Past exam of the mathematics course of the University of Cambridge / 2006 / ii / Paper 3 / 9C / Solution

The bob has $\dot{x}=\ell \cos \theta \dot{\theta }$ and $\dot{y}=at-\ell \sin \theta \dot{\theta }$. Substituting these into the Lagrangian givesThe mixed term is $\frac{d}{dt}(ma\ell t\cos \theta )-ma\ell \cos \theta$. Removing this total derivative and the time-only term leaves the equivalent Lagrangian $L_{\mathrm{eff}}=\frac{1}{2}m{\ell }^2{\dot{\theta }}^2+m(g-a)\ell \cos \theta$. The Euler-Lagrange equation becomesIn free fall, $a=g$, so $\theta ={\theta }_0+{\omega }_{0}t$: a bob initially at rest relative to the lift remains at its initial angle, and otherwise moves around the pivot with constant angular speed. Every constant angle is then an equilibrium, with no restoring force. For $a\ne g$, the equilibria are $\theta =0$ and $\theta =\pi$ modulo $2\pi$.