Question

A coin is kept at a distance of 10cm from the centre of a circular turn table if friction is 0.8 the frquency of rotation at which the coin just begins to slide

Answer 1

Hii dear,

● Answer-
f = 1.42 rps

● Explaination-
# Given-
r = 10 cm = 0.1 m
μ = 0.8

# Solution-
For a coin to just slip,
Friction force = Force due to rotation
μmg = mrω^2
ω^2 = 0.8×10/0.1
ω^2 = 80
ω = 8.94 rad/s

Frequency of rotation is calculated by
f = ω/2π
f = 8.94/(2×3.14)
f = 1.42 rps

Thus, frequency of rotation at which coin just begins to slide is 1.42 rps.

Hope that helps you...

Answer 2

Answer:

ANSWER IS 84.54

HOPE IT HELPS