Cord pulled on disk, m = 15 kg, T = 180 N

Take upward as positive for translation. The disk is homogeneous, so $I = \tfrac12 mr^2$. The forces on the disk are its weight $W=mg$ (down, at $G$) and the cord tension $T$ (up, at the rim, a distance $r$ from $G$). The cord leaves the rim tangentially, so $T$ has moment $Tr$ about $G$.

        T ^
          | A
      .---+---.
     /    G    \
     \         /
      '-------'
          | W
          v

Data

$m=15$ kg, $r=0.5$ m, $T=180$ N, $g = 9.81$ m/s²

$$W = mg = 15\times9.81 = 147.15\ \text{N}, \qquad I = \tfrac12 mr^2 = \tfrac12(15)(0.5)^2 = 1.875\ \text{kg m}^2$$

Equations of motion

$$\begin{aligned} \Sigma F_y = m\bar a &: \quad T - W = m\bar a\\ \Sigma M_G = I\alpha &: \quad T r = I\alpha \end{aligned}$$

Centre of the disk

$$\bar a = \frac{T - W}{m} = \frac{180 - 147.15}{15} = 2.19\ \text{m/s}^2\ \text{upward}$$

Angular acceleration

$$\alpha = \frac{Tr}{I} = \frac{180\times0.5}{1.875} = 48.00\ \text{rad/s}^2\ \text{(counter-clockwise, as } T \text{ acts at the right rim)}$$

Acceleration of the cord

The cord has the same acceleration as the rim point $A$ to which it is attached: the acceleration of $G$ plus the tangential acceleration $\alpha r$ of $A$ relative to $G$ (upward, since $\alpha$ is counter-clockwise and $A$ is on the right):

$$a_{cord} = \bar a_y + \alpha r = (+2.19) + 48.00\times0.5 = 26.19\ \text{m/s}^2\ \uparrow$$

Answer: (a) $\bar a = 2.19$ m/s² upward; (b) $\alpha = 48.00$ rad/s² counter-clockwise; (c) $a_{cord} = 26.19$ m/s² upward.