Aflywheel 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

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

diy

explanation: