= Solution
A common rotation changes neither the separation nor the spring energy. Its infinitesimal direction is $(1,1)$, exactly the zero-frequency <normal mode>. Thus the continuum of rotated <equilibrium points> requires that zero mode; a nonzero restoring frequency in that direction would contradict <rotational symmetry>.
The second mode has $I_1\eta_1+I_2\eta_2=0$, so it changes the relative angle while leaving the inertia-weighted common angle unchanged. It therefore carries zero perturbation in the conserved <angular momentum> associated with common motion. Indeed the coordinates
$$
q=\frac{I_1\theta+I_2\varphi}{I_1+I_2},\qquad\delta=\theta-\varphi
$$
separate the <Lagrangian> exactly:
$$
L=\frac12(I_1+I_2)\dot q^2+\frac12\frac{I_1I_2}{I_1+I_2}\dot\delta^2+\kappa\cos\delta-\text{constant}.
$$
The <cyclic coordinate> $q$ is free, while $\delta$ has restoring or inverted curvature. This also explains why all members of either equilibrium family have the same frequencies. Reflection symmetry makes the potential even in $\delta$, and the stable oscillatory mode is consistent with <time reversal in classical mechanics> through its two opposite frequency signs.
Back to article page
