A merry-go-round on a playground consists of a horizontal solid disk with a weight of 805 N and a radius of 1.47 m. A child applies a force

Question

A merry-go-round on a playground consists of a horizontal solid disk with a weight of 805 N and a radius of 1.47 m. A child applies a force 49.0 N tangentially to the edge of the disk to start it from rest. What is the kinetic energy of the merry-go-round disk (in J) after 2.95 s?

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Farah 4 years 2021-07-21T08:48:20+00:00 1 Answers 132 views 0

Answers ( )

  1. Dulcie
    0
    2021-07-21T08:49:38+00:00
    Reply

    Answer:

    255.34 J

    Explanation:

    Given,

    Weight of disk = 805 N

    radius = 1.47 m

    Force applied by the child = 49 N

    time = 2.95 s

    KE = ?

    mass of the disk

    M = \dfrac{W}{g}= \dfrac{805}{9.81} = 82.059\ Kg

    Moment of inertia of the disk

    I = \dfrac{1}{2}Mr^2

    I = \dfrac{1}{2}\times 82.059\times 1.47^2 =88.66\ kgm^2

    Torque on the child

    \tau = F \times r = 49 \times 1.47 = 72.03 Nm

    Angular acceleration

    \alpha = \dfrac{\tau}{I}=\dfrac{72.03}{88.66} = 0.812\ rad/s^2

    So, angular speed at t = 2.95 s

    \omega = \alpha t = 0.812 \times 2.95 = 2.4\ rad/s

    Now, KE of the merry go round

    KE = \dfrac{1}{2} I \omega^2 = \dfrac{1}{2}\times 88.66\times 2.4^2 = 255.34 J

    Hence, the Kinetic energy of the merry go round = 255.34 J

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