A uniform rod AB of length 2a and mass M is freely pivoted at A and is held with B vertically above A. It is then allowed to fall and when B is vertically below A it strikes a stationary particle , also of mass M,which sticks to the rod at B. The rod then turns through a further angle alpha before coming to rest. Find alpha.

Answer 1

letting the zero PE system at A, then
initial PE=mga
final PE =-mga cosAlpha-MG*2a*CosAlpha

solve for cosAlpha, setting final=initialPE

CosAlpha= 1/(-1-2)=-1/3

where alpha is measured from the vertical downward, clockwise