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Answer :
The moment of inertia of the platform-person system about the vertical axis is approximately [tex]2362 kg*m^2.[/tex]
Given:
Mass of platform (m_platform) = 103 kg
Radius of platform (r_platform) = 1.93 m
Mass of person (m_person) = 65.1 kg
First, let's calculate the moment of inertia of the platform alone using the formula for the moment of inertia of a uniform disk about its center:
[tex]\[ I_{\text{platform}} = \frac{1}{2} m_{\text{platform}} r_{\text{platform}}^2 \][/tex]
[tex]\[ I_{\text{platform}} = \frac{1}{2} \times 103 \times (1.93)^2 \][/tex]
[tex]\[ I_{\text{platform}} = 188.243 \, \text{kg*m}^2 \][/tex]
Next, we'll calculate the moment of inertia of the person about the same axis. Considering the person as a point mass, we use the formula:
[tex]\[ I_{\text{person}} = m_{\text{person}} r_{\text{platform}}^2 \][/tex]
[tex]\[ I_{\text{person}} = 65.1 \times (1.93)^2 \][/tex]
[tex]\[ I_{\text{person}} = 237.488 \, \text{kg*m}^2 \][/tex]
Now, we need to consider the parallel axis theorem to find the moment of inertia of the person relative to the center of the platform. The parallel axis theorem states that:
[tex]\[ I_{\text{platform-person}} = I_{\text{platform}} + I_{\text{person}} + m_{\text{person}} d^2 \][/tex]
Where d is the perpendicular distance between the axis through the center of mass and the axis through the point mass. In this case, d is the radius of the platform,[tex]\( r_{\text{platform}} \).[/tex]
So,
[tex]\[ I_{\text{platform-person}} = I_{\text{platform}} + I_{\text{person}} + m_{\text{person}} r_{\text{platform}}^2 \][/tex]
[tex]\[ I_{\text{platform-person}} = 188.243 + 237.488 + 65.1 \times (1.93)^2 \][/tex]
[tex]\[ I_platform-person = 2362.008 \, \text{kg*m}^2 \][/tex]
Therefore, the moment of inertia of the platform-person system about the vertical axis is approximately [tex]\( 2362 \, \text{kg*m}^2 \).[/tex]
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