Explanation:
It is given that,
Mass of the woman, m₁ = 52 kg
Angular velocity, [tex]\omega=0.47\ rev/s=2.95\ rad/s[/tex]
Mass of disk, m₂ = 118 kg
Radius of the disk, r = 3.9 m
The moment of inertia of woman which is standing at the rim of a large disk is :
[tex]I={m_1r^2}[/tex]
[tex]I={52\times 3.9^2}[/tex]
I₁ = 790.92 kg-m²
The moment of inertia of of the disk about an axis through its center is given by :
[tex]I_2=\dfrac{m_2r^2}{2}[/tex]
[tex]I_2=\dfrac{118\times (3.9)^2}{2}[/tex]
I₂ =897.39 kg-m²
Total moment of inertia of the system is given by :
[tex]I=I_1+I_2[/tex]
[tex]I=790.92+897.39[/tex]
I = 1688.31 kg-m²
The angular momentum of the system is :
[tex]L=I\times \omega[/tex]
[tex]L=1688.31 \times 2.95[/tex]
[tex]L=4980.5\ kg-m^2/s[/tex]
So, the total angular momentum of the system is 4980.5 kg-m²/s. Hence, this is the required solution.
