Question

A pulley of radius $2 \,m$ is rotated about its axis by a force $F = (20t - 5t^{2})$ newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is $10 \,kgm^{2}$, 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

Answer: (B)

$\tau = \left(20t-5t^{2}\right)2=40t -10t^{2}$
$\alpha = \frac{\tau}{I}=\frac{40t-10t^{2}}{10}=4t-t^{2}$
$\omega =\int\limits_{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=6\,s $
$\theta =\int \omega dt $
$=\int\limits_{0}^{6}\left(2t^{2}-\frac{t^{3}}{3}\right)dt $
$=\left[\frac{2t^{3}}{3}-\frac{t^{4}}{12}\right]_{0}^{6} =36 \, rad $
Number of revolution $=\frac{36}{2\pi}$ = Less than $6$