# Moment of inertia of a plate

## Homework Statement

Calculate the moment of inertia of a straight homogenous plate with mass m shaped like a square where the axis of rotation goes through the diagonal of the plate.

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       ^
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a \ | /    x
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## Homework Equations

Moment of inertia $$I=\int r^{2}dm$$

Perpendicular axis theorem $$I_{z}=I_{x}+I_{y}$$

## The Attempt at a Solution

This is what I've come up with, but I don't know if I'm right.

Being this a square, I've concluded that $$I_{x}=I_{y}$$

Using a Perpendicular axis theorem I have $$I_{z}=2I_{x}$$

I need $$I_{x}=0.5I_{z}$$

I have $$I_{z}=\frac{m*\left(a^{2}+a^{2}\right)}{12}=\frac{m*\left(a^{2}\right)}{6}$$

And then I just put it in $$I_{x}=0.5I_{z}$$ and get $$I_{x}=\frac{m*a^{2}}{12}$$

But somehow, I think I'm wrong diazona
Homework Helper
Well, why do you think you're wrong?

Also, is a the side length of the square, or is it the "half-diagonal" of the square? It's not clear from your ASCII-art diagram Here's the new picture
I hope it's better than ASCII one

http://img15.imageshack.us/img15/6522/pictureaor.jpg [Broken]

The reason why I think I'm wrong is the following. If I rotate the square on the upper picture 45° in any direction around z-axis then the x-axis no longer lies on a diagonal of the square. When I try to calculate the moment of inertia of such square plate (x-axis is the axis of rotation), I get the same solution as I get when the x-axis is on the diagonal.

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You have discovered an interesting fact, which is more than a mere coincidence.
Your working could be applied to any two perpendicular axes in the plane of the plate
through its center!

If I had to criticize, I would say it is a pity that the formula for Iz is
usually considered trickier to derive than the answer you were asked for.

David

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