Pulley on the way down

Here the rope does not slip on the pulley and hence the angular acceleration of the pulley is equal to: \ $\begin{aligned} \alpha &= \frac{a}{R} & (1) \end{aligned}$ Now, we write equations for the pulley and the block. For the block we have, $\begin{aligned} Mg - T&= Ma & (2) \\ \end{aligned}$ For the pulley, $\begin{aligned} TR &=\frac{M R^2}{2} \times \alpha & (3) \\ \end{aligned}$ Substituting value of $\alpha$ from equation (1), we get $\begin{aligned} TR &=\frac{M R^2}{2} \times \frac{a}{R} & () \\ T &=\frac{M a}{2} (4)\\ \end{aligned}$ Adding equation (2) and (4), We get $\begin{aligned} Mg - \frac{Ma}{2} &= Ma \\ a=\frac{2g}{3}\\ \end{aligned}$