A disc of radius 15 cm rotates with a speed of 33 1/2 rpm. Two coins are placed on it at 4 cm and 14 cm from its centre. ​

Question

A disc of radius 15 cm rotates with a speed of 33\(\frac13\)rpm. Two coins are placed on it at 4 cm and 14 cm from its centre. If the coefficient of friction between the coins and the disc is 0.15, which of the two coins will revolve with the disc ?

Answer 1

Data : r = 15 cm = 0.15 m,

f = 33\(\frac13\) rpm = \(\frac{100}{3\times60}\) rev/s = \(\frac59\) Hz, µ_s = 0.15, r₁

= 4 cm = 0.04 m, r₂ = 14 cm = 0.14 m

Angular speed, \(\omega\) = 2\(\pi\)f = 2 x 3.142 x \(\frac59\) = \(\frac{31.42}{9}\)

= 3.491 rad/s

To revolve with the disc without slipping, the necessary centripetal force must be less than or equal to the limiting force of static friction.

Limiting force of static friction, f_s = µ_s N = µ_s (mg) where m is the mass of the coin and N = mg is the normal force on the coin. 

∴ mω²r ≤ µ_s (mg) or ω²r ≤ µ_sg 

µ_sg = 0.15 × 9.8 = 1.47 m/s² 

For the first coin, r₁ = 0.04 m. 

∴ ω²r₁ = (3.491)² × 0.04 = 12.19 × 0.04  = 0.4876 m/s²

Since, ω²r₁ < µ_sg, this coin will revolve with the disc. For the second coin, r₂ = 0.14 m. ∴ ω²r₂ = (3.491)² × 0.14 = 12.19 × 0.14 = 1.707 m/s²

Since, ω²r₂ > µ_sg, this coin will not revolve with the disc. 

Thus, only the coin placed at 4 cm from the centre will revolve with the disc.