A) 5
B) 10
C) 0.1
D) 0.7
Correct Answer: D
Solution :
For moving on circular path without slipping centripetal force must equal to frictional force that is \[\frac{m{{v}^{2}}}{r}=\mu \,mg\] \[mr{{\omega }^{2}}=\mu \,mg\] \[r{{\omega }^{2}}=\mu \,g\] \[\therefore \] \[\omega =\sqrt{\frac{\mu g}{r}}=\sqrt{\frac{0.5\times 9.8}{10}}=0.7\,rad/s\]
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