The flywheel of a steam engine runs with a constant angular velocity of 110 rev/
ID: 1459247 • Letter: T
Question
The flywheel of a steam engine runs with a constant angular velocity of 110 rev/min. When steam is shut off, the friction of the bearings and of the air stops the wheel n = 2.9 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 55.0 rev/mm, what is the tangential component of the linear acceleration of a flywheel particle that is 30 cm from the axis of rotation? (d) What is the magnitude of the net linear acceleration of the particle in (c)?Explanation / Answer
(a)
= 110 rev/min
o = 0
t = 2.9 * 60 = 174 min
o = + *t
= - 110 / 174
= - 0.632 rev/min^2
(b)
= * t + 1/2 * *t^2
= 110*174 - 1/2 * 0.632*174^2
= 9572 rev
(c)
Tangential Component of Linear Acceleration = * R
Angular Acceleration = 0.632 rev/min^2 = (0.632 * 2 *pi)/ 60^2 rad/s^2
T = (0.632 * 2 *pi)/ 60^2 rad/s^2 * 30 * 10^-2 m
T = 3.31 * 10^-4 m/s^2
(d)
Net Linear acceleration = sqrt(T^2 + r^2 )
The radial component of acceleration, ar = v^2/r
ar = w^2 * r
Angular Velocity , w = 55.0 rev/min
w = (55 * 2 * pi)/60 rad/s
w = 5.76 rad/s
ar = w^2 * r
ar = 5.76^2 * 30 * 10^-2 m/s^2
r = 9.95 m/s^2
atotal = sqrt(9.95^2 + (3.31 * 10^-4)^2 )
atotal = 9.95 m/s^2
Magnitude of Net Linear acceleration, = 9.95 m/s^2