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Suppose we have a mass $m$ dangling from the ceiling on a vertical rope of length $\ell$ with uniform mass density $\lambda$ per unit length. The weight of the mass $mg$ is balanced exactly by the tension in the rope.

Now, suppose the rope is cut at the top of the ceiling. After a moment, the tension in the rope will be (pretty much) zero throughout and the mass will be in free fall. But, presumably the process is actually continuous, and over some period of time the tension in the rope will decrease from its initial value $T(y)$ (depending on the distance $y$ from the ceiling). How does $T(y)$ evolve over time?

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  • $\begingroup$ The drop is extremely fast, but it would probably have a gradient close to infinity $\endgroup$ – QuIcKmAtHs Dec 29 '17 at 5:47
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    $\begingroup$ Might be of interest: physics.stackexchange.com/q/56833 the behavior will be the same as for this spring / slinky, only way faster due to the much smaller elasticity $\endgroup$ – Steeven Dec 29 '17 at 7:38
  • $\begingroup$ @Steeven: Your comment is a much better answer than all the answers that have been posted. $\endgroup$ – Peter Shor Dec 29 '17 at 13:10
  • $\begingroup$ @Steeven Faster and with a lot less total displacement; so it won't look as cool. $\endgroup$ – JMac Dec 29 '17 at 13:23
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But, presumably the process is actually continuous, and over some period of time the tension in the rope will decrease from its initial value $T(y)$ (depending on the distance $y$ from the ceiling). How does $T(y)$ evolve over time?

A single value changing over time won't be a useful model. In a static or slowly evolving situation, we can model the string as massless and perfectly rigid. In this case, a single value for $T$ throughout the item is reasonable. If you continue to use this ideal model when the rope is cut, then we would consider the tension goes to zero immediately.

But if this model is insufficient, then assuming it has a single $T$ throughout is also insufficient. Instead, changes in the forces on the rope propagate from one part of the string to another at a finite speed (often very close to the speed of sound in the material). In your example, if the rope is light, then immediately after the cut, regions of the rope near the cut will have a tension near zero, while regions far from the cut will have a tension equal to $T$. Rather than a single value in the material smoothly changing over time, different portions will differ dramatically.

A sensor connected to the rope at the other end would see the force from the rope drop rapidly to zero, just later than when it was cut. The lighter the rope, the more rapid the drop.

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Let's suppose for simplicity that the cut happens instantaneously and that the rope remains straight after it is cut.

The magnitude of the tension felt by the mass is initially $mg$, since it has to balance the force due to gravity. If the rope is cut at time $0$, the information that the rupture has happened will start to travel towards the mass at velocity $v_s$, where $v_s$ is the speed of sound (longitudinal wave) in the material the rope is made of, and will reach the mass at time $t(y)=y/v_s$, where $y$ is the distance from the cut.

As a reference, the speed of a sound in nylon is $1070$ m/s according to this page. Therefore if $y=1$ m the mass will "notice" that the rupture has happened at time

$$t(1 \text{m}) = \frac{1}{1070} \text{s} \approx 10^{-3} s$$

So, from the point of view of a human, the rupture will happen almost immediately.

In this ideal case, the tension will drop instantaneously from $mg$ to $0$ at time $t(y)$. In a real case, however, the drop will happen in a finite time.

Take also a look at this plot of tension vs time during the rupture of a climbing rope, taken from this document from UIAA:

enter image description here

(On the $y$ axis the unit is $10 \cdot$ N, don't know about the horizontal axis but I'm guessing $10^{-3}$ s...these people should really learn to label their axes though!)

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  • $\begingroup$ The speed of sound? Why? You need the speed of an elastic wave. Why is that equal to the speed of sound? $\endgroup$ – Peter Shor Dec 29 '17 at 13:19
  • $\begingroup$ @PeterShor The difference is only in the terminology. Longitudinal elastic waves in a solid medium are sound waves, even if they din't always produce an audible sound. See here: farside.ph.utexas.edu/teaching/315/Waves/node29.html $\endgroup$ – valerio Dec 29 '17 at 13:42
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    $\begingroup$ @PeterShor Because the speed of sound represents any propagation of a signal through a material. You hear the sound from a knock on the other side, when the force has propagated all the way through. Sound is after all the propagated push from one particle to the other. $\endgroup$ – Steeven Dec 29 '17 at 14:23
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    $\begingroup$ But a rope isn't a single solid material; it's made up of a whole bunch of smaller strands twisted around each other. I'm sure that in a Slinky, the speed of sound is different than the speed of longitudinal elastic waves. Why shouldn't the same thing be true of a rope? $\endgroup$ – Peter Shor Dec 29 '17 at 14:28
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Consider a mass is attached to a vertical spring, dangling from the ceiling. If you suddenly detach the spring from the ceiling, the spring starts oscillating close to SH

Now, in the case of rope, because of presence of some elasticity of the rope, similar motion may be seen. Unlike spring, near SHM motion drys out very fast and practically, you don't see any motion.

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  • $\begingroup$ "Consider a mass is attached to a vertical spring, dangling from the ceiling. If you suddenly detach the spring from the ceiling, the spring starts oscillating close to SH" It would have SHM if you displaced it from equilibrium and left it attached to the ceiling. If you detached it; I'm not sure how much SHM you'd get (especially since you're now also dealing with the entire spring-mass system in an accelerating frame) (example) $\endgroup$ – JMac Dec 29 '17 at 13:22
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It would be very very hard to calculate this function.

Suppose that you're cutting the rope with a scissor and during that time, the rope will look like this:

enter image description here

Points $A$ and $B$ will have different mass densities per unit length as $B$ would be more stretched than $A$. This implies that each vertical section of the rope will have a different mass density and obviously, a different tension which would altogether make it more harder to calculate that function.

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  • $\begingroup$ "It will be very, very hard to calculate this function." That is not an answer, it's merely an opinion. If you want to say that it really is hard in a real answer, you need to find a reference that says this can't be done. What you actually need to do is write down a differential equation and solve it. I wouldn't call that "very, very hard". $\endgroup$ – Peter Shor Dec 29 '17 at 13:18
  • $\begingroup$ @PeterShor . I get your point. I'll try not to do it next time. $\endgroup$ – Aaryan Dewan Dec 29 '17 at 13:19
  • $\begingroup$ @PeterShor , perhaps it would be cool if you could help us out a little bit by writing down those differential equations yourself in form of an answer? $\endgroup$ – Aaryan Dewan Dec 29 '17 at 13:27
  • $\begingroup$ You should be able to find them in an answer to this related question. Or maybe in the references for that answer, since a rope is not an ideal spring. $\endgroup$ – Peter Shor Dec 29 '17 at 13:33

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