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Moment of inertia of a plate

  • Thread starter Lopina
  • Start date
  • #1
14
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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.

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

Homework Equations



Moment of inertia [tex]I=\int r^{2}dm[/tex]

Perpendicular axis theorem [tex]I_{z}=I_{x}+I_{y}[/tex]

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 [tex]I_{x}=I_{y}[/tex]

Using a Perpendicular axis theorem I have [tex]I_{z}=2I_{x}[/tex]

I need [tex]I_{x}=0.5I_{z}[/tex]

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

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

But somehow, I think I'm wrong :biggrin:
 

Answers and Replies

  • #2
diazona
Homework Helper
2,175
6
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 :biggrin:
 
  • #3
14
0
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.
 
Last edited by a moderator:
  • #4
181
0
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
 
Last edited:
  • #5
14
0
Ty for your help, David.
It sure brightens things up a bit for me
 

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