 Solution

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.