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By Phil Plait | September 26, 2011 2:32 pm

What happens if you hold a slinky at the top end, let it extend, and then drop it? Here’s a video setting up the problem:

OK, so what do you think? Now watch this video to get the answer:

Were you correct? I’m proud to say I figured out what would happen, and even for the correct reasons. But then I’m used to problems like this; when you write a science blog you wind up thinking about how best to explain certain phenomena, and acceleration due to gravity is a common one. All parts of the slinky are accelerated by gravity equally, but the bottom of the slinky is being supported by the top. When the top is let go, it takes time for the bottom to find out, and in that time the slinky can collapse.

Here’s another question then: what if the slinky were looser, that is, less springy? Would the same thing happen?

I’ll leave that to you to figure out. But I’ll note that at the end of the second video, they post another question answered in a third video. I got that one right, too, and again for the right reason. Maybe there’s a future in science for me.

Tip o’ the Hooke’s constant to Jeri Ryan on Google+. Yup, that Jeri Ryan.

CATEGORIZED UNDER: Cool stuff, Geekery, Science

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1. Chris P

I was wrong 😀
I thought the top and bottom would come together but the middle (centre of gravity) would be falling so they would snap together a few inches below the half way point.

Very interesting what really happens.

2. Chris P, that’s exactly what I thought. In fact, I still don’t understand what it doesn’t work that way. If the tension in the spring is holding the bottom of the slinky up, then why shouldn’t the tension pull the bottom of the slinky up past its original position?

EDIT: Ok, I think I get it. Because the tension isn’t strong enough to go against gravity before you drop it, so it isn’t going to go against gravity after you drop it, either. But the tension will continue to hold the bottom of the slinky up as the top races down to meet it.

3. John

Um. This explanation is wrong. What is actually happening is that when the slinky is let slack, it is in equilibrium between the force of gravity pulling down, and the force of the spring pulling up. The length that the slinky will slack is the exact same length that cause the entire spring to have g of tension.

When you let the top of the spring go, it experiences a force of g+g, the bottom spring experiences a force of g+(-g) and just sits there. When the spring collapses completely, it is momentarily motionless as all the spring forces cancel out, then g takes over.

The “It takes time for it to know its been dropped” thing would imply that the speed of gravity is somehow dependent on the spring constant (or type of material involved) and is significantly slower than C.

On the otherhand, adding your force vectors and understanding why the spring slacks to the length it does makes it a lot easier.

4. John

Er, — excuse me – its not momentarily motionless, its momentarily not accelerating. And only long enough for the collision to occur and sort out the momentum…short enough that there was no point in me mentioning it.

5. jaranath

I’m not sure I get the “bottom finds out” explanation. It seems to be implying that at least parts of the slinky are literally defying gravity for a while, cartoon-physics style. It seems to me that if you could calculate and animate the center of gravity of the system, you’d see it falling at the rate you’d expect.

6. John

plus the bottom of the spring is closer to the earth, so if anything it should figure out its in a gravity well faster if this weird anthropomorphic concept of gravity were true.

7. Chris

Now to put a person on the end of a slinky….

Actually that might be cool. Get enough slinkies to support a person, go up to the second story, lift the person a few feet above the ground above some padding, let the slinkies go and watch him float for a few seconds. Phil, you volunteering? It’s for science!

8. rob

another example of parts of a system needing time to “figure” out how to move is the earth-sun system. if somehow the sun disappeared, the earth wouldn’t immediately start moving in a straight line. it would continue on it’s circular orbit for 8 1/3 minutes (as if the sun were still there) until the gravitational perturbation arrived, then it would continue on a tangential course.

9. This kind of experiment makes science really fun. I had a physics prof in college who did stuff like that. Thanks for sharing!

10. @Chris #7: The person wouldn’t experience freefall, though, they’d still be subject to 1g. It’s just that the support comes from the slinky instead of the ground.

Well, okay, the freefall does come eventually.

11. John, I have done this many times. and the “It takes time for it to know it has been dropped” does not refer to the speed of gravity but the speed of waves in the slinky. In order for the bottom to “know” the top has been dropped, the information must travel from the top of the slinky to the bottom. This information is transmitted by the slinky and the fastest info can travel in a slinky is at the speed of a slinky wave.

