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Rotation dynamics pulley concept confusion plz help

  1. Jul 17, 2016 #1
    1. The problem statement, all variables and given/known data
    A light string is passed over a pulley and two masses m1 and m2 are suspended from the two free ends. The pulley is a uniform circular disc of mass M. Find the linear acceleration of the two masses. Friction may be neglected.

    2. Relevant equations
    I through center of the disc = MR^2 /2
    The answer is a= (m2-m1)g / (m1+m2+M/2)

    3. The attempt at a solution
    I drew 3 free body diagrams separately, and get T-m1g = m1a..(1)
    m2g-T=m2a...(2)
    1/2 MR^2 (a/R) = 2TR (2 tension acting both clockwise if I assume m2 is on the right hand side and it is heavier than m1) ...(3)
    Plz point out what's wrong in each equation 1,2,3, thxxxx
     
  2. jcsd
  3. Jul 17, 2016 #2
    You are wrong in 2TR because the tension direction opposite and also the rope's tension isn't same in this equation. I will add true equations in few minutes
     
  4. Jul 17, 2016 #3
    upload_2016-7-17_10-6-3.png
    \begin{equation}
    m_2.g-T_2=m_2.a
    \end{equation}
    \begin{equation}
    T_1-m_1g=m_1.a
    \end{equation}
    If i use torque
    \begin{equation}
    T_2.R-T_1.R=M.R^2.\alpha\div{2}
    \end{equation}
    \begin{equation}
    \alpha.R=a
    \end{equation}
    If there is misunderstanding in your mind, you will ask me.
     
  5. Jul 17, 2016 #4
    Thank you. I got it.
     
  6. Jul 17, 2016 #5
    You are welcome
     
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