= Solution
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 is
$$
a\dot\phi+a\dot\psi=2a\dot\theta,
\qquad\boxed{\dot\psi=2\dot\theta-\dot\phi.}
$$
This <rolling constraint inside a moving circular track> adds small-hoop rotational <energy> $Ma^2(2\dot\theta-\dot\phi)^2/2$. At quadratic order its cross term cancels the translational cross term in part a. The potential is unchanged, so
$$
\boxed{L_2=\frac12Ma^2(16\dot\theta^2+2\dot\phi^2)
-\frac12Mga(4\theta^2+\phi^2).}
$$
Static friction enforces rolling; ideal no-slip contact dissipates no <energy>.
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