Question

A disc revolves with a speed of 33 1/3 rev./min, and has a radius of 15 cm. Two coins are placed 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 coins will revolve with the record?

Answer 1

The coin can only revolve with the record when the force of friction is enough to provide the centripetal force. If this force is not enough, then the coin slips on the record.

Here, R = mg, and the centripetal force is given by mrω²

To prevent slipping, the condition should be

μ mg ≥ mrω²

or, μ g ≥ rω²    ...(i)

For first coin:

r = 4 cm = 4/100 m

v = 331/3 rev/min = {100}/{3 x 60}rev/sec

ω = 2πv = 2π x {100}/{3 x 60} = 3.49 ms^-1

rω² = 4/100 x (3.49)² = 0.49 ms^-2

and μ_g = 0.15 x 9.8 = 1.47 ms^-2

As μ_g > rω², therefore this coin will revolve with the record

For 2nd coin:

r = 14 cm = 14/100 m

ω = 3.49 s^-1

Hence, rω² = 14/100 x (3.49)² = 1.705 ms^-2

and μ_g = 1.47 ms^-2

As μ_g ≥ rω² is not satisfied, so this coin will not rotate with the record.