A 0.250-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.