The physics of flywheel launchers (like tennis ball shooters)

[cardboard_box]

TL;DR Summary

need help with a more in depth look at the physics of a flywheel launcher. for example how the surface velocity or RPM effect the final velocity while torque doesn't

in flywheel shooters, the wheel spins around with rotational kinetic energy/ angular momentum, and a ball or other projectile makes contact with its surface and therefore is accelerated. the final velocity of the projectile depends on many factors, from its own inertia, elasticity, compression, initial velocity, coefficient of friction, wheel moment of inertia and so on. but how do these variables truly work? I know that inertia will tell me how hard it is to spin the wheel or spin/move the projectile. but a lot of those other variables aren't so clear.

1. what is the effect of slippage due to friction? by looking around I saw some say that you want more slippage and some who claimed no slippage will let the ball roll (in example of a hooded FRC style shooter). how does slippage effect linear or angular velocity

2. compression, compression obviously uses energy, but what about momentum? does the ball retracting back after being released count as momentum if it is only for a short while? also, in cases like balls compression usually seems to work better then for example disks or torus', what determines the effect of compression of effiecency?

3. maybe the biggest issue is why final velocity seems so heavily influenced by RPM/tangential velocity but not at all by torque, if torque is the driving force and accelerator, wouldn't it be the main cause of velocity? if I keep torque the same and don't change the total distance the projectile is in the shooter for, energy should remain the same and so velocity should be the same (mass cant change), yet if I increase RPM alone or even decrease torque with it, the final velocity seems to go up, so where does the force come from/how is tangetial velocity influencing it?

4. somewhat regarding point 3, if due to increased velocity the flywheel now has more momentum/rotational kinetic energy to "give" to the wheel, what makes it "give" more? does it work based on % of total momentum/RKE transferred? and how (if we can at all) do we calculate the instantaneous impulses/energy transfers of a system?

5. in many of the places I looked, they treated the wheel as a single tangential force like a point mass, so to equate angular momentum of the wheel and linear velocity of the projectile. how does that work in general or if I compress the ball and so instead of a single point of contact it is now an arc.



TL;DR: my understanding of physics claims that torque should be the deciding factor of final velocity instead of RPM, in addition, I am struggling to understand the importance of friction beyond giving us a grip and letting us rotate the object around, as well as trouble understanding the momentum conversions in the system + relation and application of energy.

if anyone could help with some of these questions, I'd be grateful.

 

[Lnewqban]

Welcome!
Just imagine the extreme case of a huge torque and very small rotational speed.
The tangential velocity of the impeller is what determines the linear velocity of the projectile rather than the torque.
Torque is cosumed mainly in the impulse provided to the projectile, which mass is generally much smaller than the moment of inertia of the impeller.

 

[A.T.]

TL;DR: my understanding of physics claims that torque should be the deciding factor of final velocity instead of RPM,

That would be true If the inertia of the flywheel was negligible.

in addition, I am struggling to understand the importance of friction beyond giving us a grip and letting us rotate the object around,

A tangential force does both: accelerate linearly and angularly.

 

[cardboard_box]

Welcome!
Just imagine the extreme case of a huge torque and very small rotational speed.
The tangential velocity of the impeller is what determines the linear velocity of the projectile rather than the torque.
Torque is cosumed mainly in the impulse provided to the projectile, which mass is generally much smaller than the moment of inertia of the impeller.

hi, I get that RPM is the cause of velocity and not torque, my problem is rather understanding why its so, when normal it is force that causes acceleration and therefore velocity. also, what exactly do you mean the torque is consumed in the impulse? would that not influence the momentum and therefore the velocity?

 

[cardboard_box]

That would be true If the inertia of the flywheel was negligible.


A tangential force does both: accelerate linearly and angularly.

why would the inertia of the wheel determine whether the torque would be the deciding factor or not? isnt the inertia of the wheel simply how hard it is for the wheel to be influenced by the torque?

also the problem is where the tangential force comes from, if it comes from the torque then I am still stuck on why a higher torque doesn't result in a faster projectile velocity

 

[A.T.]

why would the inertia of the wheel determine whether the torque would be the deciding factor or not?

A very heavy flywheel can easily accelerate a ball, while free wheeling (no input torque required).

