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Homework Help: Work done along a quarter circle path

  1. Feb 18, 2006 #1
    I wasnt really sure if I should post that question here since I actually didnt understand its math part.
    this question actually already has its solution but I didnt understand this part, when I want to find the work done by a force along a quarter circle path, I define x and y with r and angle and which is for the first quarter of circle is x= r.cos(teta) and y=r.sin(teta) when it is the second quarter of the circle why the book took the x=rsin(teta) and y=r(1-cos(teta)) ?
    do you have any idea?
     
  2. jcsd
  3. Feb 18, 2006 #2

    Tide

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    The second x and y you wrote look like changes in x and y positions after integrating something over a quarter circle. Do you have any ideas why someone would do an integration for this kind of problem?
     
  4. Feb 18, 2006 #3
    actually I have my force F=(y,2x) and I need to find the Work done along that curve so I need to integrate F.dr so it becomes integral ydx + 2xdy and I need to turn x,dx,y and dy into (r) and (theta)
    there is something here but I cant see it :((
     
  5. Feb 18, 2006 #4

    lightgrav

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    you can measure theta from the positive-x axis, or from the positive y-axis
    (or anywhere else for that matter), but you have to be consistent throughout the entire solution.
    In any case, the 2x.dy results in r^2 2.sin^2(t) or r^2 2.cos^2(t).
    books will replace that 2.sin^2(t) = 1 - cos(2t) = 1 - cos^2(t) + sin^2(t) , or 2.cos^2(t) = 1 + cos(2t) ... maybe that's what they intended.
    but YOU should do the integral in a way you're comfortable with!
     
  6. Feb 18, 2006 #5

    lightgrav

    User Avatar
    Homework Helper

    you can measure theta from the positive-x axis, or from the positive y-axis
    (or anywhere else for that matter), but you have to be consistent throughout the entire solution.
    In any case, the 2x.dy results in r^2 2.sin^2(t) or r^2 2.cos^2(t).
    books will replace that 2.sin^2(t) = 1 - cos(2t) = 1 - cos^2(t) + sin^2(t) , or 2.cos^2(t) = 1 + cos(2t) ... maybe that's what they intended.
    but YOU should do the integral in a way you're comfortable with!
     
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