Slow Mo Slinky - Physics Q

[Dan_1981]

Watching this video the other day.....

http://www.youtube.com/watch?v=8UimHnsWSBc

Why does the Slinky appear to levitate and not fall at all until the top bit has met the bottom bit and then it falls to the floor?


Is this a trick of the eye due to the extreme slow motion of the film or is it because its a spring or some other physics type magic.

Or am I being pretty stupid and missing a very basic law of physics somewhere?

[tank slapper]

It's not magic, just physics. While hanging, the spring is stretched longer than its natural length due to the weight of the spring. Hooke's law describes the restoring force on the spring, and since it is being stretched longer than the natural length, the restoring force is upwards at the bottom and downwards at the top. When the spring is dropped the whole thing starts to accelerate due to gravity. At the top, since the spring force is downwards, and gravity is downwards, this part moves downwards. At the bottom, gravity is acting downwards, but the spring force is acting up.

Since the position of the bottom of the spring was reached as a result of gravity acting on the mass of the spring, we know that is the maximum extension it can have purely from its own weight and that the spring force is balancing the gravitational force. This means that when the top is released, the bottom can't go anywhere since the spring force and gravity are equal at this point. As the spring closes, the spring force is removed and there is nothing to oppose gravity and it starts to fall.

If you actually pulled the spring down further than it would go just from gravity and released both ends simultaneously, you would see the bottom end rise until the spring reached its natural length, at which point the whole things would fall.

[anonymous-user]

As above^^ The hanging springs effective CofG is in the middle, above that point it's 1G + spring force per unit of mass, and below that point it's 1G - spring force per unit mass. The CofG point does accelerate downwards at 1G, but the top bit accelerates downwards at 2G, meaning the bottom bit experiences zero G.

[Funk]

And in zero G, both ends would accelerate at each other equally, meeting in the middle.

[Nightmare]

awesome video, and without a doubt the most cogent and accurate explanation for something on pistonheads ever from TS!

[0000]

Max_Torque said:

As above^^ The hanging springs effective CofG is in the middle, above that point it's 1G + spring force per unit of mass, and below that point it's 1G - spring force per unit mass. The CofG point does accelerate downwards at 1G, but the top bit accelerates downwards at 2G, meaning the bottom bit experiences zero G.

Hmmn.

So if there was a tennis ball wedged in the bottom end of the spring how would that affect it?

[AdeTuono]

tank slapper said:

It's not magic, just physics. While hanging, the spring is stretched longer than its natural length due to the weight of the spring. Hooke's law describes the restoring force on the spring, and since it is being stretched longer than the natural length, the restoring force is upwards at the bottom and downwards at the top. When the spring is dropped the whole thing starts to accelerate due to gravity. At the top, since the spring force is downwards, and gravity is downwards, this part moves downwards. At the bottom, gravity is acting downwards, but the spring force is acting up.

Since the position of the bottom of the spring was reached as a result of gravity acting on the mass of the spring, we know that is the maximum extension it can have purely from its own weight and that the spring force is balancing the gravitational force. This means that when the top is released, the bottom can't go anywhere since the spring force and gravity are equal at this point. As the spring closes, the spring force is removed and there is nothing to oppose gravity and it starts to fall.

If you actually pulled the spring down further than it would go just from gravity and released both ends simultaneously, you would see the bottom end rise until the spring reached its natural length, at which point the whole things would fall.

Google, or off your own back? If the latter, thanks. QI didn't elaborate when it was shown the other night.

[Dan_1981]

^^^ Aha that was what I saw it on!

Thanks for the very detailed explnation above!

Much appreciated.

[tank slapper]

0000 said:

Hmmn.

So if there was a tennis ball wedged in the bottom end of the spring how would that affect it?

I haven't sat down to work it out, but I think that the same effect would occur with the difference that the spring would be stretched further due to the additional weight. The top would accelerate downwards faster than without the ball, since the restoring force would be higher but the mass of the spring at the top would be similar.

[tank slapper]

AdeTuono said:

Google, or off your own back? If the latter, thanks. QI didn't elaborate when it was shown the other night.