 

[cardboard_box]

A very heavy flywheel can easily accelerate a ball, while free wheeling (no input torque required).

yes that makes sense to me its conservesion of momentum/energy, but then where does the force come from? How can we calculate it and how it will effect the projectile? If we know the projectile and the wheel's mass + moment of inertia we are still missing the source of a force. And I dont think you can derive force simply based on initial momentum of system's objects + how they will interact.

 

[Baluncore]

I think you need to simplify the analysis by considering a two wheel launcher, rather than a single wheel and a hood. The ball will be held momentarily, between the wheels, being accelerated to the average circumferential velocity of the wheels, with the ball spin being determined by the differential velocity of the two wheels. Initial ball compression losses, will be recovered during release.

 

[cardboard_box]

I think you need to simplify the analysis by considering a two wheel launcher, rather than a single wheel and a hood. The ball will be held momentarily, between the wheels, being accelerated to the average circumferential velocity of the wheels, with the ball spin being determined by the differential velocity of the two wheels. Initial ball compression losses, will be recovered during release.

The problem is that 1. Compression costs energy, so even if it goes back to normal form after release, the momentum should still be less. Where does the momentum go and how?
2. why does it accelerates to surface velocity of the wheels? There must be a force applied and it doesn't seem to be the torque. The only thing that somewhat makes sense is if its due to friction with the wheel thats pushing it, but said friction force must come from a diffrent force pushing on it, and should have nothing to do with the surface velocity of the wheels.

 

[Baluncore]

Compression costs energy, so even if it goes back to normal form after release, the momentum should still be less.

Compression invests energy that is recovered almost immediately. With a two wheel analysis you can see how the ball is trapped, compressed, accelerated, then expands as it is released.

The only thing that somewhat makes sense is if its due to friction with the wheel thats pushing it, but said friction force must come from a diffrent force pushing on it, and should have nothing to do with the surface velocity of the wheels.

The ball contact with the wheels, is effectively a friction clutch, that locks momentarily as the ball is compressed, as it passes through the closest gap between the wheels. The ball is forced against the wheel by the restriction presented by the other wheel or the hood.

It seems you are denying either the compression of the ball, or of the hood, that is essential to locking the friction surfaces.

 

[A.T.]

... but said friction force must come from a diffrent force pushing on it...

There is no such requirement for friction. When two balls collide at an oblique angle, there is a normal force and a tangential friction force at the contact patch. No other force must be pushing either ball for that to occur.

 

[cardboard_box]

Compression invests energy that is recovered almost immediately. With a two wheel analysis you can see how the ball is trapped, compressed, accelerated, then expands as it is released.


The ball contact with the wheels, is effectively a friction clutch, that locks momentarily as the ball is compressed, as it passes through the closest gap between the wheels. The ball is forced against the wheel by the restriction presented by the other wheel or the hood.

It seems you are denying either the compression of the ball, or of the hood, that is essential to locking the friction surfaces.

but the energy isnt recovered in a very useful form, the ball pushing back to normal shape doesn't add much velocity to it, so it would make sense it would have less momentum, but then what happens in terms of momentum during compression? does the ball being compressed simply make momentum transfer less efficient? if so how do we know what will be efficient and inefficient ways of transfering momentum between 2 objects?

I get that friction is dependent on the normal force due to compression, but then why does compressing a ball more and more not really increase its final vleocity? and wouldnt the normal force between 2 rotating wheels also be due to the torque of the wheels? in which case projectile velocity should increase with torque.

 

[cardboard_box]

There is no such requirement for friction. When two balls collide at an oblique angle, there is a normal force and a tangential friction force at the contact patch. No other force must be pushing either ball for that to occur.

yes but what is the source of these forces? isn't the tangential force of the wheel simply the torque/radius?

 

[Baluncore]

but the energy isnt recovered in a very useful form, the ball pushing back to normal shape doesn't add much velocity to it, ...

Notice how compression and expansion of the ball occurs while the two contact patches between the ball and the wheels are not parallel. That accounts for the compression energy investment, and its recovery.

As the ball is compressed between the wheels, the wheels are slowed down slightly. Friction then locks the ball between the wheels while the mass of the ball is being accelerated, which slows the wheels more significantly. Then, as the wheels begin to separate, the ball expands, accelerating the wheels slightly, until the instant when contact is lost and the ball is moving independently after expansion.

