Is $H = T + U$ for a pendulum on a circle movement?

Question

I have this problem:

Obtain Hamilton's equations of motion for a plane pendulum of length $l$ with mass point $m$ whose radius of suspension rotates uniformally on the circumference of a vertical circle of radius $a$.

This is my position vector: \begin{equation} \vec{r} = (a\cos{(\omega_0 t)} + l\sin{\theta})\hat{\imath} + (a\sin{(\omega_0 t)}-l\cos{\theta})\hat{k} \end{equation} (The angle that describes the movement of the suspension point on the circle is $\omega_0 t$ for being uniform)

From this, using the definition of potential and kinetic energy, the lagrangian is: \begin{equation} L = \frac{m}{2}(a^2\omega_0^2 + 2la\omega_0\dot{\theta}(\sin{(\theta - \omega_0 t)}) + l^2\dot{\theta}^2) - mg(a\sin{(\omega_0 t)}-l\cos{\theta}) \end{equation}

Now, i tried to make my hamiltonian with the definition \begin{equation} H = p_i\dot{q_i} - L \end{equation} But... for the problem, i think that this form is useless for the Hamilton equations. Then it occurred to me to use
\begin{equation} H = T + U = \frac{m}{2}l^2\dot{\theta}^2 + mg(a\sin{(\omega_0 t)}-l\cos{\theta}) \end{equation} However, I'm not sure if I can use it because, according to me, the position vector explicitly depends on time.

Answer 1

No, you can't do that, you have to perform a Legendre transformation. Solve for $\dot\theta$, $$p_{\theta}=\frac{\partial L}{\partial\dot\theta}$$ You will get $\dot\theta(p_{\theta})$, i.e. $\dot\theta$ as a function of $p_{\theta}$. Then compute, $$H = p_{\theta}\dot{\theta}(p_{\theta}) - L(\theta,\dot\theta(p_{\theta}),t)$$ Finally, $$\dot\theta = \frac{\partial H}{\partial p_{\theta}},\quad \dot p_{\theta}=-\frac{\partial H}{\partial \theta}$$

Answer 2

This is an example of situation where the Hamiltonian $\sum \dot{q}_ip_i-L$ is not equal to $T+U$. One can see the problem immediately because $$ \sum_i \dot{q}_ip_i\ne 2T\, , $$ so this system is not a “natural” system. In fact as written $H$ is not conserved because $L$ depends explicitly in $t$.

The definition $$ H=\sum \dot{q}_ip_i-L $$ followed by the elimination of all $\dot{q}_i$ in terms of $p$ and $q$ so that $H=H(q,p)$ is correct and the starting point of the analysis. Hamilton’s equation should be use using $H(q,p)$, not $T+U$.

A related (but different) example is this question, although in the latter, $L$ does not depend on $t$ so $H$ is conserved.