= Solution
Put $I_j=m_jr_j^2$ and $K=\omega^2r_1r_2$. The <Euler-Lagrange equations> are $I_1\ddot\phi_1=K\sin\psi$ and $I_2\ddot\phi_2=-K\sin\psi$, where $\psi=\phi_2-\phi_1$. Adding shows that \b[$I_1\dot\phi_1+I_2\dot\phi_2$ is conserved]. Subtracting gives the equation of a <simple pendulum>
$$
\boxed{\ddot\psi=-k^2\sin\psi,\qquad
k^2=\omega^2r_1r_2\left(\frac1{m_1r_1^2}+\frac1{m_2r_2^2}\right).}
$$
The 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 \b[$2\pi/k$]. These are equilibria of the relative angle; a nonzero conserved total <angular momentum> permits common rotation of both particles.
Back to article page
