Past exam of the mathematics course of the University of Cambridge / 2009 / ii / Paper 3 / 9E (Classical Dynamics)

Put the cylinder's centre at the origin, with $-h/2\le z\le h/2$, $0\le r\le a$, and uniform density $\rho =M/(\pi a^{2}h)$. Symmetry makes the inertia tensor diagonal and gives $I_1=I_2$. Direct integration yieldsFor a transverse principal moment of inertia, the two contributions areTherefore $\boxed{I_1=I_2=M(a^2/4+h^2/12),I_3=M{a}^2/2}$ are the three principal moments of inertia.

Set $I=I_1=I_2$. The third of the Euler equations for a torque-free rigid body gives $I_3{\dot{\omega }}_3=0$, so ${\omega }_3$ is constant. Define $\Omega =(I_3-I){\omega }_3/I$. The first two Euler equations for a torque-free rigid body becomeso ${\omega }_1+i{\omega }_2=C{e}^{i\Omega t}$. Since the body-frame angular momentum has components $(I{\omega }_1,I{\omega }_2,I_3{\omega }_3)$, its transverse components rotate at $\Omega$ while its axial component is constant. Substitution of the principal moments of inertia givesThis is precession relative to the body axes; the angular momentum is constant in an inertial frame. If $h^2=3a^2$, the rate is zero even when ${\omega }_3\ne 0$. If the transverse components vanish, the precession cone degenerates to its axis.