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A child sits on a rotating merry-go-round, 2.1 m from its center. If the speed of the child is 2.2 m/s, what is the minimum coefficient of static friction between the child and the merry-go-round that will prevent the child from slipping?

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isabella66130

Answer:A

Explanation:

From the question, the given parameters are given.

Mass M = 30kg

Radius r = 2 m

Coefficient of static friction μ = 0.8

Coefficient of kinetic friction μ = 0.6

Kinetic friction Fk = μ × mg

Fk = 0.6 × 30 × 9.8

Fk = 176.4 N

The force acting on the merry go round is a centripetal force F.

F = MV^2/r

This force must be greater than or equal to the kinetic friction Fk. That is,

F = Fk

F = 176.4

Substitute F , M and r into the centripetal force formula above

176.4 = (30×V^2)/2

Cross multiply

352.8 = 30V^2

V^2 = 352.8/30

V = sqrt (11.76) m/s

V = 5.24 m/s

Therefore, the maximum speed of the merry go round before the child begins to slip is sqrt (12) m/s approximately

okpalawalter8

We have that the the minimum coefficient of static friction  is mathematically given as

[tex]\mu= 0.235[/tex]

Force

Generally the equation for the Force   is mathematically given as

[tex]F=n\mu\\\\Where\\\\n = mg\\\\Given\\\\Fc = m(vt2/R)\\\\Therefore\\\\\mu = vt2/gR\\\\\mu = 2.22/(9.8(2.1)) \\\\[/tex]

[tex]\mu= 0.235[/tex]

For more information on Force visit

https://brainly.com/question/26115859

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