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2 March, 02:07

A 4.2-m-diameter merry-go-round is rotating freely with an angularvelocity of 0.80 rad/s. Its total moment of inertia is1760kgm2. Four people standing on the ground, eachof mass 65 kg, suddenly step onto the edge of themerry-go-round. a. What is the angular velocity of themerry-go-round now? b. What if the people were on it initiallyand then jumped off in a radial direction (relative to themerry-go-round) ?

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  1. 2 March, 02:34
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    Answer

    given,

    mass of people = 65 kg

    number of people =

    diameter of the merry-go-round = 4.2 m

    radius = 2.1 m

    angular velocity = 0.8 rad/s

    moment of inertia = 1760 kg m²

    Using the law of conservation of angular momentum, we have

    L₁ = L₂

    I₁ω₁ = I₂ω₂

    I₁ω₁ = (I₁ + 4 m r²) ω²

    (1760 x 0.8) = (1760 + 4 x 65 x 2.12) ω²

    1408 = 2311.2 ω²

    ω² = 0.609

    ω = 0.781 rad/s

    b) If the people jump of the merry-go-round radially, they exert no torque.

    hence, it will not change the angular momentum of the merry-go-round. It will continue to move with the same ω.
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