Answer
$\omega = 0.744~rad/s$
Work Step by Step
In part (a), we found that the rotational inertia of the merry-go-round is $~~149~kg~m^2$
In part (b), we found that the angular momentum of the running child is $158~kg~m^2/s$
We can find the rotational inertia of the merry-go-round and the child:
$I = (149~kg~m^2) + (44.0~kg)(1.20~m)^2$
$I = 212.36~kg~m^2$
We can find the angular speed of the merry-go-round:
$L = I~\omega = 158~kg~m^2/s$
$\omega = \frac{158~kg~m^2/s}{I}$
$\omega = \frac{158~kg~m^2/s}{212.36~kg~m^2}$
$\omega = 0.744~rad/s$
