Question

At a playground, a 17-kg child sits on a spinning merry-go-round, as shown from above in (Figure 1).The merry-go-round completes one revolution every 6.2 s, and the child sits at a radius of r=1.8m.

Part A

What is the force of static friction acting on the child?

Express your answer to two significant figures and include appropriate units.

Part B

What is the minimum coefficient of static friction between the child and the merry-go-round to keep the child from slipping?

Answer 1

Since the child is moving at constant speed, which means acceleration will be zero, So Net force will be zero. Now Using force balance

Fnet = Ff - Fc = 0

Ff = Fc

Ff = frictional force

Fc = centripetal force = m*w^2*r

w = angular velocity = 1 rev in 6.2 sec

w = 1/6.2 rev/sec = 2*pi rad/6.2 sec

w = (2*pi/6.2) = 1.01 rad/sec

Now

Fc = 17*1.01^2*1.8 = 31.215 N = 31 N

Part B.

Since we know that

Ff = us*N = us*m*g

Ff = Fc

us*m*g = 31.215

us = coefficient of static friction

us = 31.215/(17*9.81) = 0.187

us = 0.19

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