Mechanics
The Falling Slinky: Why the Bottom Hangs in Mid-Air
The Falling Slinky is one of physics' most startling desk demonstrations. Hang a Slinky by its top, let it stretch under its own weight, and release it: the bottom coils hang motionless in mid-air, apparently ignoring gravity, while the top rushes down to meet them. Only when the collapsing coils arrive does the whole spring finally fall together. Nothing is defying gravity. The bottom simply hasn't yet received the news that the top was let go — mechanical signals travel at a finite speed, and a Slinky is slow enough to let you watch that delay unfold. Underneath the trick sits an ironclad law: the center of mass falls at g the entire time.
- Also calledSlinky drop / the "levitating" Slinky
- Governing lawa_cm = g plus finite wave speed
- Bottom hang time≈ 0.2–0.4 s (≈1 m Slinky)
- Center-of-mass accel.g ≈ 9.81 m/s² (always)
- First modeled byCalkin (1993); Cross & Wheatland (2012)
- Slinky wave speeda few m/s (steel ≈ 5000 m/s)
Interactive visualization
Press play, or step through manually. The visualization is yours to drive — try it before reading on.
Watch the 60-second explainer
A condensed visual walkthrough — narrated, captioned, under a minute.
The setup: a spring already stretched by its own weight
Hang a Slinky from one end and it does not sit at its natural length — it stretches dramatically, because each turn of the coil must support the weight of every coil beneath it. The tension at any height is set by the mass hanging below that point: T(x) = g·m_below(x). At the very top the tension equals the full weight of the spring; at the very bottom it falls to essentially zero.
Because each coil stretches in proportion to the tension it carries (Hooke's law, turn by turn), the coils are pulled wide apart at the top and packed almost touching at the bottom. That uneven spacing is the whole story in disguise: every coil hangs in equilibrium, the upward pull from the coil above exactly balancing its own weight plus the downward pull from the coil below. Nothing moves, and every part is perfectly balanced.
The mechanism: the bottom hasn't heard the news
Release the top and the collapse unfolds as a strict chain of cause and effect:
- The hand's support vanishes. The topmost coil, held up against a tension nearly equal to the whole weight, is suddenly unbalanced and accelerates downward far faster than gravity alone.
- The top coils slam onto the ones below, forming a compact bunch. Where the moving bunch meets the still-stretched spring, a sharp collapse front appears.
- That front — a compression wave — travels downward at the spring's finite signal speed, gathering coils as it goes.
- Below the front nothing has changed: the local stretch, and therefore the local tension, is exactly what it was, so those coils stay in equilibrium and hold their place.
- The bottom keeps hanging until the front finally reaches it. Only then does it begin to move, by which time the whole spring has collapsed into one clump that falls together.
The bottom is not levitating. It simply has not yet received the mechanical signal that the top was let go.
The elegant view: the center of mass never stops falling
Newton's second law applies to the Slinky as a whole. The only external force is gravity, so the center of mass must accelerate downward at exactly a_cm = g from the instant of release — no exceptions, no delay.
How can the center of mass free-fall while the bottom sits frozen? Only if the top falls faster than g to compensate — which is precisely what the collapsing coils do. So the demo illustrates a subtle truth: "the object falls at g" is a statement about its center of mass, not about every piece of it. The top plunges at many times g, the bottom holds at zero, and the mass-weighted average comes out to exactly g.
How long does it hang? A worked estimate
The hang time is set by how long the collapse front needs to travel the length of the spring. For a longitudinal wave on a spring of stiffness k, total mass M and length L, the small-amplitude speed is v ≈ L·√(k/M).
Take a typical metal Slinky: M ≈ 0.2 kg, k ≈ 0.8 N/m, stretched to L ≈ 1.2 m. Then √(k/M) = √4 = 2 s⁻¹, so v ≈ 2.4 m/s and the traversal time L/v ≈ 0.5 s — the right order of magnitude. Measured hang times for meter-scale Slinkies are typically 0.2–0.4 s. Cross-check with the top: to close the gap it must fall roughly the full length L ≈ 1.2 m in that time, an average of a few m/s — consistent with the wave speed. The real front is a nonlinear shock that moves a little faster than the small-amplitude estimate, which is why the measured time lands near the shorter end.
