A question about the tennis racket theorem with degenerate eigenvalues $I_1, I_2 , I_3$

Question

If a rigid body has a symmetry such that two of the principal moments of inertia are equals, i.e. $$I_1=I_2> I_3 \qquad{\rm or}\qquad I_1>I_2=I_3.$$ Are the rotations around the principal axes stable?

Answer 1

Repeated application of Euler's equations

$$\forall i ~\in~\mathbb{Z}_3:~~ \dot{L}_i~\equiv~ I_i \dot{\Omega}_i~=~\Omega_{i+1}(I_{i+1}-I_{i-1}) \Omega_{i-1} \tag{1}$$

leads to

$$ \begin{align}\forall i ~\in~\mathbb{Z}_3:&~~\cr I_1I_2 I_3 \ddot{\Omega}_i~&\stackrel{(1)}{=}~(I_{i+1}-I_{i-1}) \Omega_i\left\{(I_i -I_{i+1}) I_{i+1}\Omega_{i+1}^2- (I_i -I_{i-1}) I_{i-1}\Omega_{i-1}^2\right\} . \end{align}\tag{2}$$

Observation for later:

$$\forall i ~\in~\mathbb{Z}_3:~~\left( I_{i+1}=I_{i-1}\qquad \stackrel{(1)}{\Rightarrow} \qquad \Omega_{i}, L_{i}\text{ are constants}\right).\tag{3}$$

Assume that $$I_1~\geq~ I_2~ \geq~ I_3 .\tag{4}$$

There are several cases:

• Case $I_1>I_2>I_3$: Euler Eqs. (1) have only the three principal axes as equilibrium points $\dot{\vec{\Omega}}=0$.

  1. The major and the minor principal axes are stable, cf. a standard geometric argument where the intersection of an angular momentum sphere and an energy ellipsoid is a small loop, see e.g. the Phys.SE answers by Emilio Pisanty, Michael Seifert and ZeroTheHero.

  2. The intermediate axis $\vec{\Omega}\approx (0,\Omega_2,0)$ is $\color{red}{\text{unstable}}$, cf. a standard analytic argument $$\ddot{\Omega}_i~\stackrel{(2)}{\approx}~\color{red}{+}\omega^2_2 \Omega_i, \qquad i\in{1,3}, \tag{5}$$ where $$\omega_2~:=~\Omega_2\sqrt{\frac{(I_1-I_2)(I_2-I_3)}{I_1I_3}}, \tag{6}$$ see e.g. the Phys.SE answer by David Bar Moshe.

• Case $I_1=I_2>I_3$: Then $\Omega_3$ and $L_3$ are constants, cf. eq. (3). Then $$\ddot{\Omega}_i~\stackrel{(2)}{=}~-\omega^2_3 \Omega_i, \qquad i\in{1,2}, \tag{7}$$ where $$\omega_3~:=~\Omega_3\sqrt{\frac{(I_1-I_3)(I_2-I_3)}{I_1I_2}}~=~{\rm const}. \tag{8}$$ Conclusion: There is a (slow) precession of $\vec{\Omega}$ and $\vec{L}$ around the third axis with angular frequency $\omega_3$. In other words: If $\vec{\Omega}$ is close to the third axis, it will stay close; while if $\vec{\Omega}$ is close to the principal plane, it will not stay put, but precess in the principal plane.

• Case $I_1>I_2=I_3$: Then $\Omega_1$ and $L_1$ are constants, cf. eq. (3). Then $$\ddot{\Omega}_i~\stackrel{(2)}{=}~-\omega^2_1 \Omega_i, \qquad i\in{2,3}, \tag{9}$$ where $$\omega_1~:=~\Omega_1\sqrt{\frac{(I_1-I_2)(I_1-I_3)}{I_2I_3}}~=~{\rm const}. \tag{10}$$ Conclusion: There is a (slow) precession of $\vec{\Omega}$ and $\vec{L}$ around the first axis with angular frequency $\omega_1$. In other words: If $\vec{\Omega}$ is close to the first axis, it will stay close; while if $\vec{\Omega}$ is close to the principal plane, it will not stay put, but precess in the principal plane.

• Case $I_1=I_2=I_3$: $\vec{\Omega}$ and $\vec{L}$ are constants, cf. eq. (3).

Interestingly, the degenerate cases can be solved exactly with closed formulas.

[Above we have implicitly assumed that $\omega_i$ in eqs. (6), (8), and (10) are never exact zero, but strictly positive. In practice, this is true.]