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Derivation for Moment of Inertia of Rectangle rotated through center

  1. Nov 19, 2013 #1
    1. The problem statement, all variables and given/known data

    I need to know how to derive the equation for the moment of inertia of a rectangle rotated about an axis through its center. The rectangle has sides a and b. I know the equation to be (1/12)M(a2+b2), but I am having trouble deriving it. I have searched all over the internet without finding any helpful solutions.

    2. Relevant equations

    dI = ∫r2dm
    dm = λdA

    3. The attempt at a solution

    I started by taking a point mass of the rectangle to be the distance r from the center. This point would have the mass dm.

    dm would then equal λdA
    λ would equal M/A or M/ab
    dA would be the area of the small point. I'm guessing this would be dydx

    So, for dm we have (M/ab)(dydx)

    Now for r2. By pythagorean thereom, r2 = x2+y2.

    Substituting these into the equation ∫r2dm, we have (M/ab)∫(x2+y2)dydx

    I'm pretty sure this is incorrect, but even if it is correct I have no idea how to integrate it. I don't even know what limits to integrate about.
  2. jcsd
  3. Nov 19, 2013 #2
    Hi Matthew487...

    Welcome to PF!!!

    The item in red should be (M/ab)∫∫(x2+y2)dydx

    The limit of the inner integral will be values 'y' will be taking .The limit of outer integral will be values 'x' will be taking.

    Considering origin to be at the center .What should be the limits of the inner integral ?What should be the limits of the outer integral ?
  4. Nov 20, 2013 #3
    Ahhhh!! Thank You! I had never done or even heard of a double integral before. That was the problem. After quickly learning what that is, I was able to solve it. For the inside dx integral the limits are from -a/2 to a/2. The outside integral for dy is from -b/2 to b/2. Thanks a ton!
  5. Nov 20, 2013 #4
    Excellent!!! :thumbs:
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