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\begin{document}

\noindent Solution for assignment 22: Turning points of the spherical 
pendulum for a special case.\bigskip

The Energy is
\begin{eqnarray}
  E &=& \frac{m}{2}\,R^2\dot\theta^2 + 
        \frac{m}{2}\,R^2\sin^2(\theta)\dot\phi^2 + U_0\,\cos(\theta)
\end{eqnarray}
with $U_0$ given by $U_0=m\,g\,R=E/2$. The angular momentum 
\begin{eqnarray}
  M_z &=& m\,R^2\sin^2(\theta)\,\dot\phi
\end{eqnarray}
is conserved. Substituting $\dot\phi^2=M^2_z/(m^2R^4\sin^4\theta)$ the 
energy becomes
\begin{eqnarray}
  E &=& \frac{m}{2}\,R^2\dot\theta^2 + 
        \frac{M^2_z}{2\,m\,R^2\sin^2\theta} + U_0\,\cos(\theta)
\end{eqnarray}
with $M^2_z/(2\,m\,R^2)=E$. Turning points are then given by the 
solutions of
\begin{eqnarray}
  0 &=& \frac{m}{2}\,R^2\dot\theta^2\ =\ E - \frac{E}{\sin^2\theta} -
        \frac{E}{2}\,\cos\theta \\ \label{cos}
  0 &=& \sin^2(\theta) - 1 - \frac{1}{2}\,\cos\theta\,\sin^2\theta\ =\
    \cos^2(\theta)-\frac{1}{2}\,\cos\theta+\frac{1}{2}\,\cos^3\theta\,.
\end{eqnarray}
With $x=\cos\theta$
\begin{eqnarray}
  0 &=& +x\,\left(x^2 - 2\,x -1\right)
\end{eqnarray}
with the solutions $x_0=0\Rightarrow\theta_0=\pi/2$ and
\begin{eqnarray}
  x_{1,2} &=& 1\pm\sqrt{1+1}\ =\ 1\pm\sqrt{2}
\end{eqnarray}
of which only $x_2=1-\sqrt{2}$ is in the physical range of $\cos\theta$.
So, we find the turning points
\begin{eqnarray}
  \theta_{\min} &=& \cos^{-1}(0)\ =\ \frac{\pi}{2}\ = 1.5707963\dots\,,\\
  \theta_{\max} &=& \cos^{-1}(1-\sqrt{2})\ =\ 1.9978749\dots\,.
\end{eqnarray}
As an addtional task a plot of 
\begin{eqnarray}
  f(\theta) &=& \frac{E}{\sin^2\theta} + \frac{E}{2}\,\cos\theta - E
\end{eqnarray}
for, e.g., $E=1$ is instructive.
\end{document}
