Posted by roni on Saturday, November 27, 2010 at 6:14pm.
The initial kinetic energy of the bar, which is
(1/2) I wo^2 = (1/2)*M (L^2/3)*(vo/L)^2
= (1/6)*M vo^2
Must equal the increase in potential energy in the upside-down position, which is
M g L
since the center of mass raises by an amount L.
I is the moment of inertia of the rod about the pivot at the top.
M cancels out and you are left with
(1/6)vo^2 = g*L
Solve for vo
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