The flywheel of an old steam engine is a solid homogeneous metal disk of mass M

Question

The flywheel of an old steam engine is a solid homogeneous metal disk of mass M = 129 kg and radius R = 81.5 cm. The engine rotates the wheel at 483 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 = 103 N.

a) If the coefficient of kinetic friction between the pad and the flywheel is µk = 0.345, 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

mass M = 129 kg and radius R = 81.5 cm.
The engine rotates the wheel at 483 rpm.

=2*483 rad/min

=(2*483/60) rad/sec

=50.554 rad/sec

tangential speed v=R

=50.554*0.815 m/sec

=41.20151 mps

moment of inertia I= 0.5*M*R*R

=42.8425125

torque applied

=FR

=103*k*0.815=83.945*0.345 =28.961025

c)

work done by foce to bring wheel to rest is the initial kinetic energy of the wheel

since final kE=0

KE=0.5*I**

=54746.452747533

a,b) we have equations of motion

here as

=t ===>> t =/       here angular acceleration

=I =>>>

=/I =28.961025/42.8425125 =0.675988015

time taken to stop

t= / =50.554/0.675988015 = 74.78534956 sec

angle rotated

= t - (0.5)t2 equations of motion

= (50.554*74.78534956)-(0.5*0.675988015*74.78534956*74.78534956)

=1890.349280823 radians

no of rotations = /2 =301.011031978 rotations

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