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 35.5 kg.

The child grabs and clings to a bar that is 1.25 m from the center of the merry‑go‑round, causing the angular velocity of the merry‑go‑round to abruptly drop from 43.0 rpm to 17.0 rpm .

What is the moment of inertia of the merry‑go‑round with respect to its central axis?

Answer 1

36.3 kgm2

Step-by-step explanation:

According to the conservation of angular momentum, the angular momentum is preserved before and after the child grab. Since the moments of inertia increase, the angular velocity decreases.

Let I be the moment of inertia of the merry-go-round and treat the child as a point particle, his moment of inertia would be

`I_c = mr^2 = 35.5*1.25^2 = 55.5 kgm^2`

Hence the moment inertia of the child-merry-go-round system is:

I + 55.5

From here we can apply the conservation theory

`I\omega_1 = (I + 55.5)\omega_2`

`43I = (I + 55.5)17`

`43I = 17I + 943`

`26I = 943`

`I = 943/26 = 36.3kgm^2`