Rob hits it on the head. If the Sun disappeared, Earth wouldn’t find out for 8 minutes since all the information (visual and gravitational influence) would take ~8 minutes to reach us traveling at the speed of light. Another one: If you have your back turned and someone fires a starter pistol at a track meet, the information doesn’t reach you until the sound waves do. Information always travels at the speed of the respective wave.

Oh, and Phil. I believe I showed this to you many years ago at SSU and I seem to recall your first instinct turning out to be incorrect (I used one of those old Swift slinkies which tells you how long ago that was!)

12. dewoo

That is some looney toons phyiscs right there!

13. Sili

What Jeri Ryan?

I got it right(ish) for the wrong reason. But at least I learned enough to get the second one right.

The chain drop cheated me too, though …

14. Chris

No sound on the videos.

15. John

@hale-bopp – earth’s gravity does not care about the internal sturm und drang of the spring in terms of information theory. This is simply the addition of the vectors of the spring force restoring the spring (one vector up, one vector down) plus the vector of gravity down. The restorative force of the spring is greatest at the two ends.. the bottom end experiences a force upward that counteracts the force of gravity, the other end experiences a force greater than gravity. The center of gravity of the system falls at g. The moment that the spring finishes collapsing, the restorative force ends, at which point the entire spring continues free fall. (The extra speed of the top of the spring should be cancelled out or atleast reduced with some mv=mv as it collides with the momentarily motionless bottom.) And the whole thing works because gravity already “calculated” the proper length of the spring when it was let slack for this effect to happen. Now if “know it fell” is just hand waving to avoid having to introduce force vectors into this, then whatever. But any suggestion of Wile E Coyote physics (especially when it comes to gravity!) deserves a response.

The speed of sound in the metal of the spring should be incredibly fast, and the speed of the wave through the spring is a function of its tension, so as it collapses, the wave would…move slower… Plus, the top of the spring knows immediately that it is subject to 2g (or atleast 1+ n g of force where n < 1 – I have to admit I am rusty) of force…

16. CJSF

I have to say that the anthropomorphizing doesn’t really explain anything to me. The bottom of the slinky doesn’t “know” anything and doesn’t “find out” anything.

CJSF

17. Ganzy

Without watching it or reading any of above I predict both ends will come together and meet in the middle and then it will fall to the ground… Here goes..

18. Ganzy

Wow. I was wrong, and that was cool 😛

19. Cusp

It does what physics says it does .

The people in the video are Derek and Rod at Sydney Uni – Rod also does police consultancy on the physics of events (such as murders etc)

20. molybdenumfist

I passed the test, but I only got 7 of 9

21. John, I don’t think we are that far off from each other here. I agree with much of what you say. You can calculate all the forces acting on each segment of the slinky, the center of mass will fall with an acceleration of g just like you say, and the bottom will remain still until the top reaches it.

However, the speed of sound in metal is not the important part here, but the speed of a longitudinal wave in the slinky which is very different than the speed of sound. That is the speed of the wave I am referring to (and the problem is not easy since the speed of the wave depends on the tension in slinky which is different at different points in the slinky).

The laws of physics which you site are what determine how fast information can travel from the top to the bottom of the slinky just like the E&M determines how fast information can be transmitted via light.

22. rickb

Instead of a slinky, think of the idealized problem with two masses (m1 and m2) separated by a (massless, instantaneous force transmission, i.e. infinite speed of sound in the spring) spring. Before letting go all forces are zero, so k * x = m2 * g. At the instant of letting go of the top mass, m2 still doesn’t move because k*x = m2*g, but m1 is now subject to a force of m1*g + k*x or (m1+m2)*g. As the spring collapses k*x becomes smaller, so m2 starts to drop, but m1 is still subject to a force greater than gravity (and m2 conversely less as well) so m1 eventually “catches” up with m2 in a short time.

23. Peter

Gravity is pulling on the bottom of the slinky at all times. There’s no need for it to “realize” it has been dropped. It’s not like we switch on or off the gravity of the earth. There’s a constant force pulling it down (gravity) and one holding it up (hand). If it wasn’t for the tension in the slinky the bottom would have accelerated immediately. Just like it happens when a solid is dropped. However, the force in the actual slinky is fighting gravity until the moment the ends meet.

