 Solution

Put $I_j=m_j{r}_j^2$ and $K={\omega }^2{r}_1{r}_2$. The Euler-Lagrange equations are $I_1{\ddot{\varphi }}_1=K\sin \psi$ and $I_2{\ddot{\varphi }}_2=-K\sin \psi$, where $\psi ={\varphi }_2-{\varphi }_1$. Adding shows that $I_1{\dot{\varphi }}_1+I_2{\dot{\varphi }}_2$ is conserved. Subtracting gives the equation of a simple pendulumThe relative equilibria are $\psi =0$ and $\psi =\pi$ modulo $2\pi$. The relative energy ${\dot{\psi }}^2/2+k^2(1-\cos \psi )$ has a strict minimum at zero, proving stability there. Linearization gives $\ddot{\eta }=-k^2\eta$ near zero and $\ddot{\eta }=k^2\eta$ near $\pi$, so the latter is unstable. The small-oscillation period is $2\pi /k$. These are equilibria of the relative angle; a nonzero conserved total angular momentum permits common rotation of both particles.