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 = 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

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