Question
At a playground, a 20-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.
a. What is the force of static friction acting on the child?
b.What is the minimum coefficient of static friction between the child and the merry-go-round to keep the child from slipping?
a. What is the force of static friction acting on the child?
b.What is the minimum coefficient of static friction between the child and the merry-go-round to keep the child from slipping?
Answers
F = m v^2/r = m omega^2 r
m = 20
v = 2 pi r/6.2 = 2 pi(1.8/6.2)
r = 6.2
so solve for F, the mass times centripetal a
mu = Ac/9.81 = (v^2/r) /9.81
m = 20
v = 2 pi r/6.2 = 2 pi(1.8/6.2)
r = 6.2
so solve for F, the mass times centripetal a
mu = Ac/9.81 = (v^2/r) /9.81
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