Question

Constants Part A A solid, uniform cylinder with mass 8.30 kg and diameter 19.0 cm is spinning with angular velocity 205 rpm on a thin, frictionless axle that passes along the cylinder axis. You design a simple friction- brake to stop the cylinder by pressing the brake against the outer rim with a normal force. The coefficient of kinetic friction between the brake and rim is 0.345. What must the applied normal force be to bring the cylinder to rest after it has turned through 5.50 revolutions? V AED + R o 2 ? Submit Request Answer

Answer 1

I= {mr? R-D 196M = 9.5cm = 0.095 m I=(+) (8:30kg) (o095 m)? I= 0.0375 kgm² Wo2 = 205 rpm = (205)( &TT yad) -21.5 rad/s Wzzo, 0-0,25.50 rev=16:50)(27) rad B = 3406 rad we want (0-0) 10-06) (2) (3466 rad) &_=-668 rad/s2 I 2-R =- MER z = Ida -Id (00375457) - MhR=Idz nur = -(0.0375kg m2) 6-6-687ad/s3- 764 N. = (0-345 )(0.095m)