A0.250-kg wooden rod is 1.35 m long and pivots at one end. it is held horizontally and then released. part a what is the angular acceleration of the rod after it is released?

We assume that the rod's weight is evenly distributed, making its center of gravity 0.675 m from the end.
First, we calculate the moment present on the rod:
τ = F*d
τ = m*g*d
τ = 0.25 * 9.81 * 0.675
τ = 1.66

Next, in the case of rotational motion, Newton's second law is:
τ = Iα, where I is moment of inertia and α is the angular acceleration
The moment of inertia for a rod is:
I  = (ML²)/12
I = (0.25*1.35²)/12
I = 0.038

Now, we use the formula given by Newton's law:
α = τ / I
α = 1.66 / 0.038
α = 43.7 rad/s²

The angular acceleration is 43.7 radians per seconds squared.

Tons?