The flywheel of a steam engine runs with a constant angular speed of 526 rev/min
ID: 1688907 • Letter: T
Question
The flywheel of a steam engine runs with a constant angular speed of 526 rev/min (in the counterclockwise direction). When steam is shut off, the friction of the bearings stops the wheel in 2.0 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 rev/min, what is the tangential component of the linear acceleration of a flywheel particle that is 50 cm from the axis of rotation?
(d) What is the magnitude of the net linear acceleration of the particle in (c)?
6 m/s2
Explanation / Answer
Initial angular speed w = 526 rev / min final angular speed w ' = 0 time t= 2 h = 120 min (a). from the realtion w ' = w + at angular accleration a = ( w ' - w ) / t = -4.383 rev / min^ 2 (b). No.of revolutions = [ w'^ 2- w^ 2] / 2a = 31560 rev (c).angular speed w " = 75 rev / min = 75 * 2*pi/ 60 rad / s = 7.85 rad / s distance of the particle from axis of rotation r = 50 cm =0.5 m Tangential component of accleration a ' = r a = 2.1915 m / s^ 2