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Im stuck: Energy in rotational motion

  1. Oct 2, 2007 #1
    1. The problem statement, all variables and given/known data
    A uniform rope 12.0m long and with a mass 3.00kg is hanging with one end attached to a gymnasium ceiling and the other end just touching the floor. The upper end of the rope is released, and the rope falls to the floor. What is the change in the gravitational potential energy if the rope ends up flat on the floor(not coiled up) ?


    2. Relevant equations
    Fx = 0
    Fy = 0
    U = mgh1 - mgh2



    3. The attempt at a solution
    Fy = 0 = Tsin(theta) - W
    Tsin(theta) = W
    T = ((3(9.8))/(sin(theta))

    Fx = 0 = Tcos(theta)
    No other force acts on the x-axis
     
  2. jcsd
  3. Oct 2, 2007 #2

    learningphysics

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    change in gravitational potential energy = mgh2 - mgh1

    that's all you need.
     
  4. Oct 2, 2007 #3
    I thought it was inclined


    U = mgh2 - 0
    = (3.00kg)(9.8m/s^2)(12.0m)
    = 352.8 J

    thankss
     
  5. Oct 2, 2007 #4

    learningphysics

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    wait. use the center of mass... what's the height of the center of mass? you need to use 6.0m not 12.0m.
     
    Last edited: Oct 2, 2007
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