A flywheel in a motor is spinning at 590 rpm when a power failure suddenly occurs. The flywheel has mass 40.0 kg and diameter 75.0 cm . The

Question

A flywheel in a motor is spinning at 590 rpm when a power failure suddenly occurs. The flywheel has mass 40.0 kg and diameter 75.0 cm . The power is off for 30.0 s , and during this time the flywheel slows down uniformly due to friction in its axle bearings. During the time the power is off, the flywheel makes 230 complete revolutions.
a) At what rate is the flywheel spinning when the power comes back on?
b) How long after the beginning of the power failure would it have taken the flywheel to stop if the power had not come back on, and how many revolutions would the wheel have made during this time?

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Verity 3 years 2021-08-24T11:50:07+00:00 1 Answers 31 views 0

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    2021-08-24T11:51:32+00:00

    Answer:

    Explanation:

    Hello,

    Let’s get the data for this question before proceeding to solve the problems.

    Mass of flywheel = 40kg

    Speed of flywheel = 590rpm

    Diameter = 75cm , radius = diameter/ 2 = 75 / 2 = 37.5cm.

    Time = 30s = 0.5 min

    During the power off, the flywheel made 230 complete revolutions.

    ∇θ = [(ω₂ + ω₁) / 2] × t

    ∇θ = [(590 + ω₂) / 2] × 0.5

    But ∇θ = 230 revolutions

    ∇θ/t = (530 + ω₂) / 2

    230 / 0.5 = (530 + ω₂) / 2

    Solve for ω₂

    460 = 295 + 0.5ω₂

    ω₂ = 330rpm

    a)

    ω₂ = ω₁ + αt

    but α = ?

    α = (ω₂ – ω₁) / t

    α = (330 – 590) / 0.5

    α = -260 / 0.5

    α = -520rev/min

    b)

    ω₂ = ω₁ + αt

    0 = 590 +(-520)t

    520t = 590

    solve for t

    t = 590 / 520

    t = 1.13min

    60 seconds = 1min

    X seconds = 1.13min

    x = (60 × 1.13) / 1

    x = 68seconds

    ∇θ = [(ω₂ + ω₁) / 2] × t

    ∇θ = [(590 + 0) / 2] × 1.13

    ∇θ = 333.35 rev/min

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