A merry-go-round has 1 = 319 kg•m>, and is initially rotating at 2.33 rad/s. A 25.8 kg kid hops on, 1.40 m from the axis. What is the new angular velocity?

Answer 1

Let

I = 319 kgm² be the initial moment of inertia,

ω = 2.33 s^-1 be the initial angular velocity,

m = 25.8 kg be the mass of the kid,

r = 1.4 m be his distance from the axis,

L = Iω = 743.27 kgm²/s be the initial angular momentum,

L' be the new angular momentum,

I' be the new moment of inertia,

ω' be the new angular velocity.

Assuming that the kid jumps on from rest, he his contributing no angular momentum, so

L' = L = 743.27 kgm²/s

Moment of inertia is additive, therefore

I' = I + mr² = 319 kgm² + 25.8 kg × (1.4 m)² = 369.568 kgm²

ω' = L'/I' = 743.27 kgm²/s / 369.568 kgm² = 2.01 s^-1