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(D) The linear acceleration $(a)$ of the center of mass is given by Newton's second law: $a = \frac{F}{m} = \frac{20 \, N}{2 \, kg} = 10 \, m/s^2$. Th

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$\frac{1}{20}$

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(D) The linear acceleration $(a)$ of the center of mass is given by Newton's second law: $a = \frac{F}{m} = \frac{20 \, N}{2 \, kg} = 10 \, m/s^2$. The torque $(	au)$ about the center of mass is given by $	au = F 	imes d$,where $d$ is the perpendicular distance from the center of mass to the line of action of the force. Since the force $F$ is applied at the center of mass,the perpendicular distance $d = 0$. Therefore,$	au = F 	imes 0 = 0$. Using the relation $	au = I \alpha$,where $I$ is the moment of inertia and $\alpha$ is the angular acceleration: $0 = I \alpha \implies \alpha = 0 \, rad/s^2$. However,if the question implies the force is applied at the edge (tangentially) to cause rotation,then $	au = F 	imes r = 20 	imes 0.1 = 2 \, N \cdot m$. Assuming a solid disk,$I = \frac{1}{2} m r^2 = \frac{1}{2} 	imes 2 	imes (0.1)^2 = 0.01 \, kg \cdot m^2$. Then $\alpha = \frac{	au}{I} = \frac{2}{0.01} = 200 \, rad/s^2$. The ratio $\frac{a}{\alpha} = \frac{10}{200} = \frac{1}{20}$.

Find the ratio of linear acceleration $(a)$ to angular acceleration $(\alpha)$ of the center of mass $(COM)$ for the given diagram. Given: $m = 2 \, kg$,$r = 10 \, cm = 0.1 \, m$,and force $F = 20 \, N$ applied at the center.

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