= Solution
For torque-free <rotation>, <angular momentum> is fixed in inertial space, but <angular velocity> need not point along it. In principal body axes,
$$
I_1\dot\omega_1+(I_3-I_2)\omega_2\omega_3=0
$$
and cyclic permutations. A <symmetric top> with transverse moment $A$ and axial moment $C$ has constant $\omega_3$, while $\omega_1+i\omega_2$ rotates in body coordinates at signed <frequency> $(C-A)\omega_3/A$. The symmetry axis precesses around the fixed <angular momentum> in space at rate $L/A$. These body-frame and space-frame rates describe different motions and need not coincide.
\b[<Poinsot construction>] expresses the same torque-free motion geometrically. The <energy> <ellipsoid> $\sum I_i\omega_i^2=2T$ is fixed in the body. Its normal at the angular-velocity point is parallel to $\mathbf L$, and $\mathbf L\cdot\boldsymbol\omega=2T$ places that point in a fixed plane perpendicular to $\mathbf L$. The <ellipsoid> rolls on this invariable plane without slipping: the instantaneous <velocity> of its contact point is proportional to $\boldsymbol\omega\times\boldsymbol\omega=0$. The contact curve is the <polhode> on the body <ellipsoid> and the <herpolhode> on the plane.<Classical Dynamics by David Tong>[§3.4, Euler equations, stability and Poinsot construction]{ref}
For $I_1<I_2<I_3$, linearization about spin $\Omega$ on the third axis gives
$$
\ddot\omega_1=-\Omega^2\frac{(I_3-I_2)(I_3-I_1)}{I_1I_2}\omega_1.
$$
The extreme-moment axes are stable; the intermediate one is unstable. The <major-axis relaxation theorem> adds weak internal dissipation while retaining negligible external <torque>. At fixed $L$,
$$
T=\sum_i\frac{L_i^2}{2I_i}\geq\boxed{\frac{L^2}{2I_3}.}
$$
<Energy> loss favours <rotation> about the axis of largest principal moment. Here “major axis” means maximum inertia, not necessarily the body's longest geometric direction. \b[A perfectly rigid torque-free body does not dissipatively settle]; its <energy> is conserved. Distinguishing stability from relaxation also explains why minimum-inertia <rotation> can be stable in the ideal rigid model.
\b[Forced <precession>] occurs when an external <torque> changes $\mathbf L$, with $\dot{\mathbf L}=\boldsymbol\tau$. For steady <precession>, $\boldsymbol\tau=\boldsymbol\Omega_p\times\mathbf L$. A rapidly spinning, pivoted heavy <gyroscope> has $L\simeq C\omega_3$ along its symmetry axis, giving
$$
\boxed{\Omega_p\simeq\frac{Mg\ell}{C\omega_3}.}
$$
The <torque> is perpendicular to the nearly axial <angular momentum>, so it mainly changes its direction. For a steady inclined <symmetric top>, the more complete condition is
$$
A\Omega_p^2\cos\theta-C\omega_3\Omega_p+Mg\ell=0.
$$
It exhibits slow and fast <precession> branches when real solutions exist. A generic release excites <nutation> as well: inclination oscillates instead of staying fixed. Axis constraints and bearing forces must be included when specifying which <angular momentum> and pivot are being used.
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