We have the Euler equations for a rotating body as follows

$$I_1\dot\omega_1+\omega_2\omega_3(I_3-I_2)=0\\ I_2\dot\omega_2+\omega_1\omega_3(I_1-I_3)=0\\ I_3\dot\omega_3+\omega_2\omega_1(I_2-I_1)=0$$ Where $I_i$ are the moments of inertia about the $x_i$ axis, and $\omega_i$ is the angular velocity about this axis.

It can be shown (*) that if $I_1>I_2>I_3$, then objects with angular velocity very close to $\vec\omega=(0,1,0)$ are unstable. Why is this and how can I try to picture it?

I tried to picture this using a ball, but realised this is probably not a good way to visualise it, since a ball is spherically symmetric, so the moments of inertia are not distinct. Is there any visualisation or animation that could allow me to see this rotation, and possibly understand why it is unstable?

(*) In response to @SRS's comment:

I am not sure about any references, but I know how to do it: Let $\omega_1=\eta_1,\omega_3=\eta_3$ where $\eta$ is a small perturbation, and suppose $\omega_2=1+\eta_2$. Then the Euler eqns become$$I_1\dot\eta_1=(I_2-I_3)\eta_3+O(\eta^2)\tag1$$$$I_2\dot\eta_2=O(\eta^2)\tag2$$$$I_3\dot\eta_3=(I_1-I_2)\eta_1+O(\eta^2)\tag3$$Differentiate $(1)$ and sub in $(3)$ to the resulting expression$$\ddot\eta_1=\frac{(I_2-I_3)(I_1-I_2)}{I_3I_1}\eta_1$$If $I_1>I_2>I_3$, then the constant on the right hand side is positive, so the solution to this equation is an exponential (if it was any other order, then the solution would be a $\sin/\cos$). Therefore it is unstable.


To clarify, I posted this question to see other more visual ways of understanding this effect rather than solving the equations as I did above, and to see how this effect comes into play in real life. So I don't think it is a duplicate of the other questions, since they don't have answers that fit this.


3 Answers 3


There is another nice way of seeing this mathematically. It is not too hard to show that in the body frame, there are two conserved quantities: the square of the angular momentum vector $$ L^2 = L_1^2 + L_2^2 + L_3^2 $$ and the rotational kinetic energy, which works out to be $$ T = \frac{1}{2}\left( \frac{L_1^2}{I_1} + \frac{L_2^2}{I_2} + \frac{L_3^2}{I_3} \right). $$ (Note that the angular momentum $\vec{L}$ itself is not conserved in the body frame; but its square does happen to be a constant.)

We can then ask the question: For given values of $L^2$ and $T$, what are the allowed values of $\vec{L}$? It is easy to see that $L^2$ constraint means that $\vec{L}$ must lie on the surface of a sphere; and it is almost as easy to see that the $T$ constraint means that $\vec{L}$ must also lie on the surface of a given ellipsoid, with principal axes $\sqrt{2TI_1} > \sqrt{2T I_2} > \sqrt{2T I_3}$. Thus, the allowed values of $\vec{L}$ must lie on the intersection of a sphere and an ellipsoid. If we hold $L^2$ fixed and generate a bunch of these curves for various values of $T$, they look like this:

enter image description here

Note that for a given value of $L^2$, an object will have its highest possible kinetic energy when rotating around the axis with the lowest moment of inertia, and vice versa.

Suppose, then, that an object is rotating around the axis of its highest moment of inertia. If we perturb this object so that we change its energy slightly (assuming for the sake of argument that $L^2$ remains constant), we see that the vector $\vec{L}$ will now lie on a relatively small curve near its original location. Similarly, if the object is rotating around its axis of lowest inertia, $\vec{L}$ will stay relatively close to its original value when perturbed.

However, the situation is markedly different when the object is rotating about the intermediate axis initially (the third red point in the diagram above, on the "front side" of the sphere. The contours of slightly perturbed $T$ near this point do not stay near the intermediate axis; they wander all over the sphere. There is therefore nothing keeping $\vec{L}$ from wandering all over this sphere if we perturb the object slightly away from rotating about this axis; which implies that an object rotating about its intermediate axis is unstable.

