[Jaaanosik (OP)]

How is my post above different from the first stroke here?
Do you agree that the spaceship is going to be displaced after the first stroke when everything is static?
Jano

The ship is displaced, but it stops moving as soon as the balls stop moving. The barycenter hasn't moved. No net momentum is generated. As soon as the second stroke is completed, the ship has moved back into its original position.

The second stroke uses 1000 less energy.
How big the second displacement going to be?
Jano

[Kryptid]

The second stroke uses 1000 less energy.
How big the second displacement going to be?
Jano

The displacement will be the same and in the opposite direction. It'll just take much longer to complete because the balls are moving much more slowly.

[Jaaanosik (OP)]

The second stroke uses 1000 less energy.
How big the second displacement going to be?
Jano

The displacement will be the same and in the opposite direction. It'll just take much longer to complete because the balls are moving much more slowly.

Right, correct.

Let us investigate the velocity of balls/hollow cylinders when they are slowing down during the rotation.
The slowing down comes from the centripetal force.
The horizontal components of the centripetal force are constrained forces.
The initial kinetic energy goes to the rotational energy that slows down the balls/hollow cylinders and the end result is smaller centripetal force required to be imparted on the balls/cylinders when they are at the bottom part of their trajectory. This means the spaceship is continuously decreasing the momentum required to slow down the balls/cylinders, a consequence of the velocity decrease from v_L to v_B.

It is like holding 1kg in a stretched out hand, nothing is happening, no momentum change, no energy change but the constrained forces require energy. The person holding the kilo is burning energy.
Jano

[Kryptid]

Let us investigate the velocity of balls/hollow cylinders when they are slowing down during the rotation.
The slowing down comes from the centripetal force.
The horizontal components of the centripetal force are constrained forces.
The initial kinetic energy goes to the rotational energy that slows down the balls/hollow cylinders and the end result is smaller centripetal force required to be imparted on the balls/cylinders when they are at the bottom part of their trajectory. This means the spaceship is continuously decreasing the momentum required to slow down the balls/cylinders, a consequence of the velocity decrease from v_L to v_B.

I'm honestly not sure I understood this either. What is a "constrained force"? I see a v_L on your diagram, but not a v_B.

No matter how you move the balls, the barycenter of the system will not move. If the barycenter will not move, this system cannot be used for propulsion.

[Jaaanosik (OP)]

Kryptid,
The 1kg example in your hand.
F = m a = m dv/dt
F dt = m dv
There is no dv. Does it mean that the F you have to exert to maintain the potential energy is not real?
The m g is real and you have to work hard, still there is no momentum change.

The horizontal components of the centripetal force are constraint forces.
What I am trying to say that the energy 'flows' in the horizontal direction.



A - absorbers with springs
One half of energy/momentum is 'bent' to horizontal components.
Is this setup going to generate net forward momentum?
Jano

[Kryptid]

There is no dv. Does it mean that the F you have to exert to maintain the potential energy is not real?
The m g is real and you have to work hard, still there is no momentum change.

That's an artifact of that way that our anatomy works. You can put the mass on a table and the table will not expend any energy holding it up against gravity.

A - absorbers with springs
One half of energy/momentum is 'bent' to horizontal components.
Is this setup going to generate net forward momentum?

No. The balls in your most recent picture will only transfer their own forward momentum to the springs. The total momentum is unchanged.

[Jaaanosik (OP)]

...

No. The balls in your most recent picture will only transfer their own forward momentum to the springs. The total momentum is unchanged.

Kryptid,
Is all the momentum going down?
Is all the energy going to be absorbed down?
The side springs are not going to absorb any energy?
Jano

[Kryptid]

Kryptid,
Is all the momentum going down?
Is all the energy going to be absorbed down?
The side springs are not going to absorb any energy?
Jano

I'll do some calculations. I'll assume that each ball has a mass of 1 kilogram and they are travelling at 1 meter per second.downward. This gives the balls a total downward momentum of 1 kg x 1 m/s = 1 kg•m/s. We know from conservation of momentum that the downward momentum at all stages must therefore remain 1 kg•m/s. The kinetic energy of the balls is 0.5 x 1 kg x 1² m/s = 0.5 joules each. We know from conservation of energy that the total energy of the system must remain 1 joule at all stages (in the reference frame of the springs). I will assume that each spring system also weighs 1 kilogram.

