Past exam of the mathematics course of the University of Oxford / 2015 / Prelims / Mathematics IV / 4 / b / Solution

Write $c=\mathbf{k}\cdot \mathbf{e}$. Since $\mathbf{s}=\ell (\mathbf{k}-\mathbf{e})$, the total torque simplifies toTaking 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. HenceTaking the dot product of ${\dot{\mathbf{L}}}_Q=\mathbf{\tau }$ with the fixed vector $\mathbf{k}$ givesThese are conservation of the axial spin and of the vertical component of angular momentum.

For energy, the kinetic energy and potential energy areDot the equation of motion, with $\dot{n}=0$, with $\mathbf{e}\wedge \dot{\mathbf{e}}$. The cross product identity $(\mathbf{e}\wedge \mathbf{u})\cdot (\mathbf{e}\wedge \mathbf{v})=\mathbf{u}\cdot \mathbf{v}-(\mathbf{e}\cdot \mathbf{u})(\mathbf{e}\cdot \mathbf{v})$ givesIt follows thatsoThe 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.