The flywheel of an old steam engine is a solid homogeneous metal disk of mass M
ID: 2007984 • 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
braking torque = 0.8 * 130 * 0.2 = 20.8 ;
I = mr^2 /2 = 117 * ( 80 e -2 ) ^2 / 2 = 37.44 ;
= I ;
20.8 = 37.44 * ;
= 0.5555 rad /s ^2 ;
= 0 t - 1 /2 * 0.5555 t^2 ;
0 = 540 * 2 / 60 = 56.54 rad /s
^2 - 0^2 = 2 * ( -0.555 ) ;
0 ^2 - 56.54 ^2 = 2 * ( -0.555 ) ;
or = 2880.85 rad ,
2 rad = 1 rotation ; so
- 0 = t ;
0 - 56.54 = -0.555 t ;
or t = 101.873 s = 101.873 / 60 = 1.697 minutes <--- time to stop
rotations = 2880.85 / ( 2 ) = 458.50 rotations
work done = = - 20.8 * 2880.85 = - 59921.68 J