A hollow, thin-shelled cylinder of mass rolls along a horizontal surface with a translational speed of v. What percent of its total kinetic energy is translational?

Answer 1

translational=1/2 m v^2
rotational=1/2 I w^2 where w=v/r
rotaltonal=1/2 I v^2/r but I for thin shell cylinder is http://en.wikipedia.org/wiki/List_of_moments_of_inertia
rotational=1/2 mr^2*v^2/r^2=1/2 m v^2

Hmmm. Looks like 1/2 of the energy is translational.