Question

A 2.20-m radius playground merry-go-round has a mass of 101.5 kg and is rotating with an angular velocity of 0.91 rev/s. What is its angular velocity after a 38.4-kg child gets onto it by grabbing its outer edge? The child is initially at rest. Express your answer in revs/s.

Answer 1

Mass of the merry-go-round = M = 101.5 kg

Radius of the merry-go-round = R = 2.2 m

Moment of inertia of the merry-go-round = I

I = MR²/2

I = (101.5)(2.2)²/2

I = 245.63 kg.m²

Mass of the child = m = 38.4 kg

Initial angular velocity of the merry-go-round = ₁ = 0.91 rev/s = 0.91 x (2) rad/s = 5.718 rad/s

Final angular velocity of the merry-go-round = ₂

The child get on the merry-go-round by grabbing the outer edge therefore the child is at the outer edge of the merry-go-round therefore it is at a distance equal to the radius from the center.

By conservation angular momentum,

I₁ = (I + mR²)₂

(245.63)(5.718) = [245.63 + (38.4)(2.2)²]₂

₂ = 3.255 rad/s

Converting from rad/s to rev/s

₂ = 0.518 rev/s

Angular velocity of the merry-go-round after the child grabs on it = 0.518 rev/s