Question

A playground merry-go-round of radius r = 1 m has a moment of inertia of i = 240 kg*m^2. and is rotating at a rate of ω = 8 rev/min around a frictionless vertical axis. facing the axle, a 35 kg child hops onto the merry-go-round and manages to sit down on the edge. what is the new angular speed of the merry-go-round (in rev/min)?

Answer 1

When the child jumps onto the merry-go-around the moment of inertia of the system changes. If we consider the child to be point-like mass then its moment of inertia would be:
$I_{ch}=mr^2$
We get the new moment of inertia by simply adding the child's moment of inertia to the old moment of inertia.
$I_{new}=I_{old}+I_{ch}=240+35(1)^2=275 kgm^2$
Since there is no force mention we must assume that angular momentum is conserved.
$L=const.\\ L=I_{old}\omega_0=I_{new}\omega'\\ \omega'=\frac{I_{old}\omega_0}{I_{new}}$
When we plug in all the numbers we get:
$\omega'=\frac{240\cdot8}{275}=6.98 \ \frac{rev}{min}$