Discussion
A particle has Lagrangian
\[ \begin{aligned} L = \frac{1}{2}\Big[&f(\theta)\dot{\theta}^{2} + 2g(\theta)w\dot{\theta} \\ &+ w^{2}h(\theta)\Big] - V \end{aligned} \]
Then \(\frac{1}{2}\left[f(\theta)\dot{\theta}^{2} - w^{2}h(\theta)\right] + V\) represents :
Answer: A
The expression is the Hamiltonian H.
The generalised momentum is \(p_{\theta} = \partial L/\partial\dot{\theta} = f\dot{\theta} + g w\). Then
\[ \begin{aligned} H &= p_{\theta}\dot{\theta} - L \\ &= f\dot{\theta}^{2} + gw\dot{\theta} - \tfrac{1}{2}f\dot{\theta}^{2} - gw\dot{\theta} - \tfrac{1}{2}w^{2}h + V \\ &= \tfrac{1}{2}\left[f(\theta)\dot{\theta}^{2} - w^{2}h(\theta)\right] + V \end{aligned} \]
The total energy is \(T + V = \frac{1}{2}[f\dot{\theta}^{2} + 2gw\dot{\theta} + w^{2}h] + V\), which differs from H by \(gw\dot{\theta} + w^{2}h\). This is the typical situation for a system in a rotating frame (here with angular speed \(w\)): the Jacobi integral H is conserved but is not the energy. So the expression is H alone, not E, and not both.
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