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(A) The torque $	au$ applied to the pulley is $	au = F \cdot R = (20t - 5t^2) \cdot 2 = 40t - 10t^2 \ N \cdot m$.Using $	au = I \alpha$,we get $\al

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more than $3$ but less than $6$

Vedclass · Answer

(A) The torque $	au$ applied to the pulley is $	au = F \cdot R = (20t - 5t^2) \cdot 2 = 40t - 10t^2 \ N \cdot m$.Using $	au = I \alpha$,we get $\alpha = \frac{	au}{I} = \frac{40t - 10t^2}{10} = 4t - t^2 \ rad/s^2$.Since $\alpha = \frac{d\omega}{dt}$,we integrate to find angular velocity $\omega$:$\omega = \int (4t - t^2) dt = 2t^2 - \frac{t^3}{3}$.The direction of motion reverses when $\omega = 0$ (at $t > 0$):$2t^2 - \frac{t^3}{3} = 0 \Rightarrow t^2(2 - \frac{t}{3}) = 0 \Rightarrow t = 6 \ s$.Now,find the angular displacement $	heta$ by integrating $\omega$ from $t = 0$ to $t = 6$:$	heta = \int_0^6 (2t^2 - \frac{t^3}{3}) dt = [\frac{2t^3}{3} - \frac{t^4}{12}]_0^6 = \frac{2(216)}{3} - \frac{1296}{12} = 144 - 108 = 36 \ rad$.The number of rotations $n = \frac{	heta}{2\pi} = \frac{36}{2\pi} = \frac{18}{\pi} \approx \frac{18}{3.14} \approx 5.73$.Since $5.73$ is between $3$ and $6$,the correct option is $A$.

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