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June 23

Kinetic energy of symmetric top in an unusual parametrization

Suppose the orientation of the top's axis of symmetry is given by a vector $\mathbf {e}$ and its rotation about that axis is given by an angle $\phi$ . What is the kinetic energy of the top? I can do it in the case when the inertia tensor is singular (so all the mass lies on the vector $\mathbf {e}$ ) but not in the more general case. Anyone got any ideas?--Leon (talk) 08:13, 23 June 2013 (UTC)[reply]

I must be missing something, but wouldn't it just be

{\frac {1}{2}}I\phi ^{2}

where I is the moment of inertia. (I'm assuming that by angle above you mean angular speed.) Sławomir Biały (talk) 11:10, 25 June 2013 (UTC)[reply]