with uniform angular velocities in the
ratio of three to one: prove that, under certain initial con-
ditions, the path of the particle will be a closed curve, of the
same form as that represented by the equation, r =a cos 20.

44, The two ends of a smooth weightless rod are move-
able on two fixed straight wires intersecting each other at
_right angles. A particle can move on the rod and is attracted
to the point of intersection of the wires by a force varying as
the distance. Prove that if the particle have initially no
motion the angular velocity of the rod is given by an equation
of the form

ow” = n? {1 —sin? 2a cosec” 26}.
CHAPTER VII.
RADIAL AND TRANSVERSAL ACCELERATIONS.

90. Havine discussed, in the previous chapter, the use of
the components of acceleration parallel to two coordinate
axes, we now take into consideration the expressions for
radial and transversal components, leading to the equations
of motion, 7
, f¥—-7rP=P, rO4+ 2d=Q,
mP and m@ being the radial and transversal forces acting
on a particle of mass m.

For our first illustration we take the following case.
Motion of a particle in a smooth straight tube which

revolves uniformly, in a horizontal plane, about a fimed point
in the axis of the tube.

In this case the only force acting on the particle is the
pressure R of the tube, and, if w be the angular velocity, the
equations are

r—owr=0, 2mro = R.

If the particle start from the distance @ with no initial

velocity along the tube, we obtain from the first equation,

r= 5 (e*# + e-°*) =a, cosh. ot,

and from the second,
R= 2mae’ sinh «ot.
ROTATING TUBES. 105
We can also obtain the velocity and pressure in terms of
r, for the first equation gives
Pr (Cd = a’),
and therefore R=2Ime’ Jr — a.

If b be the length of the tube, the direction in which
the particle flies out is inclined to the tube at an angle 0
such that

be b

t 6 = —___ = ——— 5
si o /P-a Je —a'

If the tube revolve in a vertical plane the equations are

*— o'r =— sin wt, 2mrw = R— mg cos ot.

 

From these equations,

= aoe sin wt + Ac? + Be Chapter IL),
2 P

and R= 2mqg cos wt + 2mo* (Ae** — Be-**),

the constants being determined by initial conditions.

91. Motion of a particle in a straight tube which revolves
uniformly in a horizontal plane about a fied point at a
distance c from tts axis.

If OA (c) be the perpendicular from the fixed point on
the axis of the tube, and, P being the position of the particle,
if AP =r, the acceleration of P relative to A =7— o'r, in the
direction AP, and 27 perpendicular to AP; and the accele-
ration, wc, of the point A is wholly in the direction AO.

Hence the equations of motion are
F¥—w'r=0, m(2Qrw+o’c) = f,
and the solution is similar to that of the preceding case.
Motion of two particles,m and m’, connected together by an

inelastic string, in a straight tube revolving uniformly in a
horizontal plané about one end. is
106 ROTATING TUBES.

If r be the distance of m from the origin, J the length of
string, and 7’ its tension, the equations are

m(#— o'r) = T, m' {F- ow (r+ )}=—7,
from which we obtain

(m +m’) #— o (m+ m') r — mol = 0,

 

tthe 1 = Aevt + Be-t,

and therefore r+
m+m

92. Motion of a piece of uniform chain in a straight tube
revolving in the same manner.

 

 

Let r be the distance from O of G, the centre of gravity
of the chain, GP =p the distance from G@ of one end of an
element PQ, (6p) of the chain, m the mass per unit of length,
T the tension at P, 7'+ 67 at Q.

The equations of motion of the element are, since p is

2
independent of the time and therefore (OP) equal to #,
mép {i — wo" (r+ p)} =8T, 2mdprw = Rp,
R being the rate of pressure at P per unit of length.

Integrating the first of these equations, or, in other words,
taking the sum of the equations of motion of all the elements,
we obtain

m{G—o'r)p-a GL = T+ 0,
ROTATING TUBES, 107
Observing that 7=0 when p=—I and when p=1, 21
being the length of the chain, there results
m (¥— o'r) 21=0, or #— o'r = 0,

shewing that the motion of the centre of gravity is the same
as that of a single particle.

Taking account of this result, 7’= 4mo’ (I? — p?).

93. Motion of a bead on a smooth circular wire revolving
uniformly in its own plane about a fixed point.

 

 

If O be the fixed point, and @ the angle PCA, the accele-
rations of P in the directions PZ’ and PC, obtained by
compounding the accelerations of P relative to C with that
of C relative to O, are respectively,

a q (6+ ot) + a%csin 6, and a(6 +)? +0°c cos 8,

taking OC =c, and CP =a.

If the plane be horizontal, and & be the pressure of the
wire on the bead, the equations of motion are

ab + wcsin@=0, and m {a(6+a)*?+’c cos 6} = R.
108 CENTRAL ORBIT.

If the bead be originally attached to the wire at the
angular distance a from the line CA, and be set free, the first
of these equations leads to

a” = 20° (cos 6 — cos a),

shewing that the bead oscillates through the angle 22, and
the second determines the pressure.

CENTRAL FORCES.

94. If a particle move under the action of an attractive
force mP to a fixed centre, P being a function of r, the
equations of motion are

¢#—7rP=—P, 164+ 270=0.
From the second equation we at once obtain
76 =h,
h being a constant.

Hence t= 79= 5 pa ap if “=
p= nt gai pe
and the first equation becomes
eae Pe
dé Y
du P
or ap Ya

This equation, if the law of force be given, determines
the path, and, if the path be given, determines the law of
force.

95. If A be the area swept over by the radius vector,
A=}r6=hh,
CENTRAL ORBIT. 109

and therefore A = ht, shewing that the area is proportional
to the time, and that 4h is the area swept over in the unit of
time.

If ds be an:element of the arc of the curve, and p the
perpendicular from the centre of force upon the tangent,

$A = Lpés,
and therefore: h=2A =pi=pv,

shewing that the velocity is inversely proportional: to the
distance of the centre from the tangent to the path.

96. We have shewn,.ix. Art. (17), that the normal accele-
ration, in any curvilinear path, is equal to v°/p, where p is
the radius of curvature.

If mF be the resultant of the forces acting on a particle
m,and ¢ the inclination to the normal of the line of action of
this resultant, it follows that

2
m i = mF cos $,

and therefore, if g be the chord of curvature in direction
of the force,

v= 489.

That is, the velocity of a particle at any point of an orbit
is that which the particle would acquire if it were to move
from rest under the action of a constant force, equal to the
force at the point, through a space equal to one-fourth of the
chord of curvature in direction of the force.

97. Since — and cos =f,
P r

Ve dr

we have pre PPT
2

and therefore P= z 7 ;

an expression which is frequently useful in determining the
path for a given law of force, or the law of force for a given
path. :
110 ° CENTRAL ORBIT.

We can also obtain the same expression for P by employ-
ing the expression for the tangential acceleration.

; dv dv.
Thus —Psin =z =v | sing;
__ hd /(h\ _h'dp
therefore Pa~ oo (S)-p ae

The two expressions for P are deducible, each from the
other, by help of the equation,

i sy (S).
o dé)"

98. Another expression for the velocity is found by
utilizing the expression for the tangential acceleration; we
thus obtain

dv dr
ta? ae
and therefore y=y?—2 | Pdr,

shewing that v is a function of the distance.

Further, since
h? 1. 1 /dr\?
2 pa SS ge
- p t {a+ (G5) f>
: dé. ‘ :
it follows that r dr 8% function of the distance, and therefore

that at all points which are at the same distance from the
centre of force the angle between the radius vector and the
tangent is the same.

We now proceed to apply these formule to some particu-
lar cases.

99. To find the law of force to the focus under the action
of which a conic section can be described.
CENTRAL ORBIT. 111

Taking as the equation of the conic

cu=1+¢ cos 8,

 

eet 6. and du lak
de: ecos Vv, an det Y=
2,2 2
so that pore ate)
¢c uh6UcclUr

and therefore varies inversely as the square of the distance.

If «w be the absolute force, that is the force at the unit of
; 2 2
distance on a particle of unit mass, yu =* = * , if LZ be the

latus rectum.

To find the law of force to the centre under which a central
conic can be described.

2

Employing the equation, P= 3 a we know that in
this case,
p (? + Gb? peas *) ace a’b*,

272 2
and therefore — no =— an so that P= e Tr;
ip p ab

and therefore varies as the distance.

2

If « be the absolute force, w = ore

100. To find the law of force to the pole under which an
equiangular spiral can be described.

From the definition of the curve, p=rsina,

and therefore P= .4=35)

and the velocity =
112 CENTRAL ORBIT.

To find the law of force when a particle describes a circle
under the action of a force to a point in the circwmference.

 

In this case Pe 5 , or 7° = 2ap,
2 2 2
and therefore P=" : “ = sl =4
: Qah_ 1 i”
and the velocity Soe Sal g:

101. Motion of a particle under the action of a force to a
fixed point varying inversely as the square of the distance.

In this case, P = pu’, and we obtain
du be
rT i

the integral of which is

u— =A cos (0-4),

or u =F {1 +ec0s (0 —y)},
which is the equation of a conic section, of which e is the
2
eccentricity, and x is the semi-latus rectum.
To find the constants, let c be the initial distance, v the

velocity of projection, and 8 the inclination to ¢ of the initial
direction of motion ; then,

‘ - au :
h=ve sin 8; and, since a ~i esin (@—y),

+ =H (1 +6 0089) and “cot @=—Hesiny,

or vesin’ 8 =p (1+ecesy), v’csin @ cos 8 = — pesiny ;
CENTRAL ORBIT. 113

whence = (v’e sin? 8 — p)? + v'c* sin’ B cos’ B = pe’,
vic’sin®B  2v*c sin’ B

2 ?

pb

ve
sod Enis Does.
p—ve sin’ B

or e=1+

It follows that the conic is an ellipse, parabola, or
hyperbola, according as v? is less than, equal to, or greater
Qu
than —.
c
If 2a and 2b be the axes when the curve is an ellipse,
2

hi
—=a(l—e’);
naa (L=e)
and therefore
ve* sin® B * (= sin’B a vc’ sin? 8)

2

 

a be ie
2
or foe
c a

Since any point may be regarded as the point of projec-
tion, the velocity at the distance r is given by the equation

pale _b
ra

In the same manner, if the conic be a hyperbola, we
find that 2
2 2p + & :

r

We can also solve this question by the use of the equation
h' dp _p_
pdr to #
leading to t _ 7H +6,
por
which is an ellipse, parabola, or hyperbola, according as C is
negative, zero, or positive.

B.D. 8

Vv
114 CENTRAL ORBIT.
: Qu

If C=0, the velocity = tage :
p r

If the curve be an ellipse or hyperbola, the axes of which
are 2a and 2b, we find by comparison with the equation

 

fe 2atr
p —- r >
2 2
that w= aoe , if Z be the latus rectum, and that
AP Dye _
v= = - + .

102. The case in which the force varies as the distance
has already been dealt with in Art. (81), but we can also use-
fully employ the equation

h? dp

Pp aia
2
leading to a =C-— pr’,

which is the equation of an ellipse.
Comparing with the equation
p(a@+h—r*)=a'b’,
‘we find that C= pu (a’+ 0°), and h? = pa)’, and that

aR ane +i *) = pw. OD",
if CD be conjugate to r.
103. Motion of a particle under the action of a repulsive

force from a fixed point varying inversely as the square of the
distance from that point.

The equation of motion is

a
CENTRAL ORBIT. 115

therefore ut ii =A cos(0—y),
? 1
or Be eee
Introducing the initial conditions, we find that
2 sae 2 5
ve sin Bat =a cay Aq BS sn BoP _ osin -
2 Sn 4A? gtd
therefore e=1+ ae ene + — ;

shewing that the conic is a hyperbola.

As in the previous case,

 

h 3
— =a(e’—-1
- (—1)
egaae ve? sin? B a (7 sin? 8 4 ve ste e .
be Be
2
and therefore fama,
a ¢

Hence, any point being a point of projection, we have at
the distance r
2u

=

als

104. Path of a particle projected at the distance a with
the velocity Ju--a in the direction at right angles to the
imitial distance, and subject to the action of a central force
which on unit mass is equal to

pe {2 (a? +B”) w’ — 30°0u'}.

In this case h = ae a=,/, and therefore

= $u=2 (a 4d) uv — Babe,
116 VELOCITY IN AN ORBIT,

Multiplying by 2 S integrating and observing that when

Uu= E og we find

a’ ao
(3) = u! (au? — 1) (1— Bu’),

restoring r this becomes
dé r
dr ~ I@—r ') (® — B)
Putting 2 = 7° —b’, and oe we obtain
7 — b? = (a? — b*) cos’ (9 —y).
But r=a when 6 = 0, and therefore y = 0, and
7 = a? cos* 6+ 6 sin’ 0

is the orbit, which is the pedal of an ellipse with regard to its
centre.

VELOCITY IN AN ORBIT.

105. If the orbit be central, the velocity is given by the
equation

r=,

a
In general, whether the orbit be central or not, the
velocity is given by the equation,
v=3% Ky,
where g is the chord of curvature in the direction of the
resultant force.

For instance if the orbit be an ellipse, the force being
directed to the centre C, and if v be the velocity at P,

v= $ nOP. “CO ncDe
VELOCITY IN AN ORBIT. 117

For an ellipse, when the force is to the focus,
Gat, 2 ee
sr Ay SP UAC

For a parabola, force to the focus,

2 2
v =1 gm 4SP=-

For a hyperbola, force to the focus,

2_y & 20D? Mw, pw
=t9p- 407 spt ac:

_ For a hyperbola when the force is repulsive from the
ocus,
Patt Pe ae
2" 3p” AC AC SP

For an equiangular spiral, force to the pole,

=p hora.

For the orbit,
v= a" cos?@ +b’ sin? @, Art. (104),
> will be found that

r

‘ ali _
P= TEER rage Od 9 AE METI,

106. If a particle move in a conic section under the
ction of a force to the focus, the velocity at any point can be
scomposed into two constant velocities, one perpendicular
' the radius vector, and the other perpendicular to the aais of
1@ conte.

The sides of the triangle SPG are perpendicular to the
rection of the actual velocity, and to the directions in
iestion. Hence the velocity, and its components are in
.e ratio of the sides of the triangle. GZ being drawn per-
118 TIME IN AN ORBIT.

pendicular to SP, the component perpendicular to SP

ce eg ee eee
“SY°' PG PL 4L oh

, Art. (101).

 

 

and, since SG=e.SP, the component perpendicular to the

axis = + :

It will be noticed that the latter component is in the
direction PN, or NP, according as the body is moving
towards, or away from, the vertex.

TIME IN AN ORBIT.

107. The time of passing from one point to another of
a central orbit is in all cases determined by the equation,
7? 6 =h, that is by the fact that the area swept over by the
radius vector is proportional to the time.

Time of traversing an arc of a parabolic path, when the
force is to the focus.
TIME IN AN ORBIT. 119
The parabola being’

eee
T+cos9) 78°" 9°

dt r at 10 aé. , s 2

dé = Pua Jan? “3 = 75, ( es 5) 6 a

and therefore t=a* ff a (tom e a tan® 5) ’
B& 2° 3 2

measuring from the vertex.

we have =

108. Time in an ellipse when the force is to the centre.
The equation of the ellipse being

cos’@ sin’? @ 1

 

 

oe TR
dor or | ab
dd h abdu Jp arsin 0+ 0 cos 0”
ab sec’ 0

az ar 6? + a* tan’ 0’
and therefore the time from the end of the transverse axis
1 atan 0
= —-tan’———_..
Ju 6
The Periodic time is equal to the area of the curve divided
by 4h, and
_ mab — Qr
daby/p
It will be noticed that this is independent of the size of
the orbit.

1

109. Time in an ellipse when the force is to the focus.
120 TIME IN AN ORBIT.

Taking <= 1+ecos@ as the equation of the curve, we

 

 

 

have
i a a
dO h Juc Jp’ (k+ecos 6)"

ee I dé _ [_cos@ te _ [cos 6d

(1+ecos 6) J(1 +e cos 6)’ (1 +e cos 6)
__sin@ =A) 2... ff | a,
l+ecos@ e/|1l+ecos@ (1+ecos 6)’ ”

therefore

0-6) [ape ret ico
=~ ot | rang

and the time from the vertex is therefore, since c=a/1—e’,
i Loa i
S- {otax (4/7 *tan 8) ead ée’ sin 6)
Ju l+e 2 l+ecos@ $
The total area being crab, the periodic time

= Qrab Qrrab Qrat

ho pale) Vp
This of course can be obtained from the preceding
expression by taking @=7, and doubling the result.

110. Time in a hyperbola, when the force is to the
Focus.

The process is exactly the same, only that, e being
greater than unity, the result of the integration appears in
the form
APSES AND APSIDAL DISTANCES, 12]

e— = no |
esin 0 1 [il St maar at

(-vf -
(1+ecos 6)? I+ecosO ,/e yi ” | hn

 

 

111. Time in the orbit, r? = a? cos? 6 + B sin? 6.
In this case h=,/u, /ut = [a0

therefore /ut=4(a? +b°) 044 (a? 0’) sin 20,

2
and the periodic time = ~ (a+ m (a + 68)
a

APSES AND APSIDAL DISTANCES.

112. An apse in a central orbit is a point at which the
tangent is perpendicular to the radius vector, and the length of
the radius vector at the point ts the apsidal distance.

We have shewn that, if the central force be a function
of the distance, the velocity at any point and the inclination
of the tangent to the radius vector are also functions of the
distance, and from these facts it follows that if the motion at
any point be reversed in direction, the particle will retrace
its path in the opposite direction.

For, supposing the particle projected from P with a given
velocity v to arrive at the apse A with a velocity wu, the

value of r _ is a function of v and r, and therefore if

Bas), =] Fodder

4
and if ASP =a, a =[ FO dr.
122 APSES AND APSIDAL DISTANCES.

Now, suppose the motion reversed at A; then the values
of 6 and a are the same as before, and the orbit is therefore
retraced.

p

  

S P

We hence see that any apsidal line dimdes the orbit
symmetrically.

For on arriving at A the particle is under the same
circumstances with regard to the direction AP’ as it was
when reversed with regard to the direction AP.

Hence it is obvious that, at an apse, the radius vector has
a maximum or minimum value, and further that, in a central
orbit in which the force is a unique function of the distance,
there can only be two different apsidal distances, although
there may be any number of apses.

113. It follows that we cannot ensure the complete
description of an orbit by placing a centre of force at any
assigned point. Take for instance the case of an ellipse, and
trace its evolute.

The centre of force may be at the centre O, or at any
point of each of the four limited lines AC, DB, EA’, FB’.

If a centre of force be placed anywhere else, as at K, the
normal KP does not divide the orbit symmetrically, and
APSES AND APSIDAL DISTANCES, 123

although a particle, projected from A, may describe the arc
AP, it will not proceed in the ellipse, but will describe the
are PA turned over to the other side of KP.

 

8 P
D
Al |A
EB KE Cc
Fr
B’

The same remarks apply to an orbit of any form.

114. If the law of attraction be inversely as the nth
power of the distance, that is, if P =u", the equation of
motion is

aw ut?
t dé 6 3 +u= rw ,

leading to

2
(35) += oH aa +¢.

Hence the apsidal distances are given by the equation

ee +0,

which cannot have more than two positive roots, a result in
accordance with that of Art. (112).

U
124 APSES AND APSIDAL DISTANCES.
If however the moving particle be so projected that

i= cer and C= mae ,
du\?  ._Ac™ Su n—8
(i) Tad, tee
and the equation for the apsidal distances takes the form
(n—1) cu? = 2 (cu)"* + n — 3.
It is easily seen by taking the derived function that this

equation has two equal roots, that is, that there is really only
one apsidal distance, of length c.

we have

Taking n= 5 for example, we find that

(ae) "a

dé\? 2c"
and therefore (5) a (-r} .
Integrating we find that the equation of the orbit is
ge Qe op

e~ ted 7"

a curve which has an asymptotic circle of radius c.

Hence it appears that, when projected in the manner
described, the particle will make an infinite number of re-
volutions before arriving at the apsidal distance of its path.

If for another example we take n= 2, the equation of

the path is

(@) += 25-3
" (za) + (#-3) -0
shewing that o =0, and w= s
APSES AND APSIDAL DISTANCES. 125

or that the only possible path, under the conditions stated, is
a circle having the centre of attraction for its centre.

In fact it will be found impossible to satisfy the conditions
for any angle of projection which is not a right angle.

115. The peculiarity of the preceding case is that the
velocity of the apse is that which is suitable for circular
motion round the centre of attraction as a centre. /

For 0 = I’ {(36) + wh

dé
= anne {2 (cu)"* + n — 3}
=o when 7 =c,

116. A particle describes a nearly circular orbit about a
centre of force ; it is required to find approaimately the equa-
tion of the path and the apsidal angle.

If $(r) be the force, and if the particle be projected at
the distance c, perpendicularly to the distance with the velo-
city wf cd (c), it will describe a circle.

We shall suppose the particle projected perpendicularly

with the velocity 4/ c@ (c) at the distance c++, y being a very
small quantity.

 

We have then h? =(c + 7)’ cd (c), and if we suppose that
r=c-+ a, where a is very small,

 

_1l_1 @
7 6 Cc
a b(r)_ b(c+2) (c+2) (¢+y)?
es hae co (c)
1 Qa, ah’ (c) vy
=i(l4 e+ ee)”

if we neglect the squares of the small quantities.
126 APSES AND APSIDAL DISTANCES.

The differential equation then becomes

de of)
Bete B+ SS fam

and the ea equation of the path is

ore + Aeos{y/34 2 {) o+ah,

at the apses, - = 0, and therefore the apsidal ‘Angle is

3 + “i (c)”
Js+38
If (r) =r the apsidal angle is 3 , and if $ (r) =5 it is

equal to 7.

It will be seen that if the force vary as the nth power of
the distance n must not be less than — 3.

 

 

 

 

In other words if n is greater than — 3, a circular orbit
possesses the characteristic of stability.

If n= 3, the apsidal angle is infinite. In this case if the
particle be describing a circle about the centre of attraction
as centre, any slight divergence of path, without change of
velocity, will cause the particle to describe an equiangular
spiral.

117. Case in which $ (r)=r", where n is a positive
integer, the particle being projected from an apse at the distance
¢ with the velocity Juc™,

 

In this case the equation of motion is
du 1
dee t= Gare yi
leading to
(n+ 1) (cu)"? + 2—(n +3) (cu) _
(3) as (n+ 1) ot yt

 

=0.
RESISTING MEDIA. 127

Now if we take
F(a) = (n +1) a”? +2 — (n+ 3) v™,
we find that the equation, f(x) =0, has two roots equal to
unity, and, further, that for positive values of w, the function
J (a) is always positive.
Hence it follows, from the equation of motion, that
du
dé
or that, with the assigned conditions, a circle is the only
possible path.

=0, and r=c,

If in this case a slight disturbance of path take place, a
nearly circular orbit will be described, the apsidal angle being

a)din + 3.

118. Motion of a particle under the action of a central
Jorce, in a medium the resistance of which varies as the square
of the velocity.

Taking the transversal acceleration, we have

a ae
rd = FS as ksré,
1 ae ihe. gs
or andi (7°60) = — ks,

and therefore 76 = he-™, h being a constant; or, if p be the
perpendicular on the tangent,

ps =he-®,
Taking the normal acceleration,
# aa 5p he —iks — P) dr
= pe = Pap"

By help of the equation

: ieee al
po +(33 )
128 RADIAL AND TRANSVERSAL

this is transformed into
d*u P
qe t= Bye
119. <A particle is moving under the action of a central

force in a resisting medium ; it is required to find the resist-
ance necessary for the description of a given path.

Taking transversal and normal accelerations, and assum-
ing mR as the force of resistance,

1d 29) — Pp ee P.
mt ee a
therefore Sy 3) =— Rp, and oe 8)? =— 2Rp’s
_ tid sar
o Be~ aera (Fp

120. The two following problems will further illustrate
the use of radial and transversal components, and the appli-
cation of the principle of relative accelerations.

Two equal particles are attached to the middle point, and
to the end C, of a string ABC, the end A of which is fixed on
a smooth horizontal plane ; it is required to find the equations
of motion of the particles on the plane.

If 6 and ¢ be the inclinations of AB and BC to a fixed
line on the plane, the accelerations of B along BA and
perpendicular to it are a6 and a6, and the accelerations of C
relative to B along CB and perpendicular to it are ag* and

ad. Compounding these accelerations perpendicular and
parallel to CB, we obtain the equations of motion of 0,

ad + a8 cos ($ — 6) + a6 sin (p— 0) =0 ...... (1),
ag? + a6 cos (6 — 0) — af sin (6 — 0) = 7”......(2),
and, for the motion of B,
ab =T" sin (B10) aisvewiesicicemanene: aa (3),
a = T—T’ cos(P—O) wceecseeeeeeees (4).
ACCELERATIONS. 129

Eliminating J” we obtain,

Multiplying (1) by 1+ cos (¢ — @), and subtracting (5),
6 + (G + 6) {1 + cos (f — 6)} — (6? — &) sin (f — 4) =0,
and therefore

6+ (b+ 8){1 + cos ($ — 6)} = 0.

 

 

A

This is really the equation expressing the constancy of
the angular momentum, and could have been written down

at once.

‘Again, eliminating @ from (1) and (5),

dé’ sin (6 — 6) — ¢ cos (d — 6) — 28 =0...(6).

Multiplying (1) by 24, (6) by 26, and subtracting,
26b+466-+42 (64+ 68) cos ($— 0) —2(g*6—6"4) sin (b—0)=0,
and therefore

d? + 26° + 266 cos ( — 6) = D,

which is the equation expressing the constancy of energy,
and could have been written down at once by observing that
the velocity of C is compounded of ag perpendicular to CB,
and of a@ in the direction perpendicular to BA.

n
130 RADIAL AND TRANSVERSAL

We have given the preceding solution for the sake of
illustrating methods and principles, but so far as the present,
problem is concerned its solution is perhaps more easily
effected by employing rectangular components. The equa-
tions are then, if 2, y be co-ordinates of B, and &, of C,

m= T' cosp—Tcos0, mi =T' sing —Tsin 8,
mé = — T’ cos 4, mij = — T’ sin ¢,
with the geometrical conditions,
x=acos#, £=acos# +a cos ¢,
y=asind, n=asind+asing.
Multiplying the equations of motion by 2é, 2£, &c., and
adding, we obtain
+P + Pty =C;
we also find that :
aij — yi + Bj — nF =0,
leading to
ay — ye + En — n& =D.

These are the equations of energy and momentum, and
by means of the geometrical equations are at once expressible
in terms of @ and ¢ as before.

Of course the simplest solution consists in utilizing the
principles of energy and of angular momentum, and at once
writing down the two equations derived from these principles.

121. The same system being suspended from the end A,
it ts required to determine the small oscillations in a vertical

plane.

Neglecting the squares of small quantities, and taking @
and ¢ as the inclinations to the vertical, the acceleration of C
relative to B is aé perpendicular to CB, and that of B is ad
perpendicular to BA. If T and 7" be the tensions, the
approximate equations of motion, neglecting the product
ACCELERATIONS. 131

6 sin (b—6), and taking cos ($ — 6)=1, and sin (6 — 0)=4¢-8,
are
for C, m (ag + a) =— moh
0=—mg+T')’
and for B, = ma? =— mg + T" (p-9))
0=-mg+T-T' | :
We hence obtain 7" = mg, and T'= 2mqg;
therefore ag =g (20 — 2¢), and a6 = 9 (¢ — 28):
Multiplying by and adding .
a KAS
a($ +20) =—g @-») (6+ 2=0),

and if we assume a =A, we find that y= £ 7 2,

We hence obtain two equations of the form

 

$+ +LQ—nj(P+rO}=0,

the solution of each being of the form
$+20=A cosnt + Bsin nt, Chapter I,

shewing that the values of 6 and ¢ are each compounded of
harmonic quatitities.

EXAMPLES.

1. A heavy particle is fastened by two equal strings to
two points in a horizontal line, and then whirled round in a
vertical plane ; the velocity is such that if one of the strings
break when the particle is at its lowest point or when it is
half-way between its highest and lowest points, the particle
will continue to describe a circle; find the least distance be-
tween the points of suspension that this may be possible,
132 EXAMPLES,

2. If a body move in an ellipse under a force to a
focus, the velocity at the mean distance from the centre of
force is a mean proportional between the velocities at the
extremities of any diameter.

3. A particle describes an ellipse about a centre of force
at a vertex A. Prove that the force varies as AP+AN'*,
where WV is the foot of the perpendicular from P let fall on
the major axis.

4. Find the law, of force to the pole when the path
is the cardioid, r=a(1—cos 6); and prove that, if & be the
force at the apse, and v the velocity,

Bu = 4aF.

5. Ifa particle be describing an ellipse about a centre of
force in the centre, shew that the sum of the reciprocals of its
angular velocities about the foci is constant.

6. A particle is describing the ellipse 7=~r (1+ cos 0)
under the action of a force tending to the origin; if the velo-
city be altered at any point so as to make it describe a
parabola, shew that the vertex of the parabola lies on the
curve

r=a(l1—ecos @).

7. <A particle describes an ellipse about a central force in
the focus S, SY is a perpendicular upon the tangent to the
orbit ; shew that the angular velocity of SY is a minimum or
maximum when the particle is at the farther apse according
as the eccentricity is less or greater than 1/3.

8. A particle is acted upon by a central force which is
some function of the distance (r) and by a transverse force
equal to wr. Shew that, if the particle move completely
round any closed curve, the square of the velocity is, increased
by 4 (area of the curve).

9. A particle describes a lemniscate about a centre of
force in the node; shew that the velocity of the correspond-
ing point of the rectangular hyperbola to which the lemnis-
EXAMPLES, 133

cate is related varies as the 5th power of the radius vector of
that point.

_ 10. A point moves in a plane curve and sounds as
it moves. At a fixed point C in the plane the whole sound
produced is heard simultaneously. Shew (i) that if the point
moves uniformly, the curve is an equiangular spiral—(ii) if
the point moves as in a central orbit about O, the curve is
a reciprocal spiral.

11. Prove that the acceleration of a particle describing
an epicycloid under a centre of force at the centre of the fixed
circle varies as r/p*.

12. If at any point of a parabolic orbit about the focus,
the velocity be diminished in a given ratio, prove that the
empty foci of the elliptic orbits corresponding to different
points of change lie in a parabola.

13. An imperfectly elastic particle is under the influence
of a smooth hard gravitating sphere. Shew that (excepting
special circumstances of projection) it will perpetually describe
ares of conic sections: determine also the elements of the -
orbit described after any number of rebounds.

14. From every point of an ellipse particles are projected
in the direction of the tangent with velocities such that, when
acted on by a centre of force p/r’ in one of the foci of the
ellipse, they proceed to describe parabolas. Shew that the
directrices of these parabolas all touch one or other of two
fixed circles whose radii are equal to the major axis of the
given ellipse.

15. If v, v, be the velocities at the extremities of a
diameter of an ellipse described about the focus, and u the
velocity at either of those points when it is described about
the centre, prove that u (v, + v,) is constant.

16. Two equal and perfectly elastic particles are under
the action of the same centre of force p/r’; the one is
describing an ellipse and the other a confocal hyperbola, the
semi-major axes being (a), (a’) respectively. If they impinge
134 EXAMPLES.
then after impact they will describe two conics, cutting each
other orthogonally and of semi-major axes (a, a’), where

1] a eed,

a a@ at+a

 

the upper or lower sign being taken according as the hyper-
bola is described about the outer or inner focus.

17. OY is the perpendicular from a fixed point O on the
tangent to a curve at any point P. Ifthe curve is such that
PY is constant, and a particle describe it under the action of
a force to O, prove that the force varies as OP +O Y*.

18. A number of circles touch at a point P, and particlés
describe them under forces to a point S, on the line through
P perpendicular to the common tangent, inside all the circles,
the forces on all the particles when at P being the same:
prove that the squares of the periodic times vary as the cubes
of the radii of the circles.

19. <A curve described by a particle under the action of
a central force is such that, if at any moment the component
velocity along the radius vector be destroyed by an impulse
along the radius vector, the particle will proceed to describe
a circle: prove that the curve is a reciprocal spiral.

20. The ends of a straight tube AB, of length 4a, are
connected by an elastic string, the natural length of which is
2a; a particle is fastened to the middle point of the string,
and the tube is then made to move uniformly, in a horizontal
plane, about a fixed point O, with which it is rigidly con-
nected, and which is equidistant from its ends: determine
the motion of the particle, examining the different cases -
which may occur in the question.

21. <A particle is placed in a straight tube which revolyes
uniformly in a vertical plane about its lowest end. Supposing '
the particle to have no initial velocity relatively to the tube
and that initially the free end of the tube is vertically above
the fixed end, prove that the velocities of the particle along
the tube, when the tube returns successively to its initial
position, are proportional to
EXAMPLES. 135

22. Prove that the law of force under which the pedal
of p= (r) can be described is

2r° or dr
wie 2 ot.
(p’ p'dp
If p «x r" the law of force under which the pedal can be

2
. . . © =
described varies inversely as (distance) ”.

23. Prove that, if a particle move in a smooth tube
under the action of any forces tending to centres, the pressure
at any point of the tube will vary as

1 1d
*{o-32 5 (Pt,
p pap?)

where = is the acceleration towards any one of the centres,
and p is the radius of curvature; and hence, that the pres-
sure at any point of the tube will vary as the curvature,
whenever the orbit is such as could be described freely under
the action of each of the forces taken separately.

24, A particle moves in a plane under forces whose
: eye :
resultant in any direction is aes where ds is an element of

the line drawn in that direction from the position of the
particle ; prove that the polar equation of the orbit is

hu? (Sut u) =v ~— ae
In case V=ypu'cos 6, and the particle be projected
initially so that 9=0, h? = 2y, find the orbit and shew that
between 0=0 and 0= =e the radius vector is described with
uniform velocity. i

25. The attraction to a given point, at a distance
r, = 8u/r°+2ua’/r*, A particle is projected in a direction
making an angle tan™4 with the initial distance (a), and with
a velocity equal to that in a circle at the same distance;
determine the orbit described.
136 EXAMPLES.

26. A body is describing an ellipse of eccentricity
JZ under.a force to the focus S$: when the particle is at onc
end B of the minor axis, the centre of force is suddenly
transferred to a point S’ in BS produced such that BS’ = 4BS,
and the absolute force is doubled and becomes repulsive:
prove that the new orbit is a rectangular hyperbola.

27. If the orbit be r=asinn@, shew that the attrac-
tion is
Qa nV —-1
al - ae |:

28. Force « 2au’?+9u’+6au*. A body falls freely from
infinity towards the centre of force till its distance is a, and
then its direction is suddenly turned through the angle
cot? 4; find the orbit described. ,

29. <A particle is tied by an elastic string of length a to
a point wherein resides a repulsive force « (dist.)*: its initial
distance is a, and it is projected at right angles to this with
a velocity =./3/2 that necessary for circular motion, were the
force attractive. Ifit passes through a point of inflection at
a distance 3a/2, shew that it will come to a second apse
at the distance 3a.

30. Having given P= ypu’, and that a particle is pro-
jected from an apse at the distance c, find the orbit (1) when
the velocity of projection is ,/u/c’V2, (2) when it is /u/c’.

31. SW isa fixed line through a centre of force S; PN
is the ordinate at any point P of the path of a particle acted
on by the central force; find the force when the curve
PS always bisects the triangle PS.

 

 

32. If the path be the lemniscate the Bernoulli, the
equation to which is =a’ cos 20, prove that the square
of the force varies as the seventh power of the angular
velocity of the radius vector of the particle.

33. A point is moving in an equiangular spiral, its
acceleration always tending tuwards the pole, 8S. When
it arrives at a point P, the law of the acceleration is changed
EXAMPLES. 137

to that of the direct distance, the actual acceleration at
P being unaltered. Prove that the point will now move in
an ellipse whose axes make equal angles with SP and
the tangent to the spiral at P, and that the ratio of these

axes is tan 5 : 1, where a is the angle of the spiral.

