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Homework Help: Centroid of a semicircular arc

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

    Here, I have two ways of finding the y-coordinate centroid of a semicircular arc using polar coordinates.

    2w4he0x.jpg

    First one is considering a circle of radius a, centred at the origin. What I have done is [tex]\int ds = \int_0^{\pi} a d\theta = \pi a[/tex] and then [tex]\int y ds = \int r^2 sin \theta d\theta = a^2 \int_0^{\pi} sin \theta d\theta = 2a^2[/tex], then y-centroid is 2a/π. I'm quite sure this is the right answer.

    Second is a more roundabout method. It uses a circle with equation [tex]r=2a cos \theta[/tex]. A right angle triangle fits inside the circle and the angle between the side passing through the origin and the x-axis is theta. But I do not get a similar answer. Here my method is to use [tex]\int y ds = \int r sin\theta \sqrt{r^2 + (\frac{dr}{d\theta})^2} d\theta = \int_0^{\frac{\pi}{2}} 2a cos \theta (2a) d\theta = 4a^2[/tex], where the limits of the integral run from 0 to half pi due to how the angle is set up. (From my working, the messy square roots part becomes just 2a.) The y-centroid for the semicircle is 4a/π, which is wrong. Can someone advise me where I have gone wrong?

    The y-coordinate centroid should be the same for both as the difference between the two figures is that one is shifted a-units along the x-axis.
     

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    Last edited: Feb 1, 2009
  2. jcsd
  3. Feb 1, 2009 #2
    Sorry, I've solved this... careless... final integral should have been 2a cos theta sin theta.
     
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