Past exam of the mathematics course of the University of Cambridge / 2025 / ii / Paper 3 / 8B (Classical Dynamics)

At height $z$ from the vertex, the cone has cross-sectional area $\pi R^2{z}^2/l^2$. Hence

The parallel-axis theorem isThe displacement is along ${\mathbf{e}}_3$, so it changes the two transverse moments but not the axial one. Thus, writing the vertex moments as $I_1=I_2=I$ and $I_3=J$,and

Take $\theta$ to be the angle between ${\mathbf{e}}_3$ and the upward vertical. From the supplied angular-velocity components,ThereforeBoth $\psi$ and $\varphi$ are cyclic. Their conserved conjugate momenta areand $p_{\theta }=I\dot{\theta }$. Solving for the velocities and taking the Legendre transform gives

Fix the two conserved momenta $p_{\psi }=\lambda$ and $p_{\varphi }=\mu$. The remaining canonical variables $(\theta ,p_{\theta })$ form a two-dimensional phase space with Hamiltonianwhere

An upright spinning solution has $\theta =0$, $p_{\theta }=0$, and necessarily $\mu =\lambda$, since the vertical and body-axis angular momenta then coincide. On that momentum level,Near zero,Thus the upright configuration is a strict local minimum of the reduced Hamiltonian, and hence stable, whenHere $\lambda$ is the angular momentum about the symmetry axis. This proves stability for sufficiently large axial angular momentum.