Question

A disc revolves with a speed of 100/3 rev/min and has radius of 15 cm. Two coins are placed at 4cm and
14 cmaway 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 record?
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Answer 1

Ff = u m g = frictional force on coin

Fc = m w^2 R =  centripetal force on coin

u g = w^2 R     condition for coin to just start slipping

R = u g / w^2    distance of coin from center

w = 2 pi f = 2 * pi * 33.3 / 60 = 3.49 / s

R = .15 * 9.8 / 3.49^2 = .121 m

So only the coin at 4 cm will not slip since the required centripetal force

increases with increasing R