24. I thought that the bottom would accelerate downward, and that the upper parts of the slinky would accelerate downward faster than gravity alone would account for due to the relaxation of the stretch.

Oops!

A really cool demo! I’ll have to pass this along to Dr. Thornton of UVa Physics to include in the next National Physics Day show.

25. JupiterIsBig

I know I’m not adding much to the physics, but …
“Some people are like slinkies – basically useless, but a lot of fun when you push them down the stairs”

26. Chris Crawford

Here’s a variation on an earlier explanation that is useful for considering springs with different spring constants:

The instant that the spring is released, its center of mass begins accelerating downward at g. That center of mass acts exactly like a ball being dropped. Relative to the center of mass, the ends of the spring begin contracting at based solely on spring constant k. Both ends are accelerating towards the center. This spring just happens to have a k value such that the upper end of the spring moves downward at 2g, while the lower end accelerates at 0 g.

A spring with a higher k value would contract faster and so the lower end would initially accelerate upwards, reach minimum, and then fall normally.

A spring with a lower k value would have lower upward acceleration of the lower end, so it would appear to fall at less than g.

27. @Chris P #1: The centre of gravity does fall freely under gravity, just as you said. It will follow a smooth, steady acceleration downwards irrespective of the changing shape of the spring, including at and after the moment the spring snaps shut. But in the meantime the spring is changing shape.

For all those who don’t like the idea of the bottom “finding out” later that it’s falling – the physics of waves is the entire basis of the physics of the communication of information, and it’s probably worth getting to know it well before deciding not to like it

To get the full picture, the whole process can be derived very neatly with a bit of calculus and the basic laws of motion, and it’s a really wonderful example of the propagation of information through a system being visibly delayed.

This is how the universe works, folks! Brilliant stuff.

28. It’s actually true if you drop any object – not just a slinky. The only difference is that the speed of propagation is particularly slow through a slinky, so it’s possible to see it happen.

If you held a golf club vertically by the handle and dropped it, the bottom of the golf club wouldn’t start falling until a fraction of a second after you let go. That fraction of a second is the length of the club divided by the speed of sound.

A sound wave has to carry the news along the length of the club to inform the head of the club that it has been dropped. Until the head of the club receives the news, it stays put.

29. Robin Byron

This kind of reminds me of my last training jump at Ft Benning jump school in the 60’s, but with different physics. I was preparing to land when another jumper side slipped to a point directly below me causing my chute to collapse. Normally, not too serious a problem if you have some altitude to play with. Unfortunately, I didn’t have the altitude and free-fell about 45 feet, they said. I was told, “…get up, secure your chute and leave the DZ or be recycled(redo the entire course)”. I packed my chute in an aviator’s kit bag and hobbled off. I understand the ‘new army’ is no longer allowed to treat people that way.

30. JonF

26 comments and not a single one riffing on “Resistance is Futile”? Internet, I am disappoint.

31. I got it right, but I must admit, if no one had ever asked me the question, I might not have even considered it (as shown by the fact that I never asked the question before).

When I watch the video, it occurs to me that the spring bottom actually accelerates to the ground (when it finally gets moving) faster than it would if it was the only part of the spring that existed. The reason for that (and don’t ask me for the physics of it, because I have no idea whatsoever what that might be) is because the top of the spring has accelerated quite a bit before it actually collides with the bottom, adding acceleration to it. Besides gravity, it seems to me that the natural tension of the spring is also working on the top as it falls.

Is that correct, or is all of the tension going into keeping the bottom from falling? The tension has to have some effect.

Doesn’t it?

32. #27 – @JonF – it’s now 29.

33. rickb

@Barber Right, the potential energy of the stretched slinky gets converted into kinetic energy (along with a bit of heat and sound) so it would be going faster.

In the case of the slinky, the top of it probably eventually exceeds the speed of sound in the slinky (it goes “supersonic” so the bottom doesn’t move at all until the slinky is nearly fully relaxed.

The total energy of the system is mgh + 1/2 kx^2, so at the end the velocity of the cm should be v = 2 *sqrt(gh+1/2*k/m*x^2)

34. johnthompson

@hale-bopp Thanks for trying to keep the B.A. honest! I’ve seen this demo done several times, in front of hundreds of ph.d.s in physics, and have yet to see anyone get it right (with the correct explanation) BEFORE the fact (which isn’t to say that the B.A. couldn’t have been the first, just that I found that possibility rather unlikely).