  • 2
    $\begingroup$ I hadn't seen it like this before. I still find the diagram quite abstract, but having seen the videos about the T-bar stuff, it becomes a bit clearer - the curves follow the bigger ellipse-like lines for the unstable oscillations. Thanks! :) $\endgroup$
    – John Doe
    Apr 6, 2018 at 2:48
  • $\begingroup$ Beautiful answer. $\endgroup$
    – Bob Bee
    Apr 6, 2018 at 3:08
  • 1
    $\begingroup$ These answers - yours and ZeroTheHero's - are just beautiful. I've never thought of it in these terms before: I've just been happy to know that a stable linearization of a system about an equilibrium point means that one can find a nonzero radius ball wherein one can recast the problem as a contractive fixed point problem. But - such beauty from the global view too! $\endgroup$ Apr 6, 2018 at 4:02
  • $\begingroup$ Hey Michael, I was messing around trying to write a program to produce a plot like this. I managed to get one that looks like ZeroTheHero's plot, but I can't work out how to plot your figure. Did you compute the general form of the equation of the yellow curves, and then plot those? If so, do you remember what they were? $\endgroup$
    – John Doe
    May 3, 2018 at 10:45
  • 1
    $\begingroup$ @JohnDoe: For that version of the plot, I did actually explicitly work out the form of the curves; they won't actually fit in this comment box, though. :-) Effectively, though, you express $L_1, L_2,L_3$ in spherical coordinates, write down both surfaces in terms of spherical coordinates, set them equal to each other, use the identity $\sin^2 \theta = (1 - \cos^2 \theta)$, and then solve for $\cos^2 \theta$ as a function of $\phi$. However, if you're using Mathematica, an easier way to do it is using the MeshFunctions option, with the kinetic energy as your MeshFunction. $\endgroup$ May 3, 2018 at 15:34

There's an alternative to @MichaelSeifert method which uses angular momentum and moments of inertia: it is to deal with the vector $\vec\omega$ directly as we are interested in the evolution of this vector.

One can express the kinetic energy and length squared of $\vec L$ as \begin{align} T&=\frac{1}{2}\left(I_1\omega_1^2+I_2\omega_2^2+I_3\omega_3^2\right)\\ \vec L\cdot\vec L&=I_1^2\omega_1^2+I_2^2\omega_2^2+I_3^2\omega_3^2 \end{align} These two equations define distinct ellipsoids in $\omega$-space. As there are three components to $\vec\omega$ but only two constraints, the constraints are not enough to completely determine $\vec \omega$. Rather, the evolution of $\vec\omega$ must lie on the line intersection of these two ellipses, as illustrated by the black line in the figure. (The green ellipse is constant $\vec L\cdot \vec L$ and the yellow one constant $T$.)

enter image description here

This intersection here is roughly the shape of a flattened banana, with the long side of the banana in the $\omega_2$ direction.

Thus, during the evolution, the tip of $\vec \omega$ can move along this intersection while also keeping the energy and $\vec L\cdot\vec L$ constant. The figure illustrates the case where $$ I_1=4, \quad I_2=3/2, \quad I_3=1 $$ with $T=5/2$ and $\vert \vec L\vert=2.45$.

You can see from the figure that, if we start on the intersection at the top, the evolution of $\omega_3$ is quite restricted in range (and cannot change sign), that the evolution of $\omega_1$ is also quite restricted, but that the component $\omega_2$ is much larger (basically the entire length of the flattened banana). Hence, qualitatively speaking, the rotation about the middle axis unstable because $\omega_2$ need not stay near its initial value.

(The same argument works for the components $L_k=\omega_k I_k$ in MichaelSeifert's figure except that, in his case, $L_1$ would not change sign but $L_3$ could change sign near the hyperbolic fixed point.)

  • 3
    $\begingroup$ And this is a beautiful answer also. By the way, once you learn this, and try it in a book around its middle axis, and see it tumble around, you never forget it. The best book to tumble it with is Goldsteins Classical Mechanics $\endgroup$
    – Bob Bee
    Apr 6, 2018 at 3:10
  • $\begingroup$ These answers - yours and Michael's - are just beautiful. I've never thought of it in these terms before: I've just been happy to know that a stable linearization of a system about an equilibrium point means that one can find a nonzero radius ball wherein one can recast the problem as a contractive fixed point problem. But - such beauty from the global view too! $\endgroup$ Apr 6, 2018 at 4:02
  • 1
    $\begingroup$ This is in fact the more traditional way of doing it, and in this case the curves of intersection are called polhodes (one of my favorite physics words.) I just find it a bit easier to visualize when one of the constraint surfaces is a sphere. $\endgroup$ Apr 6, 2018 at 18:01
  • $\begingroup$ @MichaelSeifert I learned one more thing today. Thanks. $\endgroup$ Apr 6, 2018 at 20:04

My physics professor helped us visualize this with a tennis racket.

Spinning with the axis along the handle is stable.

Spinning in the plane of the racket is stable. John McEnroe flipped his rackets like that a lot.

Spinning in the other direction, like how you would swing a racket, is unstable. No matter how carefully you flip the racket, it also spins in the other direction before it comes back to your hand.

In case you don't have access to a racket to play with: https://www.youtube.com/watch?v=4dqCQqI-Gis

  • $\begingroup$ Thanks! Until I asked this question, I didn't realise that this is sometimes called the tennis racket theorem. Knowing that would have made visualising it in real life much easier $\endgroup$
    – John Doe
    Apr 6, 2018 at 18:27

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.