Answer
(a) The new angular velocity is $\frac{I_i}{I_i+mR^2}~\omega_i$
(b) $KE_{rot,i} = \frac{1}{2}I_i~\omega_i^2$
$KE_{rot,f} = \frac{1}{2}\frac{(I_i~\omega_i)^2}{I_i+mR^2}$
$L_f = I_i~\omega_i$
$L_f = I_i~\omega_i$
Work Step by Step
(a) We can find the final rotational inertia of the merry-go-round and child:
$I_f = I_i+mR^2$
We can use conservation of angular momentum to find the new angular velocity:
$L_f = L_i$
$I_f~\omega_f = I_i~\omega_i$
$\omega_f = \frac{I_i~\omega_i}{I_f}$
$\omega_f = \frac{I_i}{I_i+mR^2}~\omega_i$
The new angular velocity is $\frac{I_i}{I_i+mR^2}~\omega_i$
(b) We can find the initial rotational kinetic energy:
$KE_{rot,i} = \frac{1}{2}I_i~\omega_i^2$
We can find the final rotational kinetic energy:
$KE_{rot,f} = \frac{1}{2}I_f~\omega_f^2$
$KE_{rot,f} = \frac{1}{2}(I_i+mR^2)~(\frac{I_i}{I_i+mR^2}~\omega_i)^2$
$KE_{rot,f} = \frac{1}{2}\frac{(I_i~\omega_i)^2}{I_i+mR^2}$
We can find the initial angular momentum:
$L_f = I_i~\omega_i$
We can find the final angular momentum:
$L_f = I_f~\omega_f$
$L_f = (I_i+mR^2)(\frac{I_i}{I_i+mR^2}~\omega_i)$
$L_f = I_i~\omega_i$
