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A disc revolves with a speed of 33/3 r...

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 ?

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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).
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