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Angular acceleration of an atwood pulley

  1. Mar 14, 2017 #1
    1. The problem statement, all variables and given/known data
    An Atwood machine is a rope that passes over a pulley with a block attached to each end of the rope so that the blocks are not in contact with the floor. The frictionless axle of the pulley is oriented horizontally, and the rope is vertical save where it makes contact with the pulley. Assume the rope has no weight. The pulley is a uniform disk with a moment of inertia of 0.313 kg m² and a diameter of 0.5 meters. The first block has a mass of 10 kg, and the second block has a mass of 6 kg. Begin with the blocks at rest and at the same height. What is the angular acceleration of the pulley?

    2. Relevant equations
    net torque = Ia
    a= angular acceleration and I=moment of inertia
    m= 10kg
    M= 6 kg
    3. The attempt at a solution
    mg0.25 - Mg0.25 = Ia
    0.25g(m-M) = Ia
    0.25g*4 =0.313a
    a=31.3

    My answer is marked wrong but what's wrong with it?
     
  2. jcsd
  3. Mar 14, 2017 #2

    gneill

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    Staff: Mentor

    You haven't taken into account the inertias of the two masses; The tensions in the rope on either side of the pulley will not simply be due to the weights of the blocks. Draw free body diagrams for each component.
     
  4. Mar 15, 2017 #3
    I think I'm quite lost on this.
    I don't understand how the masses could have moments of inertia when they are not rotating. One mass will move straight down and the other will move straight up.

    How do I find the tension in the rope? I know that tension along a rope is the same throughout the rope. I think tension and the normal force are the only forces acting on the pulley. The normal force acts parallel to the lever arm so its torque is 0. But the tension force is more difficult. Is the tension acting on one point of the pulley, perhaps on either side of the pulley, or an infinite number of points because it wraps around the pulley?
    I expect that it's direction will be tangent to the pulley but where is it pointing? If the tension acts on either side of the pulley, wouldn't one have to point toward the 10kg mass and the other towards the 6kg mass in which case they cancel out. This results in no torque on the pulley which must be wrong. If it acts at an infinite number of points, I think the tensions on either side will cancel out for the same reason except for the tension at the one point at the top of the pulley, leaving one point of action. If this is correct, the direction will be horizontal and towards the 10kg mass. Tension(T) is the net force in the horizontal direction. so N -16g =0

    so using torque T(0.25) = Ia ,

    A = linear acceleration = a(0.25)

    T = 16A = 16a(0.25)

    I have two equations and two variables so I should be able to solve for them.

    16a(0.25)2 = Ia

    a cancels
    1=0.313 ..............................................ARRRGGGGGH!!!!!!

    Okay...Okay... breathe.... round 2

    If I consider only one mass at a time, weight and tension are the only forces acting on them. They will have the same magnitude of acceleration, which will be the linear acceleration (A),
    T - 10g =10(-A) and T - 6g=6A
    T = 10g - 10A
    so
    10g - 10A - 6g = 6A

    4g =16A
    A=0.25g

    A=ar so a=A/0.25
    A = g ....... wrong..... again

    I hope I have at least shown you my confusion. Where am I going wrong?
     
  5. Mar 15, 2017 #4
    If the pulley has rotational inertia, the tensions on the sections of rope on either side of the pulley are not equal. Call the tensions T1 and T2. Do a force balance on each mass, and a moment balance on the pulley. What do you get for these?
     
  6. Mar 15, 2017 #5
    How could the tension on either side of the rope not be equal? My understanding is that one property of tension is that it is always equal throughout the length of a rope.

    another question is if the pulley is frictionless then how could tension in the rope rotate the pulley?
     
  7. Mar 15, 2017 #6
    The axle of the pulley is frictionless (so the pulley is free to rotate), but the part of the surface of the pulley in contact with the rope is not frictionless. There is static friction between the rope and the surface of the pulley, which prevents the rope from sliding over the surface of the pulley. This makes it possible to have two different tensions in the two free sections of the rope. This, in turn, also produces a moment which makes the pulley rotate.
     
  8. Mar 15, 2017 #7

    gneill

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    Staff: Mentor

    Not rotational moments of inertia, linear inertia as in Newton #2: F = MA.

    upload_2017-3-15_9-2-40.png

    Using FBD's for each component, write expressions for their accelerations.
     
  9. Mar 15, 2017 #8
    T1 - 10g = 10(-A)
    T1 =10g - 10A
    A = ar = 0.25a
    T1 =10g - 2.5a
    T2 - 6g = 6A
    T2 = 6g + 6A
    T2 = 6g + 1.5a

    T1 (0.25) - T2(0.25) = Ia

    (2.5g - 0.625a) - (1.5g + 0.375a) = Ia
    g = 1.313a
    a = 7.46

    I think that's right. Please let me know if it isn't.

    I thank you for your help.
     
  10. Mar 15, 2017 #9

    gneill

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    Staff: Mentor

    It looks good numerically, but it's missing the units :wink:
     
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