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Laws of Motion

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Physics Part I

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Physics

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A disc revolves with a speed of 33 space begin inline style 1 third end style 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 coefficient of friction between the coins and the record is 0.15, which of the coins will revolve with the record?

Coin placed at 4 cm from the centre

Given, 

Mass of each coin = m

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

Frequency of revolution, ν = 100 over 3 rev/min

                                         = fraction numerator 100 over denominator left parenthesis 3 space straight x space 60 right parenthesis end fraction

                                         = open parentheses 5 over 9 close parentheses 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.

Case 1: Coin placed at 4 cm

Radius of revolution, r' = 4 cm = 0.04 m

Angular frequency, ω = 2πν

                                 = 2 x open parentheses 22 over 7 close parentheses space x space open parentheses 5 over 9 close parentheses 
                                  = 3.49 s-1

Frictional force, f = μmg

                          = 0.15 × 
m × 10

                          = 1.5
m N 

Centripetal force on the coin is, 

              Fcent. = mr'ω

                        = m × 0.04 × (3.49)

                        = 0.49 m N 

Since f > Fcent, the coin will revolve along with the record.

Case 2: 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, 

                Fcent. = mr"ω

                         = m × 0.14 × (3.49)2 

                         = 1.7m N 

Since f < Fcent., the coin will slip from the surface of the record.
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