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Pulling a puck through the hole

  1. Apr 29, 2017 #1
    1. The problem statement, all variables and given/known data
    Its a classic problem about a puck that is rotating on a frictionless tabble with a velocity v1 and radius r1. It is is connected to a string which runs through a hole a the centre of the table. The string is pulled from below until the radius decreases to r2. Find the work done by the string.

    2. Relevant equations
    I used the conservation of angular momentum to find v(r). Then as the puck is undergoing circular motion. T=mv^2/r.
    Hence W is an integral of dot product of tension and dr.

    3. The attempt at a solution
    However, I encountered a problem, how do i take the direction of r? If the direction of r is taken from the puck towards the centre, T and dr are in opposite direction.
    If r is taken from centre to the puck, T and dr are in same direction. So how do I proceed?
     
  2. jcsd
  3. Apr 29, 2017 #2

    scottdave

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    Looking at the pulling force (below the table), they point in the same direction. But the length of the string is increasing below the table. Now above the table, you are going from a larger value to a smaller value (radius decreasing). You should come up with the same answer, either way. Think about what is being done by the force.
     
  4. Apr 29, 2017 #3
    So you mean T and dr are in the same direction?
     
  5. Apr 29, 2017 #4

    haruspex

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    That does not matter. That will give you -T.dr inside the integral, but the range will be from the larger r to the smaller. That will correct the sign to give a positive result.
    By the way, there is an easier way than dealing with forces and integrating.
     
  6. Apr 29, 2017 #5

    scottdave

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    I agree with @haruspex in that it does not matter. You should come out with work done on the system by the external force. I come from the electrical background, though, where often you would make an assumption that current was in a certain direction. Often when the calculations are done, you get a negative value for current, which means that you picked the incorrect direction.
     
  7. Apr 29, 2017 #6
    Okay thank you. The problem asks it to be done both ways so I had to go through all this pain
     
  8. Apr 29, 2017 #7
    Does that thing apply in these kind of problems though? Unlike current , work can be both negative or positive.
     
  9. Apr 29, 2017 #8

    haruspex

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    I'm not sure what you mean.
    The sign of the work does matter. Calculated correctly, you should find the work done by the tension on the mass is positive. If that is not what you get, please post all your steps.
     
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