Rotational Motion Question 141
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 $10kgm^{2}$ , the number of rotations made by the pulley before its direction of motion is reversed, is:
Options:
A) less than 3
B) more than 3 but less than 6
C) more than 6 but less than 9
D) more than 9
Show Answer
Answer:
Correct Answer: B
Solution:
[b]
$\tau =(20t-5t^{2})2=40t-10t^{2}$
$\alpha =\frac{\tau }{I}=\frac{40t-10t^{2}}{10}=4t-t^{2}$
$\omega =\int\limits _{0}^{t}{\alpha dt}=2t^{2},-\frac{t^{3}}{3}$
When direction is reversed, co is zero.
So,
$2t^{2}-\frac{t^{3}}{3}=0\Rightarrow t^{3}=6t^{2}\Rightarrow t=6s$
$\theta =\int{\omega dt}$
$=\int\limits _{0}^{6}{(2t^{2}-\frac{t^{3}}{3})dt}$
$={{\frac{2t^{2}}{3}-\frac{t^{4}}{12} }^{6}} _{0}=36rad$
Number of revolution = $\frac{36}{2\pi }$= Less than 6
BETA
