= Solution
A freely rotating <rigid body> with no external <torque> conserves its inertial <angular momentum> and <kinetic energy>. In body-fixed <principal axes>, the Euler equations include
$$
I_1\dot\omega_1+(I_3-I_2)\omega_2\omega_3=0
$$
and cyclic permutations. The body can still change its orientation relative to its fixed angular-momentum vector; zero <torque> does not require every <principal axis> to remain fixed.
For a <symmetric top>, $I_1=I_2=I_\perp$, $I_3=I_\parallel$. The axial rate $\omega_3$ is constant, and $\omega_1+i\omega_2$ rotates in body coordinates with signed <frequency> $(I_\parallel-I_\perp)\omega_3/I_\perp$. In space,
$$
\mathbf L=I_\perp\boldsymbol\omega+(I_\parallel-I_\perp)\omega_3\hat{\mathbf e}_3,
\qquad
\dot{\hat{\mathbf e}}_3=\frac{\mathbf L}{I_\perp}\times\hat{\mathbf e}_3.
$$
Thus the <symmetry> axis precesses about fixed $\mathbf L$ at rate $L/I_\perp$ and constant inclination. This <body-frame and space-frame frequencies of free symmetric-top precession> distinction avoids confusing two different frame-dependent rates. A sphere has no dynamically distinguished <symmetry> axis even though a chosen body marker rotates.
For a general body, rotations about the largest and smallest principal inertias are stable to small perturbations, while rotation about the intermediate inertia is unstable. With $I_1<I_2<I_3$ and rotation $\omega_2=\Omega$, linearization gives $\ddot\omega_1=\Omega^2(I_3-I_2)(I_2-I_1)\omega_1/(I_1I_3)$, showing exponential growth. This is torque-free motion, distinct from the gravity-driven <precession> of a supported <gyroscope>.
Back to article page
