# 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?

### 3 Answers

- electron1Lv 77 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.

- billrussell42Lv 77 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

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

won't be the first time.

thanks