solution: qualitative moment of inertia

[John Denker <jsd@MONMOUTH.COM>]

At 06:46 PM 6/25/00 -0400, I posed the following puzzle:

> Suppose I take an ordinary uniform solid cube and skewer it with an axis
> that runs through an arbitrary point on one face, perhaps (X=1, Y=0.33,
> Z=pi/4), thence through the center and out the opposite
> face.  Question:  what can you say about the moment of inertia of the cube
> as it rotates about this axis?

and I gave some hints.

Some people solved the puzzle immediately -- but they didn't spoil it by
posting the details.  I have been informed (off list) that other folks
would like to see the details.  So here goes:

The way I think about it is this:

1) A vector is not just three numbers piled together;  it is a physical and
geometrical object that behaves in a definite way under
rotation.  Similarly, a tensor is not just nine numbers piled together;  it
is a physical and geometrical object that behaves in a definite way under
rotation.

2) There is such a thing as the _tensor of inertia_.

3) We know that in general an object has three principal axes of
inertia.  These are eigenvectors of the tensor of inertia.

4) Consider a cube centered on the origin and oriented so that its faces
are perpendicular to the X, Y, and Z axes.  Call this the "standard"
orientation.  It is easy to calculate the moment of inertia of the cube
about, say, the X axis.  It is obvious that the moment of inertia about
each axis (X, Y, or Z) is the same.  It is also obvious by symmetry that
each of these axes is an eigvenvector of the tensor of inertia;  the cube
won't wobble if you spin it about one of these special axes.

5) Therefore the tensor of inertia is just a multiple of the identity matrix.

6) The identity matrix is an object of very high rotational
symmetry.  There is no higher symmetry.  It is the symmetry of the sphere.

(If you want to tell the difference between a sphere and a cube, you need
something higher than a 2nd-rank tensor.)

7) Therefore the moment of inertia of the cube about _any_ axis through the
center is the same as the moment of inertia about the "standard" X-axis.