Rotation dynamics pulley concept confusion

[Clara Chung]

Homework Statement

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.

Homework Equations

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

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

 

[Safakphysics]

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

 

[Safakphysics]

\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.

 

[Clara Chung]

View attachment 103345
\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.

Thank you. I got it.

 

[Safakphysics]

You are welcome