A pulley of radius $$2 \,m$$ is rotated about its axis by a force $$F = \left( {20t - 5{t^2}} \right)$$   newton (where $$t$$ is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is $$10\,kg - {m^2}$$   the number of rotations made by the pulley before its direction of motion is reversed, is:

Solution :
$$\eqalign{ & F = 20t - 5{t^2} \cr & \therefore \alpha = \frac{{FR}}{I} = 4t - {t^2} \cr & \Rightarrow \frac{{d\omega }}{{dt}} = 4t - {t^2} \cr & \Rightarrow \int\limits_0^\omega {d\omega } = \int\limits_0^t {\left( {4t - {t^2}} \right)dt} \cr & \Rightarrow \omega = 2{t^2} - \frac{{{t^3}}}{3}\,\left( {{\text{as}}\omega = 0\,{\text{at }}t = 0,\,6s} \right) \cr & \int\limits_0^\theta {d\theta } = \int\limits_0^6 {\left( {2{t^2} - \frac{{{t^3}}}{3}} \right)} dt \cr & \Rightarrow \theta = 36\,{\text{rad}}\,\,\, \Rightarrow n = \frac{{36}}{{2\pi }} < 6 \cr} $$