A playground merry-go-round has radius 2.60 m…?

A playground merry-go-round has radius 2.60 m and moment of inertia 2300 kg*m^2 about a vertical axle through its center, and it turns with negligible friction.

(a) A child applies an 23.0 N force tangentially to the edge of the merry-go-round for 21.0 . If the merry-go-round is initially at rest, what is its angular speed after this 21.0 s interval?

w=?

(b) How much work did the child do on the merry-go-round?

W=?

(c) What is the average power supplied by the child?

P=?

PHYSICS

2 Answers

  • Torque τ = I α where α is angular acceleration

    Torque τ = F*r

    F*r = I α = I (ωf – ωi)/t

    =============================

    A)

    Given ωi = 0

    F*r = I ωf/t

    ωf = F*rt/ I = 23*2.6*21/2300 = 0.546 rad/s

    ————————————–…

    B)

    W =0.5*I* ωf² = 1150*0.546 ² = 342.8 J

    —————–

    C)

    Power Total work/time =342.8334/21 =16.3W

    =================

  • torque applied

    = 23 * 2.6

    = 59.8 Nm

    = moment of inertia * angular acc

    = 2300 *a

    a = 0.026 rad/s^2

    t =21; u=0

    v = u +at

    v = 0 + 0.026 * 21

    = 0.546 rad /s

    answer

    angular speed after this 21.0 s interval

    work the child did on the merry-go-round

    = rotational KE gained

    = 0.5 * MoI * w^2

    = 0.5 * 2300 * 0.546^2

    = 342.83 J

    answer

    average power supplied by the child

    = work the child did on the merry-go-round / time

    = 342.83 / 21

    = 16.33 W

    answer

    Source(s): my brain (Prof TBT)

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