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by parallel axes theorem the moment of inertia of the rod about the hinge is mL2/12 + m(L/4)2 = 7mL2/48.
when the rod reaches the vertical position the center of mass of the rod falls by L/4.
by conservation of energy we can write:
mgL/4 = (1/2)*I*w2 where I = moment of inertia = 7mL2/48 and w = angular velocity of the rod at the vertical position
this gives
w = (24g/7L)1/2.
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