Question

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

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

Given,
Tangential force acting on the pulley, $F=(20t-5t^2)$
Radius of pulley $\left(r\right)=$ $2\text{m}$
Moment of inertia of the pulley about axis of rotation $I=10{\text{kg-m}}^2$
$\therefore$ Torque acting on the pulley by the applied force is given by
$\tau =(20t-5t^2)2=(40t-10t^2)\text{Nm}$
From this, we find angular acceleration$\alpha =\frac{\tau }{I}=\frac{40t-10t^2}{10}=4t-t^2$
To find angular speed, we use
$\omega ={\int }_0^t\alpha dt={\int }_0^{t}4t-t^{2}dt=2t^2-\frac{t^3}{3}$
When direction is reversed, $\omega$ is zero. So
$2t^2-\frac{t^3}{3}=0\Rightarrow t^3=6t^2\Rightarrow t=6s$
Angular displacement
$\theta =\int \omega dt$
$={\int }_0^6(2t^2-\frac{t^3}{3})dt$
$={[\frac{2t^3}{3}-\frac{t^4}{12}]}_0^6=36\text{rad}$
$\text{1 rev}$= $2\pi \text{rad}$
Total number of revolution before reversal = $\frac{36}{2\pi }$
The value which is more than 3 and less than 6
Hence, the option$\left(b\right)$ is correct answer.