A playground merry-go-round of radius R = 1.20 m has a moment of inertia I = 240 kg · m2 and is rotating at 9.0 rev/min about a frictionless

Question

A playground merry-go-round of radius R = 1.20 m has a moment of inertia I = 240 kg · m2 and is rotating at 9.0 rev/min about a frictionless vertical axle. Facing the axle, a 26.0-kg child hops onto the merry-go-round and manages to sit down on the edge. What is the new angular speed of the merry-go-round?

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Minh Khuê 10 hours 2021-07-21T22:52:57+00:00 1 Answers 0 views 0

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    2021-07-21T22:54:44+00:00

    Answer:

    ω’ = 0.815 rad/s

    Explanation:

    Given,

    R = 1.20 m

    Inertia of merry-go- round= 240 kg.m²

    Rotating speed  = 9 rpm =  9\times \dfrac{2\pi}{60}

                               =0.9424 rad/s

    mass of the child, m = 26 kg

    angular speed of the merry-go-round=?

    we know

    Angular momentum, L = I ω

    Moment of inertia of the child

    I’ = m  r² = 26 x 1.2² = 37.44 kgm²

    Conservation of angular momentum

    initial angular momentum = Final angular momentum

    I ω = (I+I’)ω’

    240 x 0.9424 = (240+37.44) ω’

    226.176= 277.44 ω’

    ω’ = 0.815 rad/s

    new angular speed of the merry-go- round is equal to 0.815 rad/s

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