Can a massless rope accelerate?

Suppose I have an Atwood machine, that is, two different masses connected with an inextensible, massless rope over a pulley. Assuming no friction between the rope and the pulley, the heavier mass will accelerate towards the ground, the lighter mass will accelerate towards the pulley, and the rope will accelerate towards the heavier mass. These three accelerations will be equal in magnitude. But this makes no sense to me. Force causes acceleration. But there is no force acting on the rope. And even if there was, the acceleration of the rope would be infinite because its mass is 0. So why does the rope accelerate? And how can the magnitude of this acceleration be finite?

• Massless ropes don't exist. They're there just to make the calculations simpler. Why worry about calculations on the rope? The whole point of making them massless is so that you don't have to perform any calculations on them. – Brandon Enright Nov 23 '13 at 0:31
• Thank you, somehow it didn't occur to me that worrying about non-existent models may be a little bit pointless. – Armadillomon Nov 23 '13 at 0:53
• @Armadillomon Note that every model in physics is a "non-existent" model. An important part of physics is coming up with the simplest model that has the physics you are interested in, and making sure the simplifying approximations do not affect the answer. Physicists often ask themselves "is the exact answer to my approximate problem and approximate answer to the exact problem?" So you have a good intuition that you need to consider if the massless rope answer is consistent with the answers for very light ropes. (It is.) – Brian Moths Nov 23 '13 at 2:54

When the (inertial) mass is zero, then the acceleration can be non-zero for zero force.

This is similar, conceptually, to what has been discussed recently regarding an ideal conductor.

Consider Ohm's Law:

$$V = IR$$

Now, what if $R = 0$ as is the case with an ideal conductor?

Clearly, the voltage must be zero for any current. The current through the conductor, then, is determined by constraints external to the ideal wire, i.e., by whatever the ideal wire is connected to.

Consider Newton's 2nd Law:

$$F = ma$$

Now, what if $m= 0$ as is the case with the massless rope?

Clearly, the force must be zero for any acceleration. The acceleration, then, is determined by constraints external to the massless rope, e.g., the attached masses.

Yes, the massless rope is ideal and, thus, not physical but, there can be effectively massless ropes just as there can be effectively ideal conductors. Which is to say that, to the precision one is working to, the rope has zero mass and zero force acting on it but non-zero acceleration.

• What if the rope is in free fall? Does that mean it has an infinite acceleration? – grjj3 Dec 5 '13 at 17:51
• @grjj3, an object in free fall is inertial which is to say that an accelerometer attached to the object reads zero acceleration. From Wikipedia article "Free fall": A body in free fall experiences "0-g". – Alfred Centauri Dec 5 '13 at 18:28
• Inertial? Free fall is a motion under the influence of gravity only, meaning the object accelerates with acceleration $g$. It experiences "0-g" in the sense that it has no weight (that is, no normal force acting on it so it can actually feel itself heavy). – grjj3 Dec 6 '13 at 15:17
• @grjj3, a reference frame of a lab at rest on the Earth's surface is not inertial but is, in fact, an accelerated reference frame. An object free-falling towards the floor of the lab is inertial. This is all well known. Look up "equivalence principle". – Alfred Centauri Dec 7 '13 at 22:41
• @grjj3 An ideal rope wouldn't experience any force in a "free-fall" because there is no gravitational force acting on it because of it being massless. So, depending on the constraints, it can either be accelerating or not accelerating. However, an infinite acceleration doesn't make any sense. This is another way of saying that it is an inconsistent way of doing physics where you need to apply finite forces on massless objects. Edit: To clarify, I am answering your question in a purely Newtonian version of gravity whereas Alfred is using the more correct relativistic version of gravity. – Dvij Mankad Jul 15 at 12:45

I agree with Brandon Enright's comment. But even if there were massless ropes, if m=0 and F=0, then F=ma still would hold for any finite a.