The flywheel of an old steam engine is a solid homogeneous metal disk of mass M
ID: 1483614 • Letter: T
Question
The flywheel of an old steam engine is a solid homogeneous metal disk of mass M = 112 kg and radius R = 80 cm. The engine rotates the wheel at 590 rpm. In an emergency, to bring the engine to a stop, the flywheel is disengaged from the engine and a brake pad is applied at the edge to provide a radially inward force F = 135 N. If the coefficient of kinetic friction between the pad and the flywheel is mu_k = 0.2, how many revolutions does the flywheel make before coming to rest? How long does it take the flywheel to come to rest? Calculate the work done by the torque during this time.Explanation / Answer
a)
Moment of Inertia, I = 0.5*M*R^2
= 0.5*112*(0.8)^2
= 35.84 Kg.m^2
wi = 590 rpm
= 590*2*pi /60 rad/s
= 61.78 rad/s
Frictional force = miuk*Normal force
= 0.2*135
= 27 N
Torque due to friction = I* a
-F*R = I*a
-27*0.8 = 35.84 * a
a = - 0.603 rad/s^2
This is the angular acceleration that brings it to rest
wf = 0 rad/s since it comes to rest
use:
wf^2 = wi^2 + 2*a*d
0 = (61.78)^2 + 2* (-0.603)*d
d= 3164.8 rad
= (3164.8 ) / 2pi revolutions
= 503.7 revolutions
b)
use:
wf = wi + a*t
0 =61.78 + (-0.603)*t
t= 102.5 s
c)
Work done = T*d
= (-F*R*)*d
= (-27 N*0.8 m)*3164.8 rad
= 68360 J