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Let $C$ be a cuboid (rectangular parallelepiped) with edges of lengths $a < b < c$.

Consider an axis that passes through the centers of two opposite faces of $C$. There are three such axes, one passing through the centers of the $a$–$b$ faces, one through the centers of the $a$–$c$ faces, and one through the centers of the $b$–$c$ faces.

Someone told me many years ago that if you throw the cuboid in the air and spin it around the $a$–$b$ or the $b$–$c$ axis, rotation will be stable in the sense that the rotation will tend to return to its original axis if it is slightly perturbed. But they said that a cuboid spinning around its $a$–$c$ axis is unstable, in that any small deviation in the axis of rotation will tend to be magnified over time.

I have tried to verify this by tossing various cuboid objects, such as Zippo lighters, cell phones, and wooden blocks; it appears to be true.

My questions are:

  1. Did I describe this correctly? If not, what's the right description?
  2. What is the mathematical explanation of this phenomenon?
  3. Is there an intuitive explanation?
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You folks might want to merge the rotational-dynamics and rotational-kinematics tags if they are the same, or add an explanation if they are different. –  MJD Aug 16 '12 at 23:51
+1 I like this question. This is likely related to the kinematics of a rotating wingnut. See this question : mathoverflow.net/q/81960 and the videos linked there. –  Ryan Thorngren Aug 17 '12 at 0:35
The teacher once told us a story that in the early days of satellites, engineers deployed one by having it spin around the wrong axis and they ended up losing the satellite as a result. I do not remember when and where this happened. 60's ? May be googling this can bring it up. Will try to find out. Seem to remember he said it was a Russian satellite but not sure now. –  Nasser Aug 17 '12 at 2:07
Fair enough, that's fine. I put in a quick excerpt for each, which should help a bit. It's just something to keep in mind, that any time you see a tag without a description and you think you might be able to write one, you're free go ahead and do it. If you don't get it right, it's not a big deal because it can always be edited (and in fact if you're the first one to write the tag wiki excerpt, you get some freedom to define what it means :-P). –  David Z Aug 17 '12 at 3:57
For the Dzhanibekov effect, the tennis racquet theorem, and the intermediate axis theorem, see e.g. physics.stackexchange.com/q/17504/2451, physics.stackexchange.com/q/31475/2451 and links therein. –  Qmechanic Aug 17 '12 at 7:14
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1 Answer 1

up vote 6 down vote accepted

I like your description of this cool bit of unintuitive physics. I find the best balance of $a$ to $b$ to $c$ to cost of the object involved is best for a (boxed) pack of playing cards.

The mathematical explanation for this (see also Wikipedia) is that when considered in the principal axis frame (i.e. the frame of reference that rotates with the body and whose axes are the principal axes of inertia of the body), the motion can be described by the angular velocity $\vec{\omega}$ and the angular momentum $\vec{L}=I\vec{\omega}$, and in the absence of external torques it must conserve the magnitude of the angular momentum, $$L^2=L_1^2+L_2^2+L_3^2$$ (though not its direction since the frame is noninertial), and the rotational energy, $$E=\frac{L_1^2}{2I_1}+\frac{L_2^2}{2I_2}+\frac{L_3^2}{2I_3}.$$

The motion is then constrained to move along the intersections of an ellipsoid and a sphere:

enter image description here

These curves are closed ellipses, or nearly so, close to the axes with the smallest and largest moments of inertia, but they are locally hyperbolae close to the middle one. Hence the instability.

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Yes, I have tried it with packs of playing cards also! Thanks for reminding me. –  MJD Aug 17 '12 at 1:58
Nice! I had an inkling it had to do with Morse stuff... –  Ryan Thorngren Aug 17 '12 at 3:13
The direction of angular momentum is only not conserved in a noninertial frame, though, such as body-fixed coordinates. I think it's important to mention that. –  David Z Aug 17 '12 at 3:51
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