Past exam of the theoretical physics I course of the University of Cambridge / 2014 / Paper 1 / 1

Measure both angles from the downward vertical, as in the source diagram. With upward $y$, the cylinder centres areThe large hoop has central inertia $4M{a}^2$ and angular velocity $\dot{\theta }$, while the small hoop has no spin in this part. Translation and rotation therefore giveThus $L=T-V$. Counting each angle and velocity as first order, replacing the cosine in the velocity product by one changes only fourth-order terms. Expanding the potential and dropping its constant gives

The Euler-Lagrange equations from the quadratic Lagrangian areFor a normal mode proportional to $e^{i\omega t}$, let $\lambda ={\omega }^{2}a/g$. Nonzero amplitudes requireHenceThe corresponding amplitude ratios are $\Phi /\Theta =2\lambda /(1-\lambda )$. The lower-frequency mode has $\Phi /\Theta =2(\sqrt{2}-1)$, with the angles in phase; the higher-frequency mode has $\Phi /\Theta =-2(1+\sqrt{2})$, with opposite signs. The small hoop's spin remains zero in both modes.

Let $\psi$ be the small hoop's absolute spin angle with the same angular sign convention. Equating the two contact-point velocities cancels the large hoop centre's translation. The remaining tangential equality isThis rolling constraint inside a moving circular track adds small-hoop rotational energy $M{a}^2(2\dot{\theta }-\dot{\varphi }{)}^2/2$. At quadratic order its cross term cancels the translational cross term in part a. The potential is unchanged, soStatic friction enforces rolling; ideal no-slip contact dissipates no energy.

The new kinetic matrix is diagonal, givingFor the first mode, $\varphi =0$: the line between the centres stays vertical, both centres translate together, and $\dot{\psi }=2\dot{\theta }$. For the second, $\theta =0$: the outer hoop stays still while the inner hoop rolls within it, with $\dot{\psi }=-\dot{\varphi }$. These coordinate motions are independent only at the retained quadratic order.