Question

Three children are riding on the edge of a merry‑go‑round that has a mass of 10^5 and a radius of 1.80 m . The merry‑go‑round is spinning at 22.0 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

Answer: 22.01 rpm.

Explanation:  

If we assume no external torques are present, the angular momentum must be conserved.

L1 = L2

I1 . ω1 = I2 ω2 (1)

I1 = MR2 /2 + m1. R2 + m2. R2 + m3. R2  

I2 = MR2 /2 + m1. R2 + m3. R2

(as the 28.0 kg child moves to the center, he has no part anymore in the rotational inertia)

As I units are SI units, it is advisable to convert the angular velocity to rad/sec, as follows:

22.0 rev/min. (2π/rev) . (1min/60 sec) = 2.3 rad/sec

Solving for ω2 in (1):

ω2 = 2.31 rad/sec = 2.31 (1 rev/2 π). (60sec/1min) = 22.01 rpm