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Title: A pulley is accelerated uniformly from rest at a rate of 8 rad/s 2 . After 20 s the acceleration ... Post by: cloveb on Mar 21, 2017 A pulley is accelerated uniformly from rest at a rate of 8 rad/s 2 . After 20 s the acceleration stops
and the pulley runs at constant speed for 2 min, and then the pulley comes uniformly to rest after a further 40 s. Calculate: (a) the angular velocity after the period of acceleration, (b) the deceleration, (c) the total number of revolutions made by the pulley. Title: Re: A pulley is accelerated uniformly from rest at a rate of 8 rad/s 2 . After 20 s the acceleration ... Post by: bolbol on Mar 23, 2017 (a) Angular velocity after acceleration period,ω2 = ω1+ αt = 0 + (8)(20) = 160 rad/s
(b) ω3 = ω2 + αt from which, angular acceleration, α = (ω3 - ω2)/t = (0 - 160)/40 = - 4 rad/s2 i.e. angular deceleration is 4 rad/s2 (c) Initial angle turned through, θ1 = [(ω1 + ω2)/2]*t = [(0 + 160)/2]*(20) = 1600 rad =1600/2π rev At constant speed, angle turned through, θ2 = 160 rad/s × (2 × 60)s = 19200 rad = 19200/2π rev Angle turned through during deceleration, θ3 = [(160+ 0)/2]*(40) = 3200 rad = 3200/2π rev Hence, total number of revolutions made by the pulley = θ1+ θ2 + θ3 =1600/2π +19200/2π + 3200/2π =24000/2π = 12000/π rev or 3820 rev |