Simple Nature - Light and Matter

We can also generalize the plane-rotation equation K = (1/2)Iω 2to three dimensions as follows:K = ∑ 12 m ivi2i= 1 ∑m i (ω × r i ) · (ω × r i )2iWe want an equation involving the moment of inertia, and this hassome evident similarities to the sum we originally wrote down forthe moment of inertia. To massage it into the right shape, we needthe vector identity (A×B)·C = (B×C)·A, which we state withoutproof. We then writeK = 1 ∑m i [r i × (ω × r i )] · ω2i= 1 2 ω · ∑m i r i × (ω × r i )= 1 2 L · ωiAs a reward for all this hard work, let’s analyze the problem ofthe spinning shoe that I posed at the beginning of the chapter. Thethree rotation axes referred to there are approximately the principalaxes of the shoe. While the shoe is in the air, no external torques areacting on it, so its angular momentum vector must be constant inmagnitude and direction. Its kinetic energy is also constant. That’sin the room’s frame of reference, however. The principal axis frameis attached to the shoe, and tumbles madly along with it. In theprincipal axis frame, the kinetic energy and the magnitude of theangular momentum stay constant, but the actual direction of theangular momentum need not stay fixed (as you saw in the caseof rotation that was initially about the intermediate-length axis).Constant |L| givesL 2 x + L 2 y + L 2 z = constant .In the principal axis frame, it’s easy to solve for the componentsof ω in terms of the components of L, so we eliminate ω from theexpression 2K = L · ω, giving1I xxL 2 x + 1I yyL 2 y + 1I zzL 2 z = constant #2.The first equation is the equation of a sphere in the three dimensionalspace occupied by the angular momentum vector, whilethe second one is the equation of an ellipsoid. The top figure correspondsto the case of rotation about the shortest axis, which hasthe greatest moment of inertia element. The intersection of the two286 Chapter 4 Conservation of Angular Momentum