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.

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?

Damon · Answer

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