Three children are riding on the edge of a merry‑go‑round that has a mass of 105 kg and a radius of 1.70 m . The merry‑go‑round is spinning

Question

Three children are riding on the edge of a merry‑go‑round that has a mass of 105 kg and a radius of 1.70 m . The merry‑go‑round is spinning at 22.0 rpm . The children have masses of 22.0 , 28.0 , and 33.0 kg . If the 28.0 kg child moves to the center of the merry‑go‑round, what is the new angular velocity in revolutions per minute? Ignore friction, and assume that the merry‑go‑round can be treated as a solid disk and the children as point masses.

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Thành Đạt 6 months 2021-07-17T11:46:26+00:00 1 Answers 6 views 0

Answers ( )

  1. ngocdiep
    0
    2021-07-17T11:47:44+00:00
    Reply

    Answer:

    Explanation:

    Given that,

    Three children of masses and their position on the merry go round

    M1 = 22kg

    M2 = 28kg

    M3 = 33kg

    They are all initially riding at the edge of the merry go round

    Then, R1 = R2 = R3 = R = 1.7m

    Mass of Merry go round is

    M =105kg

    Radius of Merry go round.

    R = 1.7m

    Angular velocity of Merry go round

    ωi = 22 rpm

    If M2 = 28 is moves to center of the merry go round then R2 = 0, what is the new angular velocity ωf

    Using conservation of angular momentum

    Initial angular momentum when all the children are at the edge of the merry go round is equal to the final angular momentum when the second child moves to the center of the merry go round

    Then,

    L(initial) = L(final)

    Ii•ωi = If•ωf

    So we need to find the initial and final moment of inertia

    NOTE: merry go round is treated as a solid disk then I= ½MR²

    I(initial)=½MR²+M1•R²+M2•R²+M3•R²

    I(initial) = ½MR² + R²(M1 + M2 + M3)

    I(initial) = ½ × 105 × 1.7² + 1.7²(22 + 28 + 33)

    I(initial) = 151.725 + 1.7²(83)

    I(initial) = 391.595 kgm²

    Final moment of inertial when R2 =0

    I(final)=½MR²+M1•R²+M2•R2²+M3•R²

    Since R2 = 0

    I(final) = ½MR²+ M1•R² + M3•R²

    I(final) = ½MR² + (M1 + M3)• R²

    I(final)=½ × 105 × 1.7² + ( 22 +33)•1.7²

    I(final) = 151.725 + 158.95

    I(final) = 310.675 kgm²

    Now, applying the conservation of angular momentum

    L(initial) = L(final)

    Ii•ωi = If•ωf

    391.595 × 22 = 310.675 × ωf

    Then,

    ωf = 391.595 × 22 / 310.675

    ωf = 27.73 rpm

    So, the final angular momentum is 27.73 revolution per minute

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