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2nd moment of area + centroid

  1. Feb 19, 2010 #1

    rock.freak667

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    1. The problem statement, all variables and given/known data
    I need to get the centroid of this shape

    http://img713.imageshack.us/img713/5607/section.jpg [Broken]

    it's a semi-circular section basically. On the left most side, the distance from the origin to to where it cuts the y-axis is 'R' and the distance from the x-axis to the end of the upper arc is 'r'.

    2. Relevant equations

    [tex]I_x = \int y^2 dA[/tex]

    [tex]I_y= \int x^2 dA[/tex]

    [tex]A \bar{x}= \int xy dA[/tex]

    [tex]A \bar{y} = \int y dA[/tex]

    3. The attempt at a solution

    I just need to know if my integrals are set up properly

    The equation of the entire circle would have been x2+y2=R2.

    [tex]A= \int y dx = \int_{0} ^{r} \sqrt{R^2-x^2}[/tex]

    Assuming my 'A' is correct

    dA=y dx

    [tex] A \bar{x} = \int xy dA = \int xy^2 dx = \int_{0} ^{r} x(R^2-x^2) dx[/tex]

    my object is symmetrical about the x-axis so [itex]\bar{y} = 0[/tex]

    [tex]I_x = \int y^2 dA = \int y^3 dx = \int_{0} ^{r} (R^2-x^2)^{\frac{3}{2}}dx[/tex]

    [tex]I_y = \int x^2 dA = \int x^2 y dx = \int_{0} ^{r} x^2\sqrt{R^2-x^2)dx[/tex]

    I think my limits may be wrong, but the integrands should be correct.

    Also, if someone can just link me to a site with the centroid and second moment of area of this shape that would be helpful as I don't need to actually derive it in my report, I just need to use the result.
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Feb 20, 2010 #2

    LCKurtz

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    Typo you forgot the dx.
    These look OK.

    This one is wrong. If you set it up as this double integral

    [tex]\int_0^R\int_{-\sqrt{R^2-x^2}}^{\sqrt{R^2-x^2}} y^2\ dydx = 2\int_0^R\int_{0}^{\sqrt{R^2-x^2}} y^2\ dydx[/tex]

    and work the inner integral, you will see why.
    and the last one is OK.
     
    Last edited by a moderator: May 4, 2017
  4. Feb 20, 2010 #3

    rock.freak667

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    thanks very much, I will have to recheck the double integral one.
     
  5. Feb 20, 2010 #4

    rock.freak667

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    I need to also find the mass moment of inertia of that section if it were rotated about the x-axis.


    I assumed an elemental section such that x2+y2=R2

    dI= y2 dm =y2 ρ dV = ρ y2 (πy2 dx) =ρπy4dx.


    [tex]I = \rho \pi \int_{0} ^{r} (R^2 -x^2)dx[/tex]

    is this correct? This somehow gives me answers like 2400 kgm2 (ρ=7850 kg/m3) with the total integrand being like 0.9. The section is very small in fact and is hollow, I don't think it should be that much.
     
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