An introduction to the mathematical theory of attraction ...

146 The Potential.
also, since br is zero, Ni becomes Ni + L, where
3m
Z = 3 (G-B) { (xg-U0) + (z0 - x)}
(B -C) (y2- 2) - xy X + X },
and as the original position is one of equilibrium, Ni = O. Similar results hold
good for N2 and N3. Again, if ~ be the magnitude of the rotation which brings
the body into its actual position, a, 3, y the direction cosines of its axis, and
W the work done by the attraction of the sphere in the displacement,
W = ( (La + M + N7) dor.
Substituting their values for L, M, 1,, and remembering that 0 = ~a, 4> = aB,, = 7y, we get by integration 2 W = LO + M14 + N_. Since Ni, N2, N3 are
each zero, so also are the products xy, yz, zx; hence the point P lies on one of
the principal axes of the body, and
2 W= 3n {(B- C)(y2-z2) 02 + (C-A)(z2 -x2) 2+ (a-A-B)(X2 -y2) 2}
In a position of stable equilibrium Wmust be negative for all values of 0, p, 4.
Let A >B>; then if x= O andy = 0,
3m
2wy = - 5 2 {(B- ) 2 + (A - C) 2}.
This is always negative, and the equation of the potential ellipsoid is
(B - C) X2 + (A - C) y2 = constant, which represents a cylinder. For stable
equilibrium, the centre of the sphere must be situated on the production of the
axis of least inertia of the body.
If the body be a homogeneous ellipsoid of small ellipticity, the preceding
investigation applies even though the sphere be not distant.
5. A rigid body K having its centre of inertia fixed is attracted by a distant
immoveable homogeneous sphere; the initial position of K being given, determine the impulsive couple which must act on it in order that its axis of rotation
should be invariable.
If K be free, find the position and motion which must be given to it initially
in order that it should continue to revolve round the same axis.
If wi, a2, w3 be the angular velocities of the body round its principal axes,
its equations of motion (" Dynamics," Art. 267) are
A wi 3rn
A -(B - C) w2 W3 = N = (- B) yz, &c.;
and dt 0, provided 2 w = 3m5
and -tprovided au = /z.
dt