Rotational Motion Question 96

Question: A solid cylinder of mass 2 kg and radius 4 cm is rotating about its axis at the rate of 3 rpm. The torque required to stop after $ 2\pi $ revolutions is-

Options:

A) $ 12\times 10^{4}Nm $

B) $ 2\times 10^{6}Nm $

C) $ 2\times 10^{6}Nm $

D) $ 2\times 10^{3}Nm $

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Answer:

Correct Answer: C

Solution:

  • [c] $ \omega _2^{2}=\omega _1^{2}+2\alpha \theta $

    $ 0={{( 2\pi \frac{3}{60} )}^{2}}-2\times 2\pi (2\pi )\alpha $

    $ \alpha =\frac{9}{60\times 60\times 2}=\frac{1}{800} $

    $ \tau =I\alpha =\frac{mr^{2}}{2}\alpha $

    $ \frac{2\times 16\times 10^{4}}{2}\times \frac{1}{800}=2\times {10^{-6}} $



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