Past exam of the physics 1 paper of the University of Oxford / 2014 resit / 8

Take $\theta$ from the downward vertical. Under the conventional light-rod idealization, only the endpoint point mass contributes appreciably. Its speed is $l\dot{\theta }$, and its gravitational potential energy is $-mgl\cos \theta$. ThereforeAn additive constant in the potential energy has no effect on the Euler-Lagrange equation.

Rod-mass qualification: “thin” and “uniform” do not imply negligible mass. If the uniform rod has a finite unspecified mass $M_r$, its pivot moment of inertia is $M_r{l}^2/3$ and its centre of mass lies $l/2$ from the pivot. For a pendulum with a uniform rod and an endpoint mass, the general answer isThis formula covers either convention without inventing a rod mass. A result involving only the specified $m,l,g$ uses $M_r=0$ unless an additional relation for $M_r$ is stated.

For the general Lagrangian just obtained, the Euler-Lagrange equation givesUsing the small-angle approximation $\sin \theta \simeq \theta$, this is simple harmonic motion with ${\omega }_0^2=A/I$. ThusThe endpoint mass cancels in the light-rod limit. A concrete comparison shows why the finite rod cannot silently be included: the same specified endpoint $m$ with $M_r=m$ gives $T=2\pi \sqrt{8l/(9g)}$, distinct from the negligible-rod result. This is a physical pendulum when the rod contributes, and a simple pendulum in the light-rod limit. Both expressions require small oscillation amplitude; the full sine equation remains nonlinear.

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.