A brave child decides to grab onto an already spinning merry‑go‑round. The child is initially at rest and has a mass of 25.5 kg. The child grabs and clings to a bar that is 1.55 m from the center of the merry‑go‑round, causing the angular velocity of the merry‑go‑round to abruptly drop from 35.0 rpm to 19.0 rpm. What is the moment of inertia of the merry‑go‑round with respect to its central axis?

72.75 kg m^2

Explanation:

initial angular velocity, ω = 35 rpm

final angular velocity, ω' = 19 rpm

mass of child, m = 15.5 kg

distance from the centre, d = 1.55 m

Let the moment of inertia of the merry go round is I.

Use the concept of conservation of angular momentum

I ω = I' ω'

where I' be the moment of inertia of merry go round and child

I x 35 = (I + md^2) ω'

I x 35 = (I + 25.5 x 1.55 x 1.55) x 19

35 I = 19 I + 1164

16 I = 1164

I = 72.75 kg m^2

Thus, the moment of inertia of the merry go round is 72.75 kg m^2.