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

0.39

Step-by-step 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 (
`(mv^(2) )/(r)`)

therefore

frictional force = centripetal force

kmg =
`(mv^(2) )/(r)`

kg =
`(v^(2) )/(r)`

k =
`(v^(2) )/(r)` ÷ g

k =
`(1.1^(2) )/(0.32)` ÷ 9.8 = 0.39