= Solution
{c}
Autonomy gives conservation of <energy>, while the <cyclic coordinate> give conserved <momenta>:
$$
E=\frac A2(\dot\theta^2+\dot\phi^2\sin^2\theta)
+\frac C2(\dot\psi+\dot\phi\cos\theta)^2+mga\cos\theta,
$$
$$
p_\psi=C(\dot\psi+\dot\phi\cos\theta)=Cn,\qquad
p_\phi=A\dot\phi\sin^2\theta+Cn\cos\theta=h.
$$
Here $p_\psi$ is the <angular momentum> along the body's symmetry axis, and $p_\phi$ is its component along the fixed vertical. The time symmetry corresponds to $E$, not to an additional angle <momentum>.
With $u=\cos\theta$ and $-1<u<1$, solve
$$
\dot\phi=\frac{h-Cnu}{A(1-u^2)}.
$$
Insert this and the conserved spin rate $n$ into the <energy> equation, multiply by $A$, and isolate $\dot\theta$:
$$
\boxed{A^2\dot\theta^2
=\frac{A(2E-Cn^2-2mgau)(1-u^2)-(h-Cnu)^2}{1-u^2}
=\frac{P(u)}{1-u^2}.}
$$
For a fixed nonvertical angle, let
$$
V_{\rm eff}(u)=\frac{(h-Cnu)^2}{2A(1-u^2)}
+\frac12Cn^2+mgau.
$$
The <energy> is $A\dot\theta^2/2+V_{\rm eff}(\cos\theta)$. A constant angle in $(0,\pi)$ requires both $E=V_{\rm eff}(u_0)$ and $V_{\rm eff}'(u_0)=0$, by the reduced <Euler-Lagrange equation> and $\sin\theta_0\neq0$. Since
$$
P(u)=2A(1-u^2)(E-V_{\rm eff}(u)),
$$
both $P(u_0)=0$ and $P'(u_0)=0$, with $u_0\in(-1,1)$.
The source's unqualified final sentence omits the nonvertical restriction. A vertical axis has $u=\pm1$, where the displayed reduction divides by zero and the common root need not be interior. For example, the stationary upright top has $n=h=0$, $E=mga$, and
$$
P(u)=2Amga(1-u)^2(1+u).
$$
Its only repeated root is $u=1$, so it has no common root with its derivative in $(-1,1)$. Thus \b[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>.
Back to article page