The subtlety everyone misses: a steel rod does it too
The Slinky is not magic — it is merely slow. Every falling object behaves this way. Drop a steel rod and its bottom also stays put until an elastic wave, travelling at the speed of sound in steel (≈ 5000 m/s), arrives to report that the top was released. For a 1 m rod that delay is about 0.2 milliseconds — utterly invisible. The Slinky's compression wave crawls at a few m/s, roughly a thousand times slower, stretching that same physics into a quarter-second you can watch with your naked eye.
The common misconception is that the bottom "defies gravity," or that gravity is somehow switched off. Gravity pulls on the bottom the entire time; it is simply balanced, moment by moment, by an unchanged spring tension — the very same balance it had while hanging at rest.
History and where the idea shows up
The falling-spring problem was analysed rigorously by M. G. Calkin in "Motion of a falling spring" (American Journal of Physics, 1993), who treated the coil as a continuum and found the collapsing front. Rod Cross and Mike Wheatland refined and measured it with high-speed video in "Modeling a falling Slinky" (2012), and a widely shared Veritasium film featuring Cross made the demonstration famous. The Slinky itself was invented by naval engineer Richard James in 1943 and sold from 1945.
The same finite-signal-speed idea reaches far beyond toys. It governs how stress waves race through a suddenly-loaded crane line, how a train of railcars jerks into motion coupling by coupling, and how seismic P-waves carry the news of a rupture outward through rock. The Slinky just makes an invisible universal fact wonderfully slow.
| Part of the system | Motion during collapse | Acceleration | Why (net force) |
|---|---|---|---|
| Bottom coils | Stationary — hangs in mid-air | 0 | Spring tension up exactly balances weight; local stretch unchanged |
| Top coils | Fall faster than free-fall; coils pile up | > g (initially ≫ g) | Gravity and spring tension both pull down |
| Center of mass | Smooth free-fall from the instant of release | g ≈ 9.81 m/s² | Only external force is gravity |
| Collapse front | Travels downward, gathering coils | ≈ few m/s, roughly steady | Nonlinear compression wave (spring's signal speed) |
| Whole Slinky (after front arrives) | Falls together as one clump | g | Now a compact mass in free fall |
Frequently asked questions
Does the bottom really stay perfectly still?
To the eye and to high-speed video, yes — it stays within about a millimetre of its hanging position until the collapse front arrives. Each bottom coil remains in exact equilibrium (upward tension balancing its weight) because its local stretch has not yet changed, so there is no net force to move it.
Isn't the bottom defying gravity?
No. Gravity acts on the bottom the whole time; it is simply cancelled, instant by instant, by the unchanged spring tension pulling up. The center of mass of the whole Slinky accelerates at g throughout — the bottom's stillness is paid for by the top falling much faster than g.
How long does the bottom hang there?
For a meter-scale metal Slinky, roughly 0.2–0.4 seconds — long enough to see clearly. The time equals how long the collapse wave takes to travel down the spring, estimated by v ≈ L·√(k/M), which gives a few m/s and a few tenths of a second.
Would a solid steel rod do the same thing?
Yes, in principle. Its bottom also stays put until an elastic (sound) wave reaches it — but that wave travels near 5000 m/s, so for a 1 m rod the delay is about 0.2 milliseconds, far too brief to see. The Slinky's wave is ~1000× slower, which is why the effect becomes visible.
What makes a Slinky able to show this?
It is an extremely soft spring whose natural length is tiny compared with how far its own weight stretches it. That means a very slow longitudinal wave speed and a large collapse distance, so the finite-signal delay is stretched from microseconds to a watchable fraction of a second.
Does it matter how high above the floor you release it?
The hang effect itself does not depend on the drop height — it depends only on the spring and takes a fixed fraction of a second. You just need enough clearance that the Slinky finishes collapsing before it hits the ground, so you can watch the bottom wait for the top.