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10 December, 14:47

An airplane propeller is 2.18 m in length (from tip to tip) with mass 97.0 kg and is rotating at 2600 rpm (rev/min) about an axis through its center. You can model the propeller as a slender rod. What is its rotational kinetic energy?

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  1. 10 December, 17:41
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    1.4*10^6 J

    Explanation:

    Given that

    Length of the propeller, l = 2.18 m

    Mass of the propeller, m = 97 kg

    Speed of the propeller, w = 2600 rpm

    The formula for finding rotational Kinetic energy, K is = ½Iw²

    Where, I is the moment of Inertia, and is given as 1/12 * m * l²

    I = 1/12 * 97 * 2.18²

    I = 8.083 * 4.7524

    I = 38.41 kgm²

    w = 2600 rpm, converting to rad/s, we have

    w = 2600 * 2π rad/s

    w = 272.31 rad/s

    Now, Kinetic Energy, K is

    K = ½Iw²

    K = ½ * 38.41 * 272.31²

    K = 19.205 * 74152.7361

    K = 1424103.3 J

    K = 1.4 MJ or 1.4*10^6 J

    Thus, the rotational Kinetic Energy is 1.4*10^6 J
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