Answer:
The angular velocity is [tex]w_f = 4.503 \ rad/s[/tex]
Explanation:
From the question we are told that
  The mass of the child is  [tex]m_c = 46.2 \ kg[/tex]
  The radius of the merry go round is  [tex]r = 1.9 \ m[/tex]
   The moment of inertia of the merry go round is [tex]I_m = 130.09 \ kg \cdot m^2[/tex]
   The angular velocity of the merry-go round is  [tex]w = 2.4 \ rad/s[/tex]
    The position of the child from the center of the merry-go-round is  [tex]x = 0.779 \ m[/tex]
According to the law of angular momentum conservation
  The initial angular momentum  =  final  angular momentum
So Â
    [tex]L_i = L_f[/tex]
=> Â Â [tex]I_i w_i = I_fw_f[/tex]
Now  [tex]I_i[/tex] is the initial moment of inertia of the system which is mathematically represented as
     [tex]I_i = I_m + I_{b_1}[/tex]
Where  [tex]I_{b_i}[/tex] is the initial moment of inertia of the boy which is mathematically evaluated as
   [tex]I_{b_i} = m_c * r[/tex]
substituting values
   [tex]I_{b_i} = 46.2 * 1.9^2[/tex]
   [tex]I_{b_i} = 166.8 \ kg \cdot m^2[/tex]
Thus
  [tex]I_i =130.09 + 166.8[/tex]    Â
  [tex]I_i = 296.9 \ kg \cdot m^2[/tex]   Â
Thus Â
   [tex]I_i * w_i =L_i= 296.9 * 2.4[/tex]
    [tex]L_i = 712.5 \ kg \cdot m^2/s[/tex]
Now Â
   [tex]I_f = I_m + I_{b_f }[/tex]
Where  [tex]I_{b_f}[/tex] is the final  moment of inertia of the boy which is mathematically evaluated as
     [tex]I_{b_f} = m_c * x[/tex]
substituting values
     [tex]I_{b_f} = 46.2 * 0.779^2[/tex]
     [tex]I_{b_f} = 28.03 kg \cdot m^2[/tex]
Thus
   [tex]I_f = 130.09 + 28.03[/tex]
   [tex]I_f = 158.12 \ kg \ m^2[/tex]
Thus
   [tex]L_f = 158.12 * w_f[/tex]
Hence
   [tex]712.5 = 158.12 * w_f[/tex]
    [tex]w_f = 4.503 \ rad/s[/tex]
