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Inertia, symmetry?

  1. Mar 16, 2010 #1


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    1. The problem statement, all variables and given/known data
    I have to find the area moment of inertia about an axis 33 to the x-axis
    http://img227.imageshack.us/img227/2392/shape.jpg [Broken]

    2. Relevant equations
    [tex]I_\phi=\frac{1}{2}(I_{xx}+I_{yy})+\frac{1}{2}(I{xx}-I_{yy})cos2\phi - I_{xy}sin2\phi[/tex]

    3. The attempt at a solution

    I found Ixx, but since it is symmetrical, Ixx=Iyy, and since it is symmetrical about the x and y axes, Ixy=0

    So the I about any axis is just the 1/2(Ixx+Iyy) = Ixx.

    Is my thinking correct, or am I mistaking it?
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Mar 16, 2010 #2
    Looks ok to me.
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