34, <A particle m is projected at a distance (a) from a fixed
point, in a direction at right angles to that distance, with
a velocity ./u/2a,/2, and is acted upon by a repulsive force
mp/r® from that point; find its path, and prove that the
time from the distance a to the distance a,/2 is 2,/20°/3/u.

35, A particle moves under a force pu {3au*— 2 (a? — b*) u*},
a being > 6, and is projected from an apse at a distancea+b

with velocity ./~ + (a+b): shew that its orbit is
r=a+bcosé.

36. If the parabolic orbits of two comets intersect
the orbit of the earth, supposed circular, in the same two
given points, and if ¢,, ¢, be the times in which the comets
respectively move from one of these points to the other,
prove that

4\3
¢+yi+¢-o8=(5),

the unit of time being a year.

37. The pedal of a curve with respect to a point is
defined to be the locus of the foot of the perpendicular drawn
from the point on any tangent to the curve, and the pedal of
the pedal with respect to the same point is called the second
pedal, and so on; a particle describes the n'™ pedal freely
under the action of a force tending to that point; find the law

of force.
If the curve be a rectangular hyperbola, and the pedals be
formed with respect to its centre, prove that the n™ pedal

will be the orbit of a particle moving under the action of
6nt+1

a force varying as r*", where r is its distance from the
@

centre of force.
138 EXAMPLES,

38. A particle is projected with velocity v from the
vertex of a cycloid and describes the curve under an attraction
to a centre of force situated on the axis at a distance from the
vertex greater than the diameter of the generating circle and
less than twice that diameter; prove that the particle will be
again at an apse after a time

a 2a’ cosa+ 3asin2+ 3 sin*a cos «
o° acosa+sin a

a being the radius of the generating circle of the cycloid, and
a the apsidal angle.

 

x

39. A particle P describes a central orbit, centre
of force 8, and through P is drawn a straight line at right
angles to PS, which line touches its envelope in @: prove
that the velocity of Q

1 /d’r
o ? (Gat) 5

and is constant only when the path of P is a parabola, whose
focus is 8S.

40. <A particle is projected ata distance a with a velocity
equal to the velocity in a circle at the same distance, and at
an angle 7/4 with distance, the force being p (37° + a’r*)/r’;
determine the orbit described, and shew that the time to the
centre of force is (4— 1) a?/2/2u.

41. <A particle is projected from an apse with ve
velocity from infinity under the attraction of a force * = 5 log = ~

directed to a centre at a distance a: find the peqaittams of the
orbit described.

42. A point describes a semi-ellipse, bounded by the conju-
gate axis, and its velocity, at a distance r from the focus, is

af @=”

r (2a —7)
and f a constant acceleration; prove that the acceleration of
the point is compounded of two, each varying inversely as the

uare of the distance, one tending to the nearer focus, and
tite other from the farther focus.

i 2a being the length of the transverse axis,
EXAMPLES, 139

2:

43. If the force ata distance r be 2u (3 -5) and the
particle be projected from an apse at a distance a with a
velocity a , it will be at a distance r from the centre after a
time

1 (,, rt+J/r—-a eee
syat log sae +r Jr? —a.

44, In an orbit described under the action of a central
force, a straight line is drawn from the centre of force perpen-
dicular to the tangent and proportional to the acceleration :
if this straight line describes equat areas in equal times,
shew that the equation to the orbit is of the form

&
ade!
ye
Shew that the rectangular hyperbola is a particular case.

45. Having given P=5yu' + 8uc’u’, and that a particle
is projected from an apse at the distance ¢ with the velocity
3,/u/c, prove that the orbit is r = ¢ cos 26/3.

46. A particle acted on by the central force pw (r+ a)/r°
is projected from an apse at a distance a and with a velocity
which is to that in a circle at the same distance as 1: ,/2;
shew that the equation to the orbitis r (2 + 6°) = 2a, and that

the particle will arrive at the pole in time 7,/a*/,/84u.

47, Force co (8a°u’ — 12a7u* + 9au®—2u*). A particle
is projected at distance a with a velocity = /2/3 that in
a circle at the same distance, and in a direction making
an angle 7/4 with the radius vector; find the orbit described,
and prove that its equation is r=a (1 — tan 6).

48. Prove that there are two directions in which a
particle, acted upon by a central force varying as the distance,
can be projected from a given point, with given velocity, so as
to pass through another given ‘point.

49, The attraction to a fixed point varying inversely as
the fifth power of the distance, a particle is projected in
140 EXAMPLES.

a direction making the angle tan™ 2,/2/3 with the initial
distance c, with the velocity /17y//2c?; prove that the
orbit is :
Qr _ 3e°V? 1
C = 3¢ ov2 +1 .
50. Two heavy particles are connected by a string with-
out weight. -One particle is just dropped through:a hole ina

smooth horizontal plane and the other is projected on the
plane at right angles to the string fully stretched.

(1) Find the least velocity of projection which will
keep the particle from descending.

(2) If the velocity of projection be less than this,
determine the motion of the descending particle.

51. A particle describes an equiangular spiral under the
action of a central force to the pole in a medium in which
the resistance varies as the square of the velocity. Prove
that the distance from the pole at which the central force is
a maximum is half the distance at which the velocity is a
maximum, and that these ‘distances are independent of the
initial distance or ‘initial velocity.

52. Two equal particles P and Q, on a smooth horizontal
plane, are connected with each other by a stretched inelastic
string of length 7, which passes through a smooth ring (0)
fixed to the plane ; P being projected perpendicularly to PQ,
find the equation to its path, and shew that, when @ arrives
at (O), P will have described a right anglé about (QO) if the
initial distance of P from it be equal to J cos 7/2 4/2.

53. A rod which is extensible in accordance with Hooke’s
law is made to rotate with uniform angular velocity w, about
a line through one extremity perpendicular to its length, and
is supposed to remain straight. If a be its natural length,
m the mass of unit of length when unstretched, # its modulus
of elasticity, and / its length when rotating, shew that

_1 /&, Re
r=i,/# tan {ac zr
EXAMPLES. 141

that the stress on the axis is

E{secaw 4/7 -1},

and that the rod must break before its angular velocity is
1 NV E/mN 2a.

54, A wire in the form of a plane curve is constrained
to rotate about an axis perpendicular to its plane with
varying angular velocity. Find the motion of a bead which
slides upon it under the action of any given forces, and the
pressure on the wire.

If the wire is circular, and the axis through a point in
its circumference, and the angular velocity » uniform, shew
that the pressures on the curve at the two extremities of
the diameter perpendicular to that through the axis are
(3 + 2/2) mwa, the particle starting from rest at a point
near the axis. Draw a figure to indicate at which of these
points the pressure is the greater, and the direction of that
pressure.

55. A small smooth ring can move upon a smooth cir-
cular wire which is made to roll with uniform angular
velocity w on the outside of a horizontal circle of n times its
radius: prove that the angle through, which the ring will
have revolved with respect to the wire at a time ¢ is

{(n +1) 6—not}/(n +1) where ¢ =o sin d/(n +1).

56. Two particles P, Q, of equal mass, slide upon a
smooth endless string OPQ, which passes through a small
smooth ring at O, and lies on a smooth horizontal plane.
OP is initially equal to OQ, and the particles are projected
with equal velocities along the external bisectors of the angles
OPQ, OQP respectively ; prove that, throughout the motion,
the tension of the string varies inversely as OP.

57. A particle of mass m is attached to a fixed point by
an elastic string of natural length a, and whose coefficient of
elasticity is m. It is projected with the velocity due to half
the length of the string in a direction perpendicular to the
142 EXAMPLES,

string which is initially unstretched. Prove that the apsidal
distances of its orbit are given by

a (r?— a?) — 7 (r—ay=0.

58. A body is moving in a uniform resisting medium,
the resistance of which varies as the n™ power of the velocity,
under the action of a force varying inversely as the square of
the distance ; find the value of 1 in order that the path may
be an equiangular spiral.

59. Two centres of force pr, u'r’, tend to two fixed points
S, S', the former being attractivé and the latter repulsive.
Prove that a particle under their influence moves in such a
way that

c being a constant and @, 6’ the angles that r; r’ make
with S8’. 4

60. A smooth circular tube is fixed at one point A and -
contains a particle which is initially at rest at the opposite
extremity of the diameter through A. The tube is then
made to revolve in its own plane with a uniform angular
velocity w; shéw that the angle described by the particle
about the centré of the tube in the time ¢ is

-t 4):
4 tan (Ga :

 

61. A heavy particle is attached to a fixed point by
means of an elastic string of natural length 3a, whose coefii-
cient of elasticity is six times the weight of the particle;
when the string is at its natural length and the particle
vertically above the point of attachment the particle is pro-
jected horizontally with a velocity 3 Vag/V2; prove that the
angular velocity of the string will be constant and that the
particle will describe the curve r = a (4 — cos 6).

62. Prove that the central force tending to the pole
which causes a particle to describe an equiangular spiral in a
homogeneous medium wherein the resistance varies as the
EXAMPLES. 143

n power of the velocity (the initial velocity having been

properly chosen) varies inversely as (r)** where r is the
distance of any point from the pole. Find also the velocity
at any point in terms of r, Examine the cases where n= 2
or 1.

63. Shew that a particle may be made to describe an
epicycloid under the attraction of a circular wire whose
particles attract inversely as the cube of the distance, the
circular wire being the fixed circle of the epicycloid.

64. A heavy particle is free to move on a smooth elliptic
wire which is moveable in a vertical platie about its centre,
The wire is placed with its major axis vertical and the
particle at the highest point. The wire is then suddenly

started to rotate with a constant angular velocity af ote oh ;
where a,b are the semi-axes of the ellipse. Shew that the
particle will move relatively to the ellipse as if it were acted

upon by a force to the centre varying as the distance.

65. A particle m is projected from an apse at distance ¢
with velocity /2yc*//3. The force to the centre being
mp (r° —c'r), prove that the orbit is e+ y* =c*.

66. When the forces have a potential equal to ur? cos 0
and a particle is projected at a distance a perpendicular
to the initial line with velocity 2./u/a, then the orbit de-
scribed is

 

r=asec (v2 log tan zt").

67. A particle acted upon bya constant repulsive central
force, is projected at right angles to the initial distance with
a velocity double that which would be required in moving
from the centre of force to the point of projection; prove
that the orbit is

6 ye 1, ., /r-a
9 — tan a a 3a

 
144 EXAMPLES,

68. A smooth circular wire rotates uniformly in its own
plane, which is horizontal, about a fixed point, the distance of
which from its centre is one-third of its radius, and a bead
which can move on the wire is attached to it at the point
nearest the fixed point. If the bead be set free, prove that it
will make complete revolutions, and that, at the angular
distance sec™ 3, its pressure on the wire will vanish.

69. A particle is projected towards the origin from
infinity with any velocity, and is acted upon by a force pu* at
right angles to the radius vector; shew that it will describe
a curve of the family, u = ab* J, (8), where J, (x) is the Bessel’s

function of the n" order, and find the velocity of projection
in order that a particular curve may be described.
CHAPTER VIII.
TANGENTIAL AND NORMAL ACCELERATIONS.

122. Ir mT and mW be the forces acting on a particle of
mass ™ in directions of the tangent and normal, the equations
of motion are

os du
ms=mT, or mv — =mT,
ds
vy
and m-=mN.

We have already made a slight use of the expression for
normal acceleration in article. (96); we now proceed to
develope, somewhat at length, the utilization of these
expressions.

Motion of a heavy particle on a smooth curve in a vertical
plane.

Measuring = horizontally, and y vertically downwards, and
taking R as the normal reaction of the curve, measured
outwards, the equations of motion are

dy dy wae
my 7 = mg» aria R.

d
The first equation gives
dm (vu?) = mg (y -y');
if wu be the velocity when y=y/’,
B. D. 10
146 TANGENTIAL AND NORMAL

This is, in effect, the equation of energy, and can be
written down at once from the assumption of the truth of
the principle of energy.

The second equation determines the pressure.

123. Motion of a particle on a smooth curve under the
action of forces to fixed centres, the forces being functions of the
distances from those centres.

If 7, 7’,... be the distances of the particle from the
centres of force, and mP, mP’... the forces, the equations of
motion are

di

v dr dr’
my a= mP T+ mP +...
2 ys
a ae Sie

 

ar =mPr ag tmPr ds

From the first,
Lmet = & [mPar

this, which is the equation of energy, gives the velocity and
the second equation determines the pressure.

124, Motion of a heavy particle, placed on the outside of a
smooth circle and allowed to slide down.

 

If the particle start from the point Q, at an angular
distance a from the vertex, and v be the velocity at P,

vy’ = 2ga (cos a—cos 6),
ACCELERATIONS. 147

2
and a =mgcos@—R, R being the outward reaction of

the curve,
.. R= mg (8 cos 6 —2 cosa),

shewing that the pressure vanishes, and that the particle
flies off the curve, when

cos 9 = 2cos a

125. Motion of a heavy particle inside a smooth circular
tube in a vertical plane.

We shall suppose that the particle starts with a given
velocity u from the lowest point B.

A

P

 

B

Measuring @ upwards, the equation of energy is
4m (v* — wu’) =— mga (1 —cos 8),

and the equation for the pressure is
4
R= mg cos 6,
a
R being the pressure, inwards, of the tube on the particle.

R ut
ss m7 29 C8 — 2g +s
10—2
148 PENDULUM.

To find the height of ascent we put v=0, and to find the
point where the pressure vanishes we put R= 0.

(1) Take w?<2ga; then the highest point is given by
29a cos 6 = 2ga —u’, and the pressure never vanishes.

(2) If u’=2ga, the particle rises to C and the pressure
then vanishes.

(3) If w?>2ga and <4ga, the highest point is given by

 

 

_  uw—2ga
cos 6 = — Iga?
and the pressure vanishes, and changes sign, when
2
cos =— ™ =e
3ga

(4) If w’>4ga and < 5ga, the particle rises to A and
passes over and the pressure vanishes when

 

2
deg 2 ee
3ga
(5) If w= 5ga, the pressure vanishes at A.

‘ (6) If w?> 5ga, the pressure never changes sign.

Oscillation of a Pendulum.

126. A heavy particle, suspended by a string from a
fixed point, and oscillating in a vertical plane, forms a simple
pendulum.

Measuring @ from the vertical, and observing that if a be
the length of the string, s =a, the equation of motion is

a =—g sin 8.

If the amplitude of oscillation be very small, the approx-
imate equation is

6+26=0,
a

and therefore 6=Acos (4/2 t+a ) oa
PENDULUM. 149
This represents isochronous vibrations, the time of

a complete vibration being 2a yi F é

Finite motion of a Pendulum.

Recurring to the equation, ad =—g sin 6, and supposing
the pendulum to start at an inclination @ to the vertical,
we obtain

a? = 2g (cos 6 — cos a),

and therefore if + be the time of oscillation, from one side to
the other,

rae ei jones vi; | Jaojnmt

Panag ¢ sin ; =sin 5 sin yy, this transforms into

ravi Tuan ae
=2/2F (sin$.5)

If 6 be the angle at the time ¢ from the lowest point,
~M Qglq ./cos@—cosa g ai ae & sin’ ‘
0
nf
= Js F (sin 5%) ‘

sin? = 5 sin! =

 
150 MOTION ON

or, in the notation of Jacobi and Guderman,

vnm(y/).

and the height of the bob of the pendulum = 2asin’® #

2
wasnt wt (4/21), sind
=2asin?§ ont ( at) 2 mod. sin 5.

127. Motion of a heavy particle on the arc of a smooth
cycloid, having its vertex downwards, and axis vertical.

Measuring ¢ from the tangent at the vertex, the intrinsic
equation of a cycloid is
s= 4a sin ¢,
and the equation of motion is

§=—gsin ¢, or 8+ 2 5 =0;

1.8=A.c0s ( Lita),
4a,

shewing that the motion is an isochronous vibration, the
period of a complete vibration being

tr 4/2.
g

128. Motion of a particle in a smooth circular tube under
the action of a force to a fiaed point varying as the distance
Srom that point.

Take a as the radius of the circle, and c as the distance of
the centre of force S from the centre C.

Resolving along the tangent, the equation of motion is

© = ur cos SPL = pr

v at
SMOOTH CURVES. 151

and therefore 0° = yw {(c + a)* — 7°} = 2uac (1 — cos 6), supposin
the particle to start from A. Leepe ), Supposing

 

 

If R be the pressure on the particle, measured inwards,
2
a= mur cos CPS +R = mpPN + R,

or mal? = mp (atc cos 0) +R,
therefore 2 = mp (2c —a-— 3c¢ cos 0).

129. Motion of a particle on the arc of a smooth equi-
angular spiral under the action of a force from the pole.
If mP be the force from the pole, and & the pressure
measured inwards, we have
ne
v oF pie me Sin 4 = mP sina — R.
ds r

Tn the case in which P= y/r’ these equations give

dv p ‘ 1 1

oe ae or, =u (5-2),
if the particle start with no initial motion from the distance
a, and the second equation determines the pressure.
152 -MOTION ON

We also have
o- v COS 4 = COS % pees
and therefore the time from the starting point to the distance
b is equal to
I & Jar dr
a C084 /9u (r—a)

130. Motion of a particle on the are of a smooth hypocy-
cloid, under the action of a force to the centre varying as the
distance.

Taking BAB’ as the arc of the hypocycloid, of which A is
the vertex, measure the arc from A.

In the figure C’ being the centre of the rolling circle of
radius 6, Q is the instantaneous centre, and HP, PQ are the
tangent and normal to the hypocycloid at P.

D

 

 

 

The angle YOA being ¢, we see that
ad

a . o_ 2b¢
p+O= 5, and ae Oa or!

and p=OY=(a—-26) cos OF = (a~ 2b) cos —2F ,
SMOOTH CURVES. 153

The curve being convex to the point O,

ds__ dp _4b(a—b) ad
dp? ag a— “a —2b"

 

 

ie _ ab
and therefore s=4 a (a —b) sin an 2b"

The equation of motion is

. dp

§=— pOPcos OPY=—pPY=p ds:
% par =

or, oS gee)

shewing that the motion is oscillatory and isochronous.

A geometrical proof of the isochronism of the hypocycloid
will be found in the Principia, Book 1., Section x.

If we make the radius of the circle infinitely large, and
the quantity » infinitely small, and take ua =g, we fall upon
the case of cycloidal motion, and the above equation becomes

s+ is = 0, as in Art (127).

131. Motion of a particle sliding on a rough curve.

The equations of motion are, if BR be the pressure,
measured inwards, and pw the coefficient of friction,
du
mv ds =mT — wR,
v
m - =mN+4+ Rf.

Taking the case, for instance, of a heavy particle sliding
upwards on the arc of a curve in a vertical plane, we have

dv .
my 7 = — mg sin p — whi,

v
a

where ¢ is the inclination of the tangent to the horizon.
154 ROUGH CURVE. RESISTING MEDIUM.

dv vw ‘
Hence Seti ony HECee),

2
om oS + 2yo* = — Qgp (sin $ + cos $),
and, the intrinsic equation s=/() being given, p=/' (4),
the equation is that of Chapter II. (1).
132. Motion ofa heavy particle in a medium the resistance
of which varies as the square of the velocity.

Measuring ¢ downwards from a horizontal line, the
equations of motion are

dv : a e
07, = 9 sin b — ke’, and 3 =9 cos ¢.

The second equation shews that for a given velocity the
curvature is independent of the resistance, a theorem which
is true for motion in any resisting medium under any forces.

Eliminating v we obtain
cos of- 3p sin @ + 2kp* cos ¢ = 0,

d *) 1 :
or, =, |—])+8tan ¢.-— = 2k, leading to
dg (; ? p -
sec’ dh
=k {tan ¢ sec d + log (tan $ + sec ¢)} + C,
and this is the intrinsic equation of the path.
133. The question sometimes arises whether a given

curve can be described by a particle under the action of forces
to two or more fixed points. In such cases we obtain equations

 

2
for v - and — , and the value of v? obtained by integrating the

first must be identified with the value of v* obtained from.
the second.

If for instance the curve be an ellipse and the forces
Kw

2 to the two foci, we have

S and
TWO CENTRES OF FORCE. 155

From these equations,

pags 4
r
, , 2
d A ae :
an v = ae since p sin d AG’

134. Motion of a particle fastened to a string which is
wound round a fixed curve.

Suppose that when the string is completely wound up,
the particle is at A, and measure s and ¢ from A and the
tangent at A.

 

 

Then PQ=s, and the path of P is an involute of the
curve, its radius of curvature being PQ.

Taking mS and mN as the forces perpéndicular and
156 PARTICLE ON A STRING.

parallel to the string, and 7’ as the tension, the equations of
motion are
modv
do

o being an arc of the path, or

 

ve
=m8, i es

mode _

sdb

If there are no acting forces, v is constant, and the tension
varies inversely as PQ.

mS, m= =T+mN.

Ex. Let the curve:be a circle and the acting force a
repulsive force from the centre varying as the distance.

In this éase v a = pa,

do
so that the motion is uniformly accelerated, and
a = T—muys,

so that, if the particle start from rest at A,
T = 2mps.

135. Motion of a string or fine chain inside a smooth tube
of any shape under the action of given forces.

Taking AB as the chain, we let s represent the length
OA of the axis of the tube, measured from a fixed point 0,
and let o represent the length AP of the chain.

. Then if 7 be the tension at Pand 7+ 687 at Q, and Réc
be the pressure of the tube on the element PQ, the equations
of motion of the element, the mass of which is méc, are, ob-
serving that the velocity and the tangential acceleration of
every point of the chain are the same,

mobo .8 = 81 + méc8,
and mda : = re +m8oN + Ric,
MOTION OF A CHAIN, 157

where S and W are the tangential and normal forces per unit
of mass.

 

 

O

Taking J as the length of chain and integrating the first
equation from A to B,

l
w= Sdo,
0
and the second equation gives

- +2

R=m= —7— mW,

As a particular case consider the motion of a heavy chain
inside a smooth circular tube in a vertical plane.

Measuring from the vertical radius, and from OA, take

COA=6, and AOP = ¢.
The equation of motion of the element PQ, or madd, is
mad .ad = mads.g sin (0 + ¢) +67,
and therefore taking a as the angle subtended by the string,
mahé = T — mga cos (0 + ¢) + mgacos 6,
158 MOTION OF A CHAIN,

a

and aa6=g {cos 0 —cos (6 +a)} = 2g sin (0+5) sin 5,

i Opin * @ g
whence aad? = 29 sin 5 {eos 3 — 008 (8 + 5) :

if the end A start from C.

 

 

eae

For the rate of pressure at any point,
mad. ab? = mad}>.g cos (0 +) + TS¢ — Radd,
or Ra = mga cos (6+ 6) + T— mae.

136. Motion of a piece of fine chain inside a smooth
circular tube which is revolving in its own plane about a point
in its circwmference.

AB being the chain, let ECA =06, and ACP =¢.,

The acceleration of the point P with regard to C, in
direction of the ve is

af, (0+¢), or a8,
MOTION OF. A CHAIN, 159

and the acceleration of C is w’a in the direction CO; there-
fore the equation of motion of the element PQ is

madp {ab + wa sin (6 + ¢)} =87,

 

and, integrating from A to B, that is from ¢=0 to =a,

a0 = {cos 0 — cos (6 + a)} = 207 sin (0 + 5) sin 5°
and therefore «a6? = 0 — 40 sin 5 cos (6 + 5) ‘
the constant being determined by initial conditions.

137. The equations of motion of a free string in a plane
under the action of given forces in the plane,

If, at any instant, w and v be the tangential and normal

velocities of point P of the string, and ¢ the angular velocity
of the tangent, the accelerations of P are

io — vp, and t+ ud.
Hence the equations of motion of an element PQ are
més (%— vd) =8T + més. 8,

més (i + ud) = 72 +mos.W,
160 MOTION OF A CHAIN.

p being the radius of curvature of the string at P and mS,
mV the forces per unit of length,

or m (ig) = + mS, m (itu) == + mW,

The string being inextensible, the geometrical condition
is that the motion of Q relative to Pis ultimately perpen-
dicular to PQ, and leads to the equations, ,

(uw + 8u) cos Sp — (v + dv) sin dp =u,
86 _ (v+ Sv) cos bp + (wu + du) sin dy — 9
a 8s :
Sy being the angle between the tangents at P and Q; or
ultimately

 

du_v
dsp’ andy “ast p’

138. A particular case is that of an endless chain originally
at rest under the action of conservative forces, and the
reactions of smooth curves, and set in motion in such a
manner that each point of the chain begins to move in the
direction of the tangent at the point.

In this case the chain will retain its form.

For, if v be the velocity of each point of the chain,
the equations of motion of an element ds are,

2
0=més. P +687, mBs més. Q + =;

or 0=mP+© (P—mi), O=mQ-+e (2— mai),

leading to P+ = (Qp) = 90,

as the equation giving the form of the curve when there is
no motion, and it therefore follows that the chain retains its
form, but that the tension at each point is increased by mv’.
MOTION OF A CHAIN, 161

This question is discussed in the solutions of the Tripos

. ° Questions for 1854 by, Mr Walton and the late Bishop
Mackenzie.

Mr W. Froude, in a letter in Nature, in November 1875,
described the experimental fact that a heavy endless chain,
placed over a drum, and made to revolve rapidly, retains
unchanged the catenary form which it assumes when not
in motion; and further that if an indentation in the form
be made by a blow from a heavy hammer, the inden-
tation, if the rotation be very rapid, remains for a considerable
time.

In the latter case the tension becomes so great that the
action of gravity is unimportant, although of course, the
action of gravity will, in time, remove the indentation and
restore the catenary form.

139. The equation for determining the initial tension of
a@ string.

The problem to be considered is that of a string on the
point of motion, under the action of given forces, as for
instance a string which being in equilibrium is cut at any
point.

Let PQ (8s) be an element of the string, and més.8,
mésN the tangential and normal forces acting upon it.

Take a, 8, a+8a, @+68 as the tangential and normal
accelerations at P and Q respectively, and 7, 7+ 67 as the
initial tensions at P and Q.

Then, 56 being the angle between the tangents at P
and Q,

mos .a=87' + més. 8, and més. 8 = (T+ 67) sindd + mésN,

ds

where p is the radius of curvature.

T
or, ultimately, m2z= ae +mS, mB= a +mN,

We have also the geometrical condition that the accele-
rations of Pand Qin the direction PQ are the same; and

B. D. 11
162 MOTION OF A CHAIN.

hence, 8¢ being the angle between PQ produced and the
tangent at @, we obtain

(a + 82) cos dp — a — (8 + 68) sin 86 = 0,
or, ultimately, 7 -& =0,

and, therefore, from the mechanical equations above,

@T aS T,mN
ds* biel a a p-

If one end of the string be fixed, we have, in determining
the constants resulting from integration, to introduce the con-
dition that the tangential acceleration at the fixed point is
zero, and, if either end of the string be moveable on a fixed
curve, the condition thereby introduced is that the accele-
ration of that end of the string in the direction of the normal
to the curve is zero.

Mathematical Journal, 1864.

For example suppose that a catenary, the upper ends of
which are fixed, ts severed at the vertex.

We have s=ctand, ee ¢, N=—g cos ¢,
and therefore a8 —g cos

2 1
The equation is i =o or, transforming,

gnc 2a psig Tcosf=0.

Integrating a cos $ — 7'sin d = C,

and therefore T cos = C+.
If ¢=0, 7=0, and T’cos ¢ = Co.
MOTION OF A CHAIN. 163
If a be the value of ¢ at the upper end,
dT :
wee sing =0, when ¢=y,
which gives i ae
cosy+ysin y

For another example take the case of a catenary the ends
of which, held apart, can slide on a smooth horizontal rod, and
umagine the ends let go.

In this case, as before,
Tcos$=Ch+C’,
but the geometrical conditions are that the tangential accele-

ration of the vertex is zero, and the vertical acceleration of
each end zero.

Therefore =0, when $= 0, which gives C=0, and
T=C'sec ¢,
and asin y+ cos y=0, when ¢=y,
leading to O'=mge and T =mgc sec ¢.

The vertex has no initial acceleration, and the horizontal
acceleration of each end is g siny (1 —sin ¥).

140. A heavy chain, lying on a smooth horizontal plane,
receives tangential impulses at one or both ends; it is required
to find the impulsive tension at any point, and the direction of
the initial motion.

Taking u and v as the tangential and normal velocities of
a point P of the chain, and 7’as the impulsive tension at P,
the equations of motion of an element PQ are

mos.u = dT, mis.o= 0;
aT T
or mu=—, and mv=—;
ds p

11—2
164 MOTION OF A CHAIN,

and we have besides the equation of continuity expressing the
fact that in the limit, the velocity of Q in direction of the
tangent at P is the same as the velocity of P in that
direction.

This condition gives us

(u + du) cos 86 — (v + bv) sin 86 = u,

on dp’ dsp
We hence obtain for the impulsive tension, the equation,
rT :
ds* ~ p .

If the chain be heterogeneous, that is, if m be a variable
quantity, the equation is
aT 1ldmdT T

ds* mds ds p*

As an example take the case of a piece of chain in the
form of a portion of a catenary, bounded by a chord parallel
to its directrix, and suppose that equal tangential jerks are
applied simultaneously at its two ends.

Since s=c tan ¢, we obtain
2
cos pS? sing TAT T cosh =0,
and therefore T cos¢ = 064+C,

At the vertex, u=0, and at each end, 7 =P, if P be the
tangential jerk; and we obtain, y being the extreme
deflection,

Tcos 6 = P cosy.
Moreover meu = P cosy sin ¢, and mcv = P cosy cos ¢,
and therefore v=ucot d,

shewing that every point of the chain begins to move in the
direction parallel to the axis of the catenary.
MOTION OF A CHAIN. 165

Ex. (2). Ifa chain in the form of a quadrant of a circle,
the density at any point of which, measured from one end varies
as e°, receive a tangential jerk at the other end, we have

1dm_1
mds a’
and our equation is

&@T dT

de ap 7 = 9

the integral of which is given in Chapter II., and the geome-
trical conditions are that .

T=0, when 6 =0, and that 7’ = P when 0=4;

these conditions determine the two constants of integration.

EXAMPLES.

I. A lamina, in the form of a regular polygon, is placed
flat on a smooth horizontal plane, and fastened to the plane ;
a string, the length of which is equal to the perimeter of the
polygon, is wound round it, one end being attached to an
angular point, and the other end carrying a particle; if the
particle be projected horizontally, at right angles to the
string, find the time after which the string will be wound up
again, and its greatest and least tensions. If the lamina be
held in a vertical plane, and the particle be projected in the
same plane, with an initial velocity sufficient to keep the
string always stretched, find its velocity at the instant the
whole string becomes straight, the side of the lamina-with
which the particle is initially in contact being horizontal and
downwards.

2. A small bead is projected with any velocity along a
‘circular wire under the action of a force varying inversely as
the fifth power of the distance from a centre of force situated
in the circumference. Prove that the pressure on the wire is

constant,
166 ‘ EXAMPLES,

3. A particle is placed at the extremity of the vertical
minor axis of a smooth ellipse and is just disturbed. Shew
that if it quit the ellipse at the end-of the latus rectum the
eccentricity must satisfy the equation e°+ 5e*+ 3e"= 5.

4. Ifa particle hanging vertically by a string be pro-
jected horizontally, and rise to a point P, and there leave the
circular motion, shew that if it recommences circular motion
at Q, PQ and the tangent at P to the circle will make equal
angles with the vertical,

5. If a particle move under the action of forces F, F’,
to any number of fixed points, and if g, q’,... be the chords of
curvature of the path in directions of these forces,

2mv? = & (FQ).

6. A uniform circular ring rotates uniformly in a hori-
zontal plane about its centre. Shew that the greatest possi-
ble linear velocity of its particles is independent of the radius
of the circle and of the cross section of the ring.

Find the breaking tension in pounds per square inch
in the case of a uniform ring of radius a feet which is on the
point of breaking when it makes » revolutions a second,
the weight of a cubic inch of the material of the ring being
that of c ounces.

7. Two equal particles which repel each other with
a force = pu (distance) * are placed on the inner surface of a
smooth sphere (rad. a), and their initial distance subtends
an 422 at the centre: shew that they will perform isochro-

nous oscillations in intervals = 27ra?,/2 sin a/./u.

8. A particle descends the arc of a smooth cycloid whose
axis is vertical from the base to the vertex.

If a horizontal line through the particle meet the circle
described on the axis of the cycloid in @, the velocity of
approach of the particle to Q is constant and equal to half the
velocity of the particle at the vertex of the cycloid.

9. <A particle is placed very near the vertex of a smooth
cycloid, s=4a sin ¢, axis vertical and vertex upwards; find
EXAMPLES. 167

where the particle runs off the curve, and prove that it falls

upon the base of the cycloid at the distance (/2+,/3) a
from the centre of the base.

10. A heavy particle slides, from a cusp, down the are of
a rough cycloid, the axis of which is vertical; prove that its
velocity at the vertex will bear to the velocity at the same
point when the cycloid is smooth the ratio of

Jem p  TeR.
11. Ifa pendulum oscillate in a medium the resistance

of which varies as the velocity, prove that the oscillations are
isochronous,

12. Two imperfectly elastic particles are constrained to
move on a cycloid, whose axis is vertical; they are let fall
simultaneously from different points of the cycloid; deter-
mine the energy of the particles immediately after the
n™ collision, and the ultimate loss of energy, as the number
of collisions increases without limit.

13. Two particles are let drop from the cusp of a
cycloid down the curve at an interval of time ¢: prove that
they will meet at a time

a ¢t
am a/ 245

after the starting of the first particle, a being the radius of
the generating circle.

14. Two equal smooth circles are fixed so as to touch the
same horizontal plane, their planes being at different inclina-
tions; two small heavy beads are projected at the same
instant along these circles from their lowest points, the
velocity of each bead being that due to the height of the
highest point of the other circle above the horizontal plane ;
shew that during the motion the two beads will always be at
equal heights above the horizontal plane.

15. A particle is attached to a point in a rough plane
inclined to the horizon at an angle a, originally the string is
its natural length; prove that the particle will not oscillate
unless tana > 34, where p is. the coefficient of friction,
168 EXAMPLES,

Between what limits must yw lie in order that the particle
may come to rest at the end of its m™ semi-oscillation ?

16. A fine parabolic groove has its axis vertical and
vertex downwards, an elastic string has one extremity at-
tached to the focus, and the other to a particle in the groove;
the natural length of the string equals one-fourth the latus
rectum, and the weight of the particle is such as to stretch
the string to twice its natural length ; determine the position
of equilibrium, and shew that the time of a small oscillation

about it is 27 /2a/g.

17. If a particle, mass m, be acted upon by equal
constant forces mf in the directions of the tangent and normal
to its path, and if the resistance be mfv"/k’, prove that the
intrinsic equation of the path is

2fe
ie fe Me 1) =u? (e—1),
u being the velocity of projection.

18. Two elastic strings the natural length of each of
which is 47ra, are fastened at a point P in a circlar tube
(radius a) of small bore; the strings are stretched in opposite
directions, and their other extremities fastened to a particle
of given weight. If the plane of the tube be horizontal, and
the particle be displaced from its position of equilibrium
through an angle less than 7/2, shew that the time of an os-
cillation is independent of the extent of the displacement,

19. A heavy particle is projected upwards from the
vertex, within a smooth parabola whose axis is horizontal,
with the velocity due to a fall down the latus rectum (4a).
Investigate the subsequent motion, and shew that the particle
impinges upon the parabola again, at a distance 3a./13 from
the vertex, with a velocity that bears to the velocity of pro-
jection the ratio /5 : /2.

20. A particle describes a circular arc under the action
of a constant force not tending to the centre; shew that it
will oscillate through a quadrant.
EXAMPLES. 169.

21. A particle is moving on the convex side of a rough
equiangular spiral towards the pole, under the action of a
force to the pole =«v, where v is the velocity at distance r.
If V be the velocity at distance a, and ¢ the time of moving
from distance r to distance a, shew that, « being the coeffi-
cient of friction,

ta 1
at 22V cosa+psina y — ¢(-cosartwsinaat}

pitetana _ itp tana — =
K cOsa— msInz

where « is the angle of the spiral.

22. Having given that the normal acceleration varies as
the square of the velocity parallel to the axis of a, find the
path; and prove that if this path be described under the
action of a force parallel to the axis of y, then the whole
acceleration at any point is proportional to the velocity.

23. If the curve whose intrinsic equation, referred to a
horizontal tangent, is s* sin ¢=a’ be described by a particle
under the action of gravity, find the time of descent from any
point of the curve to the horizontal tangent.

24, A particle, mass m, moves in a smooth circular tube
of radius a, under the action of a force, wm (distance), to a
point inside the circle at a distance ¢ from its centre; if the
particle be placed very nearly at its greatest distance from
the centre of force, prove that it will pass over the quadrant
ending at its least distance in the time

Ja log (\/2 + 1)//ue.

25. <A particle P attached by a string of length a toa
fixed point ( describes a circle in one plane, under the action
of a uniform repulsive force femanating from a fixed point
O; OC being =c and >a. If V be the velocity of the
particle when at its greatest distance from O and 1 its velocity
after describing an ¢ @ from that position, shew that

v= V?—-2f {e+a— Ve +a? + 2ac cos G}.
Find also the tension of the string and shew that

y’ + 5fa.
170 EXAMPLES,

26. A heavy particle is projected in a resisting medium ;
if » be the velocity at any time, @ the inclination to the
vertical of the direction of motion, and f the retardation,
prove that

\

1 du fo
a ae

If f=’, find v in terms of ¢.

 

27. Prove that the curve possessing the property that
the product of the distances of any point on it from two fixed
points is constant, may be described with uniform velocity
under the action of two forces, each tending to one of the
fixed points, and varying as the distance from the other, the
absolute intensities of the forces being the same.

28. Prove that a particle can describe a parabola under
a repulsive force in the focus varying as the distance and
another force parallel to the axis always of three times the
magnitude of the former; and that if two equal particles
describe the same parabola under the action of these forces,

their directions of motion will always intersect on a fixed
confocal parabola. ‘

29. Find the time of a small oscillation of a particle
suspended from a point by a string of length J, when the
square of a, the angle of oscillation, is neglected ; and shew

that the time will be r (1 + ia) sf 7 if the approximation
include the square of a.

A weight is drawn up uniformly and slowly with velocity
u by means of a crane; shew that the times of small oscilla-
tions will decrease at first in arithmetical progression, the
‘common difference being m*u/2g.

30. Two equal particles, connected by a fine string, are
placed in a circular tube, to one point of which they are
attracted with a force varying inversely as the distance; one
of the particles being initially at its greatest distance from
the centre of force, and v, v' being the velocities with which
EXAMPLES. 171

they successively pass through the point whose distance from
ee yt

the centre of force is 90°, shew thate “te “=1.

31. Two equal particles are connected by a string passing
through a small hole in a smooth horizontal table, one
particle hanging down, and the other held on the table; if
the latter be projected along the table at right angles to the
string with the velocity V2gc, prove that the initial radius of
curvature of its path is 4c.

32. Within a smooth circular tube of radius a held fixed

“in a vertical plane lies a light string of length greater than

half the circumference of the tube. The string carries equal

weights at its ends, which balance within the tube, and the

string subtends at the centre an angle 2 (w—a). If they be

slightly disturbed, shew that the time of a small oscillation
is the same as that of a simple pendulum of length a sec a.

33. A particle moves under the action of a central force
which is such that the normal acceleration on the particle
is constant: find a differential equation of the first order
to the path of the particle, and shew that 7° sin30=a’ is
a particular integral.

34, A heavy particle moves on a smooth curve in a
vertical plane, the form of the curve being such that the
pressure on the curve is always m times the weight of the
particle: prove that the time of a complete revolution is

2am/a/./g (m?—1)*, and that the length of the vertical axis
of the curve is 2ma/(m? —1)*, the whole length of the curve

being ma (2m? +1)/(m?—1)4.

35. A small smooth heavy bead runs on a string
fastened at two points in the same vertical line: the string is
originally vertical and the bead in its lowest possible position,
the bead is then projected so that it proceeds to describe
a portion of an ellipse, the string being at first tight; prove
that if the string becomes slack when the bead is at the
extremity of one of the equi-conjugate diameters of the
172 EXAMPLES,

ellipse, then the free path of the bead will pass through the
other extremity of the same diameter, and the latus rectum

of its free path will be to that of the ellipse as 1: 2/2.

36. On a wire in the form of a parabola with axis
vertical and vertex downwards is a bead attached to the
focus by an elastic string whose natural length is one eighth
of the latus rectum and whose modulus is equal to the
weight of the bead. Prove that the time of a small oscillation

is 27,/a/g, where 4a is the length of the latus rectum.

37. A particle slides down the arc of a vertical parabola ,
with vertex downwards starting from rest at a height h
above a horizontal line through the vertex. Shew that the
time of descent to the vertex is equal to

VES Wea).

where E£' (k) denotes the complete second elliptic integral to
modulus k.

 

38. <A circle, centre C, is described round an internal

centre S of attractive force; shew that the force varies as
r

where SP =r, a= radius of circle, SC =.

If there be two equal centres of force S, 8’ such that
SCS’ is a straight line, and

SC=S8C =c,
br

rt roa a shew that a
particle will describe a circle, centre C and radius a (><), if
projected from a point in the line CS distant a from C with

Je @ +e)
2a (a? —c’)

39. A smooth wire is bent into the form of a circle
radius a, and rotates with uniform angular velocity w abouta

then taking each force to be

a velocity perdendicular to CS.
EXAMPLES, 178

vertical axis through the centre which makes an angle a with
the plane of the circle. If a smooth bead slide on the wire,
shew that the equation of motion of the bead along the
wire is

ad’ eos gs. 8 #8
dp 7 2@ 008" a cos = sin = —g cos asin

where s is measured from the lowest point, Hence find the
position of equilibrium of the bead, and the time of a small
oscillation about that position.

40. A heavy particle hanging by a string from a fixed
point O is projected horizontally and describes a portion of a
circle greater than a quadrant until when it arrives at a point
P the string slackens and it begins to move in a parabola:
shew that the circle is the circle of curvature of the parabola
at P, and that if OP be produced to meet the directrix the
locus of the intersection 1s a circle concentric with the given
one.

41. A heavy bead slides on a smooth fixed vertical
circular wire of radius‘a: if it be projected from the lowest
point with a velocity just sufficient to carry it to the highest,
prove that the radius through the bead will, in a time ¢,
turn through an angle

= (sin /f
2 tan (sink A a t)
42, Ifa particle move on an ellipse under a force to the
centre = mr — nr log a . where c?=a’+6’, and N be the

-pressure on the curve, p the radius of curvature, prove
Np = 2ner + constant.

'

If the velocity vanish at an extremity of the major axis
and ~ = log = , then
n c—a
Np = 2ne (r —a).

43. Snow is uniformly spread over the surfaces of a
conical pinnacle and of the hemispherical dome of a building.
174 EXAMPLES,

It begins to slide off, starting at the highest point and
clearing a path as it goes. Prove that the motion in the
two cases is the same as that of a free particle moving on the
surfaces under the action of a vertical acceleration equal
to one-fifth and one-third the acceleration of gravity re-
spectively.

44, Ifa heavy particle be projected with a given velocity
and in a given direction prove that the initial curvature of
the path is independent of the resistance. Shew that, if the
resistance = p (velocity)’, the radius of curvature (p), at the
point where the tangent is inclined at an angle ¢ to the
vertical, satisfies the equation,

sin g $f + 8p.cos $+ 2up'sing = 0;

and that, if 1/p’ be the curvature at the other point where
the inclination to the vertical is the same, the curvature at
the vertex is

Lod
2 sin’ ( + 5) :

45. <A particle is projected along a rough tube, and is
acted on by a force always bisecting the angle between the
tangent and the normal, and varying directly as the curvature
and as the sine of the angle (yr) the tangent makes with a
fixed straight line. Shew that the velocity depends only on
ar, whatever be the shape of the tube; and if it is zero when
ap =e, where the coefficient of friction is 4 cot e, shew that the
time of describing an arc s varies as

 

 

| ‘ ds
on/sin (y—e)-

46. A tube of uniform bore in the form of an equi-
angular spiral is revolving uniformly with angular velocity
in a horizontal plane about a vertical axis through its pole,
and within the tube is a smooth uniform chain of length 2/
and mass m, which is initially at rest with its middle point at
a given distance from the pole; find the space described
by the chain along the tube in a given time, and shew that
EXAMPLES. 175

the tension at any point of the chain is mw’ cos? a (J? — a”)/4l,
where «# is the arcual distance of the point from the middle
point, and a is the angle of the spiral.

47. A string of varying density slides in a smooth
cycloidal tube whose axis is vertical and vertex downwards.
Shew that if the string be let fall from any position in which
its whole length is within the tube, its centre of gravity will
reach the vertex in the same time.

48. The bob of a simple cycloidal pendulum starts from
rest at the cusp. If the string break in any subsequent
position after passing the axis of the curve, the locus of
all possible positions of the focus of the trajectory subsequently
described by the bob is given by the equation

$$ _—_—_—___—_—.

Faw oort(1— 4/24 (14/2/44,

the middle point of the base of the cycloid being taken as
origin and the length of the string being 41.

49, A string of infinite length is laid on a smooth table
in the form ofa portion of one branch of the curve r” sin n@ = a",
so that one extremity of the string is at a finite distance
from the origin of polar coordinates; to this end a tangential
impulse is applied, so that the initial direction of motion
of each point of the string and the radius vector to the point
are equally inclined to the corresponding tangent. Shew
that the impulsive tension at any point <r” and the
density of the string must

o phon (r" ooh a”) - ‘

50. A uniform string falls freely in one plane under
the influence of gravity: prove that the angular acceleration
of the tangent at any point is

ceed (=)
ma/p ds \\/p/’
where TJ is the tension at the point, p the radius of curvature,

s the arc measured from a fixed poit of the string, and m
the mass of a unit of length.
176 "EXAMPLES,

51. The equation to a curve is
(u —a) {(u—a)*— Bb? (8 +4) *} 0;

A particle m, placed on the curve, and acted upon by the
force to the origin

mht (u +38 b" (u—a)*},
is projected tangentially so that its velocity, when it arrives

at the distance a” from the pole, shall be ha; prove that its
pressure on the curve will be always zero.

52. A bead can slide on a smooth circular arc A Band is
attracted by it, the force to any point being f(r): if it
be displaced from its position of equilibrium, the time
of oscillation will be 2://2cosaf(AC), where C is the
middle point of AB, and 2a the angle AC subtends at the
centre of the circle.

53. Prove that a lemniscate can be described freely by
a particle under the action of two centres of force of equal
intensity in the foci, each varying inversely as the distance,

and that the velocity will always be equal to ./4u/3, w/r being
the acceleration of either force on the particle at a distance r,

54. A heavy particle, mass m, falls down a smooth cycloid,
whose axis is vertical and vertex upwards, in a medium
whose resistance is mv’/2c, and .the distance of the starting
point from the vertex is c; prove that the time to the cusp is

/8a (4a —c)/W/ge, 2a being the length of the axis.

55. A particle is acted on by two forces, tending to the
foci of an ellipse whose major axis is (2a), and varying
according to the law y (r° + 8a°)/8a'r’, the absolute intensities
being the same. Shew that, if it be projected along the
tangent to the ellipse with a certain velocity, then it will
continue to describe the ellipse freely, and its velocity, in any
position given by the focal distances (r, r’), will be

nr trr +r) /2/rr’,

(n) being the mean motion in the ellipse under a force p/r”
to a focus.
EXAMPLES. 177

56. In the case of a piece of uniform chain, on a smooth
horizontal table, receiving a tangential jerk at one end;

(1) Find the form of the curve if the initial direction
of motion of any point makes a constant angle with the
tangent at that point.

(2) Find the impulsive tension at any point when
the form is such that the line-mass at any point varies as the
curvature.

(8) Prove that if the impulsive tensions follow the
same law as the tensions in a non-uniform string hanging in
the same curve under the action of gravity, then

dp _dp|”
SP oP| +4,
2 ds? ds

(4) If all the particles of the chain start with equal
velocities, prove that the form must be that of a straight line
or of a catenary.

57. AB is a diameter of a horizontal circular wire, and
a particle, free to move on the wire, is repelled from A by a
constant force equal to twice its weight. Shew that if placed
at an angular distance 2a from B, it will oscillate about B in
the same time as if it were oscillating under the action of
gravity alone on the same circular wire placed with its plane
vertical through an angular distance a on each side of the
lowest point.

58, If a projectile move in a medium the resistance of
which varies as the square of the velocity, and if p, p' be the
radii of curvature of the path at two points at each of which
the direction of motion is inclined at an angle ¢ to the
horizon, and r the radius of curvature at the highest point :
shew that
2 cos’

r .

 

1 1
<= + at
Pp Pp
59. ‘Fwo points A, B, such that AB is inclined to the

horizon at an angle (G). are connected by a wire made up

B. D. . 12
178 EXAMPLES.

of two equal circular arcs of radius (a), placed so that the
tangents at the ‘extremities A,B are horizontal, while the
other ends meet at a point C in AB, so that the two arcs
have at C a common tangent. Ifa ring be projected from A,
with velocity u, it will come to rest at B, provided

ue (Qu? —1) (e** cosa — 1) + 3ue** sin a

Aga 4p * +1
p» being the coefficient of friction.

 

60. A chain, in equilibrium in one plane, receives a
tangential impulse at a given point; prove that the impulse
at any point is given by the equation

a (: “) =e
dp \pdd/ p
p being the radius of curvature, and ¢ the inclination of the
tangent to a fixed line.

’

A uniform chain hangs in equilibrium over two smooth
pegs in the same horizontal; if equal vertical impulses
be applied simultaneously to the two free ends, find the im-
pulsive tension at any point, and prove that the initial velocity
of the vertex of the catenary is to the velocity which would
be imparted to each of the straight pieces of chain, if dis-
jointed from the catenary, as 1: 1+sina, where a is the
greatest inclination of the catenary to the horizontal plane.

61. A particle is attracted to two centres of force S and
H by forces which each follow the law

et econ
where » and wp’ are the same for both centres of force. Shew
that the particle can describe a circle whose centre is midway

between S and H, if b be the length of the tangent drawn
from either centre of force to the circle.

62. A shot is fired in an atmosphere in which the
resistance varies as the cube of the velocity. If f be the
retardation when the shot is ascending at an inclination a to
EXAMPLES. 179

the horizon, f, when it is moving horizontally, and /’ when it
is descending at an inclination « to the horizon; then
1 1 _ 2cos'« = 1. 1_ 2sin a (3—2 sin*a)
fof Se OE EF g

63. <A portion of a heavy uniform string is placed on the
are of a four-cusped hypocycloid, so as to occupy the space
between two cusps, the tangents at which are horizontal and
vertical respectively, and runs off the curve at the lower
cusp; prove that the velocity which the string will have
when the whole of it has just left the curve will be the velo-
city due to nine-tenths the length of the string.

64, A fine chain of given length is contained in a smooth
circular tube which rotates uniformly in a horizontal plane
about a fixed point in the circumference; if the chain
subtend an angle 2x at the centre, and if one end be initially
fastened to the tube at the end of the diameter through the
fixed point, and be released, prove that the angular motion of
the chain relative to the tube will be given by the equation

a’ = 2w* sin a (cos 6 — cos a).

65. A piece of string in the form of part of the curve,

r=ae°'*, the density at any point of which varies as

r—™"* lies on a smooth horizontal plane, being bounded by
@=0,and 6=£8. Ifa tangential jerk be applied at the end
6 = 0, find the tangential impulse at any point, and prove that
the initial direction of motion of every point makes an angle
with the normal equal to the angle of the spiral.

66. A string is in equilibrium in the form of a circle
under the action of a central repulsive force ; if the string be
cut at any point, prove that the tension at a point, the
angular distance of which from the point of section is 0, is
instantaneously changed in the ratio

acl
ete 3/: 2,
12—2
180 EXAMPLES.

67. A heavy string, passing under and in contact with
the arc of a fixed vertical circle, centre O, has its ends
fastened to two points P, @ on the circle, each at an angular
distance a from the lowest point, the pressure at which is
zero,

If the circle be suddenly removed prove that the tension
at the lowest point is changed at once in the ratio

sing : sinha.

Examine the case in which « =.

Also determine the new tension when the ends are allowed
to be moveable on smooth rods OP, OQ at the instant the
circle is removed.

Further, ifthe ends are allowed to be moveable on smooth

rods BP, BP’ from the highest point B of the circle, the
tension at the lowest point is changed in the ratio

sin = : sinha cos — cosha sin ~
2° 2 Q*
CHAPTER IX.
MOTION IN THREE DIMENSIONS.

141. THE fundamental equation of Kinetics being that
the time-flux of the momentum of a particle, in any assigned
direction, is equal to the sum of the acting forces in that
direction, and the resulting equations for the motion of a
particle in three dimensions being given, in different forms
in Art. (52), we proceed to employ these equations in some
particular cases.

Motion of a heavy particle in contact with fixed smooth
curves or surfaces.

Measuring z vertically downwards, and taking the accele-
ration along the tangent to the path of the particle, we
obtain :

ds "7 ds’
and therefore $m (v’— wu’) = mg (2-0),

if the particle start with the velocity u from the level c.

This is in effect the equation of energy, but it must be
carefully borne in mind that, in this case, the system consists
of a particle and the earth, and that we are neglecting the
kinetic energy acquired by the earth in consequence of the
attraction between it and the particle.
182 MOTION IN THREE DIMENSIONS.

142. Motion of a heavy bead sliding down a smooth wire
tn the form of a helia with its axis vertical.

If a be the inclination of the arc of the helix to its base,
R and R’ the reactions in directions of the principal normal
and binormal, the equations of motion are

mv"

—,_ = Rh, 0= R'— mg cosa.
a sec? a

do mg sin a,

mo 7, = mg ,
From these equations, v° = 29z, if z be the height through
which the bead has fallen, and the resulting reaction is equal

to /R* + R®.

143. In general, if a particle move in free space, or in
contact with smooth curves or surfaces, the equation of
motion, obtained by taking the forces in direction of the
tangent, is

mo a = m8,

mS being the resultant of the acting forces in direction of the
tangent.

If the particle have a velocity u at the point P and a
velocity v at the point Q,

gm (v* — vu?) = | mSds,

the integral being taken from P to Q.

Now, in accordance with the definition of potential energy
in Art. 47, the change of potential energy of the system, con-
sisting of a particle in a field of force, that is, of a particle,
and attracting masses the kinetic energy of which may be
neglected, is the work which would have to be done against
the forces of the system in order to move the particle from
P to Q, and therefore, if U be the potential when the particle
is at P, and V when it is at Q,

V—U=[-msds,

the integral being taken from P to Q.
MOTION IN THREE DIMENSIONS. 183

We hence obtain
$m (? — uw’) =U-V
or £m? + V= hm’ +4 U,
which is the equation of energy.
The total energy being constant, we notice that the force

on the particle, which is the time-flux of the momentum, is
the negative space-flux of the potential energy.

In other words the force in any direction, at any point of
the field, is the rate of exhaustion, in that direction, per unit
of linear space, of the potential energy.

144, , Reaction of a smooth surface on a particle which is
moving in contact with the surface.

Let the figure be.a section of the surface by a plane per-
pendicular to the line of motion of the particle P, and let PF
be the principal normal of the path, and PG the binormal.

¢

 

The accelerations in these directions being v’/p’, and zero,
where p’ is the radius of absolute curvature of the path, it
follows that the resulting acceleration in direction of the
normal PE to the surface is v’cos¢/p’, if @ be the angle
FPE or, by Meunier’s theorem, v/p, where p is the radius
of curvature of the normal section of the surface by the
plane through the tangent to the path.
184 MOTION ON SMOOTH SURFACE.

Hence, if R be the pressure, measured inwards, and V
the acting force in the direction of the normal to the surface,

mv/p= N+ BR.
If U be the acting force in direction of the tangent PL
to the surface in the plane FPF,
mv*/p' .sin d = U,
and therefore mv’ tan 6 = pU,
an equation which determines the position of the osculating
plane of the path.

145. If no forces are in action the path is a geodesic.

For, taking the equation of motion in direction of the
binormal,
0=fRsn ¢;
therefore ¢ = 0, or the osculating plane is a normal plane.
This result is equally true if the surface be a rough sur-
face, for the same equation exists.

146. Motion of a heavy particle in a smooth surface of -
revolution the axis of which ts vertical.

 

 

A

Measuring z upwards, and employing cylindrical co-ordi-
nates, the accelerations are “s
F—7r6, 27r6+76, and Z,
MOTION ON SMOOTH SURFACE, 185

and therefore, taking the acceleration in direction of the
tangent P7' to the meridian,

sin 6 + (# — 76") cosh = —gsin ¢.
Also, there being no horizontal force perpendicular to the
plane APN, os :
7rO + 276 = 0,
and therefore 76 =h.

The equation of the surface, z= f(r), being given, these
equations determine the motion.

Taking the acceleration in direction of the normal PG,
we have, for the pressure,

m {2 cos $ — (7 — 6") sin 6} = R— mg cos ¢.
Observing that tan ¢ =dz/dr, and that

. har : ,,,had
°F de’ and 2=f' (a

the first equation becomes

50 +{f' mis + cee -20 +f" ory (5)

r*

which is the differential equation of the projection of the
path on a horizontal plane.

Multiplying by 2 = and integrating, we obtain
1 /dr\? p detec al 2
a(33) + bf OF + 3 =0- BO) 2).

If the path be a horizontal circle of radius a, we at once
obtain from the equation (1)

h? = ga’ f' (a).
Or, we can obtain this result by considering that the

acceleration to the centre, v’/a, is maintained by the action:
of gravity and the pressure of the surface.
186 MOTION ON SMOOTH SURFACE.

If we employ the principles of energy and momentum,
leading to the equations,

£m {2 +7 +H} =C—mgz, and 6 =h,
we arrive at once at the equation (2).

147. To find the apsidal angle of the projection on a
horizontal plane of the path when it is nearly circular.

When the particle i is moving in a circle, we can imagine
a slight disturbance of the motion, as for instance by the
action of a small impulse in the vertical plane through the
axis and the particle. The value of A will then be unchanged,
and if we assume r=a+», where v is a very small quantity,

and neglect (53 d i) the differential equation becomes

ner @n
a—v+ wD (a‘ + 4a®v) {f' (a) +f" (a) o} =

zs {1+ (f'a)'} apt {s+ oo} y =0,

so that the apsidal angle is

mt {fF @P/ 38+ af" (a) /f’ (a).
This shews that the motion is stable, provided.

3+ af" (a) /f" (a)

 

is positive.

148. A heavy particle is projected horizontally along the
inside of a surface of revolution ; tt is rea to o jin the
initial curvature of its path.

PG being the normal to the surface at P, let PH and PF,
in the figure of Art. (146), be the directions of the principal
normal and of the binormal of the path.

If p’ be the radius of absolute curvature, the acceleration
in the dixeptions PE is v’/p’, and therefore, if EPG =v, the
acceleration in the direction PT is v’ sin y/p’,

and therefore v’sin y/p’ =g sin ¢.
MOTION ON SMOOTH SURFACE. 187

PG is the radius of curvature of the normal section of the
surface perpendicular to the plane ANP, and therefore, by
Meunier’s theorem,

p =PGcosp=rcosy/sin d, if r= PN.
Hence v tan = gr,
A vr
ou P= singvt grt
149, Motion of a heavy particle on the surface of a
sphere. ,

In the figure of Art. (146) let G be the centre of the
sphere and take c for the radius.

The equation, r°6=h, becomes c* sin? d. 6=h, and the
equation of energy is
og? +c'sin’ 66? = C+ 2gc cos d,
: h?
252 a
or. cd seca ee ee
_ If the particle be projected horizontally, from the position
=a, with the velocity v,
h=ve sina, ¢=0, and v= (0+ 2gc cosa,
so that
. 1 2
eg’ +0" (are - 1) = 29¢ (cos @ — cos a).
_ To find the greatest and least altitudes of the particle, put
¢ = 0, then we obtain ¢ = a, or

 

v" (cos a-+ cos p) = 2gc (1 — cos’ ¢).

It is easily seen that this equation gives one value for
cos @ lying between +1 and — 1, and therefore it follows that
the whole motion of the particle is comprised between two
horizontal planes.

Taking the acceleration in the direction of the normal
GP, we find for the pressure, measuring z downwards,

mé cos } +'m ( — r6&) sin 6 = mg cos $ — R,
188 MOTION ON SMOOTH SURFACE,

leading to
= 2, U sina
R= mq cos $ + med* + m - sin’?
or R=mq (3 cos $ — 2 cosa) + mv*/c.

This is in accordance with the general result of Art. (144),
for, in this case, the square of the velocity

= v"— 2gc (cos a — cos ¢).
150. Motion of a heavy particle on the surface of a smooth
cone, the vertex being downwards, and the axis vertical.

Employing cylindrical co-ordinates, and taking the ac-
celeration along the generating line, we obtain

(* — 76) sin a+ 2 cos a =—g cos a,

or 7 —r sin’ a6? = —g sin a cos a.

Also °6=h, and putting u for , we find for the differential

equation of the projection of the path on a horizontal plane,

du - 2  gsinacosa
apes ee a .

151. Motion of a particle in a smooth plane tube which
revolves about an axis in its plane.

Take the axis of rotation as the axis of z, and ¢ as
the inclination of the normal to the axis of z.

Taking S, NV, and Tas the acting forces in directions of
the tangent and normal to the curve and perpendicular to its
plane, and R, FR’ as the reactions in the two last-named
directions, the equations of motion are

m (#—r6") cos f + mzsin d =S,
mz cos @ —m (* —r&) sing = N+ R,
m (r6 +276) =T +R.
MOTION IN A TUBE. 189

Take for instance the case of a parabolic tube with its vertex
downwards, revolving uniformly about its axis which is vertical.
We have then

S=—mgsing, N=—mg cos ¢, T= 0,

O= a, 4az=r°, and 2atan d=r,
and the first equation becomes
¥ (Aa? + 9) + ri" = (4a?w* — 2ag) r.
Integrating and supposing that initially r=c, and *=0,
7 (4a? + 1°) = 2a (2aw’ — g) (r’?—’).

If 2aw*=g, the particle will remain, in relative rest, at
whatever point it is placed initially; and according’ as
2aw* > or <g, the particle will ascend or descend.

152. Motion of a heavy bead on a smooth wire in the form
of a helix, having its axis vertical, and revolving uniformly
about tts amis.

Measuring @ from the foot of the helix, the cylindrical
co-ordinates of the bead are

a, 9+ ot, z, where z = a6 tan a,
Taking the acceleration along the tangent to the helix,
a?cosa+ Zsina=—gsina, ;
or Z=—gsin’a, and .. 2 = 2g sin? a (h—2).
This gives the vertical velocity, and the horizontal velocity
=a(6 +0) =aw +2 cota
For the motion on the arc of the helix,
s=zcoseca and .. s°=29 (h — 2).

If R, FR’ be the reactions of the wire on the bead in
directions of the principal normal and binormal of the helix,
we have, .

ma (6+) = R,

and m (@ cosa— a6 sin a) = BR’ — mg cos a.
190 MOTION IN A TUBE,

153. The motion of a bead on a tortuous wire revolving
about a fixed axis, and acted upon by any given forces, can
be similarly treated by the use of cylindrical co-ordinates,

Or we can take axes of x and y revolving with the curve,
in which case the expression for the acceleration along the
tangent to the curve will be

7 ek ee ee . dy dz
(@—w*w — 2yw) T+ (9 — wy + Abo) Fo +27,
which reduces to s-o'rZ, and if T be the sum of the

tangential forces the equation of motion is
m € -wr ar) =T.
ds

If p be the radius of absolute curvature of the curve, the
acceleration of the bead in direction of the principal normal
is equal to

- ~\ G2 4. ; dy |. dz
(@ — wa — Qyjw) p Fa + (Y¥ — w*y + 2éw) pote,

which reduces to

$ d?x d’y ; d’y . *2)
5 atp(oraty Zh) + 20 (4 ae Uae .

and the expression, when multiplied by the mass of the bead,
is equal to the sum of the forces in direction of the principal
normal, and of the reaction in that direction.

In a similar manner the reaction of the curve in direction
of its binormal can be determined.

154. Motion of a heavy particle on a smooth inclined
plane, the plane being in rigid connection with a fixed vertical
axis and revolving uniformly.

Take the line of greatest slope, drawn upwards, through
the fixed point on the plane as the axis of #, and the normal
to the plane as the axis of z. Then, referring to Art. (28),

6,=osina, 0,=0, 0,=w cosa,
MOTION ON A REVOLVING PLANE. 191

and therefore, z being zero,
U=&— wy COS 4,
v=y + wx cosa,
w= wysin a.

Taking the accelerations parallel to the axes, we have, if
BR be the reaction of the plane,

m (% — vm cos a) = — mg sina,
m (0 — wo sina + uw cos a) =0,
m (wh + vw sin a) = R — mg cos a.
The first two of these equations give
& — 2 cos ay — w" cos’ 24 = —g sina,
9 + 2 w cos aé — wy =0,

thereby determining the motion, and the third equation gives
the pressure. The integration is at once effected by aid of
the calculus of operations, for the elimination of y leads to

a (3 cost a—1) 9% +08 cos’ aw = gw’ sin a,

a linear equation with constant coefficients.

155. If the particle be constrained to move on a smooth
curve in the revolving plane, the motion is determined by
taking the resultant acceleration along the tangent to the
curve.

This leads to the equation,

(% — 2 cos ay — w? cos’ ax) + (i + 20 cos a& — wy) y

-—asine®”
= gsinaT

r § — w’ cos” ax Py 4 =— g sin a —
0 = da o wa
ds a ds"
192 MOTION ON A REVOLVING PLANE.

Take for instance the case of a bead moving on a smooth
circular wire which is made to revolve uniformly about a fixed
vertical axis through tts centre.

If 6 be the angular distance of the bead from the axis of
x, the preceding equation becomes
a6 — w*a sin? asin 0 cos 0=g sina sin 6,
and the angular motion is therefore given by

aé? = wa sin” asin? 6 — 2g sin acos 6+ C.

The radial pressure, 2, of the bead, measured inwards, is
given by the equation

ma(@? +206 cosz-+w*cos*acos’d + w* sin? 0) = R +mgsinacos6.

156. The general problem of the motion of a particle on
a smooth surface which is made to revolve about a fixed axis
can be dealt with in a similar manner by aid of the general

expressions for velocities and accelerations which are given in
Art. (28).

Taking axes rigidly connected with the revolving surface,
let the plane of za contain the fixed axis about which the
surface is revolving, and let a be the inclination of the axis
of z to this fixed axis.

We have, then,
0#,=o@sina, 0,=0, 6,= cosa,
and therefore,
w=#— yw cosa, v= %— 2 Sin a + Zw COS a,
w=2+ yo sina,

The expressions for the accelerations are
f, =u — vo cos a,
J, =U — we sin 4 + uw 008 a, °
J, = w + vo sin a,

and, if the acting forces are given, the equations of motion
can be formed. ?
PENDULUM, 193
As a particular instance consider the motion of a pendu-

jum, or, which is the same thing, of a particle inside a smooth
sphere, rotating with the earth. :

P

 

 

Try!

EP being the earth’s axis, let C be the position of relative
equilibrium of the particle.

Neglecting the size of the sphere in comparison with the
distance OF, and regarding the earth as a sphere of which #
is the centre, the direction E’Oz is defined by the considera-
tion that the resultant of the earth’s attraction in the direc-
tion O#, and of the reaction at C, is the force mw’0N in
the direction ON.

As a matter of fact it has been shewn that, if c be the
B. D. 13
194 PENDULUM.

earth’s radius, and g the acceleration due to gravity at the
equator,

2G.

289’

oc =

wo

 

and therefore w can be regarded as a very small quantity,
and the displacement HOH’ as an infinitesimal of the first
order.

Hence if mg be the pressure at C, which is the weight of
the particle,

‘o?,ON :g=EL: OL=sin¢d : cos(a+¢),
if the angle OE'N =a,
or approximately, o=w’csina cosa/g,
where a is the colatitude of the point 0.

Taking then H’O, that is the vertical at O, as the axis of
z, and O as the origin, the expressions for the velocities and
accelerations are those given above, with the exception of the
expression for the velocity parallel to the axis of y, which
now becomes,

v=4¥—Zosina+#@ cosa—wcsina.
PENDULUM. 195

The equations of motion are,
x 4
mf,=—- B=, mf,=— BE, mf,=— B% — mg.

If the particle make very small excursions from its position
of relative equilibrium, we can put z=—a, and neglect the
squares of x and y, The approximate equations are then

gets ae Re
&— 2ywcosa=——-,
ma
ji limes se 2,
m a

Sa 2s R
24yosina=—g+7,

and eliminating R we obtain finally
g

& —2y w cos a+ rac)

j+2tocosatZy=0,

157. Motion of a heavy chain inside a smooth tube which
is revolving uniformly about a vertical amis in its own plane.

Consider the motion of an element PQ, (8c) of the chain
AB. Taking the arc OA =s, and AP=a, the equation
of motion is

mba {zsin } + (7 — wr) cos } = 67 — mgéc sin ¢.

 

 

13—2
196 MOTION OF. A CHAIN IN A TUBE.

The expression Zsin@+cosp is what would be the
tangential acceleration if there were no rotation and is there-
fore equal to s, and the equation becomes

re
ee a ae :
m (8 — w’r cos ¢) 7 mg sin ¢ ;
or, analytically, « not being a function of the time,
z=ssing, 7=$cos¢,
Z=§sing + sh cos d, #=S cos f — Sp sin g,
whence Zsin @ + F#cosd =S.

Taking J as the length of chain AB, and integrating over
the length J, we obtain

Ig = 40° (r’? — 1") — 9 (2 - 2"),

7’, 2' being the co-ordinates of the end B, and rv’, 2” of the
end A. ;

All these four quantities being functions of s, we have
an equation determining the motion of the chain in the tube.

If R be the rate of pressure at P in direction of the
normal PG, per unit of length of chain, we have, by taking
the acceleration in the direction PG,

moa {Z cos $ — (# — rw’) sin g} = Rc + — mgsc cos ¢,
a2

or me + a'r sing = B+ —my cos.

For example, if the tube be circular in form, and if the are
OA subtend an angle @ at the centre, and AB an angle a,

aa0 = hoa {sin® (a+ 0) — sin? 6} —g {cos 6 — cos (a+ 8)},
and therefore,
aa6? = tw%a {sin 20 — sin 2 (a + 6)} — 2g {sin 6 — sin (a+ 6)} + C.
EXAMPLES. 197

EXAMPLES,

1. Is a railway train heavier when going east or going
west ?

__Shew that for a train weighing 180 tons, travelling 60
miles an hour in latitude 60°, the difference is about the
weight of two men.

2. <A particle of mass m is attached to one end.of an
elastic string, the other end of which is fastened to the vertex
of a smooth cone of vertical angle 22, having its axis vertical
and vertex upwards; prove that the particle can move with
a constant velocity v round the surface of the cone, and with
the string stretched to double its natural length, provided
that the modulus of elasticity > mg cos a, and that

2v" cos a < ag sin? a,

If the particle be slightly disturbed in direction of the
string, find the time of a small oscillation.

3. A point describes a rhumb line on a sphere in such a
way that its longitude increases uniformly; prove that the
resultant acceleration varies as the cosine of the latitude, and
that its direction makes with the normal an angle equal to
the latitude.

4. A material particle rests on a rough plane inclined at
a given angle to the horizon, the plane begins to rotate round
an axis perpendicular to it, with a velocity commencing from
zero and continually increasing. Determine the velocity at
which the particle will commence to move on the plane, and
the condition that the commencement of the motion is
simultaneous with that of the plane.

5. <A particle moves under the action of gravity on a
smooth surface of revolution whose axis is vertical; shew
that its path cannot be a geodesic, unless it be a meridian
section.
198 EXAMPLES.

6. A smooth right circular cylinder is placed with its
axis horizontal, and a particle moving uniformly with velocity
v along the lowest generating line receives a horizontal
impulse at right angles to this line and just sufficient to carry
it to the highest point of the cylinder. If the particle be
prevented from leaving the cylinder, shew that its subsequent
path is such that if the cylinder be developed its equation is
9

zt
vp @
’

y = 7a — 4 tane
and that the highest generating line is an asymptote to the
curve.

7. Determine the motion of a heavy particle moving in
a smooth tube in the form of a circle, which. revolves uni-
formly about a vertical diameter.

8. In the centre of a hollow sphere resides a repulsive
force. A heavy particle is projected horizontally along the
surface of the sphere from a point distant 60° from the high-
est point with a velocity due to falling through the diameter
by its weight only. Shew that it will be again moving hori-
zontally at a point whose distance from the lowest point is

tan *,/3,
9. A particle slides on a smooth helix of radius a and
angle a under the action of a force to a fixed point in the

axis equal to mu times the distance. Investigate the motion,
and prove that the pressure cannot vanish unless the greatest

velocity of the particle be Vuaseca.

10. A heavy particle moves in a smooth sphere; shew
that, if the velocity be due to the level of the centre, the
pressure on the surface will vary as the depth below the
centre.

11. Two equal particles attracting one another with a
force, the accelerating effect of which is p x distance, are
placed in two rough straight tubes at Teh angles to one
another, and the friction 1s equal to the pressure in each
tube; prove that, if they be initially at unequal distances,
EXAMPLES, 199

T
2p
that, while they are approaching the point of intersection of
the tubes, they move in the same manner as the projections
of the two extremities of a diameter of a circle upon a straight
line on which the circle rolls.

one moves for a time =~ before the other begins to move, and

12. A particle is revolving on a smooth plane about a
centre of force, the accelerating effect to the centre being
# X distance, and when the body arrives at an apse the plane
begins to revolve with an angular velocity 4/3u about the
apsidal line; shew that the subsequent orbit described on
the plane will be a portion of a parabola; and that, when the
particle leaves the plane, its velocity will be V3 x velocity at
the vertex.

13. A smooth parabolic tube whose latus rectum is 4a
rotates about its axis which is vertical, the vertex being down-
wards, with uniform angular velocity w. Find in order that
a heavy particle may be in equilibrium at any point of the tube.

If the angular velocity of the tube be greater than this
and a particle be projected down the tube from any point with
velocity just sufficient to make it reach the vertex, shew that
the equation to the projection of the subsequent path, on a
horizontal plane, is

V4a? + 7
ea a
aa 8 a4 Viet 2a

14. A heavy particle, in contact with the lower half of
the internal surface of a fixed smooth spherical shell, is
projected horizontally with velocity V, the radius through
the particle making initially an angle a with the vertical.
Find the pressure in terms of the velocity at any time and
prove that if V*cosa<2ag, the particle cannot leave the
surface, but that if V* cos a > 2ag, it may do so, and find the
additional condition necessary.

15. <A heavy bead moves along a vertical circular wire
which revolves about a vertical straight line in its own plane.
Find the time of a small oscillation, and the resistance on the
wire.
200 EXAMPLES.

16. A small bead slides on a smooth circular ring of
radius a, which is made to revolve round a vertical axis
passing through its centre with uniform angular velocity o,
the plane of the ring being inclined at a constant angle a to
a horizontal plane. Shew that the law of angular motion of
the bead on the ring is the same as that of a bead on a ring
of radius a cosec 2 revolving round a vertical diameter with.
angular velocity sin a.

17. A smooth wire, in the form of a parabola, latus
rectum J, revolves about its axis which is vertical, the vertex
being uppermost, with uniform angular velocity =/g/l; a
string, passing through a fixed ring at the focus carries, at
one end, a small ring, mass m, which slides on the wire, and
at the other end a particle, mass m’, which hangs freely.
Given the velocity, V, of the ring at the vertex, determine the
rate at which the ring describes the parabola at any point.

If m= 4m’, and 2V* =gl, prove that, at a time ¢ after the
ring has passed the vertex, the angle 6 between the two parts
of the string is given by the equation

0_ /%
cot = 4/ Be

18. Three masses m,, m,, m, are fastened to a string which
passes through a ring, and m, describes a horizontal circle as
a conical pendulum while m, and m, hang vertically.

If m, drop off, prove that the instantaneous change of
tension of the string is
gmm,
m,+m,°

19. A particle is placed between two smooth co-axial circular
cylinders of nearly equal radii, whose common axis is inclined to
the vertical, and slides down under gravity. If @ be the angular
distance of its initial position from the lowest point of the
cross-section through that position, shew that the particle
will never press the inner surface if 20<7; but if 28>,
the particle will pass from the inner to the outer surface, and
back again, and so on, when ¢, its angular distance at any point
from the lowest point of the cross-section through that point,
takes the successive values given by 3 cos 6 = 2 cos 8.
EXAMPLES, 201

20. There are two points P and Q which move so that
the line of motion of each relative to the other is always
parallel to a given direction. If the motion of P, and the
initial position of Q be given, shew how to determine the
surface on which it must move. If the orbit of P be plane,
prove that this surface is a cylinder. If the motion of P be
that of a projectile in vacuo, and the relative velocity of
P and Q constant, determine the motion of Q.

21. A smooth hollow ellipsoid of revolution is fixed with
its axis (2a) vertical, and a particle is projected from a point
in the horizontal plane through the centre and on the inside
surface with a velocity /2ga and inclination « to the horizon.
Find « in order that the greatest depth below the centre may
be 2a/3, and find in that case the greatest height reached.

22. A smooth cylinder, whose transverse section is a
cycloid generated by a circle of diameter a, is placed with its
axis horizontal, the axis of the cycloidal section being vertical
and its vertex downwards. A heavy particle is allowed to
fall from rest at any point of the surface and is attracted by
a perfectly elastic plane perpendicular to the axis of the
cylinder, with a force varying directly as the distance from the
plane, whose absolute intensity is 2g/a. Shew that the path
of the particle will be such that if the cylinder be developed
it will develope into successive portions of a parabola.

23. A smooth surface is generated by the revolution of
the curve z*y=c® about the axis of y which is vertically
downwards, and a heavy particle is projected along the sur-
face with velocity due to the depth below the horizontal
plane through the origin: prove that its path intersects all
the meridians at a constant angle.

24, A surface of revolution is such that if it be held with
its axis vertical, and a heavy particle be projected along it
with suitable velocity at any point in any direction, its path
will cut every meridian of the surface at a constant angle.
Shew that the surface may be generated by the revolution
round the axis of y of the curve

h (a? — a”) + ay =0.
202 EXAMPLES,

25. A parabolic wire, axis vertical and vertex downwards,
rotates about its axis with uniform angular velocity. A ring
slides down it under gravity ; prove that it may descend with
constant velocity.

26. A heavy string of given length is enclosed in a
smooth straight tube, which is made to revolve uniformly
about a vertical axis, so as to describe a right circular cone;
determine the motion of the string and the tension at any point.

27. A surface is of the form traced out by the revolution
of the curve z=ccosa/c about the axis of z: the surface
being placed with its axis vertical, a particle is projected
upon it in such a manner that it describes a horizontal circle
in a given time ¢. Prove that the number of possible circles
is even, except in that case in which the time of revolution
satisfies the equation

gt a
1+ dete * ©°8 J (Se 1) = 0.

28. A heavy particle moves on a curve which revolves
uniformly about a vertical axis; prove that the time of an
oscillation of the particle about a position of relative equi-
librium will be
Qa psina \

@ (=n aeate ;

p being the radius of curvature at the point of equilibrium,
a the angle made by the normal at that point with the
vertical, & the distance of the point from the axis of revolu-
tion, and w the angular velocity of the curve.

29. An anchor ring is formed by the revolution of a
circle of radius (c) about an axis in its own plane, distant (a)
from the centre of the circle. A particle is projected along
the equator, of smaller radius with velocity (v), and is acted
on by a centre of attractive force in the centre of the axis,
and equal at distance r to ur"; shew that if the particle be
slightly displaced it will continue to return to its original
path at equal angular intervals (9), where

GaSe
EXAMPLES. 203

30. <A particle constrained to move in the surface of a
smooth ellipsoid is under the attraction of an internal ellip-
soidal shell, the two surfaces being confocal ; prove that if the
particle be projected from an umbilicus with a given velocity,
it will return to the umbilicus in a time which is independent
of the direction of projection.

31. Two infinite straight lines which are at right angles
but do not meet attract according to the law of gravitation.
Prove that, if a particle be projected from the middle point
of the shortest distance between the lines in direction of the
line bisecting the angle between them, it will continue to
move in a straight line: and find the limits of the motion.
Prove also that a particle will move with uniform velocity,
under the attraction of the lines, in any smooth tube which
takes the form of the curve of intersection of a certain hyper-
bolic paraboloid with any one of a certain series of oblate
spheroids.

32. A small smooth groove is cut on the surface of a
right cone, axis vertical and vertex upwards, in such a
manner that the tangent is always inclined to the vertical
at the same angle 8. A particle slides down the groove
from rest at the vertex; shew that the time of descending a
vertical height h is equal to the time of falling freely through
a height Asec’8. Shew also that the pressure is constant
and that it makes a constant angle @ with the principal
normal to the path, such that 2 tan 6 r/cos?a — cos? B = sin a,
2a being the angle of the cone.

33. <A shot, fired at an angle of elevation (a), strikes a
point at a horizontal distance (d) from the point of projection ;
shew that if any subsequent shot falls within a small distance
(c) of the point first struck, then the direction of projection
must meet the unit sphere drawn with its centre at the point
of projection within a small ellipse whose equation when
referred to certain axes is very approximately

d? (x* cos 2a + y* cos’ a) = c’ sin” a cos” a.

34. Three particles of equal mass which attract one
another according to the law of the inverse square, are free
204 EXAMPLES.

to slide on three wires which form the edges of a prism whose
base is an equilateral triangle. If the system is slightly dis-
turbed from its position of equilibrium, prove that it executes
a small oscillation in the time 2m /a*/3m; m being the mass
of a particle and a the mutual distance of the wires.

35. A particle is free to move along a helix whose axis
is vertical, and a centre of force whose accelerating effect is
p x distance resides in the axis of the helix. The particle is
so placed as to be in equilibrium, and the centre of attraction
then begins to move vertically upwards with a velocity V;
prove that after a time ¢

usin’ a (ssin a — Vt)’ +(ssina—VY=V*,
s being the arc of the helix measured from the position of

equilibrium, and a the angle which the helix makes with the
horizon. Hence determine s in terms of ¢.

36. A circular tube of smooth bore has its centre fixed
above a rough horizontal plane and is made to roll uniformly
in contact with the plane. Shew that the motion of a particle
of unit mass within the tube is given by

ag — a0’ sin? a sin ¢ cos $ +g sin a cos p=0,
and the pressures towards the centre and perpendicular to
the plane of the tube are determined by
a (f+ cosa)? + a? sin’a sin’ $ +g sin acos ¢ = R,
2ahQ sin acos fd + a? sin a cos a cos — g cosa=S,
where © is the angular velocity of the point of contact round

the vertical and a the inclination of the plane of the tube to
the horizon.
CHAPTER X.

THE HODOGRAPH AND THE BRACHISTOCHRONE,

158. The Hodograph. If from any fixed point a straight
line be drawn parallel to the direction of motion of a moving
point and of a length proportional to the velocity of the point,
the locus of its extremity is the hodograph of the path of the
point.

Polar equation of the hodograph.

If 6 be the inclination, to any fixed direction, of the tan-
gent to the path, and if the velocity = /(8),

then r=of(0)
is the polar equation of the hodograph, ¢ being any constant.

For example, if a heavy particle slide down the arc of a
smooth vertical circle from its highest point, the hodograph is

7” = 2gc (1 — cos 6).

Again, if a particle describe an ellipse under the action of
a force to its centre, v« CD, Art. (105), and therefore the
ellipse is its own hodograph.

159. If OP and OQ represent, in direction and magni-
tude, the velocities of a particle of the times ¢ and ¢ + d¢, PQ
represents, by the triangle of velocities, the velocity imparted
during the time &, and therefore if f be the acceleration of
206 THE HODOGRAPH.

the particle, PQ is the direction of the acceleration, and its
length = f6t.
aQ

0:

Hence it follows that the tangent to the hodograph is the
direction of the acceleration, and that, if o be the arc of the
hodograph, f=, that is, the velocity in the hodograph is
equal to the acceleration of the particle.

If for instance a particle move in a plane curve under the
action of a force making a constant angle with the direction
of motion, the hodograph is an equiangular spiral.

In general, if ~, y, z be the co-ordinates of a particle in
motion, and &£ 7, € the co-ordinates of the hodograph, we
have

E=4, n=9, $=4,
and from these the equations of the hodograph can be found.

Thus, if a heavy particle slide down a smooth helix, the

axis of which is vertical,

a=acos6, y=asind, z=a0tana,

and We=V=HK 497 4+27=a' setae;
£=sin OcosaV2ga0 tana, 1 =cos A cosa V 2ga0 tan a,
and f=sinaV2ga0 tan a,

Hence & + 7° = ¢’ cot a, shewing that the hodograph is a
curve drawn on the surface of a right cone, a result which
presents itself at once from the geometry of the case.
THE HODUGKArH. 207

- 160. In the particular case of central forces, the hodo-
graph is the reciprocal polar of the path, turned through a
right angle.

S

For, if SQ= V=h/p, SQ will be the radius vector of the
hodograph, turned through a right angle. Or, which is the
same thing, the hodograph is the inverse of the pedal curve

. turned through a right angle.

Hence for a conic section described under the action of a
force to the focus, the hodograph is a circle.

If p = f(r) be the equation of a central orbit, the equation
of the hodograph is
ce ¢
7S (5):

For QE, the perpendicular on SP, is the tangent to the
path of @, and
2 2

Cae aoe
Do BY = a pe re a

In all cases of free motion under the action of parallel
forces the hodograph is obviously a straight line.
208 THE HODOGRAPH.

161. Conversely, if the hodograph and its mode of de-
scription be known, the path can be determined.

Suppose for instance the hodograph to be a helix described
with uniform velocity. We then have,

£=acosat, y=asinet, z=atan a. at,

and the integration of these equations gives the equations of
the path.

From the first two we obtain the form
(a —a) + (y— BY =e’,
so that the path is a curve on the surface of a cylinder.

162. If two particles describe the same curve, in the
same direction, under the action of the same central force,
the chord of the hodograph corresponding to their positions
at any time represents the velocity of either relative to the
other.

For instance, if two particles describe the curve,
rsin 39=a, under the action of a force to the origin,
the hodograph is a three-cusped hypocycloid. ,

Now it is a known property of this hypocycloid that
any tangent to it, bounded by the curve, is of constant
length.

Hence we infer that whenever the directions of motion
of the two particles meet on the curve the velocity of either,
relative to the other, is always the same.

The Brachistochrone.

163. The Brachistochrone is the curve along which a
particle can be guided in a given field of force from one
given point, or from one given curve or surface to another,
so as to make the transit in the least possible time. We
shall consider first some special cases and afterwards prove
some general characteristics of brachistochrones.
THE BRACHISTOCHRONE. 209

To find the brachistochrone for the case of a heavy particle
an a vertical plane from one given point to another.

Measuring y downwards from the starting-point

v = 29y,
and the expression
[2 o V1 +p ui da; is to be a minimum.
NN 2gy

Employing the ordinary processes of the Calculus of
Variations, we obtain

Vi+p op ao.
= a + 0, or aa

vy VyV1 +p 2e—y
Hence =c vers?2 — V2cy —¥° a y

is the brachistochrone, and ide represents a cycloid having
its cusp at the origin.

164, A heavy particle moves on the surface of a smooth
circular cone, axis vertical and vertex upwards ; at ts required
to find the brachistochrone from a gwen point to a given
generating line.

"a

 
210 THE BRAGHISTOCHRONE.

If A be the starting-point, VA =a and VP =r,
v® = 2g cosa (r—a@),
and the expression,

where ¢ is the azimuthal angle is to be a minimum, from
which condition we obtain

dr\* 4. g sin’ a—c(r—a)
(a) =r'sin Meter
as the differential equation of the brachistochrone.

At the limit 8, employing the boundary equation, we

dr
find that dd

horizontal at its lower extremity.

=0, and therefore that the brachistochrone is

The fact that the curve passes through the point (a, 0)
theoretically determines c, and the radius to the lowest point
is given by the positive root of the equation,

rsin? a —c(r—a)=0.
165. A particle moves under the action of a repulsive

force from a fiaed point O varying as the distance; and starts
with the velocity Vu .OA from the point A.

To find the brachistochrone to another point B, we first
observe that, at a distance r,

v= pr’,

Jet

7

dé

and therefore

 

is a minimum, the limits being constant.

This leads to r ay =C,
ds

shewing that the brachistochrone is an equiangular spiral.
THE BRACHISTOCHRONE. 211

If it be required to determine, in this case, the brachisto-
chrone from the fixed point A to a given curve, r= (0), we
have, in addition to the equation, rd@ = Cds, the boundary
condition.

Taking r,, 6, as the co-ordinates of the bounding point B
on the curve r=f(8), this condition is

1 (ar
t/ ay (2 Ne

Jr (an)

 

0

Now, if BB’ be a small arc of the given curve and
BOB = d0,, and if the dotted line BF represent part of the
brachistochrone, and B’E the slightly variated curve, meeting
in # the radius vector OB, then 6r, =— BE.

From the figure it will be seen that
BE = EL-—BL=BL cot EB’ 0 — BL cot BB’O,

and therefore or, = | t'(@)- (35) | dé.
1.

Substituting in the above equation, and observing that
d@, is an arbitrary quantity, we obtain

| r
Fos ant + l= 0, a oeene A ')
a), 70)
* See Todhunter’s Integral Calculus, Art. (358).
14—2
212 THE BRACHISTOCHRONE,

proving that the brachistochrone must intersect the given
curve at right angles.

To complete the solution, we obtain from rd@ = Cds, the
equation
r=aete, where a = OA,
and, to find r,, 0, and a, we have the above equation (A),
with the equations,
r, = f(8,), 7, = ae ote,

166. The brachistochrone for a particle moving in a given
field of force. | i

The system of the particle and the field, being a conser-
vative system, as in Art. (143), we know that the velocity
depends upon the position of the particle, and therefore, if v
be its velocity when passing through the point (a, y, 2),

v= f(xy, 2).

For the brachistochrone we have to make the expression
7 we a
[2 or [AHP ay
v v

a minimum, p and q standing for dy/dx and dz/dx.

iF

The methods of the Calculus of Variations (see Todhun-
ter’s Integral Calculus, Art. 364) lead to the equations,

d/l d/l p
Tate ays) ~ a (Torry) ~O

ed fl d/l q ?
VIED +E dels) ~ dewey ~°

which reduce to

d’y dudy , dv _
de deds t dy ~%
dz dvdz , dv_ 0

VG dds” ae
THE BRACHISTOCHRONE. 213

Multiplying by ay and . respectively, and adding the

results, we obtain the symmetrical equation

pte _dods, dv_

ds*  dsds ‘ dx

Any two of these three equations determine the brachisto-
chrone.

0.

167. The brachistochrone for a particle constrained to
move on a given smooth surface.

Taking K and L to represent the left-hand members of
the equations (4), we have in this case

fey + L8z) dx = 0;
and also, if ¢ (a, y, 2) =0 be the given surface,
ad by + bas dz = 0.

dy d.
We hence obtain the single condition
Bek,
ap dg’
dy dz

and this equation, with ¢ (a, y, 2) =0, determines the brachis-
tochrone.

168. In the case of a particle moving in a field of force,
if v be the velocity of the particle, and V the potential
energy, the equation of energy is

dmv’? +V=C.
From this equation we obtain
du dV
i oa mX,

if mX be the acting force.
214 THE BRACHISTOCHRONE.

The equations of Art. (166) now become

da  dvdx a
0 a ° Fe dg t=
2dy  ,dvdy | y_
Vag "ag to
.@2z  dvdz

Vv ae deus + Z=0.

If X, w, v be the direction cosines of the binormal we
obtain
WX + wpY+vZ=0;
and hence it follows that
the osculating plane contains the resultant of the acting forces.

Again, multiplying by the direction cosines (J, m, n) of
the principal normal, and adding, we find that

p tik + mY +nZ=0;
that is, the component of the acting force in direction of the
principal normal is equal to — mv*/p.

Now, for free motion, the force along the principal normal
is equal to mv’/p.

If then the normal force be reversed in direction, the
tangential force remaining unchanged, a free path becomes a
brachistochrone, and the converse is equally true.

In other words the forces are reflections, or images, of
each other with regard to the tangent, both in direction and
magnitude.

This theorem is due to Professor Townsend, and is given
with illustrations in Vol. xIv. of the Quarterly Journal of
Mathematics.

For instance, if a particle move in the curve, s = 4a sin d,
under the action of a force inclined to the direction of motion
at the angle 7/2+4¢, it will be found that the force is
constant.
THE BRACHISTOCHRONE. 215

The image of this case is the cycloidal brachistochrone
under the action of gravity.

For another example take the case of an ellipse described
freely under the action of forces to the two foci, each varying
inversely as the square of the distance.

The same ellipse will be a brachistochrone for repulsive
forces from the two foci, each varying inversely as the square
of the distance from the other focus.

EXAMPLES.

1. A point moves in a straight line under the action of
a force varying as the distance from a point in that line;
prove that the corresponding point in the hodograph moves
as though acted upon by a similar force.

2. One particle describes a given orbit about a centre of
force, and another particle describes the hodograph of that
orbit under the action of a force in the pole of the hodograph,
shew that the product of the accelerations of the particles at
two corresponding points of their orbits varies as the product
of the central distances of those points.

3. If Pand @ be the tangential and normal forces, and
@ the inclination of the tangent to a fixed direction, the

hodograph is
log = =| 5 a8
Sa 0G

4. A smooth elliptic tube is placed with its major axis
vertical and a particle allowed to slide down it, starting from
rest at the highest point; shew that the hodograph is given
by the equation

syack | ae
remo sin 5 jcot (; cot e).

5. Prove that the hodograph of a central orbit can
itself only be a central orbit under the action of a force to the
origin from which its radii are drawn when the central orbit
is an ellipse or hyperbola.
216 EXAMPLES,

6. A heavy particle moves on a rough curve in a
vertical plane so that the pressure on the curve is constant.
Prove that its hodograph is a conic described as about a
centre of force in the focus.

7. Ifa particle describe a lemniscate under the action of
a force to the pole, prove that the hodograph is of the form

8. If a particle move in a brachistochrone in an open
field of force the pressure on the constraining curve is 2mv"/p.

9. A rough tube in the form of a cycloid is placed with
its axis vertical and vertex upwards. A heavy particle is
projected along the tube from the vertex with a given
velocity V, find the velocity in any subsequent position.

If the coefficient of friction be tan A, and the initial
velocity be to that which would be acquired in descending
freely down the tube, supposed smooth, as sin X: 1, prove
that the hodograph is a circle. ‘

10. Find the hodograph in the cases of motion’ in a
cardioid under the action of a force to the cusp.

11, One circle rolls uniformly on the circumference of
another, on the outside; find the hodograph of a point on the
circumference of the rolling circle.

12. Find the hodograph in the cases of the motion of a
heavy particle on a smooth cycloid, the axis of which
is vertical and the vertex (1) upwards, (2) downwards.

13. A particle is moving under the action of a force
perpendicular to and proportional to the distance from the
line of zero velocity, shew that the brachistochrone is
a circle,

14. Prove that a parabola is a brachistochrone,
(1) for a constant force from the focus ;

(2) for a force from the directrix varying inversely as
the square of the distance from the directrix.
. EXAMPLES, 217

15. Prove that if the force vary inversely as the cube of
the distance from a fixed point, the brachistochrone will be
an equilateral hyperbola,

16. Shew that the parabola is brachistochronous for a
force acting perpendicularly from its axis, and varying directly
as the axial and inversely as the square of the focal distance,
the line of no velocity coinciding with the axis.

17. <A particle moves in a vertical plane in a medium
whose resistance is ku": determine the hodograph. Shew
that it will be an algebraic curve if n be an odd integer.

Defining the instantaneous parabola as the parabola that
would be described if at any instant the resistance cease to
act; shew that the vertex of such a parabola is at any instant
moving downwards at an angle tan™ (4 tan d) to the horizon,
where ¢ is the angle the particle’s path makes with the horizon.

18. Two particles are describing free paths in one plane
which are hodographs to one another; if the particles be
always at corresponding points, prove that the paths must be
conic sections, and find the nature of the forces acting on the
particles. ,

19. A body moves on a right circular cone, the velocity
varying as the nth power of the cosine of the angle of
inclination to the vertical; and the body moves along the
curve of quickest descent from one given point to another.
Shew that, if the cone be developed, the path will become a
curve such that the perpendicular on the tangent varies as
some power of the polar subtangent; and find the curves for
the cases n = 1 and n= 0.

20. If the velocity of a carriage along a road is propor-
tional to the cube of the cosine of the inclination of the
road to the horizon, determine the path of quickest ascent
from the bottom to the top of a hemispherical hill, and shew
that it consists of a spherical curve described by a point of a
great circle which on a small circle described about the

pole with a radius = together with an arc of a great circle.

6 2
How is the discontinuity introduced into this problem ?
218 EXAMPLES.

21. Prove that any curve which is a free path for a
force to a fixed centre is also a brachistochrone for an equal
force enveloping its caustic by reflexion from the fixed centre
as focus.

22. If the particle move in the brachistochrone between
two given points on a surface of revolution, prove that the
area, swept over by the radius vector on the horizontal plane
is proportional to the action.

23. A projectile moves under gravity in a uniform
medium whose resistance varies as the velocity. Prove that
the hodograph of the trajectory is a straight line, and that
the velocity of the point on the hodograph is proportional to
the horizontal velocity of the projectile.

24, Find the hodograph of an elliptic orbit described
under the action of a force to the focus, and hence prove that
the mean value, taken with regard to time, of the inverse
square of the radius vector is equal to the product of the
reciprocals of the semiaxes.

25. A particle moves freely under the action of a force
whose direction is always parallel to a fixed plane, and
describes a curve which lies on a right circular cone and crosses
the generating lines at a constant angle, prove that its
hodograph is a conic section.

26. A heavy particle is projected from a given point
along a smooth groove cut on the surface of a right circular
cone, whose axis is vertical and vertex upwards, with the
velocity due to the depth from the vertex. Prove that, if
it reach another given point not more than half way round
the cone in the least possible time, the curve of the groove
must be such as would if the cone were developed become a
parabola with the point corresponding to the vertex as focus.

27. <A particle, acted on by a central attractive force

whose accelerating effect at a distance r is Gar a being
a constant, is projected from a given point with the velocity
from infinity ; prove that the form of the groove, in which it
EXAMPLES. 219

must move in order to arrive at another given point in the
shortest possible time, is an hyperbola whose centre coincides
with the centre of force.

28. <A particle is constrained to move on a surface of
revolution under the action of forces the directions of which
pass through the axis, and which depend upon the distances
from the axis or from fixed points in it; find the differential
equation, (in r and @, or in any other form,) to the projection,
on a plane perpendicular to the axis, of the brachistochronous
path between two points on the surface, and prove that the
velocity at any point is proportional to the distance from the
axis, and to the sine of the angle between the path and the
generating curve through the point.

If the surface be a hemisphere, and the force be attractive
and vary as the distance from the axis, it may be shewn that
when the starting-point is in the rim of the surface,.the
projection is a straight line.

29, Find the differential equations of the brachistochrone
on the surface of a sphere which is rotating round a diametral
axis with uniform angular velocity.

Prove that if r and 6 be the co-ordinates of the projection
of the particle on a plane perpendicular to the axis, * will be
proportional to the resolved part of the force of constraint
perpendicular to the meridian plane of the particle at any
instant, and that under certain conditions the equation be-
tween r and 6 assumes the form

a=~r cosh mé.

30. The velocity of the current in a stream is proportional
to the distance from the bank. A man who swims at a given
rate wishes to get as far as possible down the river in a given
time ; shew that he must start from the bank at an angle
whose tangent is proportional to that time.
CHAPTER XI.
MOTION OF TWO PARTICLES ACTING ON EACH OTHER.

169. IF two particles, attracting each other, move in a
plane, we know that their centre of gravity is either at rest
or is in motion with a velocity constant in magnitude and
direction.

Having given the velocities at any instant we can find
the velocity of the centre of gravity, and by reversing this
velocity on the whole system we find the velocities of the
two particles relative to their centre of gravity, and the case
is then reduced to that of a central force.

In all cases the force of each particle on the other is
proportional to the product of the masses and a function of
the distance, and therefore by a proper choice of units is
represented by the expression mm’ ¢ (r).

If m,m’ be the masses of the particles, and u, wu’ their
initial velocities relative to their centre of gravity, these
velocities are in parallel and opposite directions, and are such
that

mu=mu'.

It is evident that the two paths about G@ are similar

curves.
TWO PARTICLES ACTING ON EACH OTHER. 221

If PQ=r, and if mm’ ¢(r) be the force exerted by each
particle on the other, then, considering the motion of P, this
force

init d ae apt,

m

 

which is a function of the distance GP, and the motion is
therefore determined: as in Chapter VII.

U, ae!

   

If the path of P with regard to G be determined in the
form, GP =f (@), the path of P relative to Q is given by the
equation,

m
mnt S(O).
Or, we can determine the path of P relative to Q by

finding the initial velocity of P relative to Q, and observing
that the acceleration of P relative to Q is

(m + m') (PQ),

which again reduces the case to that of a force to a fixed
centre. :
222 TWO PARTICLES ACTING ON EACH OTHER.

170. Motion of two particles in a plane attracting each
other with a force varying as the distance.

The force on P=mm' PQ=m (m+m’') GP, and the accelera-
tion of P in the direction PG =(m +m’) GP, hence it follows
that P describes an ellipse about G as centre in the time

2r/J/m +m’, and that, if wu be the initial velocity of P relative
to G, the semidiameter conjugate to GA =u//m +m’.
If u, uw’ be the initial velocities of P and @ relative to G,
so that mu=m’'u’, the initial velocity of P relative to Q
=utw=(m+m) u/n,
and the acceleration of P relative to Q =(m +m’) PQ, so that

the periodic time is 2m//m-+m’, and the semidiameter,
conjugate to AB, of the relative path

=u lm + mm’.
These last results are of course at once derivable by geometry
from the preceding.

171. Motion of two particles in a plane when the law of
attraction 1s that of the inverse square of the distance.

Taking the force between the particles to be
mm’/(Distance)’, the acceleration of P in the direction PG

=m'/PQY =m" /(m +m’) PG,
and hence it follows that the path of P relative to G is a
conic of which G is a focus.

This conic is a parabola, ellipse, or hyperbola according as
wu’ is equal to, less than or greater than 2m’/(m + m’)?AG.

If the relative path is an ellipse, its transverse axis 2a is
given by the equation
po ( Ss x)
~(m+mYP\AG 2a)’
and the periodic time = 27a® (m+ m')/m’*.

In the same manner if the law of attraction be that of
the inverse cube of the distance, the relative paths may be

equiangular spirals, provided the relative velocities are
properly adjusted.
EQUATIONS OF MOTION. 223

172. Ifthe two particles are not initially projected in
the same plane, we must find the velocity of the centre
of inertia of the system, and, by reversing it on each body of
the system, we shall obtain their velocities relative to the
centre of inertia. These velocities will be in parallel direc-
tions and such that mu = m/‘w’, and the plane passing through
them is the plane of the relative motion.

Hence it is seen that the actual motion consists of the
motion in the plane, while the plane, remaining parallel to
itself, moves with the centre of inertia.

173. We can also obtain these results from the equations
of motion.

For, if 2, y, 2 be the co-ordinates of one particle, a’, y’, 2’ of
the other, r the distance between them, and & the force of
each on the other, the equations of motion are,

t , t
- L-2 7 — is e-2
mit =— R——— , mij =- REL, mi=—R ae

a— a

mia =R mod mi = RY, m2 =R

 

 

Hence we obtain

 

 

é-@ _y-9 2-7 __ pmtm
e—-a@ y-y 2-2 mmr”
and, integrating, we find that
Yy-y) @-#)-@-2)Y-Y=A,
(¢—2)(a@—a') — (w@—a') (4-2) =B,
(2—2')(y-y¥) —(y—y) (@-#) = C,
A, B, and C being constants,
and .. A(e-#)+Biy-y)+C(e-2)=0,

shewing that the line joining the particles is always perpen-
dicular to the line
224 TWO HEAVY PARTICLES

174. Motion of two heavy particles, connected by an
inextensible string, and projected in any manner.

If the initial distance of the particles from each other be
less than the length of the string, there will be, after a certain
time, a jerk of the string. The velocities perpendicular to
the string will be unchanged, and the change of the velocities
in direction of the string will be determined by the considera-
tion that the momentum of the system in that direction will
not be affected by the jerk.

For the subsequent motion, since the tension of the string
produces equal and opposite momenta in any given time, it
follows that the horizontal momentum, in any direction, of
the system is constant, and that the vertical momentum
imparted to the system is the same as if there were no
string.

If then wu, v be the component horizontal velocities in
given directions, at any time, of one particle, and w’, v’ of the
other, :

mu + m'u’ and mv +m'd'

are each constant, and if w, w’ be the initial vertical velocities,
measured downwards, the vertical momentum of the system
at the time ¢ is

mu + m'w' + (m+m’) gt.

Taking then a, 8, y as the component velocities of the

centre of inertia, the horizontal resultant ./a’+ 6? is constant
in magnitude and direction, and

 

(m+ m’) y= mw + m'w' + (m+ m') gt,
so that the motion of G is the same as that of a projectile.

Further, if we imagine the velocity and acceleration of @
reversed on the system, we shall have the case of: two
particles connected with G by strings of given length, and
consequently the motion of each will be circular, and the
tension of the string will be constant. These are particular
cases of the theorems of Arts. (41) and (45).
CONNECTED BY A STRING. 225

175. We can obtain the same result by means of the
equations of motion.

If 7 be the tension and J the length of the string, these
equations are

z-2

l ?
—a 2-2
l i °

The addition of the several pairs of equations gives the
first result.

Also we find as in Art. (172) that the string is always
perpendicular to a certain fixed direction.

a— i

l

y~y
j

mé=—T

 

 

, my=—T , mé=mg—-T

 

es x
owe = T

 

 

; mf = TE, mz =mg+T

Further we have

(w@—a)P + (y-yfP+@—-2VaP,
and therefore

 

= 2) Ge 4 Re Restate =0. (a).
But from the equations of motion
a =—7(=+5)"" , &.
mm i

(@¢-#P+y-yY+(e-7P=C- Tl (=+5),

and substituting in the equation (a) we deduce that 7’ is
constant. «

176. Motion of a number of free particles, attracting each
other with forces proportional to the distance.

In this case, by a well-known theorem, the resulting
action on each particle is directed to the centre of gravity of
the system, and is proportional to the distance from it. The
particles therefore describe, relative to G, ellipses of which
@ is the centre, and in the same periodic time.

B. D.: 15
226 EXAMPLES,

EXAMPLES.

1. Two bodies attracting each other with a force which
varies inversely as the cube of the distance are projected in
parallel directions; find the condition that the relative paths
may be equiangular spirals.

2. If two particles of masses pw, w’ attract according to
the law of gravitation and be projected with velocities y, v’,
making an angle a with each other; shew that their orbits
relative to their common centre of gravity will be para-
bolas, ellipses, or hyperbolas, according as

v — Quv’' cos a pete ee)
> c

2
where ¢ is their initial distance apart.

3. Two bodies, the masses of which are m and m’, are
projected from the points A, B, and attract each other
according to the Newtonian law. The body m is projected

from A in the direction BA with a velocity i oa and

m’ is projected from B in a direction BP with a velocity

2 / ttm cos PBA;

determine completely the path of either with regard to the
other.

