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24 March, 21:29

A horizontal 649 N merry-go-round of radius 1.05 m is started from rest by a constant horizontal force of 61.3 N applied tangentially to the merry-go-round. Find the kinetic energy of the merry-goround after 3.34 s. 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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  1. 24 March, 21:32
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    The kinetic energy of the merry-go-round is 632.82 J

    Explanation:

    Given;

    weight of the merry-go-round, W = 649 N

    radius of the merry-go-round, r = 1.05 m

    applied horizontal force, F = 61.3 N

    acceleration due to gravity, g = 9.8 m/s²

    mass of merry-go-round, m = W/g

    = 649/9.8 = 66.225 kg

    moment of inertia of merry-go-round, I = ¹/₂mr²

    = ¹/₂ x 66.225 x (1.05) ²

    = 36.507 kg. m²

    Angular acceleration of the merry-go-round, α

    τ = Iα = Fr

    α = Fr / I

    Where;

    α is angular acceleration

    α = (61.3 x 1.05) / 36.507

    α = 1.763 rad/s²

    Angular velocity of the merry-go-round, ω

    ω = αt

    ω = 1.763 x 3.34

    ω = 5.888 rad/s

    Finally, the kinetic energy of the merry-go-round, K. E

    K. E = ¹/₂Iω²

    K. E = ¹/₂ x 36.507 x (5.888) ²

    K. E = 632.82 J
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