= Solution
For <torque-free precession of a symmetric top>, take principal <moments of inertia> $I_1=I_2=I_\perp$ and $I_3=I_\parallel$, with no external <torque>. In an <inertial frame> moving with its <centre of mass>, its <angular momentum> $\mathbf L$ is constant in an <inertial frame>, and its <rotational kinetic energy> is conserved.
In the principal body axes, <Euler equations for a torque-free rigid body> reduce to
$$
I_\perp\dot\omega_1+(I_\parallel-I_\perp)\omega_2\omega_3=0,\qquad
I_\perp\dot\omega_2-(I_\parallel-I_\perp)\omega_3\omega_1=0,\qquad
I_\parallel\dot\omega_3=0.
$$
Therefore $\omega_3$ is constant and
$$
\omega_1+i\omega_2=\omega_\perp e^{i(\Omega_bt+\alpha)},\qquad
\boxed{\Omega_b=\frac{I_\parallel-I_\perp}{I_\perp}\,\omega_3.}
$$
Thus the transverse <angular velocity> rotates uniformly in the body frame; its signed rate reverses between oblate and prolate inertia tensors. This body-frame wobble rate should be distinguished from the space-frame <precession> of the symmetry axis.
Let $\mathbf e_3$ be the body's symmetry axis. Since
$$
\mathbf L=I_\perp\boldsymbol\omega+(I_\parallel-I_\perp)\omega_3\mathbf e_3,
$$
its angle $\theta$ to $\mathbf L$ is constant, with $\cos\theta=I_\parallel\omega_3/L$, and
$$
\dot{\mathbf e}_3=\boldsymbol\omega\times\mathbf e_3
=\frac{\mathbf L}{I_\perp}\times\mathbf e_3.
$$
Hence the symmetry axis describes a cone about fixed $\mathbf L$ at the space-frame <precession> rate
$$
\boxed{\dot\phi=\frac L{I_\perp}.}
$$
There is no <nutation> of this angle for torque-free motion of an exactly <symmetric top>. In the usual <Euler angles>, $\omega_3=\dot\psi+\dot\phi\cos\theta$, so $\dot\psi=-\Omega_b$. Both rates describe the same motion in different axes. A freely tossed axisymmetric disc illustrates the effect; a top supported at a point under gravity is a different, externally torqued problem.
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