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hree children are riding on the edge of a merry-go-round that is 115 kg, has a 1.7-m radius, and is spinning at 24 rpm. the children have masses of 32, 34, and 40 kg. (a) if the children with masses of 34 and 40 kg move to the center of the merry-go-round, what is the new angular velocity in rpm?

Sagot :

The new angular velocity in rpm is

27.7 rpm

M = 115 kg be the mass of the merry-go-round,

r = 1.7 m be its radius,

m1 = 32 kg,

m2 = 34 kg,

m3 = 40 kg be the masses of the three children,

ω = 22 rpm be the angular velocity with all children at the edge,

ω' (to be calculated) be the angular velocity after m2 is moved to the center.

The conservation of angular momentum can be used to resolve this.

The angular momentum before and after the move must be equal.

But there are two presumptions we must make.

1) The merry-go-round has a consistent disc shape. Although it is improbable, if the merry-go-round has a more widespread. The issue gets more complex due of its convoluted shape.

2) The kids are on the merry-go-outer round's edge. Regardless of where the kids are, the issue is easily fixable.

but, since I have no information, I'll guess it's on the edge.

In general, a mass m at the edge contributes mr²ω to the angular momentum.

The rotating uniform disk contributes Mr²ω/2 to the angular momentum.

Now we can equate the total angular momentum after the move to the total

angular momentum before the move.

Mr²ω'/2 + m1r²ω' + m3r²ω' = Mr²ω/2 + m1r²ω + m2r²ω + m3r²ω

From that express

ω' = ω(M/2 + m1 + m2 + m3)/(M/2 + m1 + m3)

Substituting actual numbers

ω' = 22×(115/2 + 32 + 34 + 40)/(115/2 + 32 + 40) = 27.7 rpm

Hence, new angular velocity in rpm is 27.7 rpm

Learn more about Angular velocity here:

https://brainly.com/question/20432894

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