Asked by Jaleena
A merry-go-round makes one complete revolution in 9.1 s. An 18.8 kg child sits on the
horizontal floor of the merry-go-round 4.7 m
from the center.
Find the child’s acceleration. The acceleration of gravity is 9.8 m/s
2
horizontal floor of the merry-go-round 4.7 m
from the center.
Find the child’s acceleration. The acceleration of gravity is 9.8 m/s
2
Answers
Answered by
Damon
2 pi r = 2 (3.14)(4.7) = 29.5 meters
v = 29.5 / 9.1 = 3.25 m/s
Ac = v^2/r = 3.25^2/4.7 = 2.24 m/s^2
= .23 g
v = 29.5 / 9.1 = 3.25 m/s
Ac = v^2/r = 3.25^2/4.7 = 2.24 m/s^2
= .23 g
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