A pulley of radius 1 . 5m is rotated about its axis by a force F =...

Torque of pulley about its axis is $τ = I\alpha$, where, $I$ is moment of inertia and $\alpha$ is angular acceleration.

Torque of a force about its axis is $τ = FR$, here, $R$ is radius of pulley.

Then, \left(12t - 3{t}^{2}\right)1 . 5 = 4 . 5\alpha$</p><p>$⇒ \alpha = 4t - {t}^{2}$</p><p>Angular acceleration in terms of angular velocity is</p><p>$⇒ \alpha = \dfrac{d \omega}{d t} = 4t - {t}^{2}$</p><p>$⇒ \omega = {∫}_{0}^{t}\left(4t - {t}^{2}\right) d t$</p><p>$⇒ \omega = 2{t}^{2} - \dfrac{{t}^{3}}{3}$</p></blockquote><blockquote><p>For$\omega = 0, 2{t}^{2} - \dfrac{{t}^{3}}{3} = 0 ⇒ {t}^{2}\left(2 - \dfrac{t}{3}\right) = 0$</p><p>$⇒ t = 0\text{or}6s$</p><p>Now, using$\omega = \dfrac{d \theta}{d t}$, we have</p><p>$\dfrac{d \theta}{d t} = 2{t}^{2} - \dfrac{{t}^{3}}{3} ⇒ \theta = {∫}_{0}^{6}$\left(2{t}^{2} - \dfrac{{t}^{3}}{3}\right) d t$

$= {\left[\dfrac{2{t}^{3}}{3} - \dfrac{{t}^{4}}{12}\right]}_{0}^{6}$

$= {6}^{3}\left(\dfrac{2}{3} - \dfrac{6}{12}\right) = {6}^{3}\left(\dfrac{8 - 6}{12}\right)= \dfrac{{6}^{3}}{6} = 36$

Number of revolutions $= \dfrac{36}{2\pi} = \dfrac{18}{\pi}$

$∴ K = 18$