Asked by Bill
During take-off the angular velocity of the blades of a revving engine of engine of an air craft change from 350rev/min to 570rev/min in a time interval of 15minutes. Determine
(1).uniform angular acceleration in rad/sec
(2).angular displacement during the time interval
(1).uniform angular acceleration in rad/sec
(2).angular displacement during the time interval
Answers
Answered by
Damon
350 rev/min (2 pi rad/rev)(1min/60s) = 36.65 rad/s
570 rev/min (2 pi rad/rev)(1min/60s) = 59.69 rad/s
omega 2- omega 1 = 23.04 rad/s
delta t = 15*60 = 900 s
delta omega/delta t = alpha = 23.04/900 = .0256 rad/s^2
theta = average omega * t
= [ (36.65+59.69)/2 ] * 900 = 48.17 * 900 = 43,353 rad
570 rev/min (2 pi rad/rev)(1min/60s) = 59.69 rad/s
omega 2- omega 1 = 23.04 rad/s
delta t = 15*60 = 900 s
delta omega/delta t = alpha = 23.04/900 = .0256 rad/s^2
theta = average omega * t
= [ (36.65+59.69)/2 ] * 900 = 48.17 * 900 = 43,353 rad
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