To calculate the moment of inertia ($I$) of a door rotating about a vertical axis located 13 cm from its edge, we'll use the parallel axis theorem. Here's the step-by-step solution:

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1. Moment of inertia for a rectangular door about its center

The formula for the moment of inertia of a thin rectangular sheet about an axis parallel to one side through its center is:

$I_{\text{center}} = \frac{1}{3} M h^2$

For the door:

• $M = 20 \, \text{kg}$ (mass of the door),
• $h = 2.2 \, \text{m}$ (height of the door).

$I_{\text{center}} = \frac{1}{3} (20) (2.2)^2 = \frac{1}{3} \times 20 \times 4.84 = 32.27 \, \text{kg·m}^2$

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2. Parallel axis theorem

The door's axis of rotation is not at its center but 13 cm (0.13 m) from the edge. Using the parallel axis theorem:

$I = I_{\text{center}} + M d^2$

Where:

• $d = 0.13 \, \text{m}$ (distance from the center to the new axis).

Substitute the values:

$I = 32.27 + 20 \cdot (0.13)^2$

$I = 32.27 + 20 \cdot 0.0169 = 32.27 + 0.338$

$I = 32.608 \, \text{kg·m}^2$

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Final Answer:

The moment of inertia of the door about the vertical axis located 13 cm from its edge is:

$\boxed{32.61 \, \text{kg·m}^2}$

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Would you like further clarification or details?
Here are some related questions:

1. How does the parallel axis theorem apply to rotational dynamics?
2. What is the significance of the centroid in determining moments of inertia?
3. How would the moment of inertia change if the axis were closer to the edge?
4. How does door thickness affect the moment of inertia?
5. How is this calculation applied in real-world physics, like door hinge mechanics?

Tip: Always ensure all units are converted to standard units (meters, kilograms, etc.) before using in formulas.