Question

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?

Answer 1

Answer:

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 2

Answer:

$52.44 kgm^2$

Explanation:

First convert from revolution per minute to rad per second knowing that a revolution is 2π rad and every minute has 60 seconds

$\omega_1 = 37 rpm = 37*2\pi/60 = 3.875 rad/s$

$\omega_2 = 19 rpm = 19*2\pi/60 = 1.99 rad/s$

According to the law of momentum conservation, the total angular before and after the child jump on must be the same:

$\omega_1 I_w = \omega_2(I_w + I_c)$ (1)

where $I_w$ is the moments of inertia of the merry-go-round, which we are looking for, and $I_c$ is the moments of inertia of the child, which can be calculated if we treat him as a point mass:

$I_c = mr^2 = 34.5*1.2^2 = 49.68 kgm^2$

Therefore from eq. (1) we have:

$3.875I_w = 1.99(I_w + 49.68)$

$I_w(3.875 - 1.99) = 1.99*49.68 = 98.85$

$I_w = \frac{98.85}{3.875 - 1.99} = 52.44 kgm^2$