 

4. The co-ordinates (a, y), (a, y,), of the simultaneous
positions of two equal particles are given by the equations

x=a0 —2asin 6, wx, =a,
y=a—acos 8, ¥,=—-a+acos6;

prove that, if they move under their mutual attractions,

the law of force will be that of the inverse fifth power of the
distance.
EXAMPLES, 227

5. Two particles move under the action of their mutual
attractions, one of them being constrained to remain on a
fixed smooth wire in the form of a plane curve: if the path
of the other be an involute to this curve and the two particles
be always at corresponding points, the curve has for its
intrinsic equation

m2

S=ae 2,
where ™m is the ratio of the masses of the particles.

6. Two equal bodies attract each other with a force
varying inversely as the fifth power of the distance, and they
are projected with equal velocities, in opposite directions, at
right angles to the line joining them; prove that there are
two velocities, in the ratio of 1 : /2, for each of which the
relative orbits will be circles.

7. Two masses m, m’ are connected by an inextensible
string of length a. The extremity A to which m is attached
is compelled to move with uniform acceleration in a straight
line under the action of a force P in the straight line, and the
extremity B to which m’ is attached, is compelled to describe
a circle round A with uniform angular velocity » under the
action of a force @ perpendicular to AB, Find P and Q, and
prove that the least value of Pis

m'arw*

mf — 47 provided aw’ be < 2f.

 

8. Two smooth circular rings (of radius a) are placed in
a vertical plane with their centres in the same horizontal
line at a distance 3a. Two equal beads (of mass m) slide on
these rings and are connected by a thin elastic string, of
which the natural length is 3a and modulus of elasticity
3amg. They are held as far apart as possible and then let go.
Find when they come to rest.

In the particular case in which 7 =1, find the whole time
of the motion.

9. Two equal particles can move on a fixed smooth

circular wire and attract each other with a force varying as
15—2
228 EXAMPLES.

the distance between them. Prove that their centre of
gravity moves with uniform angular velocity, and that the
relative motion of one with respect to the other is the same
as the motion of a simple pendulum.

10. Two beads of equal mass repelling one another with
a force varying inversely as the square of the distance are
free to slide on a parabolic wire. If they are initially at the
extremities of the latus rectum, prove that if properly
projected the line joining them will always pass through the
focus of the parabola.

11. The attraction between two equal particles, each of
mass m, is wm?/r*, when r is the distance between them, and
they are projected with equal velocities on the same side of
the line (c) joining them in directions not parallel but
equally inclined to that line; prove that the path of each
will be an ellipse, parabola, or hyperbola, according as the
initial component of each velocity in direction of the line ¢
is less than, equal to, or greater than /2um/c’*.

12. Two small rings, each of mass m, which attract each
other with the force mw* x distance, are placed on smooth
wires Ox, Oy, inclined to each other at a given angle, which
commence to move in their own plane with angular velocity
w, and continue to move uniformly. Determine the motion
of the rings.
CHAPTER XII.
ENERGY AND MOMENTUM.

177. It is intended in this Chapter to illustrate the use
of the principles of momentum and energy which were laid
down in Articles (40), (48) and (48) of Chapter IV.

In many cases problems of motion are very rapidly and
easily solved by the aid of these principles, and in all the
cases to which they are wholly or partially applicable, the
problem of determining the motion of a body or a system is
reduced to the solution of equations containing differential
coefficients of the first order, or simple time-fluxes of the
co-ordinates of the system.

178. Motion of two spheres which attract each other
according to the law of nature, of given masses m and m', and
given radii a and a’, placed originally without kinetic energy
with their centres at a gwen distance c from each other.

Supposing that the configuration of zero potential energy
is when the spheres are in contact the potential energy of
the initial configuration, which is the work done in separating
the spheres,

 

& mm mma mn’
mf? dean SO
ata 7 a+a c
During the subsequent motion let u and w’ be the
velocities of the two balls when their centres are at a
distance r from each other.
230 PENDULUM.

The principles of momentum and energy give the two
equations,

 

 

mu =m'w,
' ’ ’ ,
; i mm’ mm mm mm
mu mur + Sea a wee
gmu +3 ata r ata’ c’
Dp ed
or 3 (mu? + mu”) = mm ( = 5) ;

from which u and u’ are at once determined in terms of the
distance.
179. Motion of a simple pendulum.

If a simple pendulum of length / start from the inclination
a to the vertical, the work done by gravity as the pendulum
falls to the inclination 0 is mg (J cos @ —J cosa), and the kinetic

energy is 4ml’60?,
Equating these we find that

P= 7 (cos @— cosa),

a. ft 1

dd V' 29 (Joos 8 = cos a”
If « is very small, this is approximately
dt fi 1

d~N 9 Tea

from which we obtain 6 =a cos eS ¢ tas in Art, (126).

and

180. Motion of two equal heavy particles, fastened to the
ends of a rod without weight, and oscillating in a vertical
plane inside a smooth sphere.

Taking a for the radius and 2c for the length of the rod,
the equation of energy is

2 {4ma?6"} = 2mg ./a* —c? (cos 0 — cos a),
PENDULUM. 231

6 being the inclination of the rod to the horizon,

: 2 —
or P= anes (cos 8 — cos a).

Comparing this with the first equation of the last article
we see that the length of the equivalent simple pendulum
is

e+ /a—c.

‘181. Motion of a compound pendulum, that is, of any
rigid body, or connected system of bodies, about a fixed
horizontal amis.

G being the centre of gravity of the system and GO its
distance from the axis, let @ be the inclination of GO to the
vertical at any time during the motion.

If P be the position of a particle mass m of the system,
and if OP =r, the kinetic energy of the particle m.is $mr’6”’,
for the angular velocity of OP is the same as that of OG.

Hence the equation of energy gives
> (mr°6") = Mga (cos 6 — cosa),
if OG =a, and M= the total mass.

The expression = (mr’) is called the moment of inertia of
the system about the axis, and is generally represented by the
expression Mk’, so that .

C= = (cos 0 — cosa).

Comparing this with the first equation of Art. (179), we
see that the length of the equivalent pendulum is k’~a, so
that if J be the length,

la = k’.

The point in OG at the distance J from O is called the
centre of oscillation of the system, the point O being the
centre of suspension.
232 MOMENTS OF INERTIA.

182. The evaluation of the expression & (mr’) for
particular cases is an exercise in the Integral Calculus.

In the performance of the calculation two facts are of
great utility; these are

(1) That the moment of inertia of a rigid body about any
axis is equal to the moment of inertia about a parallel axis
through the centre of gravity together with the product of the
mass by the square of the distance between those. parallel
axes;

(2) That the moment of inertia of a plane lamina about
any axis perpendicular to the plane is equal to the sum of the
moments of inertia about any two axes in the plane, per-
pendicular to each other, drawn through the foot of the axis.

For the first let r’ be the distance of any particle from the
axis, aud r its distance from the parallel axis through G.

Then if h be the distance between the axes,
= (mr) = Xam (r° + 1? — 2hr cos 6) = & (mr*) + Mh.

For the second, if x and y be the distances of any particle
of the lamina from the axes in a plane

= (mr?) = Dm (a? + y?) == (ma®) + & (my’).

For convenience afew simple results may be stated, taking
in all cases Mas the mass of the body, and Mk’ as representing
the moment of inertia.

For a straight rod of length / about an axis through one
end perpendicular to the rod, k*= 41’; and hence, if the axis
pass through the middle point, k? = 7°.

For a circular ring, about the line through its centre
perpendicular toits plane, k?’=7°. For acircular lamina about
the line through its centre perpendicular to its plane k’ = 47”.

For a parallelopiped of edges a, b, c, about the edge c,
KP =4 (a? +0’).
MOMENTA. 233

For an elliptic area about the transverse and conjugate
axes, kK? = 40°, and k’ = 4a’,
and about the line through its centre perpendicular to its plane
(k? = 4a? + 0°).
For a sphere about a diameter, k? = 2r”,

For a solid ellipsoid about the axis c,
M=1(@' +0’).

183. Hzpressions for linear and angular momenta.

Considering motion in one plane, if « and y be the
co-ordinates of a particle m of the system, the linear
momenta parallel to the axes, are &(m#) and > (my), or, if &, 7
be the co-ordinates of the centre of gravity, ME and M7.

The moments about the origin of the momenta md and my
being — méy and my#, the angular momentum of the system
is

Xm (ay — ye).

In polar co-ordinates mr6 is the part of the momentum
perpendicular to the radius vector, and therefore the angular
momentum is ;

= (mr’é).

If A be the area swept over by the radius vector,
2A = ay — ye = 7°,
so that the angular momentum is
22mA.

Again, if p be the perpendicular on the tangent to the
path of the particle m, the angular momentum of the system is

S (mip),
and, since = pds = r°d0, and p= a -y = ;

this expression at once transforms itself into either of those
preceding.
234 MOMENTA.

In the case of a single rigid body, revolving about an axis

with which it is rigidly oanncaton. 6 is the same for all the
particles, and the angular momentum is

Mk, or Mito.

184, For motion in three dimensions the expressions for
the linear and angular momenta are

% (ma), & (my), 2 (mé),
im (ys — 2y), Sn (2 — #2), Um (ay — ye).

If we take £, y, € as the co-ordinates of the centre of
gravity, and a, y, z as the co-ordinates, relative to G, of a
particle m, the angular momentum about the axis of z

= {m (E+2) (+9) —(9+y) (E+ 4)},
= M (£4 — nf) + Em (ap — ya),
since = (ma) = 0, and & (my) =

Hence it follows that the angular momentum of a system,
about any assigned axis, is the sum of the angular momentum
due to the motion of the particles of the system relative’ to
the parallel axis through the centre of gravity, and of the
angular momentum due to the mass of the system supposed
to be concentrated at, and moving with, the centre of
gravity.

It may be instructive to present the proof of this state-
ment in another form.

Let O and G be the projections of the assigned axis and of
the centre of gravity on a plane perpendicular to the axis,
GA the projection of the line of motion of G, and FK the
projection of the line: of motion of a particle.

Taking m as the mass of the particle and wu as the
component in the line FK of its velocity, and taking EG
parallel to FK, j

The angular momentum about the axis O

== (mu. OF) = (mu.O£) + % (mu. GK).
MOMENTA. 235

If & be the velocity of G in the direction GA, the first
term, which is the sum of the moments about O of a number
of momenta in lines through G, is equal to the moment of
their resultant, and therefore

= Mu. ON.

 

een

The second term
= % {m(u— cos 6) GK} + a (mGK cos 8).
But = (mGK cos 0) => (m. KL) =0,

and therefore the second term represents the angular mo-
mentum ‘due to the motion of the particles of the system
relative to the centre of gravity.

185. In the case of a single rigid body, when the motion
is entirely in two dimensions, an expression for the angular
momentum is

Mp'd + Mio,
if p and ¢ be the co-ordinates of the centre of gravity.

Take, for instance, the case of a number of rigid bodies,
rotating about the sume fied axis with given angular velocities,
and becoming suddenly, or gradually, connected together.
236 ENERGY.

In this case the total angular momentum is unchanged,
and therefore, if mk’w be the original angular momentum of
one of the bodies, and © the final angular velocity of the
system,

2.3 (mk*) = & (mk'o).

186. Expressions for kinetic energy.

For motion in two dimensions, the expression in rect-
angular co-ordinates is

2m (4° +9’),
and in polar co-ordinates,
43m (7? + 7°68).
For the case of a rigid body in motion about a fixed axis,
the kinetic energy is
£3 (mr°6") or LMBO.
For motion in three dimensions the expression for the
kinetic energy in rectangular co-ordinates, is
fim (#@ +742);
in cylindrical co-ordinates,
435m (7? + 7°@ + 2);
and in polar co-ordinates,
42m (? + °F + 7° sin? 06°).
_ Other modes of expression can also be given and will be
employed when necessary.

If 2, y, z be co-ordinates relative to the centre of gravity
the kinetic energy

=h.2{m(E+ar tty? + (b+,

= 4M (49+ 0) +42m (+7 +2).
So that the kinetic energy is the sum of the energy due to a
mass M at the centre of gravity and of the energy due to the

motions of the particles of the system relative to the centre
of gravity.
ENERGY. 237

187. In all cases in which no external forces are in
action the linear momenta and the angular momenta remain
constant.

The principle of the conservation of energy of a mechanical
system, that is to say, the assertion that the sum of the
potential and of the visible kinetic energies is constant,
applies to all those cases in which the potential energy
depends on the configuration of the system, and in which
the change of potential energy, due to a change of con-
figuration, is independent of the manner in which that
change is made.

There is no doubt that, in any system, the total energy
remains unchanged, unless extraneous force act on the
system, but, as in the case of impacts taking place between
bodies of the system, there may be an apparent loss of kinetic
energy, which is fully accounted for by the development of
invisible kinetic energy in the form of heat, or in some other
form.

Or again, kinetic energy may be created by explosions,
but in this case, the visible kinetic energy, together with the
heat added to the system, are the equivalent of the potential
energy which was lying dormant in the explosive matter
while in its quiescent condition.

Internal friction, if due to the sliding of two surfaces on
each other, is destructive of visible kinetic energy, while, on
the other hand, visible kinetic energy may be created by the
action of live things, as for instance in the case of a man
walking on a rough plank, or a rough ball, or climbing a
moveable rope.

In such cases the system is said to be not dynamically
conservative, although it is really conservative rf all the
transformations of energy be taken into the account.

* 188. If an elastic string form part of a system, the gain
of potential energy due to its extension from its natural
length is

4 (Tension) (Extension) ;
238 ENERGY.
for the work done in pulling out the string
° TH, x -
=] Tax = | * (@-1) de=y* (rl)
u U

as may also be shewn by a simple geometrical figure. In
such a case the contraction or extension of the string implies
a transformation of energy, the loss, or gain, of potential
energy due to the contraction or extension, being represented
by a change in either, or both, of the kinetic energy, and of
the potential energy due to configuration, irrespective of the
string.

nergy imparted to a system by the action of an extraneous
couple.

Any state of motion of a plane can be represented by a
state of rotation about an instantaneous centre in the plane,

 

IP

If a small displacement, 56, be made round the in-
stantaneous centre H, the work done by the force of the
couple

=P. EBs$¢—P.EAi$=P . ABSh= Gd,
G being the moment of the couple.

Hence, if @ be the total angular twist of the couple, the

work done
*

0
= i: Gdd = GO, if G be constant.

We now proceed to illustrate these general statements by
their applications to some particular cases,
ENERGY. 239

189. Motion of a heavy rod placed with its lower end in
contact with a smooth horizontal plane, and let go.

If a be the initial inclination to the vertical, and @ at any
subsequent time, the work done by gravity is
mg a (cos a — cos 8),
and the kinetic energy is
4m (7 +k6), where y = a.cos 6.
Hence we obtain

2
= (sin’6 + 4) 6’ = ag (cos a — cos @),

4. 6g cosa — cos 8
2 —_— —“ sacs: noma
- os 3sin’?A+1 °
190. Motion of a heavy rod, constrained to slide in a
vertical line, with its lower end on the curved surface of a

smooth hemisphere, the hemisphere sliding on a smooth hori-
zontal plane.

If @ be the inclination to the vertical of the radius to the
point of contact, and a its initial value, the work done by
gravity is mga (cosa —cos@),m being the mass of the rod.
If M be the mass of the hemisphere the kinetic energy of the

‘ system is

 

on eee eee ce sin 6),
7 ) (G4 :
and therefore

(m sin6 + M cos*6) @ = a (cos a — cos 8).

191. Motion of a heterogeneous sphere rocking on a rough
horizontal plane, the whole motion being parallel to one vertical
plane.

Taking O as the point of contact of the end of the radius
CG through G when it is vertical, x, y, as the co-ordinates
of G, CG =e, and the radius = a, the equation of energy is

dm {a + 9° + k°6} = mge (cos 0 — cos a),
where xz=a0—csin @, and y=a—ccos@;

@ {a? +c? — 2ac cos 6 + k?} = 2gc (cos @ — cosa).
240 ENERGY AND MOMENTUM.

192. Motion of two equal rods, AB, BC, jointed at B and
moving, on a smooth horizontal plane, about the fixed point A.

In this case the angular momentum about A and the
kinetic energy are both constant.

 

 

Taking 6 and ¢ as the inclinations to a fixed line on the
plane, the velocity of the centre of gravity, G, of BC is com-
pounded of its velocity relative to B,

that is, ag perpendicular to BC, and of the velocity of B,
which is 2a6 perpendicular to AB;

4a?
a

m a6 + mad {a + 2a cos ( — 6)} + m2a8 {2a + a cos (b — 6)};
hence the equation of momentum is
6 (48. + 2 cos h — 6) +(£+2cos $—6) = 0.
Again, the square of the velocity of G
=a? + 40°6" + 40°66 cos 6 — 0,

the angular momentum of ABis m — 6, and that of BC is

2

the kinetic energy of AB is m me

— j2
3 Be

2
and that of BC due to rotation is ms ¢? :
ENERGY AND MOMENTUM. 241

.. the equation of energy is
18 P+4¢4' +4466 cos p—0 =C",
C and C’ being constants given by the initial conditions.

These two equations determine 6 and ¢.

193. If a solid body is rotating about a fixed axis, and
is changing its shape and size, without the action of any
external force, its angular momentum remains constant, and
this determines the change of angular velocity.

If for instance a sphere rotate about a fixed diameter,
changing its size, but retaining its shape and its homogeneity,
the angular momentum, which is 2 Mr*w, remains constant.

If the radius change to 7’, the change in the kinetic
energy is

2
£M (ro —1o*) or $M (FP — 7) of

194. Motion of a heavy rod swinging about its upper
extremity which is freely jointed to a fimed support.

If 6 be the inclination of the rod to the vertical, and ¢ its
azimuthal velocity, the angular momentum about the vertical
through the fixed end and the energy are respectively,

& (mr? sin’ . 6}
and = 4S {mr + mr*sin’O . "| + Mg (a— aco 8),

if we assume that the configuration for zero energy is when the
rod is vertical and at rest.

Each of these expressions being constant, we obtain for
the determination of @ and ¢ the equations,

¢ sin’é = O sin’a,
2 a (& + g* sin’O) + g (1 — cos 6)
= 2a(o + 0’ sin’a) +9 (1 — cos a),

w and Q. being the initial values of 6 and ¢.
B.D. 16
242 ENERGY AND MOMENTUM.

Eliminating ¢ and taking the time-flux of the resulting
equation, we obtain

4a0 — 4a.0? sin‘ a cosec’@ cot 6 + 3g sin 8 = 0.
The condition that the rod may revolve uniformly, so as
to describe a right circular cone of vertical angle 22, is
4aQ7cos a = 3g.
If while thus revolving the rod receive a slight displace-
ment, not disturbing this relation, the small vibrations in 0

will be determined by putting 6=a +, when w is small in
the above equation.

The result is the approximate equation,
‘ 4a cos ap + 8g (1438 cos *a) fp =0,
so that the period of the oscillation is

4er,/a, cos a / 4/3g (1 + 3 cosa).

195. Motion of a heavy horizontal ring, fitting on a
smooth vertical cylinder, and supported by a number of verti-
cal strings fastened to its upper edge.

If the ring, when in equilibrium, be started with a given
angular velocity w, it will rise until the potential energy.
stored up by the elevation is equal to the original kinetic
energy, that is to a height A such that ,

mgh =4 maw’,
a being the radius.

As the ring rises the strings will form helices on the
surface of the cylinder, and, taking / for the length of each
string, and @ the angle through which the ring turns in

rising through the height z, we have the geometrical.
equation

P =(l—z)? + 0°6, and .*, a°06 = (1-2) 2.
The equation of energy is
£m (# + a6") = } ma’a* — mgz,
and these equations give 6 and 2 in terms of z.
ENERGY AND MOMENTUM. 243

If, instead of starting the ring with an angular velocity,
it be set in motion by a horizontal couple G, twisted through
an angle @, the energy imparted to the system is GB.

Part of this energy appears in the form of a change of
potential energy due to change of configuration, and its
amount is

mg {l— J? — a6".

The difference, G8 — mg {l—./l* — a®B}, is represented by
the kinetic energy with which the ring starts off in its new
configuration.

196. If a heavy elastic ring, in the form of a horizontal
circle, be placed on the surface of a smooth sphere and allowed
to slip down, we can determine the motion by observing that
the kinetic energy is increased by the fall, and diminished
by the extension of the rings If a be the radius of the
sphere, 27rasina the initial length of the ring, and @ be
the angular distance of each point of the ring from the
vertex at any subsequent time, the equation of energy is

ae ro _X (sin 8 — sina)
4ma’é? = mga (cos a — cos 6) 5° 27a ee
The condition that the string should just slip over the
sphere is that
Tv

6=0 when Gas:

which gives mg sinacosa=) (1 —sina)’.

197. A circular wire ring of mass M, carrying a small
bead of mass m, lies on a smooth horizontal table, and is
capable of turning about a fixed point in its circumference.
An elastic thread the natural length of which 1s less than
the diameter of the ring has one end fastened to the bead
and the other end to the fixed point. It 1s required to
determine the motion when the initial position of the thread
coincides very nearly with the diameter of the ring,

16—2
244 ENERGY AND MOMENTUM.

The angular momentum is always zero, so that if OP the
thread be inclined at an angle ¢ to the initial position of the
diameter OA, ;

M .2a°0 = mr'd,

where r = 2a cos (p + 6).

P

 

Again the equation of energy is
M206" + m (i# + 096%) = ¥ (Ba — 0? — (r — 1),

and these equations theoretically determine 6 and ¢.

When the string contracts to its natural length the kinetic
energy retains a constant value until by the motion of the
bead the string is again put into a state of tension.

198. If an elastic thread, in the form of a circle on a
smooth horizontal plane, be set rotating with a given angular
velocity, the principles of angular momentum and energy give
the equations }

mr? = maa,

dm (77? +7°6) = dmato® — = (r — a)’;

and those equations determine # and 6 in terms of r.
ENERGY AND MOMENTUM. 245

199. Motion of a wire ring on a smooth horizontal plane
produced by an insect alighting upon it and moving uniformly
round the are of the ring.

As the insect starts with a finite motion there must be an
impulsive-action, that is, of the nature of a kick, between the
insect and the ring in direction of the tangent.

Take M, m, as the masses and w as the velocity of the
insect relative to the ring.

If V, uv be the actual initial velocities, in contrary directions,
of the centre of the ring and of the insect,

MV=mv.

If be the initial angular velocity of the ring, we have
the geometrical condition,

V+ao+v=u.

Again the angular momentum about any vertical line is
equal to zero.

Taking the angular momentum about the vertical line
through the initial position of the centre of the ring,

Me’o = mua,
and these equations determine V, » and v.

 

During the subsequent motion, G remains at rest, and
the angular momentum about G is zero, so that if ¢ and @ be
246 ENERGY AND MOMENTUM.

the angles turned through by the ring and the radius to the
insect, ; ;

mPG’6 + MCG"6 — Ma’d =0,
or mO =(M+m)d.

This equation, and the geometrical condition, a (6 + é) =u,
determine @ and @..

200. Aremarkable application of the principle of energy
has been made by Professor C. Niven in discussing the
motion produced when a heavy elastic string, suspended
from one end, is cut at-any point*.

To illustrate this application we take a simple case, in
which the action of gravity has no effect.

An elastic string has one end fastened to a fixed point on a
smooth horizontal plane, and the other end is drawn out to
any assigned distance, and is then let go; it is required to
determine the subsequent motion.

Taking a as the natural length and 7 as the stretched
length of the string, the tension 7’ is ) (J—a)/a, and if dx be
the natural length of an element at the end of the string, its
potential energy, before the end is let go, which is half
tension x extension,

T° t—a\?
=45- de=3.0(=*) dx.

When the end is let go, this energy is set free, and is
immediately converted into kinetic energy.

Taking m as the mass of unit’ length, and V as the
instantaneous velocity of the end of the string, the kinetic
energy is 4mdzV”, and therefore,
ry oe

mr ma

The first element being started with this velocity the
next element will be under the same initial conditions, and

* On a case of Wave motion, by Professor C. Niven, Messenger of
Mathematics, Vol. VIII.
ENERGY AND MOMENTUM. 247

will acquire the same velocity, and all the elements in
succession will acquire the same velocity, so that a wave
like motion will be propagated from the free end to the
fixed end of the string.

To find the velocity of transmission of the wave, we
observe that when the wave is passing over the element PQ,

 

A Pg B

the tension at P is 7 and the tension at Q is zero, and

Coo in the time dé the momentum generated in PQ ‘is
dt.

But this momentum is mdz. V, and therefore

sam =s/%

so that the velocity of transmission is constant with regard
to the natural length.

Hence the time in which the wave travels from B to
A=al/ m/JX, and in this time the free end B will have:
traversed the space Vax/m/,/X which is equal to J—a.

It follows therefore that, when the wave reaches A, the
string is In its unstretched condition, and every element has
the same velocity V in the direction "BA,

If we assume that Hook’s Law holds for compression as
well as for dilatation, the kinetic energy will be gradually
converted into the potential energy of compression, which
will in its turn be reconverted into the original kinetic
energy.

If Hook’s law does not hold for compression or if the
string is so constituted that it is reduced to rest on arriving
at its natural condition, the kinetic energy, which is ap-
parently lost, is really converted into heat.

201. We can explain from first principles why it is that
the release of the end of the string instantaneously repulis 4 in
the production of a finite velocity at that end.
248 EXAMPLES.

Considering a small element at the end B, at the moment
of release, the force on the element is 7’ and its mass is mda,
so that we have a finite force acting on an infinitesimal
mass.

Now if, by means of a mental microscope, we imagine
the mass magnified so as to become a finite mass, and the
force T magnified in the same proportion, we shall have the
case of a very large force acting upon a finite mass, the result
of which is, as we know, the production of a finite velocity in
a very small time, and the smaller we take the element at
the end B, the more closely do we tend to the ultimate
form, which is the instantaneous production of velocity at
the end.

EXAMPLES.

1. A heavy particle slides down a chord of a vertical
circle to the lowest point of the circle, and then ascends the
curve; prove that in no case can it rise through more than
one-sixth of the total are.

2. A and B are two pegs very near together and in the
same horizontal line. A perfectly flexible string of length
Lis fastened at A and hangs over B, the portion between A
and B hanging down; the loose end begins to descend from
the position of equilibrium: prove that the final velocity of

the string is ,/291/3.
3. If a system of bodies initially at rest be acted on by
no forces but the attractions of its several parts, and at any

subsequent time become united as a single solid body, shew
that this body will be at rest.

4, A particle is attached to a smooth string which passes
over a rough circular arc in a vertical plane; the particle,
initially at the end of a horizontal diameter, is drawn up
with constant acceleration g/m; prove that the work expended
in drawing it to the vertex of the circle is to the work which
would be done in lifting it through the same height in the
ratio 6+44—mp : 4.

t
EXAMPLES. 249

5. The two ends of a homogeneous rod are moveable on
the arc of a fixed smooth conic, having its axis vertical and
vertex downwards; if the length of the rod be greater than
the latus-rectum, and if it be placed in a horizontal position
and slightly displaced, find the greatest kinetic energy which
it acquires.

6. A small heavy ring, moveable on a smooth circular
wire fixed in a vertical plane, is attached by an elastic string
without weight to a point at a distance below the lowest
point of the circle equal to its radius, and when the ring is at
the lowest point of the circle the string is unstretched. If
the ring be drawn along the wire till the string becomes a
tangent to the circle and then let go, determine the velocity
with which it will reach the lowest point of the circle.

7. <A particle is placed in an uniform straight tube,
which is moveable in a horizontal plane about an axis
through one end, and the mass of which is equal to that of
the particle. The tube being set in motion, shew that the
angle, which the direction of motion of the particle makes

with the axis of the tube on leaving it, lies between 5 and
tan74.

8. A cylinder whose surface is smooth, stands on a
smooth horizontal plane on one of its circular ends; find the
impulse at a given point of its surface which will cause it to
fall over.

9. The nut of a screw rests on a smooth horizontal
plane, over a hole cut so as to allow a free passage for the
screw, and the screw descends through the nut by its own
weight; determine the motion, lst, when the nut is fixed,
2nd, when it is moveable.

10. Two equal particles, connected by a stiff wire with-
out weight, are placed in a vertical circular tube, one being
at the highest point, shew that, when the other reaches the
lowest point, the velocity of each is the same as if they had
been unconnected from the beginning of the motion.
250 EXAMPLES.

11. A ring slides on a smooth wire bent into the form
of a plane vertical curve and is attached by an elastic
string to a fixed point in the plane of the curve; if it
start initially from a position in which the string is just
not stretched, prove that it will descend through a vertical
space which is a third proportional to the natural length
of the string and its extension at the lowest position, sup-
posing that the modulus of elasticity is twice the weight of
the ring, and the string is stretched throughout the motion,

12. Two equal weights W are connected by a string of
length 21, whose weight per unit of length is w, which passes
over a small pully. The system is put in motion by adding
a weight W’ at one end. Shew that when either weight has
moved through a distance a, the kinetic energy will be greater
than if the string were weightless by 4 (J — x)’ w.

13. The extremities of a uniform heavy rod of length
4a slide on the circumference of a three-cusped hypocycloid
whose plane is vertical, one of the cusps being at the
highest point of the circumscribing circle whose radius is 3a;

prove that the length of the isochronous simple pendulum
is 4a/3.

14. Four equal particles connected by inelastic strings,
forming the sides of a square, mutually repel each other with
a force w (distance); if one string be cut, then (0) the angle
either string makes with its original position is given by

& . (2—sin? 6) = 4usin 6 (2 +sin 0).

15. A number of uniformly distributed particles move
with the same velocity v in the same direction. In this
medium is placed a body of any form and such that all the
particles impinging on it adhere. Shew that, if M denote
the mass of the body at any time, and w its velocity
M(v—u) will be constant. What information can be drawr
from this fact as to the ultimate state of motion, and the
. time when this is arrived at?

16. Two equal shells are connected by a thin tube, thi
whole system having no mass, and they contain matter whicl
is transferred uniformly from one to the other in time 1
EXAMPLES, 251

When the masses in each are equal the system has an
angular velocity # communicated to it. Shew that the
curve traced out by the middle point of the tube is given by
the polar equation r = @ tanh 20/wt, where 2a is the distance
between the centres of the shells.

17. A particle of given weight is fastened to a string
which passes over a smooth circular arc in a vertical plane:
the particle initially at the extremity of a horizontal diameter
is drawn up to the vertex with constant acceleration g: prove
that the work done in drawing it up the last half is to the
work done in drawing it up the first half as

A/24+0r—4: 44,/2.

_18. Two equal particles are revolving in the same direc-
tion in the same ellipse, under the action of a force tending
to a focus; shew that, if they become rigidly connected when
they are at the extremities of a focal chord, they will after-
wards move about their centre of gravity with an angular
velocity which varies inversely as the length of the chord,
and that, wherever this takes place, the initial velocity of the
centre of gravity will be the same.

19. <A heavy elastic string, in form of a ring, is placed,
with its plane horizontal, over a smooth cone, the axis of
which is vertical; find the position of equilibrium. Also, if
the string be held in contact with the cone at its natural
length, and let go, find how far it will descend.

20. Three equal particles at rest, tied together by three
equal strings of length a, so as to form an equilateral triangle,
repel each other with a constant force f/ If one string be
cut, shew that the equation to determine the motion is

a? (3 —2 sin’6) =3/(2sin 0 — 1),
20 being the angle between the strings.

21. A uniform rod is placed with one extremity at the
middle point of the line joining two equal centres of force
attracting inversely as the square of the distance, and at
right angles to that line; find the velocity with which the
centre of the rod will reach the line.
252 EXAMPLES.

22. Two equal inelastic rods of given length, fastened
together by a smooth hinge, are placed on a vertical plane
with their other extremities at opposite sides of a hole of
given size in a smooth horizontal table. If motion be allowed
to take place, determine the condition that the rods may just
reach their position of unstable equilibrium. |

23. A cube is rotating with angular velocity w about a
diagonal, when one of its edges which does not meet that
diagonal becomes fixed; shew that the angular velocity about
this axis will be w/4/3.

24, One end of a string (length a) is attached to a
smooth circular wire of radius a, whose plane is vertical, at
one extremity of its horizontal diameter, and the other end
to one extremity of an inelastic rod of length a, the other
extremity of which is made to slide along the wire by means
of asmall ring. If the rod and string are held initially in a
horizontal position, and then abandoned to the action of
gravity, the rod will first come to rest when its middle point

is at a distance a a below the horizontal diameter.

/
80

 

25. ABCDE is a pentagon, formed of five equal uniform
rods freely jointed together. The pentagon is placed with it:
plane vertical, and angular point A uppermost; the angles £
and £ slide freely along a smooth horizontal rod. Investigate
a differential equation of the second order for the determina
tion of the inclination of BC or DE to the vertical.

26. A ring rests upon two smooth horizontal bars whicl
in the position of equilibrium subtend an angle 2a at thi
centre; shew that, if the ring be disturbed by twisting i
through a small angle about its vertical diameter, the lengt]

of the simple isochronous pendulum will be 5 cot a cosec a.

27. A fine circular tube, radius c, lies on a smooth hori
zontal plane, and contains two equal particles connected b
an elastic string in the tube, the natural length of which 1
equal to half the circumference. The particles are in contac
EXAMPLES. 253

and fastened together, the string being stretched round the
tube.

If the particles become disunited, prove that the velocity
of the tube when the string has regained its natural length is

QarrAmc
“NM (M+ 2m)’

where M, m are the masses of the tube and each particle, and
X is the modulus of elasticity.

28. If in the previous question one of the particles be
fixed in the tube, and if the tube be moveable about the
point at which the particle is fixed, prove that when the
string has regained its natural length the angular velocity of

the tube
= Pv whm
= M (2m + M)c*

29. A solid hemisphere, mass: M, is moveable on a
smooth horizontal plane, on which its flat surface rests, about
its vertical radius which is fixed. A very fine tube, the in-
side of which is smooth, is fixed on its surface, commencing
at the vertex and cutting all the meridian lines at a constant
angle a, and a particle mass m runs down this tube from the
vertex; prove that just before the particle arrives at the
horizontal plane the angular velocity w of the tube is such
that

w* (ma* + MR’) (ma? cos*a + Mk’) = 2m? ga’ sin’a,

If the tube be closed at the end on the plane, and the particle
be inelastic, what is the subsequent motion of the system ?

30. <A particle moves on a smooth parabola (of latus
rectum 4a) and is acted on at every poit by a repulsive
force from the focus which varies as the focal chord through
the point; shew that the kinetic energy of the particle
varies as
r—a

i

 

(r + 4a) (r — 2a) + 2a? log
254 EXAMPLES.

31. A particle P slides in a circular groove (radius a) ir
a horizontal plane, and is attached to a point A of the groove
by an elastic string, natural length a. Shew that the potentia
energies of P at B in the two cases when the string i:
stretched directly from .A to the particle, and when it lies ir
the groove, are in the ratio 1 : (r—1)*.

32. A thin uniform circular ring is set rotating wher
hot about its centre, and its radius diminishes in cooling by ¢
fraction proportional to the time. If no forces act, shew
that the angle through which the ring has revolved in time 1
is Qat/r, where a, © are initial radius and velocity, r the
final radius.

33. On the surface formed by the revolution of a para-
bola about its directrix is placed a uniform elastic ring. Ii
the plane of the ring be perpendicular to the directrix, shew
that the time of a small oscillation is 2 ,/Maz/T, where M is
the mass of the ring, 7’ the tension in the position of equili-
brium, and 4a the latus rectum of the parabola.

