A 4.4 m diameter merry-go-round is rotating freely with an angular velocity of 0.80 rad/s. Its total moment of inertia is 1950 kg · m2. Four

Question

A 4.4 m diameter merry-go-round is rotating freely with an angular velocity of 0.80 rad/s. Its total moment of inertia is 1950 kg · m2. Four people standing on the ground, each of 50 kg mass, suddenly step onto the edge of the merry-go-round.

What is the angular velocity of the merry-go-round now?

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Verity 4 years 2021-08-15T01:01:47+00:00 1 Answers 8 views 0

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  1. thuthuy
    0
    2021-08-15T01:03:20+00:00
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    Answer:

    Angular velocity of merry go round is 0.64 \frac{rad}{s}.

    Explanation:

    Given :

    Radius of merry go round R = \frac{D}{2} = 2.2 m

    Angular velocity of merry go round \omega = 0.8 \frac{rad}{s}

    Momentum of inertia I = 1950 Kg m^{2}

    Mass of each people M = 50 Kg

    From the formula of angular momentum conservation,

     L = I\omega

    So initial angular momentum,

    L_{i} = 1950 \times 0.8 = 1560

    Now we calculate the MOI after people jump on merry go round,

    I_{p} =\frac{1}{2} MR^{2}

    But there are four people so we have to multiply with 4 and add merry go round MOI,

    I _{tot} = 1950 + 2\times 50 \times (2.2) ^{2}

    So total MOI after people jump on is,

    I _{tot} = 2434 Kg m^{2}

    From conservation of angular momentum,

    L_{i} = L_{f}

    1560 = 2434 \times \omega_{f}

    So angular velocity is,

     \omega _{f} = \frac{1560}{2434} = 0.64 \frac{rad}{s}

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