= Solution
{c}
Write $c=\mathbf k\cdot\mathbf e$. Since $\mathbf s=\ell(\mathbf k-\mathbf e)$, the total <torque> simplifies to
$$
\boldsymbol\tau=(\kappa\ell^2-mgh)\mathbf e\wedge\mathbf k.
$$
Taking the <dot product> of the equation of motion with $\mathbf e$ gives $C\dot n=0$: both <torque> terms are perpendicular to $\mathbf e$, as are the other terms on the left. Hence
$$
\boxed{n=\text{constant}.}
$$
Taking the <dot product> of $\dot{\mathbf L}_Q=\boldsymbol\tau$ with the fixed <vector> $\mathbf k$ gives
$$
\dot p=\mathbf k\cdot\dot{\mathbf L}_Q=0,\qquad
\boxed{p=\mathbf k\cdot\mathbf L_Q=\text{constant}.}
$$
These are conservation of the axial spin and of the vertical component of <angular momentum>.
For <energy>, the <kinetic energy> and <potential energy> are
$$
T=\frac12\boldsymbol\omega\cdot\mathbf L_Q
=\frac A2|\dot{\mathbf e}|^2+\frac C2n^2,
\qquad
V=mghc+\frac\kappa2|\mathbf s|^2
=mghc+\kappa\ell^2(1-c).
$$
Dot the equation of motion, with $\dot n=0$, with $\mathbf e\wedge\dot{\mathbf e}$. The <cross product> identity $(\mathbf e\wedge\mathbf u)\cdot(\mathbf e\wedge\mathbf v)=\mathbf u\cdot\mathbf v-(\mathbf e\cdot\mathbf u)(\mathbf e\cdot\mathbf v)$ gives
$$
A\ddot{\mathbf e}\cdot\dot{\mathbf e}
=(\kappa\ell^2-mgh)\mathbf k\cdot\dot{\mathbf e}.
$$
It follows that
$$
\dot T=(\kappa\ell^2-mgh)\dot c,\qquad
\dot V=(mgh-\kappa\ell^2)\dot c,
$$
so
$$
\boxed{E=T+V=\frac12\boldsymbol\omega\cdot\mathbf L_Q+mgh\mathbf k\cdot\mathbf e+\frac\kappa2|\mathbf s|^2=\text{constant}.}
$$
The applied <force> is conservative even if $\kappa$ is negative: its <potential energy> is $\kappa|\mathbf s|^2/2$. This is a <symmetric top with a quadratic alignment potential>.
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