Question 16.SP.4: A cord is wrapped around a homogeneous disk of radius r = 0.......

Equations of Motion. We assume that the components \overline{ \mathbf{a}}_x \text { and } \overline{ \mathbf{a}}_y of the acceleration of the center are directed, respectively, to the right and upward and that the angular acceleration of the disk is counterclockwise. The external forces acting on the disk consist of the weight W and the force T exerted by the cord. This system is equivalent to the system of the effective forces, which consists of a vector of components m \overline{ \mathbf{a} }_x \text { and } m \overline{ \mathbf{a} }_y attached at G and a couple \bar{I} \alpha. We write

\begin{array}{rlrlrl}\stackrel{+}{\rightarrow} \Sigma F_x=\Sigma\left(F_x\right)_{\text{eff}}: & 0 =m \bar{a}_x & \overline{ \mathbf{a} }_x=0 \\+\uparrow \Sigma F_y=\Sigma\left(F_y\right)_{ \text{eff} }: & T-W =m \bar{a}_y \\&\bar{a}_y =\frac{T-W}{m}\end{array}

Since T = 180 N, m = 15 kg, and W = (15 kg)(9.81 m/s²) = 147.1 N, we have

\bar{a}_y=\frac{180  N -147.1  N }{15\text{ kg}}=+2.19  m / s ^2 \quad \overline{ \mathbf{a} }_y=2.19  m / s ^2 \uparrow

\alpha=-\frac{2 T}{m r}=-\frac{2(180  N )}{(15 \text{ kg} )(0.5  m )}=-48.0\text{ rad}/ s ^2

α = 48.0 rad/s² ↓

Acceleration of Cord. Since the acceleration of the cord is equal to the tangential component of the acceleration of point A on the disk, we write

\begin{aligned}\mathbf{a} _{\text {cord }} & =\left( \mathbf{a} _A\right)_t=\overline{ \mathbf{a} }+\left( \mathbf{a} _{A / G}\right)_t \\& =\left[2.19  m / s ^2 \uparrow\right]+\left[(0.5  m )\left(48\text{ rad}/ s ^2\right) \uparrow\right]\end{aligned}

\mathbf{a} _{\text {cord }}=26.2 m / s ^2 \uparrow