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

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.