35. Mike

That’s all good and dandy except that the idiots called it “slow motion video”. It’s not. It’s high-speed.

36. RwFlynn

I got the first one right, but the second one threw me off. I had thought that the greater potential energy would actually change something, but now I know better! 😀

Edit: Well, I should say it did change something, but not the thing that mattered!

37. VinceRN

I saw this done years ago, but when I guessed back then I got it wrong. Since it’s posted here, I showed it to my 7 year old daughter, and she got it right, though she couldn’t articulate why and didn’t totally understand the explanation.

After reading everyone’s dueling explanations I now understand this less than I did before. The explanation I remember from before is pretty close to what they guy in the video said.

38. Morgan Grigg

What was the answer to the second question? What will happen to the slinky with a tennis ball on the bottom?

39. rickb

I just realized I screwed up my analysis of the potential energy since I didn’t take into account conservation of momentum. Analysis of problem by disregarding the force details and just using conservation laws only works if you take into account all of them. The potential energy in the spring doesn’t change the velocity of the center of mass. It does still add energy to the impact when it hits the ground, of course, heat, sound and a little bit of KE to the earth.

40. wildcardjack

I was right, but I’ve spent too much time dealing with springs in dynamic systems.

41. JB of Brisbane

“Slinkies – can’t explain that!” 😉

42. jennyxyzzy

Interesting. Clearly there are dynamic aspects to this experiment (the time for the information to travel down the spring) because if the spring was able to be fully modelled by Hooke’s Law, a static description of springs, the bottom would in fact start moving before the spring has completely collapsed. Here’s my reasoning:

At the initial state, the tension in the spring is equal to the force of gravity on the spring (because the spring is motionless).

From Hooke’s Law, we know that the tension is linearly related to the extension of the spring. So as the spring collapses, the tension force gets smaller in a linear fashion, until the spring is fully collapsed, at which point the acceleration experienced by the bottom of the spring would be equal to gravity. However, halfway through the collapsing process, the bottom of the spring should already be accelerating at half gravity. The fact that this is not the case, as demonstrated in the video, shows that Hooke’s Law is not sufficient to describe the behaviour of the spring in this case.

Actually, this is quite a nice demonstration of how a scientific theory can be ‘correct’, but still gets replaced. Hooke’s Law is indeed correct for a static system – you can accurately predict the length of a spring’s extension by measuring the force being applied to it in a steady state manner. But Hooke’s Law is not enough to correctly describe the system in the experiment, so we use another description of the spring – as a medium for wave transmission. Those calculations give the correct answer – the bottom of the spring won’t move until the spring has fully collapsed.

But, if we look at that film in high resolution / very slow motion, I’m willing to bet that we do see some motion of the bottom of the spring – and that would be due to the fact that our models of springs as wave mediums are not exact either – for example they don’t take into account innaccuracies in the manufacturing of the spring, or the non-constant gravitational field (gravity will be stronger on the bottom of the spring than on the top – ok, not a big deal for this experiment, but if you’re building a sky-hook, you had better understand this effect in detail) which means that our perfectly balanced gravity and tension in the experiment are not actually perfectly balanced. Maybe the bottom was actually oscillating, or maybe it was falling, or even being pulled upwards.

43. Bill3

@Chris Crawford – You wrote: “A spring with a higher k value would contract faster and so the lower end would initially accelerate upwards, reach minimum, and then fall normally.

A spring with a lower k value would have lower upward acceleration of the lower end, so it would appear to fall at less than g”

With the given experiment, holding one end of the spring and dangling the other, this is not true. The higher k value spring would stretch less at equilibrium before the drop, but behave the same after being dropped. The lower k spring would similarly stretch more prior, but behave the same when dropped.

44. db26

The bottom of the slinky is simply falling at the exact same rate as the slinky tension is wanting it to be pulled towards the top. Very slight, but so is gravity. Before the slinky is dropped, the bottom is already in perfect balance of tension versus gravity. It would stay there forever until the energy/mass from the top slammed into it.

