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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?

Qmechanic
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Gabriel Sandoval
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1 Answers1

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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 $$\vec{\Omega}~=~\begin{pmatrix}0\cr \Omega^{(0)}_2\cr 0\end{pmatrix} ~+~{\cal O}(\epsilon), \qquad |\epsilon|~\ll~ \Omega^{(0)}_2 ,\tag{5}$$ is $\color{red}{\text{unstable}},$ cf. a standard analytic argument $$\ddot{\Omega}_i~\stackrel{(2)+(5)}{=}~\color{red}{+}\omega^2_2 \Omega_i~+~{\cal O}(\epsilon^2), \qquad i~\in~\{1,3\}, \tag{6}$$ where $$\omega_2~:=~\Omega^{(0)}_2\sqrt{\frac{(I_1-I_2)(I_2-I_3)}{I_1I_3}}, \tag{7}$$ see e.g. the Phys.SE answer by David Bar Moshe. The solution to the 2nd order ODE (6) is $$ \Omega_i(t)~=~\sum_{\pm} A_i^{\pm}\exp(\pm\omega_2t)~+~{\cal O}(\epsilon^2), \qquad i~\in~\{1,3\}, \tag{8}$$ with 4 integration constants $A_1^{\pm},A_3^{\pm}\in\mathbb{R}$. The solution (8) generically blows up exponentially except if both constants $A_1^+=0=A_3^+$ are exactly zero, which is unlikely.

      Moreover, we know that $(\Omega_1,\Omega_3)$ is constrained to move on the intersection of the surfaces of constant energy $E$ and constant angular momentum square $\vec{L}^2$, which is a hyperbola (line) in the $(\Omega_1,\Omega_3)$ plane $${\cal O}(\epsilon^2)~=~\underbrace{\left(\frac{I_1}{I_2}-1\right)}_{>0}I_1\Omega_1^2 - \underbrace{\left(1-\frac{I_3}{I_2}\right)}_{>0}I_3\Omega_3^2~=~\frac{\vec{L}^2}{I_2}-2E ~=~{\rm constant}, \tag{9}$$ if the constant is non-zero (zero), respectively. One may show that if the constant (9) is non-zero, then $A_1^+$ and $A_3^+$ cannot both be zero, i.e. the solution blows up.

  • Case $I_1=I_2>I_3$: Then $\Omega_3$ and $L_3$ are constants, cf. eq. (3). Then $$\ddot{\Omega}_i~\stackrel{(2)}{=}~-\omega^2_3 \Omega_i, \qquad i~\in~\{1,2\}, \tag{10}$$ where $$\omega_3~:=~\Omega_3\sqrt{\frac{(I_1-I_3)(I_2-I_3)}{I_1I_2}}~=~{\rm const}. \tag{11}$$ 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 $$\ddot{\Omega}_i~\stackrel{(2)}{=}~-\omega^2_1 \Omega_i, \qquad i~\in~\{2,3\}, \tag{12}$$ where $$\omega_1~:=~\Omega_1\sqrt{\frac{(I_1-I_2)(I_1-I_3)}{I_2I_3}}~=~{\rm const}. \tag{13}$$ 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).

[Above we have implicitly assumed that $\omega_i$ in eqs. (7), (11), and (13) are never exact zero, but strictly positive. In practice, this is true.]

Interestingly, the degenerate cases can be solved exactly with relatively simple closed formulas. See also this related Phys.SE post.

Qmechanic
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