4
$\begingroup$

In short: Do bouncing balls keep bouncing forever? If not, does it have to do more with energy than velocity?

Learner.org: Energy

If energy is conserved, why do bouncing balls, pendulums, and other cyclic systems eventually stop moving? The energy doesn’t disappear but rather it is transferred to the surroundings, where it is unavailable to keep the cycle going. The result is that most objects moving in cycles eventually move less and less and then finally stop.

But that wouldn't make sense mathematically: although the limit converges to zero, it doesn't fully ever approach it. Do the mathematical laws break down at the micro physical level much like they do in general relativity? Does the ball keep bouncing up and down but reaching some height in space $\epsilon > 0$?

$\endgroup$
11
$\begingroup$

Every simple equation we use to model a system is based on assumptions, and these assumptions often cease to be accurate after a certain amount of time. Given the intuition you may have developed from the simple versions of pendulums, bouncing balls, etc., it's natural that you'd think that they'd go on forever. Really, we have to consider what actually happens in these systems and how long our assumptions are valid.

The pendulum is easier. A perfect pendulum will swing forever. What is it that makes a pendulum non-perfect? Friction can come from the air around it and the pivot it swings on. Also, the swinging of the pendulum pulls its pivot side-to-side, transferring a little bit of energy into the system it's hanging from. Each of these effects can be characterized, but doing that requires some assumptions. Let's just worry about the friction. Kinetic friction is the frictional force between two moving surfaces. If the surfaces aren't moving any longer, the surfaces 'settle in' a bit more and don't give way so easily. If you move slowly enough, however, you can get a little bit of both. This is what makes cellos to sound, hinges to squeak, and tires to screech. Static friction will kill your pendulum.

As for the bouncing ball, what is it that makes it bounce? When the ball hits the ground, it deforms and snaps back into place, launching itself into the air. But the ball doesn't return all of the energy; some of it ripples around the ball while it's in the air. Watch this video for a great example of that. Eventually, the impact will be enough to deform the ball a little, but the 'snapping back' won't be enough to launch it into the air. Even more likely, the ball will hit at such a time that its vibrations and its motion cancel out, like when you jump on a trampoline right after someone next to you and steal their jump. Physics is all about making complex things simple enough to understand, but in the end we only want to understand them well enough that we can make them complex again.

Good question, Jossie.

$\endgroup$
2
$\begingroup$

Just an addition to the other good answer: while dissipation reduces the amplitude it is not the cause for the breakdown of the underlying model. Even with dissipation you could have in theory an ever shrinking oscillation amplitude.

But we can answer the second part of the question "Do the mathematical laws break down at the micro physical level much like they do in general relativity?" easily:

Yes, the mathematical model is based on Newtonian physics. At smaller and smaller amplitudes quantum effects dominate and a completely different approach is required. The oscillation amplitude can be described by the Heisenberg principle and will never be exactly zero.

$\endgroup$
  • $\begingroup$ If we're going this route, it should be noted that at a certain point, the amplitude will be smaller than the de broglie wavelength of an atom in the pendulum or of the pendulum itself, even, and at this point, the oscillation will no longer mean anything. $\endgroup$ – Jerry Schirmer Aug 29 '13 at 18:25
0
$\begingroup$

The height of bounce will asymptote to zero, so also the time taken for the bounce will get smaller and smaller and also asymptote to zero. So as time moves past this point, the ball will have stopped bouncing.

$\endgroup$
  • 1
    $\begingroup$ Asymptote to zero ? I have heard tend to zero, but I think asymptote to zero isn't the correct way to say "tend to zero". Asymptote is defined for curves as a line whose distance with the curve becomes zero as we move towards infinity, so how can a line simply tend to zero without a curve into consideration. $\endgroup$ – Mitchell Jun 20 '17 at 14:03

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.