Frictional conversion of spin to rolling ($\Delta K=-\frac{MI}{2(I+M{a}^2)}(v_0-a{\omega }_0{)}^2$)

For a circular rigid body of radius a, mass M and axial moment of inertia I on a fixed horizontal surface, choose positive spin so that rolling means $v=a\omega$. During purely horizontal frictional relaxation, an impulse J changes $Mv$ by J and $I\omega$ by $-aJ$. Hence $I\omega +Mav$ is constant. At final rolling without slipping, $v_f=(Ia{\omega }_0+M{a}^2{v}_0)/(I+M{a}^2)$. Subtracting initial from final translational-plus-rotational kinetic energy gives the title formula. The coefficient of friction controls the relaxation time, but not this loss once the no-slip state is reached. Vertical landing losses and rolling resistance are excluded.