Question

In a playground, there is a small merry-go-round of radius $1.20m$ and mass $180kgs$. Its radius of gyrationis $91.0cm$. A chikd of mass $44.0kg$ runs at the intiially stationary merry-go-round and then jumps on. Neglect friction between the bearings and the shaft of the merry-go-round. Calculate the rotational inertia of the merry-go-round about its axis of rotation,

Answer 1

No external torques act on the system consisting of the child and the merry-go-round,
so the total angular momentum of the system is conserved.
An object moving a straight line has angular momentum about any point that is not
on the line. The magnitude of the angular momentum of the child the center of the merry-go-round is given by Eq. $mvR$, where $R$ is the radius of the merry-go-round.
In terms of the radius of gyration $k$, the rotational inertia of the merry-go-round is
$I=M{k}^2$. With $M=180kg$ and $k=0.91m$, we obtain
$I=\left(180kg\right)(0.910m{)}^2=149kg.m^2$.