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?

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

b) returns freshwater to the earth.

c) returns freshwater to the earth.c) occurs after condensation.

d) returns freshwater to the earth.c) occurs after condensation.d) roughly equal to the amount of evaporation.

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the displacement produced by the body.  

the amount of force subjected to the body.  

the angle between the direction of force and displacement