Answer:
w = 25.05 rad / s
, α = 0.7807 rad / s²
, θ = 1972.75
Explanation:
This is a kinematic rotation exercise, let's start by looking for the acceleration when the engine is off
θ = w₀ t - ½ α t²
α = (w₀t - θ) 2/t²
let's reduce the magnitudes to the SI system
w₀ = 530 rev / min (2pi rad / 1 rev) (1 min / 60 s) = 55.5 rad / s
θ = 250 rev (2pi rad / 1 rev) = 1570.8 rad
let's calculate the angular acceleration
α = (55.5 39 - 1570.8) 2/39²
α = 0.7807 rad / s²
having the acceleration we can calculate the final speed
w = w₀ - ∝ t
w = 55.5 - 0.7807 39
w = 25.05 rad / s
the time to stop w = 0
0 = wo - alpha t
t = wo / alpha
t = 55.5 / 0.7807
t = 71.09 s
the angle traveled
w² = w₀⁹ - 2 α θ
w = 0
θ = w₀² / 2α
let's calculate
θ = 55.5 2 / (2 0.7807)
θ = 1972.75
