 Solution

Autonomy gives conservation of energy, while the cyclic coordinate give conserved momenta:Here $p_{\psi }$ is the angular momentum along the body's symmetry axis, and $p_{\varphi }$ is its component along the fixed vertical. The time symmetry corresponds to $E$, not to an additional angle momentum.

With $u=\cos \theta$ and $-1<u<1$, solveInsert this and the conserved spin rate $n$ into the energy equation, multiply by $A$, and isolate $\dot{\theta }$:

For a fixed nonvertical angle, letThe energy is $A{\dot{\theta }}^2/2+V_{\mathrm{eff}}(\cos \theta )$. A constant angle in $(0,\pi )$ requires both $E=V_{\mathrm{eff}}(u_0)$ and $V_{\mathrm{eff}}^{\prime }(u_0)=0$, by the reduced Euler-Lagrange equation and $\sin {\theta }_0\ne 0$. Sinceboth $P(u_0)=0$ and $P^{\prime }(u_0)=0$, with $u_0\in (-1,1)$.

The source's unqualified final sentence omits the nonvertical restriction. A vertical axis has $u=\pm 1$, where the displayed reduction divides by zero and the common root need not be interior. For example, the stationary upright top has $n=h=0$, $E=mga$, andIts only repeated root is $u=1$, so it has no common root with its derivative in $(-1,1)$. Thus the interior double-root conclusion holds for nonvertical steady precession. The vertical configurations require separate endpoint treatment. This is nonvertical qualification in the steady-top double-root condition.