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



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