Re: moment of inertia of a cube

"John C. Polasek" wrote:

On Fri, 30 Mar 2007 13:44:22 -0800, Uncle Al <UncleAl0@xxxxxxxxxxxxx>
wrote:

"John C. Polasek" wrote:

On 29 Mar 2007 22:05:37 -0700, bob@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.
It would be interesting to see how you calculated that fact. But it is
true and can be proved trivially.

J is a 2d rank tensor for a cube J11=J22=J33 so we J is equal to the
scalar j x the identity tensor I.

The similarity transform for a 2d rank tensor T is A*TA
where A and A* are cosine rotation matrices for angle a and -a. Thus
for any rotation a about (any) axis (compound or principal is OK)

J' = A*JA = A*IA = A*A= I = J

Thus the rotations have no effect on the cube tensor.

John Polasek

What about the right circular cylinder of height=(radius)[sqrt(3)]?
Get funky!

Same answer but fruitier: .5Ma^2 x Identity matrix. Who'd athunk it?
John Polassek

We did, though not through the front door. Petitjean's quantitative
geometric parity divergence normalized measure CHI depends upon a mass
distribution's moments of inertia. If CHI->1 that chiral distribution
must have three identical, indistinguishable moments of inertia in all
Eulerian angle orientations of the principle axis passing through the
center of mass.

Right prisms also work. Take that cylinder, shave its sides into an
orthogonal regular hexagon, and measure the radius "properly." All
three moments of inertia are indistinguishable as long as the
principle axis passes through the center of mass.

One cute thingie was finding classes of chiral mass distributions that
met this criterion but had CHI->0 with increasing radius. Uncle Al
isn't telling.

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

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