Question

A flywheel in a motor is spinning at 530 rpm when a power failure suddenly occurs. The flywheel has mass 40.0kg and diameter 75.0cm . The power is off for 39.0 s , and during this time the flywheel slows down uniformly due to friction in its axle bearings. During the time the power is off, the flywheel makes 250 complete revolutions. At what rate is the flywheel spinning when the power comes back on? How long after the beginning of the power failure would it have taken the flywheel to stop if the power had not come back on, and how many revolutions would the wheel have made during this time?

Answer 1

w = 25.05 rad / s , α = 0.7807 rad / s² , θ = 1972.75

Step-by-step 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