= Symmetric top with a quadratic alignment potential
{title2=$\kappa\ell^2=mgh+A\Omega^2\cos\alpha-Cn\Omega$}
Consider a <symmetric top> with fixed vertex $Q$, axial <unit vector> $\mathbf e$, <principal moments of inertia> $A,A,C$, and centre of mass at $h\mathbf e$. Let a <force> $\kappa\ell(\mathbf k-\mathbf e)$ act at $\ell\mathbf e$, where $\mathbf k$ is the upward vertical <unit vector>. Together with <gravity>, it produces <torque> $(\kappa\ell^2-mgh)\mathbf e\times\mathbf k$ and <potential energy> $mgh\mathbf k\cdot\mathbf e+\kappa\ell^2(1-\mathbf k\cdot\mathbf e)$.
The <axial angular velocity of a symmetric top> $n$, vertical <angular momentum> $\mathbf k\cdot\mathbf L$ and total <mechanical energy> are conserved. For steady <precession>, $\dot{\mathbf e}=\Omega\mathbf k\times\mathbf e$ and $\mathbf k\cdot\mathbf e=\cos\alpha$. Substituting $\mathbf e\times\ddot{\mathbf e}=\Omega^2\cos\alpha\,\mathbf e\times\mathbf k$ into $A\mathbf e\times\ddot{\mathbf e}+Cn\dot{\mathbf e}=(\kappa\ell^2-mgh)\mathbf e\times\mathbf k$ gives the condition in the title for nonvertical inclination. If the total <angular velocity> itself is $\Omega\mathbf k$, then $n=\Omega\cos\alpha$, and the condition becomes $\kappa\ell^2=mgh+(A-C)\Omega^2\cos\alpha$.
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