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In a playground, there is a small merry-go-round of radius $$1.20\ m$$ and mass $$180\ kgs$$. Its radius of gyrationis $$91.0\ cm$$. A chikd of mass $$44.0\ kg$$ 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,


Solution

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 
$$I=Mk^2$$. With $$M=180\ kg$$ and $$k=0.91\ m$$, we obtain
$$I=(180\ kg)(0.910\ m)^2=149\ kg.m^2$$.

Physics

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