This question is taken from the book "Classical Dynamics of Particles and Systems" - Marion, problem 7.24. The problem is about a pendulum that is set into motion, it's length varies at a constant rate $$\frac{dl}{dt} = - \alpha.$$ The reader is asked to compute the Lagrangian, the Hamiltonian and discuss conservation of energy. Computing the Lagrangian we get: $$ L = T - U = \frac{1}{2}m(\dot{l}^2 + l^2\dot{\theta}^2) + mgl\cos{\theta} = \frac{1}{2}m(\alpha^2 + l^2\dot{\theta}^2) + mgl\cos{\theta}. $$ Now, we can find the generalized momenta both for $\theta$ and $l$: $$p_{\theta} = \frac{\partial L}{ \partial \dot{\theta}} = ml^2\dot{\theta}$$ $$p_{l} = \frac{\partial L}{ \partial \dot{l}} = m\dot{l} = -m\alpha.$$ Now that we have the generalized momenta we can write the Hamiltonian following: $$H = \sum_i p_i\dot{q_i} - L = \frac{p_{\theta}^2}{2m l^2} + \frac{p_{l}^2}{2m}- mgl\cos{\theta}.$$ But, according to the book, the Hamiltonian is actually equal to: $$H = \sum_i p_i\dot{q_i} - L = \frac{p_{\theta}^2}{2m l^2} - \frac{p_{l}^2}{2m}- mgl\cos{\theta}.$$ This result basically ignores the term $p_l\dot{l}$. Why?

  • $\begingroup$ I don’t think that l is a generalized coordinate, so pl is not correct $\endgroup$
    – Eli
    Sep 16, 2018 at 21:18
  • 1
    $\begingroup$ Why wouldn't it be? Considering that we need to know l to specify the position of the pendulum mass. $\endgroup$
    – prm
    Sep 16, 2018 at 22:24

1 Answer 1



  1. The original Lagrangian system has 2 variables $(\ell,\theta)$ and 1 constraint $\ell=\ell_0-\alpha t$.

  2. By inserting the constraint we can rewrite this as a reduced Lagrangian system with 1 variable $\theta$ and 0 constraints.

  3. The corresponding reduced Hamiltonian system then has variables $(\theta,p_{\theta})$ with Hamiltonian $H=p_{\theta}\dot{\theta}-L$.


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