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Circular motion Work Energy problem

  1. Oct 26, 2012 #1

    AmK

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    A small bead of mass m is moving on a smooth circular wire (radius R) under the action of a force
    F directed towards a point P at a distance R/2 from the centre .What should be the minimum velocity of the bead at the point where it is closest to P so that it may complete the circle.

    I worked it out as follows
    The bead will complete the circle if it is able to just reach the point diametrically opposite to the nearest point to P because after that a component of F will act along the tangential direction and accelerate the bead.
    Using conservation of energy 1/2mv^2 + work done by force F >= 0

    But i couldn't calculate the work done by the force F.
    i know W = ∫F.dr so i tried to integrate but i couldnt find the angle between the force and displacement for the integral.

    Please give some idea on how to work out this integral.
    Also,can it be done without the integral ?
     
    Last edited: Oct 26, 2012
  2. jcsd
  3. Oct 26, 2012 #2

    tiny-tim

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    Welcome to PF!

    Hi AmK! Welcome to PF! :smile:
    If the bead is at X, and if the centre is O,

    then the angle is the angle between PX and the tangent, or 90° minus OXP :wink:
     
  4. Oct 26, 2012 #3

    AmK

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    Thanks Tiny-Tim
    the angle is what u said but for the integration i have to relate it to that angle to the angle subtended at the centre since the displacement would be in terms of that and i wasnt able to relate them
     
  5. Oct 26, 2012 #4

    haruspex

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    Consider the triangle OPX. You know∠XOP, distance OP, distance OX. Can you write down an equation giving you ∠XPO? Can you find ∠PXO from there?
     
  6. Oct 27, 2012 #5

    AmK

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    Cosine formula for length PX and then sine formula for ∠ PXO
    That would make the integral very complex and i dont think i could solve that
     
  7. Oct 27, 2012 #6

    tiny-tim

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    Hi AmK! :smile:
    After a substitution, isn't it one of the standard integrals?

    Show us what you get. :smile:
     
  8. Oct 27, 2012 #7

    AmK

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    I think i got it.
    i'll denote the length by l then ∠oxp by β and ∠xop by θ
    then l/sinθ = R/2sinβ
    ∫F.dr = ∫F*sinβ*Rdθ
    sinβ = (R/l)*sinθ
    ∫F*sinβ*Rdθ = ∫F*(R/l)*sinθ*Rdθ
    l=√R^2 + R^2/4 - R^2cosθ
    and then i can sub cosθ=t and range of t will be from 1 to -1
    Thanks guys
     
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