A uniform disk of radius 0.2 m and mass m = 16 kg is mounted on a nearly frictionless axle. A string is wrapped tightly around the disk and

Question

A uniform disk of radius 0.2 m and mass m = 16 kg is mounted on a nearly frictionless axle. A string is wrapped tightly around the disk and pulled with a constant force of F = 1 N. After a while the disk has reached an angular speed of !0 = 1.2 rad/s. What is its angular speed 1.5 s later?

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Thu Thảo 4 years 2021-08-17T03:23:24+00:00 2 Answers 28 views 0

Answers ( )

  1. trucchi
    0
    2021-08-17T03:24:53+00:00
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    Answer:

    ω = 2.13rad/s

    Explanation:

    Given

    F = 1N = force applied, m = 16kg, R = 0.2m

    ωo = 1.2rad/s

    t = 1.5s

    ω = ?

    Where

    α = angular acceleration

    ωo = initial angular speed

    And ω = angular speed at a later time t

    For a rigid body rotating about an axis

    τ = FR= I×α

    I = 1/2×MR² for a uniform disk

    FR = (1/2×mR²)×α

    = 1/2×mR²×(ω – ωo)/t

    Therefore

    FR= 1/2×mR²×(ω – ωo)/t

    2Ft/mR = ω – ωo

    ω = 2Ft/mR + ωo

    ω = 2×1×1.5/(16×0.2) + 1.2 = 2.13 rad/s

  2. haidang
    0
    2021-08-17T03:25:18+00:00
    Reply

    Answer:

    0.684 rad/s

    Explanation:

    The moment of inertia of a uniform disk is assumed = 0.5MR² where m is mass and R is radius

    I, moment of inertia = 0.5 × 16 × (0.2 m)² = 0.32 kgm²

    Torque of the force = F × R = 1 × 0.2 = 0.2 Nm

    angular speed = moment of inertia I × angular speed ω

    new angular speed  = Iω + (Torque × time) = (0.32 kgm²× 1.2rad/s) + ( 0.2 Nm × 1.5 s) = 0.684 rad/s

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