45. Chris P:

I thought the top and bottom would come together but the middle (centre of gravity) would be falling so they would snap together a few inches below the half way point.

Think about it. The bottom is stationary because the force of gravity down is exactly balanced by the force of tension up. In order for the bottom to be pulled up, the tension would need to increase. Releasing the top of the Slinky decreases the amount of tension on the Slinky as a whole, as your hand is no longer applying the necessary upward force to keep the entire Slinky above the ground. However, it takes time for that force to propagate. As the top of the Slinky falls, the bottom of the Slinky still “feels” the same amount of tension as before, at least for some time.

I’d like to see this added to the experiment…

Hold some other object at the same height as the top of the Slinky. Then, release both objects at the same time. How do the two objects compare as they fall? Does the (top of) the Slinky fall slower, faster, or the same, as the “free” object?

(My prediction is that the top of the Slinky will fall faster than the “free” object, for reasons I will post later.)

46. When the top is let go, it takes time for the bottom to find out

Hey, whatever happened to the Bad Astronomy website’s take on “don’t anthropomorphize inanimate objects”? (You know, things like “the car wants to go in a straight line” and the like.)

47. Messier Tidy Upper

Wow. Interesting question, awesome slow-motion photography.

I guessed the slinky’s ends would come together for the first question, guessed the tennis ball would just fall down for the second.

The gravity-defying aspect there – no I didn’t expect. Trippy!

48. Stan9fromouterspace

Obviously a conspiracy. The slinky fell into its own footprint.

49. Aubri

It wouldn’t matter how springy the spring is, you would always get the same result. (Otherwise we’re assuming that particular slinky was tuned to 9.8 m/s!) A heftier spring would require less stretch to counter gravity, but would thus take proportionally less time to transmit the wave down its length. A theoretical slinky with no stretch at all would be an inelastic string or rod, which would begin falling immediately.

@Ken B
The center of mass for the slinky system must begin falling immediately, with the top accelerating at gravity + spring tension. If you painted the center coil, I’ll bet you would find that it falls at the same rate as the other object.
If you idealize the slinky into two masses with a massless spring in the middle, the top should accelerate at 2G*. (In reality, the situation is probably way, way more complicated since the weight is spread along the spring.) When the top and bottom collide, it’s like a Newton’s Cradle — the top half gives its excess “spring acceleration” to the bottom so the whole thing is suddenly falling at the “right” speed… less whatever energy was expended to noise and such, I suppose. It would be interesting to test how much velocity is lost that way relative to a control object.
If you were to attach extra weight to the bottom, I suppose the top would have to accelerate down faster to balance it. Which makes sense, because the extra weight would mean extra stretch, which means more tension.

*Edit: assuming the end masses are equal.

50. Daniel

I too am not totally convinced by the information travel time idea. Maybe it’s a part of it, but I don’t think it can do without accounting for the dynamic equalibrium of forces experienced by the bottom of the slinky between tension and gravity.

Also, is it just me or does the tennisball actually seem to be falling (much slower than the normal rate, but still falling) in the third video?

51. Aubri

@Daniel:
The tennis ball was bouncing slightly before release. It continued to happily do so until the rest of the slinky arrived and said “HI WE’RE FALLING NOW!”

Information travel time is one way to approach it, but like most of physics, you can approach the problem in many ways and get the same answer. A force diagram would give you the same result.

52. Thought experiment…

You’re at the carnival, and get on one of those centripetal (or is is centrifugal?) force rides, which hold you against the outer wall of a spinning room as they drop the floor out from under you. You hold the same Slinky+TennisBall such that, in your spinning frame of reference, the tennis ball is now stationary.

Release the “top” of the Slinky.

Plot the path of the Slinky+TennisBall from your point of reference, as well as from the reference of an outside observer. Will the inside-the-ride observer still see the tennis ball remain motionless until the “top” of the Slinky reaches it?

Who said calculus and differential equations have no use?

53. Chris A.

I contend that the bottom of the slinky did _not_ remain stationary until the top hit it. Instead, it begins accelerating downward as soon as the slinky is released (well, plus the time for the wave to propagate from top to bottom). It’s just that the net force on the bottom (gravity minus tension) is really close to zero (i.e. the tension is decreasing very slowly) until the time when the top crashes down onto it, so the acceleration of the bottom is really close to zero until that time.

