Post by: Jaaanosik on 04/04/2020 23:16:19
Would it be possible to built an inertial propulsion system for a spaceship?
Inertial meaning that nothing is ejected from the spaceship and the spaceship is not interacting with any outside fields.
Energy would be used only on moving something/anything inside of the spaceship to generate an acceleration.
Is it doable?
Jano
Post by: Kryptid on 04/04/2020 23:23:36
Post by: Janus on 04/04/2020 23:33:49
Post by: Jaaanosik on 05/04/2020 00:59:23
So many views and only two opinions?
Jano
Post by: Kryptid on 05/04/2020 01:02:10
Come on guys, why so serious :)
So many views and only two opinions?
Jano
Opinions can't change the laws of physics.
Post by: evan_au on 05/04/2020 01:56:53
Post by: Jaaanosik on 05/04/2020 03:45:17
For a third opinion, see: https://www.thenakedscientists.com/forum/index.php?topic=79146.msg598377#msg598377Yes,
I've seen that. That's the reason I said no interaction with outside fields.
Just something happening inside the ship itself,
Jano
Post by: Jaaanosik on 05/04/2020 03:52:57
Kryptid,Come on guys, why so serious :)
So many views and only two opinions?
Jano
Opinions can't change the laws of physics.
the solution does not change the laws of physics.
It strives in them!
We'll have a vote if works at the end.
All,
Imagine you are a broker in Las Vegas.
What odds you would give me that I can present something that moves the ship ahead?
100:1?
1000:1?
Come on guys, this is for entertainment in this tough times, let us cheer up. :)
Let's here opinions,
Jano
Post by: Kryptid on 05/04/2020 04:02:26
Kryptid,
the solution does not change the laws of physics.
It strives in them!
So how do you propose to create momentum out of nowhere without violating conservation of momentum?
Post by: Jaaanosik on 05/04/2020 04:10:24
What do you mean out of nowhere?Kryptid,
the solution does not change the laws of physics.
It strives in them!
So how do you propose to create momentum out of nowhere without violating conservation of momentum?
I said there would be an energy source and it would do something inside the ship.
No ejection of a mass out is allowed though,
Jano
Post by: Kryptid on 05/04/2020 04:12:39
What do you mean out of nowhere?
I said there would be an energy source and it would do something inside the ship.
No ejection of a mass out is allowed though,
Jano
Energy is not momentum.
Post by: Jaaanosik on 05/04/2020 04:14:34
Energy is not momentum.But it can be used to generate momentum,
Jano
Post by: Kryptid on 05/04/2020 04:28:28
But it can be used to generate momentum,
Jano
No, it can't. The net momentum of a system is a conserved quantity.
Post by: syhprum on 05/04/2020 09:21:46
Post by: Janus on 05/04/2020 16:52:56
Not net momentum. Momentum is a vector, it has both magnitude and direction. You can take a certain store of potential energy and convert it to kinetic energy, Energy is a scalar, so all you have to worry about is conserving magnitude in terms of conserving energy. So a compressed spring will have a stored energy of E = 1/2kx^2, where k is the spring constant and x the amount the spring was compressed. This can be converted into kinetic energy as mv^2/2. meaning a mass m moving at velocity v. Energy is conserved.Energy is not momentum.But it can be used to generate momentum,
Jano
Momentum also in conserved. Momentum is mv. Now in the energy equation, v is squared, thus the direction of the velocity, which is itself a vector, doesn't matter. v^2 and (-v)^2 produce the same answer.
However with momentum it isn't, thus mv is not the same as m(-v), and, in fact, they are opposites.
Like energy, momentum is conserved. Our mass and compressed spring start with no velocity, so the total momentum is m(0) = 0. After the spring has converted its stored energy into kinetic energy of mass m, the total momentum must still be 0.
The only way to do this to divide m in two where m1+m2 = m and then m1v1 + m2v2 = 0 Since m cannot be negative, either v1 or v2 must be negative, Meaning m1 and m2 must be sent in opposite directions. The only way to get m1 moving to the right at v1, is to have m2 moving to the left at v2 such that the sum of the momenta add up to zero.
