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24 April, 02:38

A 65.0-kg woman stands at the rim of a horizontal turntable having a moment of inertia of 480 kg · m2 and a radius of 2.00 m. The turntable is initially at rest and is free to rotate about a frictionless vertical axle through its center. The woman then starts walking around the rim clockwise (as viewed from above the system) at a constant speed of 1.50 m/s relative to the Earth.

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  1. 24 April, 03:02
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    A 65.0-kg woman stands at the rim of a horizontal turntable having a moment of inertia of 480 kg · m2 and a radius of 2.00 m. The turntable is initially at rest and is free to rotate about a frictionless vertical axle through its center. The woman then starts walking around the rim clockwise (as viewed from above the system) at a constant speed of 1.50 m/s relative to the Earth.

    (a) In what direction and with what angular speed does the turntable rotate?

    (b) How much work does the woman do to set herself and the turntable into motion?

    The answers are

    (a) 0.40625 rad/s ccw

    (b) 112.73 J

    Explanation:

    To solve the question we have

    Mass of woman = 65. kg

    Moment of inertia of the turntable = 480 kg·m²

    Radius of turntable = 2.00 m

    Speed of woman = - 1.50 m/s, clockwise motion

    Sum of anguar momentum = constant

    Therefore Inital total angular momentum = Finat toatal momentum

    M₁ₐ, M₁ₓ = initial and final momentum of turn table and M₂ₐ, M₂ₓ for the woman

    Therefore

    M₁ₐ + M₂ₐ = M₁ₓ + M₂ₓ

    Since the momentum = Inertia, I multiplied by angular velocity, ω, we have

    I₁*ω₁ = - I₂*ω₂ and

    ω₁ = - I₂*ω₂:I₁

    Which gives m·r²*v/r:I₁ where ω = v/r and I = m·r²

    =-m·r·v:l₁, Substituting the values for the mass, radius, the speed and the moment of inertia of the turntable gives

    = - 65*2*-1.5:480 = 0.40625 rad/s ccw

    The work done can be found from

    Total work = change in kinetic ennergy

    = Initial Kinetic energy of the system less final kinetic energy of the system

    = Kinetic energy of the woman + Kinetic energy of the turntable - 0

    1/2*I₁*ω₁² + 1/2*m*v₂²

    = 1/2*480 kg·m² * (0.40625 rad/s) ² + 1/2*65 kg * (-1.5 m/s) ²

    = 112.734375 kg·m/s ≈ 112.73 J
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