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\begin{center}
{\bf PHY 4936 -- Solution 32}\\\vspace {2mm} 
\end{center}
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\noindent
The energy is
$$ E\ =\ \frac{1}{2}MR^2\dot\theta^2\left[\frac{3}{2} 
      - \frac{8}{9\pi}\cos\theta\right] 
      + MgR\left[1-\frac{4}{9\pi}\cos\theta\right]. $$
Solving for $v^2=R^2\dot\theta^2$ gives
$$ v^2\ =\ R^2\dot\theta^2\ =\ \frac{2\,[E-MgR\,(1-4\cos\theta/
                       (9\pi)]}{M\,[3/2 - 8\cos\theta/(9\pi)]} $$

\noindent
A. $E\ =\ MgR$:
\begin{equation}
  v^2\ =\ \frac{8gR\cos\theta/(9\pi)}{3/2-8\cos\theta/(9\pi)}\,,
        \qquad v\ =\ \pm \sqrt{v^2}\,. 
\end{equation}
\bigskip

\noindent
B. $E\ =\ MgR\,[1-4/(9\pi\sqrt{2})]$:
\begin{equation}
  v^2\ =\ \frac{8gR\,[\cos\theta/(9\pi)-1/(9\pi\sqrt{2})]} 
  {3/2-8\cos\theta/(9\pi)}\,, \qquad v\ =\ \pm \sqrt{v^2}\,. 
\end{equation}
\bigskip

\noindent
With $g=9.81\,$[m/s] and $R=1\,$[m] plots of $v(\theta)$ for (1) and 
(2) are given in the figure.

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