A) \[\frac{2\pi }{15}N-m\]
B) \[\frac{\pi }{12}N-m\]
C) \[\frac{\pi }{15}N-m\]
D) \[\frac{\pi }{18}N-m\]
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 wheels rotation is |
| \[\tau =I\,\alpha =I\left( \frac{{{\omega }_{0}}-\omega }{t} \right)\] |
| \[\therefore \] \[\tau =\frac{2\times 2\pi \times 60}{60\times 60}=\frac{\pi }{15}N-m\] |
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