 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,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.