= Solution
Measure both angles from the downward vertical, as in the source diagram. With upward $y$, the cylinder centres are
$$
\mathbf R=(2a\sin\theta,-2a\cos\theta),\qquad
\mathbf r=\mathbf R+(a\sin\phi,-a\cos\phi).
$$
The large hoop has central inertia $4Ma^2$ and <angular velocity> $\dot\theta$, while the small hoop has no spin in this part. <Translation> and <rotation> therefore give
$$
T=\frac12Ma^2[12\dot\theta^2+\dot\phi^2+
4\cos(\theta-\phi)\dot\theta\dot\phi],
\qquad V=-Mga(4\cos\theta+\cos\phi).
$$
Thus $L=T-V$. 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
$$
\boxed{L_2=\frac12Ma^2(12\dot\theta^2+\dot\phi^2+4\dot\theta\dot\phi)
-\frac12Mga(4\theta^2+\phi^2).}
$$
Back to article page
