Friction/heat kinetic energy - reactive force

[haruspex]

I'm a bit late to this thread, but from reading your (@TonyCross) references to reactive force and some of your examples I get the impression you are confusing the force exerted by the wood, squash court wall, whatever, at the time of the impact with what happens later.

As a squash ball compresses, the lost KE is going into internal energy of the gas, deformation of the rubber, and some heat. The reaction force from the wall is equal and opposite to the force the ball exerts on the wall.
As the ball rebounds, the forces between the wall and the ball are smaller than at the corresponding point during the compression phase.
This is an example of hysteresis. If you were to plot a graph of force against degree of compression you would get a closed curve. The area inside the curve represents the energy lost as heat and increased internal energy of the gas. The raised internal energy also gets lost as heat as the ball cools.

Similarly with the nail in the wood. As the nail moves into the wood, the action and reaction are equal and opposite. But in this case there is negligible rebound.

 

[sophiecentaur]

Reactive force is important as the object in motion applies a force to the object it collides with this is known as reactive force, my question relates directly to the reactive force.

You want to distinguish between the Force and the Reactive force. But, for a start, they will be equal and in opposite directions. But also where is the Kinetic Energy actually dissipated? The Power (rate of dissipation) will be the Force times the relative velocities of the pads and the discs. That energy will be shared between the 'fixed' pads and the 'moving' disc (or whichever other pair you are discussion). How would you (would you want to?) distinguish between the work done on the pads and on the discs? The same magnitudes of force ('reactive ' or 'other') and velocities are involved for each. Each will get hot although the pad will heat up quicker because of its thermal mass and also the pad material will be designed to be more lossy. But what if you had steel pads and a steel disc? It's much more than just trying to find a dichotomy between the two forces involved.

 

[TonyCross]

I'm a bit late to this thread, but from reading your (@TonyCross) references to reactive force and some of your examples I get the impression you are confusing the force exerted by the wood, squash court wall, whatever, at the time of the impact with what happens later.

As a squash ball compresses, the lost KE is going into internal energy of the gas, deformation of the rubber, and some heat. The reaction force from the wall is equal and opposite to the force the ball exerts on the wall.
As the ball rebounds, the forces between the wall and the ball are smaller than at the corresponding point during the compression phase.
This is an example of hysteresis. If you were to plot a graph of force against degree of compression you would get a closed curve. The area inside the curve represents the energy lost as heat and increased internal energy of the gas. The raised internal energy also gets lost as heat as the ball cools.

Similarly with the nail in the wood. As the nail moves into the wood, the action and reaction are equal and opposite. But in this case there is negligible rebound.

Hi,
Thanks for your reply, re: the squash ball example, may I ask the following question, if we know the mass of the ball (25gram) and the velocity of the ball (50 m/s) just before it impacts the wall and we then have a total impact which may last let's say 10 milliseconds let's say for example that the initial compression phase of the ball takes 8 milliseconds and the rebound takes 2 milliseconds we could perhaps plot Force(y)/t(x) on a graph to show the unknown maximum value of F. Using the Impulse formula {I=F.delta t} Do you think that any heat generated by the ball in this example would detract from the area of this graph?

This is around about way of asking is the total impact force/reaction force with the wall less, the hotter the ball gets?

 

[A.T.]

This is around about way of asking is the total impact force/reaction force with the wall less, the hotter the ball gets?

If more heat is generated, the ball will have less kinetic energy so the total transmitted impulse is less. But this doesn't determine the force, which also depends on the duration of the collision.

 

[Dale]

This is around about way of asking is the total impact force/reaction force with the wall less, the hotter the ball gets?

For your specific example, yes, but this is not a general rule. It is easy to come up with alternative examples that violate the rule.

Remember, heat is a transfer of energy. Impulse/force is a transfer of momentum. Those are two separate quantities. They can be transferred independently.

 

[haruspex]

is the total impact force/reaction force with the wall less, the hotter the ball gets?

First, let's get the terms clear.

There is a reaction force from the wall all the time the two are in contact, and at each instant it is equal and opposite to the force the ball exerts on the wall.
I think you mean the rebound forces, i.e. the force pair (action and reaction) between ball and wall during the ball's expansion phase.

By "impact force/reaction force" I presume you mean the normal force generally, i.e. not separating compression phase and expansion phase.

Next, how are you defining total force here? The integral wrt time will give the change in momentum, while wrt to displacement will give change in KE.

