A solid cylinder of mass 2 kg and radius 4 cm rotating about its axis at the rate of 3 rpm. The torque required to stop after 2π revolutions is

Given : Mass M=2 kg , Radius R=4 cm  Initial angular speed ω0=3rpm=3×2π60rad/s=π10rad/s  We know that, ω2=ω02+2αθ ⇒ 0=π102+2×α×2π×2π ⇒ α=-1800rad/s2  Moment of inertia of a solid cylinder,  I=MR22=2×410022=16104  Torque τ=Iα=16104×-1800=-2×10-6Nm