A horizontal 790-N merry-go-round of radius 1.60 m is started from rest by a constant horizontal force of 45 N applied tangentially to the m

Question

A horizontal 790-N merry-go-round of radius 1.60 m is started from rest by a constant horizontal force of 45 N applied tangentially to the merry-go-round. Find the kinetic energy of the merry-go-round after 4.0 s. (Assume it is a solid cylinder. Also assume the force is applied at the outside edge.)

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Vodka 3 years 2021-08-26T05:19:21+00:00 1 Answers 25 views 0

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  1. minhkhoi
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    2021-08-26T05:20:43+00:00
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    Answer:

    404.3 J

    Explanation:

    Given that

    Weight of the merry go round = 790 N

    Radius if the merry go round = 1.6 m

    Horizontal force applied = 45 N

    Time taken = 4 s

    To find the mass of the merry go round, we divide the weight by acceleration due to gravity. Thus,

    m = F/g

    m = 790 / 9.8

    m = 80.6 kg

    We know that the moment of inertia is given as

    I = ½mr², on substitution, we have

    I = ½ * 80.6 * 1.6²

    I = 103.17 kgm²

    Torque = Force applied * radius, so

    τ = 45 * 1.6

    τ = 72 Nm

    To get the angular acceleration, we have,

    α = τ / I

    α = 72 / 103.17

    α = 0.70 rad/s²

    Then, the angular velocity is

    ω = α * t

    ω = 0.7 * 4

    ω = 2.8 rad/s

    Finally, to get the Kinetic Energy, we have

    K.E = ½ * Iω², on substituting, we get

    K.E = ½ * 103.17 * 2.8²

    K.E = 404.3 J

    Therefore, the kinetic energy is 404.3 J

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