54. John Nouveaux

I want to see Adam, Jamie and crew Mythbusters-size (Mythbustersenate?) this experiment!

55. Isaac

Yeah, Phil, I got ’em both right, too, but I don’t imagine that there is a future in science for me.

56. physicsman

hey that’s Derek my physics lecturer from UTS!

57. Mal

If I dropped a golf club in the presence of gravity but in a vacuum there could be no sound wave that would “send the message” but I’m sure the golf club head would still drop. In any case if the handle dropped momentarily before the club head there must be shortening of the club for this to happen and the golf club surely doesn’t shorten.
The tension forcing the spring to collapse is high at the top because it has the entire weight of the spring opening it in the first place. The lower we go in the spring the lesser the extending force is. Right at the bottom of the spring there is virtually no tension. This is interesting but not the reason for the phenomena.
The top of the spring is collapsing at a rate faster than gravity. It collapses in a concertina fashion (like the twin towers did) which has the effect that the lowermost leading edge of the new compressed mass of the spring has the effect of being a new suspension point maintaining the tension in the open spring below it. The forces below the compressed mass remain in equilibrium and thus the lowermost spring edge doesn’t move until the compressed mass hits it.

58. Nikola

Silly question: why the tip to Jeri Ryan? I can not find this mentioned on her G+ public posts?

59. Moose

The explanation was… less than satisfying.

When the slinky is extended, the force of gravity on the bottom of the slinky stretching it out is the same as the force being applied by the hand holding up the top. In balance. The gravity on the whole slinky is also balanced by the spring action trying to pull it back together.

When you let go, the balance changes. The system’s center of gravity falls at the expected 9.8m/ss. The top end and bottom end are also (each) closing at 9.8m/ss. The bottom end is stationary because it’s falling and climbing at the same acceleration, resulting in a zero vector. The top end is falling at 2G. You could see this by holding and dropping a tennis ball beside the slinky at the half-way point.

Here’s the thing: the tennis ball adds weight to the bottom of the slinky. The center of gravity is thus lower in the system. The tennis ball (and you can see this in the third video) is actually falling, just slowly. The top of the slinky will also be falling at slightly faster than 2G. Again, dropping a second tennis ball at the half way point of the slinky will show this.

60. Buzz Parsec

@57 Mal – The “sound” wave people are discussing doesn’t propagate through air. It propagates through the solid metal of the golf club shaft. So doing the experiment in a vacuum wouldn’t change anything, except eliminate the confounding effects of air resistance.

If we aver return to the Moon, we should definitely do this experiment. No air to confuse things and everything happens much slower in 1/6 g.

BTW, the golf club is stretched (very slightly) under the force of gravity, and the shaft does exert a spring force up on the head, which causes the club to shorten when released. This is the cumulative stretching of the bonds between the atoms in the solid structure of the club, and probably amounts to a few nanometers under Earth’s gravity, but it is very important as it is what allows the club to not break when it hits a golf ball, nor to fall apart when you swing it.

I think your description of the falling spring is correct, but you left out one little bit that might make it clearer… At the very bottom of the spring, the tension is zero, but the mass being supported by that tension is also zero. As you move up the spring, the tension is greater, and so is the mass supported by the tension, in perfect balance so the net force (gravity plus tension) is zero. This is proven by the fact that nothing is moving. (If the forces weren’t in balance, the spring would oscillate, but the experimenter has deliberately damped out the oscillations before dropping the slinky.)

BTW, about the speed of sound in the slinky, there are several different wave effects, that may propagate at different speeds. There is the speed of sound in the metal the slinky is made of, which is probably much higher than the speed of sound in air, and which I think doesn’t enter into this problem. (But I’m not sure.) Then there are the speeds of transverse and compressional waves in the slinky, which depend on the spring constant and the tension. If you stretch out a slinky (have a friend hold the other end about 10-20 feet away), you can easily make it wiggle up and down or side to side. These a the transverse waves. You can also make compressional waves. Bunch up a few coils at one end then release them, and the bunch of close-together coils will propagate to the other end and reflect back. The speed of this is the compressional wave speed. This is what’s relevant to the falling slinky experiment. I’m not sure how or if the various speeds are related to each other. All three probably depend on the detailed structure of the metal (or plastic) in the slinky.

