### Q.37:- A disc revolves with a speed of 100 / 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 co-efficient of friction between the coins and the record is 0.15, which of the coins will revolve with the record?

**Answer:-**

Mass of each coin = *m*

Radius of the disc, *r* = 15 cm = 0.15 m

Frequency of revolution, *ν* = 100 / 3 rev/min = 100 / (3 × 60) = 5 / 9 rev/s

Coefficient of friction, *μ* = 0.15

In the given situation, the coin having a force of friction greater than or equal to the centripetal force provided by the rotation of the disc will revolve with the disc. If this is not the case, then the coin will slip from the disc.

__Coin placed at 4 cm____:__

Radius of revolution, *r*‘ = 4 cm = 0.04 m

Angular frequency, *ω* = 2π*ν = *2 × (22/7) × (5/9) = 3.49 s^{-1}

Frictional force, *f* = *μ**m*g = 0.15 × *m* × 10 = 1.5*m* N

Centripetal force on the coin:

*F*_{cent.} = mr’ω^{2}

= m × 0.04 × (3.49)^{2}

= 0.49m N

Since* f >* F_{cent}, the coin will revolve along with the record.

__Coin placed at 14 cm:__

Radius, r” = 14 cm = 0.14 m

Angular frequency, *ω* = 3.49 s^{–1}

Frictional force, *f*‘ = 1.5*m* N

Centripetal force is given as:

*F*_{cent.} = mr”ω^{2}

= *m* × 0.14 × (3.49)^{2}

= 1.7*m* N

Since *f* < *F*_{cent.}, the coin will slip from the surface of the record.