Answer to Question #16043 in Mechanics | Relativity for Ashlee

Question

A merry-go-round makes one complete revolution in 8.35 s. A 25.04 kg child sits on the
horizontal floor of the merry-go-round 2.13 m
from the center.
What minimum coefficient of static friction
is necessary to keep the child from slipping?
The acceleration of gravity is 9.8 m/s^2.

Answer 1

& a = r*ω² = 2.13*(2π/8.35)² = 1,206 m/s²
µmin = a/g = 1,206 /9.8 = 0,123