 Solution

Allowing azimuthal motion makes the endpoint speed satisfy $v^2=l^2({\dot{\theta }}^2+{\sin }^2\theta {\dot{\phi }}^2)$. The light-rod spherical pendulum therefore hasThe azimuth $\phi$ is a cyclic coordinate, since the Lagrangian is independent of it. Its conjugate canonical momentum is the conserved vertical angular momentum:For a finite uniform rod, replace $m{l}^2$ by $I$ and $mgl$ by $A$ from i. The same azimuthal symmetry then gives $p_{\phi }=I{\sin }^2\theta \dot{\phi }$.

The Hamiltonian is the Legendre transform of this Lagrangian. With $p_{\theta }=I\dot{\theta }$,There is no explicit time dependence, so $dH/dt=\partial H/\partial t=0$. Here $H$ also equals the total mechanical energy: the constraint is stationary, the kinetic energy is quadratic in generalized velocities, and gravity is conservative. The apparent singularity of the momentum expression at $\sin \theta =0$ is a spherical-coordinate singularity; trajectories reaching a pole must be described consistently, and nonzero $J_z$ prevents reaching it at finite energy.