 Solution

Take $\theta$ as the inclination of the symmetry axis to the upward vertical, $\varphi$ as its azimuth, and $\psi$ as rotation of the body about that axis relative to the line of nodes. The kinetic energy and potential energy areThe cyclic coordinates $\psi ,\varphi$ and time independence give three conserved quantities:Their meanings are the component of angular momentum along the body axis, the component along the vertical, and the total energy.

For a vanishingly small perturbation of the upright spinning state, $L\simeq Cn$, $p_{\varphi }\simeq L$, and $E\simeq L^2/(2C)+mg\ell$. Eliminating $\dot{\varphi }$ gives the limiting effective potentialIts quadratic change near zero is $\{L^2/(8A)-mg\ell /2\}{\theta }^2$. Thus if $C^2{n}^2>4mg\ell A$, zero is a strict local minimum and a sufficiently small perturbation remains near it.

In the unstable case, the limiting energy equation becomesAt the nonzero turning point the bracket vanishes, yieldingThe approximation is the small-perturbation limit, not an assertion that arbitrary disturbances preserve the upright state's exact conserved quantities.