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Work done by a force going around a circle ?

  1. May 4, 2015 #1
    I have a problem with question 4 part (iv) which is on here http://www.mei.org.uk/files/papers/m208ju_ail7.pdf

    Basically, I used Ep(initial) + Ek(initial) + W(Forward) = Ep(Final) + Ek(final) + W(Resistive) equation, where Ep is the potential energy, Ek the kinetic energy and W is the work done.
    This finalized to mgh + 1/2mu2 = 1/2mv2 + Fs (since there is no forward work done and h = 0 at the lowest point)

    I worked values of the equation:
    mgh = 0.15 x 9.8 x 0.4679
    1/2mu2 = 1/2 x 0.15 x 2.52
    1/2mv2 = 1/2 x 0.15 x v2 <-- need to find v

    Fs is the problem; The mark scheme says that Fs = 0.6 x 40/360 x 2pi x 2

    I understand that it must be that 0.6 acts over 40/360 of the circle, and then the distance from A to the lowest point is 2m as I've previously worked out, so this is the s. What I don't understand is 2pi, and whether the 2pi relates to the distance or the force... Could anybody explain how Fs, the resistive work done, can be equal to that and maybe give an equation for working out work done when it acts in circular motion?

    Thank you for reading and please if you could help me with my question I would really appreciate it...
     
  2. jcsd
  3. May 4, 2015 #2

    pasmith

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    The distance you want is that travelled by the sphere from B to the lowest point, which is 40/360 of the circumference of a circle of radius 2m.

     
  4. May 4, 2015 #3
    Can anyone explain 2pi?
     
  5. May 4, 2015 #4

    haruspex

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    pasmith explained already:
    Look at the part after "40/360".
     
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