Past exam of the mathematics course of the University of Cambridge / 2007 / ii / Paper 4 / 9C / b

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.