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Homework Help: Product of inertia of a quadrant

  1. Apr 7, 2010 #1
    1. The problem statement, all variables and given/known data

    Find the product moment of inertia of the quadrant about the x-y axis. Circle diameter is 30mm

    [PLAIN]http://img132.imageshack.us/img132/2382/58644163.png [Broken]

    2. Relevant equations

    the centroid of the quadrant is located at ( 4r/3[tex]\pi[/tex] , 4r/3[tex]\pi[/tex] )

    and in general, Ixy = [tex]\int(x y dA)[/tex] equation 1

    and Ixy = Ix'y' + A.dx.dy (using the parallel axis theorm) equation 2

    3. The attempt at a solution

    Ixy = Ix'y' + A.dx.dy

    where A.dx.dy = ( 1/4 [tex]\pi[/tex] 302 ) . (35) . (60)

    My difficulty is at the Ix'y' part. I know that for symetrical objects, Ix'y' is 0.

    And since the x'y' axis is not an axis of symmetry for the quadrant, then Ix'y' is not zero.

    I think I should use equation 1 and use integration to find the the product moment of inertia about the centroidal axis. However I am stuck with deriving this integral form.

    Any help is appreciated. thanks
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Apr 7, 2010 #2


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    are you sure that is the integral you want to calcuate, I'm not too sure what the x-y axis means

    I would have thought it might mean around the axis point out of the page
    [tex] \int \int (x^2+y^2)dA [/tex]
    if that were the case, you could consider the MoI of a whoel circle, divide by 4 then use parallel axis theorem... but its been a while since I've done these
  4. Apr 7, 2010 #3


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    yes, I'd just integrate

    [tex]I_{xy}=\int_A xy dA[/tex]

    and then transform to polar coordinates x=rcosθ and y=rsinθ, dA=rdrdθ.

    You can make the boundaries for θ and r easily and then apply the parallel axis theorem.

    EDIT: I am sorry,your radius is constant so the transforms would just be x=Rcosθ and y=Rcosθ with dA being an area of an elemental section of angle dθ
  5. Apr 7, 2010 #4
    i just found the answer from an ebook to be (11/200)x(r^4) (the product moment of inertia of a quadrant about its centroidal axis)

    honestly what I really wanted was the final answer and not the derivation.

    however I still like to see the derivation of it just for curiosity. I can't manage to do it on my own..
    Last edited: Apr 7, 2010
  6. Apr 7, 2010 #5


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    how about finding the MoI f a full circle around an axis through its centre, the take 1/4 and use parallel axis theorem as detailed before?
  7. Apr 8, 2010 #6
    the result quoted from the ebook was the moment of inertia about the centroidal axis, and not the product of inertia.

    so i'm stuck again ...

    If I use the equation

    Ixy = Ix'y' + A.dx.dy

    to find the product of inertia with respect to the xy axis of a whole circle, the equation becomes Ixy = A.dx.dy (since Ix'y' is zero since its an axis of symmetry).

    Then divide the obtained answer of Ixy by 4, so as to get Ixy for the quadrant, and then use the parallel axis theorm to calculate the product of intertia with respect to the new axis passing through the centroid of the quadrant, will it work this way?

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