34. Two particles m, m’ slide in a smooth circular groove
of radius a, whose plane is vertical, and are connected by'a
weightless rod whose length subtends an angle a at the
centre of the circle; find an expression for the velocity o!
either particle at any instant, under given initial conditions
‘and shew that the length of the simple equivalent pendulum
for finite oscillations is

m’ sina
m+’ cos a”

m+m

* con B+ mi’ cos (a= Ay’ Where tan A=

35. A fine string is placed in a smooth cycloidal tube o:
same length with openends. The tube is placed in a vertical
plane with base of cycloid horizontal and vertex upwards
The string being slightly disturbed, investigate the motior
completely, findmg the velocity at any subsequent time
Shew that when the string leaves the tube the velocity is
that due to a fall from rest through eight-thirds of the
diameter of the generating circle and that this takes place

after a time Ja/g cos ,/2/3 from commencement of motion.
EXAMPLES, 255

36. A circular disc rolls down a rough curve in a vertical
plane ; if the initial and final positions of the centre of the
disc be given, prove that, when the time of motion is the
least possible, the curve is an involute of a cycloid.

37. AB is a vertical rod and a weight C of mass m is
suspended from A by a weightless rod of length J, while a
weight of mass M is connected to C bya second rod of length
I, and can slide freely on AB. Ifthe whole system be made
to rotate about AB with uniform angular velocity w, and 0
be the angle which AC makes with the vertical when there
is relative equilibrium, shew that

cos d= 5 (142%),
™m

wl

and that if the system be slightly disturbed the time of a
small oscillation is

Qer (m? (4M +m) oP — 4M (2M + mg"
@ mol? — m (2M + m)*g" :

38. Two centres of force of equal strength, one attractive
and the other repulsive, are placed at two points, S and H,
the law of force being that of the inverse square of the
distance. Shew that a particle if placed anywhere on the
plane bisecting SH at right angles will oscillate in a semi-
ellipse of which § and # are the foci.

39. On a smooth wire bent into the form of an ellipse,
a small ring is placed, which is attached to the foci by two
elastic strings of the same material, whose natural lengths
are equal to the shortest focal distance. Prove that, if the
ring be placed at a very small distance from an extremity of
the major axis, the velocity at any point varies as the dis-
tance from the major axis, and that the ratio of the pressures
on the curve at the extremities of the axes is b (1 —e): a.

40. A particle is suspended so as to oscillate in a cycloid
whose vertex is at the lowest point; if it begin to move from
a point distant a from the lowest point measured along the
curve, and the medium in which it moves gives a small
resistance kv? tothe acceleration, prove that before it next
256 EXAMPLES.

comes to rest energy has been dissipated which is 8ka/3 of its
original value.

41. A tube in the form of a plane curve

(+ b*) y+a°=0
rotates freely about the axis of y which is vertical and
measured upwards; the mass of the tube is M and its radius
of gyration k; a heavy particle of mass P is capable of sliding
in the tube, and the velocity of rotation is such that P is in
equilibrium with respect to the tube: shew that its equi-
librium will be stable or unstable according as Pb’ is greater
or less than Mik’.

42. A heavy circular disc rests on a rough horizontal
plane at A, a light elastic string is fastened at A, and, lying
partly in a small groove in the rim of the disc, is just long
enough to reach a point B vertically above A where also it is
fastened. If the disc be drawn aside in its own plane and
let go, find the velocity of its centre when passing through
its initial position.

43. Three equal rods OA, OB, OC, of length a are
jointed freely at O. The extremities A, B and C are con-
nected by strings of length a and the system is placed upon a
smooth table with A, B and C in contact with the table.
The string AB is then cut; shew that, when the angle
between OA and OB is 20, ‘

96 — 144 sin? + 64 sin*@ + 13 tan’6 + 9 tan‘d a)

3—4 sin’? (a

= Hg (,/2- oa)
a 3 2

 

44, Four equal uniform rods, jointed freely so as to form
a rhombus, are laid upon a smooth horizontal plane; the
system is acted on by a repulsive force, tending from the
centre of the rhombus, and varying inversely as the square of
the distance: prove that, if 2a be the initial value of an
acute angle of the rhombus, 26 the angle between the rods
containing this angle at the time ¢,

dy. 0, /r @ a T a
(=) varies as log tan stn (7-5) cot 5 cot (TS).
EXAMPLES. 257

45. A perfectly rough rod is gently placed with one end
upon another rod of equal mass and in the same vertical
plane, moving with the velocity ./2gc on a smooth table. If
the initial inclination of the first rod to the horizon be a, and
its length 2a, shew that it will just rise to a vertical position
if

2a (1 —sin a) (5 + 3 cos’a) = 8¢ sin’a.

46, Six equal uniform rods are jointed together, so as to
form a hexagon; every particle of each of two opposite rods
repels every particle of the other rod with a force varying
inversely as the square of the distance; prove that, if the
rods be arranged in the form of a regular hexagon, and left
to themselves, then if @ be the angle between either of the
repelling rods, and one of the adjacent rods, after a time t,

(2 +3 cos*@) & varies as (1 + 4 sin’O)? — 2 sin 0 —2

(1+ 4sin?6)? +1
tyes gain 8

47. Anendless string passes round the rim of an elliptic
disc and a smooth peg in a horizontal plane in which the
disc is revolving with uniform angular velocity w about the
peg, the major axis produced always passing through the
peg. The distance between the foci is 2c, that between the
centre and the peg is a, and the radius of gyration round a
vertical axis through the centre is k. Shew that when a
slight disturbance is given, small oscillations take place,
whose periods are 27/n, where

nh? (a? — 0”) = c’a? (a? +k’).

Tripos, June, 1882.

B.D. 17
CHAPTER XIII.
EQUATIONS OF MOTION.

202. We have endeavoured in the preceding chapter to
illustrate the application of the principles of the conservation
of momenta, and of the conservation of energy, leading at
once to differential equations of the first order for the
determination of the motions of bodies.

We now proceed to consider the more general cases in
which, by the action of external forces, momenta, or energy,
or both, are imparted to, or taken from, a system, and also to
explain the method of determining the actions and reactions
between the bodies of a system.

It has been shewn in Chapter Iv., as a result of Newton’s
Laws of Motion, that the aggregate of the time-fluxes of
momenta of the bodies of a system is the exact equivalent of
the aggregate of the acting forces.

The phrase ‘effective forces’ is usually applied to what
we have called the time-fluxes of momenta, and, in this
language, the system of effective forces is the exact equiva-
lent of the system of acting forces.

If impulses be applied to a system, the aggregate of the
changes of momenta of the particles of the system is the
exact equivalent of the applied impulses,
EQUATIONS OF MOTION. 259

208. Recapitulating from Chapter Iv., we observe that,
for the determination of the motion of a body or of a system,
we have, as a result of the laws of motion, the following
principles :

(1) The time-flux of the linear momentum of a system,
in any assigned direction, is equal to the sum of the forces
acting in that direction.

(2) The time-flux of the angular momentum of a
system, about any assigned axis, is equal to the sum of the
moments, about that axis, of the acting forces.

(3) The change, in any assigned direction, of the mo-
mentum of a system is equal to the sum of the applied
impulses.

(4) The change of the angular momentum of a system,

about any assigned axis, is equal to the sum of the moments,
about that axis, of the applied impulses.

204, The linear momenta, parallel to the coordinate
axes, of a system are ~

= (ma), & (my), > (me),
and the angular momenta, about these axes, are
2m (yz— zy), Lm (ze — 22), Lm (ay — ya).

Hence it follows that the mathematical forms of the
statements (1) and (2) are as follows, .

(1) ¥ (mi) =X, (mj) = Y, 3 (mi) =Z,
(2) Sm(yz—2i) =L, Sm(2# — 22) =M, Sm (xy — ye) =N.

Or these equations can be obtained, as in Art. 53, from
the principle, Art. 46, that the system of the time-fluxes of
momenta of the several particles of a system is the exact
equivalent of the system of acting forces, and therefore that
the sums of the components of the one system and their
moments about the axes are the same as those of the other

system.
17—2
260 - EQUATIONS OF MOTION.

Tn the case in which impulses are applied to a system,
the mathematical forms of the statements (3) and (4) are

(3) Sm(u'—u)=P, Sm(v'-v)=Q, Ym (w’—w) =R,
(4) Sm fy (w’—w) —2(o'-v } = G,
im {z(w’ —u) —a (w’—w)} =H,
Sin {ee (o' —0) —y(w'—w) =K,
where P, Q, R, are the sums of the applied impulses in

directions of the axes, and G, H, K the moments about the
axes of those impulses,

205. Independence of the motions of the centre af gravity
and of the system relative to the centre of gravity.

If & 7, € be the coordinates of G, the centre of gravity,
the equations (1) become

EX (m)=X, 2 (m)=Y, $3 (m)=Z,

shewing that the motion of G is the same as if the whole
mass were concentrated into a particle at that point, with
all the extraneous forces acting upon it.

Again, if we take a, y, 2 as the coordinates, relative to the
centre of gravity, of a particle m of the system, its accelera-
tions are #+ & +4, 2+ and, if L, M, N’ be the moments
about the axes through G, of the acting forces, the equations
(2) are replaced by

Xm {y (2+ 6) —2(9 + %)} =L, &e,
or, since & (my) = 0, and & (mz) = 0,
Sm (yz— 27) = L, &e.,
shewing that the motion relative to the centre of gravity is
independent of the motion of the centre of gravity itself.

206. In the case of impulses being applied to the
system, if the velocities of the centre of gravity change from
a, 6, c to a’, b,c’, the equations (3) become

M(a'—a)=P, M(B -b)=Q, M('-c)=R,
If being the mass of the whole system.
MAN ON SPHERE. 261

Again, taking the instantaneous position of the centre of
gravity as the origin, and taking u, v, w as the velocities of
m relative to the centre of gravity, so that the actual
velocities of m-are u+a,uv+b, w+c, the first of the equations
(4) becomes

Sm {y (w' + —w—e)—2(v +b'-v —b)} = &,
os Sn {y (w' —w) —2 (v' — 0)} = G.

These results shew that the changes of motion of the
centre of gravity, and the changes of angular momentum
about axes through the centre of gravity, due to impulsive
actions, are independent of each other.

207. To illustrate at once the use of these principles,
consider the case of

A man walking on a large rough sphere, so as to make
the sphere roll in a straight line on a horizontal plane, the
man keeping himself at a constant angular distance (a) from
the highest point of the sphere.

Take M, m as the masses of the sphere and the man, and
let w be the angular velocity of the sphere at any instant.

The velocity of the man is aw parallel to the plane, and
the time-flux of his momentum is therefore maw.

For the sphere the time-fluxes of the linear momentum,
and of the angular momentum about the centre of gravity
are

Mae and M20’,

Hence taking moments about the horizontal axis, perpen-
dicular to the motion, through the point of contact, we
obtain

Mo’ + M20’ + mash = mga sin a,
or ® (7Ma + 5mh) = 5mg sin a,
where h is the constant height, above the plane, of the centre
of gravity of the man.

We may also solve this problem by finding the rate of
change of the angular momentum, at any time, about the
262 MAN ON SPHERE,

axis fixed in space which passes through the point of contact
at that time.

To do this we observe that, at any time, the angular
momentum of the system about the point of contact is equal
to

£ Mao + mah.

After the lapse of an interval 6¢ of time, the angular
velocity is » + dw, and the linear velocity of the centre of the
sphere and of the man is a (w+ do).

Observing that the distances from the assigned axis of
the lines of acceleration of the centres are the same as at the
time t, the angular momentum at the time ¢ + 6¢ is

Ma? (w + 8) + 2Ma? (w + 8) + ma (w+ bw) h,
and the rate of change is therefore
2Ma’o + maho,
which is equal to the moment mga sin a of the acting forces.
If the reactions and frictions be required, we observe that
the time-fluxes of the linear momenta of the sphere and the

man are
Mae and mas,
so that, if 2” be the friction between the sphere and plane,
and R, F the reaction and friction between the man and the
sphere,
Mao + mao =F’
maw = Rsina— F cosa
0=Reosa+Fsina—mg,

and these equations determine F, F” and R.

208. This system is not, from a purely mechanical point
of view, a conservative system.

The work done by friction upon the sphere is not equal
and opposite to the work done by friction upon the man, and
the energy of motion at any time is due to the difference
between the amounts of work done upon the two bodies.

Thus if the figure represent two consecutive positions,
P being the point of contact, the horizontal distance
MAN ON SPHERE, 263

PP'= 86, and if P’C'Q=80, the point P of the sphere is
carried to Q.

L\Q

 

Cc’

 

 

 

Hence, if P’L, QN be perpendiculars on the tangent at P,
the work done by friction upon the sphere is /’. PN, and the
work done upon the man is f’.(—PL). The sum of these
=F. LN = Fa8é, and the work done upon the system

z | Fad6.

But, the acceleration of the man in the direction of the
tangent at P being a® cos a, it follows that

Mae cosa = mg sina — F,
and therefore the work done
= | (mga sin a —ma’o cos a) dé.

This is transformed into energy of motion, or kinetic

energy, the measure of which is
$M.2a’o’ +4 ma’o’.
Equating these two expressions, we find that
ta’o ($M+m+mcosa) =mgasina. 8,

and, taking the time-fluxes of each side of this equation, we
obtain the same equation for @ as before.

The fact that kinetic energy is produced and increased is
264 CHAIN ON PULLEY.

a proof that potential energy is being lost somewhere. The
explanation is that the man’s power of exerting himself is
diminishing; he is getting tired, or, in other words, the man
is a machine which has acquired potential energy by being
wound up, and is running down.

209. A heavy chain passes over a rough pulley, moveable
about a horizontal axis through its centre, and the portions:
hanging down, which are of different lengths, are coiled up and
held close to the ends of the horizontal diameter ; if both coils
be let go at the same instant what ts the subsequent motion ?

The coils will fall freely, each leaving behind a straightened
portion, and, at the time ¢, each will have fallen through the
space 4gt’.

Take a for the radius of the pulley, b and ¢ for the lengths
of the portions coiled up, b being greater than c. Then if 0
be the turn of the pulley in the time ¢, the lengths 0’, ¢’ of
the pieces in the coils at that time will be

b— (4gt? — a6) and c — (49? + a6).

The velocity of each point of the straight pieces and of
the portion in contact with the pulley will be a6, and the
velocity of each coil will be gt.

Let Mk’ be the moment of inertia of the pulley, m the
mass of the whole chain, and J its length, so that

 

l=b+c+ 7a.
Then the angular momentum of the system will be
Mi6+™ (g# + ma) a+ gta — ne a,
or Mio + 7 (g@? + a) a6 + mgt —c+2a8).

Taking the time-flux of this expression and equating it
to the moment of the acting forces which is

mb’ _ me’
ge
or mI" (b —¢ +206),

i
SLIDING ROD. 265

we obtain the equation
2 oe 2. =
{ue 4 (ot +a) 8 4. mga bp x0),
from which, by integration and the introduction of initial
conditions, 6 can be determined.

 

210. Motion of a heavy rod sliding between a smooth
vertical plane and a smooth horizontal plane in a vertical
plane perpendicular to both.

The angular velocity in any position is at once deter-
mined by the principle of energy, but the question is intro-
duced for the sake of further illustrating the meaning of the
phrase angular momentum. ° ;

 

|
oo
2
of

oe
-
oer

 

 

 

If AB be the rod, and # the instantaneous centre of the
motion, the velocity of Gis aw, and the angular momentum
about # is ma’o + mk’w, or 4 ma’.

After a time 6¢, the rod having turned through an angle
60, the perpendicular distance from # of the line of motion
of G’ is 2a.cos 89 —a, which to the first order of infinitesimals

is equal to a.
Hence the new anguldr momentum about F is
ma? (w + Sw) + mk? (w + So),

and therefore the rate of change of the angular momentum
about EF is 4ma’# which is equal to mga sin 0;

; y+ 39 . a __ 39 =
G=7 sin 8, and 0 = 9, (cose cos @).
266 FIXED AXIS.

This is one view of the case, and another is to observe
that aw? and aw are the accelerations of G in the direction
GO and perpendicular to it, so that the time-fluxes of
momenta are maw’, maw, and mk’s, and the moments of
these about £ are equal to mgasin 0.

The reactions R, #' in the directions AH, BE are given
by the equations,

R= ¢ (mab cos 6) = ma (6 cos 6 — 6° sin 6),

R—mg=4 (-mad sin 0) =— ma (6 sin 0 + 6° cos 6).
It will be seen from these equations that R’ vanishes and
changes sign when
3 cos 0 = 2 cos a,
shewing that the rod will then leave the vertical plane.
If it should be required to take moments about 0, it will

be seen that the angular momentum is maw — mk’o and
therefore that the equation of motion is

2ma’o = mga sin 6 + R'2a cos 6 — R2a sin 6,

but then the preceding equations must be employed for the
elimination of R and £&’.

211. Motion of a rigid body about a fined axis.

If w be the angular velocity of the body at any instant,
its angular momentum is M (k’ +h’) w, h being the distance,
OG, of the centre of gravity from the axis and Mk’ the
moment of inertia about the line through G parallel to the
axis.

Hence if V be the moment of the acting forces,
M(? +h’) o=N,

or, if @ be the inclination of the plane through G and the
axis to a fixed plane through the axis,

Me +h) O=N.
FIXED AXIS. 267

If r be the distance of a particle m from the axis, its
accelerations in the directions of r and perpendicular to it
are respectively — wr and ra.

Taking the fixed axis as the axis of z and OG as the axis

of x, the components of these parallel to # and y are
— o's — oy and — wy + ox.

Hence the time-fluxes of the momenta and their moments

about the axes are respectively
2m (— wa — ay), Um (— wy + wz), 0,

and Ym (w°yz— wz), Xm (—w'xz— wyz), ZS {me (a? +4},
or, — Mah, Moh, o,
and Dw’ — Eo, — Ew — Do, M (kh? +1’) @,
where D and £ represent the quantities, Smyz, and Ymzz.

The first three of these expressions are equal to the sums
of the acting forces and of the reactionary stresses of the axis,
and the next three are equal to the sums of the moments
about the axes, of the same quantities*.

Suppose that the body is connected with the axis at two
points at distances c and c’ from O, and that U, V, W, and
U’, V', W’ are the stresses upon the axis at these points.

Then, if X, Y, Z, L, M, N be the sums and moments of
the acting forces, we shall have the equations

—~Mo*h=X-U-U’,
Moh=Y-V-—V’,
o=Z—-W-W',
Dow — Es = L — Ve —V'e’,
— Eo? — Da = M+ Uc+ U'e'.

212. Impulses applied to a body in motion about a fixed
ants.

* It will be seen in the next chapter that these expressions can be
also obtained by first writing the expressions for the angular momenta and
then employing the general expressions given in that chapter for the rates
of change of the angular momenta,
268 FIXED AXIS.

If w'—w be the change of angular velocity due to the
application of impulses, and K the moment of the impulses,
the change of angular momentum is given by the equation

UM (i? + h’) (o’ —@) = K.
Taking the axes as in the last article, and observing that
m (#’ —w)r is the change of momentum of the particle m in
the direction perpendicular to r, of which the components

are —m(o'—@)y, and m(w’—) #, we find that the sums
and moments of the changes of momenta are respectively

o, M (w' — w)h, 0,
and —E(#'—w), -—D(w'-—o), UM (k’ +2’) (o’ —@).

Equating these expressions to the sums and moments of
the applied impulses, and of the reactionary stresses, we can
calculate the latter quantities.

The right-hand members of the equations in the preceding
article will be the expressions for the sums and moments, if
the symbols employed be supposed to represent the applied
impulses and the impulsive stresses on the axis at two
points. .

213. We have defined, in Art. 181, the centre of oscilla-
tion of a compound pendulum.

We can shew that the centres of oscillation and suspension
£ and O are convertible.

For OL. OG=04 +2,
whence OG. EG=R,
shewing that O and EF are convertible.

214, Centre of percussion.

If a single impulse can be applied to a rigid body which
is capable of motion about a fixed axis, so as to produce no
impulsive stress on the axis, the line of the impulse is called
the line of percussion, and the point in which it meets the
plane through G perpendicular to the axis is called the
centre of percussion.
‘ILLUSTRATIONS. 269

If w be the angular velocity produced the sums and
moments of the changes of momenta will be

0, Moh, 0, — Ho, — Do, M (kh +h)o;
OG being the axis of 2, and Oz the fixed axis as in Art. 211.
The single impulse P must therefore be in the direction
of the axis of y, and, if &, be the coordinates of the point
in which its line of action meets the plane wy, we must have
— Ew =— Pi, —-Do=0, M(h +2’) w= PE

Hence it follows, as a necessary condition for the existence
of a centre of percussion, that D, or & (myz), must vanish,
and that, if there is a centre of percussion, its distance from
the axis is the same as that of the centre of oscillation.

When £, or & (mz), vanishes, the centres of oscillation
and percussion are coincident.

215. Motion of two heavy rods AB, BC jointed at B, and
swinging in a vertical plane about the end A which 1s jointed
to a fixed horizontal axis. ,

If 6 and ¢ be the inclinations of AB and BC to the
vertical, the angular momentum of the system about A, 2a
and 2b being the lengths of the rods, is

4q? . 9 OG. er
m 3 O+m x > +m'bd {b + 2a cos (6 —4)}
+ m'2a6 {2a +d cos (db — 6)},
and the time-flux of this expression
=— mga sin 0 — m'g (2a sin 6 + bsin @).

Next, the accelerations of G being compounded of bf and

bf? perpendicular and parallel to GB, and of 2a8 and 2a6"
perpendicular and parallel to BA, we obtain by taking
moments about B for the rod BC, and dividing by m’

0b + 5 $+ 2a b cos (p— 6)
+ 2a6°b sin (b — 0) =—gb sin ¢,
or. 4bh + 2a [6 cos (b — 6) + sin ($ — 6)} =—gsing.
270 ILLUSTRATIONS.

Or for a second equation we might have expressed the
constancy of the energy which is

3ma?O? + 4m'd'd? + 5 {B'? + 4a°6? + 4abd6 cos (f — 6)}
— mga cos 6 — mg (2a cos 6 + b cos ¢).

In either case we obtain two equations for the determina-
tion of @ and ¢.

The horizontal and vertical components of the momentum
of the system are respectively
maé cos 0 + m'bd cos ¢ + m’2a8 cos 6,
and ma sin 0 + mb sin b + m’2a0 sin 8,
The time-fiux of the first of these is the horizontal com-
ponent of the stress at A, and the time-flux of the second

increased by the weight of the rods is the vertical com-
ponent.

In the same manner the stress at B is determined by
writing down the horizontal and vertical components of the
momentum of the rod BC.

216. Motion of a heavy sphere, the centre of gravity of
which is eccentric, on a smooth horizontal plane.

If the sphere have no initial kinetic energy, the centre of
gravity G moves in a vertical line GO, and if OG =y, the
equation of moments about C or about any point in the
vertical line CP through the point of contact P is, with the
notation of Art. 189,

mijc sin 0 + mk*6 = — mge sin 8,
where y =a—ccos 6.
Substituting, multiplying by 26 and integrating we obtain
6 (k? +c? sin?0) = C+ 29c cos 6,
which is the equation of energy.
The reaction, &, at P is given by the equation
R- mg = mi.
MOTION OF LAMINA. 271

If the plane be rough so that the sphere Tolls, the equation
of moments about P is
mie + méy + mije sin 0 = — mg.c sin 6,
where 2=a0—csin#, and y=a—ccos6,
making the substitutions and integrating we obtain the result
of Art. 189,

The horizontal and vertical reactions at P are given by
the equations mé=—F, mij= R— mg.

217. Motion of a plane lamina, of any given shape, on a
smooth plane, when a given point of the lamina is made to
move in a given manner, and the lamina is besides acted upon
by known forces.

If A be the given point, and G the centre of inertia of the
lamina, take 0 as the inclination of AG to a fixed line in the
plane, so that 6 is the angular velocity of the lamina.

If f and f’ be the accelerations of the point A in the
direction AG and perpendicular to it, which are known
functions of the position of .A, or of the time, the accelera-

tions of G in the same directions are f— a6’, and f’ +a,
and the time-flux of the angular momentum about G is

MO.

Hence, if V be the moment about A of the acting forces,

M(f' +06) a+ MO =N,

the equation which determines the angular motion of AG.

If P, Q be the requisite constraining stresses at A, and
X, Y the resultant forces, in the direction AG and perpen-
dicular to it,

M(f-a6)=X+P, and M(f’ +a6)=Y+Q.

218. Motion of a plane lamina, bounded by a curve,

rolling on a fiaed curve under the action of given forces.

We have first to solve the kinematical question of the
time-flux of the angular momentum about the point of
contact.
272 “MOTION OF LAMINA.

The angular momentum being the sum of the angular
momenta due to the rotation and the motion of the centre of
gravity, the first part is Mk’w. :

&
Gg’

AY

For the second, taking as the angular velocity when the
point P of the rolling curve is in contact with the point A of
the fixed curve, and @ as the consecutive point of contact, the
angular momentum at A = Mr’a, if

AG=PG =r,
In the consecutive position, when the motion of G' is”
perpendicular to QG’, the angular momentum about A
= M (r+ 6r) (w+ 80) 7,

remembering that AP is an infinitesimal of the second
order, and the difference

= Mr (réo + wor).

Hence the time-flux of the angular momentum due to
the motion of G

= Mr’ + Morr,
MOTION OF LAMINA, 273

and consequently the equation of motion is.
M (+7) 0+ Morr =L,
L being the moment about P of the acting forces.

If p, p' be the radii of curvature at P of the rolling curve
and the fixed curve, and if the are AQ=6s,
oot = as + ae i
pp

and, if @ be the angle between the line AG and the normal
at A,

 

,
err =— or sin 6s = — wr sin 8 = 3
p

so that the equation of motion takes the form

 

Mk +9°) @— Mo'r sin 0—PP— = 1.
pt p

If the fixed curve be a straight line

pee,
p

and the equation is then
M (k? +7") a — Mo’pr sin 0 = L.

219. The result of the preceding article may be other-
wise obtained.

In Art. 25, it is shewn that the acceleration of the point
P, when at A, in direction of the normal at A is
o'pp'/(p +p’).
The acceleration of G relative to A, in the direction

perpendicular to AG js rw, and therefore the actual accelera-
tion of G perpendicular to AG is

ro — opp’ sin 6/(p +p’).
Hence, equating momenta about A,

Mia + Mr {re — opp’ sin O/(p + p’)} = L.
B. D. 18
274 MOTION OF LAMINA.

220. The investigation of the two preceding articles is
the infinitesimal case of the following general statement.

@ @f

 

 

If Q be the linear momentum of a body, in motion in one
plane, when P is the instantaneous centre, and Q’ at a
subsequent time when P’ is the instantaneous centre, G
being the centre of inertia, the change of the angular
momentum about P is

Mi? (o' — w) + Q' (r’ + PP’ cos 4) — Qr.
221. Asa particular case consider the motion of a heavy
uniform circular disc of radius ¢ rolling on the curve,

s=of (), starting from the highest point, from which s and
are measured.

In this case, 9=0,

1 1 " Se r
and o= ff + ate (9) ¢=(1+f'9) ¢,
and the equation of motion becomes

30° ; Pah ia IS 5
MEL +f GIb+S" () 4b} =Uge sin $.
Suppose the curve to be a cycloid, s=csin ¢;

then cos f—sin § f= 3.2 sin$,
MOTION OF LAMINA. 275

the integral of which is

ihe ohn. aie
3cp cost 5 = 2g(1 cos 5)

 

and therefore ait (1 — cost $ ) .
3c 2
To find the pressure we have the equation,
ew M
e. g cos p— Rh,

shewing that & vanishes, and therefore that the disc flies off,
when 5 cos ¢ = 3.

At the instant of flying off, w” = 249/25c, and the velocity
of the centre of the disc is 2/6gc/5.

222. Change of motion produced in a lamina, moving in
any manner in its plane, produced by its impact on a rough
curve.

We suppose the roughness to be so great that there is no
sliding, and we have simply to express the fact that the
angular momentum round the point of contact is unchanged.

P being the point, let v be the component perpendicular
to GP of the velocity of G, w and w’ the angular velocities
just before and after the impact; then

M (kh +17") 0 = Mk’o + Mor.

' Suppose for example that the disc of the preceding
article, just after flying off the cycloidal arc impinges on a
fixed rough peg just beneath its lowest point.

In this case
So «se ‘
7? =5 o+c’a cos ¢,
from which we obtain 150’ = 11.
The disc will then turn round the peg as a fixed point,
and the equations of motion will determine the angle
through which it turns before leaving the peg.

 

18— 2
276 BREAKING OF RODS.

223. In general if a rigid body, or system of any kind,
be in motion, and if a straight line of the system suddenly
become fixed, the angular momentum of the system about
the axis is unchanged.

In the case of a single rigid body this at once determines
the angular velocity about the axis.

Thus, if Q be the linear momentum of a rigid body
perpendicular to the axis which becomes fixed, p the distance
from the axis of its centre of inertia G, and H the angular
momentum about the line through G parallel to the axis,
the angular velocity is given by the equation

Mio = Qp + H,

Mi? being the moment of inertia about the axis which
becomes fixed.

224, Tendency of a rod in motion to break at any assigned
point.

Imagine a rod of small section and of any shape, its axis
however being a plane curve, to be in motion in that plane
under the action of forces in the plane.

Taking any cross section, through any point P of the axis,
the stress at this section, that is, the action and reaction
between the two parts of the rod separated by the cross
section, may be represented by two forces Z'and NV at P in
directions of the tangent and normal to the axis, and a couple
G in the plane of the axis.

The velocities and accelerations of the various points of
the axis having been previously determined, the quantities
L, N, and G can be found by writing down the equations of
motion of either of the two parts of the rod, including 7, N,
and G amongst the acting forces.

Now a rod may break in three ways; the internal
molecular forces may not be sufficient to withstand the force
T in direction of the tangent, or they may give way in
direction of the normal, or the moment about P of the mole-
cular forces may be overpowered by the couple G.
BREAKING OF RODS. 277

In other words the rod may break by tearing, by shearing,
or by snapping, and the quantities 7, NV, and G are, respec-
tively, the measures of the tendencies to break in these three
ways.

To illustrate, take the case of a heavy straight rod swing-
ing in a vertical plane about one end, and examine the
tendency to break at the middle point.

Writing down the equations of motion of the lower half
of the rod, and taking ras the distance from the axis-of a
point in the rod, we obtain

2a 6dr 5 3 2a dr
[ m 578 = N— ging sin 0, [ m5

a

76? = T — kg cos 0,

2 .
[omer @-a) d= G — mga sin 8,

where 6 and @ are known functions of 6, and these equations
determine the values of 7, NV, and G at the middle point of
the rod.

225. If arod have its state of motion suddenly changed
by impulsive action, impulsive stresses are created at all
points of the rod, and the method of determining them is the
same as in the previous case.

If for instance the free end of the swinging rod, supposed
inelastic, be suddenly stopped by impinging against a fixed
surface, and if 7, N, and G then represent impulses, the
equations are

2 ‘
T= 0, [m= N-P
i 2a

2a .
| — mF r(r—a)b= G —4Pa,

a
where P, the impulse at the free end, is given by the equation,
mab = BP.
226. Determination of initial stresses, and initial accelera-

tions, when some of the constraints of a system, previously in
equilibrium, are removed.
278 INITIAL STRESSES.

In such cases the equations of motion should be written
down for the configuration ofthe system at the instant of
release from constraint,

These equations, in combination with the kinematical
relations of the system, will be sufficient for the determination
of the required stresses.

The equations are simplified by the fact that the linear
and angular velocities are zero, so that radial and transversal
accelerations take the forms 7 and 76, and normal accelerations
are evanescent*.

Exampies. (1) wo rods AB, CD, of lengths 2a and
2b, are connected by equal strings AC, BD, of length c, and the
system 1s supported, with the rods horizontal, by a fixed hori-
zontal axis through the middle point of AB; if one string AC
be cut, at is required to find the initial tension of the other.

Let w, w, and w be the initial angular accelerations, in
the directions figured, of the two rods and the string.

/)
wo

A

  

Sp) manne neneen ened

 

The vertical acceleration of G, the centre of gravity of CD,
= ba’ + cw” cos a+ aw,
and its horizontal acceleration = cw” sin a.

Hence the equations of motion are

a
mx @ = Ta sina,

* Some illustrations of the method of finding initial stresses are given in
an article in the Mathematical Messenger for 1866.
INITIAL CURVATURES. 279

m’ (bw' + co” cosa + aw) = m'g— T'sina,

m'cw” sina = T cos a,
Bg.
ms wo = T.bsina,

which determine the tension and the angular accelerations.

(2) A heavy rod, of length 2a, is supported against a
smooth fixed sphere by a horizontal string fastened to its upper
end A, and also to the highest point of the sphere; if the
string be cut it is required to find the initial pressure on the
sphere.

If « be the angular distance of the vertex from P the
point of contact, it will be found that PG =asin’a. Observ-
ing that the acceleration of P is wholly tangential, and taking
as the initial angular acceleration of the rod, it follows
that wPG is the acceleration of G perpendicular to the rod,
and therefore taking moments about P
a
3
We have also moP G=mgcosa—R,

and we obtain R(1+3sin*a) =mg cosa.

m= o + mPG’o =,mgP G cos a.

227. Determination of the initial radit of curvature of
the paths of assigned points of a system, when the system is set
in motion in any given manner, or, being in a state of equi-
librium, has some of tts constraints removed.

If the velocity and direction of motion of an assigned
point of the system be known, the expression for the normal
acceleration, v’/p, determines the curvature, 1/p, for the
acceleration of the point in the direction of the normal to its
path is obtainable from the equations of motion of the
system.

EXAMPLES. (1) Two particles, m and pu, are connected by
a string passing over a smooth fixed horizontal rail, and, the
portions of string, of lengths a and b, being vertical, the
particles are projected horizontally, in opposite directions
perpendicular to the ratl.
280 INITIAL CURVATURES,

The initial equations of motion are
m (# — 76°) = mg — T, m (rb + 276) =0,
ub — pd) =n9 —T, w (pb + 264) = 0.
If u and v be the velocities of projection, then, initially,
r=a, p=b, r=0, p=0, aG=u, bp =2,
and therefore 6=0, d =0,
uw vw
m(i#-F)=mg-F, »(6-F)=ng- 7,
and these equations, with the equation *+ p= 0, determine

*, p, and T.

The initial radii of curvature R, R’ of the paths of m and |
p are given by the equations
uw

vw
m= T—mg, mm = T — ug,

assuming the concavities to be upwards.

If Tis less than either mg, or yg, the concavity in that
case will be downwards.

(2) A rod AB is moveable in a vertical plane about the
end A, and a string BC carries a heavy particle at C; the
particle is held in a given position in the vertical plane through
the rod, and is projected in the direction perpendicular to BC’
in the vertical plane.

If 6 and ¢'‘be the inclinations of AB and BC to the
vertical at the moment of projection, the initial equations of
motion are

ue b+ mbp {b + 2a cos (f — 6)} — mb? 2a sin (d — 0)

+ m2a6 {2a + b cos (¢ — 6)} + m2a6’ b sin (¢ — 6)
= Mga sin 0+ mg (2asin 6 + bsin ¢),
and b+ 2a6 cos (f — 6) + 2a6* sin (6 — 0) =—gsin ¢,
where 2a and 6 are the lengths of AB and BC.
INITIAL CURVATURES. 281

Initially, if u be the velocity of projection,
6=0, and u=dd, :
and the preceding equations determine @ and ¢.

The acceleration of the particle in the direction CB is
initially bd? — 2a0 sin (f — 9), and this is equal to u*/p if p be
the initial radius of curvature of the path of C.

228. Ifasystem have initially no motion, and. we wish
to find the initial curvature of the path of any assigned
point of the system, we must first find the initial direction
of motion, and then, observing the small displacements which
take place in a very short time, we can sometimes obtain the
curvature by an immediate application of the Newtonian
expression for the diameter of curvature, viz. (arc)” + perpen-
dicular subtense. Sometimes however it is necessary to take
the analytical expression, in Cartesian or polar co-ordinates,
or in some other system, and to expand, in ascending powers
of the time, the various terms contained in these expressions.
An illustration of each case will be sufficient to explain the
methods.

(1) Two rods, AB, BC, freely jointed together at B, and
moveable about the end A, are held in a horizontal position so
as to form a straight line ABC, and are then let go;
it is required to find the initial curvature of the path
of C.

