Four 50-g point masses are at the corners of a square with 20-cm sides.What is the moment of inertia of this system about an axis perpendicular to the plane of the square and passing through its center? A)1.0 $\times$ 10^-3 kg · m² B)4.0 $\times$ 10^-3 kg · m² C)2.0 $\times$ 10^-3 kg · m² D)8.0 $\times$ 10^-3 kg · m² E)2.8 $\times$ 10^-3 kg · m²

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Quizplus · Accepted Answer

The moment of inertia for point masses is calculated as I = Σmr². Each mass is 50 g (0.05 kg) and the distance from the center to each corner is √(0.1² + 0.1²) = √0.02 m. Thus, I = 4 * 0.05 * 0.02 = 4.0 × 10⁻³ kg·m².

Four 50-G Point Masses Are at the Corners of a Square