Question

# A body moves along circular path of radius 10 m and the coefficient of friction is 0.5. What should be its angular velocity in rad/sec if it is not to slip from the surface? $$\displaystyle \left( g=9.8{ m }/{ { s }^{ 2 } } \right)$$

A
0.1
B
0.7
C
5
D
10

Solution

## The correct option is B 0.7Since, the centripetal force is supplied by the frictional force, therefore$$\displaystyle \mu mg=\frac { { mv }^{ 2 } }{ r } mr{ \omega }^{ 2 }$$$$\displaystyle \Rightarrow \quad { \mu }_{ g }=r{ \omega }^{ 2 }\quad \Rightarrow \quad 0.5\times 9.8=10\times { \omega }^{ 2 }$$$$\displaystyle \Rightarrow \quad { \omega }^{ 2 }=0.7\quad rad/sec$$$$\displaystyle$$$$\displaystyle$$Physics

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