 Solution

For steady precession at rate $\Omega$ about the fixed vertical unit vector $\mathbf{k}$,Consequently $\mathbf{e}\wedge \ddot{\mathbf{e}}={\Omega }^2\cos \alpha \mathbf{e}\wedge \mathbf{k}$ and $\dot{\mathbf{e}}=-\Omega \mathbf{e}\wedge \mathbf{k}$. Since the axial angular velocity of a symmetric top $n$ is constant, substitution in the equation of motion givesAs $0<\alpha <\pi /2$, the cross product can be cancelled. For a specified axial spin, the required value isThe corresponding total angular velocity isIf the stated angular velocity $\Omega \mathbf{k}$ is the total angular velocity of the top, then $n=\Omega \cos \alpha$ and the requested value specializes toThis specialization permits a direct check: ${\mathbf{L}}_Q=A\Omega \mathbf{k}+(C-A)\Omega \cos \alpha \mathbf{e}$, whose time derivative is $(A-C){\Omega }^2\cos \alpha \mathbf{e}\wedge \mathbf{k}$, exactly the required torque. The first formula keeps the separate axial spin explicit when $\Omega$ is understood as the rate of precession.