# How do I squeeze a $\theta(t)$ and $\varphi(t)$ out of this?

A ball attached to a fixed-length massless rod swings about under gravity. Mathematically:

$$L=T-U=\frac{MR^2}{2}(\sin^2(\theta)\dot{\varphi}^2+\dot{\theta}^2)+MgR \cos(\theta)$$

$$\frac{d}{dt}\left(\frac{\partial L}{\partial \dot{\theta}}\right)=\frac{\partial L}{\partial \theta}$$ $$MR^2 \ddot{\theta}=MR^2\sin(\theta)\cos(\theta)\dot{\varphi}^2-MgR \sin(\theta)$$ $$MR^2 \ddot{\theta}=\frac{MR^2}{2}\sin(2\theta)\dot{\varphi}^2-MgR \sin(\theta)\tag{1}$$ $$\frac{d}{dt}\left(\frac{\partial L}{\partial \dot{\varphi}}\right)=\frac{\partial L}{\partial \varphi}$$ $$\frac{d}{dt}(MR^2 \sin^2(\theta) \dot{\varphi})=0$$ $$MR^2( \sin^2(\theta) \ddot{\varphi}+2\sin(\theta)\cos(\theta) \dot{\varphi}\dot{\theta})=0$$ $$MR^2( \sin^2(\theta) \ddot{\varphi}+\sin(2\theta) \dot{\varphi}\dot{\theta})=0\tag{2}.$$

Any ideas as to the domestication of these equations? Any approximative tricks?

Edit: I forget that it's useful if I post some of my own insights to aid answerers

1. $\sin(\theta)=\theta+o(\theta^3)$, so approximate $\sin(\theta) \approx \theta$
2. For cosine, it's not so easy, because $cos(\theta)=1+o(\theta^2)$ doesn't hold for as long, so perhaps $cos(\theta) \approx 1-\frac{1}{2}\theta^2$ could work, but there's enough trouble already with the equations being nonlinear before adding a squared term.
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Those look like equations of motion to me. Are you looking for a closed-form solution? – Rahul Jan 27 '13 at 19:52
Edited title: I wondered whether explicit $\theta$ and $\varphi$ would be found as functions of $t$. – Alyosha Jan 27 '13 at 19:54
Your equation is invariant under translation of $t$ and $\phi$. You will get two constant of motions, one for the "angular momentum" of $\phi$ and one for energy. At the end, you can express $t$ as inverse function of some integral over $\theta$. In similar problem without the $\phi$ term, I think $\theta(t)$ can be expressed in terms of elliptic functions but I'm not 100% sure. Hope this helps. – achille hui Jan 27 '13 at 21:58
If you're assuming $\theta\ll1$, then this is just the motion of a particle in a $2$D plane under a centripetal force proportional to $\theta$, which has a classical closed-form solution as an ellipse. – Rahul Jan 27 '13 at 22:02
Hmm.. the physics of the equation is wrong. If the $\dot{\varphi}^2$ term really represent kinetic energy associated with motion in $\varphi$ direction, then the coefficient in front of it should be $\sin(\theta)^2$, not $\sin(\theta)$. – achille hui Jan 27 '13 at 23:01

There is no closed form solution for this, unless you assume $\theta \ll1$. In that case you get from the second equation that: $$\theta\dot{\phi}=C$$ This really doesn't help you unless you make some further assumptions, namely that either $\dot{\phi}$ or $\dot{\theta}$ are equal to 0. I've seen solutions that try to perturb the system around one of these assumptions and then use a Taylor series to get an approximate solution.