A) \[\frac{2\pi }{15}\text{N-m}\]
B) \[\frac{\pi }{12}\text{N-m}\]
C) \[\frac{\pi }{15}\text{N-m}\]
D) \[\frac{\pi }{18}\text{N-m}\]
Correct Answer: C
Solution :
Given: \[I=2kg-{{m}^{2}},\,\,{{\omega }_{0}}=\frac{60}{60}\times 2\pi \,rad/s,\] \[\omega =0,\,\,t=60\,s\] The torque required to stop the wheel?s rotation is \[\tau =I\,\alpha =I\left( \frac{{{\omega }_{0}}-\omega }{t} \right)\] \[\therefore \] \[\tau =\frac{2\times 2\pi \times 60}{60\times 60}=\frac{\pi }{15}\text{N-m}\]
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