Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2014 / Paper 1 / 9

For a variation $q\mapsto q+ϵ\eta$ with $\eta (t_0)=\eta (t_1)=0$, the first variation isThe endpoint term vanishes. Since the interior variation is arbitrary, the fundamental lemma of the calculus of variations gives the Euler-Lagrange equationAlong such a trajectory,If $L$ has no explicit time dependence, the first integral is thereforeIts negative is the equivalent Beltrami form. For the usual time-independent mechanical Lagrangian $L=T-V$ with quadratic velocity dependence, this conserved quantity is the total mechanical energy $T+V$.

Use the angle orientation in the PDF: $M$ has coordinates $(R\sin \theta ,-R\cos \theta )$ relative to the hoop centre. The wrapped string segment has length $R(\pi /2-\theta )$. As $\theta$ increases, it releases length $R\theta$, so the suspended mass moves down by $R\theta$. Both speeds have magnitude $R|\dot{\theta }|$.

Thus the hoop with a suspended counterweight hasand $L=T-V$. The Euler-Lagrange equation givesAt equilibrium,The positive curvature $V^{\prime \prime }({\theta }_0)=MgR\cos {\theta }_0$ establishes stability. Put $\theta ={\theta }_0+\delta$ and linearize. The equilibrium force cancels, leavingHenceFor $M\gg m$, ${\theta }_0\simeq m/M$ and $\omega \to \sqrt{g/R}$, the small-oscillation frequency of a simple pendulum of length $R$. The linearization assumes an amplitude small enough to remain inside the stated angular range and a taut string throughout.