Question

In: Physics

A 50.0 kg child stands at the rim of a merry-go-round of radius 1.50 m, rotating...

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?

Expert Solution

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

genius_generous answered 2 years ago

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