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

Answer 1

here, at equilibrium condition,

frictional force = centrifugal force

$\mu N$ = $m\omega^2r$

where $\mu$ is coefficient of friction, N is normal reaction acts between coin and disc, $\omega$ is angular frequency of disc, r is the distance between coin and axis of rotation.

given, $\omega=3.5rev/s=7\pi rad/s$
N = weight of coin = mg

so, $\mu mg=m\omega^2r$

or,$\mu g=\omega^2r$

or, $\mu$ × 10 = (7π)² × 1.25 × 10^-2

$\mu=0.6$

Answer 2

Answer:

Hence answer (D)0.6

Option :d