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 = 85.3 cm. The engine rotates the wheel at 417 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 = 193 N.

a) If the coefficient of kinetic friction between the pad and the flywheel is ?k = 0.231, 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

a)F=193N

So

=193.706kJf(friction)=0.231F=44.583N

Now torque about center

T=I*(ang.acc)

So

44.583*R=(MR^2/2)*(|ang.acc|)

So

|ang.acc|=0.81033 rad/s^2 (this is actually retardation)

So

v=u+at

So

0=417*2pi/60-0.81033t

So

t=53.89 sec (time taken to stop)

v^2-u^2=2as

So

s=angle=(417*2pi/60)^2/(2*0.81033)=1176.623 radians =187.265 rotations

W=F*r*s=44.583*0.853*(187.265*2pi)=44.746kJ

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