Re: moment of inertia of a cube

On Mar 30, 2:13 pm, Andy Resnick <andy.resn...@xxxxxxxxxxx> 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?

I've read a few of the responses, but I am wondering what question you
are really asking- the moment of inertia of *any object*, being defined
as: I = Integral(r^2 dm) is always defined in terms of a coordinate
origin, which is taken to be on the axis of rotation (except for the
parallel axis theorem). So, unless your mass distribution falls off as
1/r^2 irrespective of your coordinate origin (something clearly
unphysical), there will be a dependence on where the axis of rotation is.

Or am I missing something? After all, it's friday afternoon, and I spent
the morning fighting with my microscope...

I think he is asserting is that for axes passing through the center of
the cube the moment of inertia is independent of the orientation of
the axis. I don't know if I ever knew this, but it's certainly
plausible: for some purposes cubic symmetry is as good as spherical
symmetry.

For any substance with cubic symmetry, for example, any property
represented by a rank-2 tensor is represented by a tensor proportional
to the unit tensor. Here would seem to be an application of this
general principle, with the tensor identified with the moment of
inertia tensor -- though I have to think about this more clearly.

It's probably possible to understand the property of the moment of
inertia by breaking the cube up into a union of sets of points, each
of which independently has the desired property. No doubt this idea
will gel in a moment or two. ;-)

.

Relevant Pages

  • Re: moment of inertia of a cube
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  • Re: moment of inertia of a cube
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