 Solution

Let ${\delta }_0={\theta }_0-{\phi }_0$ be $0$ or $\pi$, and write the angular displacements as ${\eta }_1,{\eta }_2$. To quadratic order the restoring potential energy is $\kappa \cos {\delta }_0({\eta }_1-{\eta }_2{)}^2/2$. The linearized normal mode equation isThe determinant gives ${\omega }^2[{\omega }^2{I}_1{I}_2-\kappa \cos {\delta }_0(I_1+I_2)]=0$. Thus the normal modes of two spring-coupled coaxial rotors areFor the aligned family, the relative mode oscillates with positive frequencyFor the antipodal family, ${\omega }_{\mathrm{rel}}^2=-{\Omega }^2$, so that displacement has solutions $e^{\pm \Omega t}$ and the equilibrium is unstable.

The common zero mode has ${\eta }_1={\eta }_2=C+Dt$. It describes a neutral displacement or uniform common rotation, not a restoring oscillation. Thus the aligned family is stable with respect to relative motion and has orbital stability modulo common rotations. A particular static point is not strictly Lyapunov stable against unrestricted small velocity perturbations: arbitrarily small nonzero $J$ eventually moves the common angle far from that point. With $J=0$, the common angle is fixed and the aligned relative equilibrium point is stable. This distinguishes the symmetry-induced drift from the exponential instability of the antipodal family.