PHYSICS HIGH SCHOOL

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?

Answers

Answer 1

Answer:

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 ( $(mv^(2) )/(r)$ )

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

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