The flywheel of a steam engine runs with a constant angular speed of 363 rev/min
ID: 1999686 • Letter: T
Question
The flywheel of a steam engine runs with a constant angular speed of 363 rev/min (in the counterclockwise direction). When steam is shut off, the friction of the bearings stops the wheel in 2.0 h.
a) How many revolutions does the wheel make before stopping?
b) At the instant the flywheel is turning at 75 rev/min, what is the tangential component of the linear acceleration of a flywheel particle that is 50 cm from the axis of rotation?
c) What is the magnitude of the net linear acceleration of the particle in (b)?
Explanation / Answer
(a) w = 363/60 rev /sec
no. of revolutions = average angular velocity*time = (363/(2*60))*2*3600 = 21780
(b) w=w0-ft
0=363*2pi/60-f*2*3600
f= 5.2796*10^-3 rad/sec^2
tangential acceleration = r*f = 0.5*5.2796*10^-3= 2.6398*10^-3 m/sec^2
we are assuming constant decelration; hence, 75rpmm will not affect tangentiasl acceleration.
(c) axial racceleration = rw^2 = 0.5*(2pi*75/60)^2= 30.8425 m/sec^2
net linear acceleration = sqrt[30.8425^2+(2.6398*10^-3)^2] = 30.8425 m/sec^2 Ans.