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Re: [Phys-L] intermediate axis theorem

• From: John Denker <jsd@av8n.com>
• Date: Wed, 26 Dec 2012 19:41:09 -0700

On 12/26/2012 07:26 AM, Carl Mungan wrote:
I'm looking for an intuitive argument for the intermediate-axis
theorem (IAT), rather than a formal derivation via Euler's
equations.

Gack! In my previous answer, I violated an important principle:
When talking about stability, it pays to say explicitly "stable
with respect to WHAT?"

Web searches indicate that I am not the first person to make
this mistake in the context of this problem.

So, there are two separate versions of this problem to be
considered ... plus an interesting interaction between the
two versions.

A) The easy version applies to an isolated _almost_ rigid
system. There is a tiny bit of internal dissipation. The
dissipation arises from the fact that the system is not 100%
rigid.

The assumptions guarantee that the system has constant
angular momentum ... but not necessarily constant rotational
energy. Some of the energy might be dissipated as heat.

This can be analyzed in terms of basic 3rd-grade physics.
As surely as balls roll downhill and are stable at the
bottom of the hill, this system is stable (in the
thermodynamic sense) when rotating around the axis with
the largest angular momentum.

B) The more challenging case, which was probably the one CM
intended, assumes constant rotational energy as well as
constant angular momentum.

Here we are interested in stability in the Lyapunov sense.
That is, given nearby initial conditions, do we get nearby
trajectories in the long term?

The intuitive way to handle this is to think about it in
angular momentum space. The contours of constant energy
are ellipsoids in this space. Let's choose one such
contour, i.e. let's choose a definite energy.

If the norm of the angular momentum is as small as possible
(consistent with the chosen energy) the angular momentum must
be at the short axis of the ellipsoid. If the norm is merely
/near/ the smallest possible value, the angular momentum must
be near the short axis. The geometry of spheres intersecting
with an ellipsoid guarantees it.

Similarly, if the norm of the angular momentum is at or near
the largest possible value, the angular momentum vector must
be at or near the long axis of the ellipsoid.

Now (!) consider the intermediate case. There are lots of
places the angular momentum vector could wander relative to
the ellipsoid. (In any inertial frame the angular momentum
vector doesn't wander at all, but the ellipsoid does, as
we re-orient the object. Alternatively, in a frame comoving
with the body, the ellipsoid stays fixed but the angular
momentum wanders.)

So, in the intermediate case I haven't yet proved that it
will tumble, but I have explained why it could tumble (and
could not tumble in the other two cases). Proving that it
*will* tumble might require a more detailed analysis.

I tried googling for images of the intersection of spheres
with an ellipsoid. Here is the crucial intermediate case:

It should be trivial to cobble up more such images using x3d
aka VRML. I can easily visualize such things in my mind's
eye, but when explaining it to somebody else it helps to
have the picture.

C) In the real world, the object always has some internal
friction. If the object is spinning on the "big" axis, it
is unconditionally stable, so the friction doesn't matter.
Interestingly, if it is spinning smoothly on the "small"
axis, the friction might act only slowly. This axis is
therefore in some sense metastable. In the intermediate
case, the system is wobbling like crazy, causing lots of
internal friction, so it will rather quickly decay to the
lowest-energy state.

This case (C) is a weird combination of the previous two
cases.