A brave child decides to grab onto an already spinning merry‑go‑round. The child is initially at rest and has a mass of 34.5 kg. 34.5 kg. The child grabs and clings to a bar that is 1.20 m 1.20 m from the center of the merry‑go‑round, causing the angular velocity of the merry‑go‑round to abruptly drop from 37.0 rpm 37.0 rpm to 19.0 rpm . 19.0 rpm. What is the moment of inertia of the merry‑go‑round with respect to its central axis?

the moment of inertia of the merry go round is 38.04 kg.m²

Explanation:

We are given;

Initial angular velocity; ω_1 = 37 rpm

Final angular velocity; ω_2 = 19 rpm

mass of child; m = 15.5 kg

distance from the centre; r = 1.55 m

Now, let the moment of inertia of the merry go round be I.

Using the principle of conservation of angular momentum, we have;

I_1 = I_2

Thus,

Iω_1 = I'ω_2

where I' is the moment of inertia of the merry go round and child which is given as I' = mr²

Thus,

I x 37 = ( I + mr²)19

37I = ( I + (15.5 x 1.55²))19

37I = 19I + 684.7125

37I - 19 I = 684.7125

18I = 684.7125

I = 684.7125/18

I = 38.04 kg.m²

Thus, the moment of inertia of the merry go round is 38.04 kg.m²

answer would be d.) i hope this

explanation: