 Solution

Inertia and principal directions. About the centre of mass, the inertia tensor isIt relates angular momentum and angular velocity by $\mathbf{J}=\mathsf{I}\mathbf{\omega }$ and gives rotational kinetic energy $T=\mathbf{\omega }\cdot \mathsf{I}\mathbf{\omega }/2$. 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 $I_1,I_2,I_3$.

Derivatives in a rotating frame. Write $\mathbf{J}={\sum }_i{J}_i{\mathbf{e}}_i$ in a body-fixed orthonormal basis. Each basis vector rotates according to $(d{\mathbf{e}}_i/dt{)}_0=\mathbf{\omega }\times {\mathbf{e}}_i$. Differentiating the components and basis separately proves the rotating-frame derivative formula,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.

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 $J_i=I_i{\omega }_i$:These are Euler's equations in the principal body frame.

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 areFor ${\omega }_1=a{e}^{pt}$ and ${\omega }_2=b{e}^{pt}$, eliminate $a,b$ fromThis givesIts units are inverse square seconds. Positive $p^2$ gives a growing real exponential; negative $p^2$ gives bounded oscillatory perturbations.