Question

A pulley of radius $1.5m$ is rotated about its axis by a force $F=\left(12t-3t^2\right)N$ applied tangentially (while $t$ is measured in seconds). If moment of inertia of the pulley about its axis of rotation is $4.5kg{m}^2$, 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

Answer: 18

$\tau =I\alpha \Rightarrow \left(12t-3t^2\right)1.5=4.5\alpha$
$\Rightarrow \alpha =4t-t^2$
$\Rightarrow \frac{d\omega }{dt}=4t-t^2\Rightarrow \omega ={\int }_0^t\left(4t-t^2\right)dt$
$\Rightarrow \omega =2t^2-t^3/3$
For $\omega =0=2t^2-\frac{t^3}{3}\Rightarrow t^2\left(2-\frac{t}{3}\right)=0$
$\Rightarrow t=0,6$
$\frac{d\theta }{dt}=2t^2-\frac{t^3}{3}\Rightarrow \theta =\underset{0}{\overset{6}{\int }}\left(2t^2-\frac{t^3}{3}\right)dt$
$={\left[\frac{2t^3}{3}-\frac{t^4}{12}\right]}_0^6$
$=6^3\left(\frac{2}{3}-\frac{6}{12}\right)=6^3\left(\frac{8-6}{12}\right)$
$=\frac{6^3}{6}=36$
No. of revolutions $=\frac{36}{2\pi }=\frac{18}{\pi }$
$\therefore K=18$