According to the intermediate axis theorem, an object rotating about its intermediate axis with a very slight perturbation will undergo periodic flips in its orientation in the absence of external forces. Initially, almost all the angular momentum is directed along the intermediate axis. However, during the flip, some of the angular momentum must be transferred to the other two principal axes to allow the flip to occur. As far as my understanding goes, angular momentum conservation says that in the absence of external torque, both the magnitude and direction of angular momentum must be conserved. According to simulations I've created, the magnitude of angular momentum is conserved, but the direction changes during a flip as described above. How does this mesh with angular momentum conservation?

Edit to clarify: By slight perturbation, I mean that the object begins rotating about an axis that is ever so slightly tilted from its intermediate axis. There is no torque at any point during the simulation. What I mean by angular momentum not being conserved is this: Let's say the intermediate axis is initially along the y-axis. Then $\omega_y$ is on the order of 1, while $\omega_x$ and $\omega_z$ are on the order of 1e-5. Likewise, $L_x=I_x\omega_x$ and $L_z=I_z\omega_z$ are tiny compared to $L_y$. Yet, at fixed intervals, the object will rapidly flip its orientation. After the flip, angular momentum is the same as before, yet during the flip, the object must rotate about the x and/or z axes with high $\omega_x$ and/or $\omega_z$. This has been observed in experiments and computer simulations and is the part that I believe violates angular momentum conservation, not its rotation after the flip.

  • $\begingroup$ "a slight perturbation" mens a slight force, so "absence of external forces" is not true, the angular momentum is almost conserved. $\endgroup$
    – trula
    Jul 13 at 18:25
  • $\begingroup$ By slight perturbation, I mean that the initial axis of rotation is slightly off from the intermediate axis. There is no force. $\endgroup$
    – Leon
    Jul 13 at 18:42
  • $\begingroup$ The angular momentum vector $~\vec L$ is not conserved . $\vec L \cdot \vec L$ is conserved $\endgroup$
    – Eli
    Jul 13 at 20:02
  • $\begingroup$ Why isn't 𝐿⃗ conserved? Conservation of angular momentum requires both magnitude and direction. $\endgroup$
    – Leon
    Jul 13 at 20:54

4 Answers 4


The angular and linear momenta are both conserved. This is the reason why a stable rotation about a non-easy and non-hard axis is not possible.

A continuous rotation about any other axis requires periodic torque to be maintained.

This is because when a body spins about such an axis, its angular moment will be not aligned with its rotation axis. As a result, the rotational axis will precess around the angular momentum vector.

This short clip nicely demonstrates this behavior for a rotational axis that is almost along the intermediate axis. The body reverses its own sense of rotation, but it also flips over at the same time, conserving its angular momentum.

  • $\begingroup$ I don't understand what you mean by a non-easy or non-hard axis. Here is the Wikipedia article on the intermediate axis theorem (en.wikipedia.org/wiki/Tennis_racket_theorem). There is no external force required. This experiment has been done in zero gravity, and it occurs there without any torque. $\endgroup$
    – Leon
    Jul 13 at 18:44
  • $\begingroup$ I mean the axes with the highest and lowest moment of rotational inertia. These are the "hardest" and "easiest" to make an object spin, hence these names. $\endgroup$
    – tobalt
    Jul 13 at 18:45
  • $\begingroup$ This doesn't answer my question. If angular momentum is conserved, how can the axis of rotation shift as occurs for, as you say, any non-easy or non-hard axis? Angular momentum is parallel to the axis of rotation. $\endgroup$
    – Leon
    Jul 13 at 18:48
  • 2
    $\begingroup$ @Leon Angular momentum is parallel to the axis of rotation. exactly not. 😊 Angular momentum is the product of the axis of rotation and the moments of rotational inertia. If these moments differ, then the rotation and angular momentum are in general non-parallel. $\endgroup$
    – tobalt
    Jul 13 at 18:53
  • $\begingroup$ Even so, can you say that the object seen at this timestamp doesn't change the direction of its angular momentum during the flip? It clearly changes its axis of rotation in a drastic way, and it's not just a precessing either. $\endgroup$
    – Leon
    Jul 13 at 19:02

First off: recommendation of the following two resources:

2018 Article by Nicholas Mecholsky Analytic formula for the geometric phase of an Asymmetric top

2020 Youtube video by David Brown The Dzhanibekov effect, equations and simulations

The video by David Brown is particularly clarifying.

David Brown has set up simulation of the most symmetric case where there are still three different moments of inertia:

Screenshot of the type of asymmetric top used in Physics Unsimplified Youtube video

Screenshot from the first video (of 2 videos) about the intermediate axis theorem

The implementation of the simulation makes the following clear: when the object is rotating there are internal stresses. If the struts would be flexible they would flex, dissipating kinetic energy. (In the equations for the simulation the object is for simplicity treated as perfectly rigid, of course.)

