= Solution
The new kinetic matrix is diagonal, giving
$$
\ddot\theta+\frac{g}{4a}\theta=0,
\qquad\ddot\phi+\frac{g}{2a}\phi=0,
\qquad
\boxed{\omega_1^2=\frac g{4a},\quad\omega_2^2=\frac g{2a}.}
$$
For the first mode, $\phi=0$: 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 $\dot\psi=-\dot\phi$. These coordinate motions are independent only at the retained quadratic order.
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