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Moment Inertia of a cube

by Cosmossos
Tags: cube, inertia, moment
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Cosmossos
#1
Jan12-10, 08:22 AM
P: 100
To which of the two cubes has a larger moment of inertia?
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I think it's the right one, is it correct?
How can I explain that without using the parallel axis theorem?
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berkeman
#2
Jan12-10, 12:10 PM
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Quote Quote by Cosmossos View Post
To which of the two cubes has a larger moment of inertia?
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I think it's the right one, is it correct?
How can I explain that without using the parallel axis theorem?
Why do you say the right one? Are you familiar with the relevant equation for calculating the moment of inertia?
Cosmossos
#3
Jan12-10, 01:01 PM
P: 100
what relevant equation?

I think it's the right one because We know that the minimal moment of inertia is throw the principal axes that goes throw the center of mass. in the right one , the rotation isn't throw the principal axes . there is also the following theorem :

The moment of inertia about an arbitrary axis is equal to the
moment of inertia about a parallel axis passing through the
center of mass plus the moment of inertia of the body about
the arbitrary axis, taken as if all of the mass M of the body
were at the center of mass.

Am I wrong?

berkeman
#4
Jan12-10, 04:05 PM
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Moment Inertia of a cube

Quote Quote by Cosmossos View Post
what relevant equation?

I think it's the right one because We know that the minimal moment of inertia is throw the principal axes that goes throw the center of mass. in the right one , the rotation isn't throw the principal axes . there is also the following theorem :

The moment of inertia about an arbitrary axis is equal to the
moment of inertia about a parallel axis passing through the
center of mass plus the moment of inertia of the body about
the arbitrary axis, taken as if all of the mass M of the body
were at the center of mass.

Am I wrong?
There may be a shortcut way to tell which has a higher moment of inertia, but for me, I'd need to calculate it. I'd use the standard definition of the Mmoment of inertia, and evaluate thge integral for the diagonal case. I don't think you can use the parallel axis theorm, since the two axes are not parallel.

I'd do the 2-D case first, to see if it offered some intuition. That is, the moment of inertia for a flat rectangular sheet, with the axes going straight versus diagonal.
vela
#5
Jan12-10, 04:58 PM
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I think they'll turn out to be equal. Try computing the moment of inertia tensor.


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