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9 June, 11:15

A 125-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 0.700 rev/s in 2.00 s? (State the magnitude of the force.)

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  1. 9 June, 11:32
    0
    201.6 N

    Explanation:

    m = mass of disk shaped merry-go-round = 125 kg

    r = radius of the disk = 1.50 m

    w₀ = Initial angular speed = 0 rad/s

    w = final angular speed = 0.700 rev/s = (0.700) (2π) rad/s = 4.296 rad/s

    t = time interval = 2 s

    α = Angular acceleration

    Using the equation

    w = w₀ + α t

    4.296 = 0 + 2α

    α = 2.15 rad/s²

    I = moment of inertia of merry-go-round

    Moment of inertia of merry-go-round is given as

    I = (0.5) m r² = (0.5) (125) (1.50) ² = 140.625 kgm²

    F = constant force applied

    Torque equation for the merry-go-round is given as

    r F = I α

    (1.50) F = (140.625) (2.15)

    F = 201.6 N
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