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Question:

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 of 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)

0.5

0.3

0.7

0.6

Solution:

3.5 rev/second
1 rev → 2π rad
3.5 rev → 2π × 3.5 rad → w = 7π rad/sec
μmg = mv²/r
1.25 cm
μmg = m(rw)²/r
μmg = mrw²
μ = rw²/g = 1.25 × 10⁻² × (7 × 22/7)² / 10 = 1.25 × 10⁻² × 44²/10 = 0.6