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Variable Length Pendulum

  1. Dec 16, 2007 #1
    [SOLVED] Variable Length Pendulum

    Hi, everybody! I'm not sure about my solution to the problem.
    1. The problem statement, all variables and given/known data

    Consider a simple pendulum with variable length r(t). The suspension point remains fixed.
    (i)Write the Lagrangian for the pendulum and the equations of motion.
    (ii)Write the Hamiltonian for the system.
    (iii)Calculate the total mechanical energy of the system.

    2. Relevant equations

    Lagrangian: L=T-V
    Euler-Lagrange equation: [tex]\frac{d~}{dt} \ \left( \, \frac{\partial L}{\partial \dot{q}_i} \, \right) \ - \ \frac{\partial L}{\partial q_i} \ = \ 0[/tex]
    Legendre transform of the Lagrangian: [tex]H\left(q_j,p_j,t\right) = \sum_i \dot{q}_i p_i - L(q_j,\dot{q}_j,t)[/tex]

    3. The attempt at a solution

    (i) I think the system has only one degree of freedom, the angle [tex]\theta[/tex]. The kinetic energy of the system is [tex]T=\frac{m}{2}\left( \dot{r}^2+r^2 \dot{\theta}^2\right)[/tex]. The potential potential energy is [tex]V=-mgr \cos\theta [/tex].
    [tex]\Rightarrow L=\frac{m}{2}\left( \dot{r}^2+r^2 \dot{\theta}^2\right)+mgr \cos\theta[/tex].
    Using the Euler-Lagrange equation for [tex]q=\theta: 2mr\dot{r}\dot{\theta}+mr^2 \ddot{\theta}-mgr\sin\theta =0 [/tex]
    (ii)With the generalized momenta [tex]p_{\theta}=\frac{\partial L}{\partial \dot{\theta}}=mr^2 \dot{\theta} \Leftrightarrow \dot{\theta}=\frac{p_{\theta}}{mr^2}[/tex] I can write the Hamiltonian [tex]H=p_\theta \dot{\theta} -L=\frac{p_\theta}{2mr^2}-\frac{m}{2}\dot{r}^2-mgr\cos\theta[/tex]. The Hamiltonian is not conserved since r=r(t) is time dependent.
    (iii) The total energy of the system is [tex]E=T+V=\frac{p_\theta}{2mr^2}+\frac{m}{2}\dot{r}^2-mgr\cos\theta[/tex]
    Last edited: Dec 16, 2007
  2. jcsd
  3. Dec 19, 2007 #2
    Math seems fine, but [itex]r[/itex] and [itex]\theta[/itex] are both changing with time. How exactly are you defining the degree of freedom?
  4. Dec 19, 2007 #3


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    r is a fixed function of time r(t). You can't predict its values from the dynamics.
  5. Dec 19, 2007 #4
    First I was confused about [tex]H(\theta,p_{\theta})\neq E[/tex], but after reading some books this seems to be a classic example of this type of problem. If I understand this correct r(t) is a holonomic constraint.
    Anyway the solution seems to be correct. Thanks for your responses.
  6. Dec 19, 2007 #5


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    What happens is that the system is not isolated, there is some extrenal force dictating how r changes with time. In that case H will not be E. The same thing happens, for example, when there is something rotating at a constant angular velocity, which means there is an extrenal force acting and forcing the rotation.
  7. Dec 23, 2007 #6
    Ok, so its essentially given.
  8. Dec 24, 2007 #7


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    Right. It could also as Zoidberg pointed out, could be considered as a constraint.
  9. Jul 17, 2010 #8


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    Re: [SOLVED] Variable Length Pendulum

    Shouldn't pΘ appear on the second power here?

    (off: why can't I write latex formulas? The preview shows a completely different formulas from that I write down. E.g. how [tex]p_\theta[/tex] looks like?)

    edit: wow! now it displays good. Only the preview was wrong.
    Last edited: Jul 17, 2010
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