Re: moment of inertia of a cube
- From: John C. Polasek <jpolasek@xxxxxxxxxx>
- Date: Fri, 30 Mar 2007 21:10:13 -0500
On Fri, 30 Mar 2007 13:44:22 -0800, Uncle Al <UncleAl0@xxxxxxxxxxxxx>
wrote:
"John C. Polasek" wrote:Same answer but fruitier: .5Ma^2 x Identity matrix. Who'd athunk it?
On 29 Mar 2007 22:05:37 -0700, bob@xxxxxxxxxxxxxx wrote:
I calculated today that the moment of inertia of a cube does notIt would be interesting to see how you calculated that fact. But it is
depend on the axis of rotation. Is there a real intuitive way to see
this?
Thank you.
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!
John Polassek
.
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