Blocks over a massive pulley

Given: $m_A = 8$ kg, $m_B = 4$ kg, pulley $M = 10$ kg, $r = 0.2$ m. Since $m_A > m_B$, A moves down and B moves up. Pulley moment of inertia:

$$I = \tfrac12 Mr^2 = \tfrac12(10)(0.2)^2 = 0.20\ \text{kg·m}^2$$

        ____
       /    \   pulley (M, r)
  T_A |      | T_B
      |      |
     [A]    [B]
   (down)   (up)

The rope does not slip, so $a = \alpha r$. The tensions in the two parts differ because the pulley has inertia.

Equations

Block A (downward positive): $m_Ag - T_A = m_A a$ (1)

Block B (upward positive): $T_B - m_Bg = m_B a$ (2)

Pulley (moments about the axle): $(T_A - T_B)\,r = I\alpha = \tfrac12 Mr^2\,\dfrac{a}{r}$

$$T_A - T_B = \tfrac12 Ma = 5a \quad (3)$$

Adding (1), (2) and (3): $(m_A - m_B)g = (m_A + m_B + \tfrac12 M)\,a$

Acceleration

$$a = \frac{(8 - 4)(9.81)}{8 + 4 + 5} = \frac{39.24}{17} = 2.308\ \text{m/s}^2$$

Angular acceleration

$$\alpha = \frac{a}{r} = \frac{2.308}{0.2} = 11.54\ \text{rad/s}^2$$

The pulley turns in the sense that moves A's side down (counter-clockwise in the sketch, where A is on the left).

Tensions

$$T_A = m_A(g - a) = 8(9.81 - 2.308) = 60.01\ \text{N}$$ $$T_B = m_B(g + a) = 4(9.81 + 2.308) = 48.47\ \text{N}$$

Check: $T_A - T_B = 11.54$ N $= 5a$. It agrees.

Answer: a = 2.31 m/s²; α = 11.54 rad/s²; T_A = 60.0 N; T_B = 48.5 N.