A coin is kept at a distance of 10 cm from the centre of a circular turntable. 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 begin to slip.
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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 = 80
ω = 8.94 rad/s
Angular frequency is calculated by
f = ω/2π
f = 8.94/(2×3.14)
f = 1.42 rps
Thus, angular frequency at which coin just begins to slip is 1.42 rps.
Hope that helps you...
● 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 = 80
ω = 8.94 rad/s
Angular frequency is calculated by
f = ω/2π
f = 8.94/(2×3.14)
f = 1.42 rps
Thus, angular frequency at which coin just begins to slip is 1.42 rps.
Hope that helps you...
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