Question

In a playground, there is a small merry-go-round of radius $1.20m$ and mass $180kgs$. Its radius of gyrations $91.0cm$. A child of mass $44.0kg$ runs at the initially stationary merry-go-round and then jumps on. Neglect friction between the bearings and the shaft of the merry-go-round. Calculate the angular speed of the merry-go-round and child after the child has jumped onto the merry-go-round.

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. $11-21,mvR$, where $R$ is the radius of the merry-go-round.
The initial angular momentum is given by $L_f=L_{child}=mvR$
; the final angular momentum is given by $L_f=(I+m{R}^2)w$, where $w$ is the final common angilar velocity of the merry-go-round and child. Thus $mvR=(I+m{R}^2)w$ and
$w=\frac{mvR}{I+m{R}^2}=\frac{158kg.m^2/s}{149kg.m^2+\left(44.0kg\right)(1.20m{)}^2}=0.744rod/s$.
Note:The child initially had an angular velocity of
$w=\frac{v}{R}=\frac{3.00m/s}{1.20m}=2.5rod/s$.
After he jumped onto the merry-go-round, the rotational inertia of the system (merry-go-round+child)increases, so the angular velocity decreases.