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AriAstronomer Mar14-12 10:01 AM

Moment of inertia tensor
 
1 Attachment(s)
1. The problem statement, all variables and given/known data
Find the moment of inertia tensor of the plate attached below


2. Relevant equations
σ = area density


3. The attempt at a solution
So the main problem I'm having is solving for Ixx and Iyy:
1) Ixx = σ∫∫(y^2 + z^2)dydz, since there are no dz components, I don't see how you end up with 1/3Ma^2. But iIf you approach it like this:
2) Ixx = σ∫∫(y^2 + z^2)dA = (M/a^2)∫(y^2 + z^2)(ady) = 1/3Ma^2, you get the right answer (where after integration you fill in z=0), and it makes sense, but is there a way to get there from 1)?

Also, my prof seemed really confused about how the moment of inertia tensor components translate to rotations about different axes. I had a hunch that Ixx would translate to the rotation about the x-axis, Iyy about the y-axis, Izz about the z-axis, Ixy about some diagonal rotation in the xyplane, Ixz about some diagonal rotation in the xzplane, etc. Is there any truth to this? My prof said it doesn't translate that way, but didn't seem sure. I want some closure.

Thanks,
Ari

tiny-tim Mar14-12 01:25 PM

Re: Moment of inertia tensor
 
Hi Ari! :smile:
Quote:

Quote by AriAstronomer (Post 3814803)
So the main problem I'm having is solving for Ixx and Iyy:
1) Ixx = σ∫∫(y^2 + z^2)dydz, since there are no dz components, I don't see how you end up with 1/3Ma^2.
is there a way to get there from 1)?

Yes, this is an "infinitely thin" plate (in the z-direction), so as the thickness zmax -> 0, σ∫z2dz -> 0 (if we're keeping M, and therefore σzmax, constant) :wink:
Quote:

Also, my prof seemed really confused about how the moment of inertia tensor components translate to rotations about different axes. I had a hunch that Ixx would translate to the rotation about the x-axis, Iyy about the y-axis, Izz about the z-axis, Ixy about some diagonal rotation in the xyplane, Ixz about some diagonal rotation in the xzplane, etc. Is there any truth to this? My prof said it doesn't translate that way, but didn't seem sure. I want some closure.
So long as your z y z axes are axes of symmetry of the body ("principal axes"), the tensor will be diagonal, and the moment of inertia for each axis will give you the ratio between angular momentum and angular velocity for that axis.

For any other axes, the tensor isn't actually any use unless the tensor is for part of a composite body, and the axis is a principal axis of the whole body (but not that one part).

In no case does Ixy etc have any physical significance (other than as part of a calculation).


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