I didn't need google, but it was familiar from looking at sprung systems from studying mech. engineering.

[Baron Greenback]

Still love the one trillion frames per second movie of light going through the coke bottle! Just thinking of the storage needed is mind boggling.

[Alfanatic]

tank slapper said:

0000 said:

Hmmn.

So if there was a tennis ball wedged in the bottom end of the spring how would that affect it?

I haven't sat down to work it out, but I think that the same effect would occur with the difference that the spring would be stretched further due to the additional weight. The top would accelerate downwards faster than without the ball, since the restoring force would be higher but the mass of the spring at the top would be similar.

Yeah I would expect it would do the same thing, the ball would remain still. When it's extended the tension in the spring and gravity have reached equilibrium. The extra mass of the tennis ball means that the gravitational force is stronger, so the spring extends further before equilibrium is reached. From that point on the whole system would act the same as the spring by itself?

[0000]

Tennis ball within the slinky: https://www.youtube.com/watch?v=oKb2tCtpvNU

[mrmr96]

Baron Greenback said:

Still love the one trillion frames per second movie of light going through the coke bottle! Just thinking of the storage needed is mind boggling.

That wasn't "filmed" as such. It's an animation of multiple shots taken from multiple experiments.

"Direct recording of light is impossible at that speed, so the camera takes millions of repeated scans to recreate each image."
http://www.bbc.co.uk/news/technology-16163931

Not sure how that translates into storage required, since it says it still needs millions of scans. Can you shed any light on it?

[Dr Jekyll]

Is this effect related to the way vehicle suspension works? I know the go faster brigade get very worked up about the perils of 'unsprung weight' and I assumed it was because the suspension would tend to compress from both ends instead of forcing the wheels back onto the ground.

[mrmr96]

Dr Jekyll said:

Is this effect related to the way vehicle suspension works? I know the go faster brigade get very worked up about the perils of 'unsprung weight' and I assumed it was because the suspension would tend to compress from both ends instead of forcing the wheels back onto the ground.

Not really, at least I don't think so. The unsprung mass thing is to do with keeping the tyre on the road. When you hit a bump the wheel lifts up, but then you want it to come back down as soon as possible, and not jump (aka remain off the road after the bump.) When the car hits a bump the suspension compresses and the car stays more or less where it is. The spring then pushes the tyre back onto the road and the less un-sprung mass the faster this can happen. (Force = Mass x Acceleration. So for a given force, that of the spring, the lower mass leads to higher acceleration.) This is different from the slinky, because that's a spring under tension which is lifting the bottom of it up as the top falls, whereas the suspension is a spring in compression.

[Alfanatic]

Dr Jekyll said:

Is this effect related to the way vehicle suspension works? I know the go faster brigade get very worked up about the perils of 'unsprung weight' and I assumed it was because the suspension would tend to compress from both ends instead of forcing the wheels back onto the ground.

The heavier the wheel assembly is, the more inertia it has, so it needs more force to accelerate it up or down. It needs a stronger spring / damper to cope, which makes ride quality worse, and it also responds less well to sudden deflections (bumps). Over a speedbump, for example, once the speedbump has accelerated the wheel upwards, more force would be put into accelerating the whole vehicle upwards as well through the stronger spring, and once over the bump, the spring has to work harder to force the wheel back down fast enough to keep in contact with the road. The vehicle has much more mass, therefore much more inertia, so it's the wheel assembly that does most of the moving. The lighter this assembly is, the easier it is to control that movement.

[Dr Jekyll]

So does this mean it's the unsprung mass itself that matters, rather than proportion of sprung to unsprung?

[Alfanatic]

I've always understood, from what I've read, that it's the ratios (proportions) that matter. I've not understood why. I guess that the spring is trying to move the sprung mass upwards with the same force it's trying to move the wheel downwards. If the either the sprung or unsprung masses increase, so the spring's strength needs to increase as well. I suppose then, if there is 25kg of wheel, brakes and suspension arms hanging off a 250 corner of a car, the spring will not have a steady foundation from which to be pushing the wheel back down?

Edited by Alfanatic on Thursday 3rd January 10:56