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AIEEE 2011 Solved Paper

Section: Physics

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Question : 37 of 91

Marks: +1, -0

A pulley of radius

2\text{ m}

is rotated about its axis by a force

F=(20t-5t^2)

t

is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is

10\text{ kg}-\text{m}^2

the number of rotation made by the pulley before its direction of motion is reversed, is:

more than 3 but less than 6
more than 6 but less than 9
more than 9
less than 3

Solution:

Given

F=20t-5t^2, R=2\text{ m}

and

I=10\text{ kg m}^2

Torque applied on pulley

\tau=FR

\; \therefore \alpha=\; \frac{FR}{I} [\; \text{ as } \; \tau=I\alpha]

\; \Rightarrow \alpha=\; \frac{(20t-5t^2) \times 2}{10}

\; \Rightarrow \alpha=4t-t^2

\; \Rightarrow \; \frac{d\omega}{dt}=4t-t^2

\; \Rightarrow \; \int\limits_{0}^{\omega} d\omega = \int\limits_{0}^{t} (4t-t^2) dt

\; \Rightarrow \omega=2t^2-\; \frac{r^3}{3}

\;(\; \text{ At } \; t=0,6\,\text{s}\,\omega=0)

\; \omega=\; \frac{d\theta}{dt}=2t^2-\; \frac{t^3}{3}

\; \int\limits_{0}^{\theta} d\theta = \int\limits_{0}^{6} \left(2t^2-\; \frac{r^3}{3}\right) dt

\Rightarrow \theta=36\,\text{rad} \Rightarrow n=\; \frac{36}{2\pi} < 6

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