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A disc rotates about its axis of symmetry in a horizontal plane at a steady rate of 3.5 revolutions per second. A coin placed at a distance 1.25 cm from the axis of rotation remains at rest on the disc. The coefficient of friction between the coin and the disc is (g = 10 m/s²)

1. 0.5
2. 0.7
3. 0.3
4. 0.6

Correct answer: 0.6

Solution

omega = 3.5 rev/s = 7*pi = 21.99 rad/s. The coin stays put when friction supplies the centripetal force: mu*g >= omega^2*r, so mu = omega^2*r/g = (21.99^2 * 0.0125)/10 = 0.60.

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