We have computed the inertia tensor for a cube with the origin at a corner and the axes along the edges.
Lets solve the equation to find the principal axes.
(The moment of inertia will be
with this definition of
.)
To facilitate the solution of the cubic equation, lets get some zeros in this determinant by subtracting the second row from
the first and the third.
If we let
, we can write the determinant.
So there are three roots,
,
, and
; corresponding to three values for
of
,
, and
.
Knowing that two equal moments of inertia along principal axes indicate a symmetry, lets find the axis for the
third value
.
One can see that all three of these equations are solved for the three components of
being equal.
So this
principal axis is the diagonal of the cube starting from the origin and going toward
.
The other roots just give the same equation three times
.
There are many possible solutions. The equation defines a plane perpendicular to the cube diagonal.
We can choose and orthogonal pair of axes in that plane.
Jim Branson
2012-10-21