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12 January, 17:03

A 160-kg merry-go-round in the shape of a uniform, solid, horizontal disk of radius 1.50 m is set in motion by wrapping a rope about the rim of the disk and pulling on the rope. What constant force would have to be exerted on the rope to bring the merry-go-round from rest to an angular speed of 0.600 rev/s in 2.00 s? (State the magnitude of the force.)

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  1. 12 January, 17:54
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    F = 453.6 N

    Explanation:

    We apply Newton's second law to the merry-go-round:

    ∑Ft = m*at Formula (1

    Where:

    ∑Ft : algebraic sum of tangential forces (N)

    m: merry-go-round mass (Kg)

    at: tangential acceleration (m/s²)

    Kinematics equations of the merry-go-round

    ω = ω₀ + α*t Formula (2)

    at = α*R Formula (3)

    Where:

    α : angular acceleration (rad/s²)

    t : time interval (s)

    ω₀ : initial angular speed (rad/s)

    ωf : final angular speed (rad/s)

    R : radius of the circular path (m)

    Data

    m=160-kg

    R = 1.50 m

    ω₀=0

    ωf = 0.600 rev/s

    1 rev = 2π rad

    ωf = 0.6*2π rad/s = 3.77 rad/s

    t = 2 s

    Calculation of angular acceleration of the merry-go-round

    We replace data in the formula (2)

    ω = ω₀ + α*t

    3.77 = 0 + α * (2)

    α = 3.77 / 2

    α = 1.89 rad/s²

    Calculation of tangential acceleration

    We replace data in the formula (3)

    at = α*R

    at = (1.89) * (1.5)

    at = 2.835 m/s²

    Constant force (F) that would have to be exerted on the rope to bring for the merry-go-round from rest to an angular speed of 0.600 rev/s in 2.00 s:

    We replace data in the formula (1)

    F = m*at

    F = (160 kg) * (2.835 m/s²)

    F = 453.6 N
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