Post by: A-wal on 05/04/2020 20:25:17
A propulsion system that's entirely internal can have no external effect but wanting the ship to accelerate is itself an effect that's external to the ship so you're asking if you can exert an external influence without exerting an external influence.
So no, no you can't.
Post by: Jaaanosik on 05/04/2020 20:39:57
Not net momentum. Momentum is a vector, it has both magnitude and direction. You can take a certain store of potential energy and convert it to kinetic energy, Energy is a scalar, so all you have to worry about is conserving magnitude in terms of conserving energy. So a compressed spring will have a stored energy of E = 1/2kx^2, where k is the spring constant and x the amount the spring was compressed. This can be converted into kinetic energy as mv^2/2. meaning a mass m moving at velocity v. Energy is conserved.Energy is not momentum.But it can be used to generate momentum,
Jano
Momentum also in conserved. Momentum is mv. Now in the energy equation, v is squared, thus the direction of the velocity, which is itself a vector, doesn't matter. v^2 and (-v)^2 produce the same answer.
However with momentum it isn't, thus mv is not the same as m(-v), and, in fact, they are opposites.
Like energy, momentum is conserved. Our mass and compressed spring start with no velocity, so the total momentum is m(0) = 0. After the spring has converted its stored energy into kinetic energy of mass m, the total momentum must still be 0.
The only way to do this to divide m in two where m1+m2 = m and then m1v1 + m2v2 = 0 Since m cannot be negative, either v1 or v2 must be negative, Meaning m1 and m2 must be sent in opposite directions. The only way to get m1 moving to the right at v1, is to have m2 moving to the left at v2 such that the sum of the momenta add up to zero.
Janus,
agreed, this is all good.
Having said that I am not going to break the physics laws, I am going to bend them, literally.
Before going further a statement/disclaimer:
If this is going to be proven as a viable propulsion system I claim prior art/design and it is my wish this design belongs to public domain/all human kind.
Jano
(https://i.imgur.com/c5Q33VW.png)
Let us discuss.
Is this going to work?
Post by: Kryptid on 05/04/2020 20:47:14
It won't work.
Post by: Janus on 05/04/2020 23:06:24
All youv'e done here is fail to account for all the momentum transfers between balls and tubes. You will get no net gain of momentum with this set up.Not net momentum. Momentum is a vector, it has both magnitude and direction. You can take a certain store of potential energy and convert it to kinetic energy, Energy is a scalar, so all you have to worry about is conserving magnitude in terms of conserving energy. So a compressed spring will have a stored energy of E = 1/2kx^2, where k is the spring constant and x the amount the spring was compressed. This can be converted into kinetic energy as mv^2/2. meaning a mass m moving at velocity v. Energy is conserved.Energy is not momentum.But it can be used to generate momentum,
Jano
Momentum also in conserved. Momentum is mv. Now in the energy equation, v is squared, thus the direction of the velocity, which is itself a vector, doesn't matter. v^2 and (-v)^2 produce the same answer.
However with momentum it isn't, thus mv is not the same as m(-v), and, in fact, they are opposites.
Like energy, momentum is conserved. Our mass and compressed spring start with no velocity, so the total momentum is m(0) = 0. After the spring has converted its stored energy into kinetic energy of mass m, the total momentum must still be 0.
The only way to do this to divide m in two where m1+m2 = m and then m1v1 + m2v2 = 0 Since m cannot be negative, either v1 or v2 must be negative, Meaning m1 and m2 must be sent in opposite directions. The only way to get m1 moving to the right at v1, is to have m2 moving to the left at v2 such that the sum of the momenta add up to zero.
Janus,
agreed, this is all good.
Having said that I am not going to break the physics laws, I am going to bend them, literally.
Before going further a statement/disclaimer:
If this is going to be proven as a viable propulsion system I claim prior art/design and it is my wish this design belongs to public domain/all human kind.
Jano
(https://i.imgur.com/c5Q33VW.png)
Let us discuss.
Is this going to work?
I always amazes me that people believe that they are the first person to think up such an idea. I can't tell you how many times I've seen the same type of idea tossed around. The reason we don't already have devices that use this type of arrangement is that they just don't work.
