\begin{problem*}{10.59}
A uniform solid disk and a uniform hoop are placed side by side at the
top of an incline of height $h$.  \Part{a} If they are released from
rest and roll without slipping, which object reaches the bottom
first?  \Part{b} Verify your answer by calculating their speeds when
they reach the bottom in terms of $h$.
\end{problem*}

\begin{solution}
\Part{a}
The disk reaches the bottom first, because it has a lower moment of
inertia, so there is less energy wasted on rotational kinetic energy
and more left over for translational kinetic energy.

\Part{b}
Conserving energy
\begin{align}
  E_i = mgh &= E_f = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2
    = \frac{1}{2}mv^2 + \frac{1}{2}I\p({\frac{v}{r}})^2
    = \frac{1}{2}\p({m+\frac{I}{r^2}})v^2 \\
  v^2 &= \frac{2mgh}{m+\frac{I}{r^2}} \\
  v &= \sqrt{\frac{2mgh}{m+\frac{I}{r^2}}} \;.
\end{align}

For a disk, $I=\frac{1}{2}mr^2$, so
\begin{equation}
  v = \sqrt{\frac{2mgh}{m+\frac{1}{2}m}}
    = \sqrt{\frac{2gh}{\frac{3}{2}}}
    = \ans{\sqrt{\frac{4gh}{3}}} \;.
\end{equation}

For a hoop, $I=mr^2$, so
\begin{equation}
  v = \sqrt{\frac{2mgh}{m+m}}
    = \sqrt{\frac{2gh}{2}}
    = \ans{\sqrt{gh}} \;.
\end{equation}

As we claimed in \Part{a},
\begin{equation}
  v_d = \sqrt{\frac{4gh}{3}} > \sqrt{gh} = v_h \;.
\end{equation}
\end{solution}
