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 · Accepted Answer

force = mass * centripetal acc = m v^2/r c = 2 pi r = 2 pi * 1.8 v = (c/6.2) meters/second so F = 20 * (c/6.2)^2/1.8 when F = mu * m g, we slip mu = Ac/g= (c/6.2)^2/(1.8*9.81)