Let uw, m be the masses and 2a, 2b the lengths of AB
and BO, then, if w and ’ be the initial angular accelerations
of AB and BC, it can be shewn, by taking moments about A
for the system and about B for the rod BC, and combining
the equations, that

a @

Now supposing that, after a short time ¢,@ and ¢ are
the inclinations of 4B and BC to the horizontal, it follows
that the horizontal and vertical displacements of C are

2a + 2b'— 2a cos 6 — 2b cos ¢, and 2a sin @ + 2bsin ¢,
or, approximately, ;

ab? + bg? and 2a0 + 2b4.
282 INITIAL CURVATURES.

The initial tangent to the path of C is vertical, and
consequently

5 ~ ., 4 (a8 +b)"
< 2o = Limit of oo = we .

The first approximations to @ and ¢ are
0 = ot’, 6=ho't’,
and we hence obtain

i EE
2ab--b (m+ 2u)* +ap""

 

(2) A plane lamina, which is moveable about a horizontal
axis on its plane, is inclined to the horizontal at an angle a,
and a heavy particle m is placed upon tt at a distance c from
the axis and below the ams; it is required to determine the
initial curvature of the path of the particle.

If @ be the inclination at any time, the equations of
motion are
Mk’6 + mr (r6 + 276) = mgr cos 6,
¢—7r6=gsin 0.
The expression for the radius of curvature is
(r°6* + 5)8
PO + 27°6 — r (6% — 76)’
and we have to expand in powers of ¢, the various terms of

this expression, which can be effected by Maclaurin’s
Theorem.

Taking m? to represent Mk’ + mc’, we obtain, from the
equations of motion,

pe Shay x gcocos a
F=gsina, 6, = E

»  g'ecos? Qc? ~ g' 8in a COSa 7c
=f F ae Fr) 6 = (1-+).

, #,=0, 6,=0,

 

 
VIRTUAL WORK. 283

Now G=8 E+... PHPEL
. as se s, wee. 3
and 6 —7O= Birt #8) BF ene

and therefore
G2+0d2)!
C03 + 2128, 2 (Bs — 78)

 

Po =

and, making the requisite substitutions, we obtain p, in terms
of a, J, and ¢.

If a=0, it will be found that
3c 3inc®

Po= FB Ue
the negative sign shewing that the concavity of the initial
path is outwards-from the axis.

229. Application of the principle of virtual work.

If at any instant the geometrical configuration of a
system be contemplated, and if a geometrical displacement
be imagined, the virtual work of the time-fluxes of momenta,
or of the effective forces, and of the acting forces will be the
same,

We must however include in the phrase acting forces
any internal forces such as the tensions of elastic strings,
or sliding frictions, by means of which. work is done on the
system.

Considering a single rigid body, i.e. a material system of
invariable form, if #’ be the time-flux of the linear momentum
in any direction, and és the displacement of the centre of
gravity in that direction; and if K be the time-flux of the
angular momentum about an assigned axis through the
centre of gravity, and 5¢ the angular displacement about the
axis, then the corresponding portions of the virtual work

are Fés and Ké¢.

(1) Consider for example the case of the two rods in
Art. 215. ‘
284 VIRTUAL WORK.

Taking « and y as the horizontal and vertical co-ordinates
of G, we obtain

2 2
m = 680 +m’ hdd + m'#d2x + m'ijdy = m'gdy — mga sin 88.

Now «=2asin6+bsin ¢, and y= 2acos 6 +b cos ¢,

from which dx and éy are obtained in terms of 5@ and 8d,
and observing that 8? and $¢ are arbitrary quantities and
independent of each other, their coefficients must each
vanish, and we thus obtain the equations for the determina-,
tion of 6 and ¢.

ExampiLe. (2) Motion of an extensible circular ring,
placed horizontally over a smooth surface of revolution the aats
of which ts vertical.

Let s represent the distance along a meridian are from a
fixed level to the ring, 7 its radius and z the depth of its
plane. The accelerations of any point of the ring down the
meridian are and perpendicular to it are § and s’/p, and there-
fore, if we imagine a displacement by slightly shifting the
ring downwards on the surface the equation of virtual work
is, m being the mass of the ring,

ms8s = mgdz — TS (207),

or mi= mg — 2D S,
an equation which can also be obtained by considering the
meridional motion of an element of the ring.

Observing that Z=2(r—a)/a, and that r=f (z);
this equation determines the acceleration along any
meridian.

(3) Four equal rods, of length 2a, are jointed together in
the form of a square OADB, and suspended from the joint
O, the square form being maintained by a string OD;

of ae string be cut it 1s required to find the change of stress
at O.

_ Take as the expression for the initial angular accelera-
tion of each rod; then if G and H be the centres of gravity
VIRTUAL WORK. 285

of OA and AD, the initial vertical accelerations of G and H

are aw/,/2 and 3aw/,/2 ;

and the initial horizontal accelerations are each awn! 2, also

the time-flux of the angular momentum of each rod is
mew/3. .

_ If we imagine D pulled through a small space, so as to
displace each rod angularly through the small angle @, the
linear displacements of G and H are, vertically, «6//2,
3a6/,/2, and horizontally each is a0/,/2; also the vertical dis-
placement of the centre of gravity K of the system is a0,/2.

Hence we obtain

awd 9a?wd vod awd =

at 2m = + dm SS + dm <= dng /2,
from which 10a = 39/,/2, and therefore the acceleration of
K is 39/5, shewing that the pressure on the point of support
is instantaneously diminished by three-fifths of the weight of
the system.

If P be the horizontal stress at D, it can be determined
by giving the rod AD a small arbitrary twist, 0, about 4,
breaking the connection at D.

2m

 

 

 

The equation of virtual work will be

& 38maw a0 maw af ae
= 50 — oe ae ee eee —— + Paé 2,
ME ae gale
awh! _ _ mg
from’ which P = io°

If Q and R be the actions at A upon OA in the directions
DA and OA, these quantities may be found by giving DA a
twist about D, and OA a twist about O, breaking in each
case the connection at A.

The equations obtained are
maw 6 3mao a maw ad

aé
es — .—— = R?2 6- Fe
3 oo we ge
286 VIRTUAL WORK,

ao maw a0 maw ad aé

al o° fe ere 2. 6— ==;
mee 8 i
and from these we find that
R=3,mgJ2 and Q=y5mg,/2.

In solving this question our object has been to illustrate
the use of the principle of work, but the same result may be
obtained from the initial equations of motion of the rods,

which are 4maw mga

 

3 ~/e — 2aQ,
ate gh mab Vd _ pp
J/2 J2 * f2 fh ,
mara Pa
3 noes
EXAMPLES.

1. A smooth sphere is at rest on a smooth horizontal
plane, and an equal sphere is placed gently upon it, so as to be
in contact very nearly at the highest point; prove that the
centre of the upper sphere will describe a portion of the arc
of an ellipse, and that when @ is the inclination to the
vertical of the line of centres,

a6? (1 + sin? 6) = 2g (1 — cos 6)...
Shew that the spheres will part company when
cos 0=,/3—1.
2. Two equal smooth spheres rest in contact on a
smooth horizontal plane and another sphere is placed between
them, so that motion ensues. Find the pressure between

them at any time in terms of the inclination to the horizon
of the lines joining their centres.
EXAMPLES, 287

3. If two weights be suspended by a weightless string,
which passes over a rough circular cylinder of given mass,
the space described by either in any time is independent of
the radius of the cylinder.

4, Two equal smooth spheres are placed one upon the
other and both in contact with a smooth vertical wall.
If the lower one just leave the wall, prove that they will
separate when the line joining their centres is inclined to the
horizon at an angle cos’ (2), the motion being supposed to
take place in a vertical plane.

5. Four equal rods are jointed together so as to form a
square ABCD, and the system is suspended from the point
A, the square form being maintained by a string connecting
A and C. Find the tension of the string.

‘If the string be cut, prove that during the subsequent
motion,
aG (1 + 8sin’ 6) = 3g (cos g- =) ;
2a being the length of each rod, and @ the inclination to the
vertical.

6. Three equal smooth balls are kept in contact with
each other on a smooth horizontal plane by a string passing
round them, and a fourth equal ball rests upon the three; if
the string be cut, what is the initial change of pressure
between the upper ball and each of the lower ones?

7. Four equal smooth inelastic circular discs, of radius
a, are placed in one plane with their centres at the four
corners of a square of which each side is 2a. They attract
one another with a force varying as the distance. <A blow
being given to one of them in the line of one of the diagonals
of the square, investigate the whole of the subsequent
motion,

8. A solid sphere resting on a smooth horizontal plane,
is suddenly divided into two equal parts by a vertical plane
through its centre. It is required to determine the initial
horizontal pressure between the two parts and the initial
vertical re-action of the plane.
288 EXAMPLES.

9. The middle point of a uniform rod is fixed midway
between two centres of force, which attract with a force
varying inversely as the square of the distance. Prove that
the time of a small oscillation is

om (a —c*) J M/V3pa0,
where J/ is the mass of the rod, 2c its length, 2a the distance

between the centres of force, and wé2/r* the attraction on an
element 5z of the rod at a distance r.

10. A rod of given length is formed into the quadrantal
arc of a circle, and is made to rotate about an axis through
one end perpendicular to its plane.

Supposing the are to become suddenly fixed to its axis,
find the measure of the tendency to break off; and shew
that, if the rod were formed into a semi-circular arc, the
tendencies to break off in the two cases would be compared
by the ratio 477 —8 : m.

11. A rough cylinder rests on a horizontal plane. Find
the least velocity of a second cylinder of given larger radius,
which will, after impinging upon it, pass over it.

12. Ifa bullet of mass m be fired with velocity u per-
pendicular to the face of a block of wood of mass M, placed
on a smooth horizontal plane, and remains just imbedded,
prove that the angular velocity acquired by the block is

mbu
(M +m) k? + m (a? + 6*)’

where a is the distance of the centre of gravity of the block
from the face struck, and 6 is the distance of the point struck
from the foot of the perpendicular drawn from the centre of
gravity of the block to the face.

13. A uniform rod of length 2a is rotating, in a vertical
plane, about its middle point, which is fixed, with an angular
velocity /67g/a. At the instant the rod is horizontal, the
ascending end is struck by a ball of equal mass, which was
dropped from a height 37a; and when it is next horizontal,

 
EXAMPLES, 289

the same extremity is struck by a second equal ball similarly
dropped. The elasticity being perfect, determine the subse-
quent motion of the rod and balls,

14. A wire in the form of the portion of the curve
r=a(1+cos@) cut off by the initial line rotates about the
origin with angular velocity w; shew that the tendency to
break at the point 26=7 is measured by 12 V2mo’a*/5,
where m is the mass of a unit of length of the wire.

15. A-rod rests horizontally upon two supports; if one
support be suddenly withdrawn, find an equation to deter- '
mine where the initial strain on the rod is greatest.

16. A rough cylinder of radius a loaded so that its
centre of gravity is at a distance A from its axis is placed on
a board of » times its mass, which can move on a smooth
horizontal plane. Find the time of an oscillation when the
system is slightly disturbed from its position of stable
equilibrium, and prove that if 2 be the length of the simple
equivalent pendulum

(n+1) (k—7) =n (ah)?

17. <A circular disc is suspended from a fixed point by
four equal strings symmetrically attached to its rim, and
inclined at an angle a to the vertical. If two of the strings
be cut, shew that the initial tension of each of the others is
to the weight of the disc in the ratio

cosa : 1+5cos’a.

18. A circular ring, mass M and radius a, lies on a
smooth horizontal plane, and a fly, mass m, alighting upon it
starts off and crawls round the ring, with a velocity v, which
is uniform relative to the ring. Prove that the angular

velocity of the ring = (4 Imya ‘

19. If the ring in the previous question be vertical and
moveable about its centre of gravity, and if the fly start off
and as before move uniformly relative to the ring, find its

B.D. ‘ 19
290 EXAMPLES.

angular velocity in any position, supposing the fly to start
from the lowest point. Also find the least ratio of the
masses in order that the fly may ever be at the highest point
of the ring.

20. A flat square board ABCD, moveable about its
centre on a smooth horizontal plane, has attached to A a
fine string AP (=AB) which carries a patticle P; P is
projected horizontally from B at right angles to AB. If the
mass of the board is nine times that of the particle, find the
angular velocities of the string and board when AP and AD
are in the same straight line.

21. A heavy circular disc is supported from a point
whose height above it is equal to the radius by three equal
strings attached to the circumference at equal intervals.
One of the strings being cut, shew that the tension of the
others is immediately diminished in the ratio of 2 : 3.

22. If one end of a rigid rod of length 6a be made to
move in the circumference of a circle of radius a with the
uniform angular velocity » round the centre; shew that at
the time #¢, the angle between the rod and the radius of the
circle to that end is given by the equation

wt
o_ 3
cot a e",
the rod being initially in a radius produced.

23. Two rods AB, BC connected by a hinge at B are
in motion on a smooth horizontal plane, the end A being
fixed. If, initially, AB has no angular velocity, that of BC
being w, shew that when BC has no angular velocity that of
AB will be bw/2a and the angle between the rods will then

be cos™ {2 (26 — a)/3b}, 2a and 26 being the lengths of the _ .

rods, which are supposed equal in mass.

24. A sphere falling vertically impinges on a fixed
inclined plane, the sphere having an angular velocity about
EXAMPLES. 291

a horizontal axis parallel to the plane. If it be reflected ~
vertically, shew that its angular velocity is

(5e — 9e) w/(5 — 9e),

e being the modulus of elasticity.

_ 25. A smooth spherical shell of mass M rests on an
inclined plane, being fastened to a point of the plane by a
string; a particle m rests inside the sphere; prove that if
the string be cut, the ratio of the initial pressure between
the sphere and the particle to that between the sphere and
the plane is

m (1 — sina cos a)
(M+m)cosa ’

where a is the inclination of the plane to the horizon.

26. The ends of a straight rod are moveable on two
smooth fixed rods, intersecting each other at right angles; if
the rod be set in motion, prove that when @ is its inclination
to either fixed rod, the measure of the tendency to break
at any point is proportional to sin 20.

27. Two equal rods AB, AC are jointed together and
rest symmetrically over a smooth sphere; the junction of
the rods at A being severed, what is the initial pressure of
each rod on the sphere ?

Suppose the sphere divided by a vertical plane through
A, perpendicular to the plane of the rods, and imagine the
left-hand hemisphere to be suddenly annihilated; it is
required to determine the initial action at A.

28. P and Q are two points in a uniform rod equidistant
from its centre. The rod can move freely about a hinge at P.
The hinge-is constrained to move up and down in a vertical
line. If the motion be such that Q moves in a horizontal
line, determine the velocity when the rod has any given
inclination, the rod being supposed to start from rest in a
horizontal position. ;

19—2

\
292 EXAMPLES.

In the case in which the whole length of the rod =,/3. PQ,
shew that the time of a complete oscillation

= (2n)* (14)? JPQ/2g.
In this case also find the equation to the hodograph of
the middle point of the rod:

29. A number (n) of equal uniform rods, AA,, A,A

A,A,, &¢c., are jointed together at their ends A,, A,,... and
the end A of the first rod is attached to a fixed point. The
rods are held so as to form a straight line AA,...4,, the end
A, being free, and the supports are simultaneously removed.
Prove that if w be the weight of a rod, and a= 15°, the

initial action at A, is equal to
w(—4)" sin” a cos” a — cos* a. sin” a
2 (8) ° sin” a + cos” a

30. AB, BC, CD are three equal uniform rods freely
jointed together and moveable about the extremity A;
the rods fall from a horizontal position of rest: prove that
the radius of curvature of the initial path of the extremity
D of the further rod is 44, a, where a is the length of each
rod. Prove also that the initial stresses at. C, B and A are
in the ratio of 1, 4 and 15,

31. An arc of a circle is placed in its position of equi-
librium in a vertical plane resting on a perfectly rough
horizontal plane and slightly disturbed in the former plane;
shew that the square of the time of oscillation varies as
azsina, a being the radius of the circle, and 2a the angle
subtended at its centre by the arc.

 

32. A ball is projected from a point in a perfectly rough
horizontal plane, without any rotation; if the coefficient of
frictional elasticity be 2/5, prove that the horizontal velocity

of the ball, after the n™ rebound, will be e +(—2)* o. where
w is the horizontal velocity of projection.

33. Two equal rods, connected by a hinge, which allows
them to move in a vertical plane, rotate uniformly about a
vertical axis through the hinge; and a string, whose length
EXAMPLES, 293

is double that of either rod, is fastened to their extremities,
and supports a weight at its middle point. Determine the
angular velocity when in the position of relative equilibrium
the rods and the string form a square; and supposing the
weight slightly displaced in a vertical direction, find the
time of a small oscillation.

34, A smooth plane of mass M is freely moveable about
a horizontal axis lying within it and passing through its
centre of gravity, the radius of gyration of the plane about
the axis being & The plane being inclined at an angle a to
the vertical, a sphere of mass m is placed gently upon it. If
initially the centre of the sphere be in a vertical through
the axis of the plane, and if h be its initial height above
that axis, shew that the angle ¢, which the initial direction
of motion of the centre makes with the vertical, is given by

(Mk? + mh*) tan 6 = MP’ tan a.

35. A uniform beam of mass M and length 2a can turn
round a fixed horizontal axis at one end; to the other end of
the beam a string of length / is attached, and at the end of
the string is a particle of mass m. Determine the relation
that must hold in order that, during a small oscillation of
the system, the inclination of the string to the vertical may
be twice that of the beam.

36. If a given power P raise a given weight Q through
a given space in the least possible time, on a wheel and axle,
the masses of which are M and JM’ respectively, prove that
the ratio of their radii must be

Q ig Q 20+’
PtN Pt opr t*
37. A uniform heavy beam of length 2c is supported in

a horizontal position by means of two strings, without weight,
each of length 6, which are fastened to its ends, the other
ends of the strings being fixed; in equilibrium each of the
strings makes an angle a with the horizon: find the time of
a small oscillation when the system is slightly displaced in
the vertical plane in which it is situated, the strings not
being slackened.
294 EXAMPLES.

38. A circular ring hangs in a vertical plane on two
pegs. If one be removed, prove that, P,, P, being the
instantaneous pressures on the other peg, calculated on the
supposition that the ring is (1) smooth, (2) rough,

P2: PR:s4&:44+tan*a,

where a is the angle which the line drawn from the centre of
the ring to the peg makes with the vertical.

39. A heavy bar is suspended in a horizontal position by
two equal and parallel vertical strings attached to its ends, it
is then set swinging so that the strings move ‘in vertical
planes perpendicular to the bar, if one string breaks when
the rod is in its lowest position, prove that the tension of the
other string is instantaneously diminished by one half its
value.

If the second string be cut when the bar is vertical, prove
that the subsequent rotation will be uniform and round a
horizontal axis fixed in direction; but if the second string
be cut at any other time, the vertical plane containing the
bar will rotate with an angular velocity varying as sec’, and
6 will increase at a rate varying as Va+6 sec’ 0, where @ is
the inclination to the horizon and a, 6 are constant.

40. A chain of mass m and length / hangs in equilibrium
over a smooth pulley, an insect of mass M alights gently at
one end and begins crawling up with uniform relative velocity
V; shew that the velocity with which the chain leaves the

pulley will be
M? 2, UM+im _,)t
laren? ++ m gt}

41. A man walks on a large rough ball so as to make
the ball roll straight up an inclined plane of inclination a,
keeping himself at an angular distance 8 from the highest
point of the ball; prove that the acceleration of the man is

sin 8—2sina
ee 10cos’$ (a+ 8)’
taking the man and ball to have equal weights.

 
EXAMPLES. 295

42. A light uniform lamina in the form of a regular
trapezoid is suspended by one of the parallel edges, and a
weight mg is uniformly distributed over the opposite edge ;
supposing the lamina to be elastic only in the direction of
the breadth, find the position of equilibrium,

Shew that the time of a small oscillation is

ml (log a — log 8)
am / on (a—=b)”
when 2a and 26 are the lengths of the parallel edges, J is the
breadth of the lamina when unstretched, and » the modulus
of elasticity.

43. The extremities of a uniform heavy rod of length 2c
slide on a smooth wire in the form of the parabola #’—4ay=0,
the axis of the parabola being vertical, and c>2a. If the
rod be slightly displaced from its position of stable equi-
librium, prove that the time of a small oscillation is

9 Qac B
™ 139 (¢—2a)) *

44, Four equal uniform rods are jointed together so as
to form a square ABCD, and the system is suspended from
the joint A, the square form being maintained by an elastic
string joining A and C.

Find the tension of the string, and, the modulus of
elasticity being twice the weight of one of the rods, prove
that, if C be slightly depressed, the length of the simple
isochronous pendulum will be 540/12. If, when there is
equilibrium, the string be cut, prove that the initial pressure
at C is equal to one-tenth of the weight of one of the rods,
and that the initial acceleration of C is equal to 69/5.

45. A cylindrical mass of snow rolls down an inclined
plane covered with snow of uniform depth e; gathering up
all the snow it rolls over, and always remaining circular.
Prove that it will move with uniform acceleration g sin a/5 if
initially when its radius is a it be started with velocity
a2: sin a/5¢, where a is the inclination of the plane to the
horizon.
296 EXAMPLES.

46. Three particles A,B, C are connected by two strings
AB, AC and placed in a line on a smooth table. The
extreme particles B and C are then projected at right angles
to the strings with velocities u, v. Prove that the initial
curvatures of the paths of the extreme particles are respec-
tively

v v wu

uw
(q+ m) — +O 5 a (p +m) 5 +P
(p+q+m) wu (p+q+m)v

m, p, ¢ being the masses of the particles, and a, b the lengths
of the strings.

 

>

47. Two particles of masses m,, m, are tied to the
extremities of a string that passes through a bead of mass VM,
the whole is placed on a smooth table with m,, m, at the
acute angles and M at the right angle of a right-angled
triangle, if the particles at the ends be projected with
velocities w,, w,, perpendicular to the respective strings, then
shew that the initial radii of curvature of their paths are p,, p,

2 2

U, U,

2 2 _ Pa
where MU Melly Gy
A fe UD

—+ 7+
m <M m,

48. A circular disc of mass M, radius a, and moment of
inertia about the centre MK’, is spinning with angular
velocity © and impinges normally with any velocity on a
rough rod resting on a smooth plane. Shew that the angular
velocity immediately after impact becomes reduced to

(M+m) K?
ma’ + (M+m) ke
49. Two uniform rods AB, BC of masses m, m’ freely
jointed at B lie-upon a smooth horizontal table and AB is

struck perpendicular to its length at a point between A and
B; shew that the point B will begin to move in a direction

ate ym) cot al; a being

‘ : 1 1
making with BC an angle tan | Aa

the angle between the rods.
EXAMPLES, 297

50. A spherical shell of mass m, whose outer surface is
rough and of radius a, has its inner surface smooth and of
radius 6; a particle of mass m moves inside while the shell
rolls on a rough table, shew that if the exctrsions of the
particle be a on either side of the vertical, then

[AL (a? + i) + ma? sin?6] &
= = LM (a? + k*) + ma’ [cos @ — cos a].

51. A smooth massless rod HM of length 1+2a turns
freely about a hinge at one end H. A string of length J is
fastened at the end M and also at a point S in a horizontal
line with Hand at a distance 2a,/2 from H. A smooth ring
of mass m is slipped over the rod and string at Mand moved
up the rod until the string is tight and the rod horizontal;
it is then allowed to fall, find the velocity of the ring at any
instant before it slips off the rod. Shew that the tension of
the string when the ring has fallen through a vertical height

. _ mg 3a’t+ 27?
y is T= 5. . a ay
mg being the weight of the ring.

52. A smooth thin spherical shell of mass Jf and radius
a rests on a smooth inclined plane by means of an elastic
string which is attached to the sphere and toa peg at the
same distance from the plane as the centre of the sphere and
a particle of mass m rests on the inner surface of the shell.
In the position of equilibrium the string is parallel to the
plane, find the times of oscillation of' the system when it is
slightly displaced in a vertical plane and prove that the arc
traversed by the particle and the distance traversed by the
centre of the shell from their positions of equilibrium can
always be equal if

Mg + mg (1+ cos a) = ra (1 + cos a)/e,
where A is the coefficient of elasticity of the string, and ¢ its
natural length.

53. A uniform circular disc moving in any way is placed
gently upon a rough horizontal plane. Assuming that the
298 EXAMPLES,

friction between any element of the disc and the plane varies
as the relative velocity and is in a direction opposite to it,
find the motion of the disc, and shew that if w and » be the
velocity of the centre and the angular velocity about it at any
instant, wo, =u,w@, where u, and @, are the initial values of u
and a.

54, A number (n) of equal uniform rods 4,C,B,, A,C,B.,,...
are placed on a smooth horizontal plane so that the end
A, of the 2nd is in contact with the middle point C, of
the first, the end A, in contact with C,... and the angles
C\A,B, C,A,B,... are each equal to 6, so that the figure
A,C,C,C,...C,, is a portion of a regular polygon. At the end
A, an impulse P is applied inwards in the direction making

an angle 7/2 —@ with A,C,. Prove that the impulse between
the r” and r+1]", supposing them smooth and rigid,

= P (B"a” = a"Q")/(8" — a"),
where 4 and £ are the roots of the equation

2 — (28sec 643 cos 6)z+1=0.
CHAPTER XIV.
MOTION IN THREE DIMENSIONS.

230. WE now proceed to consider the motion of asystem
referred to three rectangular axes, either fixed, or moving in
a given manner.

As in Art. (28) we employ 6@,, 6,, and @, to represent the
angular velocities of the system of axes.

Taking @,, w,, and w, as the angular velocities, at any
instant, of a rigid body about the axes, it follows as
in Art. (28) that the angular accelerations are respectively

@,— 0, 7 00,,
@, — 0,9, T o,9,,
©, —0,0,+ 0,0,

From the definition of the linear momenta and the
angular momenta of a system it follows that these
quantities are vectors and are subject to the parallelo-
grammic law.

Let p,, p,, p, represent the linear momenta of a system

in the directions of the axes, and h,, h,, 4, the angular
momenta of the system about those axes.

Then it follows, as in Art. 28, that, if we take OL, Ol,
and OWN to represent either the quantities p,, p,, p, or the
quantities A,, h,, h,, the rates of change of these quantities
300 EQUATIONS OF MOTION.

are, on the same scale, the velocities of the point of which OL,
OM, ON are co-ordinates, and are therefore respectively

Py, — pO, + POs iy * h,O, + iO,
Ps = ps0, + p,9, and h, — h,O, a h,9,
Ps —p9, + pA, h, i hO, + h,O,

The equations of motion of the system are obtained by
equating these expressions to the components of the acting
forces and of the acting couples.

The equations of motion, in the forms thus obtained, were
first given by Mr R. B. Hayward, F.R.S., of St John’s
College, Cambridge.

They are contained in a paper published in 1856, in
Part L, Vol. X., of the Cambridge Philosophical Transactions.

231. If a, y, z be the co-ordinates of a particle m of
the system, and if u, v, w be the component velocities of
the particle,

p, = mu, p,= Lm, p, = Imw;
h, = Xm (wy — v2), h, = =m (uz —we), h, = Xm (ve — uy).

The total motion of the system at the instant in
question is thus represented by three linear momenta in
the directions of the axes and three angular momenta about
those axes.

These are equivalent to a single linear momentum and a
single angular momentum.

232. Ifthe origin be not a fixed point, the expressions
for the rates of change of the linear momenta are unaffected,
but the expressions for the rates of change of angular
momenta will require modification.

Let «, 8, y be the component velocities of the origin, and
suppose the axes to have no rotation.
EQUATIONS’ OF MOTION. 301

Since h, =m (wy — v2),

the angular momentum, at the time ¢+ 6¢, about the axis Ow,
fixed in space °

= Xm {(w + dw) (y + dy + BSt) — (v + bv) (2 + 82 + ydt)}
= h, ab oh, cle (p,8 — Py) ot,

and, subtracting A, and dividing by 6t, we obtain the
additional term

PB —- Py

so that the complete expression for the time-flux, about the
instantaneous position of the axis, of the angular mo-
mentum is

h, ey h,0, a 10, + p8 — pyy.
If the origin be the centre of gravity of the system, the

expressions for angular momenta and their rates of change
are those of Art. (230).

It will be seen that the terms p,8—p,y of the previous
article disappear in this case, for

p, = MB, and p, = My.

233. Motion of a rigid body about a fixed point.
In this case
U=20, — YO. V= LW, — 20,, W= YO, > LW,,
and therefore
h, = Xm (y? +24) @, —¥ (may) wo, — 5 (may) 0,
and, if we represent the three moments of inertia by
A, B, C and the three products of inertia by D, #, F
we have
h, = Ao, — Fo, — Eo,,
h,= Bo, — Do, — Fo,,
h,= Co, — Ew, — Do,
If the expressions D, EH’, F all vanish the axes are said to

be principal axes; if two vanish, the corresponding axis is a
principal axis.
302 SPHERE ON PLANE.

In the case of a sphere, or a solid bounded by any
regular polyhedron, when the centre is the origin, D, H, F all
vanish, and A, B, C are all the same, so that the angular
momenta take the forms .

Aw,, Ao,, Ao,

In the case of a solid of revolution, or of a regular
pyramid, the axis of which is one of the axes, D, H, F all
vanish, and the angular momenta.are

Aw,, Aw,, Co,.

In the case of a plane lamina, when one axis is perpen-
dicular to its plane, D and £ vanish, and the angular
momenta are

Ao, —Fw,, Bo,— Fw,, (A + B) a,.

234. If the axis of z be fixed in space, the time-
fluxes of angular momenta about the instantaneous positions
of the axes are I . é

h,—h,O,, h, +h,O,, hg

It may be instructive to obtain these expressions directly.
Thus, at the time ¢ + 6, the angular momenta about Oz’, Oy’,
the consecutive positions of Om, Oy, being h,+ dh,, h, + 8h,,
it follows that the angular momenta about Ox and Oy are
respectively

(h, + 8h,) cos 0,5¢ — (h, + 8h,) sin 6,8¢,
(h, + 5h,) cos ,6t + (h, + 6h,) sin 6,64,

and, subtracting h, and h, and dividing by 6, we obtain, in
the limit, the expressions given above.

The general expressions of Art, 230 may be obtained in a
similar manner.

235. We are now in a position to solve some problems,
and we commence with the motion of a sphere on a rough
plane, under the action of forces the resultant of which passes
through the centre of the sphere.

Referring to fixed axes the linear momenta are ma, my,
and the time-fluxes of the angular momenta are

Aw, Aw, Aw, where A = 2me’/5.
FIXED AXIS. 303

Assuming X, Yas the forces, and taking moments about
the lines in the plane through the point of contact parallel to
x and y, we obtain

—mijc+ Ao, =— Ye, miic+ Aa, = Xe.
We have also the geometrical conditions,
é—cw,=0, y+ cw, =0,
and we hence obtain
mi =iX, my =hY.

If the frictional reactions be required, they are given by
the equations

mé&=F+X, my=G+Y,
so that F=—-2X and G=-32Y.

If the plane be made to revolve uniformly, with the
angular velocity Q about the axis of 2, the equations of
motion are the same, but the geometrical conditions are

é—cw, =— Oy, y+cw, = On.
If in this case there be no forces in action, the elimination
of w, and w, leads to the equations
Ti+ 20y=0, 77 —20%=0,
or,if Tnr=20, +n? (e@—a)=0, +n? (y—b) =0,
where a and 6 are constants.

Integrating these equations and eliminating the time, we
shall find that the path of the centre of the sphere is an
ellipse, of which the point (a, 0) is the centre.

236. To illustrate the use of two moving axes, consider
the motion of a rigid body about a fixed axis.

Taking, as in Art. 211, the line OG as the axis of w and
the fixed axis as the axis of z,
w=— Yo, v= 20, w=),
and therefore,

h,=— Eo, h,=—- Dw, h,=Co.
30+ ROTATING DISC.

Hence h, —h,0,= — Eo + Do’,
h,+h,0, =—Do— Eo’,
and equating these expressions to the moments of the

acting forces, we obtain, as in Art. 211, the stresses on
the axis.

237. A circular disc, the plane of which is vertical, and
centre fixed, is rotating about a horizontal amis through its
centre perpendicular to its plane, which axis is itself rotating
Jreely in the horizontal plane through the centre, and an
insect crawls in a given manner on the disc.

Taking the figure of Art. (247), let ZC be the plane of
disc, OC being a given radius of the disc.

The equations of motion are obtained by observing that
the angular momentum about Oz is constant, and that the
time-flux of the angular momentum about OF’ is equal to the
moment about OF of the weight of the insect.

Let p be the distance of the insect from O, and ¢ the
angular distance of p from OC measured in the direction CZ,
so that p and ¢ are known functions of the time.

Putting ZC = 6, and XZC'= wy, we obtain
Mk'yp + mp’ sin’ (0+ $) p=C.
» 2, be the angular momenta about OK and OF,
h, = mp" sin (0+) cos (8 + $),
h, = 2Mk’O + mp’ (6 + ).
Hence we obtain, since h,+h, is the time-flux of the
angular momentum about OF,
2MKO + mp? (6+46)-+ 2mpp (0+46)-+mpp? sin (4+ ) cos (0+¢)
= mgp sin (0 + $),
and 6 and are determined by these equations.

If h

238. Motion of a heavy sphere on the interior rough
surface of a vertical cylinder.

The figure being a section by the horizontal plane through
the centre of the sphere, take the axis (3) through C vertically
upwards and measure z upwards.
SPHERE ON CYLINDER, 305

The accelerations of C in the directions (1) (2) and (8) are
—(a—c) $*, (a—c) , 2.

In this case, since h, = Aa,,h, = Aw, and 0,= 4, the time-
fluxes of the angular momenta about the instantaneous
positions of the lines (1), (2), and (3) are

Ao, — Aa, , Aw, + Ao, d, Ao,

 

Tv

_ Taking moments about CP, PT, and the vertical through
P, we obtain
0, —o,6 = 0, Aw, + Aw,d + mez=—mge, Aw,—m (a—c) ch=0.

Expressing the fact that the point P of the sphere has
no velocity, the geometrical conditions are

(a—c)p + cw, = 0, z—cw, =0,
we at once see that ¢ and a, are constant, that co, = dz, and
therefore that 47+ Ad¢’z+mc’z is constant, or that if 2¢7=7n’,
Z+nv7z=C,

shewing that the ball rolls up and down, between two fixed
levels, in the time 7/n.

If the cylinder, instead of being fixed, be made to revolve
with a constant angular velocity about a vertical generating
line through the point O in the figure, the angular velocity
of the line CH is +; and the accelerations of C in the
directions (1), (2) and (3) are, putting b for a—c,

aw® cos d —b (w +b)’, bf — aw’ sin ¢, 2.
B. D. 20
306 SPHERE ON CONE.

Hence, taking moments about the same axes as before,
we obtain,

@,— @, (o +) =0, $ cw, + 3CO, (o+ d) +Z=- 9,
2 cw, —b G+ aw’ sin = 0.
The geometrical conditions are that the velocities of the

point P of the sphere and of the point P of the cylinder are
the same,

cw, +b (w +) — aw cos 6 = aw—awcos d
or co, +b (w+¢) = aa,
and %—cw, = 0.
Eliminating , we find that
7b = 5aw* sin ¢,
an equation which determines the angular motion of the
centre of the sphere relative to the cylinder. —

239. Motion of a heavy sphere on the interior rough
surface of a cone having tts axis vertical and vertex
downwards..

The figure being a section of the system by the vertical
plane through the axis of the cone and the centre of the
sphere, the accelerations of C are

F—r¢’, rot 2rd, 2,
and the time-fluxes of angular momenta are the same as in
the preceding case.
The geometrical conditions are

TF — CW, SIN A= Oe eeeeeee eens (1)
rh + cw, Sin a+ Co, COS4=O.eeeeseeeeee (2)
B= Oil, COR = Oo. sa sidaaisac'vonain re (3).

