A rope is wound around a hollow cylinder of mass 3 kg and radius 40 cm. What is the angular acceleration of the cylinder if the rope is pulled with a force of 30 N ?

m=3 kg,r=40 cm=40×10-2 m,F=30 NMoment of inertia of hollow cylinder about its axis =mr2=3 kg×(0.4)2 m2=0.48 kg m2The torque is given by, τ=Iαwhere I =  moment of inertia, α= angular acceleration  In the given case, τ=rF, as the force is acting  perpendicularly to the radial vector. ∴  α=τI=Frmr2=Fmr=303×40×10-2=30×1003×40 α=25rads-2