I get that friction is dependent on the normal force due to compression, but then why does compressing a ball more and more not really increase its final velocity?

Because once the compression is sufficient to lock the frictional contact, the ball surface will be travelling at the circumferential speed of the wheel. More compression force cannot change that situation.

 

[A.T.]

When two balls collide at an oblique angle, there is a normal force and a tangential friction force at the contact patch.

yes but what is the source of these forces?

The objects deform slightly due to their relative motion and resist that deformation.

 

[cardboard_box]

Notice how compression and expansion of the ball occurs while the two contact patches between the ball and the wheels are not parallel. That accounts for the compression energy investment, and its recovery.

As the ball is compressed between the wheels, the wheels are slowed down slightly. Friction then locks the ball between the wheels while the mass of the ball is being accelerated, which slows the wheels more significantly. Then, as the wheels begin to separate, the ball expands, accelerating the wheels slightly, until the instant when contact is lost and the ball is moving independently after expansion.


Because once the compression is sufficient to lock the frictional contact, the ball surface will be travelling at the circumferential speed of the wheel. More compression force cannot change that situation.

how do you calculate or relate the momentum loss due compression? momentum should be a vector quantity so how does a ball being compressed and decompressed to the same dimensions on both sides result in less momentum in the ball while also less final momentum in the wheel?

do you have any sources where I can read more in depth about friction accelerating a "locked" object to surface velocity? my intuition still says that its the force of the wheel that should accelerate, because if we for example had 2 cubes and we move one so that it "rubs" and touches the other one and the friction between the 2 is enough to prevent any sliding, then wouldn't it be the force of the first moving cube that determine the acceleration of the other one? the velocity of the 2 fuzed cubes should drop from when it was only one cube moving due to conservation of momentum, does it slow down due to friction + unmoving cube's inertia? if so is there a way to calculate it with cubes or with the wheel and projectile? also, what happens if the wheel or cube doesn't have enough friction to simply cancel any slippage, what happens if we move from static to kinetic friction? since kinetic friction can't be higher then static friction does it mean our projectile shouldn't be able to move at all even though real life examples show it can move with some slippage?

and finaly, in a lot of the examples I saw they specificaly assume the ball isn't being compressed at all and so is touching the wheel on a single point mass pushing tangentialy, does that mean in those examples it shouldnt work at all because without any normal force we wouldnt have any friction?

 

[cardboard_box]

The objects deform slightly due to their relative motion and resist that deformation.

im sorry, I am not sure i am following, the objects deform due to being compressed and resist that compression, but isnt the compression coming from the torque of the wheel pushing the projectile against the hood? if we didnt have sufficient enough torque we wouldn't be able to compress it. since friction force is dependent on the normal force wouldn't it mean that final velocity is based on torque + compression? which makes no sense to me since compression takes wheel energy and usually makes the final velocity slower and increasing torque even without decreasing RPM doesn't seem to increase final velocity.

 

[Baluncore]

how do you calculate or relate the momentum loss due compression? momentum should be a vector quantity so how does a ball being compressed and decompressed to the same dimensions on both sides result in less momentum in the ball while also less final momentum in the wheel?

Compression and expansion of the ball are not changes in momentum, they are changes in energy. As the wheels come together, compressing the ball, the force of compression multiplied by the reduction in wheel separation, increases the potential energy of the elastic ball. It also slows the wheels slightly.
That process is then reversed.

Ball momentum is irrelevant. The ball velocity, and the kinetic energy, are increased.

 

[cardboard_box]

Compression and expansion of the ball are not changes in momentum, they are changes in energy. As the wheels come together, compressing the ball, the force of compression multiplied by the reduction in wheel separation, increases the potential energy of the elastic ball. It also slows the wheels slightly.
That process is then reversed.

Ball momentum is irrelevant. The ball velocity, and the kinetic energy, are increased.

but why would it have any effect on ball velocity? velocity is a vector quantity and so the ball compressing and decompressing in the Z axis shouldn't increase its X axis velocity. if compression doesn't cause change in a ball's momentum it can't increase its velocity or lower wheel velocity, and if does cause change in momentum then how? the ball's compression shouldn't add more velocity, but it takes velocity from the wheel. so if there is an effect of compression on momentum how exactly does it happen?

