Suppose that the rod is held at rest horizontally and then released. (throughout the remainder of this problem, your answer may include the symbol i, the moment of inertia of the assembly, whether or not you have answered the first part correctly.) what is the angular acceleration α of the rod immediately after it is released?

α = 3/2 g / L

Explanation:

Let's analyze this exercise a bit, the bar is held at one end, so that we can turn, therefore, let's write Newton's second law for rotational movement

    τ = I α

Where ta is the torque, I the moment of inertia and angular acceleration. Let's calculate the torque, for which we set a reference system with the origin at the fixed end of the bar, in that case the only external force is the weight that is applied at the center of the bar, the forces at the point turning has zero distance to the origin

    τ = W  L/2

    W L/2 = I α

The moment of inertia of a bar attached at one end is

    I = 1/3 m L²

    m g L/2 = (1/3 m L²) α

     g/2 = 1/3 L α

    α = 3/2 g / L

This is the angular acceleration of the system.