The flywheel of an old steam engine is a solid homogeneous metal disk of mass M
ID: 2261468 • Letter: T
Question
The flywheel of an old steam engine is a solid homogeneous metal disk of mass M = 131 kg and radius R = 80.5 cm. The engine rotates the wheel at 457 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 = 177 N.
a) If the coefficient of kinetic friction between the pad and the flywheel is ?k = 0.573, how many revolutions does the flywheel make before coming to rest?
b) How long does it take for the flywheel to come to rest?
c) Calculate the work done by the torque during this time.
Explanation / Answer
m = 131 kg, R = 0.805 m
w1 = 457 rpm = 457*2*pi/60 = 47.83 rad/s
a) torque on the disk, T = mue*F*R = 0.573*177*0.805 = 81.644 N.m
momenti fo inertia of the disk, I = 0.5*M*R^2 = 42.45 kg.m^2
we know, T = I*alfa
alfa = T/I = 81.644/42.45 = -1.92 rad/s^2
negative sign represents the disk is decelleratng.
w2^2-w1^2 = 2*alfa*theta
theta = (w2^2-w1^2)/(2*alfa)
here, w2 = 0.
theta = (0-47.83^2)/(-2*1.92) = 595.8 radians = 94.87 revolutions
b)
w2 = w1+alfa*t
0 = 47.83 - 1.92*t
==> t = 47.83/1.92 = 24.9 s
c)work done by torque, W = -T*theta = -81.644*595.8 = -48643.5 J