A disc revovles with a speed of `33 (1)/(3) rev//min` and has a radius of 15 cm Two coins A and B are palaced at 4 cm and 14 cm away from the center of the record If the coefficient of friction between the coins and the record is 0.5 which of the coins will revolve with the record ?

Options

A. A
B. B
C. Both A and B
D. Neither A nor B

Answer

A

Explanation

Here, `mu=0.15`.
`v=33(1)/(3)rpm=(100)/(3)rpm=(100)/(3xx60)rps=(5)/(9)rps`
`therefore omega=2piv=2xx(22)/(7)xx(5)/(9)=(220)/(63)rad s^(-1)`
The coin will revolve with the disc. If the force of friciton is enousght to provide the necessary centripetal force.
`i.e. (mv^(2))/(r) le mumg`
`As v=r omega therefore m omega^(2)r le mumg`
For the given `mu and omega`, the condition is `r le (mug)/(omega^(2))...(i)`
`As(mug)/(omega^(2)) =(0.15xx10)/(((220)/(63))^(2))approx12cm`
For coin, r=4cm The above condition is satisfied , therefore coin A will revolve with the disc.
For coin B, r=14cm The above condition is not satisfied, therefore coin B will not reolve with the disc.
NOTE: We have nothing to do with the radius of the disc.