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19 April, 20:09

A merry-go-round rotates from rest with an angular acceleration of 1.04 rad/s2. How long does it take to rotate through (a) the first 1.59 rev and (b) the next 1.59 rev

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  1. 19 April, 20:12
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    (a) 4.38 s.

    (b) 1.817 s

    Explanation:

    (a)

    Using

    θ = ω₀t + 1/2αt² ... Equation 1

    Where θ = number of revolution, t = time, α = angular acceleration, ω₀ = angular velocity.

    Given: θ = 1.59 rev = 1.59*2π = 9.992 rad, ω₀ = 0 rad/s, α = 1.04 rad/s².

    Substitute into equation 1

    9.992 = 0 (t) + 1/2 (1.04) (t²)

    t² = (2*9.992) / 1.04

    t² = 19.984/1.04

    t = √ (19.215)

    t = 4.38 s.

    (b)

    also using

    θ = ω₀t + 1/2αt² ... Equation 1

    Given: θ = 3.18 rev = 3.18*2π = 19.97 rad, ω₀ = 0 rad/s, α = 1.04 rad/s².

    Substitute into equation 1

    19.97 = 0 (t) + 1/2 (1.04) (t²)

    t² = 19.97*2/1.04

    t = √ (38.40)

    t = 6.197 s

    The time require = 6.197-4.38 = 1.817 s
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