Past exam of the physics B course of the University of Cambridge / 2014 / Paper 2 / Section A / A4

For slow gyroscopic precession, the gravitational torque turns the spin angular momentum: $Mg\ell \sin \theta =\Omega I_3\omega \sin \theta$, with $I_3=M{R}^2/2$. ThusThe inclination cancels at this order. More precisely, uniform precession of a heavy symmetric top satisfiesIts slow root gives $T_p\simeq 1.57\mathrm{s}$. About $1.6\mathrm{s}$ is the slow-precession answer; arbitrary initial conditions can also produce nutation rather than exact steady precession.