Question

3. After leaving a helicopter, you notice the rotors motion changes from 300 rev/min to 225 rev/min in 1 minute. Al find the average acceleratio this period. B) Assuming the acceleration remains constant, how long will it take the rotor to stop from 225 rev/min? And C) how many revolutions will the rotor undergo while stopping from 225 rev/min?

Answer 1

The angular acceleration and revolution

wo= 300 əv min W, = 225 der min aj Arena Angulars Acceleration a=W-wo 2= 225-360 = - 75 dev & ming 2 x = - -7sav x 25 sad x (min) ² min) 2 xv (6o seej 2 2= -0.13/_dad. seca at W2=0 then wz= c + Lt o= 225or- 75 av I min min) t = 225 = 3 min 75 t = 3 min = 180sec

o = Witt L &t? 0 = 225x3 - 4X75X6332 o = 675 - 33703 lo = 337.5 ev /