= Stopping angle under constant resisting torque
{title2=$\Delta\phi=I\omega_0^2/(2\tau_f)$}
A rigid body's axial <moment of inertia> $I$ and a constant resisting <torque> $\tau_f>0$ give constant <angular acceleration> $-\tau_f/I$. Hence $t_{\rm stop}=I\omega_0/\tau_f$ and the <angular displacement> is the displayed expression. Equivalently, the lost <rotational kinetic energy> $I\omega_0^2/2$ equals the resisting work $\tau_f\Delta\phi$. The total number of revolutions is $\Delta\phi/(2\pi)$; the number of whole revolutions completed is its <floor function>.
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