Past exam of the mathematics course of the University of Cambridge / 2007 / ii / Paper 4 / 9C (Classical Dynamics)

Assume the spheroid is homogeneous, as implicit in the stated moments. Put $s=\sqrt{1-e^2}$ and map a uniform solid sphere by $(X_1,X_2,X_3)\mapsto (X_1,X_2,s{X}_3)$, preserving total mass by adjusting density. For the sphere, symmetry and its moment $2M{a}^2/5$ give $\int X_i^{2}dm=M{a}^2/5$ for each coordinate. The transformed squared-coordinate integrals are consequently $M{a}^2/5,M{a}^2/5,s^{2}M{a}^2/5$. Reflection symmetry makes the off-diagonal inertia entries vanish. ThusThe original PDF gives the correct solid-sphere factor $2/5$, rather than the converted TeX's $2/3$.

Put $I=I_1=I_2$. The third of the Euler equations for a torque-free rigid body gives ${\dot{\omega }}_3=0$. The first two become ${\dot{\omega }}_1=-\Omega {\omega }_2$, ${\dot{\omega }}_2=\Omega {\omega }_1$, whereTherefore ${\omega }_1+i{\omega }_2=C{e}^{i\Omega t}$. In body coordinates, $L_1+i{L}_2=IC{e}^{i\Omega t}$ and $L_3=I_3{\omega }_3$ is constant. The angular momentum precesses about the body symmetry axis, with periodFor positive ${\omega }_3$ this is the printed expression. In the inertial frame the torque-free angular momentum is constant; the precession here is its changing representation in the rotating body frame. If $e=0$ or ${\omega }_3=0$, this body-frame precession rate is zero.