TITLE: Doesn't rotational KE of a rolling marble change if there is no friction to provide torque?
QUESTION [1 upvotes]: The question arise from the following situation:
A marble at the border of a uniform bowl begins rolling within it from rest. There is enough friction in the first half the bowl for the marble to not slip, but there's no friction in the other half. Find the height reached by the marble, measured from the bottom of the bowl.
I read the solution, but I can't undestand the following:
When marbel arives to the bottom, it has both translational an rotational KE. The solution says when marble stops at the maximun height, its rotational KE is still the same as when it was at the bottom (because there's no friction to provide torque). Does it mean that marble is still rolling? Could anyone explain why would it be wrong if I propose that RKE is zero at the maximun height?

REPLY [2 votes]: Here's the marble in its starting position:

During its descent on the side with friction, potential energy $U$ is converted to kinetic energy $K$. Assuming there's enough friction for rolling without slipping ($v=\omega R$) then kinetic energy is partitioned into translational and rotational energy, so:
$$U=K$$
$$mgy=\frac12mv^2+\frac12I\omega^2$$
$$mgy=\frac12mv^2+\frac12\frac{I}{R^2}v^2$$
If $I$ is known, like in the case of a marble, $v$ can be calculated from that equation. For example for a solid sphere like a marble, $I=\frac25 mR^2$, so:
$$v^2=\frac{10}{7}gy$$
Now when the marble starts travelling upwards in the friction-free zone, its rotational state of motion remains unaltered, as per Newton's law, because no friction also means no torque that can affect $\omega$. Effectively, rotational kinetic energy remains constant during the climb.
The kinetic energy available for conversion to potential energy is thus only:
$$K_{trans}=mgy-\frac12\frac{I}{R^2}v^2$$
This energy is converted back to potential energy, acc.:
$$mgy_2=mgy-\frac12\frac{I}{R^2}v^2$$
So:
$$y_2=y-\frac12\frac{I}{mgR^2}v^2$$
Reworked with the above we get:
$$y_2=\frac57 y$$
Note that there's no loss of energy: part of the initial kinetic energy remains owned by the marble due to its rotation.
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