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Rope pulling cylinder - final velocity?

  1. Feb 6, 2013 #1
    1. The problem statement, all variables and given/known data

    A massless rope is wrapped several times around a solid cylinder of radius 20cm and mass 20kg, which is at rest on a horizontal surface. Someone pulls 1m of the rope with a constant force of 100N, setting the cylinder in motion. Assuming the rope neither stretches nor slips and the cylinder rolls without slipping, what is the final angular velocity and speed at its surface?

    2. Relevant equations

    [itex] I = MR^2/2 [/itex]



    3. The attempt at a solution

    I assumed it would be simple energy conservation. The work done on the cylinder in pulling is equal to its total energy afterwards.

    [itex] E = Work = F \times s = 100 \times 1 = 100J [/itex]

    Then equating to the cylinder energy...

    [itex] 100 = 0.5Iw^2 + 0.5mv^2 [/itex]

    Using [itex] v = wR [/itex] where v is the surface speed and subbing in I, I get;

    [itex] 100 = 0.75Mv^2 [/itex]

    If I re-arrange this I get v = 2.58ms^-1, but my answer says 3.65ms^-1.

    Can anyone get the answer they provide? Where is the mistake in my working?
     
    Last edited: Feb 6, 2013
  2. jcsd
  3. Feb 6, 2013 #2

    Doc Al

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    Don't assume that the speed of the rope is the same as the speed of the cylinder's center of mass.
     
  4. Feb 6, 2013 #3
    Thanks for the reply, but I don't see where I've made that assumption anywhere. I don't incorporate the speed of the rope in my answer, and the only relation between the pulling of the rope and my final answer is in the work done - which doesn't use the rope's speed at all.
     
  5. Feb 6, 2013 #4

    Doc Al

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    They wanted the speed "at its surface", by which I thought they meant the speed of the rope and the surface it touches. You calculated the speed of the center.

    Either way, that won't explain the answer. Let me think about it. (Did you provide the entire problem word for word?)
     
  6. Feb 6, 2013 #5
    My mistake. There was one crucial thing I missed, everything else is word for word;

     
  7. Feb 6, 2013 #6

    Doc Al

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    In that case I agree with your answer, which is the speed of the surface with respect to the center.

    Is this from a textbook? If so, let me know which book/problem, in case I have a copy.
     
  8. Feb 6, 2013 #7
    It's a one-word answer to a physics synopsis practice paper of mine. In all possibility it might be wrong, and I'll mention it to my example class demonstrator next week.

    Thanks for the help.
     
  9. Feb 6, 2013 #8

    haruspex

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    Be very careful interpreting that. What do you think it means?
     
  10. Feb 9, 2013 #9
    Hmm, so I think I understand what you might be getting at. The rope is being pulled 1m, but in doing the "pulling" the cylinder is moving towards us. So we actually pull 1m of rope, but we've been pulling for a distance of longer than 1m to keep up with the fact that the cylinder is coming towards us at the same time. It comes down to finding the actual distance we've pulled, I'll work it out later when I get time. Out of curiousity, have you been able to get the right answer?
     
  11. Feb 9, 2013 #10

    Doc Al

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    Ah, good point.

    I interpreted the problem to be that someone standing still pulled one meter of rope with the given force or pulled the rope one meter along the ground with that force. But they probably meant that they pulled until one meter of rope unwound from the cylinder. Quite a difference, of course.
     
  12. Feb 9, 2013 #11

    Doc Al

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    FYI, that does produce the given answer. Thanks again to haruspex for being on the ball.
     
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