Answer:
The angular momentum is  [tex]L = 8440.32 \ kg \cdot m^2 \cdot s^{-1}[/tex]
Explanation:
From the question we are told that
  The mass of the woman is  [tex]m = 50 \ kg[/tex]
   The angular  speed of the rim is  [tex]w = 0.80 \ rev/s = 0.8 * [\frac{2 \pi}{1} ] = 5.024 \ rad \cdot s^{-1}[/tex]
   The mass of the disk is  [tex]m_d = 110 \ kg[/tex]
    The radius of the disk is [tex]r_d = 4.0 \ m[/tex]
The moment of inertia of the disk is mathematically represented as
    [tex]I_D = \frac{1}{2} m_d r^2_d[/tex]
substituting values
     [tex]I_D = \frac{1}{2} * 110 * 4^2[/tex]
     [tex]I_D = 880 \ kg \cdot m^2[/tex]
The moment of inertia of the woman is Â
     [tex]I_w = m * r_d^2[/tex]
substituting values
    [tex]I_w = 50 * 4^2[/tex]
    [tex]I_w =800\ kg[/tex]
The moment of inertia of the system (the woman + the large disk ) is Â
    [tex]I_t = I_w + I_D[/tex]
substituting values Â
   [tex]I_t = 880 +800[/tex]
   [tex]I_t =1680 \ kg \cdot m^2[/tex]
The angular momentum of the system is
   [tex]L = I_t w[/tex]
substituting values Â
   [tex]L = 1680 * 5.024[/tex]
   [tex]L = 8440.32 \ kg \cdot m^2 \cdot s^{-1}[/tex]
