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

Answer:

$330kgm^2$

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=$330kgm^2$