A 5m ladder is leaning against a wall. If the bottom of the ladder is pulled along the ground away from the wall at a constant rate of 0.40m/s, how fast will the top of the ladder be moving down the wall when its bottom is 3m away from the wall?(This is a version of the problem I found today at math.stackexchange.)
The problem is straightforward as stated (the answer is 0.3 m/s), but my colleague wanted to know what I thought, as a physicist, about the fact that the speed of the top point of the ladder exceeds any finite value as this point nears the ground. I can see from a different math.stackexchange page that this is a frequently asked question, so I thought I'd post some thoughts about it here.
I added a reversed phase of the motion just because I think it aids the eye to see the "bounce."
The equations I used to generate this figure were
x(t) = t
y(t) = (1 − t2)½
0 ≤ t ≤ 1
These equations are the non-dimensionalized versions of
X(T) = vT
Y(T) = (L2 - v2T2)½
0 ≤ T ≤ L/v
The functions x(t) and y(t) are differentiable on the interval (0, 1), which means that points F and W both have well-defined speeds during this interval. However, y(t) has no derivative at t = 1, not even a left-sided one; this means that the speed of point W doesn't exist at the moment of time when the ladder is horizontal.
People usually say that the speed of point W "becomes infinite" at t = 1…but if "the speed of point W" means |dy/dt|, then saying that the speed is infinite is saying that dy/dt = ∞, and I used to discourage my beginning Calculus students from writing such things, because I wanted them to appreciate that dy/dt is defined by a limit, and limits are numbers that either exist or don't.
(It's certainly true that the speed of point W exceeds any finite value as t approaches 1, and that's enough to justify asking for a physicist's take.)
The coordinates of the CM point are ½(x(t), y(t)). As the ladder moves, the CM point traces a circle of radius ½. This isn't uniform circular motion, however, because the CM point isn't moving at constant speed.
The next animation shows the horizontal and vertical locations of the CM point.
The CM point has zero horizontal acceleration during the motion. But the CM point has a downward vertical acceleration that increases without bound throughout the motion.
From the Second Law, Fnet, cm = Mtotacm, we conclude that during the time when the ladder is dropping, there is zero net horizontal force on the ladder, while the net vertical force on the ladder is downward, increasing without bound as the ladder reaches the ground.
So I think the main takeaway from physics is that no finite force is capable of making a ladder move this way.
To analyze the problem any further, I would want to specify the coupling between the ladder and the wall and floor. For instance, on the wall there could be a freely sliding ring, allowing point W to move vertically without resistance, but constraining point W so that it can't move horizontally. That version of the problem is considered in this article, again leading to the conclusion that the constraint forces cannot be finite. This had to happen, given the motion of the CM point during the stipulated motion.
The ladder problem reminds me of a class of problems treated by philosophers of science, called supertasks. A supertask is an infinite sequence of operations conducted in a finite amount of time. One question is whether supertasks ever happen, or are ever possible. Zeno's Paradox could be thought of as a question about whether motion itself is a supertask. You can read about supertasks on Wikipedia.
After Zeno, the most famous of a supertask is the example of Thomson's lamp; here is a paper about that paradox by John Earman and John D. Norton. I learned about some of Norton's work while writing my paper on the Law of Inertia and determinism in Newton's Laws.