[Kryptid]

what happens when two torques of the opposite rotations of two wheels cancel out?

Then there is no net angular momentum. What did you think would happen?

[Jaaanosik (OP)]

what happens when two torques of the opposite rotations of two wheels cancel out?

Then there is no net angular momentum. What did you think would happen?

Kryptid,
the balls lost the velocity/energy, the wheels are spinning, the velocity/energy went to the wheels rotation and the rotation is not slowing down the spaceship, the same way as the rotation of CMGs on the ISS is not slowing down the space station.
Do you understand that you are going against the CMG experiment?
Jano

[Kryptid]

Kryptid,
the balls lost the velocity/energy, the wheels are spinning, the velocity/energy went to the wheels rotation and the rotation is not slowing down the spaceship, the same way as the rotation of CMGs on the ISS is not slowing down the space station.

The scenarios are completely different. A gyroscope spun by a motor can only impart angular momentum to a spacecraft at most. And that's only for the spacecraft: the gyroscope's angular momentum will perfectly counter the spacecraft's angular momentum such that there is no net change in angular momentum in the system.

In your scenario, where the wheel is spun by throwing a ball at it, the same is not true. There is an exchange of both linear momentum and angular momentum between the ball and the wheel. That doesn't occur when a motor spins a gyroscope.

Do you understand that you are going against the CMG experiment?

No, I'm not. I explained the difference.

And since you still seem to be trying to argue that angular momentum and linear momentum can be interconverted, I'm going to post this yet again: https://van.physics.illinois.edu/qa/listing.php?id=24173&t=is-linear-momentum-converted-to-angular-momentum

The two conservation laws- linear and angular momentum- are absolutely separate. Neither one can be converted to the other.

So in accordance with the above quote, you're trying to break conservation laws.

[Jaaanosik (OP)]

Kryptid,
we can play this game for a long time:

https://ccrma.stanford.edu/~jos/pasp/Relation_Angular_Linear_Momentum.html





Jano

[Kryptid]

Yes, there is a relationship between the two. There is also a relationship between electric charge and energy. Electric potential energy is a function of total electric charge and distance. Does that mean that energy and electric charge can be interconverted? No, it doesn't. Electric charge and energy both have their own conservation laws. Likewise, linear momentum and angular momentum, although related, have their own conservation laws as well. Nothing you have posted contradicts that.

[Jaaanosik (OP)]

Kryptid,
I'll tell you what.
This does not go anywhere.
I am going to find a software for animation and I'll prepare a video.
We'll talk then.
It is going to be a few weeks though.
... stay tuned, ... to be continued,
Jano

[Jaaanosik (OP)]

All,
before I'll post the video I am going to show that:

Energy rules!!!

Energy is superior to momentum in showing what is going on.
This is a challenge to all the mathematicians/physicists that want to defend the theorem saying linear momentum has nothing to do with the angular momentum.

The kinetic energy got converted to potential energy.
Where is the conservation of the linear momentum?

This is not a new theory, this is the TRUTH!
The truth is that linear momentum got converted to increased tension, higher vibrations of spring atoms.
The vibrations have angular momentum deep down so the angular momentum got bigger in expense of the linear momentum.
Jano

[Kryptid]

Energy is superior to momentum in showing what is going on.

They are both equally important because both of them are equally conserved.

This is a challenge to all the mathematicians/physicists that want to defend the theorem saying linear momentum has nothing to do with the angular momentum.

Nobody said that they have nothing to do with each other. They are, indeed, related. What was stated was that they have separate conservation laws. Both must be conserved independently of the other.

Where is the conservation of the linear momentum?

The linear momentum of the system started out at zero because the linear momentum of the ball on the left canceled out the linear momentum of the ball on the right. Momentum is a vector. If two objects have a linear momentum that is equal in magnitude and opposite in direction, then the total linear momentum is zero. When both balls come to a stop after compressing the spring, the linear momentum is still zero. So there is no violation of conservation of linear momentum.

