Why isn't tension equal throughout?

Before rotation is introduced into the course, Atwood machines were considered to have equal tension throughout. In this particular problem however, the pulley did have a significant body which rotated about an axis. In captions the authors stated that the tensions may not be equal even though it is on the same rope, an indeed the tensions are different numerically.

Why can we assume equal tensions when the pulley is negligible but not when it is significant?

• I do not understand what you are asking here. What is "the course", what is "this particular problem", who are "the authors"? Oct 26, 2016 at 3:34
• That is besides the point, but the textbook contains two problems from different chapters. In the chapter of just translational forces, the Atwood machine had a pulley with no significance other than just to hold the two objects. Oct 26, 2016 at 3:42
• Continuation: In that case the tension of the string holding the blocks was considered equal. In the chapter of rotational dynamics, the authors asked about the acceleration of the system. To solve for this, one needs to not assume that the tension forces on the two blocks are equal, otherwise the answer would have been incorrect. My question is why can we assume equal tension forces for the negligible pulley but not for the one with rotational dynamics. After all, it is the same string. Oct 26, 2016 at 3:42

It is all a matter of progression.

When one starts studying Mechanics it is in terms of point masses, massless and inextensible strings, frictionless and massless pulleys, etc..
This is done to produce some foundations on which to build.
Strings and pulleys are just devices for transferring forces from one place to another and changing the directions of forces.

A step forward is to introduce rotational dynamics where bodies are no longer treated as point masses and pulleys suddenly become nearer to those that can be found in the real world.

It appears that you have taken such a step.

If you have a pulley of moment of inertia $I_{\rm C}$ about its centre of mass then to produce an angular acceleration $\alpha$ of the pulley requires a torque about the centre of mass of the pulley of $\tau = I_{\rm C} \;\alpha$.
Note that if the pulley is massless then it has no moment of inertia and so requires no torque to accelerate it.

The torque is applied using a string which can still be assumed massless and inextensible but now communicates with the pulley via a static frictional force.

This now means that the tension in the string is no longer constant as is shown in the solution to the capstan problem. (However note that there are no accelerations in the derivation).

You normally do not worry about the detail of the interaction between the pulley and the string what interests you is the fact that the tensions at either end of the string are difference and it is that difference which enable the string to apply a torque on the pulley.

In some cases you could assume that the moment of inertia of the pulley is so small as to make the very little difference to the tension in the string which is mirroring the assumption that the pulley is massless.

I wonder what assumptions (simplifications) NASA made when designing their pulley system?

For a rope to move on a pulley, the pulley needs to rotate. Now when we use an ideal pulley, the force required to rotate the pulley thought to be zero so there is no work done by the rope to rotate the pulley. When we use a real pulley with mass and a moment of inertia, we need to do some work to get the pulley to rotate and that usually comes through as friction from the rope. But because of newton's third law the reaction force of the friction on the rope can lower or raise the tension of the rope from one side to the other.