Data : r = 15 cm = 0.15 m,
f = 33\(\frac13\) rpm = \(\frac{100}{3\times60}\) rev/s = \(\frac59\) Hz, µs = 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, fs = µs N = µs (mg) where m is the mass of the coin and N = mg is the normal force on the coin.
∴ mω2r ≤ µs (mg) or ω2r ≤ µsg
µsg = 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 < µsg, 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 > µsg, this coin will not revolve with the disc.
Thus, only the coin placed at 4 cm from the centre will revolve with the disc.