Re: moment of inertia of a cube

On Mar 30, 12:06 pm, Uncle Al <Uncle...@xxxxxxxxxxxxx> wrote:
b...@xxxxxxxxxxxxxx wrote:

I calculated today that the moment of inertia of a cube does not
depend on the axis of rotation. Is there a real intuitive way to see
this?

Thank you.

We trivially locate the center of mass through which all relevant
rotation axes must pass. By what means can we distinguish any of the
eight corners or six faces to create a unique rotation axis? A point,
planes, and rotation axes of symmetry of the cube show the corners and
faces are inertially indistinguishable. Small stuff.

Does a rotation axis through a corner and the center of mass give
three identical orthogonal moments of inertia as does a rotation axis
through a face center and the center of mass? Do all 0.5(pi)
steradians of unique rotation axes give identical triplets of moments
of inertia? (Any axis through the top hemisphere goes through the
bottom hemisphere - mirror plane. Then chop out redundancy from the
other two mirror planes.)

How 'bout a tetrahedron with rotation axes passing through its center
of mass?

How 'bout a right circular cylinder whose height is (radius)[sqrt(3)],
with rotation axes passing through its center of mass? Do all unique
axes of rotation give orthogonal identical triplet moments of
inertia? The cylinder is not a Platonic solid nor does it have a
minimum surface/volume ratio (that would be height equal to
diameter). Symmetry again severely cuts down the possible unique
orientations of the first rotation axis.

Are we having fun yet? "8^)

--
Uncle Alhttp://www.mazepath.com/uncleal/
(Toxic URL! Unsafe for children and most mammals)http://www.mazepath.com/uncleal/lajos.htm#a2

You are learning to use third transform in solution. Very good.

"Axes" as third location. Very good.

.

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