A pulley of radius $$1.5 \text{ m}$$ is rotated about its axis by a force $$F = (12t - 3t^2) \text{ N}$$ 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 ______.

We are given a pulley of radius $$r = 1.5 \text{ m}$$, a tangential force $$F = (12t - 3t^2) \text{ N}$$, and moment of inertia $$I = 4.5 \text{ kg m}^2$$. We need to find $$K$$ where the number of rotations before the direction reverses is $$\dfrac{K}{\pi}$$.

$$\tau = F \times r = (12t - 3t^2) \times 1.5 = 18t - 4.5t^2$$

$$\alpha = \frac{\tau}{I} = \frac{18t - 4.5t^2}{4.5} = 4t - t^2$$

Since the pulley starts from rest ($$\omega = 0$$ at $$t = 0$$):

$$\omega = \int_0^t \alpha \, dt = \int_0^t (4t - t^2) \, dt = 2t^2 - \frac{t^3}{3}$$

The direction reverses when $$\omega = 0$$ (after the pulley has started rotating):

$$2t^2 - \frac{t^3}{3} = 0$$

$$t^2\left(2 - \frac{t}{3}\right) = 0$$

$$t = 0$$ or $$t = 6 \text{ s}$$

So the direction reverses at $$t = 6 \text{ s}$$.

$$\theta = \int_0^6 \omega \, dt = \int_0^6 \left(2t^2 - \frac{t^3}{3}\right) dt$$

$$\theta = \left[\frac{2t^3}{3} - \frac{t^4}{12}\right]_0^6$$

$$\theta = \frac{2(216)}{3} - \frac{1296}{12} = 144 - 108 = 36 \text{ rad}$$

$$\text{Number of rotations} = \frac{\theta}{2\pi} = \frac{36}{2\pi} = \frac{18}{\pi}$$

Comparing with $$\dfrac{K}{\pi}$$, we get $$K = 18$$.