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Motion of

The kinetic energy of the motion of M relatively to 0, its a rigid body pivoted on centre of inertia, is ( 281) one of its principal axes mount

3 (* + Bp° + °); ed on a gimballed and (S 280) its whole kinetic energy is obtained by adding the bowl.

kinetic energy of a material point equal to its mass moving with
the velocity of its centre of inertia. This latter part of the
kinetic energy of M is most simply taken into account by sup-
posing n to include a material point equal to M placed at 0);
and using the previous notation k, e, f for radii of gyration of n
on the understanding that n now includes this addition. Hence
for the present example, with the preceding notation G, D, we
T'=} {(G + D cos' 0) + nk* 0}
+ A (0 sin $ - 4 sin 6 cos 6) + B (è cos 0 + 4 sin 6 sin )

+C (j cos 0 + 0)}. Rigid body From this the three equations of motion are easily written down. rotating freely; referred to

By putting G = 0, D = 0, and k = 0, we have the case of the the v, 4,0 co-ordinates motion of a free rigid body relatively to its centre of inertia. ($ 101).

By putting B = A we fall on a case which includes gyroscopes Gyroscopes and gyrostats of every variety; and have the following much and &yrostats. simplified formula :

T=} {[G+ A + (D A) cos’O] + (nik® + A) 8 + C (4 cos 0 + 0)"},


T = }{(E+F cos? ) je +(nko + A) Q* + C (cos 0 + 0)"}, if we put E = G + A, and F= D-A.

Gyroscopic pendulum.

Example (D).-Gyroscopic pendulum.-A rigid body, P, is attached to one axis of a universal flexure joint (§ 109), of which the other is held fixed, and a second body, Q, is supported on P by a fixed axis, in line with, or parallel to, the first-mentioned arm of the joint. For simplicity, we shall suppose Q to be kinetically symmetrical about its bearing axis, and OB to be a principal axis of an ideal rigid body, PQ, composed of P and a mass so distributed along the bearing axis of the actual body Q as to have the same centre of inertia and the same moments of inertia round axes perpendicular to it. Let AO be the fixed arm, O the joint, OB the movable arm bearing the body P, and coinciding with, or parallel to, the axis of Q. Let BOA' = 0; let 6 be the angle which the plane AOB makes with a fixed plane of reference, Gyroscopic

pendulum. through 0 A, chosen so as to contain a second principal axis of the imagined rigid body, PQ, when OB is placed in line with A0; and let

I y be the angle between a plane of reference in Q through its axis of symmetry and the plane of the two principal axes of PQ already mentioned. These three co-ordinates (0, 0, 4)

B clearly specify the configuration of the system at any time, t. Let the moments of inertia of the imagined rigid body PQ, round its principal axis OB, the other principal axis referred to above, and the remaining one, be denoted by A, B, C respectively; and let A' be the moment of inertia of Qround its bearing axis.

We have seen ($ 109) that, with the kind of joint we have supposed at 0, every possible motion of a body rigidly connected with OB, is resolvable into a rotation round 01, the line bisecting the angle AOB, and a rotation round the line through ( perpendicular to the plane AOB. The angular velocity of the latter is è, according to our present notation. The former would give to any point in OB the same absolute velocity by rotation round 01, that it has by rotation with angular velocity 4 round AA'; and is therefore equal to

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This may be resolved into 20 sino 10 = (1 - cos ) round OB, and 20 sin 10 cos 0= sin round the perpendicular to OB, in plane AOB. Again, in virtue of the symmetrical character of the joint with reference to the line 01, the angle d, as defined above, will be equal to the angle between the plane of the two first-mentioned principal axes of body P, and the plane AOB. Hence the axis of the angular velocity • sin 0, is inclined to the principal axis of moment B at an angle equal to . Resolving therefore this angular velocity, and è, into components round the axes of B and C, we find, for the whole component angular velocities of the imagined rigid body PQ, round these axes, o sin 6 cos $+ sin , and - sin o sin $ + cos , respectively. The whole kinetic energy, T, is composed of that of the imagined rigid body PQ, and that of Q about axes through its centre of


Gyroscopio pendulum.

inertia : we therefore have 2T=A(1-сos 6)* $*+B ($sin 6 cos$+ėsino)+C($sin 6 sin$- İcoso)

+ ' {-(1-сos 6)). d

dT Hence A' {- (1 - cos ()}, = 0,

dy dT

A (1 – cos 6)* $ + B (o sin 0 cos 0 + 0 sin ) sin 6 cos

+C(sin sin 0-0 cos) sinsin $-A'{4-4(1- cos )}(1-cos), dT

B (ď sin 0 cos 0 + 0 sin ®) (sin sin 0 - 0 cos ®) do

+C(sin sin 0-0 cos) (osin coso+isino), dT

= B (sin cos$+) sin )sin -C(sin 6 sin • -, cos ) cos o de

and A (1 – cos 6) sin 08° + B cos 0 cos 00 (sin 0 cos 0+sin o)

+Ccos 6 sin ( sin sin p-, cos 6)-A' sin 60 {-(1- cos 6) }.

