# Rolling on a track in vertical plane

A sphere of mass m and radius r is released from rest at point A on a track in vertical plane. The track is rough enough to support rolling between A and B and from B onwards it is smooth. My thinking: From A to B it will roll and friction will do no work so mechanical energy at A = mechanical energy at B. After B since no friction is present and it going on an incline it cannot roll. Hence the height reached after B is proportional to it's translational kinetic energy only.

My question: If it cannot roll after B what will happen to it's rotational kinetic energy?

Note: The above question is not a homework question, it is a doubt that I have while solving the homework question.

## 1 Answer

Nothing will happen to the rotational Kinetic energy,since there is no friction there is no torque about the center of mass of the object and hence its angular velocity about center of mass must remain constant. You said,

"It cannot roll between B and C"

This is partially true,it can roll but cannot be in a state of pure rolling. Usually an object requires frictional force to attain angular velocity sufficient enough for the "pure rolling" state($$v_{com}=\omega*r$$),but here when the ball leaves point B it has a certain amount of angular velocity so it will rotate about the center of mass until an external torque is given to oppose this velocity.

So, rotational kinetic energy is constant between B and C. i.e it will keep rotating with the same angular velocity at point C as it had in point B.