A heavy uniform bar is carried by two men on their shoulders. The weight of the bar is W If one man lets it fall from the end carried try him, the weight experienced by the other at the instant the rod is falling is

See fig.First of all we consider the acceleration of C.M. Here, the bar turns about A with an angular acceleration α after withdrawing the support of one man.Hence Iα=Wl2 or  α=Wl2IFor a bar,  I=Ml2/3∴α=mgl×32×Ml2=3g2lThe linear acceleration of C.M.a=l2×α=l2×3g2l=3g4From figure, W - R = m amg−R=M×(3g/4)R=mg4=W4