These internal forces are continuously relocating momentum from one part of the asymmetric top to another. The orientation and magnitude of the global angular momentum is constant (since there is no external torque). The continuous internal relocation of momentum gives the pattern of continuous change of the orientation of the angular velocity vector with respect to the asymmetric top.

David Brown points that there are only two circumstances where (internal) relocation of momentum does not occur: when the rotation is exactly along the axis of largest moment of inertia, and when the rotation is exactly along the axis of smallest moment of inertia.

In all other circumstances internal relocation of momentum does occur. That is: when the rotation is just a little bit off the axis of largest/smallest moment of inertia there is continuous relocation of momentum too, it's just less vivid.

Rigid pendulum

Pendulum with a rigid rod

David Brown offers the following comparison: a pendulum with a rigid rod. When the pendulum amplitude is very small the motion is to a close approximation harmonic oscillation. The motion pattern of an asymetric top that is rotating just slightly off the axis of smallest/largest moment of inertia looks like just a slight wobble, analogous to a pendulum swinging with small amplitude.

The other end of the spectrum is that the pendulum is released from a close to completely inverted position. Let me refer to that as being released at 175 degrees away from hanging vertically down. The pendulum will then swing back and fort between 175 degrees and minus 175 degrees. Visually that looks as if the pendulum is lingering in the inverted orientation, with an apparently sudden swing to the other inverted orientation.

Being released from close to completely inverted is a release with the potential energy of the pendulum maxed out. (As in: you cannot release with more inital potential energy than that.)

In the case of the asymmetric top:
Setting up the initial rotation state very close to the axis of intermediate moment of inertia is like the case of releasing an inverted pendulum from close to completely inverted.

Dissipation of kinetic energy

I think it is also very instructive to consider what happens when the system does have significant dissipation of kinetic energy. (Here I mean with dissipation of kinetic energy: internal dissipation due to flexing of the structure, not dissipation due to friction from something external.)

Angular velocity around the axis of largest moment of inertia has the masses furthest away from the center of rotation. So: for the same angular momentum the rotation rate will be the slowest.

In the presence of dissipation of kinetic energy: If the initial state of rotation is that the angular velocity vector is close to the axis of intermediate moment of inertia then the asymmetric top will over time proceed to a state where the rotation is along the axis of largest moment of inertia.

The initial state of rotation being close to the intermediate moment of inertia is in a sense a maxed out state.

There is an end state such that all the kinetic energy that can dissipate has dissipated. That end state is rotation around the axis of largest moment of inertia. (There is still kinetic energy at that point, but it has no opportunity to dissipate.)

An initial state of rotation close to the axis of intermedite moment of inertia is a state that is the most charged with additional kinetic energy, as compared to state of rotation around the axis of largest moment of inertia.


The intermediate axis theorem only states that this rotation is unstable, i.e., IF there is a perturbation, the axis will become another. The perturbation produces a slight non-conservation of angular momentum (it will be produced by an external force and, therefore, we do not expect strict conservation in that situation).

However, if the perturbation the alteration of the angular momentum will also be small and there will be a quasi-conservation. Although the angular momentum will be almost the same, the axis of rotation will perform a precession around the direction of the angular momentum. Although the axis will be radically different, the energy and angular momentum will be almost the same as without the existence of the perturbation.


The moment of inertia of an axis is denoted as $I$
Energy of rotation about an axis is $E = Iw^2 / 2$
Angular momentum about an axis is $L = Iw$
Each axis has a characteristic ratio $L^2/E = 2I$

Given an object that has axes with three different moments of inertia $I_1 < I_2 < I_3$, and no way dissipate rotational energy :

If you initiate a spin around the axis with the largest moment of inertia, then you have provided AM and Energy that meets $L^2/E = 2I_3$ . This ratio is too high to transfer any momentum or energy to an axis with a lower ratio, while conserving both.

If you initiate a spin around the axis with the smallest moment of inertia, then you have provided AM and Energy that meets $L^2/E = 2I_1$ . This ratio is too low to transfer any momentum or energy to an axis with a higher ratio, while conserving both.

If you provided 99% $I_3$, and 1% $I_2$, it could never transfer any more or less .

But if you initiate 99% $I_2$, and 1% $I_3$, then you have provided AM and Energy with an intermediate ratio. The object can find endless combinations of $I_1, I_2, I_3$ that can satisfy conservation of AM and Energy. It is free to follow whatever path $\vec F=m\vec a$ takes .

It is nearly impossible to set up a purely intermediate axis spin, and it can transfer to $I_1$ and $I_3$ simultaneously, thus the transfer, thus the flip. Conserved all the way, always.


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