A 4.2-m-diameter merry-go-round is rotating freely with an angular velocity of 0.80 rad????s. Its total moment of inertia is 1360 kg ???? m2

Question

A 4.2-m-diameter merry-go-round is rotating freely with an angular velocity of 0.80 rad????s. Its total moment of inertia is 1360 kg ???? m2. Four people standing on the ground, each of mass 65 kg, suddenly step onto the edge of the merry-go-round. (a) What is the angular velocity of the merry-go-round now? (b) What if the people were on it initially and then jumped off in a radial direction (relative to the merry-go-round)

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Neala 6 months 2021-07-18T00:01:41+00:00 1 Answers 5 views 0

Answers ( )

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    2021-07-18T00:02:57+00:00

    Answer:

    A) 0.44 rad/s

    B) 3.091 rad/s

    Explanation:

    Initial angular momentum must be equal to final angular momentum.

    W1 = 0.80 rad/s

    I = 1360 kgm^2

    Initial moment Iw = 0.8 x 1360 = 1088 rad-kg-m^2/s

    For second case

    I = (65 x 4)4.2 + 1360 = 2452 kg-m2-rad/s

    New momentum = 2452 x W2

    Where w2 = final angular momentum

    = 2452W2

    Equating both moment,

    1088 = 2452W2

    W2 = 0.44 rad/s0

    If they were on it before,

    Initial momentum

    I = 1360 – (60 x 4)4.2

    = 352 rad-kg-m^2/s

    352W1 = 1088

    W1 = 3.091 rad/s

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