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Lilit [14]
3 years ago
15

A 5.0-m radius playground merry-go-round with a moment of inertia of 2000 kg · m 2 is rotating freely with an angular speed of 1

.0 rad/s. Two people, each having a mass of 60 kg, are standing right outside the edge of the merry-go-round and step on it with negligible speed. What is the angular speed of the merry-go-round right after the two people have stepped on?
Physics
1 answer:
lidiya [134]3 years ago
5 0

Answer:

angular speed = 0.4 rad/s

Explanation:

given data

radius = 5 m

moment of inertia = 2000 kg-m²

angular speed = 1.0 rad/s

mass = 60 kg

to find out

angular speed

solution

Rotational momentum of merry-go-round = I?

we get here momentum that is express as  

momentum = 2000 × 1

momentum = 2000 kg-m²/s

and  

Inertia of people will be here as

Inertia of people = mr² = 60 × 5²

Inertia of people = 1500 kg-m²

so Inertia of people for two people  

1500 × 2 = 3000

and

now conserving angular momentum(ω)

moment of inertia × angular speed = ( momentum + Inertia of people ) angular momentum  

2000 ×  1 = (2000 + 3000 ) ω

solve we get now  

ω = 0.4 rad/s  

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if you need any clarification or more explanation pls do mention on the comment section.i would like to help more

Hope this helps and if it does mark as branliest answer thx

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