Question

A playground merry-go-round of radius r = 2.00 m has a moment of inertia = 250 ⋅ , and is rotating at 10.0 rev/min about a frictionless vertical axle. Facing the axle, a 20.0-kg child hops onto the merry-go-round, and manages to sit down on the edge. What is the new moment of inertia of the merry-go-round?

Answer 1

`330kgm^2`

Step-by-step explanation:

We are given that

Radius,r=2 m

Moment of inertia,I=250
`kgm^2`

Angular velocity,
`\omega_1=10 rev/min`

Mass of child,m=20 kg

We have to find the new moment of inertia of the merry go round.

New moment of inertia ,
`I'=mr^2+I`

Using the formula

`I'=20(2)^2+250`

`I'=80+250=330kgm^2`

Hence, the new moment of inertia of the merry -go-round=
`330kgm^2`