question 28 of chapter 10 rotation and rolling...///////////////////////////////

 = 100 rev/min = 5/8 rev/s = 10/3 rad/s
 = 10 rev = 20  rad, r = 0.2 m
After 10 revolutions the wheel will come to rest by a tangential force
Therefore the angular deacceleration produced by the force =  = 2/2
Therefore the torque by which the wheel will come to an rest = Icm × 
 F × r = Icm ×   F × 0.2 = 1/2 mr2 × [(10/3)2 / (2 × 20)]
 F = 1/2 × 10 × 0.2 × 100 2 / (9 × 2 × 20)
= 5 / 18 = 15.7/18 = 0.87 N.