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Atwood's Problem (pulley has mass)

  1. Apr 26, 2010 #1
    1. The problem statement, all variables and given/known data

    In the Atwood's machine shown below the masses of the blocks are ma = 3kg and mb = 2kg. The moment of inertia of this frictionless pulley about its axis is Ic = .6 kg m2 and its radius is Rc = .2m. Note that there is no slipping between the rope and the pulley.

    Find the accelerations of the blocks and the angular acceleration of the pulley

    http://commons.wikimedia.org/wiki/File:Atwoodmachine.gif
    this is the closest diagram i could find... the distance of the right block (I'll name it block 2 and the left block 1) above the ground is h.

    I thought of two ways to do this problem, but I wanted to try and use Newton's laws.
    I solved it using conservation of energy:
    EPot f - EPot i + Kfinal - Kinitial = 0
    so we know that:

    Kinitial = 0 (since its not moving yet)
    EPot f = m1(g)(h)
    EPot i = m2(g)(h)
    Kfinal = (1/2)m2v2 + (1/2)m1v2 + (1/2) Iw2

    then using vf2= vi2+2ah
    again vii = 0, so we can now replace every vf2 by 2ah

    at this point you have:
    m1(g)(h) - m2(g)(h) + (1/2)(ma)(2ah) + (1/2)(m1)(2ah) + (1/2)(I)(2ah/R2)

    Now cancel out all the h's and get:
    m1(g) - m2(g) + (1/2)(ma)(2a) + (1/2)(m1)(2a) + (1/2)(I)(2a/R2)

    After solving for a I got a = .49 m/s2 of the blocks (I'm not concerned with the angular acceleration right now)

    I think my answer is right but..
    Is there any way to use Newton's Laws... Maybe use the following?
    m2(g) - T2 = m2(a)
    T1 - m1(g) = m1(a)
    Or does this only apply for massless pulleys?
    I tried using it, and I get a different Answer... Also if someone could check my answer itd be nice
     
  2. jcsd
  3. Apr 27, 2010 #2
    M1g - T1 = M1a
    T2 - M1g = M2a
    torque (t) = T1R - T2R = Ia/r

    eliminate T1,T2

    M1g -M2g - (T1-T2) = (M1 + M2)a

    T1 - T2 = Ia/r^2

    M1g - M2g - Ia/r^2 = (M1 + M2)a

    a = (M1 - M2)g/(M1 + M2 + I/r^2 )

    massles pulley implies I = 0

    your answer 0.49 is correct using this approach
     
  4. Apr 27, 2010 #3
    Thanks for the help, I get it now... Just another question though... When we consider the pulley to massless does it mean that the tensions are equal? Since T1-T2=Ia/r2 Or am I jumping to a false conclusion? Thanks again
     
  5. Apr 27, 2010 #4
    True, if we consider the pulley massless T1 = T2. The difference is due to the fact that there is a friction force that turns the pulley.
     
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