 

[A.T.]

... if we didnt have sufficient enough torque we wouldn't be able to compress it. ...

As already explained, if the flywheel has enough inertia, then it can launch the ball even while free wheeling, without any torque applied by the motor. It will slow down a bit, so a torque from the motor is needed to speed it up again between subsequent launches.

 

[Baluncore]

the ball compressing and decompressing in the Z axis shouldn't increase its X axis velocity.

But the wheel surfaces are not parallel at the entry and the exit. As the ball moves toward the point of release, the Z force provides a torque to the wheels, that accelerates the wheels, and therefore increases the velocity of the ball, up to the moment it is released.

 

[cardboard_box]

But the wheel surfaces are not parallel at the entry and the exit. As the ball moves toward the point of release, the Z force provides a torque to the wheels, that accelerates the wheels, and therefore increases the velocity of the ball, up to the moment it is released.

I am not sure I understand, in a linear shooter (2 wheels on each side) the ball's compression is due to the wheels and (i think) their torque, I don't see how the compression would accelerate the wheels or why it would matter if it already de-accelerated them when they had to compress the projectile. the wheels don't allow for the ball to de-compress until it's outside the shooter, and the force the compressed projectile applies to the wheels shouldn't necessarily accelerate them back, and if it does it then best case scenario it simply accelerates them back to the velocity they were before having to compress it, otherwise it breaks conservation of momentum/energy.

 

[cardboard_box]

As already explained, if the flywheel has enough inertia, then it can launch the ball even while free wheeling, without any torque applied by the motor. It will slow down a bit, so a torque from the motor is needed to speed it up again between subsequent launches.

yes that I know, the problem is that even though we know for a fact this happens, we don't know where the force comes from. something must be applying a force on the projectile, otherwise there is no impulse and no change in momentum. if it isn't motor torque then what is it? simply knowing the momentum of a system at start isn't enough to know the end momentum of each subsystem, we need to find the impulse which relays on a force I dont understand and a time that also complicates things since in theory it should mean a lower speed flywheel would have more time to apply the force and so greater final velocity.

basically if we know the momentum of object 1 and 2 how do we figure out the impulse they will cause on each other without looking at each of their momentum post-impulse?

 

[A.T.]

simply knowing the momentum of a system at start isn't enough to know the end momentum of each subsystem,

For simple elastic collisions that is enough. But not for this machine, with unknown energy dissipation and more complex interaction.

However, if friction is known/assumed to be high enough, such that contact surfaces end up at equal velocities, then you can use this constraint in the solution.

 

[cardboard_box]

For simple elastic collisions that is enough. But not for this machine, with unknown energy dissipation and more complex interaction.

However, if friction is known/assumed to be high enough, such that contact surfaces end up at equal velocities, then you can use this constraint in the solution.

you are speaking of energy loss to outside the system, I don't really know how to relate it to momentum. and even if we work with KE instead it isn't easy, since we don't know what the energy distribution between the 2 subsystems will be. whether we use energy or momentum the problem still remains that we don't know how much momentum or energy it transfers without looking at end results. with momentum I still don't get the source of the force and the time is counter intuitive since it will mean a slower wheel should have either more or the same final velocity of a faster one. energy is based on distance and since the distance the ball is in contact with the wheel doesn't change it would mean velocity is determined by the unknown force alone.
Another problem is why friction accelerates a ball to the same speed as the wheels surface velocity, even though it should keep the same system-wide momentum and overall de-accelerate, if we drop a ball into a flywheel for a 2 cm or 2 seconds contact or 4 cm/4 seconds contact shouldn't their velocities be different? how do we know when/how the projectile will equalize to wheel velocity? and that also ignores rolling/rotating.

what exactly happens to the projectile inside the shooter on the more instantaneous level, what is the accelerating force and how does it behave in relation to other variables?

 

[A.T.]

what exactly happens to the projectile inside the shooter on the more instantaneous level, what is the accelerating force and how does it behave in relation to other variables?

To investigate this you could use high-speed cameras or a dynamic finite element simulation. Even simple collisions can get arbitrarily complex if you want to know what happens in detail.

