Question

skepta distance of 10 cm from the centre of a circular turn table if the coefficient of static friction between the table and the coin is 0.8 find the frequency of rotation of the disc at which the coin will just be begin to sleep​

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...