= Frictional conversion of spin to rolling
{title2=$\Delta K=-\frac{MI}{2(I+Ma^2)}(v_0-a\omega_0)^2$}
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>, $v_f=(Ia\omega_0+Ma^2v_0)/(I+Ma^2)$. 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.
Back to article page
