Question

Three children are riding on the edge of a merry-go-round that is 202 kg, has a 1.60 m radius, and is spinning at 15.3 rpm. The children have masses of 17.4, 28.5, and 32.8 kg. If the child who has a mass of 32.8 kg moves to the center of the merry-go-round, what is the new angular velocity in rpm?

Answer 1

Use conservation of angular momentum: I_i ω_i = I_f ω_f, so ω_f = ω_i (I_i / I_f).

Moments of inertia:
- Merry-go-round (solid disk): I_disk = (1/2) M R^2 = 0.5(202)(1.60^2) = 258.56 kg·m^2.
- Children at edge: I_children_initial = (17.4+28.5+32.8)R^2 = 78.7(1.60^2) = 201.472 kg·m^2.
Thus I_i = 258.56 + 201.472 = 460.032 kg·m^2.

After the 32.8 kg child moves to center, her contribution is zero:
- I_children_final = (17.4+28.5)R^2 = 45.9(1.60^2) = 117.504 kg·m^2.
So I_f = 258.56 + 117.504 = 376.064 kg·m^2.

ω_f = 15.3 rpm × (460.032 / 376.064) = 15.3 × 1.22333... ≈ 18.7 rpm.

Answer: ≈ 18.7 rpm (about 18.72 rpm).