Past exam of the mathematics course of the University of Cambridge / 2019 / ii / Paper 4 / 15E (Classical Dynamics)

A Lagrange top is a rigid body that is symmetric about a principal axis, has a point on that axis fixed in space, and has its center of mass on the same axis while gravity acts uniformly. Here $I_1=I_2$ is the transverse principal moment of inertia about the fixed point, $I_3$ is the moment about the symmetry axis, $M$ is the total mass, and $l$ is the distance from the fixed point to the center of mass.

The Euler angles for a symmetric top use $\theta$ for inclination, $\varphi$ for precession, and $\psi$ for spin about the body axis. Since $\psi$ is a cyclic coordinate, its generalized momentumis conserved. The coordinate $\varphi$ is also cyclic, sois a second integral. Finally, the Lagrangian has no explicit time dependence, and conservation of energy from time-translation invariance gives the independent integral

For steady precession set $\theta$ constant and $\dot{\varphi }=\Omega$ constant. The $\theta$ Euler-Lagrange equation, with $0<\theta <\pi /2$, reduces after division by $\sin \theta$ toThis quadratic has a real precession rate precisely when its quadratic discriminant is nonnegative. Hence Steady precession of a Lagrange top is possible if and only if

Put $z=x_3$. The body is a solid of revolution with disk radiusIts volume and uniform mass density are

By the continuous center of mass formula, the distance of the center of mass from the pointed origin is

For a disk of radius $R$, $\int r^{2}dA=\pi R^4/2$. Therefore the axial moment isFor the transverse $x_1$-axis, $\int x_2^{2}dA=\pi R^4/4$, while every slice also contributes $z^2$ times its mass. HenceBy axial symmetry $I_2=I_1$. These values agree with the moment of inertia of a pointed cubic-profile solid of revolution.