= Solution
\b[Inertia is a <tensor>.] For a <rigid body>, distances between material points are fixed. Write each position relative to the <centre of mass> as $\mathbf r$, so its rotational <velocity> is $\boldsymbol\omega\times\mathbf r$. Integrating $\mathbf r\times(\boldsymbol\omega\times\mathbf r)$ gives <angular momentum>
$$
L_i=I_{ij}\omega_j,\qquad
I_{ij}=\int(r^2\delta_{ij}-r_ir_j)\,dm.
$$
The <rotational kinetic energy> is
$$
T_{\mathrm{rot}}=\frac12\boldsymbol\omega\cdot\mathbf I\boldsymbol\omega.
$$
The <inertia tensor> is real and symmetric, so <orthogonal> <principal axes> diagonalize it. <Angular momentum> is parallel to <angular velocity> only along a <principal axis>, or within a subspace of equal principal moments. For translation plus <rotation> about the <centre of mass>, $T=Mv_{\mathrm{CM}}^2/2+T_{\mathrm{rot}}$. Moving the reference to a parallel axis adds $M(d^2\mathbf1-\mathbf d\mathbf d^{\mathsf T})$ to the <tensor>, the <tensor> form of the parallel-axis theorem.
\b[Euler's equations.] The external <torque> about the <centre of mass> obeys $\boldsymbol\tau=(d\mathbf L/dt)_{\mathrm{space}}$. In body axes, the transport rule gives
$$
\boldsymbol\tau=\left(\frac{d\mathbf L}{dt}\right)_{\mathrm{body}}+\boldsymbol\omega\times\mathbf L.
$$
The principal moments are constant in that frame, hence
$$
\boxed{I_1\dot\omega_1+(I_3-I_2)\omega_2\omega_3=\tau_1,}
$$
$$
I_2\dot\omega_2+(I_1-I_3)\omega_3\omega_1=\tau_2,\qquad
I_3\dot\omega_3+(I_2-I_1)\omega_1\omega_2=\tau_3.
$$
These are <Euler's equations (rigid body dynamics)>. With zero <torque>, spatial <angular momentum> and <kinetic energy> are conserved, although the body components generally vary. For distinct moments, small perturbations about <rotation> around the largest or smallest principal moment oscillate, whereas those about the intermediate moment grow. This is the intermediate-axis instability.
\b[A lamina.] Choose the lamina's plane as $z=0$. Then $I_{xz}=I_{yz}=0$, so the normal is a <principal axis>, and
$$
\boxed{I_z=\int(x^2+y^2)dm=I_x+I_y.}
$$
This perpendicular-axis theorem applies to a planar <mass> distribution; it is not a general identity for a three-dimensional body. Rotate the in-plane axes to diagonalize the remaining symmetric two-by-two block. For a uniform rectangular lamina with sides $a,b$ and <mass> $M$, its central principal moments are $I_x=Mb^2/12$, $I_y=Ma^2/12$ and $I_z=M(a^2+b^2)/12$. If $a>b$, then $I_x<I_y<I_z$: torque-free <rotation> about the long in-plane axis or the normal is stable, while that about the short in-plane axis is unstable. Coincident moments in a square require the degenerate, rather than strict intermediate-axis, case. Under a constraint to rotate only about the normal, the motion reduces to the scalar equation $\tau_z=I_z\dot\omega_z$.
\b[A <symmetric top> without <torque>.] Let $I_1=I_2=I_\perp$ and $I_3=I_\parallel$. Euler's equations reduce to
$$
\dot\omega_3=0,\qquad
\dot\omega_1=-\nu\omega_2,\qquad
\dot\omega_2=\nu\omega_1,
\qquad \nu=\frac{I_\parallel-I_\perp}{I_\perp}\omega_3.
$$
Thus $\omega_1+i\omega_2$ rotates at signed body-frame <frequency> $\nu$, while the axial component stays fixed. The symmetry axis has constant angle to fixed spatial <angular momentum>. Indeed $\mathbf L=I_\perp\boldsymbol\omega+(I_\parallel-I_\perp)\omega_3\mathbf e_3$, so
$$
\dot{\mathbf e}_3=\boldsymbol\omega\times\mathbf e_3
=\frac{\mathbf L}{I_\perp}\times\mathbf e_3.
$$
It precesses in space about $\mathbf L$ with angular speed $L/I_\perp$. This spatial <precession> and the body-frame <frequency> $\nu$ describe different motions and should not be identified.
\b[A heavy top.] If a <symmetric top> is pivoted a distance $\ell$ from its <centre of mass>, gravity gives <torque> $\mathbf r_{\mathrm{CM}}\times M\mathbf g$. Here the relevant moments are about the fixed pivot. The axial <angular momentum> and the vertical angular-momentum component are conserved, because gravity supplies neither an axial <torque> nor a <torque> about the vertical. Using tilt $\theta$, <precession> angle $\phi$ and <spin> angle $\psi$, its <Lagrangian> is
$$
\mathcal L=\frac12I_\perp(\dot\theta^2+\dot\phi^2\sin^2\theta)
+\frac12I_\parallel(\dot\psi+\dot\phi\cos\theta)^2-Mg\ell\cos\theta.
$$
Eliminating the two conserved cyclic momenta leaves one-dimensional motion in an effective tilt potential; bounded oscillation in tilt is <nutation>. Steady <precession> at $\dot\phi=\Omega$ and fixed tilt satisfies
$$
I_\perp\Omega^2\cos\theta-L_3\Omega+Mg\ell=0.
$$
For a rapidly spinning top on the slow branch, the first term is small and
$$
\boxed{\Omega\simeq\frac{Mg\ell}{L_3}=\frac{Mg\ell}{I_\parallel\omega_3}.}
$$
Gravity then changes the direction of <angular momentum> rather than simply tipping the top straight down. The approximation fails near conditions where <nutation> or the fast-precession branch matters; the full equations retain those possibilities.
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