ω = 100 rev/min = 5/8 rev/s = 10π/3 rad/s
θ = 10 rev = 20 π 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 = α = ω2/2θ
Therefore the torque by which the wheel will come to an rest = Icm × α