Taking moments about PH, PC, and the line through P
perpendicular to the plane of the figure we obtain

(Ao, — Aw,¢) sin a+ Ad, cos a—m (rb + 27) o= 0...(4)
(Ao, — Aw, d) cos a— Ad, SNA =Ovieeceeserrereeees (5)
Ao, + Aw, p+ m (i — 7g") csina + mic cosa =— mge cosa...(6).
SURFACE OF REVOLUTION. 307

From (4) and (2) we find that rd + 27'6=0, or that r°d=h,
From (4) and (5) it follows that #,=0, or that a, =n.

(3)
H

N (1)

 

0

Lastly from (6), with the aid of (1), (2), (3), and the
preceding results, we find, if we write w for 1/r, the equation,

du  2+5sin"a ,gsinacosa ,cncosa
dd? eee

240. The general problem of the motion of a sphere on
any surface of revolution may be treated in the same manner,
or we may employ three moving axes.

Taking the axes as in the figure, and taking wu, v, w as the
velocities of C in the directions (1), (2), and (8), the geo-
metrical conditions are

u—co,=0, v+co,=0, w=0.

Since 0,=¢ cos 6, 6,=6, 0,=¢sin 8,

&— v0, + wO,= cw, + co, sin 0,

d—w0, + uO, = — ca, + ca, sin 8,

w — uO, + v0,= — cw,0 — cw, p cos 9,
which are the accelerations of C.

 
308 SURFACES OF REVOLUTION.

Hence, if L, M@, N be the moments of the acting forces
about PT, about the line through P perpendicular to the
plane AGP, and about PO, the equations of motion are

A (o, — @,g sind + w,6) — me (— cw, + cw, sin 0) = L,...,
A (&, — ag cos 0 +, sin 8) + me (cm, + 00h sind) = M,...ii,
A (@, — 0,6 +, cos 0) = N.. iii.

 

 

It will be seen that C moves on a parallel surface of which
A is the vertex, and if AN=z and CN =r, the relation
between r and z is known, so that z=/(r).
Hence if s be the are AC
co, =u=s=— pb,
if p be the radius of curvature of AC at C.
We also have cw, =—v=— rd.
If for example the surface be a sphere, and gravity the
only force in action,
L=0, N=0, and M=— mge cos 0.
Also, in this case,
cw, = — a6, cw, = — acos 64,
and therefore, from equation iii,
®,=0, or o, =n.
SPHERE ON DISC. 309

These values of w,, w,, o, being substituted in equation
1, we obtain

a cos Op — 2a sin 6 6b = 3 end, °
and therefore

cos"0g = C +2 = sin 0.

Substituting in equation ii and integrating we obtain a
differential equation of the first order for 0.

This last however is more easily obtained from the
equation of energy which is

a6” + a? cos* 04 + 20 (w,? +o," + ,°) = D + 2ga.cos 6,
‘9 con. ,\* 10g
or 0 + s00%9 (C+? sind) =H+ >~ cos 6,
a 7a
C and £ being constants determined by initial conditions.
241. Motion of a rough sphere on the surface of a flat
disc, which is moveable on a smooth horizontal plane, the
upper surface of the disc being perfectly rough.
wt

7 NV ax

FO i.

(2)

 

 

 

We shall suppose that the centre of gravity of the system
has no motion; this will be the case if the disc be initially at
310 SPHERE ON DISC.

rest, and if the sphere, in a state of rotation about a diameter,
be placed gently upon the disc.

In the figure O is the projection on a horizontal plane of
the centre of gravity of the system, Z of the centre of gravity
of the disc, and C of the centre of the sphere; Ox, Oy are
fixed directions.

Taking £, 7, and a, y, as the co-ordinates of H and 0,
measured in opposite directions, and Q as the angular velocity
of the disc, the geometrical conditions are

& — c0,= — E—O (yt), veceeceeeeseennees (1)
YtCo, =—AEA(GHE). creccrcceccseceees (2).

Taking moments for the motion of the sphere, about the
horizontal tangent lines parallel to the axes,

mic+mk'o,=0, mijc—mk*o, =0,
and therefore #+2cw,=A, y—2cw,=B....... (3), (4).
The angular momentum of the system about any
assigned vertical line is constant, and if the sphere have

initially no rotation about the vertical diameter this constant
is zero.

Taking moments about the vertical line, fixed in space,
through which JF is passing,

MK*O,= ma (y +) — my (@+ &),
or MRP =m (+ M) (bY — Yb) ciececcceceves (5).
We have besides ma = ME, and my = My, and we thus

have seven equations to determine the seven unknown
quantities.

If the original axis of rotation of the sphere be above the
line Ox and parallel to it, so that initially

o,=n, and wo, =0,
we obtain &+2co,=0, y—2co, = B.
The elimination of w, and », leads to
24 (M+ m) + 5Mé = — 2yO, (iM +m),
2y (M +m) + 5My — 5MB= 220 (M+ m),
ROTATING ROD. 311

and, substituting for Q its value from (5), these equations
take the forms

(a+ by’) & =bay y, (a+ ba’) y = bay é+e,
where a, 6 and c are constants.

The integration of the first of these equations gives
a+ by’ = Cx’,

shewing that the path in space of the centre of the sphere is

a hyperbola, (a result given in the Tripos Examination,
Jan. 1882).

It should be mentioned that nm is the initial angular
velocity of the sphere just after having been placed in contact
with the disc.

‘If w be the angular velocity of the sphere about the
diameter parallel to the axis of # before the contact, and if X
be the angular velocity of the disc immediately after the
impact, m and A are determined by the equation

mk'n — myc = mk*w,

combined with the preceding equations (1), (2), and (5) in
their initial forms.

242. Motion of a heavy rod AB, the ends of which shide
on a fixed vertical rod OB, and a horizontal rod OA, which 1s
made to revolve uniformly.

If r be the distance from G, in the direction GA, of a
point P of the rod, the accelerations f, f’ of the point in the
2 2

directions LP and MNP, are - PL—o'PL, and ae PN,
where PL, PN are the perpendiculars upon OB and OA, so
that PL =(a+r)cos 0, and PN= (a—1) sin @.

Taking moments about the line through £ perpendicular
to the plane OAB, we obtain

: mo {(a+r) fsin 0—(a—7r) f’ cos 6} = mga cos 0.

—a
312 ROTATING ROD.

Substituting for f and /’ their values, and integrating,
this reduces to
oe ie 3g
2 a I
0+’ sin 8 cos 6 = he 08

We have solved this question by an appeal to first princi-
ples, but it may be instructive to indicate the method of
dealing with it by the aid of the expressions for angular

momenta,

 

 

 

 

 

()
B B
¢ F,
y : P $
wv 4

 

Taking for axes the line GA, and the lines through G

perpendicular to and in the plane OAB,
6,=—asin6, 6,=6, 0,=acos8,

and h,=0, h,=mk’o,, h, = mk’o,.

The acceleration of G in the plane OAB, perpendicular
to OG =a +asin@cos@.0%. Art. 15,
and the rate of change of the angular momentum about the
second axis =mk’o, + mk’w, w sin 6.

Hence the equation of moments about the line through Z
perpendicular to the plane is
2

a? a ‘
3 tm ww sin @=— mga cos 6;

and, observing that , = 6 and that w, = cos 0, this reduces
to the equation previously obtained.

ma? (6 + sin 6 cos Ow") +m
SOLID OF REVOLUTION. 313

If the system instead of being made to revolve uniformly

be set in motion and left to itself, we shall have, taking ¢ for

the azimuthal motion, and neglecting the inertia of the rods
OA, OB,

6+ gtsin 0 cos 6 = — 22 cos 6,
with the additional equation, derived from the fact that the
angular momentum about OB is constant,

c mo (a+r) cos’ 0.46 = C.

—a

243, Steady motion of a heavy body in the form of a solid
of revolution, rotating uniformly about its axis, one point O of
which is fixed while the amis has a constant inclination to the
vertical (a), and a constant azimuthal motion (Q).

(3)

 

 

®

Taking moving axes as in the figure, the second axis
being perpendicular to the plane of the paper,

6,=—QOsina, 0,=0, 0,= 0 cosa,
314 SOLID OF REVOLUTION.

also o,=0 and o,=—Q sina, and, taking moments about
the axes, we obtain

Ao, =0, Co, Asina+ Ao, XO cosa=mgasina, Co, =0.
Putting n for w, this gives
COn — AQ? cos a = mga,
as the condition for steady motion.

Ifa=— , COn = mga, so that if the angular velocity n be

imparted to the body about OC, when OC is _ horizontal,
and OC be then started with the angular velocity mga/Cn
about Oz, the axis OC will continue to revolve in a horizontal
plane.

_ In the general case, when the motion is not steady, take
ap as the azimuthal motion, so that

6,=—wsin 0, 0,=6, 6,= cos 8,
and therefore
Ad, — Aap cos 6+ Cw,6 = 0,
Aw, + Co, sin 6+ Aw, cos 6 = mgasin 6,
Ca, — Aw,6 — Ao, sin 0 = 0.

Now o,=—sin 0, and w,=6, and we find that , is a
constant (n) and that

— Ap sin 6 — 2AW4 cos 0 + Cnb =0,
AG + Cn sin 6 — Ary’ sin 6 cos 6 = mgasin 6,
two equations which completely determine the motion.

If the motion be very nearly steady, the small oscillations
are determined by putting

6=a+¢, and p=N+4+y,

and neglecting the squares and products of the small quanti-
ties @ and x.

These results are roughly illustrated by observing the
SOLID OF REVOLUTION. 315

motion of an ordinary spinning top on a rough horizontal
plane.

The most complete illustration is obtained by means of a
Gyroscope.

K

 

 

 

A heavy brass solid ring is moveable about the diameter
OC of a circular brass framework, and, holding this frame-
work, a rapid rotation can be imparted to the ring. This
can be effected by looping a string to a peg at H, winding up
the string and pulling it out sharply.

This being done, the point O may be held by a string,
and OC placed at any inclination to the vertical.

Or, if a small peg be fixed at O underneath the rim of
OC, this may be placed on the cup K of a fixed vertical _
stand KZ, and then the axis OC may be seen to revolve |
horizontally.

244, Steady motion, about a fixed point O, of a rigid
body so constituted that two of the principal moments of inertia
at the point O are equal.

In this case, of which the preceding is a particular case,
the constraining couple is equal to
(COn — AQ? cos a) sin a.
If Cn = AQ cos a, this expression vanishes, and the steady
motion continues without the action of any extraneous force.
316 CASE OF TWO EQUAL MOMENTS.

245. Motion of a rigid body about a fixed point O under
the action of no extraneous forces, the body being so constituted
that A and B are equal at the point O.

The body is supposed to be set in motion in a given
manner, or by means of given impulses.

The moments about the principal axes of the given
impulses are the initial values of the angular momenta

Aa,, Aw,, Ca,

There being no forces in action the angular momenta
about any fixed lines through the point O remain unchanged
and therefore the resultant angular momentum is a constant
quantity, H, and the axis of resultant angular momentum is
a fixed line.

This line, which is called the invariable line, we shall take
for Oz.

Take OC as the axis of (3), and, for axes of (1) and (2)
the line in the plane zOC perpendicular to OC, and the line
perpendicular to the plane 200.

Then if y be the inclination of the plane 200 to a fixed
plane through Oz,

6,=—ysin 6, 6,=6, 0,= cos 8,

and the equations of motion are

A@, — Ao,rp cos 6 + 00,0 = 0, cescceccsseeeesees (i)
Ao, + Cor sin 0+ Aw, cos 0 = 0, oe (ii)
Co, — Aw,6 — Aor sin 0=0. veces (iii).

We have also o,=— sin 9, and o, = 6, and therefore,
from (iii), it follows that @, is constant.

Now, Hcos 6=Cw,, and therefore. it follows that @ is
constant,

From equation (i), we find that w, is constant and there-
fore that vf is constant.

Further since — Hsin 0 = Aw,, we obtain Ay = H.
CASE OF TWO EQUAL MOMENTS. 317

_ The condition, Cw, sin 8+ Aa, cos 0 = 0, derived from (ii),
simply expresses the known fact that the angular momentum

- about the line in the plane zOC which is perpendicular to Oz
is zero,

The conclusion is that the angular velocity about OC is
constant, and that the plane zOC revolves with uniform
angular velocity H/A about the axis of resultant angular
momentum Oz.

246. What has been proved of motion about a fixed
point is equally true of the motion of a rigid body relative
to its centre of gravity.

This can be easily illustrated by tossing into the air a
solid body of any symmetrical shape, such as a piece of wood
in the form of a circular cylinder, or in the form of a cir-
cular cone, or any regular prism or regular pyramid, taking
care to give the body rather a rapid rotation about its axis.

In all such cases, in whatever manner the body may be
thrown up, its axis will be seen to describe uniformly a right
circular cone about a line through the centre of gravity the
direction of which remains unchanged.

This line is the axis of the resultant angular momentum
originally imparted to the body.

If the Gyroscope described in Art. (243) be mounted in
a fixed framework so as to give the diameter OC free motion
about the fixed point G, we obtain an apparatus for directly
demonstrating the fact that the earth has a motion of
rotation independently of its motion of translation.

The ring being set rotating with great rapidity, so as to
continue rotating for some hours, it will be seen that its
position relative to the room in which the machine is
situated gradually changes; and, as we know that the
direction of the axis of rotation of the ring does not change,
it follows that the earth itself is in a state of rotation.

247. Euler's Equations. In the case of the motion of a
single rigid body about a fixed point or about its centre of
318 EULER'S EQUATIONS.

gravity, if we take for our moving axes three straight lines

which are always coincident with three given lines in the
body,

0,=0,, 0,=0,, 0,=o%
If further we take these three lines to be principal axes,
h,=Aa,, h,=Bo,, h,=Ca,,
and the equations of motion take the forms
Ao, —(B- C) 0,0, = L,
Bo,—(C— A) o,0,=M,
Co,—(A—B) oo, =N,
which are Euler’s Equations.

Geometrical equations connecting the angular motions about
three moving axes with the motion of the body referred to lines
Jixed in space.

If OA, OB, OC be three lines fixed in the body, and OX,
OY, OZ three lines fixed in space, and if these lines end on
the surface of a sphere, the position of the body will be
completely determined by the quantities , 0, ¢, if

v= CZK, 0=Z0, =ACE.

Zz
Z

 

 

Moreover the motion is completely determined by the
rotation @ about OF, the rotation y about OZ and by the
°

IMPULSES. 319

rate of separation d of the plane OCA in the body from the
moving plane ZCE.

This system must be equivalent to the system of angular
rotations w,, ,, and w, about OA, OB and OC.

Expressing this equivalence we obtain,
o, =6cos FA +> cos ZA = 6sin d — sin 0 cos ¢,
w,=6 cos FB+ cos ZB=6cos$+sin O sin d,
w, = cos 0+ d.

These equations may be employed in the discussion of
the general problem of the motion of a rigid body about a
fixed point, :

For this however we refer the student to Poinsot’s
Nouvelle Théorie de ta Rotation des corps solides, to Dr
Routh’s Rigid Dynamics, and other treatises.

248. Impulses. We have given in Arts. (200) and (201),
the principles which determine the effects of impulses, and
the equations for the calculation of those effects.

If smooth inelastic bodies impinge on each other, it must
be carefully borne in mind, that the velocities, immediately
after impact, of the points of the bodies in contact with each
other, are the same in the direction of the common normal
to their surfaces. y

If however two perfectly rough inelastic bodies impinge
‘on each other, the geometrical condition is that the velocities
of the points of contact are the same in any direction.

If when a system is in motion, a straight line in the
system is suddenly fixed, the impulses called into action
have no moment about the line, and consequently the
angular momentum about it remains unchanged.

If a point of the system be suddenly fixed, the change
of motion is determined by the fact that the angular
momentum round any straight line whatever, through the
point, remains unchanged.
320 ILLUSTRATIONS OF WORK.

In the case of a single rigid body, if u, v, w be the
velocities of the centre of gravity, and @,, @,, @, the angular
velocities about principal axes through the centre of gravity,
the changes of these quantities due to impulses whose com-
ponents are P, @, R, and moments, about axes through the
centre of gravity, G, H, K are given by the equations

M(u'—u) =P, M (v1) =Q, Mw'—w) =,
A (w,’—0,)=G, B(w,'—o,) = H, C (a, —@,) = K.

The solution of problems involving impulses may some-
times be facilitated by the use of the principle of virtual
velocities.

For any imagined geometrical displacement the virtual
moment of the changes of linear and angular momenta is
equal to the virtual moment of the applied impulses.

It may be well to notice that if a couple be displaced
about a line parallel to its plane the virtual work is zero, so
that if a couple is displaced about an axis not perpendicular
to its plane, all that is necessary is to find the component of
the couple about the axis of displacement.

The solutions of the two following problems will serve as
further illustrations of the application of the principles of
momentum and energy.

249. A system, consisting of four equal rods forming a
square ABCD, having universal joints at A, B, C and D, is
rotating freely with an angular velocity n about the line EF
joining the middle points of BC' and DA; it is required to
determine the changes of motion produced ‘when the point A
ais suddenly fixed.

In order to mark directions take the axis of z per-
pendicular to the plane of the square, and let @,, @,, @,, @, be
the angular velocities of AB, BC, CD, DA immediately after
A is fixed.

Expressing the fact that the angular momentum of the
ILLUSTRATIONS OF WORK. 321

rods AB, BC about the straight line AC is unchanged, we
obtain

 

 

 

 

 

an an
Prag 2 geen
< at % at 2m) rc 53 v2 3y2?
or 5@, +@, = —2n.
D BR a
Me ! Pd
igs ie
‘ i x
MK \ ra
Faas pee eae I a
oe "ee
co ' ah
Bg i “sy
c Ei B
y

Further the angular momenta of ADC about AC, and
of BCD about BD are unchanged, and therefore

264 a(o,+20,) ao, _ an, an

+R 2 ~ 3/2 So fe”

a 4 H% an an

@ (0, + 20,) 5 + 5s ta (0, + 20,) vat 3/2 = 3/72 +»
or 5o,+ 0, = 2n,

3, + 20, + 20, + 30, = 2n.

Further we have the geometrical condition obtained by
equating the two expressions for the velocity of C which is

2aw, + 2aw, = 2aw, + 2aw,

or @, +0, =O, +0,
322 ILLUSTRATIONS OF WORK.

From these equations we find that

Ee ey

If P, Q, R be the impulses at B, C, D, we obtain, by
taking moments about A, B, C, for the rods AB, BC, CD,

Po, + ao, + a’n = 2aP,
a’ (w, + 20,) +k* (w, — n) = 2aQ,.
a? (w, + 20,) —a’n — kw, = 2ak,
and therefore Q = 0, 14P = na, 14R = — na.

Finally the impulse at A which is the change of linear
momentum of the system

= a0, + (ao, + 2aw,) + (aw, + 2aw,) + aw, = $ na.

250. A cube, the edges of which are twelve equal uniform
rods hinged together, is hung up by one corner, the cube form
being maintained by a string joining this corner with the
lowest corner. Itis required to find the initial change of stress
at the point of support when the string is cut.

é. EE

 

 

 

 

B

The corner O being the point of support and the diagonal
ILLUSTRATIONS OF WORK. 323

OD vertical, it is clear that the initial angular accelerations
of OA, OB, OC will be respectively in the planes AOD,
BOD, COD and will be equal to each other; and further
that the angular accelerations of all the other rods will
be the same and will be, respectively, in parallel planes,

If w represent this initial angular acceleration, 2a, which
we shall call 2f will be the linear acceleration of A in the
direction AD, of B in direction BD, and of C in direction CD.

Taking accelerations parallel to OA, OB, and OC, we
obtain the following forms, where K and JL are the centres of
the rods CE, ED.

The accelerations of Cc are

IV 2, fr/2, 9,
the accelerations of K relative to C are
. 0, v2, f/./2,
of E relative to C,
0, f2, f./2,

and of Z relative to E,
S/V2, 0, f//2.

Therefore the actual accelerations of K are

Iv/2, 3/2, fl’ 2,
8f)/2, 2fr/2, 8f]r/2.

Now suppose a displacement made by slightly increasing
the length of OD, so as to turn every rod through a small
angle 0.

The displacements of the various points follow the law of
the accelerations and are of the same forms, replacing o
by @.

Observing that there are six rods under the same condi-
tions as CEH, three under the same conditions as FD, and
three other rods OA, OB, OC, the equation of virtual
work is

6mfab (2 + $+ 4) + 3mfad (3 +8 + ¥)
2 2
+3m a. wf + 9m i w8 = 12mgabn/6,
21—2

and of Z
324 ILLUSTRATIONS OF WORK.

for the displacement of D is the resultant of the displace-
ments of A, B, and C.

From this equation we obtain
25aw = 3g 1/6,

and since the acceleration of D is 2aw /6, it follows that the
acceleration of G is aw 1/6 and is therefore 189/25.

Hence it follows that the diminution of stress at O is
18/25 of the total weight of the system.

The initial stresses at the several joints can be obtained
by giving independent displacements to the.several rods,
breaking the connections at the different joints.

251. The endeavour of this chapter has been to explain
the use of moving axes, and to illustrate the solution of
problems in three dimensions by the direct application of
fundamental conceptions and of equations derived imme-
diately from these conceptions.

It is hoped that these illustrations will serve as an
introduction to a very extensive field of thought, a field
which has been worked extensively by many mathematicians
with great power and success.

For the more general problems of the motions of solid
bodies on surfaces, for the great problems of Precession and
Nutation, and for the utilization and application of systems
of generalized coordinates, the student will consult such
English works as the Natural Philosophy of Sir William
Thomson and Professor Tait, Dr Routh’s Rigid Dynamics,
and the Treatise on Generalised Coordinates by Dr H. W.
Watson and Mr S. H. Burbury.

The student who has time at his disposal will study the
great works of Laplace and Lagrange, and those works, with
the English Treatises to which we have referred, and the
treatises and papers of Jacobi, Laurent, Bertrand, Resal,
Mathieu and numerous others, offer a vast area for considera-
tion, and suggest unlimited fields of research.
EXAMPLES. 825

EXAMPLES.

1. Two equal smooth circles are fixed so as to touch the
same horizontal plane, their planes being at different inclina-
tions; two small heavy beads are projected at the same
instant along these circles from their lowest points, the
velocity of each bead being that due to the height of the
highest point of the other circle above the horizontal plane,
shew that during the motion the two beads will always be at
equal heights above the horizontal plane.

2. A sphere is projected horizontally on an inclined
plane, the surface of which is perfectly rough; shew that its
centre will describe a parabola.

3. An imperfectly elastic rough sphere is projected
obliquely without rotation against a fixed plane; if 1,7’, be
the angles of incidence and reflection, A the coefficient of
elasticity for direct impact, and p the ratio of the tangential
forces of restitution and compression, prove that

29=5—T7rtanv coti.

4. Two particles of masses m, 2m are fixed to the ends
of a weightless rod of length 2a which is freely moveable
about its middle point. Prove that if @ be the inclination of
the rod to the vertical when the particles are moving with
uniform angular velocity o,

30a cos 0 = g.

5. A solid rectangular parallelepiped with edges of
length a, b, c, is acted on by instantaneous couples with axes
parallel to these edges and of moments proportional to
p:q:7; shew that the direction cosines of the instantaneous
axis of rotation are in the ratio

Oe a
P+ C+a° a+b

 
326 EXAMPLES.

6. If an octant of an ellipsoid bounded by three principal
planes be rotating about the axis a with angular velocity ,
and if this axis suddenly become free, and the axis 0 fixed,

hew that th lar velocity is "2
shew that the new angular velocity is Fee

7. <A ring, mass VM, is placed on a smooth table, and is
held fast while a circular disc of smaller radius and mass m,
is made to run round the inside of it, the coefficient of friction
being infinite. The ring being now set free determine com-
pletely the motion and the paths traced out by the centres of
the ring and disc respectively.

8. ‘There are n discs placed on a smooth table, with their
edges in contact, and their centres in a straight line, the
edges of the discs being perfectly rough. The first disc has
communicated to it a velocity u in a direction at right angles
to the line of centres. Prove that the velocity and angular
velocity of the pth disc are given by

= = O{(e+ a) (s=ey" + (« — a) (3) u,
(« = a)" + (« a ay )

and a, # are the radius and radius of gyration of each disc.

 

 

 

where

9. Two equal rods AB, BC are jointed at one extremity
B of each, and the other end A of one is fixed: if C be held
in such a position that ABC is a right angle and AC hori-
zontal, prove that when C is suddenly let go the initial
pressure at B will be + of the weight of either rod and hori-
zontal.

10. A sphere is moving at a given moment on an
imperfectly rough horizontal table with velocity V, and at
the same time an angular velocity round some horizontal
diameter, the angle between the direction of V and the axis
of O being a: prove that the centre will describe a parabola

if ak? O? + OV sin a (a? — hk’) -aV*? =0.
EXAMPLES. 827

11. One end of a heavy rod rests on a horizontal plane
and against the foot of a vertical wall, the other end rests
against a parallel vertical wall, all the surfaces being smooth.
Shew that if it slips down, the angle @ through which it
turns round the common normal to the vertical walls is given
by the equation,

d¢* 6 :
+ (1 +3 c0s%g J =~ Feptin gd +6

where 2a is the length of the rod, and 2b the distance
between the walls. .

12. Asmooth plate inclined at an angle ¢ to the horizon
is made to rotate about a vertical axis AB with uniform
angular velocity w. A rod of mass m is compelled by guides
to be always vertical, and at a distance r from AB, while it
rests with one end in contact with the plate, sliding up and
down as the latter rotates. Shew that, if the rod be initially
in its lowest position, the pressure on the plate at the end of
the time ¢ will be

m (g cos @ + rw’ cos wt sin d) sec’¢.
13. A square is revolving, with angular velocity ©, round
a diagonal which is vertical, the highest point of which

is fixed; if suddenly one of the angles adjacent to the fixed
point become fixed, shew that the square will just revolve

round the fixed side, if 0? = st

sures at the two fixed points are in the ratio 3:2; a being
the length of a side of the square.

, and the impulsive pres-

14. A smooth homogeneous cone is placed within a
hollow sphere circumscribing it, and set rotating about its
axis of figure with a given angular velocity; determine the
motion.

15. <A plane mirror whose thickness may be neglected
and whose centre of gravity is fixed, moves under the action.
of no forces so that its invariable plane is horizontal ; prove
that the image of any luminous point below the mirror and
328 EXAMPLES.

on the invariable line moves on the surface of a sphere so
that its zenith distance (@) and azimuth (yr) are connected by
the equation
dnp 2
—* =m +n cosec

dt a.

16. A sphere of radius c on a smooth horizontal plane is
placed in contact with a rough vertical plane, which is made
to revolve with angular velocity w about a vertical axis in
itself: if @ be the initial distance of the point of contact from
the axis, and r the distance at the time ¢, prove that

2r = (a + en/Z) VE" + (a — cn/Z) VI"
Also shew that the ratio of the friction to the pressure -
approximates, as ¢ increases indefinitely to 1 : /35.

17. A rough plane is made to revolve uniformly, with
angular velocity «, about a horizontal line in itself, and a
sphere is projected so as to move upon it, determine the
motion ; and if, when the plane is horizontal, the centre of
the sphere be vertically above the axis of revolution, and be
moving parallel to it, prove that the contact will cease when
the plane has revolved through an angle @ given by the
equation

11 cos 0 aw” ee
. = oO ers 12 5 (eve +e ev),

18. A frame consists of four equal uniform rods loosely
jointed at their ends so as to form a square, and one of the
rods carries a light ring fastened to it at its middle point.
The frame moves with uniform velocity on a table. All
kinds of friction being neglected, prove that when a vertical
bolt is shot through the ring the frame will be brought
absolutely to rest.

 

19. A solid cube is in motion about any angular point
which is fixed; if @,, ,,@, be the angular velocities about
the three edges through the fixed angular point, and if no
extraneous forces be in action, prove that w,+,+ 0, is
constant, and that ,’+ w,’ + @,’ is also constant.
EXAMPLES. 329

20. A circular wire ring of radius a rolls on a rough
horizontal plane, so that its plane maintains a constant in-
clination (a) to the vertical; if w be the angular velocity of
a ring, and © the azimuthal motion of its centre, prove
that \

daw, cos a — a? sin a cosa = 2g sin a.

.21. A hollow cone, the internal surface of which is
perfectly rough, is fixed in a position in which its axis is
inclined at an angle a to the vertical, and a solid cone of
smaller vertical angle is placed inside, its vertex coinciding
with the vertex of the fixed cone, and allowed to perform
small oscillations; prove that the length of the simple
isochronous pendulum is

4X? sin (8 — 7)
3A sinasin’y ’

2a and 2y being the vertical angles, h the height of the
moving cone, and & its radius of gyration about a generating
line.

22. A vertical hollow infinitely rough cylinder is move-
able about its axis. A sphere is projected horizontally in
contact with the cylinder. Shew that the cylinder will move

eduring the subsequent motion with a constant angular
velocity, and find its magnitude, having given V the velocity
of projection of the sphere before it touched the surface, a, m
the radius and mass of the sphere, and 6, M those of the
cylinder.

23. A sphere is. pressed between two perfectly rough
parallel boards which are made to revolve with the uniform
angular velocities Q and ’ about fixed axes A and B per-
pendicular to their planes. Prove that the centre of the
sphere describes a circle about an axis which is in the same
plane as A and B, and whose distances from these axes are
inversely proportional to 2 and 0”.

Shew that, when A and B coincide, the points of contact
will trace on the sphere small circles, the tangents of whose
330 EXAMPLES.

radii will be oe. , a being the radius of the sphere, and
a+ a
c that of the circle described by its centre.

24, If a rectangular parallelopiped (edges 2a, 2a, 2b)
move freely about its centre of gravity under no forces:
shew that its angular velocity about one principal axis is
constant and about the other principal axes is periodic, the
period being to the period of revolution about the first-
mentioned principal axis as 6?+ a" : b’— a’.

25. A system of two bodies is acted on by no forces, and
moving, one about a fixed axis, the other about an axis
fixed in the first, and parallel to the former axis; shew
under what circumstances the two bodies will revolve about
the fixed axis as a single body, and determine when disturbed
from such a. condition, whether they will oscillate or not,
and the time of a small oscillation. Prove that if the bodies
be two equal rods freely jointed at a common extremity and
moveable about the other extremity of one of them, the time

of a small oscillation = /7g x time of a complete revolution.

26. A body in the form of a solid of revolution has a
point in its axis attached by a universal joint and a string of
length 7 to a fixed point; an angular velocity » about its
axis of figure is given to the body, and nm is the angular
velocity of the centre of gravity about the vertical, @ the ‘
angle which the string, @ that which the axis of the body
makes with the vertical, a the distance between the centre of
gravity and the point of suspension; ( being the moment of
inertia about the axis of figure. Shew that when the motion
is steady ,

(A -C) nsin $ (w — n cos ¢) cos 6 = Mga sin (¢ — @)
w (Jsin 8+ asin $) = tan 0.
27. A ring rests upon two smooth horizontal bars which
in the position of equilibrium subtend an angle 2a at the

centre; shew that, if the ring be disturbed by twisting it
through a small angle about its vertical diameter, the length

of the simple isochronous pendulum will be 5 cot a cosec a.
EXAMPLES. 331

28. A smooth circular tube, of radius a, has its centre
fixed at a height h above a fixed horizontal plane, which it
touches, a particle of mass m is placed within it at a distance
c below the horizontal plane through the centre of the tube
which becomes z at the time ¢, prove that, if the moment of
inertia of the tube about a diameter be nma?

(5) _ 29 (2-0) (h? — 2’) (nh? + a? — 2°)
dt/ a (n+l) h?—-# :
If the particle be initially in the horizontal diameter, prove

‘that the initial pressure between the tube and particle will
be less than if the ring were fixed, in the ration : n+1.

 

29. A sphere is rolling on the rough surface of a
cylinder, the cross section of which is the curve

8r=aV2&V?;

prove that, if there be no forces, the path of the point of
contact becomes, when the cylinder is developed into a plane,
a curve of the form,

-y = (a+ Bx) cos (tog “) + (y + 62) sin (log *) :

30. A sphere moves under the action of gravity on the
inside of a rough cylindrical surface, of which the generating
lines are inclined at an angle a to the horizon, and the
transverse section perpendicular to the generating lines is a
cycloid with its vertex at the lowest generating line.

The sphere is projected initially with a velocity V along
the generating line at which the curvatures of the sphere
and cycloid are equal.

Prove that the motion will be comprised within a length
2a

male v5 COS a

successive instants of the sphere reaching the original

generating line is 4r a Se

5g cos a

the generating circle of the cycloid.

 

of the cylinder, and that the time between

, where a is the radius of
332 EXAMPLES,

31. A perfectly rough circular disc, of which the centre
is fixed, is constrained to roll with its edge on a horizontal
plane, so that the point of contact moves with angular
velocity n.

A sphere is placed on the disc in contact with a horizontal
diameter, and in a state of relative rest. Obtain the
equations of motion, and shew that the sphere moves down
the disc as if that were at rest and the sphere a smooth
heavy particle, and that the trace of the point of contact on
the disc, referred to axes in its plane, that of y being
horizontal, is the curve

2/2e cosa —? [2a COS a:
yes fer ne oes

g ,

the disc being inclined at an angle a to the vertical, and ¢
being the initial value of y.

32. Ata point P of the earth’s surface a sphere has its
centre O fixed-and by means of pegs at the extremity C of a
diameter, the diameter OC is compelled to move in the
meridian at P; if originally OC be parallel to the axis of
the earth and the sphere have angular velocity about OC;
prove that if the sphere be disturbed OC will oscillate in the
meridian in a time

Qa

Jnw
where » is the angular velocity of the earth about its
axis.

33. <A solid of revolution has a point O in the circum-
ference of its base fixed and receives a blow in a direction
passing through G its centre of gravity and parallel to the
tangent line to the base at O. If the principal moments of
inertia at G in directions perpendicular to’ the blow are
proportional to those about parallel axes through O, prove
that the instantaneous axis and OG are equally inclined to
the plane of the base.
EXAMPLES. 333

34. A heavy, uniform, and inextensible string is in
equilibrium in the form of a horizontal ring on a smooth
sphere ; prove that, if it be cut at a point A, the initial change
of tension at a point P will be to the weight of the string in
the ratio ;

cos h (cos a) : 2a cot a cos h (a cosa),

a being the angular distance of the string from the vertex of
the sphere, and w—@ the angle subtended at the centre of
the ring by the are PA. Tripos, Jan. 1883.

35. A rough heavy sphere, radius‘e, rolls on a fixed
rough surface, of the form generated by the revolution of a
circle, radius 0, about a vertical axis in its own plane, distant
a from the centre, a being greater than b+c. Prove that if,
at the time ¢, @ be the angle through which the plane
through the vertical axis and centre of the sphere ‘has
turned, @ the inclination to the vertical of the common
normal, w, the angular velocity about that common normal, a
and \ the initial values of 6 and ¢, and if 6 and a, be initially
zero, and l=b++¢,

ca, = abd,
PO + (a —Isin 6) ¢? — 2 (a—Jsin a) + 30,”

= 1Pgl (cos a — cos @),

= \@- isin 6)’ “ash + 2aw, (a —/sin 0) = 0.

36. A perfectly rough plane, inclined at a fixed angle
to the vertical, rotates about a vertical line with uniform
angular velocity; shew that the path of a sphere which is
placed upon it is given by two equations of the forms

jtar+by=0, é-ayt+ a=,

the origini being the point where the vertical line meets the
plane, and the axis of y being the straight line in the plane
which is always horizontal.
334 EXAMPLES,

37. Two equal spheres attracting each other, the force
varying as the distance, are rolled upon a perfectly rough
horizontal plane. Prove that they will describe ellipses
about each other in the periodic time

Dit ef oe

10."

If the plane revolve with a uniform angular velocity w
about a vertical axis, prove that their centre of gravity will

move in a circle with uniform angular velocity 2 and that

their relative orbits’ will be such that each will appear to the
other to describe a circle with uniform angular velocity

10u o ow
Vr ee

while the centre of that circle moves with uniform angular
velocity

in another circle.

 

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April 1884.
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