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(A) The torque $	au$ is given by $	au = F \cdot R = I \alpha$.Given $F = (20t - 5t^2) \ N$,$R = 2 \ m$,and $I = 10 \ kg \ m^2$.$(20t - 5t^2) \cdot 2

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$5$

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(A) The torque $	au$ is given by $	au = F \cdot R = I \alpha$.Given $F = (20t - 5t^2) \ N$,$R = 2 \ m$,and $I = 10 \ kg \ m^2$.$(20t - 5t^2) \cdot 2 = 10 \alpha \implies \alpha = 4t - t^2$.Integrating $\alpha$ with respect to time to find angular velocity $\omega$:$\omega = \int (4t - t^2) dt = 2t^2 - \frac{t^3}{3}$.Integrating $\omega$ with respect to time to find angular displacement $	heta$:$	heta = \int (2t^2 - \frac{t^3}{3}) dt = \frac{2t^3}{3} - \frac{t^4}{12}$.The direction of motion reverses when $\omega = 0$:$2t^2 - \frac{t^3}{3} = 0 \implies t^2(2 - \frac{t}{3}) = 0 \implies t = 6 \ s$.Calculating angular displacement at $t = 6 \ s$:$	heta = \frac{2(6)^3}{3} - \frac{6^4}{12} = \frac{2 \cdot 216}{3} - \frac{1296}{12} = 144 - 108 = 36 \ rad$.Number of rotations $N = \frac{	heta}{2\pi} = \frac{36}{2\pi} = \frac{18}{\pi} \approx 5.73$.

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