Post by: Petrochemicals on 06/04/2020 01:47:15
Is not gravity a worthy outside field ? If you cannot interact with an outside field you cannot go anywhere because nothing exists, there would not be a consept of outside, just like there is not a consept of outside the universe in a massive way as that would be part of the universe, thus inside it.For a third opinion, see: https://www.thenakedscientists.com/forum/index.php?topic=79146.msg598377#msg598377Yes,
I've seen that. That's the reason I said no interaction with outside fields.
Just something happening inside the ship itself,
Jano
Post by: Jaaanosik on 06/04/2020 03:03:12
...Janus,
All youv'e done here is fail to account for all the momentum transfers between balls and tubes. You will get no net gain of momentum with this set up.
I always amazes me that people believe that they are the first person to think up such an idea. I can't tell you how many times I've seen the same type of idea tossed around. The reason we don't already have devices that use this type of arrangement is that they just don't work.
Please, have a look at the second stroke.
That's the case where the net momentum is 0 because it is linear.
The first stroke has the original momentum changed to angular momentum of the balls and the momentum that hits sideways.
Therefore not 100% of the original momentum is applied in the linear way and that's the reason forward net momentum is generated,
Jano
Post by: Kryptid on 06/04/2020 03:40:18
That's the case where the net momentum is 0 because it is linear.
You're wrong. Newton's third law states that for every action there is an equal and opposite reaction. If the balls are being redirected by the curved paths towards the front of the device, then that means a force is being applied to the balls to change their path. In turn, that means the balls are producing an equal and opposite force against the curved paths (and therefore the device as a whole). This second stroke is equal and opposite to the first stroke and thus produces an equal and opposite force. There is no net movement.
Post by: Jaaanosik on 06/04/2020 04:45:08
Kryptid,That's the case where the net momentum is 0 because it is linear.
You're wrong. Newton's third law states that for every action there is an equal and opposite reaction. If the balls are being redirected by the curved paths towards the front of the device, then that means a force is being applied to the balls to change their path. In turn, that means the balls are producing an equal and opposite force against the curved paths (and therefore the device as a whole). This second stroke is equal and opposite to the first stroke and thus produces an equal and opposite force. There is no net movement.
your first stroke analysis is wrong. The forces change direction at the back of the ship.
It is better to analyze momentum or energy.
If you just think about the angular momentum of the balls. Where does it come from?
The amount of the angular momentum is going to be missing in equalization of the momentum from the front where the first stroke started. Don't you think?
Jano
Post by: Jaaanosik on 06/04/2020 05:25:12
if the balls fly out with 10m/s at the beginning of the first stroke.
What would be the velocity when they hit the side of the spaceship? It is not going to be 0m/s for sure.
If it is 5m/s then how much momentum went to slowing down the ship?
Still, remember that some linear momentum went to the angular momentum of the balls.
10m/s = v for slowing down momentum + v for angular momentum + 5m/s
10m/s > v for slowing down momentum
... we have a net momentum forward,
Jano
Post by: Kryptid on 06/04/2020 05:36:51
Kryptid,
your first stroke analysis is wrong. The forces change direction at the back of the ship.
And?
If you just think about the angular momentum of the balls. Where does it come from?
The amount of the angular momentum is going to be missing in equalization of the momentum from the front where the first stroke started. Don't you think?
There is no net angular momentum. The balls end up spinning in opposite directions, so they cancel each other out.
What would be the velocity when they hit the side of the spaceship?
Depends on the reference frame. Is this the frame of an outside observer or the frame of the machine?
The very act of firing the balls backwards will push the rest of the assembly forward with momentum equal and opposite to that of the balls. As the balls begin to push against the curved track, their backwards momentum slowly starts to cancel out the forward momentum of the machine as a whole. Once they reach the wall, their backwards momentum is gone and so is the forward momentum of the machine. The sideways momentum of the two balls cancel each other out because they are moving in opposite directions.
There is no net movement.
Post by: Jaaanosik on 06/04/2020 15:47:19
...
All youv'e done here is fail to account for all the momentum transfers between balls and tubes. You will get no net gain of momentum with this set up.
I always amazes me that people believe that they are the first person to think up such an idea. I can't tell you how many times I've seen the same type of idea tossed around. The reason we don't already have devices that use this type of arrangement is that they just don't work.
