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COM of a half circle

  • Thread starter Physgeek64
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  • #1
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Homework Statement



Find the centre of mass of a semi-circle

Homework Equations



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

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 :)
 

Answers and Replies

  • #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.
 
  • #3
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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.
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....
 
  • #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:
  • #5
NascentOxygen
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R‧dΘ is the arc length, so it's a sloping distance perpendicular to neither axis.
 
  • #6
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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.
 
  • #7
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Actuallly it does give the correct answer. Could you show at least your calculation ?

View attachment 101612

But: at first you mentioned vertical rectangles.
And you had already defined ##\theta##; why change it ?
Hi- sorry if I was not being very clear. This is what I had meant :) but I can't see my mistake
 

Attachments

  • #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 .....
 
  • #9
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The rectangular strip has its long side of length 2‧R‧cosθ
but its dimension perpendicular to this is .....
Oh okay so it should be ##dy= Rcos \theta d \theta ## in place of ##R d \theta## ? :)
 
  • #10
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The rectangular strip has its long side of length 2‧R‧cosθ
but its dimension perpendicular to this is .....
And it comes out with that correction! Thank you so much :)
 
  • #11
NascentOxygen
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So you get the answer you quote in post #1 ?
 
  • #12
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