Answer :

Given:

Speed of revolution of disc, v = 33.3 rev/min = 0.55 rev/sec


Radius of Disc = 15 cm = 0.15 m


Distance of coin 1 from centre, r = 4 cm = 0.04 m


Distance of coin 2 from centre, R = 14 cm = 0.14 m


Coefficient of friction, � = 0.15


To decide which coin will rotate, we shall calculate the net force on each coin,


For coin 1, at 4 cm distance


Radius of path = 0.04 m


Angular frequency, ω = 2πv


ω = 2× 3.14× 0.55 rev/sec


ω = 3.49 s-1


Frictional force, f = �× m× g


f = 0.15× m Kg× 10 ms-2


f = 1.5 m N


Centrifugal force on the coin:


F = m× r× ω2


F = m× 0.04m × 3.492


F = 0.49m N


Since force of friction is larger than the centrifugal force, the coin will not slide and will revolve along the record.


For coin 2, at 14 cm distance


Radius of path = 0.14 m


Angular frequency, ω = 2πv


ω = 2× 3.14× 0.55 rev/sec


ω = 3.49 s-1


Frictional force, f = �× m× g


f = 0.15× m Kg× 10 ms-2


f = 1.5 m N


Centrifugal force on the coin:


F = m× r× ω2


F = m× 0.14m × 3.492


F = 1.7m N


Since force of friction is smaller than the centrifugal force, the coin will slip on the surface of record.


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