Not sure what you are asking re "the hotter the ball gets". During early play the ball gets hotter, increasing its internal pressure and improving its bounce. The duration reduces, so the forces increase.
If you are asking how the amount of the temperature increase during a single bounce correlates with the magnitude of the forces, yes, I think that a cold ball will bounce less efficiently and so heat up more, and the forces will be less. But I see no general principle here applicable to other contexts; it is just a result of the characteristics of a squash ball. Is that what you had in mind in post #55, @Dale ?

we could perhaps plot Force(y)/t(x) on a graph to show the unknown maximum value of F. Using the Impulse formula {I=F.delta t}

That formula will only give you the average force over each time interval. You could model it as a spring to find the curves for each phase (one against time, one against displacement). As it goes from compression phase to expansion phase, the force will drop fairly suddenly, so the force/displacement curve will return via a lower path.

Do you think that any heat generated by the ball in this example would detract from the area of this graph?

The area bounded by the whole curve in the force/displacement graph shows the KE lost. I'm not a thermodynamicist, but I believe the right description is that during compression KE is turned into internal energy, some of the rubber, but mostly of the gas in the ball. During expansion, most of that turns back into KE. What remains as internal energy will dissipate out over time as heat.

 

[TonyCross]

If more heat is generated, the ball will have less kinetic energy so the total transmitted impulse is less. But this doesn't determine the force, which also depends on the duration of the collision.

I'm sorry I still cannot understand why a loss of kinetic energy doesn't determine the force, surely more velocity more force, less velocity less force.
I agree it depends on the duration of the collision, however Force in this example is not a fixed value it peaks at a certain value, during the upramp to this maximum Force point of the graph, kinetic energy is being lost in the form of velocity (can't be mass) due to the heating effect on the ball. The total Force is the total area contained under the curve on the graph. My contention is that the graphed Force area must be less due to the loss of kinetic energy in the heating process.
Consider a bicycle pump, you apply a force/time to the handle this causes the tyre to inflate, however it also
causes the pump to experience an increase in temperature, the force used on the pump must surely be divided into 2 parts - air into the tyre and some of the energy must cause the heating of the pump. If the pump didn't heat up, the tyre would be easier to inflate. Hence this does determine the force.

 

[TonyCross]

First, let's get the terms clear.

There is a reaction force from the wall all the time the two are in contact, and at each instant it is equal and opposite to the force the ball exerts on the wall.
I think you mean the rebound forces, i.e. the force pair (action and reaction) between ball and wall during the ball's expansion phase.

By "impact force/reaction force" I presume you mean the normal force generally, i.e. not separating compression phase and expansion phase.

Next, how are you defining total force here? The integral wrt time will give the change in momentum, while wrt to displacement will give change in KE.

Not sure what you are asking re "the hotter the ball gets". During early play the ball gets hotter, increasing its internal pressure and improving its bounce. The duration reduces, so the forces increase.
If you are asking how the amount of the temperature increase during a single bounce correlates with the magnitude of the forces, yes, I think that a cold ball will bounce less efficiently and so heat up more, and the forces will be less. But I see no general principle here applicable to other contexts; it is just a result of the characteristics of a squash ball. Is that what you had in mind in post #55, @Dale ?

That formula will only give you the average force over each time interval. You could model it as a spring to find the curves for each phase (one against time, one against displacement). As it goes from compression phase to expansion phase, the force will drop fairly suddenly, so the force/displacement curve will return via a lower path.

The area bounded by the whole curve in the force/displacement graph shows the KE lost. I'm not a thermodynamicist, but I believe the right description is that during compression KE is turned into internal energy, some of the rubber, but mostly of the gas in the ball. During expansion, most of that turns back into KE. What remains as internal energy will dissipate out over time as heat.

Hi,
Sorry if I am not making myself clear.
Thanks for the answers, I was only considering a single event, the instance when the cold ball makes contact with the wall, where the ball compresses and heats up then expands and rebounds.
I didn't want to consider the gas in the ball or the effect of gravity or the angle the ball hits the wall, just one question if the ball did not heat up would the ball hit the wall harder? Or perhaps the only result would be a loss in Ke on the rebound? My thought is that because the heating takes place during the compression phase, or the impact phase, this loss of Ke of the ball would impact on the force the ball contacts the wall, hence hitting the wall softer.

 

[Dale]

I see no general principle here applicable to other contexts; it is just a result of the characteristics of a squash ball. Is that what you had in mind in post #55, @Dale ?

Yes, that is what I meant.

 

[Dale]

just one question if the ball did not heat up would the ball hit the wall harder?

