Past exam of the mathematics course of the University of Cambridge / 2019 / ii / Paper 4 / 8E (Classical Dynamics)

Let $e_1,e_2,e_3$ be a principal body frame, let $I_1,I_2,I_3$ be the corresponding principal moments of inertia, and write the angular velocity as $\mathbf{\omega }={\sum }_i{\omega }_i{e}_i$. The body-frame components of the angular momentum areNo external torque acts, so $\mathbf{L}$ is constant in the fixed space frame. Since a body-fixed basis vector satisfies ${\dot{e}}_i=\mathbf{\omega }\times e_i$,Taking body-frame components gives the Euler equations for a torque-free rigid body:

Multiply the three Euler equations for a torque-free rigid body by ${\omega }_1,{\omega }_2,{\omega }_3$ and add. The cubic terms cancel, giving conservation of the rotational kinetic energyInstead multiply the equations by $I_1{\omega }_1,I_2{\omega }_2,I_3{\omega }_3$ and add. The cubic terms again cancel, givingthe squared magnitude of the conserved angular momentum. For an asymmetric body these two quadratic forms are not proportional, so the torque-free rigid-body invariants are independent.

First consider rotation predominantly about the third principal axis, writing ${\omega }_3=\Omega +O({\omega }_1^2+{\omega }_2^2)$. To first order the third Euler equation gives $\dot{\Omega }=0$, while the transverse components satisfyThereforeand the same equation holds for ${\omega }_2$.

Order the principal moments as $I_1<I_2<I_3$. The coefficient above is negative, so perturbations about axis three oscillate and remain bounded. Cyclically applying the same linear stability of principal-axis rotation calculation about axis one givesso that rotation is also stable. About axis two, however,which has an exponentially growing solution. Thus the intermediate axis theorem gives