Past exam of the mathematics course of the University of Cambridge / 2025 / ii / Paper 4 / 15B (Classical Dynamics)

The rotational kinetic energy and squared angular momentum are the torque-free rigid-body invariantsandDifferentiating the energy and inserting the Euler equations givesSimilarly,

Setting ${\omega }_1={\omega }_3=0$ makes all three Euler equations stationary. The invariants then giveso the two rotations arewhere $L=\sqrt{L^2}$.

Linearize about $(0,\Omega ,0)$, where $\Omega$ is either of these values. Writing the transverse perturbations as $(\xi ,\zeta )=(\delta {\omega }_1,\delta {\omega }_3)$ givesConsequentlyThe transverse eigenvalues are thereforeone of which is positive. Both rotations are linearly unstable, which is the intermediate axis theorem.

Subtracting $I_2$ times the energy identity from the angular-momentum identity, under $L^2=2E{I}_2$, givesCombining this withyieldsand

The second Euler equation now gives, after squaring or choosing one orientation of the orbit,Indeed, the factor $I_3-I_1$ in $I_2{\dot{\omega }}_2=(I_3-I_1){\omega }_3{\omega }_1$ cancels the denominator from the product of the two preceding expressions.

For the plus sign, separation givesAfter choosing the time origin $t_0=0$,For example, a compatible choice of the other components isAs $t$ runs from $-\infty$ to $+\infty$, the solution approaches the rotations ${\omega }_2=-\mu$ and ${\omega }_2=+\mu$. It is the intermediate-axis separatrix, a heteroclinic trajectory that exhibits the instability found in part (b).