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19 July, 09:57

A playground merry-go-round of radius R = 1.40 m has a moment of inertia I = 245 kg · m2 and is rotating at 9.0 rev/min about a frictionless vertical axle. Facing the axle, a 24.0-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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  1. 19 July, 10:50
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    Answer: 11.65rpm

    Explanation:

    Moment of inertia of the child = mr²

    Inertia = 24 * 1.4 * 1.4 = 47.04kgm²

    Converting rpm to rad/s we have

    12rpm = 12 * π/30

    12rpm = 37.7/30

    12rpm = 1.257 rad/s

    Angular momentum = Iω

    (245 * 1.257) + (47.04 * 0) = 355kgm²

    355 = (245 + 47.04) * ω

    355 = 292.04ω

    ω = 355/292.04

    ω = 1.22 rad/s

    1.22 rad/s = 1.22 * 9.55 = 11.65 rpm

    Thus, the new angular speed of the merry go round is 11.65rpm
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