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Find Acceleration

  1. Jul 12, 2009 #1
    1. The problem statement, all variables and given/known data
    A pulley (in the form of a uniform disk)
    with mass 79 kg and a radius 14 cm is at-
    tached to the ceiling in a uniform gravita-
    tional field and rotates with no friction about
    its pivot. The masses are connected by a
    massless inextensible cord.
    Determine the acceleration of the mass
    33 kg. The acceleration of gravity is 9.8 m/s2 .
    Assume up is positive. Answer in units of
    m/s2.

    2. Relevant equations
    Net Torque = Moment of Inertia *Angular Acceleration
    Torque = Force * Radius



    3. The attempt at a solution

    I attempted this:
    I*Alpha = net toque
    1/2MR^2*(A/R) = (56-33)*9.8*R
    M*A=23*9.8*2
    A=(23Kg*9.8m/s/s*2)/(79Kg)

    Unfortunately that didnt work...
    Thanks for any help.
     
  2. jcsd
  3. Jul 12, 2009 #2

    Redbelly98

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    If I understand the situation correctly, the rope tensions are not simply equal to the weights of the masses.

    Maybe try applying F=mA to each mass, and see if that helps.
     
  4. Jul 12, 2009 #3
    Youre right, I tried just using the tensions and such and that didnt work either. Thanks for your help, and Ill try that.
     
  5. Jul 12, 2009 #4

    Redbelly98

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    Oh, also note that the 2 rope sections can have different tension.
     
  6. Jul 12, 2009 #5
    Not only can they, but they do. As one mass is accelerating up, and the other down.
     
  7. Jul 12, 2009 #6
    I messed up typing it originally.
    The pulley has two masses attached. One of mass 56 kg on one side of the pulley and one of mass 33 kg on the other side.
     
  8. Aug 3, 2009 #7

    ideasrule

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    Don't forget to write the rotational equation of motion for the pulley too!
     
  9. Aug 4, 2009 #8
    Just on a conceptual note, from this it does not follow that the tension is different on each side of the pulley. If it were a mass-less pulley, one mass would be accelerating up, and the other down, but the tension in the rope would still be the same throughout.
     
  10. Aug 5, 2009 #9

    rl.bhat

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    1/2MR^2*(A/R) = (56-33)*9.8*R
    This equation should by
    1/2MR^2*(a/R) =[ T1- T2]*R...(1)
    You have to write two more equations.
    56*g - T1 = 56*a ....(2)
    T2 - 33*g = 33*a.....(3)
    From these three equations, you can get the required result.
     
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