The physics of a falling Slinky

Professor Ted Bunn (no relation), who still gets email about a Black Holes FAQ he wrote over fifteen years ago, recently blogged about what happens when you suspend a Slinky vertically, then let go.

No wait, come back.

In Part the First, he provides an alternate (but equivalent) explanation for the phenomenon.

Allen Downey on his wonderfully-titled blog Probably Overthinking It started in with the equations, which inspired Ted to write Part the Second, which contains with lots of math, but there are also charts.

Okay, fine, I don't blame you for leaving.

Tags

1. Matt says:

The way I look at this is that there are two forces in action in this system: Tension and Gravity.

At the top of the spring, gravity is pulling the spring downwards, and tension is pulling the spring towards the centre of mass of the spring – i.e. also downwards. When the spring is released we should therefore expect the top of the spring to accellerate towards the ground.

At the bottom of the spring, gravity is also pulling the spring downwards, and tension is also pulling the spring towards the centre of mass of the spring – but in this case that's upwards.

In this case the tension of the spring is derived from the fact that the spring is suspended under gravity. Before the spring is dropped, it is stationary, which means that the tensile force upwards at the bottom of the spring is exactly equal and opposite to g, since the spring is stationary but still under gravity before it is dropped.

So when the spring is dropped, the forces of tension continues to oppose the force of gravity. The interesting thing is that (as your second link points out) this isn't entirely flat because the spring's tension isn't uniformly distributed, but to get an obviously non-uniform distribution of tension you'd need a really weird spring that is huge, falls over a massive distance and has very weak tension. In the case of a slinky you won't notice the fact that there's an integral rather than a straightforward multiple in there.

The spring therefore accellerates towards the ground at more than g at the top, and at effectively zero accelleration at the bottom until the centre of mass catches up with the bottom of the spring.

2. Anonymous Coward says:

Also interesting is what happens if you let a spring fall vertically, letting both ends go. The gravitational force at the bottom is ever so slightly higher than at the top, so the spring stretches and that increases the effect. The spring will eventually reach an equilibrium between its tension and gravity – unless it hits the surface. To prevent that, it's horizontal velocity must be sufficient for the spring to remain in orbit.

3. Ken Hagan says:

@AC 11:11am: The difference in gravitational force with height depends on the difference in distance to the centre of the Earth, so it is less than one part per million per metre. If you can measure the length of a metre-long spring to better than 1 micron accuracy and build a mechanical device that releases both ends within a microsecond or so of each other, I'll believe what you say about the spring stretching.

4. alegr1 says:

@Anon:

It's not gravity difference, it's air drag.

5. Anonymous Coward says:

Ken, either get yourself a bigger planet or more patience. For example, the moons of Jupiter are in some sense springs: if you stretch them they'll bounce back, and they live in equilibrium with Jupiter's gravity. Some live in a dynamic equilibrium where they're rotating, but the stretch-axis always points in the same direction, heating them up by churning the entire moon.

Alegr, I wasn't talking about air drag. (Otherwise I wouldn't have been talking about the orbiting spring, and besides air drag isn't really the subject we're discussing at the moment. Maybe I should have started with ‘imagine a heavy massive object without atmosphere blah blah blah’ but I thought we were smarter than that.)