(https://i.imgur.com/HQiG5V0.png)
Hi Janus,
If we would shoot just the balls on the left side then the conservation of the momentum would start to rotate the spaceship.
We shoot the balls at the same time so the horizontal components cancel out.
They are still there though.
This is the reason why the backward component is not equal to 100% of the original momentum.
There is going to be a delta front net momentum, plus some momentum goes to the rotation of the balls,
Jano
Post by: Jaaanosik on 06/04/2020 15:56:04
There could be a spaceship with this type of propulsion.
The spaceship could spread its sails, collect the interstellar hydrogen, burn it for the propulsion,
Jano
Post by: Janus on 06/04/2020 17:41:55
Wrong. Momentum is still full conserved. Just because you don't see it, doesn't mean that it doesn't occur. Your system will not, and can not, produce net momentum, no matter how hard you want to believe otherwise. The fact that by your analysis, it does, just means that your analysis is incomplete....
All youv'e done here is fail to account for all the momentum transfers between balls and tubes. You will get no net gain of momentum with this set up.
I always amazes me that people believe that they are the first person to think up such an idea. I can't tell you how many times I've seen the same type of idea tossed around. The reason we don't already have devices that use this type of arrangement is that they just don't work.
(https://i.imgur.com/HQiG5V0.png)
Hi Janus,
If we would shoot just the balls on the left side then the conservation of the momentum would start to rotate the spaceship.
We shoot the balls at the same time so the horizontal components cancel out.
They are still there though.
This is the reason why the backward component is not equal to 100% of the original momentum.
There is going to be a delta front net momentum, plus some momentum goes to the rotation of the balls,
Jano
Post by: Jaaanosik on 06/04/2020 18:20:42
...Janus,
Wrong. Momentum is still full conserved. Just because you don't see it, doesn't mean that it doesn't occur. Your system will not, and can not, produce net momentum, no matter how hard you want to believe otherwise. The fact that by your analysis, it does, just means that your analysis is incomplete.
Well, let's talk about conservation of energy.
The 2 balls have the weight of the spaceship, so we can ignore the mass.
The beginning of the first stroke, balls fly out at v and the spaceship -v, agreed?
At the end of the first stroke, the balls stopped the ship to zero for no net gain and they have additional energy to hit the side walls and the balls have additional rotational energy.
Where is the conservation of energy in this?
What do I miss?
Jano
Post by: Kryptid on 06/04/2020 20:58:55
Janus,
Well, let's talk about conservation of energy.
The 2 balls have the weight of the spaceship, so we can ignore the mass.
The beginning of the first stroke, balls fly out at v and the spaceship -v, agreed?
At the end of the first stroke, the balls stopped the ship to zero for no net gain and they have additional energy to hit the side walls and the balls have additional rotational energy.
Where is the conservation of energy in this?
What do I miss?
Jano
We can use the principles of conservation of momentum and conservation of energy to figure this out.
We start off with the balls being equal in mass to the rest of the ship. So when they are sent backwards, an equal amount of kinetic energy is received by both the balls and the ship. Thus, the balls are sent backwards at -v and the ship is sent forward at +v. The momentum of the balls is equal and opposite to that of the ship, thus giving the ship/ball system a total momentum of zero (equal to when it was at rest).
Now, the balls encounter the curved paths. Since the curved paths are attached to the ship, they are moving forward at a velocity +v. This extra velocity component accelerates the balls as they change direction. In turn, the ship is decelerated as kinetic energy is transferred from the ship to the balls via the curved paths. Once the balls have completed the curve and hit the wall, they have gained all of the kinetic energy that was once a part of the ship, thus bringing the ship's forward velocity down to zero. The rotational energy of the balls comes from part of the existing kinetic energy, so that energy is not new.
The balls themselves are, just before hitting the wall, traveling at a right angle to the ship's initial heading. So the balls no longer have a backwards momentum component either. Since the balls no longer have rearward momentum and the ship no longer has forward momentum, then the total momentum at the end of the cycle is still zero. The total energy of the system has also remained constant because all of the kinetic energy that was once in the ship has been transferred to the balls. If the collision of the balls with the walls is inelastic, then all of the kinetic energy in the balls is converted into heat energy.
