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Arisa [49]
3 years ago
9

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

he 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?
Physics
1 answer:
pashok25 [27]3 years ago
6 0

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

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{\mathfrak{\underline{\purple{\:\:\: Given:-\:\:\:}}}} \\ \\

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\\

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