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(D) The torque applied is $	au = rF = 2(20t - 5t^2) = 40t - 10t^2\, N\cdot m$.Using $	au = I\alpha$,we have $40t - 10t^2 = 10\alpha$,so $\alpha = 4t

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more than $3$ but less than $6$

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(D) The torque applied is $	au = rF = 2(20t - 5t^2) = 40t - 10t^2\, N\cdot m$.Using $	au = I\alpha$,we have $40t - 10t^2 = 10\alpha$,so $\alpha = 4t - t^2\, rad/s^2$.Integrating $\alpha$ with respect to time to find angular velocity $\omega$ (assuming $\omega_0 = 0$): $\omega = \int (4t - t^2) dt = 2t^2 - \frac{t^3}{3}$.The direction of motion reverses when $\omega = 0$,so $2t^2 - \frac{t^3}{3} = 0$,which gives $t^2(2 - \frac{t}{3}) = 0$. Thus,$t = 6\, s$.To find the angular displacement $	heta$,we integrate $\omega$: $	heta = \int_0^6 (2t^2 - \frac{t^3}{3}) dt = [\frac{2t^3}{3} - \frac{t^4}{12}]_0^6$.$	heta = \frac{2(216)}{3} - \frac{1296}{12} = 144 - 108 = 36\, rad$.The number of rotations is $N = \frac{	heta}{2\pi} = \frac{36}{2\pi} \approx \frac{36}{6.28} \approx 5.73$.Since $5.73$ is between $3$ and $6$,the correct option is $D$.

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