 Solution

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 $\mathbf{\omega }\times \mathbf{r}$. Integrating $\mathbf{r}\times (\mathbf{\omega }\times \mathbf{r})$ gives angular momentumThe rotational kinetic energy isThe 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, $T=M{v}_{\mathrm{CM}}^2/2+T_{\mathrm{rot}}$. Moving the reference to a parallel axis adds $M(d^{2}1-\mathbf{d}{\mathbf{d}}^{\mathsf{T}})$ to the tensor, the tensor form of the parallel-axis theorem.

Euler's equations. The external torque about the centre of mass obeys $\mathbf{\tau }=(d\mathbf{L}/dt{)}_{\mathrm{space}}$. In body axes, the transport rule givesThe principal moments are constant in that frame, henceThese are Euler's equations. 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.

A lamina. Choose the lamina's plane as $z=0$. Then $I_{xz}=I_{yz}=0$, so the normal is a principal axis, andThis 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 $I_x=M{b}^2/12$, $I_y=M{a}^2/12$ 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 ${\tau }_z=I_z{\dot{\omega }}_z$.

A symmetric top without torque. Let $I_1=I_2=I_{\perp }$ and $I_3=I_{\parallel }$. Euler's equations reduce toThus ${\omega }_1+i{\omega }_2$ 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 $\mathbf{L}=I_{\perp }\mathbf{\omega }+(I_{\parallel }-I_{\perp }){\omega }_3{\mathbf{e}}_3$, soIt 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.

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 $\varphi$ and spin angle $\psi$, its Lagrangian isEliminating the two conserved cyclic momenta leaves one-dimensional motion in an effective tilt potential; bounded oscillation in tilt is nutation. Steady precession at $\dot{\varphi }=\Omega$ and fixed tilt satisfiesFor a rapidly spinning top on the slow branch, the first term is small andGravity 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.