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30 August, 16:36

A horizontal 810-N merry-go-round of radius 1.40 m is started from rest by a constant horizontal force of 55 N applied tangentially to the merry-go-round. Find the kinetic energy of the merry-go-round after 2.0 s. (Assume it is a solid cylinder. Also assume the force is applied at the outside edge.)

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  1. 30 August, 16:54
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    52.218 J

    Explanation:

    Mass of the merry go round m = weight / g

    = 810 / 9.8

    = 82.65 kg

    radius r = 1.40 m

    torque applied τ = r * F

    = 1.40 * 55

    = 77 N - m

    Moment of inertia (of cylinder) I = (1/2) * m * r²

    = 0.5 * 82.65 * 1.96

    = 80.997 kg-m²

    angular acceleration α = τ / I

    = 77 / 119.43

    = 0.6447 rad/s²

    According to first equation in angular motion final angular velocity ω = ω0 + α * t

    ω = 0 + 0.6447 * 2.0

    = 1.2894 rad/s

    Hence k. e. after 2 seconds K = (1/2) * I * ω²

    = 0.5 * 80.997 * 1.2894

    = 52.218 J
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