Question 16.SP.5: A cord is wrapped around a homogeneous disk with a radius of...
A cord is wrapped around a homogeneous disk with a radius of r = 0.5 m and a mass of m = 15 kg. If the cord is pulled upward with a force T of magnitude 180 N, determine (a) the acceleration of the center of the disk, (b) the angular acceleration of the disk, (c) the acceleration of the cord.
STRATEGY: Since you have forces and are interested in determining accelerations, use Newton’s second law.
MODELING: Choose the disk and the cord as your system. Assume that the \overline{\mathbf{a}}_x components 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 (Fig. 1). A free-body diagram and kinetic diagram for this system are shown in Fig. 2. The external forces acting on the disk consist of its weight W and the force T exerted by the cord.
ANALYSIS:Equations of Motion. Applying Newton’s second law in the x and y directions gives
\stackrel{+}{\rightarrow} \Sigma F_x=m \bar{a}_x: \quad 0=m \bar{a}_x \overline{\mathbf{a}}_x=\mathbf{0}
+\uparrow \Sigma F_y=m \bar{a}_y: \quad T – W=m \bar{a}_y
\bar{a}_y=\frac{T – W}{m}
Since T = 180 N, m = 15 kg, and W = (15 kg)(9.81 m/s²) = 147.1 N, you have
\bar{a}_y=\frac{180 \mathrm{~N}-147.1 \mathrm{~N}}{15 \mathrm{~kg}}=+2.19 \mathrm{~m} / \mathrm{s}^2 \quad \overline{\mathrm{a}}_y=2.19 \mathrm{~m} / \mathrm{s}^2 \uparrow
Now taking moments about the center of gravity, you get
+ \circlearrowleft \Sigma M_G=\bar{I} \alpha: -\operatorname{Tr}=\bar{I} \alpha
-T r=\left(\frac{1}{2} m r^2\right) \alpha
\begin{array}{r}\alpha=-\frac{2 T}{m r}=-\frac{2(180 \mathrm{~N})}{(15 \mathrm{~kg})(0.5 \mathrm{~m})}=-48.0 \mathrm{rad} / \mathrm{s}^2 \\\alpha=48.0 \mathrm{rad} / \mathrm{s}^2 \circlearrowright \end{array}
Acceleration of Cord. The acceleration of the cord is equal to the tangential component of the acceleration of point A on the disk, so you have (Fig. 3)
\begin{aligned}\mathbf{a}_{\text {oord }}=\left(\mathbf{a}_A\right)_t &=\overline{\mathbf{a}}+\left(\mathrm{a}_{A / G}\right)_t \\&=\left[2.19 \mathrm{~m} / \mathrm{s}^2 \uparrow\right]+\left[(0.5 \mathrm{~m})\left(48 \mathrm{rad} / \mathrm{s}^2\right) \uparrow\right]\end{aligned}
\mathrm{a}_{\text {cord }}=26.2 \mathrm{~m} / \mathrm{s}^2 \uparrow
REFLECT and THINK: The angular acceleration is clockwise, as we would expect. A similar analysis would apply in many practical situations, such as pulling wire off a spool or paper off a roll. In such cases, you would need to be sure that the tension pulling on the disk is not larger than the tensile strength of the material.
