A pulley of radius 1.5,m is rotated about its | vedclass

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(A) The torque applied is $	au = F \cdot r = (12t - 3t^2) \cdot 1.5 = 18t - 4.5t^2$.Using $	au = I\alpha$,we have $18t - 4.5t^2 = 4.5\alpha$,which g

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$18$

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(A) The torque applied is $	au = F \cdot r = (12t - 3t^2) \cdot 1.5 = 18t - 4.5t^2$.Using $	au = I\alpha$,we have $18t - 4.5t^2 = 4.5\alpha$,which gives $\alpha = 4t - t^2$.Since $\alpha = \frac{d\omega}{dt}$,we integrate to find angular velocity: $\omega = \int (4t - t^2) dt = 2t^2 - \frac{t^3}{3}$.The motion reverses when $\omega = 0$,so $t^2(2 - \frac{t}{3}) = 0$,giving $t = 6\,s$.The angular displacement $	heta$ is $\int_{0}^{6} (2t^2 - \frac{t^3}{3}) dt = [\frac{2t^3}{3} - \frac{t^4}{12}]_{0}^{6} = \frac{2(216)}{3} - \frac{1296}{12} = 144 - 108 = 36\,rad$.The number of rotations is $N = \frac{	heta}{2\pi} = \frac{36}{2\pi} = \frac{18}{\pi}$.Comparing with $\frac{K}{\pi}$,we get $K = 18$.

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