= Solution
The <azimuth> $\phi$ is a <cyclic coordinate>, since $\partial\mathcal L/\partial\phi=0$. Its <Euler-Lagrange equation> gives conservation of the vertical component of <angular momentum>:
$$
\boxed{h=p_\phi=\frac{\partial\mathcal L}{\partial\dot\phi}=ml^2\sin^2\theta\,\dot\phi=\text{constant}.}
$$
This $h$ is the actual <angular momentum>, not <specific angular momentum>. The Lagrangian also has no explicit <time> dependence. The conserved <energy> is
$$
\boxed{E=\dot\theta\frac{\partial\mathcal L}{\partial\dot\theta}+\dot\phi\frac{\partial\mathcal L}{\partial\dot\phi}-\mathcal L
=\frac12ml^2(\dot\theta^2+\sin^2\theta\dot\phi^2)-mgl\cos\theta.}
$$
Along solutions, $\dot E=-\partial\mathcal L/\partial t=0$. These are the conserved quantities associated with rotational <symmetry> about the vertical and <time-translation symmetry>, respectively.
Back to article page
