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3 years ago

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A 50.0-kg child stands at the rim of a merry-go-round of radius 2.00 m, rotating in such a way that it makes one revolution in 2

.09 s. What minimum coefficient of static friction between her feet and the floor of the merry-go-round is required to keep her in the circular path

1 answer:

Vaselesa [24]3 years ago

7 0

Answer:

2.01

Explanation:

First, we need to find the centripetal acceleration.

We're given that the merry go round rotates 1 revolution in 2.09 seconds. Converting to rpm, we know that it rotates 30 revolution per minute

Now this speed gotten in rpm will be converted to m/s, to ease the calculation

30 rpm = πdN/60 m/s

30 rpm = (3.142 * 4 * 30)/60

30 rpm = 377.04/60

30 rpm = 6.284 m/s

a(c) = v²/r

a(c) = 6.284²/2

a(c) = 39.49 / 2

a(c) = 19.74 m/s²

F = ma

F = 50 * 19.74

F = 987 N

Also, Normal Force, F(n) =

F(n) = mg

F(n) = 50 * 9.81

F(n) = 490.5

We then use this to find the coefficient of static friction, μ

μ = F/F(n)

μ = 987 / 490.5

μ = 2.01

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$Q = k *\int\limits^{2*\pi}_0 {\frac{\pi r^4}{4}} \, d\phi$

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