What Is the Moment of Inertia of a Square Plate About Its Diagonal?

[Lopina]

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.

       ^
       |y
       |
      /|\
     / | \ a
-------|------>
   a \ | /    x
      \|/
       |
       |

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]

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

 

[Lopina]

Here's the new picture
I hope it's better than ASCII one

http://img15.imageshack.us/img15/6522/pictureaor.jpg

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.

 

[davieddy]

Your answer is correct, (and well worked out).
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

 

[Lopina]

Ty for your help, David.
It sure brightens things up a bit for me