Rotational Motion Question 55

Question: A wheel having moment of inertia $ 2kg-m^{2} $ about its vertical axis, rotates at the rate of 60 rpm about this axis. The torque which can stop the wheel’s rotation in one minute would be:

[AIPMT (S) 2004]

Options:

A) $ \frac{2\pi }{15}N-m $

B) $ \frac{\pi }{12}N-m $

C) $ \frac{\pi }{15}N-m $

D) $ \frac{\pi }{18}N-m $

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Answer:

Correct Answer: C

Solution:

  • Given: $ I=2kg-m^{2},{\omega_0}=\frac{60}{60}\times 2\pi rad/s, $ $ \omega =0,t=60s $

    The torque required to stop the wheel’s rotation is $ \tau =I\alpha =I( \frac{{\omega_0}-\omega }{t} ) $

    $ \therefore $ $ \tau =\frac{2\times 2\pi \times 60}{60\times 60}=\frac{\pi }{15}N-m $



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