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Centroid problem

  1. Aug 31, 2009 #1
    1. The problem statement, all variables and given/known data

    A blade on an industrial fan has the configuration of a semicircle attached to a trapezoid (see figure). Find the centroid of the blade.

    2. Relevant equations

    3. The attempt at a solution

    My plan was to solve for the centroid of the trapezoid, then of the semi circle, then average the two. I got the wrong answer, however. Here is my work:


    [tex]M_{x} = 0[/tex] (since this is the axis of symmetry)
    [tex]M_{y} = \int^{6}_{0} 2x(\frac{1}{6}x + 1)dx = 60[/tex]
    [tex]A_{trap} = \frac{(4+2)(6)}{2}[/tex]

    [tex]\overline{x} = \frac{60}{18} = \frac{10}{3}[/tex]
    [tex]\overline{y} = 0[/tex]


    [tex]M_{x} = 0[/tex]
    [tex]M_{y} = \int^{2}_{-2} \frac{\sqrt{4-y^{2}}+6}{2}(\sqrt{4-y^{2}}+6) dy = \frac{4(9\pi + 58)}{3}[/tex]
    [tex]A_{circle} = \frac{\pir^{2}}{2} = 2\pi[/tex]

    [tex]\overline{x} = \frac{\frac{4(9\pi + 58)}{3}}{2\pi} = \frac{2(9\pi + 58)}{3\pi}= [/tex]
    [tex]\overline{y} = 0[/tex]


    [tex]\frac{\frac{10}{3} + \frac{2(9\pi + 58)}{3\pi}}{2} = \frac{2(7\pi + 29)}{3\pi}[/tex]
    [tex]\overline{y} = 0[/tex]

    Book Answer: [tex](\overline{x}, \overline{y}) = (\frac{2(9\pi + 49)}{3(\pi +9)}, 0)[/tex]

    Attached Files:

  2. jcsd
  3. Sep 1, 2009 #2


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    Science Advisor

    The average of the centroids of two figures is NOT, in general, the centroid of their union. You need to use a "weighted" average, multiplying both numbers by the area of each figure, then dividing by the sum of the areas, not 2.
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