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Pulleys moving down with constant speed

  1. Nov 19, 2011 #1
    1. The problem statement, all variables and given/known data
    Please refer diagram attached.
    The pulleys are moving down with constant speed u each. Pulleys are light , string ideal. What is the speed with the mass moves up?






    2. Relevant equations



    3. The attempt at a solution
    Now, I got the answer as ucosΘ, obviously. But the answer is u/cosΘ.
    the solution too is given. But I feel that the latter is wrong as the velocity goes to to infinity as the mass goes upwards. Please tell me which is correct.
     

    Attached Files:

  2. jcsd
  3. Nov 20, 2011 #2

    gneill

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    Why don't you explain your reasoning for the 'obvious' result, u*cos(θ)?
     
  4. Nov 20, 2011 #3

    ehild

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    If the half-length of the rope between the pulleys is L, the mass is at depth y=L*cos(Θ), and dL/dt = -u. The rate of change of y is dy/dt=d(L*cos(Θ))/dt. Both L and Θ are changing as the rope is pulled while x = LsinΘ stays constant. If you take these into account you will get the correct result, that the mass raises wit u/cosΘ.

    ehild
     
  5. Nov 20, 2011 #4
    Well, the string section 'r' should move with speed u only as it is a light pulley. Then the vertical component is ucosθ.
     
  6. Nov 20, 2011 #5
    But then is my interpretation of decreasing vertical velocity wrong?
     
  7. Nov 20, 2011 #6

    ehild

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    It is wrong. And you did not prove it.

    ehild
     
  8. Nov 20, 2011 #7

    gneill

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    The problem with this analysis (as tempting as it is) is that the rope end at D will NOT move along the line DA (or DB), as it is constrained by its mirror-image partner. While your length r may be getting shorter, point D has no velocity along r. So trying to treat dr/dt = u as a velocity component for point D is not right.
     
  9. Nov 20, 2011 #8

    gneill

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    The vertical velocity will decrease, but not for the reason you think! As the mass rises and the angle θ approaches 90°, the tension in the rope required to maintain the rope speed u will head to infinity. IF the rope speed could be maintained at u (or any positive real value greater than zero) then the velocity of the mass would go infinite, but rope speed cannot be so maintained by any real apparatus.
     
  10. Nov 20, 2011 #9

    Doc Al

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    The rate at which the length of segment DA changes will equal u. You need to relate the rate of change of segment DC to the rate of change of DA. How are those lengths related?
     
  11. Nov 20, 2011 #10
    OK, I see now. Using Pythagoras theorem we can relate the legths as
    [itex]DA^{2}[/itex]=[itex]DC^{2}[/itex]+[itex]AC^{2}[/itex]
    then differentiating,
    [itex]2DA\frac{dDA}{dt}[/itex]=[itex]0+2AC\frac{dDC}{dt}[/itex]
    And finally the answer will come u/cosθ. Right?
     
    Last edited: Nov 20, 2011
  12. Nov 20, 2011 #11

    ehild

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    If you replace AC with DC in the second equation it will be right.

    ehild
     
  13. Nov 21, 2011 #12
    Yes, thank you.
     
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