# Equations of motion for a spherical pendulum? [closed]

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?

• Just to be clear is the position of the mass at $\vec{r} = (L\cos\phi\sin\theta, L\sin\phi\sin\theta, -L\cos\theta)$ and gravity acting in the -z direction? – John Alexiou Dec 22 '13 at 0:03
• Hints: 1. Go to the Routhian formulation. 2. The $\theta=0$ singularity stems from the centrifugal potential. – Qmechanic Nov 20 '19 at 21:42

• 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!
• But if $\theta$ goes near zero, $\dot{\phi}$ will also grow to compensate, which is what the conservation law is telling you. Furthermore, if your angular momentum there is zero initially, then it will always be zero. That pretty much tells you either you start in the $\theta = 0$ regime or that $\dot{\phi} = 0$ always, depending on your initial conditions. – webb Nov 21 '13 at 23:33