Timing a bouncing ping-pong ball

[Xlotic]

When I throw a ping-pong ball as free fall(not in projectile motion), the time is same from t0 to t1 and from t1 to t2. How can this be possible? Or is there any necessary assumption to prove it?

 

[kuruman]

Are you describing the results of an experiment that you performed? If so, have you done a calculation? Remember that you cannot ignore air resistance when you have ping-pong balls in free fall.

You may wish to look at this study.

 

[Lnewqban]

Welcome, Xlotic!
It seems to me that the ball acts like a projectile among those times, being subjected to almost the same mean aerodynamic drag in the way up and in the way down (some drag-degraded mean velocity here).

 

[sophiecentaur]

Welcome, Xlotic!
It seems to me that the ball acts like a projectile among those times, being subjected to almost the same mean aerodynamic drag in the way up and in the way down (some drag-degraded mean velocity here).

That all depends on the height you are discussing. Air resistance on the way will reduce the KE of the ball going up and down so the |velocity| when it next hits the floor will be less than when it leaves. I remember a project one of my students did with bouncing pingpong balls and air resistance was measurable for trajectory heights of greater than 3 or 4m (>one room height ) and he did the test in a stair well of a building with four floors (tiled concrete floors). Iirc, he used a TV camera to look at speeds so as to offset the effect of varying coefficient of restitution.
This link shows measured values of actual terminal velocity of a pingpong ball. It was reached at heights of 12m.

 

[kuruman]

This link shows measured values of actual terminal velocity of a pingpong ball. It was reached at heights of 12m.

Ahem ... there is a missing link. Can you provide it?

 

[Vanadium 50]

.. there is a missing link

You mean his avatar? It's been that way for a while. <rimshot>

the time is same from t0 to t1 and from t1 to t2. How can this be possible

Take a ping-pong ball and bounce it. Listen. Does it go

tock-----tock-----tock-----tock-----tock-----tock-----tock-----tock-----tock-----tock-----tock

or does it go

tock---------------tock----------tock-----tock----tock---tock--tock-tock-tocktocktocktocktock

?

 

[sophiecentaur]

Ahem ... there is a missing link. Can you provide it?

Sorry, here it is. I realize he's not actually on our line but he's a nice looking (Neanderthal) guy so what the?

 

[kuruman]

Sorry, here it is. I realize he's not actually on our line but he's a nice looking (Neanderthal) guy so what the?
View attachment 280869

Thanks for the link, he looks nice. Can he play the piano?

 

[rcgldr]

For a ping pong ball, the time decreases between each bounce. On a ping pong table, just before it stops bouncing, you can hear an increasing frequency of the bounces.

 

[sophiecentaur]

For a ping pong ball, the time decreases between each bounce. On a ping pong table, just before it stops bouncing, you can hear an increasing frequency of the bounces.

But that (at the speeds involved) will be largely due to hysteresis as the ball flexes, I think.

 

[cjl]

It seems to me that very few people have looked at the diagram here. The question is not about the duration of the first bounce vs the second, the question is about the time from first contact to apex compared to apex to second contact. These two times should be very similar (differing only due to air resistance).

Xlotic: why would you think this wouldn't be the case? You seem to treat this as a surprising result, so I'd be curious to hear your thought process here so I can better address it with my answer.

 

[Halc]

These two times should be very similar (differing only due to air resistance).

But air resistance is there, else all the bounces would be much closer to the same height (differing only to energy losses to inelasticity of the bounces).
So t0 to t1 has to be shorter since the ball is moving faster on that leg on average than the t1 to t2 points. If the average speeds were the same, then no energy is lost to air resistance, which would be a pretty amazing trick for a ping pong ball.
So I ask:

When I throw a ping-pong ball as free fall(not in projectile motion), the time is same from t0 to t1 and from t1 to t2. How can this be possible?

What makes you assert that the two intervals in question are the same length? It's not like it's trivial to measure.

 

[Vanadium 50]

It's not like it's trivial to measure.

Take a ping-pong ball and bounce it. Listen. Does it go

tock-----tock-----tock-----tock-----tock-----tock-----tock-----tock-----tock-----tock-----tock

or does it go

tock---------------tock----------tock-----tock----tock---tock--tock-tock-tocktocktocktocktock

It's fairly trivial to measure.

 

[Halc]

Listen.

No tock sound at the apex events, so not going to get the answer this way.
Re-read cjl's post just above since you seem to be answering a different question.

 

[Vanadium 50]

No tock sound at the apex events

Sure, but if the tocks come closer together, the apex events must as well.

 

[Ibix]

Sure, but if the tocks come closer together, the apex events must as well.

Yeah - but the question seems to be about comparing tock-apex and apex-tock times, which are equal for a given bounce if air resistance can be neglected.

 

[sophiecentaur]

Sure, but if the tocks come closer together, the apex events must as well.

Yet the tocks do not tell you when the apex events occur; certainly not mid-way between tocks, except in the limit where the velocity through the sir approaches zero.
Detecting the apex events is certainly harder than detecting the tocks - the old peak / zero crossing problem.

 

[A.T.]

Sure, but if the tocks come closer together, the apex events must as well.

The question is not about the time between apex events, but how each apex event divides the flight phase. If you do it in vacuum it will be in the middle.

 

[cjl]

But air resistance is there, else all the bounces would be much closer to the same height (differing only to energy losses to inelasticity of the bounces).
So t0 to t1 has to be shorter since the ball is moving faster on that leg on average than the t1 to t2 points. If the average speeds were the same, then no energy is lost to air resistance, which would be a pretty amazing trick for a ping pong ball.
So I ask:
What makes you assert that the two intervals in question are the same length? It's not like it's trivial to measure.

Some air resistance is there, but I would bet that the majority of energy loss is happening due to the dissipation during bounces, not dissipation due to air resistance, at least for short bounces. Certainly I would expect a slight asymmetry in the direction you indicated, but I bet it'd actually be pretty hard to measure, at least for shorter bounces.

 

[jbriggs444]

Some air resistance is there, but I would bet that the majority of energy loss is happening due to the dissipation during bounces, not dissipation due to air resistance, at least for short bounces. Certainly I would expect a slight asymmetry in the direction you indicated, but I bet it'd actually be pretty hard to measure, at least for shorter bounces.

Suppose that we wanted to experimentally resolve this.

The first step is to model the phenomenon. We have two competing explanations. So we need to work through two competing models.

Suppose that the dominant energy loss is due to surface impact. Assume that the coefficient of restitution is roughly constant for low impact velocities. Then we would expect energy to decay exponentially as a function of the number of impacts. If energy decays exponentially then velocity also decays exponentially. The delay from one impact to the next scales with velocity, so we are looking at a decaying geometric series. The total delay until the ping pong ball stops generating "tocks" will be bounded (the sum of a decaying geometric series).

If we plot inter-tock time versus tock number, we should get a decaying exponential and a finite time of last tock.

Suppose instead that the dominant energy loss is due to air resistance. If the resistance is linear than we have what is roughly a decaying exponential. But this time it is a function of elapsed time, not a function of number of bounces. This means that the sequence of "tocks" will continue indefinitely.

If we plot inter-tock time versus elapsed time, we should get a decaying exponential and an unbounded time of last tock. [If its quadratic air resistance then we get yet another decay curve]

Do the experiment, plot both ways and see if the data fits a predicted curve.