A long plying record revolves with a speed of `33 (1)/(3) "rev min"^(-1)`, and has a radius of 15 cm. Two coins are placed at 4 cm and 14 cm away from the centre of the record. If the coefficient of friction between the coins and the record is 0.15, which of the two coins will revolve with the record ? Take `g = 9.8 ms^(-2)`.

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).