= Solution
\b[Inertia and principal directions.] About the centre of mass, the <inertia tensor> is
$$
I_{ij}=\int(r^2\delta_{ij}-r_ir_j)\,dm.
$$
It relates <angular momentum> and <angular velocity> by $\mathbf J=\mathsf I\boldsymbol\omega$ and gives rotational <kinetic energy> $T=\boldsymbol\omega\cdot\mathsf I\boldsymbol\omega/2$. The <principal axes> are orthogonal <eigenvector> directions of this real symmetric <tensor>; in that basis it is diagonal, with entries the <principal moments of inertia> $I_1,I_2,I_3$.
\b[Derivatives in a rotating frame.] Write $\mathbf J=\sum_iJ_i\mathbf e_i$ in a body-fixed orthonormal basis. Each basis vector rotates according to $(d\mathbf e_i/dt)_0=\boldsymbol\omega\times\mathbf e_i$. Differentiating the components and basis separately proves the <rotating-frame derivative formula>,
$$
\boxed{\left.\frac{d\mathbf J}{dt}\right|_0
=\left.\frac{d\mathbf J}{dt}\right|_{\rm body}
+\boldsymbol\omega\times\mathbf J.}
$$
The body derivative differentiates the scalar components in that basis. A fixed pivot can replace the centre of mass when the corresponding angular-momentum balance is valid.
\b[Euler equations.] The inertial angular-momentum equation is $d\mathbf J/dt=\mathbf G$. Since the principal moments are constant in the body frame, substitute $J_i=I_i\omega_i$:
$$
\boxed{\begin{aligned}
G_1&=I_1\dot\omega_1+(I_3-I_2)\omega_2\omega_3,\\
G_2&=I_2\dot\omega_2+(I_1-I_3)\omega_3\omega_1,\\
G_3&=I_3\dot\omega_3+(I_2-I_1)\omega_1\omega_2.
\end{aligned}}
$$
These are <Euler's equations (rigid body dynamics)> in the <principal body frame>.
\b[Linear perturbation of free rotation.] After the impulse, $\mathbf G=0$. Let $\omega_3=\Omega$ to leading order and $\omega_1,\omega_2$ be small. The third Euler equation changes $\omega_3$ only at quadratic order, so the linear transverse equations are
$$
I_1\dot\omega_1=(I_2-I_3)\Omega\omega_2,\qquad
I_2\dot\omega_2=(I_3-I_1)\Omega\omega_1.
$$
For $\omega_1=ae^{pt}$ and $\omega_2=be^{pt}$, eliminate $a,b$ from
$$
I_1pa=(I_2-I_3)\Omega b,\qquad
I_2pb=(I_3-I_1)\Omega a.
$$
This gives
$$
\boxed{p^2=\Omega^2\frac{(I_2-I_3)(I_3-I_1)}{I_1I_2}.}
$$
Its units are inverse square seconds. Positive $p^2$ gives a growing real exponential; negative $p^2$ gives bounded oscillatory perturbations.
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