= Solution
For $\alpha>1$, $\theta=0$ is an unstable stationary point and the interior minimum satisfies $u=\alpha^{-1/2}$. Hence
$$
\boxed{\theta_{\min}=2\arccos\alpha^{-1/4},\qquad
V_{\min}=-C^2(\sqrt\alpha-1)^2.}
$$
The potential then diverges at $\pi$. For $\epsilon=\omega_0/C$, the turning point is specified exactly by the smaller quadratic root
$$
u_{\rm turn}=\frac{1+\alpha+\epsilon^2-\sqrt{(1+\alpha+\epsilon^2)^2-4\alpha}}{2\alpha},\qquad
\theta_{\max}=2\arccos\sqrt{u_{\rm turn}}.
$$
For small positive $\omega_0$, this approaches $\boxed{\theta_{\max}\simeq2\arccos\alpha^{-1/2}}$, 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 period
$$
\boxed{T_\theta=2\int_0^{\theta_{\max}}\frac{d\theta}{\sqrt{\omega_0^2-V(\theta)}}.}
$$
Near 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 $\omega_0\to0$. 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 <S7/2/c>.
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