Question

Three children are riding on the edge of a merry‑go‑round that has a mass of 105 kg and a radius of 1.60 m . The merry‑go‑round is spinning at 16.0 rpm. The children have masses of 22.0, 28.0, and 33.0 kg. If the 28.0 kg child moves to the center of the merry‑go‑round, what is the new angular velocity in revolutions per minute? Ignore friction, and assume that the merry‑go‑round can be treated as a solid disk and the children as point masses.

Answer 1

Given : m = 105 kg ; m₁ = 22 kg ; m₂ = 28 kg ; m₃ = 33 kg , r = 1.60 m ; N= 16 rpm

Solution :

As this is a closed system so the net angular momentum will be conserved.

L_initial = L_final

I_ii = I_ff

Initially the system consists of merry go round and three children so

I_i = I_M+ I₁+I₂+I₃

Now if the second child moves to the center of the merry go round , then

I_f = I_M + I₁ + I₃

i.e.

(I_M + I₁+I₂+I₃)_i = (I_M+I₁+I₃)_f

.............(1)

Moment of inertia of merry go round is given as : I_M = (1/2)mr²

i.e. I_M = (1/2)(105)(1.60)² = 134.4 kg m²

Moment of inertia of childs are given as :

I₁ = (22kg)(1.60 m)² = 56.32 kg m²

I₂ = (28 kg)(1.60 m)² = 71.68 kg m²

I₃ = (33kg)(1.60 m)² = 84.48 kg m²

Now putting values in equation (1)to get final.angular velocity as :

= 20.17 rpm

Answer : 20.17 rpm