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Moment of inertia for a rhombus

  1. Nov 30, 2006 #1
    Hi, this is the question:

    Give the moment of inertia for a rhombus with sides of length c and mass m, about an axis that is parallel to the plane of the rhombus and goes from one corner to the opposite corner.

    I have set up the integral:

    I=(m/c^2)*2* int(from x=0 to x=c) int(from y=x to y=-x+c*sqrt2)x^2dydx.

    where I have used the y-axis as the axis of rotation.

    I don't know if this is the correct integral. In particular I suspect of the 2 multiplying the first integral. The thing is that this 2 would be correct if I were just finding the area of the rhombus, but since i have introduced the
    x^2 that doesn't have anything to do with the area, then I don't know if it works here.
     
  2. jcsd
  3. Nov 30, 2006 #2
    Hmm, I don't know why you have a two multiplying it.

    Did you use this as your moment?
    [tex]I = \iint(x^2 + y^2) \rho (x,y) dA[/tex]

    I see the bounds to going from x=c to x=(x-c)c and y=0 to y=c/sqrt(2).
     
    Last edited: Nov 30, 2006
  4. Nov 30, 2006 #3
    No. I'm not sure I understand the bounds that u give for x.
     
  5. Nov 30, 2006 #4
    Oh sorry the bounds for x that i give in the expresion is wrong. The correct bounds for x are 0<=x<=c/sqrt2.
     
  6. Nov 30, 2006 #5

    OlderDan

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    I think you have the wrong axis. The axis is in the plane of the rhombus
     
  7. Nov 30, 2006 #6

    OlderDan

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    Don't you need to know the angle of the rhombus?
     
  8. Nov 30, 2006 #7
    I put the 2 in there because I gave those boundaries for x. The actual area of the rhombus is the double of what those boundaries in the integrals represent. But again, i'm not sure that is right. Help please.
     
  9. Nov 30, 2006 #8
    Sure. Once the rhombus is cut in half by the axis, there are four 45 degree angles and two 90. But why do u ask?
     
  10. Nov 30, 2006 #9
    Oops, you're right. You don't need the angles of the rhombus if all the sides are equal, i.e. lozenge.
     
  11. Nov 30, 2006 #10

    OlderDan

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    You do need the angle. A rhombus has 4 equal sides and can have any smaller angle. I take it from the other post that this is actually a square.

     
  12. Nov 30, 2006 #11
    yes that's right but since i put the y-axis as the axis of rotation then the square won't 'seat' on a base but on a corner by convential methods anyways.
     
  13. Nov 30, 2006 #12
    I think the original integral is correct, but OlderDan is right and you do technically need an angle because assuming the small angle is 45 is a little sketchy.
     
  14. Nov 30, 2006 #13
    I still don't understand ur question about the angle.
     
  15. Nov 30, 2006 #14
    ok great, minds, but ill wait for the other opinions too. I have a bad feeling about that integral.
     
  16. Nov 30, 2006 #15

    OlderDan

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    OK.. I have the picture now. There are actually 4 equal contributions from the 4 triangles in the 4 quadrants. The lower right quadrant boundary is y = x - c/sqrt(2). The upper right is bounded by y = -x + c/sqrt(2). I think your sqrt(2) is in the wrong place.

    Added: and your x is from 0 to c/sqrt(2) for one of the 4 triangles.
     
    Last edited: Nov 30, 2006
  17. Nov 30, 2006 #16
    But then does it mean that I have to multiply by four the double integral?
     
  18. Nov 30, 2006 #17

    OlderDan

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    You can do it that way.. 4 times the integral over 1 of the 4 triangles. See my previous note for the added comment about the x interval.
     
  19. Nov 30, 2006 #18
    If so the result should be the same from the original integral (with a corrected x boundary that I mentioned) the only difference is that your origin is at the center of the rhombus but in mine the origin is at
    (-c/sqrt2, -c/sqrt2) from ur origin. But if u actually do both integrals in this way, (with the 2 or the 4) u get different results. Why?
     
  20. Nov 30, 2006 #19
    sorry my origin is only (0,-c/sqrt2) from yours. But what I last mentioned holds ground.
     
  21. Nov 30, 2006 #20

    OlderDan

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    Ah. You were close in the first place. Sorry

    If you are integrating over x², the origin has to be on the y axis. If you put the corner at the origin, your limits should be x from - c/sqrt(2) to + c/sqrt(2) and y from x to -x + c*sqrt(2), so it was just your x limit that was off. Then you can do 2 times the integral with x from 0 to c/sqrt(2) as you intended, or 4 times the integral using those x limits and y from x to c/sqrt(2) (the lower right of the 4 triangles)
     
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