This is not a new theory, this is the TRUTH!
The truth is that linear momentum got converted to increased tension, higher vibrations of spring atoms.
The vibrations have angular momentum deep down so the angular momentum got bigger in expense of the linear momentum.
Jano

The total angular momentum is zero both before the balls hit the spring and after they compress the spring. There is no violation of conservation of angular momentum either.

[Jaaanosik (OP)]

...
The linear momentum of the system started out at zero because the linear momentum of the ball on the left canceled out the linear momentum of the ball on the right. Momentum is a vector. If two objects have a linear momentum that is equal in magnitude and opposite in direction, then the total linear momentum is zero. When both balls come to a stop after compressing the spring, the linear momentum is still zero. So there is no violation of conservation of linear momentum.
...

Kryptid,
why it took so long to understand this?
The balls have zero momentum at the end.
Hitting the side walls cannot stop the spaceship!
Jano

[Kryptid]

Kryptid,
why it took so long to understand this?

It didn't. I knew this from the beginning.

The balls have zero momentum at the end.

Yes, because they transferred their linear momentum to the rails that they are riding on.

Hitting the side walls cannot stop the spaceship!

Hitting the walls isn't what is stopping the spaceship. It's the transfer of linear momentum to the rails which are attached to the spaceship.

[Jaaanosik (OP)]

Kryptid,
why it took so long to understand this?

It didn't. I knew this from the beginning.

The balls have zero momentum at the end.

Yes, because they transferred their linear momentum to the rails that they are riding on.

Hitting the side walls cannot stop the spaceship!

Hitting the walls isn't what is stopping the spaceship. It's the transfer of linear momentum to the rails which are attached to the spaceship.

Kryptid,
the spaceship is stopped after ds in the forward direction.
The second part of the U turn is not there to bring the spaceship back.
New balls push the spaceship forward again.
Stop after ds, new balls push again, stop , push, ....
There is nothing bringing the spaceship back!!!
There is a net forward momentum over 'longer' dt,
Jano

[Kryptid]

Kryptid,
the spaceship is stopped after ds in the forward direction.
The second part of the U turn is not there to bring the spaceship back.
New balls push the spaceship forward again.
Stop after ds, new balls push again, stop , push, ....
There is nothing bringing the spaceship back!!!
There is a net forward momentum over 'longer' dt,
Jano

I don't know what you mean by "ds" or "dt".

Let's take this back to the very beginning before anything in the ship starts moving. The linear momentum of the ship and all of its components is zero. The angular momentum of the ship and all of its components is also zero. If you are going to argue that a system with zero linear momentum and zero angular momentum can end up with non-zero momentum, then you are breaking conservation of momentum. Arguing that linear momentum and angular momentum can be interconverted will not solve this problem. If there is zero linear momentum, then you couldn't get non-zero angular momentum of out it. Likewise, if there is zero angular momentum, you can't get non-zero linear momentum out of it.

You cannot make this ship have net momentum without violating conservation of momentum.

[Jaaanosik (OP)]

Let say the spaceship moved 10m upwards/forward in the barycenter frame after initial shooting of the balls.
The one quarter rotation (90 degrees) stopped the spaceship in the barycenter frame.
The balls stopped moving in the backward direction as well, they do not have any backward velocity, only sideways velocity.
The balls stop hitting the side walls.
What is going to bring the spaceship back 10m?
Jano

[Kryptid]

What is going to bring the spaceship back 10m?

Nothing. The ship can move 10 meters upward just fine so long as the balls also move downward 10 meters in order to keep the barycenter in the same place.

[Jaaanosik (OP)]

What is going to bring the spaceship back 10m?

Nothing. The ship can move 10 meters upward just fine so long as the balls also move downward 10 meters in order to keep the barycenter in the same place.

