A thin-walled sphere rolls along the floor. What is the ratio of its translational kinetic energy to its rotational kinetic energy through its center of mass?

Answer 1

rotational:

omega = v/r = angular velocity
I = (2/3) m r^2

Ke = (1/2)I omega^2
= (1/2)m (2/3)mr^2 v^2/r^2
= (1/2)m (2/3)v^2
but
translational Ke = (1/2) m v^2
so
2/3

Answer 2

sorry 3/2 the way it is phrased