Question

A pulley of radius 2 m is rotated about its axis by a force $F=(20t-5r^2)$ newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis fo rotation is 10 $kg{m}^2,$ the number of rotations made by the pulley before its direction of motion is reversed, is:

• A. less than 3
• B. more than 3 but less than 6
• C. more than 6 but less than 9
• D. more than 9

Answer 1

The correct option is B more than 3 but less than 6

$\tau =(20t-5t^2)2=40t-10t^2$
$\alpha =\frac{\tau }{I}=\frac{40t-10t^2}{10}=4t-t^2$
$\omega ={\int }_0^t\alpha dt=2t^2-\frac{t^3}{3}$
When direction is reversed, $\omega$ is zero. So
$2t^2-\frac{t^3}{3}=0\Rightarrow t^3=6t^2\Rightarrow t=6s$
$\theta =\int \omega dt={\int }_0^6(2t^2-\frac{t^3}{3})dt={[\frac{2t^3}{3}-\frac{t^4}{12}]}_0^6=36rad$
Number of revolution = $\frac{36}{2\pi }$ = Less than 6