= Solution
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,
$$
V\simeq\frac{C^2(1-\alpha)}4\theta^2.
$$
For $\omega_0/C\ll1$, away from the marginal case $\alpha=1$, this gives
$$
\boxed{\theta_{\max}\simeq\frac{2\omega_0}{C\sqrt{1-\alpha}},\qquad
\Omega=\frac C2\sqrt{1-\alpha}.}
$$
The 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>.
\Image[/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-spinning-top-effective-potential.png]
{title=The vertical spinning top changes from a stable minimum to an unstable maximum}
{description=Dimensionless effective potentials are shown for alpha equal to 0.5 and 2. The second panel marks the off-axis minimum and the zero-energy turning angle approached by a near-separatrix excursion.}
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