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clairez93
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Homework Statement
A blade on an industrial fan has the configuration of a semicircle attached to a trapezoid (see figure). Find the centroid of the blade.
Homework Equations
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:
Trapezoid
[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]
Circle
[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]Averaging:
[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]