Solution to 1986 Problem 76

The initial kinetic energy of the ball is Mgh. The final kinetic energy is
\begin{align*}\frac{1}{2} I \omega^2 + \frac{1}{2}M v^2\end{align*}
v = \omega R and for a hoop I = M R^2, so the final kinetic energy is
\begin{align*}\frac{1}{2} M R^2 \omega^2 + \frac{1}{2}M \left(\omega R\right)^2 = M R^2 \omega^2\end{align*}
\begin{align*}\omega = \sqrt{g h /R^2}\end{align*}
The angular momentum about the center of mass is
\begin{align*}L = I \omega = \boxed{M R \sqrt{g h}}\end{align*}
Therefore, answer (A) is correct.

return to the 1986 problem list

return to homepage

Please send questions or comments to where X = physgre.