Discussion
The Lagrangian of a particle of mass \(m\) in spherical polar co-ordinates is given by :
\[ \begin{aligned} L &= \tfrac{1}{2}m(\dot{r}^{2} + r^{2}\dot{\theta}^{2} + r^{2}\sin^{2}\theta\,\dot{\phi}^{2}) \\ &\quad - mgl\cos\theta \end{aligned} \]
The quantity that is conserved is :
Answer: C
The conserved quantity is \(\partial L/\partial\dot{\phi}\).
\(\phi\) does not appear in L, only \(\dot{\phi}\) does, so \(\phi\) is a cyclic coordinate. Lagrange's equation for it gives
\[ \begin{aligned} \frac{d}{dt}\frac{\partial L}{\partial\dot{\phi}} &= \frac{\partial L}{\partial\phi} \\ &= 0 \end{aligned} \]
so \(p_{\phi} = mr^{2}\sin^{2}\theta\,\dot{\phi}\), the angular momentum about the vertical axis, is constant.
\(r\) and \(\theta\) both appear explicitly in L (through \(r^{2}\dot{\theta}^{2}\), \(\sin^{2}\theta\) and \(\cos\theta\)), so \(\partial L/\partial\dot{r}\) and \(\partial L/\partial\dot{\theta}\) are not conserved, and \(\partial L/\partial r\) is a generalised force, not a momentum.
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