Question

A solid, uniform cylinder with mass 8.25 kg and diameter 10.0 cm is spinning with angular velocity 245 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 0.344.

Answer 1

Hi buddy!

Given data :

Mass of cylinder = 8.45 kg

diameter of cylinder =10.0 cm

radius of cylinder = r= 5 cm =0.05 m

moment of inertia of solid cylinder = I =  

Initial angular velocity = wi

=, where f = frequency

=21.991 rad/s

initial rotational kinetic energy=

let the normal force be F

The coefficient of kinetic friction = 0.343

friction force = f = normal force =0.343 F

torque of friction =T =rf= 0.05*0.343 F=0.01715F Nm

angular displacement=theta=2pi N=2pi*5.25=32.9867 rad

work done by torque=T*angular displacement = T theta=0.01715F*32.9867=0.5663F

work done by torque=initial rotational kinetic energy=2.5389 J

0.5653F=2.5389

=4.4912 N

thus applied normal force be to bring the cylinder to rest is 4.4912 N