A disc revolves with a speed of 33 1/3rev/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 ?

A disc revolves with a speed of 33 1/3rev/min. 3 and has radius of 15

1.	A disc revolves with a speed of 33 1/3rev/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 ?
Answer» Solution :For the coin to revolve with the DISC, the force of friction should be enough to provide thenecessary centripital force `i.e (mv^2)/( r) lt= mu MG ` . Now `v = r OMEGA` , where ` omega = (2pi)/( T)` is the angular FREQUENCY of the disc.For a given `mu` and `omega` , the condition is `r lt= mu g// omega^2 ` .The condition is SATISFIED by the nearer coin ( 4 cm from the centre).