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14 October, 00:39

A uniform rod rotates in a horizontal plane about a vertical axis through one end. The rod is 3.46 m long, weighs 12.8 N, and rotates at 226 rev/min. Calculate (a) its rotational inertia about the axis of rotation and (b) the magnitude of its angular momentum about that axis.

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  1. 14 October, 02:03
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    a. Rotational inertia: 5.21kgm²

    b. Magnitude of it's angular momentum: 123.32kgm²/s

    Explanation:

    Length of the rod = 3.46m

    Weight of the rod = 12.8 N

    Angular velocity of the rod = 226 rev/min

    a. Rotational Inertia (I) about its axis

    The formula for rotational inertia =

    I = (1/12*m*L²) + m * (L : 2) ²

    Where L = length of the rod

    m = mass of the rod

    Mass of the rod is calculated by dividing the weight of the rod with the acceleration due to gravity.

    Acceleration due to gravity = 9.81m/s²

    Mass of the rod = 12.8N / 9.81m/s²

    Mass of the rod = 1.305kg

    Rotational Inertia =

    (1/12 * 1.305 * 3.46²) + 1.305 (3.46:2) ²

    Rotational Inertia = 1.3019115 + 3.9057345

    Rotational Inertia = 5.207646kgm²

    Approximately = 5.21kgm²

    b. The magnitude of the rod's angular momentum about the rotational axis is calculated as

    Rotational Inertia about its axis * angular speed of the rod.

    Angular speed of the rod is calculated as = (Angular velocity of the rod * 2π) / 60

    = (226*2π) / 60

    = 23.67 rad/s

    Rotational Inertia = 5.21kgm²

    The magnitude of the rod's angular momentum about the rotational axis

    = 5.21kgm² * 23.67 rad/s

    = 123.3207kgm²/s

    Approximately = 123.32kgm²/s
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