Both momentum and energy are conserved.
Post by: Jaaanosik on 07/04/2020 02:51:23
Thanks, good response.
Question:
(https://i.imgur.com/xGP1rpH.png)
If a ball is sliding on ice and it hits the red carpet.
The friction at the bottom will cause the ball to roll.
The roll decreases the velocity v and this decreases the forward kinetic energy that is available to slow down the spaceship.
Quote
The rotational energy of the balls comes from part of the existing kinetic energy, so that energy is not new.Isn't this energy going to be missed?
Jano
Post by: Kryptid on 07/04/2020 04:28:00
Isn't this energy going to be missed?
No, because the same force that is slowing the ball's forward kinetic energy will act in an equal and opposite manner on the carpet. Remember, the red carpet is attached to the ship, so any forces applied to the carpet will also be applied to the ship. If a force of X slows down the ball, the same force of X will slow down the ship by the same amount (since you specified that their masses are equal).
Post by: Jaaanosik on 07/04/2020 15:20:23
here is the spaceship, the ball mass is the same as the mass of the spaceship.
If the ball is pushed down on the left side the ball and the spaceship will rotate around the barycenter.
The spaceship does not move anywhere when v_L = v_B and the v is constant all around the rotation.
If v_B < v_L then we have a forward (up in the figure) net momentum gain, considering the mechanism is coupled with symmetrical wheel/ball in opposite direction.
Here is a little secret, it is better to use hollow cylinders instead of balls for the original design where cylinder v_L kinetic energy is transformed to cylinder rotational energy at the bottom of the spaceship because more kinetic energy will be absorbed by the hollow cylinder and v_L, v_B delta will be bigger.
Jano
(https://i.imgur.com/L7CdpX0.png)
Post by: Kryptid on 07/04/2020 15:24:59
Post by: Jaaanosik on 07/04/2020 18:40:11
Unfortunately, I don't really understand your diagram or set-up.Kryptid,
that's very cool you said that.
Does this help?
The barycenter is important for the inertial propulsion system,
Jano
Post by: Kryptid on 07/04/2020 21:19:19
Kryptid,
that's very cool you said that.
Does this help?
The barycenter is important for the inertial propulsion system,
Jano
I know what a barycenter is. What is causing the balls to move in the diagram? Where is the spaceship at?
I can say this much: if you think that this will cause a net change in momentum, then you are trying to break conservation of momentum. The laws of physics do not allow that. For this reason, I'm considering this a "new theory" and thus moving it to that board.
Post by: Jaaanosik on 07/04/2020 22:41:39
...
I know what a barycenter is. What is causing the balls to move in the diagram? Where is the spaceship at?
I can say this much: if you think that this will cause a net change in momentum, then you are trying to break conservation of momentum. The laws of physics do not allow that. For this reason, I'm considering this a "new theory" and thus moving it to that board.
(https://i.imgur.com/hEOzDrR.png)
Kryptid,
The point A is an axle through the spaceship, balls rotate. The L ball mass plus R ball mass is equal to the spaceship mass.
The position A, also the geometrical center of the spaceship, was at B when the balls were horizontal.
When the L/R balls get pushed down at the same time the spaceship moves up/forward.
The point B is the barycenter.
The balls get stopped at the bottom, the dotted position, two balls - one behind the other.
Now, nothing moves, everything is steady. The new dotted position is the 'displaced' spaceship that got moved ahead.
Agreed?
Jano
Post by: Kryptid on 07/04/2020 23:16:56
Kryptid,
The point A is an axle through the spaceship, balls rotate. The L ball mass plus R ball mass is equal to the spaceship mass.
The position A, also the geometrical center of the spaceship, was at B when the balls were horizontal.
When the L/R balls get pushed down at the same time the spaceship moves up/forward.
The point B is the barycenter.
The balls get stopped at the bottom, the dotted position, two balls - one behind the other.
Now, nothing moves, everything is steady. The new dotted position is the 'displaced' spaceship that got moved ahead.
Agreed?
Jano
As long as you agree that the barycenter didn't move, sure.
Post by: Jaaanosik on 07/04/2020 23:44:21
...
As long as you agree that the barycenter didn't move, sure.
