Normal modes of two spring-coupled coaxial rotors (${\omega }_{\mathrm{rel}}^2=kaA(1/I_1+1/I_2)\cos {\delta }_0$)

For point masses on concentric horizontal rings in parallel planes, set $I_1=m{a}^2$, $I_2=M{A}^2$, and $\delta =\theta -\phi$. A spring of zero natural length gives $V=\text{constant}-kaA\cos \delta$. The inertia-weighted angle $q=(I_1\theta +I_2\phi )/(I_1+I_2)$ separates the Lagrangian into a free common rotor with inertia $I_1+I_2$ and a relative rotor with inertia $I_r=I_1{I}_2/(I_1+I_2)$. Static equilibria have ${\delta }_0=0$ or $\pi$. Their normal modes are the common displacement $(1,1)$ with zero frequency and the relative displacement $(I_2,-I_1)$ with the displayed squared frequency. The aligned family has a restoring relative mode; the antipodal family has an exponentially growing one. The zero mode is required by rotational symmetry. Small nonzero total angular momentum produces common drift, so stability of the aligned family modulo rotations should be distinguished from Lyapunov stability of one fixed equilibrium point.