The flywheel of an old steam engine is a solid homogeneous metal disk of mass M
ID: 1264718 • Letter: T
Question
The flywheel of an old steam engine is a solid homogeneous metal disk of mass M = 125 kg and radius R = 86.7 cm. The engine rotates the wheel at 477 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 = 181 N.
a) If the coefficient of kinetic friction between the pad and the flywheel is ?k = 0.445, how many revolutions does the flywheel make before coming to rest? Tries 0/99 b) How long does it take for the flywheel to come to rest? Tries 0/99
c) Calculate the work done by the torque during this time. Tries 0/99
Explanation / Answer
braking torque = 0.867 * 125 * 0.445 = 48.22 ;
I = mr^2 /2 = 125 * ( 86.7 e -2 ) ^2 / 2 = 46.9 ;
? = I ? ;
48.22 = 46.9 * ? ;
? = 1.028 rad /s ^2 ;
? = ?0 t - 1 /2 * 1.028 t^2 ;
?0 = 477 * 2 ? / 60 = 49.92 rad /s
?^2 - ?0^2 = 2 * ( -1.028 ) ? ;
0 ^2 - 49.92 ^2 = 2 * ( -1.028 ) ? ;
or ? = 1212.06 rad ,
2? rad = 1 rotation ; so
? - ?0 = ?t ;
0 - 49.92 = -1.029t ;
or t = 48.51 s = 48.51 / 60 = 0.8085 minutes <--- time to stop
rotations = 1212.06/ ( 2? ) = 193 rotations
work done = ? ? = - 48.22 * 1212.06 = - 58445.53 J