Question

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 co-efficient of friction between the coins and the record is 0.15, which of the coins will revolve with the record ?

Answer 1

Given that the speed of revolution of the disc is 33 1 3  rpm, the radius of the disc is 15 cm, the distance of the two coins from the center of the record is 4 cm and 14 cm respectively, and the coefficient of friction between the coins and the record is 0.15.

The equation to determine the angular velocity of the disc is,

ω=2πf

Substitute the values in the above equation.

ω=2π( 33 1 3 rpm ) =2π( ( 100 3 rpm ) ( 1 rps 60 rpm ) ) = 10 9 π  rad/ s 2

The equation to determine the maximum radius of rotation of the coin is,

F f = F c mr ω 2 =μmg r= μg ω 2

Here, the frictional force is F f , the centripetal force is F c , the mass of the coin is m, the radius of rotation of the coin is r, the angular velocity of rotation is ω and the kinetic friction is μ.

Substitute the values in the above equation.

r= ( 0.15 )( 10 ) ( 10 9 π ) 2 =0.12×100 cm =12 cm

The maximum radius of rotation of the coin placed is 12 cm.

Thus, the coin at 4 cm would revolve with the record.