# Precession of angular velocity about the body-fixed axis

My textbook mentions that under force-free motion of a symmetric top, its angular velocity vector $\overrightarrow \omega$ precesses about the $z$-axis of the body-fixed coordinate system. This seems impossible to me. Assuming that the axis of symmetry of the top is the $z$-axis, how can $\overrightarrow \omega$ point in any direction other than the $z$-axis? It's got to rotate about the $z$-axis and hence point along it. What am I missing?

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The general derivation is done on the assumption that the symmetry axis of the top is not vertical. It is, however, interesting to ask what happens when it is. –  dmckee Sep 20 '12 at 4:25
@Joebevo (possibly off topic): no reason for typesetting physics to be a pain, which it will be if you have to type \overrightarrow all the time. Try \vec{x} instead. It also formats better :) –  Chris White Sep 20 '12 at 5:53

You can spin a symmetric top around axes other than the symmetry axis: your text is considering the general case of an arbitrary rotation, and how the motion evolves in that case.

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You're right. I read too quickly and ended up answering a different question. Oops. –  Chris White Sep 20 '12 at 6:29
@ChrisWhite: no problem. I beat my brains out trying to understand Feynman's wobbling plate, so I was tuned to this scenario. –  Art Brown Sep 20 '12 at 6:34
I think that it is too confusing. A top by definition spins around its body axis, otherwise it is not a top. If there exists a force, gravity for example, then omega, which will be pointing in the z direction of the body coordinate may, if given an impulse, precess around the perpendicular defined by gravity. If there is no force, as the question states, why would there be a precession, about what axis? –  anna v Sep 20 '12 at 11:44
@annav: The defining characteristic of symmetric top is that it has two equal principal moments of inertia. A plate, for example, qualifies. And the angular velocity need not be parallel to the symmetry axis. One famous example is Feynman's wobbling plate; see: physics.stackexchange.com/questions/15082/… . In such a situation, the angular momentum vector is not parallel to the angular velocity. With no applied torque, angular momentum is constant, but not angular velocity. –  Art Brown Sep 20 '12 at 16:11
oops, within the body frame the angular momentum will also appear to precess. The other thing is that when the texts talk about angular velocity with respect to the "body frame" they actually mean a fixed frame that's instantaneously aligned with the body axes. Yes, it's confusing. –  Art Brown Sep 20 '12 at 17:29

Think of a single particle, an electron, moving in the x y plane and a magnetic field perpendicular to its direction of motion, the z direction.

The electron will trace a circle, the angular velocity is associated with the electron and not the z axis passing through the center of its circle. Just the z direction. As the electron could be described as precessing so one could describe the omega associated with it as precessing . I agree it is a confusing terminology for a rigid body.

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