A pulley of radius $1.5$ m is rotated about its axis by a force $F = (12t - 3t^2)$ N applied tangentially (while $t$ is measured in seconds). If moment of inertia of the pulley about its axis of rotation is $4.5\ \text{kg m}^2$, the number of rotations made by the pulley before its direction of motion is reversed, will be $\dfrac K\pi$. The value of $K$ is ______ .
Numerical value type. Enter your answer.
Answer: 18
$$\alpha = \frac{FR}{I} = \frac{1.5(12t - 3t^2)}{4.5} = 4t - t^2$$
Starting from rest:
$$\omega = 2t^2 - \frac{t^3}{3}$$
The direction reverses when $\omega$ returns to zero: $t^2\left(2 - \dfrac t3\right) = 0 \Rightarrow t = 6$ s.
$$\theta = \int_0^6\left(2t^2 - \frac{t^3}{3}\right)dt = \frac{2(216)}{3} - \frac{1296}{12} = 144 - 108 = 36\ \text{rad}$$
Rotations $= \dfrac{36}{2\pi} = \dfrac{18}{\pi}$, so $K = 18$.
Solution by Sreeraj P, M.Sc Physics