The instantaneous angular velocity of the rod is 0, so the centripetal acceleration of each point is 0, and there is only tangential acceleration.

The linear density σ=p/(gL) and the acceleration a=α. X at each point along the x direction, the inertia force dq = a.dm = α.x.σ. Dx of each point (so it can be seen that the inertia force is distributed along the triangle of the rod).

What is the resultant force of inertia force? q =∫(0-L)dQ =∫(0-L)α.σ. x . dx =(0-L)α.σ。 x^2/2=P.L.α/(2g)

If the inertia force is distributed along the triangle of the rod, the line of action of the resultant force should pass through the center of the triangle, so L'=2L/3.

* You can also use static and dynamic methods to formulate the equilibrium equation: ∑MA=0, Q.L'=P(L/2), or you can get:? L'=2L/3