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Calculate the rotational inertia of a meter stick, with mass 0.390 kg, about an axis perpendicular to the stick and located at the 21.1 cm m
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Answers ( )
Answer:
0.0651 kgm²
Explanation:
Parameters given:
Mass of meter stick, m = 0.39 kg
Distance of axis from center of meter stick, D = 50 – 21.1 cm = 28.9 cm = 0.289 m
The moment of inertia of a thin rod about an axis that is perpendicular to it and is located at a particular distance, D, from its center is given as:
I = 1/12 * m * L² + m * D²
I = (1/12 * 0.39 * 1²) + (0.39 * 0.289²)
I = 0.325 + 0.326
I = 0.0651 kgm²