Answer:
[tex]330kgm^2[/tex]
Explanation:
We are given that
Radius,r=2 m
Moment of inertia,I=250[tex]kgm^2[/tex]
Angular velocity,[tex]\omega_1=10 rev/min[/tex]
Mass of child,m=20 kg
We have to find the new moment of inertia of the merry go round.
New moment of inertia ,[tex]I'=mr^2+I[/tex]
Using the formula
[tex]I'=20(2)^2+250[/tex]
[tex]I'=80+250=330kgm^2[/tex]
Hence, the new moment of inertia of the merry -go-round=[tex]330kgm^2[/tex]
