= 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 $-aJ_f$. Thus $I\omega+Mav$ is conserved. At the end $v_f=a\omega_f$, giving
$$
\boxed{v_f=\frac{v_0+a\omega_0}{2},\qquad \omega_f=\frac{\omega_0+v_0/a}{2}.}
$$
Because $I=Ma^2$, the final total <kinetic energy> is $K_f=Mv_f^2$. Subtraction yields
$$
\boxed{\Delta K=-\frac M4(v_0-a\omega_0)^2=-\frac{M(R-r)^2\omega_0^2}{16}.}
$$
The 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.
\Image[/past-exam-of-the-physics-1-paper-of-the-university-of-oxford/2016-resit-drum-rolling-transition.png]
{title=Axle rolling, the formal impulse height and friction-driven conversion to rolling on the outer drum}
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