Question

A solid ball of mass 2 kg, rolls down a hill that is 6 meters high.  What is the rotational KE at the bottom of the hill?

Answer 1

This is a little complicated, it is rolling, and translating .

total energy=mgh
let translational velocity be v.
then w=v/r
translational KE=1/2 m v^2=1/2 m w^2r^2
rotational KE= 1/2 I w^2=1/2*2/5 mr^2*w^2
= 1/5*mv^2
but total energy=sum of above
mgh=(1/5+1/2)mv^2=0.7 mv^2
solving for v
v^2=gh/.7
and then rotational KE= .2*m*v^2

check all that.