Yahoo Answers: Answers and Comments for A coin lies on the top of a turntable at a distance R from the center? [Physics]
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From Reham
enUS
Sat, 25 May 2019 13:47:48 +0000
3
Yahoo Answers: Answers and Comments for A coin lies on the top of a turntable at a distance R from the center? [Physics]
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https://answers.yahoo.com/question/index?qid=20190525134748AAIfK2D
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From oubaas: m*g*μ ≥ m*V^2/r
mass m cross
μ ≥ V^2/(r*g)
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https://answers.yahoo.com/question/index?qid=20190525134748AAIfK2D
Sun, 26 May 2019 06:38:27 +0000
m*g*μ ≥ m*V^2/r
mass m cross
μ ≥ V^2/(r*g)

From billrussell42: Centripetal force F = mV²/R
which is force of ...
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https://answers.yahoo.com/question/index?qid=20190525134748AAIfK2D
Sat, 25 May 2019 16:10:47 +0000
Centripetal force F = mV²/R
which is force of friction
weight of the coin is mg
coefficient of static friction = friction force/weight
µ = mg / mV²/R = gR/V²
edit, correction
µ = (mV²/R) / mg = V²/gR
Centripetal force f = mV²/r = mrω²
ω is angular velocity in radians/sec
1 radian/sec = 9.55 rev/min
m is mass in kg
r is radius of circle in meters
V is the tangental velocity in m/s = ωr
f is in Newtons

From electron1: When the turntable is rotating, there is centr...
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Sat, 25 May 2019 21:03:42 +0000
When the turntable is rotating, there is centripetal force on the coin,
Fc = m * v^2 ÷ R
To prevent the coin from sliding off of the turntable, the friction force must be equal to the centripetal force,
Ff = μ * m * g
μ * m * g = m * v^2 ÷ R
μ = v^2 ÷ (g * R
I hope this is helpful for you.