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.
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?
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1 answer