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Here $(I_1,I_2,I_3)=(20,17,5)M{a}^2/12$, so $I_1>I_2>I_3$ and $e_2$ is the intermediate axis. Eliminating the other components from $L^2=2I_{2}E$ and the conservation of energy relation gives, with $J=2E-I_2{\omega }_2^2$,Thus the second Euler equations for a torque-free rigid body reduces toThe sign is minus the sign of ${\omega }_1{\omega }_3$; it is constant along a nontrivial branch. Let $\Omega =\sqrt{24E/(17M{a}^2)}$. Since ${\omega }_2(0)=0$, integration yields ${\omega }_2(t)=\pm \Omega \tanh (3\Omega t/5)$, with the same sign as in the scalar equation. The remaining components areTheir initial squares are fixed by the preceding algebraic formulas with $J=2E$; their initial signs determine the branch. Hence $\boxed{\omega (t)⟶\pm \Omega e_2}$ as $t\to \infty$. This trajectory is an intermediate-axis separatrix: approaching intermediate-axis rotation on this special conserved-energy level does not make that rotation stable to generic perturbations. If $E=0$, all components vanish identically.

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).