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Math Help - Center of mass

  1. #1
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    Center of mass

    Find the center of mass of the thin semicircular region of consant density,sigma=4 bounded by the x-axis and the curve y=(121-x^2)^1/2
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  2. #2
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    The center of mass is simply finding the the mass of the object = density * volume and finding the displacement of the mass to the x-axis and y-axis to find the center of mass, which is usually pair (x',y')
    However, This link can explain better than I can

    Pauls Online Notes : Calculus II - Center of Mass
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  3. #3
    MHF Contributor chisigma's Avatar
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    The computation of the coordinates of the 'center of mass' of a thin region A with density \sigma (x,y) is perfomed is succesive steps...

    a) first is computed the 'whole mass' as...

    M= \int \int_{A} \sigma (x,y)\cdot dx\cdot dy (1)

    b) then the coordinates of the 'center of mass' are computed as...

    x_{g} = \frac{1}{M} \cdot \int \int_{A} x\cdot \sigma(x,y)\cdot dx\cdot dy

    y_{g} = \frac{1}{M} \cdot \int \int_{A} y\cdot \sigma(x,y)\cdot dx\cdot dy (2)

    In our case \sigma(x,y) is constant so that it can be \sigma (x,y)=1 and A is an half circle of radious r and also for it we can suppose r=1. Now we can compute first ...

    M= \int_{-1}^{1} \int_{0}^{\sqrt{1-x^{2}}} dx\cdot dy = \int_{-1}^{1} |y|_{0}^{\sqrt{1-x^{2}}} \cdot dx =

    = \int_{-1}^{1} \sqrt{1-x^{2}}\cdot dx = \frac{1}{2}\cdot |x\cdot \sqrt{1-x^{2}} + \sin^{-1} x|_{-1}^{1} = \frac{\pi}{2} (3)

    ... and then...

    x_{g} = \frac{2}{\pi}\cdot \int_{-1}^{1} \int_{0}^{\sqrt{1-x^{2}}} x\cdot dx\cdot dy = \frac{2}{\pi}\cdot \int_{-1}^{1} x\cdot \sqrt{1-x^{2}}\cdot dx = 0 (4)

    y_{g} = \frac{2}{\pi}\cdot \int_{-1}^{1} \int_{0}^{\sqrt{1-x^{2}}} y\cdot dx\cdot dy = \frac{2}{\pi}\cdot \int_{-1}^{1} |\frac{y^{2}}{2}|_{0}^{\sqrt{1-x^{2}}} \cdot dx =

    = \frac{1}{\pi}\cdot \int_{-1}^{1} (1-x^{2})\cdot dx = \frac{1}{\pi}\cdot |x - \frac{x^{3}}{3}|_{-1}^{1} = \frac{4}{3\pi} (5)

    Kind regards

    \chi \sigma
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