Question

The flywheel of a steam engine runs with a constant angular velocity of 150 rev/min. When steam is shut off, the friction of the bearings and of the air stops the wheel in 1.9 h. (a) What is the constant angular acceleration, in revolutions per minute-squared, of the wheel during the slowdown? (b) How many revolutions does the wheel make before stopping? (c) At the instant the flywheel is turning at 75.0 rev/min, what is the tangential component of the linear acceleration of a flywheel particle that is 33 cm from the axis of rotation? (d) What is the magnitude of the net linear acceleration of the particle in (c)?

Answer 1

a)

Initial angualar speed ?₁ = 150 rev/ minute
Final angular speed ?₂= 0
Time = 1.9 hours.
Angualar accleration =? = ?? / ?t = 0 -150 / 114 min = -1.31 rev/ minute. = -1.31/60 = 0.022 rev/s^2
? = 2?*0.022 = 0.137 rad/s^2

b)

Initial angualar speed ?₁ = 150 rev/ minute
Final angular speed ?₂= 0  

Average angular speed = 150 /2 = Total no of revolutions /Time
Total no of revolutions = 150 /2 * time = [150 /2]* 114 = 8550 revoluttons.

c)

Linear acceleration a = r ? = 0.33*0.137 = 0.045 m/s^2

d)

Normal accelertion = r?^2
net acceleration =? [(r?^2)^2 + (r ?)^2] = r?[?^4 + ?^2]
? = 75 rev/min = (2?*75 / 60) = 7.85 rad /sec
0.33*? {(7.85}⁴ + (0.045)² = 20.33 m/s^2