To start off with, I will assume different starting circumstances where the balls are moving downward at 0.5 meters per second and the springs are moving upwards at 0.5 meters per second. When the balls contact the springs, they bring each other to a halt along the up-down axis. However, the kinetic energy can't just disappear. If none of it is dissipated as heat, then all of it goes into moving the balls towards each other along the left-right axis. Since they are travelling in opposite directions, their momentum cancels out and the total momentum of the system is still zero.

The reason I used that particular set-up was because of relativity. If we were in the reference frame of the springs, the scenario above would look identical to your set-up. That is, it would look to us like we weren't moving and that the balls were moving twice as fast instead. So we know that, from the reference frame of the springs, it looks as if the balls are moving along the left-right axis after the collision but not along the up-down axis. This means that, for an outside observer, the balls must always look as if they are moving left-right relative to the springs. So if the springs end up moving downward in some reference frame after the collision, the balls must have a component of their momentum moving downward as well. Their up-down velocity must be equal to whatever the up-down velocity of the springs is.

So when we start off, the balls are moving downwards at a velocity of 1 m/s while the springs are stationary. Then the collision happens. Some of the downward momentum will be carried by the springs now, but (and this is important) not all of it will. The balls will still carry some downward momentum. To an outside observer, they are moving both downward and towards the center at the same time (both balls now follow a diagonal path). In order for the downward momentum to remain 1 kg•m/s while a total mass of 2 kilograms is now carrying that momentum, the downward component of the velocity must be 1 kg•m/s / 2 kilograms = 0.5 m/s. So both the springs and the balls are moving downward with a velocity of 0.5 m/s. That is equivalent to a total kinetic energy of 0.5 x 2 x 0.5² m/s = 0.5 joules. That's only half of our energy budget, so that is okay.

The remaining 0.5 joules must therefore be in the form of the inward kinetic energy of the balls moving towards each other. That is an inward velocity of 1 m/s for each ball. The total momentum and energy of the system is conserved.

[Jaaanosik (OP)]

Thank you Kryptid,
I appreciate your post.

It is detailed and I need to read it multiple times.
Can we stick to barycenter reference frame? This would be the frame after the initial separation where balls and the ship have velocities of the same magnitude but in the opposite directions.
I guess we want to establish if the box (the spaceship) has a net forward momentum.

Would this be the outside observer that you mentioned as well?




Let us assume the absorber shafts have friction less slider connection to the deflection plate.
The shafts of the absorbers move after the initial contact.
I just want to point out that even thought the horizontal components go in opposite direction and they cancel out, each horizontal shaft moves in the horizontal direction.
Do you agree that this momentum to the side will reduce the momentum down?
In other words, less downward momentum is available to slow down the ship.
Jano

[Kryptid]

Can we stick to barycenter reference frame?

The barycenter frame would be the one where the balls look like they are moving downward and the springs are moving upward at the same velocity. So that particular scenario is described in my last post.

Do you agree that this momentum to the side will reduce the momentum down?

No. The upward movement of the springs completely cancels out the downward movement of the balls upon impact. The total momentum before the collision is zero according to the barycenter, and it remains zero after the collision.

[Jaaanosik (OP)]

Can we stick to barycenter reference frame?

The barycenter frame would be the one where the balls look like they are moving downward and the springs are moving upward at the same velocity. So that particular scenario is described in my last post.

Do you agree that this momentum to the side will reduce the momentum down?

No. The upward movement of the springs completely cancels out the downward movement of the balls upon impact. The total momentum before the collision is zero according to the barycenter, and it remains zero after the collision.

Kryptid,
did we make some extra energy on the side absorbers springs?
Jano

[Kryptid]

did we make some extra energy on the side absorbers springs?

No, the total energy remains the same. The kinetic energy is only stored in the springs temporarily. Then it is all sent into the balls. Initially, the balls were moving downward and the springs were moving upward. After the collision, the springs have stopped moving upward and the balls are now moving inward. The kinetic energy that was once in the springs is now added to the kinetic energy of the balls. The total is the same. It has only been distributed differently.

[Jaaanosik (OP)]

did we make some extra energy on the side absorbers springs?

No, the total energy remains the same. The kinetic energy is only stored in the springs temporarily. Then it is all sent into the balls. Initially, the balls were moving downward and the springs were moving upward. After the collision, the springs have stopped moving upward and the balls are now moving inward. The kinetic energy that was once in the springs is now added to the kinetic energy of the balls. The total is the same. It has only been distributed differently.

