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, \[I=2\,kg-{{m}^{2}},{{\omega }_{0}}=\frac{60}{60}\times 2\pi \,rad/s,\] \[\omega =0,\,t=60\,s\] The torque required to stop the wheels rotation is \[\tau =I\alpha =I\left( \frac{{{\omega }_{0}}-\omega }{t} \right)\] \[\tau =\frac{2\times 2\pi \times 60}{60\times 60}=\frac{\pi }{15}N-m\]
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