 Solution

Let $R$ be the center of mass and $g$ the uniform gravitational acceleration. The gravitational torque about $R$ is

Choose body-fixed principal axes of inertia. The inertial derivative of the angular momentum isFor torque-free motion $N=0$ and $L_i=I_i{\omega }_i$, giving the Euler equations for a torque-free rigid body

For a symmetric top, let $I_1=I_2=I$ and let $I_3$ be the axial moment. Then ${\omega }_3$ is constant andThus ${\omega }_1^2+{\omega }_2^2$ 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.