Question

A disc revolves with a speed of 33 1/3 rev./min, and has a radius of 15 cm. Two coins are placed at 4 cm and 14 cm away from the centre of the record. If the coefficient of friction between the coins and the record is 0.15. Which of the coins will revolve with the record?

Answer 1

Hii dear,

# Answer- coin placed at 4 cm will revolve with record

# Explanation-
# Given-
r = 15 cm = 0.15 m
μ = 0.15
f = 100/3 rpm = 5/9 Hz

# Solution-
Here, frictional force is given by
F = μmg
F = 0.15×10×m
F = 1.5 N

Also angular frequency is
ω = 2πf
ω = 2×3.14×5/9
ω = 3.49 rad/s

a) When coin placed at 4 cm,
r = 4 cm = 0.04 m
Fc = mrω^2
Fc = m×0.04×(3.49)^2
Fc = 0.49 N
Here, Fc < F Thus coin will revolve with the record.

b) When coin placed at 14 cm,
r = 14 cm = 0.14 m
Fc = mrω^2
Fc = m×0.14×(3.49)^2
Fc = 1.71 N
Here, Fc > F Thus coin will slip off from the record.

Hope that helps you...

Answer 2

I hope u get the answer