A disc revolves with a speed of 33 `1/3` rev/min. 3 and has radius of 15 cm. Two coins are placed at 4 cm and 14 cm away from the centre of the record. If the co-efficient of friction between the coins and the record is 0.15 which of the coins will revolve with the record ?

Explanation

Radius of disc r= 15 cm

Frequency of roation of disc
`= (100)/( 3)`
`=(5)/(9) ` Revolution `//` second
Angular frequency `epsilon = 2 pi f`
co - efficient of friction between disc and coin
centripetal force `F_(c ) = (mv^(2))/( r ) = mrepsilon^(2)`
in order that coin do not slip `f le F_(c )`
For First A coin
and `mug = 0.15 xx 9.8 = 1.47`
Here `mu g (1.47 ) gt pi ^(2) = (0.48)`
`pi epsilon^(2) =0.14 xx ((220)/( 63))^(2)`
Here `mu g (1.47 ) le (r epsilon ^(2) (1.70)`
Coin at 4 cm from disc will revolve with disc