Tension in whirling rope 
The question was to find the tension in a rope at a distance r from one end which is being rotated at a constant angular speed. Why is the tension at x=L zero?  Shouldn't the part of the rope towards the axis pull the rope which is at the end and thus provide tension? 
 A: For a small portion $\Delta r$ of the rope at the far end there is only one force acting, the tension force, $T$, from the rope on the 'inner' side of the portion. So applying Newton's second law to the portion (whose mass is $\frac{m}{L}\Delta r) $:$$T=\frac{m}{L}\Delta r\ L \omega^2=m\ \Delta r\ \omega^2$$ As $\Delta r$ approaches zero, $T$ must approach zero, because $m \omega^2$ is constant. 
A: I think the confusion lies in working in an accelerated frame. Your question, which is basically, "shouldn't the rope be pulling the end [in a circle] thereby providing tension" sounds right--you have to keep that end going round and round, which requires force, which means tension.
Now if we just consider a hanging rope in a uniform gravitational field, this concern doesn't happen. The end of the rope is holding no weight and thus has no tension. Of course, the maximum tensions is at the top, where the entire weight of the rope is suspended.
So how does the spinning rope differ from a rope in a non-uniform (linear) gravitational field? It doesn't. If we just consider that static case, you should be able to convince yourself that the tension at the end is zero.
A: I think this is more subtle than JEB's answer lets on.  Tension is a pulling force on an object, and a rope can be thought of as a bunch of objects (atoms) linked together by forces that we agree to call tension, even though we could name those forces if we wanted to get down into the weeds.  A simple model, in fact, of a string would be a series of equally spaced masses with massless strings between them. The forces keeping the particle spacing are what we have agreed to call tension (we could also call the forces keeping the particles themselves together tension, and in some ways this would be correct with our labeling of tension, but it might be easiest to think of these particles as unbreakable).
Each particle is being pulled by tension.  For both the spinning rope and the hanging rope example JEB gives, there MUST be a tension force acting on the the last particle or it will be lost and no longer be part of the rope. If you have built this rope made of masses and tension providing pieces, then the tension in the piece between the last two particles is and must be non-zero.  The tension at the absolute end of the rope, then, is the tension within a particle.  If we could spin the rope fast enough that the last particle could fall apart into two pieces, say, then even that "tension" is non-zero if the particle stays together.  At the very end, where no matter how hard you spin nothing else comes off of the rope, at that point the tension is zero as there is no force required to keep that part on.  You can think of it as a last connector in the simple model that has nothing attached at one end and is just flopping around.  It doesn't really make sense as a connection from something to nothing is really no connection, but the equation is right that it takes no tension if you want to attach nothing to the end of the rope and spin that nothing in a circle.   