Yes. For that one question the answer is yes.

Just be aware that this is not a general principle and it cannot be applied in other contexts.

 

[TonyCross]

Yes. For that one question the answer is yes.

Just be aware that this is not a general principle and it cannot be applied in other contexts.

Thanks very much, I really must phrase my questions better in future...(:

 

[haruspex]

if the ball did not heat up would the ball hit the wall harder?

Again, you need to define hitting harder.
The peak force could be exactly the same, indeed, the whole force profile during compression could be the same, but the change in momentum will be greater since the forces during expansion will be greater.
Consider instead a spring mechanism. In one scenario the springs rebounds perfectly elastically; in another, a latch operates at maximum compression to inhibit any rebound.

 

[jbriggs444]

initial compression phase of the ball takes 8 milliseconds and the rebound takes 2 milliseconds

Those numbers suggest a rebound that is four times as fast as the impact. Is the ball coated with explosives?

 

[A.T.]

..., surely more velocity more force, less velocity less force.

No. You can have a very soft but elastic ball, that will have a lot of rebound but low peak force, because the impulse is spread out over a long time.

 

[haruspex]

I modeled it as a spring with two different constants, a higher one during compression and a lower during relaxation. As you can see, the rebound takes longer.

Read the next chart as clockwise around the loop. The enclosed area represents the net KE lost.

And for completeness:

 

[Physics guy]

Take a hammer and nail, I move the hammer at a constant speed with a certain mass this is kinetic energy or potential energy, it hits the nail with a Force (Newtons) the action of the nail has to be the reactive force causing
the nail to move into the wood with the transferred kinetic energy, or am I wrong.

First of all before giving my opinion you seem to be a heck lot confused with the concept of force and energy. Maybe because you think a little too much about it. Force and energy are concepts which help us understand and study the motion around us. They are compatible with each other but each helps us understand motion in a different way.

Energy can be of any type but one thing that we know for sure is that the total energy is conserved for an isolated system no matter in what situation. This has been strongly established. Kinetic energy is the energy due to the motion of the body and potential energy is the energy due to the position of a body with respect to another body. In ideal situations when there are no dissipative forces we can conclude that the total mechanical energy (sum of kinetic and potential energy) is conserved. The question of hammering a nail if you want to study it by forces then we apply an external force on the nail which causes the nail to move and it hits the wall bit stops because the wall exerts force causing the nail to stop (note that this has nothing to do with the reaction force due to the force of the nail on the wall). If we were to study it in terms of energy then we use the energy stored in the body to be converted to kinetic energy of the hammer which inturn transfers the energy to the nail causing it to enter the wall. But the dissipative forces exerted by the wall cause the kinetic energy to be converted to any other form of energy about which we are not concerned. I don't understand what potential energy has to do in this situation.

 

[TonyCross]

I modeled it as a spring with two different constants, a higher one during compression and a lower during relaxation. As you can see, the rebound takes longer.
View attachment 280488
Read the next chart as clockwise around the loop. The enclosed area represents the net KE lost.

View attachment 280489

And for completeness:
View attachment 280491

I have attempted to use your graph with some imaginary velocities, does this seem correct?

 

[A.T.]

I have attempted to use your graph with some imaginary velocities, does this seem correct?

To much imaginary stuff there, especially that "make it a triangle for the sake of ease" part.

Why would the rebound take longer, if the rebound force is the same as the compression force? You have now more impulse transmitted on the rebound, than on the compression. Also, if rebound force is the same as the compression force, then you have no energy dissipation, so why is the final speed smaller than the initial speed.

 

[TonyCross]

To much imaginary stuff there, especially that "make it a triangle for the sake of ease" part.

Why would the rebound take longer, if the rebound force is the same as the compression force? You have now more impulse transmitted on the rebound, than on the compression. Also, if rebound force is the same as the compression force, then you have no energy dissipation, so why is the final speed smaller than the initial speed.

All good points how would you approach the problem?

 

[A.T.]

All good points how would you approach the problem?

What is the question?

 

[Dale]

All good points how would you approach the problem?

I like the approach by @haruspex

 

[TonyCross]

I like the approach by @haruspex

So do I...

 

[Dale]

So do I...

OK, so just use that and don’t change it.

 

[TonyCross]

OK, so just use that and don’t change it.

Thanks Dale, you actually answered my initial question earlier in the thread, haruspex made a really good examination of the physics. I think we can now call this question well and truly answered. Thanks to everyone who gave their input to the discussion.