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I am trying to solve for the equations of motion to simulate a pendulum. I decided to use the spherical coordinates. The Lagrange equation is:

enter image description here

where

  • L = length of the rope
  • ϕ= angle of the projection of the rope on x-y plane with x-axis
  • θ = angle with the z- axis

I solved these equations: enter image description here and enter image description here

and I got

enter image description here

and $$ \frac{ d}{ dt}(mL^2sin^2θ\dot\phi) = 0 $$ 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 dose not make sense to me because it goes to infinity when θ goes to 0. Any ideas on what I am doing wrong.

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All seems good. Just a question, where is $ \dot \theta$ in the last expression? –  Anuar Nov 21 '13 at 21:35
    
thanks I added it. But still i cant use this as this means that as my angle goes towards 0 the simulation blows up which should not happen. –  harsha Nov 21 '13 at 21:48
    
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? –  ja72 Dec 22 '13 at 0:03

1 Answer 1

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!

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but this still dose not address the fact that as θ approaches 0 the value gets very large. –  harsha Nov 21 '13 at 23:03
1  
No, it doesn't. Zero times very large is still zero. –  Emilio Pisanty Nov 21 '13 at 23:33
1  
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

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