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Moment of inertia of a plate

  1. May 9, 2009 #1
    1. The problem statement, all variables and given/known data
    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 (Text):

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

    2. Relevant equations

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

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

    3. 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:
     
  2. jcsd
  3. May 9, 2009 #2

    diazona

    User Avatar
    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 :biggrin:
     
  4. May 10, 2009 #3
    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: May 4, 2017
  5. May 10, 2009 #4
    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: May 10, 2009
  6. May 10, 2009 #5
    Ty for your help, David.
    It sure brightens things up a bit for me
     
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