So the gain is 10m forward.
The second stroke does not change the 10m forward gain because there is no rotation involved.
Anything pulled back after shooting the balls from the back is gained back up front.
We have the balls back up front with the net gain 10m forward,
Jano

[Kryptid]

The second stroke does not change the 10m

It has to if you want to avoid violating conservation of momentum.

Anything pulled back after shooting the balls from the back is gained back up front.
We have the balls back up front with the net gain 10m forward,

The very act of bringing the balls back to the front must bring the rest of the ship back by an equal amount due to Newton's third law. A force must be applied on the balls to move them to the front. An equal and opposite force must therefore be present as well. Whatever it is that is experiencing that equal and opposite force will then move backward by the same amount.

Going along what I was saying before, the angular momentum and linear momentum are both zero before the engine starts. Then the first stroke happens. The balls move down the rails and stop once they hit the walls. So what is the angular momentum present after the balls stop? It must be zero. So, what must the overall linear momentum be at the end of the stroke in order for the total momentum at the end of the stroke to match the total momentum before the stroke?

[Jaaanosik (OP)]

The second stroke does not change the 10m

It has to if you want to avoid violating conservation of momentum.

Anything pulled back after shooting the balls from the back is gained back up front.
We have the balls back up front with the net gain 10m forward,

The very act of bringing the balls back to the front must bring the rest of the ship back by an equal amount due to Newton's third law. A force must be applied on the balls to move them to the front. An equal and opposite force must therefore be present as well. Whatever it is that is experiencing that equal and opposite force will then move backward by the same amount.

Going along what I was saying before, the angular momentum and linear momentum are both zero before the engine starts. Then the first stroke happens. The balls move down the rails and stop once they hit the walls. So what is the angular momentum present after the balls stop? It must be zero. So, what must the overall linear momentum be at the end of the stroke in order for the total momentum at the end of the stroke to match the total momentum before the stroke?

We need to remember conservation of energy as well.
Not all of the kinetic energy is used to stop the spaceship.
Part of the energy is in the rotation of the balls.
I do not mean just big circle rotation but physical rotation of balls itself.

Never mind, let's defer this discussion,
Jano

[Kryptid]

We need to remember conservation of energy as well.

I haven't forgotten about it.

Not all of the kinetic energy is used to stop the spaceship.

Correct. Just as much is used to stop the balls as it is to stop the ship.

Part of the energy is in the rotation of the balls.
I do not mean just big circle rotation but physical rotation of balls itself.

I'm well aware of that. It doesn't make any difference. In the barycenter's frame, half of the rotational energy in the balls came from the kinetic energy that was already in the balls. The other half came from the kinetic energy of the ship.

Might you know the answer to this question?

So, what must the overall linear momentum be at the end of the stroke in order for the total momentum at the end of the stroke to match the total momentum before the stroke?

0 angular momentum + 0 linear momentum = 0 angular momentum + ? linear momentum. What is the value of the question mark?

[Jaaanosik (OP)]

Kryptid,
I got a hand on Blender sw to create a video.
Here is the script what's going to be in it.



CMG - four rotating wheels in a tetrahedron configuration.
If a torque is applied to CMG - software control will regulate the speed of the wheels so the CMG as a system is NOT going to rotate.
CMGs are free in space. They are not attached to the space ship. The red lines are strings.
The spaceship will start to reel in.
CMGs do not rotate, only the spaceship moves forward in the CMG/background frame.
CMGs shut down prior to hitting the back of the spaceship.
Mass of the CMG on ISS is multiple times less than the mass of the ISS. CMG still works.
We can assume the same thing. CMG mass is smaller than the rest of the spaceship.

Do we have a net forward momentum in the background frame or not?
Jano

[Kryptid]

Do we have a net forward momentum in the background frame or not?

I'll answer your question once you've answered mine:

Might you know the answer to this question?

So, what must the overall linear momentum be at the end of the stroke in order for the total momentum at the end of the stroke to match the total momentum before the stroke?

0 angular momentum + 0 linear momentum = 0 angular momentum + ? linear momentum. What is the value of the question mark?