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