= Solution
Let $R$ be the <center of mass> and $g$ the uniform gravitational acceleration. The gravitational <torque> about $R$ is
$$
\sum_i(r_i-R)\times m_i g
=\left(\sum_i m_i(r_i-R)\right)\times g=0.
$$
Choose body-fixed <principal body frame>[principal axes of inertia]. The inertial derivative of the <angular momentum> is
$$
N=\left(\frac{dL}{dt}\right)_{\rm body}+\omega\times L.
$$
For torque-free motion $N=0$ and $L_i=I_i\omega_i$, giving the <Euler equations for a torque-free rigid body>
$$
\begin{aligned}
I_1\dot\omega_1+(I_3-I_2)\omega_2\omega_3&=0,\\
I_2\dot\omega_2+(I_1-I_3)\omega_3\omega_1&=0,\\
I_3\dot\omega_3+(I_2-I_1)\omega_1\omega_2&=0.
\end{aligned}
$$
For a <symmetric top>, let $I_1=I_2=I$ and let $I_3$ be the axial moment. Then $\omega_3$ is constant and
$$
\dot\omega_1=-\Omega\omega_2,
\qquad
\dot\omega_2=\Omega\omega_1,
\qquad
\Omega=\frac{I_3-I}{I}\omega_3.
$$
Thus $\omega_1^2+\omega_2^2$ is constant and the transverse angular-velocity vector rotates uniformly about the body symmetry axis. Its rate is $|\Omega|$; it precesses in the positive axial sense when $(I_3-I)\omega_3>0$ and in the opposite sense when this product is negative.
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