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Acceleration, pulley

  1. Nov 16, 2007 #1
    1. The problem statement, all variables and given/known data
    A string passing over a pulley has a 3.8-kg mass hanging from one end and a 3.15-kg mass hanging from the other end. The pulley is a uniform solid cylinder of radius 4.0 cm and mass .80 kg.
    (a) If the bearings of the pulley were frictionless, what would be the acceleration of the two masses?
    (b) In fact, it is found that if the heavier mass is given a downward speed of .2 m/s, it comes to rest in 6.2 s. What is the average frictional torque acting on the pulley?


    2. Relevant equations
    I= .5*M*R^2
    For mass 1: F_1 - (m_1g) = m_1a
    For mass 2: m_2g - F_2 = m_2a


    3. The attempt at a solution
    The inertia for the pulley is simple. .5*.8*.04^2 = .00064

    Now, rearranging the first two equations:
    a= (F_1 - m_1g)/(m_1)
    a=(m_2g-F_2)/(m_2)

    Since the magnitude of the acceleration has to be the same for both masses...

    And now is where I am stuck. I have 3 unknowns (F_1, F_2, a), and the Inertia equation times alpha can be equivalent to (F_2 - F_1)*r. But, that introduces alpha into the mix, which would be another unknown.

    What am I missing? Any help is greatly appreciated!
     
  2. jcsd
  3. Nov 16, 2007 #2
    I belive you can simply make this problem easier by elimitating one unknown either a, or alpha since there is a relationship between the two:

    a = alpha * r (Not only the linear acceleration of the two masses will have the same magnitude but the one of the pulley as well since they are all in the same system)
    By using this equation should be a little easier to find the acceleration since you would have three equations and three unknowns. (a, F2 and F1) and you would get another eqatuion : a = (F2 - F1)r^2/I in addition to the ones you already mentioned

    I hope this helps!
     
  4. Nov 16, 2007 #3

    Doc Al

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    Staff: Mentor

    You are missing the simple relationship between alpha and the linear acceleration.
     
  5. Nov 16, 2007 #4
    Thank you both Doc Al and Hells Kitchen. The relationship between the alpha and linear accel. was floating around in my mind during my post, but wasn't really certain how to go about it. Sometimes I get so sucked into the problem that I lose focus and forget simple relationships that are the keys to success. Thanks again!
     
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