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------------------------------------------------ A woman with a mass of 52.0 kg is standing on the rim of a large disk that is rotating at an angular velocity of [tex]0.470 \, \text{rev/s}[/tex] about an axis through its center. The disk has a mass of 118 kg and a radius of 3.90 m.

Calculate the magnitude of the total angular momentum of the woman-plus-disk system. (Assume that you can treat the woman as a point.)

Answer :

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.