 

[Baluncore]

The ball is being compressed from first contact with the wheels, until it is half-way through. The ball is expanding from when it is half-way through, until it loses contact with the wheels. It is symmetrical.

... the wheels don't allow for the ball to de-compress until it's outside the shooter, ...

It is symmetrical. Are you saying the ball is also being compressed before it enters and contacts the wheels?

 

[cardboard_box]

The ball is being compressed from first contact with the wheels, until it is half-way through. The ball is expanding from when it is half-way through, until it loses contact with the wheels. It is symmetrical.

It is symmetrical. Are you saying the ball is being compressed before it enters and contacts the wheels?

no you are correct, my mistake. but the issue still remains that even if it does de-compress and while doing so it does accelerate the wheels, why would it matter? at that point the projectile barely touches the wheels and their velocity can only go back to the amount they lost to compress it. it can somewhat explain where the momentum "goes" in compression but that is assuming the projectile decompression perfectly pushes the wheels, which I don't know if it will do so. also, how does that work in a hooded shooter? there I highly doubt the de-compression will perfectly push the wheels.

 

[Baluncore]

There are terminology problems here. You need to provide an annotated diagram or picture of the device you are discussing.

 

[cardboard_box]

There are terminology problems here. You need to provide an annotated diagram or picture of the device you are discussing.

sure thing, here are some videos/pictures of hooded and linear flywheel shooters:

linear:

hooded:

notes: these are for ball only, but for linear shooters you are more then welcomed to treat it as a torus or disk as well (for pictures/videos on that just look up FRC 2024 and you will find tons of flywheel shooters)

hope this is helpful.

 

[Baluncore]

also, how does that work in a hooded shooter? there I highly doubt the de-compression will perfectly push the wheels.

The profile of the hood will determine the rate of compression and the rate of expansion, and they must balance.

After the wheel has pulled in the ball, the ball, as it exits, will push against the wheel and the hood. Since the gap widens towards the exit, that must return the compressive energy to the wheel.

The gas inside the ball will be heated by compression, it will then cool by expansion as it exits, before there has been time for significant heat loss. Perfection is impossible, there will be minor energy losses from the gas to the ball, and from the initial sliding friction to the ball.

 

[cardboard_box]

The gas inside the ball will be heated by compression, it will then cool by expansion as it exits, before there has been time for significant heat loss. Perfection is impossible, there will be minor energy losses from the gas to the ball, and from the initial sliding friction to the ball.

wait, I'm not sure I follow, what gas inside a ball? what if the ball is solid or has air or something else inside it?

 

[Baluncore]

wait, I'm not sure I follow, what gas inside a ball? what if the ball is solid or has air or something else inside it?

Air is a mixture of different gasses. All the balls I see in the launcher videos are typically air-filled. Gasses exhibit a thermal inefficiency when they are compressed and expanded, heated and cooled.

The ball may be made from a closed cell foam, in which case it will be as lossy as a gas filled ball. If the ball was filled with only one species of gas, it would probably be dry nitrogen, to avoid chemical reactions with the envelope during hot impacts.

If the ball is solid, maybe a polymer, then it will have elastic properties, and will probably bounce more efficiently than a gas filled ball.

A loaded, or a weighted ball, could be liquid filled, or a bean bag. I have trouble thinking of a sport that is played with a weighted ball, or why it would need a launcher. Weighted balls are more like weapons, likely to cause injuries.

 

[Lnewqban]

hi, I get that RPM is the cause of velocity and not torque, my problem is rather understanding why its so, when normal it is force that causes acceleration and therefore velocity. also, what exactly do you mean the torque is consumed in the impulse? would that not influence the momentum and therefore the velocity?

Yes, you are correct.
Certain amount of energy flows from the launcher to the projectile.
If the launcher is not fed with replenishing energy (like from an electric motor), and keeps giving impulse to several projectiles consecutively, it will simultaneously slow down more, and more and will deliver weaker and weaker torque, after each launch.

 

[cardboard_box]

A loaded, or a weighted ball, could be liquid filled, or a bean bag. I have trouble thinking of a sport that is played with a weighted ball, or why it would need a launcher. Weighted balls are more like weapons, likely to cause injuries.

it is used in tennis pitching machines to practice.