The turntable rotates very slowly at 60 rpm, hence the coin does not slide off as the static friction force is greater than the centripetal force.
Missing data from the question:
Static coefficients of friction (μs) = 0.8
Kinetic coefficients of friction (μk) = 0.5
There are two forces works on the coin, the friction force and the centripetal force. The coin will start to slide off if the centripetal force is equal to the maximum static friction force.
Let:
Fs = static friction force
Fc = centripetal force
Fs = μs . mg
Fc = m . v²/R = m . ω²R
Where:
m = coin's mass
g = gravitational acceleration = 10 m/s²
ω = angular velocity
Parameters given:
R = 15 cm = 15 . 10⁻² m
m = 5 gram = 5 . 10⁻³ kg
ω = 60 rpm = 60 . 2π/60 = 2π rad/s
Hence,
Fs = μs . mg = 0.8 . 5 . 10⁻³ . 10 = 4 . 10⁻² N
Fc = m . ω²R = 5 . 10⁻³ . (2π)² . 15. 10⁻² = 2.96 .10⁻² N
Since Fs > Fc, hence the coin does not slide off.
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