Equations of motion for a spherical pendulum? I am trying to solve for the equations of motion to simulate a spherical pendulum. I decided to use the spherical coordinates. The Lagrange equation is,
$$
L=T-V=\frac{1}{2}m\left(L\dot\theta\right)^2+\frac{1}{2}m\left(L\sin\theta\dot\phi\right)^2-\left(-mgL\cos\theta\right),
$$
where $L$ is the length of the rope, $ϕ$ is the angle of the projection of the rope on $x$-$y$ plane with $x$-axis and $θ$ is the angle with the $-z$-axis
I solved these equations: 
\begin{align}
\frac{\mathrm d}{\mathrm dt}\left(\frac{\partial L}{\partial\dot\theta}\right)-\frac{\partial L}{\partial\theta}&=0, \\
\frac{\mathrm d}{\mathrm dt}\left(\frac{\partial L}{\partial\dot\phi}\right)-\frac{\partial L}{\partial\phi}&=0,
\end{align}
and I got 
$$
\ddot\theta=\sin\theta\cos\theta\dot\phi^2-\frac{g}{L}\sin\theta
$$
and
$$
\frac{\mathrm d}{\mathrm dt}(mL^2\sin^2θ\dot\phi) = 0
$$
This seems like the change in angular momentum is conserved. But when I solve it more  
$$
\ddot\phi = -2\dot\phi\dot\theta\cot\theta
$$
This does not make sense to me because it goes to infinity when θ goes to 0.
Any ideas on what I am doing wrong?
 A: The cases with zero and nonzero angular momentum should be treated separately.


*

*If $\theta$ ever crosses zero then the angular momentum at that time is zero. By the conservation law it means that the angular momentum vanishes at all times. This implies that $\phi$ is constant and all its derivatives vanish. (That means that $\ddot\phi\propto\dot \phi\cot(\theta)$ doesn't blow up, because both sides vanish.) In this case you're back to the planar case, and you should solve it as such, or treat the crossings from positive to negative $\theta$ in a careful way.

*On the other hand, if the angular momentum is nonzero - as must happen if $\dot\phi$ is ever nonzero - then the particle can never cross the pole, and your equation is perfectly well defined.
The general scheme for solving this is to find that $\ell=\sin^2\theta\ \dot\phi$ is conserved, and to forget about $\phi$ temporarily. Substituting this in your other equation you get a single second-order equation in $\theta$; once you solve this you automatically get $\phi$ from integrating $\ell/\sin^2\theta$. The equation for $\theta$, however, radically changes character depending on whether $\ell$ is zero or not: if it's not, then an angular momentum barrier will appear that stops $\theta$ from ever reaching zero. Try it out!
