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Moment of inertia (double integral)

  1. Apr 12, 2016 #1
    1. The problem statement, all variables and given/known data
    sqZ9I.jpg
    Determine the moment of inertia of the shaded area about the x axis.



    2. Relevant equations
    Ix=y^2dA

    3. The attempt at a solution
    Okey so I now get how to do this the standard method. But I want to know if the method I tried is correct aswell or where my mistake lies.

    My attempt:
    I look at the portion above x-axis and then multiple it with 2.
    ty9ELmz.png
    So I tried to use the method that I have been taught, a small element that touches the graph like in the picture.
    dA= (Rcos (a)- X)dady
    since x^2+y^2=R^2, x= sqrt (R^2-y^2)
    so I(x)= y^2(Rcos(a)-sqrt (R^2 - y^2))dady
    First I integrate this to da with the limits 0 to 1/2a (since the angle goes from 0 to 1/2a)
    and then that answer I integrate to dy with limits 0 to Rsin(a/2) (since y goes from 0 to Rsin(a/2)

    So basically after the first integration I get:
    y^2Rsin(a/2)-(a/2)(y^2)sqrt(R^2-y^2)dy

    and then when I integrate this I get this very long and quite complicated equation but is it correct?
    I know now the simpler way, but I find it important that I understand it.
     
  2. jcsd
  3. Apr 12, 2016 #2

    andrewkirk

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    No, it's not correct. The error arises here:
    The equation x^2+y^2=R^2 only applies to points on the curved line, and the X in your equation dA= (Rcos (a)- X)dady is the x coordinate of a point on the diagonal straight line - not on the curved line.
     
  4. Apr 13, 2016 #3
    Ah okey I think I get it. So it would have worked if I had the equation of the straight line and implented that? instead of x^2+y^2=R^2

    Can I use x= y/tan(a/2) ?

    Owh and by the way, I have not had double integrals before, but how do you know if you should integrate towards da or dy first? How to figure out the order of integration?
     
    Last edited: Apr 13, 2016
  5. Apr 13, 2016 #4

    andrewkirk

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    Yes, that should work.
    In most cases it is possible to switch the order of integration. You just need to be careful to adjust the integration limits accordingly when you do so. So you can integrate in either order. Choose whichever one gives the easiest looking integration to do.
     
  6. Apr 13, 2016 #5

    Thanks, you been a great help =D
     
  7. Apr 13, 2016 #6
    So if I get it straight the integral will be come:
    dA= (Rcosα- y/tan (α/2))dy
    and Rcosα= x = √(r^2-y^2)
    so dA= (√(r^2-y^2) - y/tan (α/2)) dy
    and Ix= y^2dA so I(x)= (y ^2√(r^2-y^2) - y^3/tan (α/2)) dy and then integrate this from 0 to Rsin(α/2).

    I thought at first it would be dA= (Rcosα- y/tan (α/2))dαdy, but I had no justification why in this situation dα would be there. Because in order to get the area of the rectangle element you don't multiply anything with a small angle.
    But so is: I(x)= (y^2√(r^2-y^2) - y^3/tan (α/2)) dy and then integrate this from 0 to Rsin(α/2)

    When I use standard integral for (y^2√(r^2-y^2)dy, I get a fairly complicated equation.
    I know the answer should be R^4/8(α-sinα), but I don't know if the answer for 2* I(x) = 2* (y^2√(r^2-y^2) - y^3/tan (α/2)) dy from 0 to Rsin(α/2) simplifies to that.
     
  8. Apr 13, 2016 #7
    So I think dA= (Rcosα- y/tan (α/2))dαdy is wrong aswell, the dα seems random and I believe should not be in there. But since cosα is a variable related to dy, I need to rewrite it. So the correct integral to me seems: dA= (√(r^2-y^2) - y/tan (α/2)) dy ?
     
  9. Apr 13, 2016 #8

    andrewkirk

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    Yes, that looks right.
     
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