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For nonvertical uniform precession of a heavy symmetric top, write $\dot{\varphi }=\Omega$ and fix $\theta$. The $\theta$ Euler-Lagrange equation reduces toFor $\cos \theta \ne 0$ its solutions areFor $0<\theta <\pi /2$, uniform precession therefore requires $|{\omega }_3|\ge 2\sqrt{I_{1}Mgl\cos \theta }/I_3$. For $\pi /2<\theta <\pi$, the discriminant is positive for every spin, including zero. At $\theta =\pi /2$, the condition is $\Omega =Mgl/(I_3{\omega }_3)$, requiring nonzero spin. Once this condition holds, choose $L_z=I_1\Omega {\sin }^2\theta +L_3\cos \theta$ and $\dot{\psi }={\omega }_3-\Omega \cos \theta$; both cyclic equations are then satisfied. The exactly vertical configurations are also solutions for arbitrary spin, but the azimuthal member of the Euler angles is degenerate there, so their existence is separate from this nonvertical discriminant condition.

Put $L_z=L_3(1-{ϵ}^2/2)$ and seek $\theta =O(ϵ)$. Using $\cos \theta =1-{\theta }^2/2+O({\theta }^4)$ and ${\sin }^2\theta ={\theta }^2+O({\theta }^4)$, the $\theta$-dependent part of the effective potential isIt has a local minimum precisely whenwhich is the gyroscopic stabilization of an inverted symmetric top condition. Differentiating the displayed approximation givesand henceTaking $L_3>0$ without loss of orientation, the steady precession rate isso