Reham asked in Science & MathematicsPhysics · 7 months ago

# A coin lies on the top of a turntable at a distance R from the center?

A coin lies on the top of a turntable at a distance R from the center. The table rotates at a

constant speed v. What is the minimum coefficient of static friction that can prevent the

coin from sliding of the table?

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• oubaas
Lv 7
7 months ago

m*g*μ ≥ m*V^2/r

mass m cross

μ ≥ V^2/(r*g)

• 7 months ago

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.

• 7 months ago

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