Question

A coin is placed 32 cm from the center of a horizontal turntable, initially at rest. The turntable then begins to rotate. When the speed of the coin is 110 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

Answer:

0.39

Explanation:

distance from the center (r) = 32 cm = 0.32 m

speed of the coin (v) = 110 cm/s = 1.1 m/s

acceleration due to gravity (g) = 980 m/s^{2} = 9.8 m/s^{2}

find the coefficient of static friction (k) between the coin and the turn table

frictional force = kmg

before the table begins to move, the frictional force balances the centripetal force ($\frac{mv^{2} }{r}$)

therefore

frictional force = centripetal force

kmg = $\frac{mv^{2} }{r}$

kg = $\frac{v^{2} }{r}$

k = $\frac{v^{2} }{r}$ ÷ g

k = $\frac{1.1^{2} }{0.32}$ ÷ 9.8 = 0.39