= Normal modes of two spring-coupled coaxial rotors
{title2=$\omega_{\rm rel}^2=kaA(1/I_1+1/I_2)\cos\delta_0$}
For point masses on concentric horizontal rings in parallel planes, set $I_1=ma^2$, $I_2=MA^2$, and $\delta=\theta-\varphi$. A spring of zero natural length gives $V=\text{constant}-kaA\cos\delta$. The inertia-weighted angle $q=(I_1\theta+I_2\varphi)/(I_1+I_2)$ separates the <Lagrangian> into a free common rotor with inertia $I_1+I_2$ and a relative rotor with inertia $I_r=I_1I_2/(I_1+I_2)$. 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.
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