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

For the coin to revolve with the disc, the force of friction should be enough to provide the necessary 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).