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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 is 0.15, which of the coins will revolve with 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 mumg`. Now, `v=romega`, where `omega =(2pi)/T` is the ANGULAR frequency of the disc. For a given `MU` and `omega`, the condition is `r le (mug)/omega^(2)` The condition is SATISFIED by the nearer coin (4 cm from the centre). |
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