Cord on disk, m = 15 kg, T = 200 N

Principle of impulse and momentum for plane motion of a rigid body: the momenta of all particles of the body at time $t_1$, plus the impulses of the external forces acting between $t_1$ and $t_2$, equal the momenta at time $t_2$. The momenta of the particles reduce to a linear momentum vector $m\bar{\vec v}$ at the mass centre and an angular momentum $\bar I\omega$ about it:

$$\begin{aligned} m\bar{\vec v}_1 + \sum\int \vec F\,dt &= m\bar{\vec v}_2\\ \bar I\omega_1 + \sum\int M_G\,dt &= \bar I\omega_2 \end{aligned}$$

The moment equation may also be written about a fixed point $O$ (or about any point, with the appropriate momentum terms): $H_{O1} + \sum\int M_O\,dt = H_{O2}$. In scalar form the principle gives three equations (two for the linear part, one for the rotational part). It is the best method when forces are given as functions of time, or when a time interval (rather than a distance) is involved, as in impact problems.

 (momenta)1   +   (impulses)   =   (momenta)2

   m v1  .--.         F dt          m v2  .--.
   ---> (G) I w1  +   ------>  =    ---> (G) I w2
        '--'                             '--'

Numerical part

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, $T=200$ 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

$$\bar a = \frac{T - W}{m} = \frac{200 - 147.15}{15} = 3.52\ \text{m/s}^2\ \text{upward}$$

Angular acceleration

$$\alpha = \frac{Tr}{I} = \frac{200\times0.5}{1.875} = 53.33\ \text{rad/s}^2\ \text{(counter-clockwise, as } T \text{ acts at the right rim)}$$

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):

$$a_{cord} = \bar a_y + \alpha r = (+3.52) + 53.33\times0.5 = 30.19\ \text{m/s}^2\ \uparrow$$

Answer: (a) $\bar a = 3.52$ m/s² upward; (b) $\alpha = 53.33$ rad/s² counter-clockwise; (c) $a_{cord} = 30.19$ m/s² upward.