Starting from rest a centrifuge accelerates uniformly at the rate of 40 "rad"//s"^2. What is the number of revolutions made by it when its angular velocity becomes 1800 rev/min ?

Starting from rest a centrifuge accelerates uniformly at the rate of 4

1.	Starting from rest a centrifuge accelerates uniformly at the rate of 40 "rad"//s"^2. What is the number of revolutions made by it when its angular velocity becomes 1800 rev/min ?
Answer» Solution :`THETA = OMEGA^(2)//2alpha = (60pi)^(2)//2 xx 40 = 44.368 RAD, N = 443.68//2pi = 71 `rev