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Pulley problem (Rotation)

  1. Mar 23, 2008 #1
    [SOLVED] Pulley problem (Rotation)

    Mass m1 = 9.07 kg sits on a frictionless, horizontal surface. A massless string is tied to m1, passes over a pulley (a solid disk of mass Mp = 5.42 kg and radius Rp = 37.9 cm), and is tied to m2 = 6.94 kg hanging in space.

    Find the acceleration of the masses.

    It seems I'm having some trouble with this problem. I'm not sure if my algebra is right or if I am doing the wrong thing with the moment of Inertia. I was hoping someone could help me.

    torque = I [tex]\alpha[/tex]
    T = ma
    a = R[tex]\alpha[/tex]


    My attempt is as follows:

    The question assumes that there are non-slip conditions and it starts from rest.

    1 . First, I find the pulley's torque

    T1 = ma
    T2 = mg - ma
    torque = I [tex]\alpha[/tex]
    r T = I [tex]\alpha[/tex]
    T2 R - T1 R = I [tex]\alpha[/tex]

    2. Second, I find the acceleration.

    a = R [tex]\alpha[/tex]
    a / R = [tex]\alpha[/tex]

    3. Third, I solve for a.

    T2 R - T1 R = I[tex]\alpha[/tex]
    R (T2 - T1) = I (a/R)
    T2 - T1 = I (a/R) / R
    T2 - T1 = I a / R[tex]^{2}[/tex]
    m2g-m2a - m1a = I a/R[tex]^{2}[/tex]
    m2g = m2a + m1a + I a/R[tex]^{2}[/tex]
    m2g = a (m2 + m1 + I /R[tex]^{2}[/tex]
    m2g / m2 + m1 + I / R[tex]^{2}[/tex] = a
    (6.49)(9.81) / (I/R[tex]^{2}[/tex]) + 9.07 +6.94 = a
    I = mass of pulley * radius[tex]^{2}[/tex]
    I = (5.42 kg)(.379 m)[tex]^{2}[/tex]
    I = .7785

    (6.49)(9.81) / (.7785/.379[tex]^{2}[/tex]) + 9.07+6.94 = a
    63.667 / 5.420 + 9.07 + 6.94 =
    63.667/21.43 = 2.971 m/s[tex]^{2}[/tex]
     
    Last edited: Mar 23, 2008
  2. jcsd
  3. Mar 24, 2008 #2
    It's OK except for the moment of inertia of a disk. it's I = (1/2)mR^2.
    I would also use more parentheses and write m2g / (m2 + m1 + I / R^2) instead of
    m2g / m2 + m1 + I / R^2
     
  4. Mar 24, 2008 #3
    yeah i'm sorry i had to edit it a few times since some of the subscripts were displayed oddly. It worked out it thanks a lot for your help.
     
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