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 wheel's rotation is \[\tau =I\alpha =I\left( \frac{{{\omega }_{0}}-\omega }{t} \right)\] \[\therefore \] \[\tau =\frac{2\times 2\pi \times 60}{60\times 60}=\frac{\pi }{15}\,N-m\]
You need to login to perform this action.
You will be redirected in
3 sec
