1. Given data

• Radius of pulley: $r=1.5\,\text{m}$
• Applied tangential force: $F(t)=12t-3t^2$
• Moment of inertia: $I=4.5\,\text{kg m}^2$

We need the angular displacement made before the direction of motion reverses.

2. Torque and angular acceleration

Since the force is tangential, torque is $\tau(t)=rF(t)=1.5(12t-3t^2)$ $\tau(t)=18t-4.5t^2$

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

3. Angular velocity as a function of time

Assume the pulley starts from rest, so $\omega(0)=0$

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

Integrating, $\omega(t)=\int (4t-t^2)dt = 2t^2-\frac{t^3}{3}+C$

Using $\omega(0)=0$, we get $C=0$. Hence, $\omega(t)=2t^2-\frac{t^3}{3}$

4. Time when direction reverses

Direction reverses when angular velocity becomes zero again after being positive.

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

Thus, $t=0 \quad \text{or} \quad t=6\,\text{s}$

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

5. Angular displacement till reversal

Angular displacement is $\theta=\int_0^6 \omega(t)\,dt$

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

Integrate: $\theta=\left(\frac{2t^3}{3}-\frac{t^4}{12}\right)_0^6$

At $t=6$, $\theta=\frac{2(6)^3}{3}-\frac{(6)^4}{12}$ $\theta=\frac{2\cdot 216}{3}-\frac{1296}{12}$ $\theta=144-108=36\,\text{rad}$

6. Number of rotations

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

Given that this is $\frac{K}{\pi}$, we get $K=18$

7. Comparison with stored answer

Stored correct answer: $18$

Our derived answer is also $18$.

So the stored answer is correct.