Now let a couple, G, act on the body Q, in a plane perpendi-
cular to its axis, and let L, M, N act on P, in the plane perpen-
dicular to OB, in the plane A’OB, and in the plane through OB
perpendicular to the diagram. If y is kept constant, and $
varied, the couple G will do or resist work in simple addition
with L. Hence, resolving L +G and N into components round
01, and perpendicular to it, rejecting the latter, and remembering
that 2 sin 100 is the angular velocity round 01, we have
Q=2 sin 30{-(L+G) sin }O+Ncos 10}={-(L+G)(1-cos()+N sin 6}.
Also, obviously

V = G, M = M. Using these several expressions in Lagrange's general equations (24), we have the equations of motion of the system. They will be of great use to us later, when we shall consider several particular cases of remarkable interest and of very great importance.

Example of varying relation without constraint (rotating axes).

Example (E).- Motion of a free particle referred to rotating axes. Let x, y, z be the co-ordinates of a moving particle referred to axes rotating with a constant or varying angular velocity round the axis 02. Let x,, Y., z, be its co-ordinates referred to the same axis, O2, and two axes 0X,, OY,, fixed in the plane per

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pendicular to it. We have

Example of

varying , = x cos a – y sin a, y = x sin a + y cosa;


without a cos 0-2 sin - (sin a + 2 cos a) i, j, = etc. constraint


axes). where

a, the angle 8,0X, must be considered as a given function of t. Hence

T = {m {i* + jo + 32 + 2 (xy ) å + (aco + yo) ä'},

= m (i - ya), m (j + ca), = mi,


= m (+ ää?),
= m (- cà + yaʼ), da




d d
at da

= m (ä – ja -- ), āt dij

d IT

= m (i + că + ää),

and hence the equations of motion are

m (ä - 2ja – xå? – ) = X, m (y + 2.că ya® + xä) = Y, = 2, X, Y, Z denoting simply the components of the force on the particle, parallel to the moving axes at any instant. In this example t enters into the relation between fixed rectangular axes and the co-ordinate system to which the motion is referred; but there is no constraint. The next is given as an example of varying, or kinetic, constraint.

Example (F).--A particle, influenced by any forces, and at- Example of

varying tached to one end of a string of which the other is moved with any relation constant or varying velocity in a straight line. Let o be the kinetic

constraint. inclination of the string at time t, to the given straight line, and • the angle between two planes through this line, one containing the string at any instant, and the other fixed. These two coordinates (0, 0) specify the position, P, of the particle at any instant, the length of the string being a given constant, a, and the distance OE, of its other end E, from a fixed point, 0, of the line in which it is moved, being a given function of t, which we shall denote by u. Let x, y, z be the co-ordinates of the particle referred to three fixed rectangular axes. Choosing OX as the given straight line, and YOX the fixed plane from which is measured,

we have

x = x + a cos 0, y = a sin 0 cos $, z = a sin o sin ,

d = - a sin 0);

Example of varying relation due to kinetic constraint.

and for y, z we have the same expressions as in Example (A). Hence

T=T + } m (i' - 2 iba sin 6) where I denotes the same as the T of Example (A), with 3 = 0, and r = a. Hence, denoting as there, by G and H the two components of the force on the particle, perpendicular to EP, respectively in the plane of 8 and perpendicular to it, we find, for the two required equations of motion,

d (sin' 10) m{a (ö – sin 0 cos 0$) – sin 0 ü} = G, and ma


dt These show that the motion is the same as if E were fixed, and a force equal to – were applied to the particle in a direction parallel to EX; a result that might have been arrived at at once by superimposing on the whole system an acceleration equal and opposite to that of E, to effect which on P the force – is required.

Example (F'). Any case of varying relations such that in 318 (27) the coefficients (4,4), (4,4)... are independent of t. Let T denote the quadratic part, L the linear part, and K [as in § 318 (27)] the constant part of T in respect to the velocity components, so that

T = } {(4,4) 49 + 2 (4,0) 40 +(4,0) +
L = (4) 4 + (0) $+ ..

K = (4, , 0, ...)
where (4,4), (4, p), (4,0) ... denote functions of the co-ordi-
nates without t, and (y), (), ..., (4, 0, 0, ...) functions of the
co-ordinates and, may be also, of t; and
T=T + L + K........

....(6). dK We have


Hence the contribution from K to the first member of the y-

equation of motion is simply - Again we have


(4) ;

d d
dt dy dy


d (4) 6 + etc. +

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