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

the answer is b i think