Question

A coin is kept at distance of 10cm from the centre of a circular turn table. If  = 0.8 , the frequency of rotation at which the coin just
begins to slip is (g = 9.8ms−2 )

Answer 1

Answer: 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...

Explanation: MARK AS BRAINLIEST....