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13 February, 15:40

Find the moment of inertia about each of the following axes for a rod that is 0.360 {cm} in diameter and 1.70 {m} long, with a mass of 5.00*10-2 {kg}.1. About an axis perpendicular to the rod and passing through its center. (in I=kg*m^2) 2. About an axis perpendicular to the rod and passing through one end. 3. About an axis along the length of the rod.

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  1. 13 February, 15:43
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    The complete question is;

    Find the moment of inertia about each of the following axes for a rod that is 0.36 cm in diameter and 1.70m long, with a mass of 5.00 * 10 ^ (-2) kg.

    A) About an axis perpendicular to the rod and passing through its center in kg. m²

    B) About an axis perpendicular to the rod and passing through one end in kg. m²

    C) About an axis along the length of the rod in kg. m²

    Answer:

    A) I = 0.012 kg. m²

    B) I = 0.048 kg. m²

    C) I = 8.1 * 10^ (-8) kg. m²

    Explanation:

    We are given;

    Diameter = 0.36 cm = 0.36 * 10^ (-2) m

    Length; L = 1.7m

    Mass; m = 5 * 10^ (-2) kg

    A) For an axis perpendicular to the rod and passing through its center, the formula for the moment of inertia is;

    I = mL²/12

    I = (5 * 10^ (-2) * 1.7²) / 12

    I = 0.012 kg. m²

    B) For an axis perpendicular to the rod and passing through one end, the formula for the moment of inertia is;

    I = mL²/3

    So,

    I = (5 * 10^ (-2) * 1.7²) / 3

    I = 0.048 kg. m²

    C) For an axis along the length of the rod, the formula for the moment of inertia is; I = mr²/2

    We have diameter = 0.36 * 10^ (-2) m, thus radius; r = (0.36 * 10^ (-2)) / 2 = 0.18 * 10^ (-2) m

    I = (5 * 10^ (-2) * (0.18 * 10^ (-2)) ^2) / 2

    I = 8.1 * 10^ (-8) kg. m²
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