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12 March, 05:40

How much work is required to uniformly accelerate a merry-go-round of mass 1490 kg and a radius of 7.80 m from rest to a rotational rate of 1 revolution per 8.33 s? Model the merry-go-round as a solid cylinder (I = ½ MR2). How much power is required to accelerate the merry-go-round to that rate in 8.33 s?

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Answers (2)
  1. 12 March, 06:52
    0
    Given:

    Time, t = 8.33 s

    Mass, M = 1490 kg

    Angular velocity, w = 1 rev in 8.33 s

    = 0.12 rev/s

    Converting rev/s to rad/s,

    1 rev = 2pi rad

    0.12 rev/s * 2pi rad/1 rev

    = 0.754 rad/s

    Radius, r = 7.8 m

    Inertia, I = 1/2 * M * r^2

    = 1/2 * 1490 * 7.8^2

    = 45325 kgm^2

    Kinetic energy, Uk = 1/2 * I * w^2

    = 1/2 * 45325 * 0.754^2

    = 12884.5 J

    = 12.8 kJ

    Power = energy/time

    Kinetic energy, Uk = energy

    Power = 12884.5/8.33

    = 1546.725 W

    = 1.55 kW.
  2. 12 March, 07:10
    0
    W = 12,884.221 J

    Explanation:

    I = ½ MR² (given)

    I = 0.5 * 1490 kg * (7.8 m) ²

    I = 45,325.8 Kg m²

    and ω = 2π / T = 2 * 3.14 / 8.33 s

    ω = 0.754 radians/s

    now W = Δ K. E. = 1/2 Iω² = 0.5 * 45,325.8 Kg m² * (0.754 radians/s) ²

    W = 12,884.221 J
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