# Torque on part of one-dimentional pivoting rigid rod

A one-dimentional rigid rod, at rest and initially horizontal, is performing a rotational motion around fixed point O. I painted the last part of length l/3 with different color to focus on that. Given that $$I_O = 1/3ml^2$$, at $$t = 0$$, rod OA has a rotational acceleration $$\theta ''= 3g/2l$$ and AD part a center of mass acceleration $$a_K = 5g/4$$. To have an acceleration greater than $$g$$ , OA part must exert a force F to AD at point A downwards. But then, the sum of torques acting at AD around center of mass K tend to rotate AD counterclockwise. How do you explain that? Before the rod OD is released from an equilibrium horizontal position, there is an upward force $$\frac13 mg$$ acting on AD which keeps it from falling down, together with an anticlockwise torque which keeps it from rotating about an imaginary joint at A. If the only external force on section AD is its weight then these forces and torques come from the section OA.