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Problems with work and F*dr

  1. Jan 25, 2006 #1
    A semi-circle is given by the function [tex]y=R cos(\frac{x\pi}{2R})[/tex]. An attractive force F at [tex]x = R[/tex] and [tex]y = 0[/tex] such that [tex]F(x, y) = F(R, 0) = 0 [/tex] and [tex]F(x, y) = F(-R, 0) = F_{0} [/tex]. This force along the semi-circle, relative to its source, is given by [tex]F(\theta) = F_{0} sin(\theta/2)[/tex]. What amount of work is done against the force when a particle moves along the circle from [tex]\theta = 0[/tex] to [tex]\theta = \pi[/tex].
    To find the work I have the equation
    [tex]W = - \int^{\pi}_{0} F dr[/tex]
    For the force I have the equation
    [tex]F= F_{0} sin(\theta/2)[/tex]
    For the distance from the particle to the orgin of the force I have
    [tex]2 R sin(\theta/2)[/tex]
    so for dr I have
    [tex]2 R sin(\theta/2) d\theta[/tex]
    Putting this all tegether I get
    [tex]W = - \int^{\pi}_{0} F_{0} sin(\theta/2) 2 R sin(\theta/2) d\theta[/tex]
    This rearanges to
    [tex]W = -F_{0} R \int^{\pi}_{0} \left[1-cos(\theta) \right] d\theta[/tex]
    Integrating I get
    [tex]W = -F_{0} R \left[ \theta-sin(\theta) \right]^{\pi}_{0}[/tex]
    This works out to
    [tex]W = -F_{0} R \pi[/tex]
    The answer I am suppose to get is
    [tex]W = -F_{0} R[/tex]
    What did I do wrong?
     
    Last edited: Jan 26, 2006
  2. jcsd
  3. Jan 25, 2006 #2

    Gokul43201

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    The question doesn't make very much sense. Could you please reproduce the question exactly as it appears in the text, and also provide the name of the text ?
     
  4. Jan 26, 2006 #3
    I rewrote the problem a little bit. I'm afraid it's still a little unclear with out the diagram though. This question is not from a text book.
     
  5. Jan 26, 2006 #4

    Gokul43201

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    I think the mistake you are making is in interpreting what dr refers to. This is an elemental distance traveled by the particle (along the semicircular path) - it is not an element of the distance of the particle from the origin. Also keep in mind that the angle between F and dr changes with the motion of the particle.
     
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