Question

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?

Answer 1

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

Step-by-step 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