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1 July, 00:24

A flywheel accelerates for 5 seconds at 2 rad/s2 from a speed of 20 rpm. Determine the total number of revolutions of the flywheel during the period of its acceleration. a. 5.65 b. 8.43 c. 723 d. 6.86

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  1. 1 July, 04:00
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    option (a)

    Explanation:

    t = 5 sec, α = 2 rad/s^2, f0 = 20 rpm = 20 / 60 rps

    Use second equation of motion for rotational motion

    θ = ω0 x t + 1/2 α t^2

    θ = 2 x 3.14 x 5 x 20 / 60 + 0.5 x 2 x 5 x 5

    θ = 10.47 + 25 = 35.47 rad

    Number of revolution = 35.47 / (2 x 3.14) = 5.65
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