JEE Main 2022 — Rotational Motion Question with Solution

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

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

Then, $\left(12t-3t^2\right)1.5=4.5\alpha$ 

$\Rightarrow \alpha =4t-t^2$ 

Angular acceleration in terms of angular velocity is

 $\Rightarrow \alpha =\frac{d\omega }{dt}=4t-t^2$

$\Rightarrow \omega ={\int }_0^t\left(4t-t^2\right)dt$

$\Rightarrow \omega =2t^2-\frac{t^3}{3}$

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

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

Now, using $\omega =\frac{d\theta }{dt}$, we have

$$\begin{gathered} \frac{d\theta }{dt}=2t^2-\frac{t^3}{3} \\ \Rightarrow \theta ={\int }_0^6\left(2t^2-\frac{t^3}{3}\right)dt \end{gathered}$$

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

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

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

$\therefore K=18$