A disc revolves with a speed of 33`/3` rev/min, 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 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 be disc, the force of friction should be enough to provide the necessary centripetal force. i.e., `(mv^(2))/r lemumg`. Now `v=romega`, where `omega=(2pi)/T` is the angular frequency of the disc. Fow a given `mu and omega`, the condition is `rle(mug)/omega^(2)`
The condition is satisfied by the nearer coin (4cm from the centre).