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If the motion is confined to one plane, that of r, 0, we have

= 0, and therefore H= 0, and the two equations of motion

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These equations might have been written down at once in terms of the second law of motion from the kinematical investigation of

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are the components of acceleration along and perpendicular to the radius-vector, when the motion of a point in a plane is expressed according to polar co-ordinates, r, 0.

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problem.

The same equations, with instead of 0, are obtained from the polar equations in three dimensions by putting 0 = 1, which implies that G = 0, and confines the motion to the plane (r, 4). Example (B).-Two particles are connected by a string; one Dynamical of them, m, moves in any way on a smooth horizontal plane, and the string, passing through a smooth infinitely small aperture in this plane, bears the other particle m', hanging vertically downwards, and only moving in this vertical line: (the string remaining always stretched in any practical illustration, but, in the problem, being of course supposed capable of transmitting negative tension with its two parts straight.) Let be the whole length of the string, r that of the part of it from m to the aperture in the plane, and let be the angle between the direction of r and a fixed line in the plane. We have

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Also, there being no other external force than gm', the weight of the second particle,

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Case of stable equilibrium due to motion.

Examples continued; C (a), folding door.

The motion of m' is of course that of a particle influenced only by a force towards a fixed centre; but the law of this force, P (the tension of the string), is remarkable. To find it we have (§ 32), P=m(-+r). But, by the equations of the motion,

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where (according to the usual notation) denotes the moment of momentum of the motion, being an arbitrary constant of integration. Hence

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The particular case of projection which gives m a circular motion and leaves m' at rest is interesting, inasmuch as (§ 350, below) the motion of m is stable, and therefore m' is in stable equilibrium.

Example (C).-A rigid body m is supported on a fixed axis, and another rigid body n is supported on the first, by another axis; the motion round each axis being perfectly free.

Case (a).-The second axis parallel to the first. At any time, t, let and be the inclinations of a fixed plane through the first axis to the plane of it and the second axis, and to a plane through the second axis and the centre of inertia of the second body. These two co-ordinates, 4, 4, it is clear, completely specify the configuration of the system. Now let a be the distance of the second axis from the first, and b that of the centre of inertia of the second body from the second axis. The velocity of the second axis will be ap; and the velocity of the centre of inertia of the second body will be the resultant of two velocities

ap, and by,

in lines inclined to one another at an angle equal to 4-, and its square will therefore be equal to

a2¿a + 2ab↓↓ cos (¥ − $) + b2¿3.

Hence, if m and n denote the masses, j the radius of gyration of the first body about the fixed axis, and k that of the second

body about a parallel axis through its centre of inertia ; we have, Examples according to $ 280, 281,

T = { {mj3 ¿2 + n [a2¿2 + 2ab$$ cos (4 –− $) + b2¿a + k2 4o]}.

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= nab cos (4 − $) $+n(b2+k2) 4 ;

nab sin (-) $4.

The most general supposition we can make as to the applied forces, is equivalent to assuming a couple, , to act on the first body, and a couple, , on the second, each in a plane perpendicular to the axes; and these are obviously what the generalized components of stress become in this particular co-ordinate system, 4, 4. Hence the equations of motion are

(−)]

(mj2 + na2) ÿ + nab d [4 cos (4-4)] – nab sin (4-4) ¿↓ = ®,

d

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nab [$ cos (4-4)]

dt

dt

- $) $4

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If there is no other applied force than gravity, and if, as we may suppose without losing generality, the two axes are horizontal, the potential energy of the system will be

gmh (1 − cos p) + gn {a [1 − cos (p + A)] + b [1 − cos (↓ + A)]}, the distance of the centre of inertia of the first body from the fixed axis being denoted by h, the inclination of the plane through the fixed axis and the centre of inertia of the first body, to the plane of the two axes, being denoted by A, and the fixed plane being so taken that = 0 when the former plane is vertical. By differentiating this, with reference to 4 and 4, we therefore have

- Þ = gmh sin & + gna sin (p + 4), − ¥ = gnb sin († + A).

We shall examine this case in some detail later, in connexion with the interference of vibrations, a subject of much importance in physical science.

When there are no applied or intrinsic working forces, we have = 0 and 0: or, if there are mutual forces between the two bodies, but no forces applied from without, ☀ + ¥ = 0. In

continued; C (a), folding door.

Examples of the use of Lagrange's generalized equations of motion.

C (b). Motion of governing

masses in Watt's centrifugal governor: also of gimballed

compassbowl.

either of these cases we have the following first integral:-
:-

(mj3 + na3) $ + m'ab cos (4 − $) ($ + ¿) + n (b2 + k3) 4 = C ;
obtained by adding the two equations of motion and integrating.
This, which clearly expresses the constancy of the whole moment of
momentum, gives & and in terms of (-) and (4-4). Using
these in the integral equation of energy, provided the mutual forces
are functions of -, we have a single equation between
d (4-4)
(4-4), and constants, and thus the full solution of
the problem is reduced to quadratures. [It is worked out fully
below, as Sub-example G,.]

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Case (b). The second axis perpendicular to the first. For simplicity suppose the pivoted axis of the second body, n, to be a principal axis relatively [§ 282 Def. (2)] to the point, N, in which it is cut by a plane perpendicular to it through the fixed axis of the first body, m. Let NE and NF ben's two other principal axes. Denote now by

h the distance from N to m's fixed axis;

k, e, f the radii of gyration of n round its three principal axes through N;

j the radius of gyration of m round its fixed axis;

the inclination of NE to m's fixed axis;

the inclination of the plane parallel to n's pivoted axis through m's fixed axis, to a fixed plane through the latter.

Remarking that the component angular velocities of n round NE and NF are cos and i sin 0, we find immediately

or,

T = {} {[mj2 + n (h3 + e2 cos2 0 +ƒ3 sin3 0)] ¿a + nk2 02}, if we put

mj + n(l +f*) = G, n (ẻ - f*) = D;

-

T = {{(G+D cos2 0) ¿2 + nk2 02}.

The farther working out of this case we leave as a simple but most interesting exercise for the student. We may return to it later, as its application to the theory of centrifugal chronometric regulators is very important.

a rigid body

one of its

axes mount

Example (C'). Take the case C (b) and mount a third body M Motion of upon an axis OC fixed relatively to n in any position parallel to pivoted on NE. Suppose for simplicity O to be the centre of inertia of M principal and OC one of its principal axes; and let OA, OB be its two ed on a gimballed other principal axes relative to 0. The notation being in other bowl. respects the same as in Example C (b), denote now farther by A, B, C the moments of inertia of M round OA, OB, OC; & the angle between the plane AOC and the plane through the fixed axis of m perpendicular to the pivoted axis of n; w, p, σ the component angular velocities of M round QA, OB, OC.

In the annexed diagram, taken from § 101 above, ZCZ' is a

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circle of unit radius having its centre at O and its plane parallel
to the fixed axis of m and perpendicular to the pivoted axis
of n.

The component velocities of C in the direction of the arc ZC
and perpendicular to it are and sin; and the component
angular velocity of the plane ZCZ' round OC is cos 0. Hence

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