A disc rotates about its axis of symmetry in a hoizontal 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 d

Question

A disc rotates about its axis of symmetry in a hoizontal 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 / s 2 )

TestHub · Accepted Answer

Using, μ m g = r m v 2 ​ = m r ω 2 ω = 2 πn = 2 π × 3.5 = 7 π rad / sec Radius, r = 1.25 cm = 1.25 × 1 0 − 2 m Coefficient of friction, μ = ? μ m g = r m ( r ω ) 2 ​ ( ∵ v = r ω ) ⇒ μ = g r ω 2 ​ = 10 1.25 × 1 0 − 2 × ( 7 × 7 22 ​ ) 2 ​ = 10 1.25 × 1 0 − 2 × 2 2 2 ​ = 0.6 ​

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