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COM of a half circle

  1. Jun 3, 2016 #1
    1. The problem statement, all variables and given/known data

    Find the centre of mass of a semi-circle

    2. Relevant equations

    ##y_{cm}=\frac{1}{M} \int y dm ##

    3. The attempt at a solution
    So ## y= R cos \theta ## where theta is measured from the vertical, and the base of the semi-circle is along the horizontal

    Now apparently from here you can change coordinated to polar coordinates and replace ##dm## with ## \rho r dr d \theta## to obtain the correct answer of ##\frac {4}{3 \pi}## But I'm confused as to why you can't also replace ##dm## with ##2 \rho R sin \theta R d \theta## to split to 'dm' segment into rows perpendicular to the horizontal base?

    Many thanks :)
     
  2. jcsd
  3. Jun 3, 2016 #2

    BvU

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    With ##R\cos\theta## the height of such a slice, what would you have for the width ? I would guess ##\ R\cos\theta\ d\theta\ ## And the center of mass of such a slice would be at ##{R\over 2} \cos\theta##. It might work. Make a drawing to check what you are doing.
     
  4. Jun 3, 2016 #3
    Hi- I did draw it out, but unfortunately cannot scan it in. The best way I can describe what Im doing is by saying that I'm calculating it in the same way I would the moment of inertia by dividing the semicircle into rectangles of length ## 2 R sin \theta## (or ## sin \theta## depending on how you define your angles) and width ##Rd \theta## It does not appear to give the correct answer....
     
  5. Jun 3, 2016 #4

    BvU

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    Actuallly it does give the correct answer. Could you show at least your calculation ?

    upload_2016-6-4_0-32-14.png

    But: at first you mentioned vertical rectangles.
    And you had already defined ##\theta##; why change it ?
     
    Last edited: Jun 3, 2016
  6. Jun 3, 2016 #5

    NascentOxygen

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    R‧dΘ is the arc length, so it's a sloping distance perpendicular to neither axis.
     
  7. Jun 3, 2016 #6
    I think it would be better to set this up in cartesian coordinates, and then to convert to polar coordinates in doing the trig substitution as part of the integration. If y is the distance above the base, the area of the differential "rectangle" between y and y + dy is ##2\sqrt{R^2-(R-y)^2}dy##. The moment of the area about the base is ##2y\sqrt{R^2-(R-y)^2}dy##. This can be integrated from y = 0 to y = R to get the total moment of the semi-circle.
     
  8. Jun 5, 2016 #7
    Hi- sorry if I was not being very clear. This is what I had meant :) but I can't see my mistake
     

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  9. Jun 5, 2016 #8

    NascentOxygen

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    The rectangular strip has its long side of length 2‧R‧cosθ
    but its dimension perpendicular to this is .....
     
  10. Jun 5, 2016 #9
    Oh okay so it should be ##dy= Rcos \theta d \theta ## in place of ##R d \theta## ? :)
     
  11. Jun 5, 2016 #10
    And it comes out with that correction! Thank you so much :)
     
  12. Jun 5, 2016 #11

    NascentOxygen

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    So you get the answer you quote in post #1 ?
     
  13. Jun 5, 2016 #12
    Yes- thank you
     
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