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Answer please! A pulley of radius 2 m is rotated about its axis by a force F = (20t - 5t2) newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m2 , the number of rotati

A pulley of radius 2 m is rotated about its axis by a force F = (20t - 5t2) newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m2 , the number of rotations made by the pulley before its direction of motion if reversed, is

Option 1)

less than 3

Option 2)

more than 3 but less than 6

Option 3)

more than 6 but less than 9

Option 4)

more than 9

Answers (1)
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As we learnt in

Analogue of second law of motion for pure rotation -

\vec{\tau }=I\, \alpha

- wherein

Torque equation can be applied only about two points

(i) centre of motion.

(ii) point which has zero velocity/acceleration.

 

 \int \tau{}.d\theta=0

\tau{}=(20t-t^{2})\times 2=40t-10t^{2}

I\alpha=40t-10t^{2}

\alpha=\frac{40t-10t^{2}}{10}=4t-10t^{2}

\alpha=\frac{d\omega}{dt}

d\omega{}=\alpha.dt

=>\int d\omega=\int_{0}^{t}(4t-t^{2})dt

=4\int_{0}^{t}t\ d t-\int_{0}^{t}t^{2}dt

\omega=2t^{2}-\frac{t^{3}}{3}

\omega is zero at  2t^{2}-\frac{t^{}3}{3}=0

\Rightarrow\ \;2t^{2}=\frac{t^{3}}{3}

t = 6 sec   

\theta=\int_{0}^{6} {\omega}dt

\Rightarrow\ \;\int_{0}^{6} \left ( 2t^{2}-\frac {t^{3}}{3} \right ) dt=36\ rad

No. of revolution \frac{36}{2\pi} less than 6

Ans. 4


Option 1)

less than 3

This an incorrect option.

Option 2)

more than 3 but less than 6

This is the correct option.

Option 3)

more than 6 but less than 9

This an incorrect option.

Option 4)

more than 9

This an incorrect option.

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