Wikipedia:Reference desk/Archives/Mathematics/2013 June 23

= 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)


 * 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)