# Problem: A playground merry-go-round of radius R = 2.00 m has a moment of inertia I = 250 kg•m2 and is rotating at 10.0 rev/min about a frictionless, vertical axle. Facing the axle, a 25.0-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?

###### FREE Expert Solution

We have jumped onto a rotating system along the radius. Adding mass changes the moment of inertia of the merry-go-round. The angular momentum of the system is conserved.

Conservation of angular momentum:

$\overline{){{\mathbf{L}}}_{{\mathbf{1}}}{\mathbf{=}}{{\mathbf{L}}}_{{\mathbf{2}}}}$

Angular momentum:

$\overline{){\mathbf{L}}{\mathbf{=}}{\mathbf{I}}{\mathbf{\omega }}}$ where I is the moment of inertia and ω is the angular speed.

Moment of inertia of a point mass:

$\overline{){\mathbf{I}}{\mathbf{=}}{\mathbf{m}}{{\mathbf{r}}}^{{\mathbf{2}}}}$

91% (221 ratings) ###### Problem Details

A playground merry-go-round of radius R = 2.00 m has a moment of inertia I = 250 kg•m2 and is rotating at 10.0 rev/min about a frictionless, vertical axle. Facing the axle, a 25.0-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?