Answer
The merry-go-round is rotating at an angular speed of $0.61~rev/s$
The rotational kinetic energy decreased by $663~J$
Work Step by Step
We can find the initial rotational inertia of the merry-go-round:
$I_0 = \frac{1}{2}MR^2$
$I_0 = \frac{1}{2}(160~kg)(2.0~m)^2$
$I_0 = 320~kg~m^2$
We can find the final rotational inertia of the merry-go-round and child:
$I_f = \frac{1}{2}MR^2+mR^2$
$I_f = \frac{1}{2}(160~kg)(2.0~m)^2+(180~N/9.80~m/s^2)(2.0~m)^2$
$I_f = 393.5~kg~m^2$
We can use conservation of angular momentum to find the new angular speed:
$L_f = L_0$
$I_f~\omega_f = I_0~\omega_0$
$\omega_f = \frac{I_0~\omega_0}{I_f}$
$\omega_f = \frac{(320~kg~m^2)(0.75~rev/s)}{393.5~kg~m^2}$
$\omega_f = 0.61~rev/s$
The merry-go-round is rotating at an angular speed of $0.61~rev/s$
We can find the change in rotational kinetic energy:
$\Delta KE_{rot} = \frac{1}{2}I_f~\omega_f^2 - \frac{1}{2}I_0~\omega_0^2$
$\Delta KE_{rot} = \frac{1}{2}(393.5~kg~m^2)[(0.61~rev/s)(2\pi~rad/rev)]^2 - \frac{1}{2}(320~kg~kg~m^2)[(0.75~rev/s)(2\pi~rad/rev)]^2$
$\Delta KE_{rot} = -663~J$
The rotational kinetic energy decreased by $663~J$.
