In: Physics
A 50.0 kg child stands at the rim of a merry-go-round of radius 1.50 m, rotating with an angular speed of 2.50 rad/s.
(a) What is the child's centripetal acceleration?
m/s2
(b) What is the minimum force between her feet and the floor of the
merry-go-round that is required to keep her in the circular
path?
N
(c) What minimum coefficient of static friction is required?
Is the answer you found reasonable? In other words, is she likely
to be able to stay on the merry-go-round?
Mass of child,m=50 kg
Radius of merry go round,R=1.50 m
Angular speed,=2.50 rad/s
Part a.
child's centripetal acceleration=2R=2.5021.50=9.375 m/s2
Part b.
minimum force between her feet and the floor of the merry-go-round that is required to keep her in the circular path,F=m2R=502.5021.50=468.75 N
Part c.
For this
frictional force=minimum force between her feet and the floor of the merry-go-round that is required to keep her in the circular path
So
mg=m2R
So
=2R/g=(2.5021.50)/9.8=0.957