Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

enter image description here

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.
share|improve this question
    
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
1  
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
show 3 more comments

1 Answer

up vote 2 down vote accepted

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.

As one of my physics professors told us:

All problems that can be solved exactly will be taught to you in your undergraduate degree. From then on, it's all approximations.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.