Re: moment of inertia of a cube
- From: Uncle Al <UncleAl0@xxxxxxxxxxxxx>
- Date: Fri, 30 Mar 2007 13:44:22 -0800
"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 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!
--
Uncle Al
http://www.mazepath.com/uncleal/
(Toxic URL! Unsafe for children and most mammals)
http://www.mazepath.com/uncleal/lajos.htm#a2
.
- Follow-Ups:
- Re: moment of inertia of a cube
- From: John C . Polasek
- Re: moment of inertia of a cube
- From: Androcles
- Re: moment of inertia of a cube
- References:
- moment of inertia of a cube
- From: bob
- Re: moment of inertia of a cube
- From: John C . Polasek
- moment of inertia of a cube
- Prev by Date: Re: Entanglement.
- Next by Date: Probabilistic-Statistical Physics 4.1: Quantum Probability in Propaganda in Physics Against Probability and Statistics
- Previous by thread: Re: moment of inertia of a cube
- Next by thread: Re: moment of inertia of a cube
- Index(es):
Relevant Pages
|
