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

Question

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 mark. (Treat the stick as a thin rod.)

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Neala 3 years 2021-08-14T04:37:38+00:00 1 Answers 27 views 0

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  1. minhkhang
    0
    2021-08-14T04:39:04+00:00
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    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

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