Past exam of the mathematics course of the University of Cambridge / 2006 / ii / Paper 3 / 9C (Classical Dynamics)

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 $2\pi$.

Near $\theta =0$, the linearized equation is $\ddot{\theta }+(g-a)\theta /\ell =0$. When $a<g$ this is an harmonic oscillator, so the downward configuration is stable. Writing $\theta =\pi +\epsilon$ instead gives $\ddot{\epsilon }-(g-a)\epsilon /\ell =0$, with an exponentially growing solution. Downward is stable; upward is unstable.

When $a>g$, the signs reverse. The downward configuration has an exponentially growing disturbance, whereas near $\theta =\pi$ one obtains $\ddot{\epsilon }+(a-g)\epsilon /\ell =0$. Upward is stable; downward is unstable. The effective gravity in the descending lift points upward in this case.