Answer
The angular speed of the merry-go-round just after the ball is caught is $ 0.07 \frac{rad}{s} $
Work Step by Step
We can solve this problem by using the conservation of Angular Momentum:
$L_i=L_f$
$ I_i\omega_i + L_{ball} = I_f\omega_f$
The angulur momentum of the ball is given by:
$L_{ball} = mvr\sin(90-\phi) $
$L_{ball} = (1\ kg)(12 \frac{m}{s})(2\ m)\sin53$
$L_{ball} = 19.17 kg\frac{m^2}{s}$
The moment of inertia $I_f$ after the ball is caught is given by:
$I_f = I_{wheel} + I_{child} + I_{ball} $
$I_f = (150 \ kg \ m^2) + m_{child}R^2 + m_{ball}R^2 $
$I_f = (150 \ kg \ m^2) + (30\ kg)(2\ m)^2 + (1\ kg)(2\ m)^2$
$I_f = 274 \ kg \ m^2$
Simplifying the original equation yields:
The initial angular velocity of the merry-go-round is 0, so: $\omega_i = 0 $
$I_i(0) + L_{ball} = I_i\omega$
$\omega_f =\frac{L_{ball}}{I_f} $
$\omega_f =\frac{(19.17\ kg\ m^2/s)}{(274 \ kg \ m^2)} $
The final speed of the merry-go-round is $0.07\ \frac{rad}{s} $
