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## Homework Statement

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]

## Homework Equations

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

and in general, I

_{xy}= [tex]\int(x y dA)[/tex]

*equation 1*

and I

_{xy}= I

_{x'y'}+ A.dx.dy (using the parallel axis theorm)

*equation 2*

## The Attempt at a Solution

I

_{xy}= I

_{x'y'}+ A.dx.dy

where A.dx.dy = ( 1/4 [tex]\pi[/tex] 30

^{2}) . (35) . (60)

My difficulty is at the I

_{x'y'}part. I know that for symetrical objects, I

_{x'y'}is 0.

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

_{x'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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