A small coin is kept at the rim of a horizontal circular disc which is set into rotation about vertical axis passing through its centre . If radius of the disc is 10 cm and `mu_(s) = (3)/(7)` , then the angular speed at which the coin will just slip off at

Options

A. `5 rad s ^(-1)`
B. `6.54 rad s^(-1)`
C. `10 rad s^(-1)`
D. `4.9 rad s^(-1)`

Answer

B

Explanation

For just slip off the coin
`mumg=mromega^(2)rArrmug=romega^(2)rArromega^(2)=(mug)/(r)`
`rArr" "omega=sqrt((mug)/(r))=sqrt((3)/(7)xx(10)/(10xx10^(-2)))=6.54rads^(-1)`