Physics
posted by Theresa .
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.

letting the zero PE system at A, then
initial PE=mga
final PE =mga cosAlphaMG*2a*CosAlpha
solve for cosAlpha, setting final=initialPE
CosAlpha= 1/(12)=1/3
where alpha is measured from the vertical downward, clockwise
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