How to Handle Real Pulley in a Tension Problem

This is our old favorite - a couple masses tied together, string going over a pulley. In the good ol' days, the pulley was frictionless, and we simply could say that the tensions at the two ends of the (massless) string were the same, and set up F = ma for the system to get the acceleration. Worked out well.

But, in reality the pulley accelerates, too. How do we handle this? This means there has to be a net torque on the pulley in order to cause an angular acceleration. The only way for this to happen is if the tensions on the sides of the pulley, due to the hanging masses, are different. Here we see how to deal with this new situation, and apply F = ma on the two blocks, and torque = I*alpha on the pulley. We will assume there is no slipping between the string and the pulley, so we can relate linear motion of the blocks to the rotational motion of the pulley. Check it out.

How to do Rotational Motion for a rotating, falling bar - NON-constant acceleration

Check out an example of a NON-constant angular acceleration problem, where a bar starting in static equilibrium (up = down, cw = ccw) goes into non-equilibrium and accelerates. You can see how torque = I*(alpha) gives us the angular acceleration of the bar at any given angle it has rotated through, and also how to use rotational energy and energy conservation to determine the angular speed at any given angle.

One thing to keep in mind as far as linear acceleration and linear speed is that each point of the bar has different values for these quantities, as determined by a = R*alpha and v = R*omega. Check it out...