Question

A pulley of radius $1.5 \,m$ is rotated about its axis by a force $F =\left(12 t -3 t ^2\right) N$ applied tangentially (while $t$ is measured in seconds). If moment of inertia of the pulley about its axis of rotation is $4.5\, kg \,m ^2$, the number of rotations made by the pulley before its direction of motion is reversed, will be $\frac{ K }{\pi}$. The value of $K$ is _____

Answer 1

$ \tau= I \alpha \Rightarrow\left(12 t -3 t ^2\right) 1.5=4.5 \alpha$
$\Rightarrow \alpha=4 t - t ^2 $
$ \Rightarrow \frac{ d \omega}{ dt }=4 t - t ^2 \Rightarrow \omega=\int_0^{ t }\left(4 t - t ^2\right) dt$
$ \Rightarrow \omega=2 t ^2- t ^3 / 3 $
For $ \omega=0=2 t ^2-\frac{ t ^3}{3} \Rightarrow t ^2\left(2-\frac{ t }{3}\right)=0 $
$ \Rightarrow t =0,6 $
$ \frac{ d \theta}{ dt }=2 t ^2-\frac{ t ^3}{3} \Rightarrow \theta=\int\limits_0^6\left(2 t ^2-\frac{ t ^3}{3}\right) dt $
$ =\left[\frac{2 t^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) $
$=\frac{6^3}{6}=36$
No. of revolutions $=\frac{36}{2 \pi}=\frac{18}{\pi}$
$\therefore K =18$