61. Buzz Parsec

Moose @59, We need a bigger budget and better experimental apparatus! Some kind of stand to hold the slinky at the top, and at least three platforms to hold tennis balls at the top, the bottom and a third at the middle (center of gravity) of the spring, and a mechanism to release them all at the same time. The three tennis balls should stay the same distance apart as they fall. The center one should stay right next to the central coil of the slinky (which should be marked to make it easier to follow.) There should be a scale (horizontal lines) on the structure to calibrate it. Finally, the entire structure should be in a vacuum chamber, or preferably on the Moon, to eliminate air resistance on the tennis balls and slinky.

Anyone want to apply for a grant?

62. cxevalo

this clip is very short so double click the play button to pause it before the end’

it was shot well into civil twilight so the quality is not so hot.

the charting was done with ‘tracker’ http://www.cabrillo.edu/~dbrown/tracker/
i made one pass marking the top if the slinky and the center of the classic Newtonian falling body i.e. an apple. I was suprised how nice the plots came out. The slinky is the top with the apple under it.

63. Dwayne

Arrrghh, you people!

This is not the ideal massless spring of your high school and/or college freshman level physics class supporting a single mass/weight at the bottom! This is a real world Slinky! It has mass all along the spring, and in this demonstration the weight of that mass is what is stretching the spring.

So, when it comes to the tension in the spring, there is a gradient all along its length. The highest tension, then, is near the top where it is supporting the most weight and the lowest tension is near the bottom where it is supporting the least weight.

Think about it! It is this initial gradient in the tension that causes the observed effect when the top of the slinky is released – when the small differential mass at the top begins accelerating downward with the force of its own weight plus the force due to the tension in the spring just below it (which should be equal to the weight of all the Slinky below it).

So, the tension in any section of the Slinky below the descending pancaking top doesn’t decrease until it becomes part of the pancaking section itself. In other words, it’s just business as usual for anything below the pancaking section – as can be easily seen in the video.

…oops…

I overlooked Mal’s post. Sorry.

Oh, well, I’ll post this anyway.

64. Matt B.

In #22, the masses need to be equal, otherwise you can get the drop that’s noticeable in the tennis ball (despite the claim in the third video that it didn’t move).

The center of mass of the slinky is not precisely halfway between its ends, though; since the upper half has more weight to hold up, it’s more extended.

It would be interesting to put the tennis ball at the top and see what the bottom does.

65. Jeff

#5 is right. Look at Eulers 1st law of motion instead of Newton’s 2nd, the 2 laws are the same except that Eulers says that linear forces are equal to the mass times the acceleration on the center of mass, and Newtons 1st law. According to Newtons’s 1st law, the particles would stay stationary unless acted on by another force by an outside force, and Euler’s law which states that gravity is acting on the center of mass, then we can say that a specific particles will not accelerate unless the only way for the center of mass to fall would be for that particle to accelerate.

66. anon

Here’s the explanation I understood:

The bottom of the slinky, when the slinky is still held up, experiences one g down and one g up (in order for it to stay unmoving, the sole downward force, g, must be equal and opposite to the sole upward force, in this case the force of the spring). The top must then experience 2g down and 2g up (The spring holds up the bottom with one g, which means the spring must pull down on the top with that same force, g, which adds up to 2g of total downward force once you add in gravity. The upward force on the spring by the hand must balance this 2g downward force in order to keep the slinky from accelerating.). Immediately upon release, the top experiences only that 2g of downward force, accelerating it downwards. The bottom, however, is still being pulled up with that same g of spring force from the beginning, which counteracts the constant one g of force down on it due to gravity. Thus, the bottom does not accelerate until the top of the slinky finally bumps into it from above.

What I don’t get about that explanation, however, is this:

Perhaps this does not occur because of some momentum-related issue (where the spring force, F=kx, does not change from its initial value when x was equal to the unstretched length), or perhaps it arises out of the distribution of mass (the mass is not distributed solely on the top and bottom, but rather continuously and relatively evenly over the slinky’s length). The explanation could also lie (lay?) in the slinky–perhaps the slinky is not simplistic enough to be modeled with F=kx.

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