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
(https://i.imgur.com/c5Q33VW.png)
Post by: Kryptid on 07/04/2020 23:51:50
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.
Post by: Jaaanosik on 07/04/2020 23:58:47
The second stroke uses 1000 less energy.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.
How big the second displacement going to be?
Jano
Post by: Kryptid on 08/04/2020 00:02:07
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.
Post by: Jaaanosik on 08/04/2020 01:54:23
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
Post by: Kryptid on 08/04/2020 05:58:27
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 vL on your diagram, but not a vB.
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.
Post by: Jaaanosik on 08/04/2020 16:37:29
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.
(https://i.imgur.com/W8iLdPw.png)
A - absorbers with springs
One half of energy/momentum is 'bent' to horizontal components.
Is this setup going to generate net forward momentum?
Jano
Post by: Kryptid on 08/04/2020 16:42:59
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.
Post by: Jaaanosik on 08/04/2020 16:49:53
...
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
Post by: Kryptid on 08/04/2020 21:40:52
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.
Post by: Jaaanosik on 08/04/2020 22:49:09
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?
(https://i.imgur.com/5eWlD7a.png)
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
Post by: Kryptid on 08/04/2020 22:53:10
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.
Post by: Jaaanosik on 08/04/2020 22:57:35
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
Post by: Kryptid on 08/04/2020 23:00:37
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.
Post by: Jaaanosik on 08/04/2020 23:09:14
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
Post by: Kryptid on 08/04/2020 23:13:36
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.
Post by: Jaaanosik on 08/04/2020 23:18:25
Is energy in the side springs going to be missing from the slowing down the ship?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.
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
Post by: Kryptid on 08/04/2020 23:56:44
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.
Post by: Jaaanosik on 09/04/2020 00:59:18
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.
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
Post by: Kryptid on 09/04/2020 01:10:56
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.
Post by: Jaaanosik on 09/04/2020 01:31:51
Is horizontal component slowing down the ship?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.
What work is done to slow down the spaceship with the horizontal component?
Jano
Post by: Kryptid on 09/04/2020 01:38:29
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.
Post by: Jaaanosik on 09/04/2020 01:44:35
No, the horizontal component is not going to slow down the spaceship.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.
No work is done to slow down the spaceship.
How do we know?
Because work is a dot product of F and s.
The force is 90 degrees to the trajectory of the spaceship, no work is done.
F*s = 0 (as a dot product)
Jano
Post by: Jaaanosik on 09/04/2020 02:13:09
When the ball pushes the horizontal spring shaft then it is doing a work.
The trajectory s is horizontal and the inertia of the ball is pushing the shaft - the ball is doing work F*s.
The ball is losing its kinetic energy, transferred to the horizontal spring potential energy,
... but it is NOT slowing down the ship!
Jano
Post by: Kryptid on 09/04/2020 05:57:34
No, the horizontal component is not going to slow down the spaceship.
No work is done to slow down the spaceship.
How do we know?
Because work is a dot product of F and s.
The force is 90 degrees to the trajectory of the spaceship, no work is done.
F*s = 0 (as a dot product)
Jano
If no work is being done to slow down the ship, then no work is being done to slow down the balls either. Remember, just as much energy entering the springs is coming from the ship as is coming from the balls. The fact that kinetic energy is being removed from the balls and the ship to compress the springs means that both must be slowed down.
Now to continue along the line.
When the ball pushes the horizontal spring shaft then it is doing a work.
And when the ship is pushing the spring against the ball, it is doing work as well.
Quote
The trajectory s is horizontal and the inertia of the ball is pushing the shaft - the ball is doing work F*s.
The ball is losing its kinetic energy, transferred to the horizontal spring potential energy,
... but it is NOT slowing down the ship!
If the balls are losing kinetic energy, then the ship is too. Remember, the situation is perfectly symmetrical. From the reference frame of the balls, the balls are sitting still and the springs are moving towards them. From the balls' point of view, the kinetic energy of the ship-spring system is what is compressing the springs because the balls have no kinetic energy in their own reference frame. From the reference frame of the barycenter, energy from both the balls and the ship are being used to compress the springs.
Post by: Jaaanosik on 09/04/2020 16:11:33
Kryptid,No, the horizontal component is not going to slow down the spaceship.
