6
$\begingroup$

Consider an elastic ball is bounced off a hard flat surface. I would like to reconcile two different answers to the question "how does the contact time between the ball and surface depend on the speed of the ball?"

The first solution, which was the accepted solution here, is that the contact time is independent of the speed. It depends on the ball's diameter and the elasticity and density of its material. The signal for the ball to turn around is a compression wave that travels through the ball. The medium is non-dispersive.

The second solution, which seems to be more correct, is based on Hertzian contact mechanics. It claims that the bounce time is inversely proportional to the fifth root of the collision speed.

But what is the flaw in the first analysis? Is the speed of a compression wave through a sphere homogeneous material dispersive? Is the relevant distance that the wave travels equal to the amplitude of the indentation, not the diameter of the ball? This seems to be what the authors of the second solution are asserting, but I don't see how this is justified.

$\endgroup$
2
+50
$\begingroup$

I believe the source of discrepancy in the second analysis comes in the line where they claim the time of impact is given by

$$\tau_{imp} \text{~}\frac{h_{max}}{v} $$

since that ignores the argument that the impact time must require the shock wave to travel to the other end of the ball and back again. When $h<<d$, then for values $v < \frac{h}{d}c$ where $c$ is the speed of sound in the ball, this equation is valid. But once it is greater, the time is too short for the entire ball to "know" that it is bouncing. The authors' argument becomes invalid at this point.

I think there are three different regimes:

First - treat the system as a mass + linear spring. Regardless of impact velocity, impact time will be the same. This works for very slow impacts, where deformation is very small but the natural frequency $\omega=\sqrt{\frac{k}{m}}$ is such that the shock wave of the impact would bounce many times - a quasi-static situation

Second - add the nonlinear behavior that results from the deformation of the ball: as the ball deforms more, it will have a higher spring constant. This affects the time - it will get shorter. This is shown by the $\tau\text{~}v^{-\frac15}$ relationship.

Third - the situation where the stiffness of the ball is such that the ball remains in contact while the pressure wave travels to the other end of the ball and back again. Once again, you have no dependence on velocity. You can think of this as a situation where the mass of the ball is not known at the time of impact, because parts of the ball have not yet been informed (by the pressure wave) that the ball is bouncing and so they can't yet contribute to the dynamics.

Which of these is the best regime depends on the actual properties of the ball: size, density, modulus - and to some extent, velocity.

$\endgroup$
  • $\begingroup$ Two questions: What about the properties of that "hard flat surface" which reflects the ball? 2nd, a elastic wave originating from the point of contact in the ball would be reflected to that point of contact? I doubt that. A wave originating from the center of the ball would be reflected back on its origin. $\endgroup$ – Georg Jan 15 '15 at 20:01
  • $\begingroup$ @Georg - since the question stated "hard surface" I decided to leave that aspect alone and focus just on properties of the ball. Yes, strictly speaking the above is "more valid" for a long thin rod than for a ball, but the basic principle stands: if part of the ball is not yet aware that a collision has occurred, that mass cannot participate in the equation of motion. The timing of the release of the ball is broadly governed by the round trip time of sound across the diameter - it might be slightly less but much more than $h/v$. $\endgroup$ – Floris Jan 15 '15 at 21:02
  • 1
    $\begingroup$ For inspiration on how to treat problems like this I look at these high speed frames i.stack.imgur.com/QLmjI.jpg $\endgroup$ – ja72 Jun 14 '16 at 14:52
  • $\begingroup$ There is also the consideration when the ball is not very stiff that there will be multiple subimpacts as the pressure waves bounces back and forth, so the concept of "contact time" becomes undefined. $\endgroup$ – ja72 Jun 14 '16 at 15:41
  • $\begingroup$ @ja72 are you saying that in the "not very stiff" case the ball would lose contact from time to time? I don't think so - the reflected wave will have a lower negative pressure than the (increasing) contact pressure that will be present when the wave returns. $\endgroup$ – Floris Jun 14 '16 at 20:37
0
$\begingroup$

I think you need to look at this problem using the concept of "Coefficient of Restitution". You can probably derive the equation using that concept.

For the period of deformation, the equation would be: $$ M*V = -\int_0^T Pdt\, $$ The restitution equation would be very similar. It's simply the impulse-momentum concept.

The compression of a ball as it strikes a wall is similar to a spring being compressed. The above equation clearly shows that the time of compression is dependent upon the mass and velocity of the ball as well as some average force P exerted by the wall. But I think P could also be expressed as P(t) or P(x).

The only way that time would not be dependent upon velocity is if P would always be equal to V. They would cancel. It doesn't make any logical sense that time would always equal mass. Time must be dependent upon velocity.

$\endgroup$

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.