 Double-root criterion for steady precession of a symmetric top

After fixing the two cyclic angular momenta of a heavy symmetric top, its inclination has an effective potential $U(u)$ with $u=\cos \theta$. Write the reduced energy equation in the form $P(u)=2A(E-U(u))(1-u^2)$. A steady nonvertical inclination has $E=U(u_0)$ and $U^{\prime }(u_0)=0$, the latter coming from the reduced second-order Euler-Lagrange equation. Therefore $P(u_0)=P^{\prime }(u_0)=0$.

This proves the double-root condition without dividing by a velocity that vanishes in the very solution under study. A simple turning point has $P(u_0)=0$ but nonzero derivative, and is not a steady inclination. Exactly vertical spinning corresponds to $u=\pm 1$, where the inclination coordinate and its reduced denominator are singular; these endpoint configurations require a separate treatment and do not supply an interior double root.