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Zoidberg

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**[SOLVED] Variable Length Pendulum**

Hi, everybody! I'm not sure about my solution to the problem.

## Homework Statement

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.

## Homework 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]

## 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]

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