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8 October, 14:26

A flywheel turns through 24 rev as it slows from an angular speed of 1.2 rad/s to a stop.

(a) Assuming a constant angular acceleration, find the time for it to come to rest. s (b) What is its angular acceleration? rad/s² (c) How much time is required for it to complete the first 12 of the 24 revolutions? s

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  1. 8 October, 14:50
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    a) 251.36 s b) - 4.77 * 10⁻³ rad/s² c) 73.64 s

    Explanation:

    Using equation of angular motion

    θ = (ω₀ + ω₁) t/2

    2θ = (ω₁ + ω₀) t where ω₀ is the initial angular speed = 1.2 rad/s, ω₁ is the final angular speed = 0 and t is the time

    2 * 24 * 2π = (0 + 1.2) t

    t = 301.632 / 1.2 = 251.36 s

    b) the angular acceleration can be calculated by:

    ω₁ = ω₀ + αt

    ω₁ - ω₀ = αt

    0 - 1.2 = 251.36 α

    α = - 1.2 / 251.36 = - 4.77 * 10⁻³ rad/s²

    c) the amount of time required to complete the first 12 of the 24 revolutions

    θ = ω₀t + 0.5αt²

    12 * 2π = 1.2t - 0.00239t²

    75.408 = 1.2t - 0.00239t²

    - 0.00239t² + 1.2t - 75.408 = 0

    multiply by - 1 both sides

    0.00239t² - 1.2t + 75.408 = 0

    using quadratic formula to solve for t

    -b ±√ (b² - 4ac) / 2a

    1.2 ±√ (1.44 - 0.721) / (2*0.00239)

    1.2 ± √0.719 / 0.00478

    1.2 - 0.848 / 0.00478 or 1.2 + 0.848 / 0.00478

    0.352 / 0.00478 or 2.048 / 0.00478

    t = 73.22 s or 428.5

    t = 73.64 s
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