PhysicsLAB Resource Lesson
Rotational Dynamics: Pulleys

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We are now ready to put torque and an object's rotational inertia together into a formulation for rotational dynamics:
 
net F = ma
net τ = Iα
 
We will begin our investigation with pulleys and yo-yo's. Note that we will also be incorporating our knowledge of freebody diagrams, systems of equations, as well as both linear and angular kinematics. Where we previously assumed that our pulleys were "massless and frictionless," we will now allow our pulleys to have mass, and hence rotational inertia. In most cases our pulleys will still be frictionless; that is, there will be no loss of energy to heat.
 
Our pulley's inertia will be evidenced by the fact that it will now rotate as the rope moves across its circumference. We will no longer be able to assume that the tension on both sides of the rope are equal. Our ropes no longer "slip across the pulley" but "grab and rotate the pulley." Our ropes could conceivably "snap" if their tensile strength was insufficient to support the motion of the weights hanging from their ends.
 
Since the rope rotates without slipping along the edge of the pulley, we can use our equations for s = rθ, v = rω, and a = rα to relate the motion of a point along the rim of the pulley (aka, the rope and subsequently the masses on the end of the rope) to the pulley's rotational motion.
 
 
Stationary Pulleys - Pure Rotational Motion
 
Now consider the Atwood machine shown at the right. It is an example of pure rotational motion; that is, the center of gravity of the pulley does not translate up/down or to the left/right.
 
The equations of motion for an Atwood machine that has a pulley with rotational inertia are:
For the smaller mass:
net F = ma
T1 - mg = ma

For the larger mass:
net F = ma
Mg - T2 = Ma

For the pulley:
net τ = ICMα
(T2 - T1)r = Iα
In this case net torque is calculated by finding the product of (T2 - T1)r where T2 is the tension in the direction of the pulley's rotation (towards the larger mass) and T1 is the tension in the cord on the other side of the pulley. The radius of the pulley is r. Once again, remember that T1 and T2 can only be equal if the pulley is "massless and frictionless."
 
The equation a = rα will allow you to relate the linear acceleration of the hanging masses with the angular acceleration of the pulley.
 
 
Falling Yo-Yo's - Rotational and Translational Motion
 
Now consider a "perfect" yo-yo which is released from rest. This will be an example of both rotary motion and translational motion.
 
The equations of motion for this simple yo-yo as it begins falling are:
 
net F = ma
mg - T = ma

net τ = Iα
Tr = Iα
 
Note that the weight of the yo-yo does not produce a torque because its line of action passes through the pivot point, or center of rotation. These two equations, along with the relationship a = rα, will allow you to solve for the pulley's angular acceleration.
 
The yo-yo will fall straight down since there are only vertical forces acting on its mass.
 



 
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