A horizontal 694 N merry-go-round of radius 1.96 m is started from rest by a constant horizontal force of 61.5 N applied tangentially to the

Question

A horizontal 694 N merry-go-round of radius 1.96 m is started from rest by a constant horizontal force of 61.5 N applied tangentially to the merry-go-round. Find the kinetic energy of the merry-go round after 3.89 is. The acceleration of gravity is 9.8 m/s 2 . Assume the merry-go-round is a solid cylinder. Answer in units of J.

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Hưng Khoa 4 years 2021-08-22T16:39:14+00:00 2 Answers 22 views 0

Answers ( )

  1. thienthanh
    0
    2021-08-22T16:40:58+00:00
    Reply

    Answer:

    808.19J

    Explanation:

    Image attached gives a proper explanation

  2. minhkhue
    0
    2021-08-22T16:41:12+00:00
    Reply

    Answer: 808.24625J

    Explanation:

    Moment of inertia is the physical quantity, that expresses the tendency of a body to resist angular acceleration. It determines the Torque needed for a desired angular acceleration, about a rotational axis

    I = 1/2mr²

    I = 1/2(W/g)r²

    I = 1/2 * 694/9.8 * 1.96²

    I = 1/2 * 272.048

    I = 136.024kgm²

    Also,

    Iα = Fr

    136.024α = 61.5*1.96

    136.024α = 120.54

    α = 120.54/136.024

    α = 0.8862 rad/s²

    Angular Velocity,

    ω = αt

    ω = 0.8862*3.89

    ω = 3.4473 rad/s

    K = 1/2Iω²

    K = 1/2*136.024*3.4473²

    K = 1/2 * 1616.4925

    K = 808.24625J

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