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

RE=1/2 *I w^2=.5*v^2/r^2
But I=2/5 mr^2

so TE/Re= (1/2 mv^2)/(Mr^2*v^2/5)=2.5

check