a meterstick pivots freely from one end. if its released from a horizontal position, find its angular velocity when it passes through the vertical. treat the stick as a uniform thin road.
The easy way is to use conservation of energy.
At the vertical position, the potential energy decrease,
M g L/2,
equals the kinetic energy increase, which is
(1/2) I w^2 = (1/6)ML^2*w^2
gL/2 = L^2 w^2/6
w^2 = 3 g/L
w = sqrt(3g/L) = 5.42 rad/s
(since L = 1 meter)