 Solution

Fixed-angular-momentum form. Eliminating $\dot{\varphi }$ using the conserved angular momentum $h$ givesEquivalently, the polar motion has effective potentialFor the steady conical pendulum at ${\theta }_0$, $\ddot{\theta }=0$, soThe sign of $h$ agrees with the sign of $\Omega$; substituting $h=m{l}^2{\sin }^2{\theta }_0\Omega$ verifies the constant-angle result in part (d).

Period of small polar oscillations. Put $\theta ={\theta }_0+\alpha (t)$, with $|\alpha |\ll 1$ and $h$ held fixed. This is the appropriate small nutation about a conical pendulum; keeping $\dot{\varphi }$ fixed instead would violate the angular momentum constraint during the oscillation. Ifthen $f({\theta }_0)=0$ andThe linearization is $\ddot{\alpha }+{\omega }_{\theta }^2\alpha =0$, givingSince $0<{\theta }_0<\pi /2$, the coefficient is positive and these small polar oscillations are stable. To first order, $\dot{\varphi }=\Omega (1-2\alpha \cot {\theta }_0)$, making the compensating change in azimuthal angular velocity explicit. The limit ${\omega }_{\theta }/|\Omega |\to 2$ as ${\theta }_0\to 0$ concerns radial modulation of a small nonzero circular orbit; the exactly planar $h=0$ pendulum is a different, degenerate limit of these polar coordinates.