• Support PF! Buy your school textbooks, materials and every day products Here!

Product of inertia of a quadrant

  • Thread starter mattu
  • Start date
  • #1
3
0

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, 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
 
Last edited by a moderator:

Answers and Replies

  • #2
lanedance
Homework Helper
3,304
2
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
 
  • #3
rock.freak667
Homework Helper
6,230
31

Homework 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

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
 
  • #4
3
0
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:
  • #5
lanedance
Homework Helper
3,304
2
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?
 
  • #6
3
0
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?

thanks..
 

Related Threads on Product of inertia of a quadrant

  • Last Post
2
Replies
48
Views
15K
  • Last Post
Replies
4
Views
1K
Replies
2
Views
5K
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
5
Views
664
  • Last Post
Replies
2
Views
3K
  • Last Post
Replies
1
Views
10K
  • Last Post
Replies
1
Views
2K
Replies
1
Views
716
Replies
10
Views
1K
Top