Question

1.A merry-go-round makes one complete revo-
lution in 12.2 s. A 49.4 kg child sits on the
horizontal floor of the merry-go-round 2.3 m
from the center.
Find the child’s acceleration. The accelera-
tion of gravity is 9.8 m/s2 .
Answer in units of m/s2

2.Find the horizontal force of friction that acts
on the child.
Answer in units of N

3.What minimum coefficient of static friction is
necessary to keep the child from slipping?

Answer 1

1. Speed V = (2.3 m)*2 pi/12.2 s
= 1.185 nm/s
acceleration = V^2/R = 0.61 m/s^2

2. M*a = M V^2/R = ____ N

3. Ustat*M*g = M*V^2/R
Ustat = V^2/(g*R)