Q37 of 40 Page 109

A disc revolves with a speed of 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 co-efficient of friction between the coins and the record is 0.15, which of the coins will revolve with the record?

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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