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Frictional torque sign

  1. Dec 27, 2014 #1
    1. The problem statement, all variables and given/known data

    A string passing over a pulley has a 3.80 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 0.04 m and mass 0.80 kg.
    It is found that if the heavier mass is give a downward speed of 0.20 m/s, it comes to rest in 6.2 s. What is the average frictional torque acting on the pulley?

    2. Relevant equations

    T = F*r
    F = ma
    T = I* alpha

    3. The attempt at a solution

    Newton's second law for masses:

    FTA - mag= ma*a (the heavier one)
    mb*g -FTB = mb*a

    For the pulley:

    FTB*rpulley(ccw-positive) - FTA*rpulley(cw-negative) + (+-_____ )= I*alpha

    The pulley is rotating ccw, so I thought I had to write minus friction because the friction must be opposite of rotation direction, but then saw this and got really confused.

    Determining the sign of frictional torque in this equation, should it be positive or negative?
  2. jcsd
  3. Dec 27, 2014 #2


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    Assuming the heavier mass is on the right and the lighter on the left, and given that the heavier one is pushed down and accelerates up, which way is the pulley rotating before it stops?
  4. Dec 27, 2014 #3
    I think no matter what the acceleration is, for your question, it rotates clockwise as long as the heavy mass has that downward velocity. (until v = 0)
    In my calculations, I assumed that the heavy mass was on the left.
    Last edited: Dec 27, 2014
  5. Dec 27, 2014 #4


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    Heavy on left moving down implies ccw pulley rotation ...yes , no , maybe? Watch signage it is not easy.
  6. Dec 27, 2014 #5
    Heavy on left (F_T1) and let ccw be positive
    (FT1-FT2)*r - Tfrictional = I* alpha

    Since rotation and acceleration are in different directions for the pulley, I should've written minus I*alpha for my setup, right?
  7. Dec 27, 2014 #6


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    Yes. So for clarity, your last equation using A and B instead of 1 and 2 is???
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