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7 February, 13:41

A disk-shaped merry-go-round of radius 2.93 m and mass 135 kg rotates freely with an angular speed of 0.621 rev/s. A 62.4 kg person running tangential to the rim of the merry-go-round at 3.21 m/s jumps onto its rim and holds on. Before jumping on the merry-go-round, the person was moving in the same direction as the merry-go-round's rim. What is the final angular speed of the merry-go-round? rad/s

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  1. 7 February, 13:47
    0
    2.55 rad/sec

    Explanation:

    Use conservation of angular momentum

    The merry go round has

    I = ½mr²

    I = ½ * 135 * 2.93²

    I = 579 kgm²

    the person has

    I = mr²

    I = 62.4 * 2.93²

    I = 536 kgm²

    Converting 0.621 rev/sec to rad/sec we have 0.621 * 2π

    0.621 * 2 * 3.14 rad/sec

    3.9 rad/sec

    for the person v/r = w

    w = 3.21 / 2.93

    w = 1.10 rad/sec

    So

    579 * 3.9 + 536 * 1.1 = (579+536) * w

    2258 + 589.6 = 1115w

    2847.6 = 1115w

    solve for w

    w = 2847.6 / 1115

    w = 2.55 rad/sec

    Thus, the final angular speed of the merry go round is 2.55 rad/sec
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