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A disc revolves with a speed of 33(1)/(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 coefficient of friction between the coins and the record ? |
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Answer» Solution :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 )le 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 le (mu G)/(omega^(2))`. The condition is SATISFIED by the nearer coin (4 CM from the centre). |
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