The flywheel of an old steam engine is a solid homogeneous metal disk of mass M
ID: 1696347 • Letter: T
Question
The flywheel of an old steam engine is a solid homogeneous metal disk of mass M = 118 kgand radius R = 80 cm.The engine rotates the wheel at 460 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 = 130 N.If the coefficient of kinetic friction between the pad and the flywheel is µk = 0.2,how many revolutions does the flywheel make before coming to rest?How long does it take for the flywheel to come to rest?
Calculate the work done by the torque during this time.
Explanation / Answer
The angular speed of the wheel is i = 460 rpm
= 460*2/60 rad/s
= 15.33 rad/s
The radially inward force F = 130 N
This will cause inward torque given by in = 130 N * 0.8 m
in = 104 Nm
The frictional torque is given by f =( mu)mg
= 0.2*118 kg * 9.8 m/s^2
= 231.28 Nm
The net torque acting on the wheel is = in - f
I = -127.28 Nm
[MR^2/2] = -127.28 Nm
The angular acceleration of the wheel is = -3.37 rad/s^2
Using the equations of circular motion
f^2-i^2 = 2(f-i)
-(15.33)^2 = 2*-3.37*(f-i)
(f-i) = 343.78 rad
(f-i) = 54.74 revolutions
f = i+t
0 = 15.33 + (-3.37)t
t = 4.55 s
The work done by the torque in this time is
W = (f-i)
W = 127.28*343.78
W = 4.375*10^4 J