Past exam of the mathematics course of the University of Cambridge / 2005 / ii / Paper 1 / 15C (Classical Dynamics)

Consider a smooth variation $q^a(t)+ϵ{\eta }^a(t)$ with ${\eta }^a(t_1)={\eta }^a(t_2)=0$. Differentiating the action at zero and integrating the velocity term by parts givesThe boundary term vanishes because the endpoints are fixed. Independent compactly supported variations and the fundamental lemma of the calculus of variations make every bracket zero. Thus the Euler-Lagrange equations areThe required principle is stationarity of the action; an extremal need not be a strict minimum.

Measure $\theta$ from the downward vertical, so the bob is at $(x+l\sin \theta ,-l\cos \theta )$. Its squared speed is ${\dot{x}}^2+2l\dot{x}\dot{\theta }\cos \theta +l^2{\dot{\theta }}^2$. Up to a constant potential term, the Lagrangian isThe Euler-Lagrange equations giveAt the downward stable equilibrium, linearization gives $(M+m)\ddot{x}+ml\ddot{\theta }=0$ and $l\ddot{\theta }+\ddot{x}+g\theta =0$. Eliminating $\ddot{x}$ yieldsThere is also a zero-frequency translational mode because the rail fixes no preferred pivot position.

If an external force imposes $\ddot{x}=A$ constantly, the bob's angular equation becomes $l\ddot{\theta }+A\cos \theta +g\sin \theta =0$. Its stable equilibrium therefore hason the branch near the downward vertical. The displaced bob leans opposite the acceleration. The solution differing by $\pi$ is unstable, corresponding to alignment opposite the effective gravitational field in the accelerating frame.