= Solution
Choose body-fixed <principal axes> through a fixed point or the centre of <mass>, with principal moments $I_1,I_2,I_3$. <Angular momentum> has components $L_i=I_i\omega_i$. The inertial <derivative> is the body <derivative> plus $\boldsymbol\omega\times\mathbf L$, so <Euler's equations (rigid body dynamics)> read
$$
\boxed{I_1\dot\omega_1+(I_3-I_2)\omega_2\omega_3=\tau_1,\quad
I_2\dot\omega_2+(I_1-I_3)\omega_3\omega_1=\tau_2,\quad
I_3\dot\omega_3+(I_2-I_1)\omega_1\omega_2=\tau_3.}
$$
They avoid differentiating the rotating <inertia tensor> in laboratory coordinates. <Torques> must refer to the chosen point, and the displayed fixed moments require a <rigid body>.
For zero <torque>, laboratory <angular momentum> is constant, while its components can change in the body frame. Both $L^2$ and rotational <kinetic energy> $T=\tfrac12\sum I_i\omega_i^2$ are conserved. The intersections of their constant-value surfaces constrain the motion. For distinct $I_1<I_2<I_3$, steady rotation about either extreme <principal axis> is stable against small perturbations, whereas rotation about the intermediate axis is unstable. For example perturbing rotation $\Omega$ about axis 2 gives
$$
\ddot{\delta\omega_1}=
\Omega^2\frac{(I_2-I_3)(I_1-I_2)}{I_1I_3}\,\delta\omega_1,
$$
whose coefficient is positive. A symmetric body has additional simplifications and can precess. \b[Conservation of laboratory <angular momentum> does not imply constant body-frame <angular velocity>.]
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