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

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

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.

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.