Posted by **Theresa ** on Saturday, May 26, 2012 at 6:14am.

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.

- Physics -
**bobpursley**, Saturday, May 26, 2012 at 1:29pm
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

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