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A coin is placed 29 cm from the center of a horizontal turntable, initially at rest. The turntable then begins to rotate. When the speed of the coin is 120 cm/s (rotating at a constant rate), the coin just begins to slip. The acceleration of gravity is 980 cm/s^2 . What is the coefficient of static friction between the coin and the turntable?

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Karabo99

Answer:

The coefficient of static friction between the coin and the turntable is 0.51

Explanation:

at the time of the slip:

centripetal force = frictional force

mv^2/r = x*m*980

v^2/r = 980x

x = v^2/980r

  = [(120)^2]/[980*29]

  = 0.51

Therefore, The coefficient of static friction between the coin and the turntable is 0.51

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