You probably remember the Atwood's machine, which is two masses connected by a rope over a pulley. Previously, we considered the pulley to be massless. Now let's take into account that the pulley has a mass and that the rope does not slip over the pulley. What is the acceleration? First we use Newton's second law to describe the motion of each mass. Let's assume that the mass on the left is going down and the mass on the right is going up. Choosing up to be in the positive direction and down to be negative, which equation corresponds to Newton's second law for m1 on the left? ... Which equation corresponds to Newton's second law for m2 on the right? Notice that because the pulley has mass, the tension is not the same throughout the rope as it was in the massless pulley case. Thus with Newton's second law for each mass, we see that there are two equations but three unknowns: the two different tensions and the acceleration. Another equation that can be applied in this situation is Newton's second law for rotation, which states the sum of the torques is equal to the moment of inertia times the angular acceleration. Looking closely at the pulley, the torque is provided by the rope on either side of the pulley. Since the torques act in the opposite direction, whichever one is largest will cause the wheel to spin one way or the other. Assuming the tension in the rope makes a ninety-degree angle with the pulley at the contact point, which expression gives the sum of the torques? We can relate the angular acceleration to the linear acceleration using which expression? What is the moment of inertia of the pulley, which you can take here to be a solid cylinder with mass M and radius r? We now have five equations that we can use to solve the problem. Substituting for the moment of inertia and the angular acceleration finally gives us three equations and three unknowns. We can use this system of equations to find the acceleration. What is the final answer for the acceleration? When the pulley has a mass the acceleration is lower than when it is massless since some of the energy now goes into the rotating pulley. The difference appears in the denominator, which now includes the pulley's mass divided by two.

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