Past exam of the short options paper shared by Prelims and Parts A and B of the physics course of the University of Oxford / 2018 / S7 / 2

Differentiation givesThe angles $\varphi$ and $\psi$ are cyclic, so their canonical momenta are conserved. The Lagrangian is autonomous, so the total energy is also conserved. The momentum $p_{\theta }$ is not generally constant because gravity and the angular kinetic terms depend on $\theta$.

At the initially vertical configuration, the two conserved momenta coincide:The apparent $0/0$ angular term at $\theta =0$ is interpreted by its regular limit with these equal momenta. Substitution into energy conservation, followed by $1-\cos \theta =2{\sin }^2(\theta /2)$, givesWriting $C=I_3\omega /I_1$ yieldsThe zero of the potential at the vertical configuration includes the initially specified transverse kinetic energy separately as ${\omega }_0^2$.

Use the conserved $p_{\varphi }=p_{\psi }=I_3\omega$ in the canonical-momentum equation. For a nonzero tilt,Cancelling with the half-angle identities givesIts regular limit at the north pole is $C/2$, although the azimuth itself is not a uniquely defined coordinate at that pole.

Put $u={\cos }^2(\theta /2)$. Then $V/C^2=u^{-1}-1-\alpha +\alpha u$. For $\alpha <1$ it increases from zero at the north pole to infinity as $\theta \to \pi$ and has no interior minimum. The vertical orientation is stable. Near zero,For ${\omega }_0/C\ll 1$, away from the marginal case $\alpha =1$, this givesThe locally signed tilt oscillates at angular frequency $\Omega$, while the azimuth precesses approximately at $C/2$. The nonnegative polar coordinate is $\theta \simeq {\theta }_{\max }|\sin \Omega t|$, so its successive pole-to-pole excursions have period $\pi /\Omega$ and angular repetition frequency $2\Omega$. Stating this convention avoids confusing a signed small displacement with the polar-angle magnitude. The full axis direction combines that excursion with precession.

For $\alpha >1$, $\theta =0$ is an unstable stationary point and the interior minimum satisfies $u={\alpha }^{-1/2}$. HenceThe potential then diverges at $\pi$. For $ϵ={\omega }_0/C$, the turning point is specified exactly by the smaller quadratic rootFor small positive ${\omega }_0$, this approaches $\boxed{{\theta }_{\max }\simeq 2\arccos {\alpha }^{-1/2}}$, a finite excursion even when the initial impulse is arbitrarily small. An excursion from the north pole to the outer turning angle and back has periodNear the unstable pole, $V\simeq -{\Omega }_u^2{\theta }^2$ with ${\Omega }_u=C\sqrt{\alpha -1}/2$. The near-pole contribution gives $T_{\theta }\sim (2/{\Omega }_u)\log (\mathrm{constant}\times C/{\omega }_0)$, so the period diverges logarithmically as ${\omega }_0\to 0$. The exactly unperturbed top can remain at the unstable vertical orientation; the limiting excursions spend ever longer close to it. The direction also precesses according to part c.