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Consider the problem of a rigid body rotating in 3D space under gravity with one point fixed. I am particularly curious about the equilibrium state where the body is spinning at a constant angular velocity $\omega$ along the vertical axis (the $z$-axis). As a consequence of conservation of angular momentum, in this situation, the center of mass must lie on the vertical axis. (Edit: There might be a mistake. I don't know whether the angular momentum must be constant for the angular velocity to be cosntant and along the $z$-axis.)

Denote the moments of inertia as $I_1$, $I_2$, and $I_3$. Note that in the equilibrium state above, the center of mass need not lie on a principal axis of inertia. (Edit: There may be a mistake. It seems the center of mass must lie on a principle axis if both the angular velocity and angular momentum are constant.)

One special case of this problem is the Lagrange top, which stands for the case $I_1 = I_2$ and the center of mass lies on the principal axis of $I_3$. This is also the case for a familiar toy top. From our experience, a spinning top is stable in the sense that it won't fall down (in a short time), while a top at rest will always fall down immediately.

We can, in fact, prove the stability of the spinning (Lagrange) top rigorously. Since the motion of the Lagrange top is integrable, we can solve its ODE of dynamics directly. We have the following stability result: there is a threshold $\omega_0 > 0$ such that the equilibrium will be stable if $\omega > \omega_0$ and unstable if $\omega < \omega_0$. A reference for this would be Goldstein's Classical Mechanics, though I think it is not rigorous in mathematics.

The word "stability" used above does not mean "Lyapunov stability", since even a slight perturbation of angular velocity $\omega$ will result in a large deviation of the rotation angle in the long run. By "stable," it actually means that the position deviation and velocity of the center of mass will remain small as long as the perturbation is small (so the line crossing the center of mass and the fixed point will also be close to the $z$-axis). However, in the Lagrange top case, this "stability" can be transformed into Lyapunov stability of an equilibrium of some reduced ODE.

My question is, what can we say about the stability of the equilibrium (spinning at a constant speed around the $z$-axis) for general rigid bodies? For example, when the moments of inertia are all different and the center of mass does not lie on any principal axis. (Edit: Actually I am most concerned with the case of a "modified" Lagrange top: the center of mass lies on the principal axis of $I_3$, which is also the $z$-axis, but $I_1$ is not necessarily equal to $I_2$. Imagine a toy top with the shape of ellipse.)

The Dihanibekov effect (for a reference, see The "Dzhanibekov effect" - an exercise in mechanics or fiction? Explain mathematically a video from a space station) tells us this question is not as simple as the Lagrange top case.

To elaborate on this, consider another integrable special case, the "Euler top", which means the fixed point is exactly at the center of mass (so gravity has no effect). Suppose $I_1 > I_2 > I_3$, we have the following result concerning Euler's equation: spinning around the principal axis of $I_1$ or $I_3$ is a Lyapunov stable equilibrium, while spinning around the principal axis of $I_2$ is unstable.

The Euler top shows there are cases where the equilibrium would always be unstable, no matter how big $\omega$ is.

Background: A month ago, I started to study the dynamics of spinning tops, concentrating on the equilibrium question because I thought it would be a good chance to practice qualitative theory of ODEs. However, I was wrong; this question is much more difficult than I thought. We cannot determine the stability of equilibrium of Hamiltonian system by linearization, since it would be impossible for all eigenvalues of its Jacobian matrix to have negative real parts. I could only solve the special cases where the equation is integrable. I have no idea how to deal with the general, nonintegrable case.

I wish to see an elegant, qualitative examination of the equilibrium of a spinning top. It would be better if we can learn new techniques for determining the stability of Hamiltonian equilibriums.

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  • $\begingroup$ This is all very standard. The trajectories (in 3D phase space of 3 angular speeds about 3 principle axis) lie on the intersection of 2 ellipsoids that are surfaces of constant angular momentum and energy. See video here (starting at 13 min mark): youtube.com/watch?v=DPu6hb2HN_8 $\endgroup$ Commented Mar 19 at 18:17
  • $\begingroup$ Regarding stability: since it is Hamiltonian system, each equilibria will either be a center or saddle type. Center is neutrally stable leading to precession, and saddle is unstable and will lead to tumbling $\endgroup$ Commented Mar 19 at 18:18
  • $\begingroup$ @PiyushGrover The video describes the motion of Euler top, which is an integrable special case. In the general case the energy and angular momentum won't be constant. I doubt the method of ellipsoids still applies (even in the Lagrange top case). $\endgroup$ Commented Mar 20 at 7:39
  • $\begingroup$ @PiyushGrover A equilibirum that is center in the linearized system may not also be a center in the original system. I do not know any specific Hamiltonian system example of this, but this article (scholarpedia.org/article/Stability_of_Hamiltonian_equilibria) says spectral stability does not imply Lyapunov stability in Hamiltonian systems. $\endgroup$ Commented Mar 20 at 7:43
  • $\begingroup$ Yes the example is special case but you mentioned special cases in your question, e.g., Euler top/ Dihanibekov effect can occur in this special case. Re: center not being center in nonlinear system, well the whole phase space portrait is plotted in that video, so you can exactly see what happens. It remains center. $\endgroup$ Commented Mar 20 at 12:00

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