 Solution

Define the classical axial spin by the axial angular velocity of a symmetric top,Since $\mathbf{e}$ is a unit vector, $\mathbf{e}\cdot \dot{\mathbf{e}}=0$. The vector triple product identity applied to $\dot{\mathbf{e}}=\mathbf{\omega }\wedge \mathbf{e}$ givesThe inertia tensor multiplies the component perpendicular to $\mathbf{e}$ by $A$ and the component along $\mathbf{e}$ by $C$. Therefore the angular momentum about $Q$ isIts time derivative isbecause $\dot{\mathbf{e}}\wedge \dot{\mathbf{e}}=0$.

The gravitational torque about $Q$ is $h\mathbf{e}\wedge (-mg\mathbf{k})$, and the applied torque is $\ell \mathbf{e}\wedge (\kappa \mathbf{s})$. The support force acts at $Q$, so has zero torque about $Q$. The angular momentum equation is consequently