Kryptid,
imagine a mainspring - spiral torsion spring in the absorbers. They wind-up and stay locked. No energy goes back to balls.
What's then?
Do we have an extra energy on the side?
Jano

[Kryptid]

Kryptid,
imagine a mainspring - spiral torsion spring in the absorbers. They wind-up and stay locked. No energy goes back to balls.
What's then?
Do we have an extra energy on the side?
Jano

In that case, the balls effectively "stick" to the springs as the springs are pushed downward. The total downward momentum is the same and the total downward velocity would be the same. The only difference is that the energy that would have been put into pushing the balls inward is now in the form of potential energy stored inside the locked springs.

[Jaaanosik (OP)]

Kryptid,
imagine a mainspring - spiral torsion spring in the absorbers. They wind-up and stay locked. No energy goes back to balls.
What's then?
Do we have an extra energy on the side?
Jano

In that case, the balls effectively "stick" to the springs as the springs are pushed downward. The total downward momentum is the same and the total downward velocity would be the same. The only difference is that the energy that would have been put into pushing the balls inward is now in the form of potential energy stored inside the locked springs.

Is energy in the side springs going to be missing from the slowing down the ship?
100% of the energy released at the top.
Let us say 30% in the side springs.
Is the 70% going to stop the spaceship?
Jano

[Kryptid]

100% of the energy released at the top.

It can't be. Some mechanism aboard the ship is going to have to launch the balls. When that happens, the ship must be propelled in the opposite direction of the balls due to Newton's third law. If the ship weighs as much as the balls, then 50% of the energy will be in the ship's upward movement (and therefore the springs' upward movement) and the 50% will be in the ball's downward movement. When the balls collide with the springs, both the balls and the springs (plus the ship) come to a stop. The energy that would have been used to propel the balls inward is now stored as potential energy in the side springs.

[Jaaanosik (OP)]

100% of the energy released at the top.

It can't be. Some mechanism aboard the ship is going to have to launch the balls. When that happens, the ship must be propelled in the opposite direction of the balls due to Newton's third law. If the ship weighs as much as the balls, then 50% of the energy will be in the ship's upward movement (and therefore the springs' upward movement) and the 50% will be in the ball's downward movement. When the balls collide with the springs, both the balls and the springs (plus the ship) come to a stop. The energy that would have been used to propel the balls inward is now stored as potential energy in the side springs.

Kryptid,
if you say 50% that's fine. The energy is frame dependent. This is for the barycenter reference frame.
The balls have kinetic energy - either 50% in the barycenter frame or 100% in the spaceship frame.

Let us go with the 50% - 50%.
If the balls will push the side springs and the balls use 40% of the energy to wind-up the springs of 50% total then we have a 20% - 30% ratio of the 50% total.
Is 30% of the energy going to stop the spaceship released with 50% in the barycenter frame?
Jano

[Kryptid]

Let us go with the 50% - 50%.
If the balls will push the side springs and the balls use 40% of the energy to wind-up the springs of 50% total then we have a 20% - 30% ratio of the 50% total.
Is 30% of the energy going to stop the spaceship released with 50% in the barycenter frame?
Jano

Even if 100% of the energy could be absorbed by the springs it will all still come to a complete stop. The very act of compressing the springs requires a force. If enough force was imposed on the balls to bring them to a complete stop, then an equal and opposite force must act on the springs (and thus the ship). Since the ship has the same mass as the balls, that amount of force will stop it. Basically, you are taking all of the kinetic energy in the system as a whole and storing it in the springs.

[Jaaanosik (OP)]

Let us go with the 50% - 50%.
If the balls will push the side springs and the balls use 40% of the energy to wind-up the springs of 50% total then we have a 20% - 30% ratio of the 50% total.
Is 30% of the energy going to stop the spaceship released with 50% in the barycenter frame?
Jano

Even if 100% of the energy could be absorbed by the springs it will all still come to a complete stop. The very act of compressing the springs requires a force. If enough force was imposed on the balls to bring them to a complete stop, then an equal and opposite force must act on the springs (and thus the ship). Since the ship has the same mass as the balls, that amount of force will stop it. Basically, you are taking all of the kinetic energy in the system as a whole and storing it in the springs.

Is horizontal component slowing down the ship?
What work is done to slow down the spaceship with the horizontal component?
Jano

[Kryptid]

Is horizontal component slowing down the ship?

Yes. All of the kinetic energy possessed by the ship and balls is stored in the springs at the end of the interaction. Keep in mind that the springs are technically being pressed from both the bottom (due to the ship's velocity) and the top (due to the ball's velocity). An equal amount of energy is being absorbed from both the ship and the balls in the process because both the balls and the ship have the same mass.