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7 August, 05:30

A propeller is modeled as five identical uniform rods extending radially from its axis. The length and mass of each rod are 0.827 m and 2.89 kg, respectively. When the propeller rotates at 507 rpm (revolutions per minute), what is its rotational kinetic energy K

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  1. 7 August, 09:00
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    3241.35J

    Explanation:

    No. Of rods = 5

    Mass = 2.89kg

    Length (L) = 0.827m

    W = 507rpm

    Kinetic energy of rotation = ½I*ω²

    For each rod, the moment of inertia (I) = ML² / 3

    I = ML² / 3

    I = [2.89 * (0.827) ²] / 3

    I = 1.367 / 3 = 0.46kgm²

    ω = 507 rev/min. Convert rev/min to rev/sec.

    507 * 2Πrads/60s = 53.09rad/s

    ω = 53.09rad/s

    k. e = ½ I * ω²

    K. E = ½ * 0.46 * (53.09) ²

    K. E = 648.27.

    But there five (5) rods, so kinetic energy is equal to

    K. E = 5 * 648.27 = 3241.35J
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