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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
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B
0.7
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C
5
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D
10
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Solution

The correct option is B 0.7
Since, 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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