The flywheel of an old steam engine is a solid homogeneous metal disk of mass M
ID: 2007801 • Letter: T
Question
The flywheel of an old steam engine is a solid homogeneous metal disk of mass M = 117 kgand radius R = 80 cm.The engine rotates the wheel at 540 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?____ revolutions
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 = 540 rpm
= 540*2/60 rad/s
= 56.52 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 = -127.28Nm
[MR^2/2] = -127.28 Nm
The angular acceleration of the wheel is = -3.39 rad/s^2
Using the equations of circular motion
f^2-i^2 = 2(f-i)
-(56.22)^2 = 2*-3.39*(f-i)
(f-i) = 466.1 rad
(f-i) = 74.21 revolutions
f = i+t
0 = 56.52 + (-3.39)t
t = 16.67s
The work done by the torque in this time is
W = (f-i)
W = 127.28*466.1 rad
W = 5.93*10^4 J