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Question

A disc revolves with a speed of 3313rev/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 coin revolves with the record? Answer 1.00 for the coin placed at 4 cm, 2.00 for the coin placed at 14 cm and 3.00 for both coins.

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Solution

Given
angular frequency ω=3313 rev/min=1003×160 rev/s
ω=2×227×1003×160=3.5 rad/s
For coin to revolve with the record, the force of friction must be enough to provide necessary centripetal force,
i.e.,
mv2Rμmgmω2Rμmg
Rμgω20.15×10(3.5)2=0.12 m or 12 cm
The coin should be maximum 12 cm from the centre to revolve with the disc. Hence, the coin placed at 4 cm from centre will revolve with the disc.

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