 Solution

For the frictional conversion of spin to rolling, consider the horizontal slipping stage on the rough surface, whose contact radius is now $a$, and neglect extra vertical impact or height-change energies at the transition. A friction impulse $J_f$ changes $Mv$ by $J_f$ and clockwise spin $I\omega$ by $-a{J}_f$. Thus $I\omega +Mav$ is conserved. At the end $v_f=a{\omega }_f$, givingBecause $I=M{a}^2$, the final total kinetic energy is $K_f=M{v}_f^2$. Subtraction yieldsThe result is independent of the friction coefficient, provided sufficient contact time allows rolling without slipping to be regained. Initially $v_0=b{\omega }_0<a{\omega }_0$, so the contact point slips backwards. Friction acts forwards and can increase translational kinetic energy, while decreasing rotational kinetic energy by more. A finite vertical drop or inelastic landing would require additional geometric information; the boxed expression describes the horizontal rolling-conversion loss.