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

If the coin is to revolve with the record, then the force of friction must be enough to provide the necessary centripetal force.
`therefore` mr`omega^(2)` `le` `mu_s`mg or r `le` `mu_s`mg / m`omega^(2)` or r`le` `mu_s`g / `omega^(2)`
frequency = 33(1/3)rpm = 100 / 3 rom = 100 / `3 xx 60 rps`
The problem in which centripetal force is obtained from force of friction , start with the following equation,
m r`omega^(2)` `le` `mu_s`mg
`omega` = 2`pi` xx 100 / `3 xx 60 rad s^(-1)` = 10/9 `pi` `rad s^(-1)`
`mu_s`g / `omega^(2)` = 0.15 `xx `10 / `((10 / 9) pi) ^(2)` m = 0.12 m = 12 cm
The codition (r `le`12 cm ) is satisfied by the coin placed at 4 cm from the centre of the record. So, the coin at 4 cm will receive with the record.