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.

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.

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.

Anyone want to apply for a grant?

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.

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.

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?

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.

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?

@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.

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!

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.)

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.)

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.