Flywheel Physics Question

I apologize in advance for anything I say that is not “proper” from a physics standpoint. Feel free to correct me, it is my weak area for sure. Here is the question:

If a single flywheel system is spinning at 8 m/s (tangential velocity) and a ball is fed through it, in theory it would leave the flywheel also going 8 m/s correct? I know that is unrealistic, but in a perfect world.

So my question is, if I have a double flywheel system and both of those flywheels are spinning at 8 m/s, will the ball in theory leave at 16 m/s or still 8 m/s? Basically, will more flywheels result in a lower overall speed required to launch balls? (from a perfect world physics standpoint)

Thanks.

Edit: Sorry, accidentally posted this under the general forum…any mod can feel free to move it if they want.

It would still be traveling at 8m/s, not 16m/s.

The idea behind 2 flywheels = lower speed (I believe) is that with just one flywheel the ball will exit the flywheel spinning quickly but actually traveling forward at < 8 m/s. Not sure how bad this is, though.

I thought single wheel has the ball leaving at half the velocity, 4m/s in your example.

Yes, that’s what I thought, too.

We’ve been beating the physics of flywheels to death on this thread:

https://vexforum.com/t/physics-of-the-flywheel-launcher/29357/1

But, really, we’ve only scratched the surface. There’s plenty left to discuss, I think, especially when it concerns the variability of the ball squishiness, mass, etc.

If by speed you mean rotational speed of the wheels, then the short answer is: yes. But, of course, there is a minimum tangential velocity that must be maintained in order to provide the balls with enough initial velocity to make it to the target. More wheels provide more kinetic energy, which means that the wheels don’t slow down as much when the ball is accelerated by the wheels.

The single wheel launcher must spin faster not only because of geometric reasons (see the diagram) but also for kinetic energy reasons, otherwise it will slow down too much to launch the ball at its intended target. In the single wheel launcher, part of the kinetic energy is sucked up in getting the ball to spin (which might help it in other ways, however).

Good, I’m not going crazy. That’s also what we observed at the last meeting and why I’m working on a single flywheel design that runs at 4000 rpm.

To quote one of my kids: Eww may gad! :eek:

But, really, we’ve only scratched the surface. There’s plenty left to discuss, I think, especially when it concerns the variability of the ball squishiness, mass, etc.

If by speed you mean rotational speed of the wheels, then the short answer is: yes. But, of course, there is a minimum tangential velocity that must be maintained in order to provide the balls with enough initial velocity to make it to the target. More wheels provide more kinetic energy, which means that the wheels don’t slow down as much when the ball is accelerated by the wheels.

The single wheel launcher must spin faster not only because of geometric reasons (see the diagram) but also for kinetic energy reasons, otherwise it will slow down too much to launch the ball at its intended target. In the single wheel launcher, part of the kinetic energy is sucked up in getting the ball to spin (which might help it in other ways, however).

I thought I understood what you were saying until I went back and read the thread you linked and came across this:

hold on, I if my wheel is 10 inches in circumference, and spinning at 600 RPM
then the edge of the wheel would move at 100 inches per second, and then the ball would leave at 100 inces per second in a perfect world right?

This is what I’m confused about as well. Based one what people are saying it would leave at 1/2 that speed?

We tested a flywheel with 1680 theoretical rpms and THAT was scary. I can’t imagine 4000 rpm. I’m not walking anywhere near these robots! :eek:

hopefully if anything goes bad, the only thing that will come flying off is soft rubber

One side of the ball would be accelerated to 100 inches per second (forgetting about all the things that will slow it down, friction etc.), however, the other side of the ball is stationary. See the diagram that FullMetalMentor posted.

We ran one last weekend with that theoretical speed but not enough power (2 motors). We probably achieved 3000 rpm, wasn’t too bad.

how is one side of the wheel stationary, the way I have this planned is that there will be a ramp that goes 180 degrees around a wheel, and the ball will definitely be moving along that ramp, so over that 180 degree distance wont the ball be accelerated to the speed of the wheel?

I learned this concept in my physics class this year and it threw me off too. Think of it this way: when a bicycle wheel is rolling along the ground, relative to the ground the bottom of the wheel is not moving, correct? If the bottom of the wheel was moving relative to the ground then the wheel would be slipping.

The center of the wheel is moving forward at velocity V m/s, relative to the ground. Because the bottom of the wheel is technically moving backwards relative to the center of the wheel, the bottom has a tangential velocity of 0 m/s relative to the ground. Because the top of the wheel is moving forwards relative to the center of the wheel (moving at V m/s), the top of the wheel has a tangential velocity of 2V m/s relative to the ground. This explains why the top spokes of a wheel are blurred when the bike is moving and we can see the bottom spokes clearly.

In our situation the bicycle wheel is the ball being launched. In a single-wheel flywheel, the point of contact between the ball and the flywheel launcher is the top of wheel in the bicycle wheel example and the point of contact between the ball and the surface it rolls along is the bottom of the bicycle wheel. In a two-wheel flywheel, both sides of the ball are accelerated at the relatively the same rate instead of just one side.

Does this clear up any confusion? This is definitely not an intuitive concept.

I think that diagram is suppose to show the ball in an of “instant” of time. The flywheel would be at the top of the diagram, and the bottom of the diagram would be the wall of whatever kind of aiming system you use. In that instant of time, the bottom of the ball would be “stuck” to the wall and therefore not moving linearly. But the center of the ball would be moving linearly at v. The flywheel (top of the ball) would need to have a tangential velocity of 2v in order for the geometry to work out so the center of the ball is moving at v. Although the bottom of the ball is “stuck” to the wall in one instant of time, the ball would certainly be rolling along from one instant in time to another. But then the ball would be touching a different part of the wall and a different part of the flywheel as it went rolling along with its center moving at velocity v. This is why the single flywheel imparts lots of spin to the ball. Am I making any sense? Did that help at all?

how could all parts of the ball NOT be moving at the same speed? aren’t all parts of the ball tied together physically?

It’s because it’s not just translation, it’s also rotation. That’s why the top is moving 2v, because the whole ball is moving forward while it’s rotating.

Here’s a diagram that helped me. The vectors for rotation and the vectors for translation add up to create that diagram posted above:

The bottom velocity vectors cancel out (v - v = 0). The top velocity vectors add up together to 2v (v + v = 2v). Hope that helps

With a single wheel launcher, the ball is made to spin. It’s the spinning aspect that freaks out people’s perspective of what’s happening linearly. With a different kind of mechanism than what we’re talking about there, you could actually end up with a linear velocity and a spin rate that causes the bottom of the ball to appear to be moving backwards for an instant.

Ok I get it now, the diagram explained it well for me, so pretty much that extra velocity it lost in the fact that you are also acceleration the ball to rotate at V, and then to also translate at V, so you need to input 2V, right?