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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).

A useful readout of a tumble is the alignment of the intermediate principal axis ${\hat{e}}_2$ with the conserved angular momentum direction $\hat{L}$. In the principal body frame, write $L_i=I_i{\omega }_i$ and $L=|\mathbf{L}|$. ThenA change from $c_2\simeq +1$ to $c_2\simeq -1$ means that the directed body axis has turned nearly $180^{\circ }$ relative to $\mathbf{L}$. It does not mean that $\mathbf{L}$ reversed in the fixed space frame. The unsigned line through that principal axis hides this distinction, so a marked end matters in an actual demonstration.

For the normalized intermediate-axis model with $\mathbf{\omega }(0)=(0.01,1,0)$, the rotational kinetic energy and squared angular momentum are $E=1.00005$ and $L^2=4.0001$. Hence $L^2-2E{I}_2=-0.0001$: this initial condition lies close to, but off, the intermediate-axis separatrix. An independent calculation evolves ${\dot{\mathbf{L}}}_{\mathrm{body}}={\mathbf{L}}_{\mathrm{body}}\times I^{-1}{\mathbf{L}}_{\mathrm{body}}$ and a rotation matrix taking body components to space components. The intermediate principal axis crosses $c_2=0$ at normalized times approximately $10.3773$ and $31.1319$, alternating long nearly aligned intervals with quicker reversals.

Both calculations use step size $0.005$ without rescaling either invariant. Over $0\le t\le 50$, the tumbling calculation changes rotational kinetic energy and squared angular momentum by less than $4\times 10^{-13}$ relative to their initial values. Reconstructing $\mathbf{L}$ in the fixed space frame gives relative error below $3\times 10^{-12}$. Halving the step changes the body components and rotation matrix entries by less than $3\times 10^{-10}$. These checks distinguish physical tumbling from numerical drift. On the exact intermediate-axis separatrix, approach to an intermediate-axis rotation takes infinite time; the recurrent off-separatrix example here is different.