What happens when a ball stops bouncing?

Question

If I were to drop a bouncy ball onto a surface, each successive bounce will be lower in height as energy is dissipated. Eventually, however, the ball will cease to bounce and will remain in contact with the ground. What happens during that small moment of time that is the transition period?

Here's my line of thought so far:

Let's say, for example, that the ball could be modeled as an "ideal" bouncy ball. Each time it hits the floor, it rebounds with one-half of the impact velocity, the other half of which is absorbed and dissipated by various things. Will this ideal bouncy ball ever stop bouncing? The answer, as far as I can tell, is yes (well, maybe). Since each bounce has half of the initial velocity of the previous bounce, the time between each bounce also halves. The first bounce might take $1$ second, the second takes $0.5$ seconds, the third $0.25$ seconds, etc. This infinite series converges to a finite amount of time, but with an infinite number of bounces.

Since it's obviously not possible for an actual ball to bounce an infinite number of times in finite time, it is clear that the above model doesn't work. Eventually, I decided that the problem must be in the fact that the model doesn't take into account the fact that ball spends time compressing and decompressing while in contact with the floor.

Even once the ball touches the ground, the center of gravity continues to move downward as the ball compresses. Soon, the ball expands, and the center of gravity moves upward. Lets $h_1$ be the height of the center of gravity when the ball is uncompressed, and $h_2$ the height of the center of gravity of the ball when compressed, and $D = h_1 - h_2$. Eventually, the height of the bounce will be smaller than $D$, so that center of gravity simply vibrates between those two locations, and the ball never leaves the ground.

Answer 1

There is a thresh-hold below which the ball will not bounce determined by the mass of the ball and gravitational acceleration (force of gravity on the ball in other words)

In order for the ball to bounce (meaning physically lose contact with the surface of the ground) the compressive force created by the impulse of impact (the point at which velocity is downwards but acceleration upwards until the ball reaches its point of maximum deformation and velocity is 0) must exceed the force of gravity on the ball, so the ball will continue to bounce so long as this condition is true.

However, once the point is reached where the ball achieves maximal deformation but is unable to accelerate upwards in excess of gravity, it will cease to bounce. All subsequent release of energy will be entirely other forms (deformation, vibration, etc) so it isn't exactly like a pendulum, nor is it a perfect Zeno's paradox scenario.

Answer 2

If the ball and surface are ideal rigid bodies, then it would simply bounce an infinite number of times in a finite period of time, as you say. There is no problem with that. (Though "half-velocity" implies sqrt(half) time for successive bounce - but it still gives us a geometric series.) Here is a full analysis.

For a real ball, it depends on what you mean by "bounce". For me, bounces are those things separated by local minima in the ball's gravitational potential energy, or equivalently, by local minima in the height of the ball's center of mass.

In this sense of the word, the ball is bouncing to a lower minimum ("energy floor") than an ideal ball would be bouncing, and is in contact with the floor during a part of its bounce. Moreover the minimum ("energy floor") to which the ball bounces is itself rising gradually as the ball compresses less on each successive bounce. However, the asymptote to which this "energy floor" converges is below that of an ideal ball because the ball compresses slightly under its own weight even when not bouncing. The (non-zero) difference between this asymptote and that of an ideal ball, forms a threshold (let us call it "escape energy") that must be attained by the ball in order to loose contact with the floor.

Initially, the ball would spend most of its time in free fall, approximating the Zeno behavior already described. However, once the bounce energy diminishes to below the "escape energy" defined above, it is no longer sufficient to raise the ball's center of mass above its natural radius, so the ball no longer looses contact with the ground, but simply bobs up and down in the center-of-mass sense. During this regime, I believe the system exhibits second-order linear (or at least linearizable) dynamic behavior, like a mass sitting on a spring, since the corrective force is now proportional to displacement in the infinitesimal neighborhood of equilibrium.

Now any system linearizable to a second order system (like a pendulum) never stops oscillating because its period converges to some non-zero constant. It just swings less and less, but it never stops swinging. So it is with the ball. The "bobbing" decays exponentially over an infinite amount of time or until other disturbances such as impacts from air molecules drown out motion.

EDIT: I hadn't read your question properly before jumping in and now I see you said pretty much the same thing. Oh well.. guess I'll just leave it up.

Answer 3

There is no contradiction, because after some time, the period of each bounce tends to zero, and consequently the frequency of jumps tends to infinity. This can be easily checked when drop a ball partially inelastic: as more  time passes the ball bounces faster and faster to a height less and less, until you no longer perceives the movement (but it exists as time is less that the limit of infinite sum, with frequency and amplitude to infinity tends to zero). So mathematically it is possible to bounce ball infinite times in a finite time. After a time equal to the limit of infinite series, the ball should stop. Until then and excuse me for English ... I'm still learning. Hope this helps.