No work is done to slow down the spaceship.
How do we know?
Because work is a dot product of F and s.
The force is 90 degrees to the trajectory of the spaceship, no work is done.
F*s = 0 (as a dot product)
Jano
If no work is being done to slow down the ship, then no work is being done to slow down the balls either. Remember, just as much energy entering the springs is coming from the ship as is coming from the balls. The fact that kinetic energy is being removed from the balls and the ship to compress the springs means that both must be slowed down.Now to continue along the line.
When the ball pushes the horizontal spring shaft then it is doing a work.
And when the ship is pushing the spring against the ball, it is doing work as well.QuoteThe trajectory s is horizontal and the inertia of the ball is pushing the shaft - the ball is doing work F*s.
The ball is losing its kinetic energy, transferred to the horizontal spring potential energy,
... but it is NOT slowing down the ship!
If the balls are losing kinetic energy, then the ship is too. Remember, the situation is perfectly symmetrical. From the reference frame of the balls, the balls are sitting still and the springs are moving towards them. From the balls' point of view, the kinetic energy of the ship-spring system is what is compressing the springs because the balls have no kinetic energy in their own reference frame. From the reference frame of the barycenter, energy from both the balls and the ship are being used to compress the springs.
The balls start to rotate in the barycenter frame.
The balls velocity component down is getting smaller because of this rotation in the barycenter frame.
The spaceship velocity component up is not getting smaller because the spaceship does not rotate in the barycenter frame.
(https://i.imgur.com/eF2dbsp.png)
If you look at the figure. The 'ideal' absorber is stiff, does not bend, just provides deflection.
When the deflection happens the spring is wind-up and locked.
If we pair the left absorber with the same type absorber on the other side of the spaceship then the spaceship will not rotate.
The spaceship stays inertial and the balls start to rotate in the barycenter frame.
Why?
Because, the work done on the pair of springs in the figure above is the same as in the figure below.
If the spaceship system with balls is moving upwards the motion of the balls have no effect on the upward velocity.
The constraint forces, 90 degrees to the velocity.
(https://i.imgur.com/fEif763.png)
Going back to the deflected ball.
The short time after the deflection the ball is still free-flying but on a different trajectory.
The work done on the spring did not change the spaceship velocity/momentum in the barycenter frame.
The deflected free-flying ball has smaller velocity/momentum in the downward direction because the velocity vector direction change does not come free, the angular momentum does not come free.
I hope this helps,
Jano
Post by: Kryptid on 09/04/2020 16:22:48
The balls velocity component down is getting smaller because of this rotation in the barycenter frame.
No, it's getting smaller because it's meeting resistance. The act of compressing a spring creates a force against whatever is compressing it. So you can just as easily redesign the system where the balls are prevented from rotating and you get the same result.
The spaceship velocity component up is not getting smaller because the spaceship does not rotate in the barycenter frame.
Again, the resistance caused by the spring is what causes the ship to slow, not rotation.
The work done on the spring did not change the spaceship velocity/momentum in the barycenter frame.
Yes it did, for reasons I've mentioned.
I hope this helps,
It doesn't because it's wrong. You can't break the law of conservation of momentum.
Post by: Jaaanosik on 09/04/2020 16:57:12
Kryptid,
The spaceship moves upwards/forward.
Are you saying that the horizontal motion will slow down the spaceship?
Are you saying that winding-up the springs on the absorbers will slow down the spaceship?
Jano
Post by: Kryptid on 09/04/2020 20:46:59
(https://i.imgur.com/fEif763.png)
Kryptid,
The spaceship moves upwards/forward.
Are you saying that the horizontal motion will slow down the spaceship?
Are you saying that winding-up the springs on the absorbers will slow down the spaceship?
Jano
In this most recent image of yours, there is no upward movement. The balls are moving to the side. So it isn't akin to the previous version. In your earlier version, the spring paddles are at a 45 degree angle to the trajectory of the balls. This means that both a horizontal and a vertical component to the force will be present when the ball hits the paddle. The horizontal component will go into compressing the spring, whereas the vertical component will be carried through the paddle and spring to the wall of the ship where the spring is attached. It is this vertical component that slows down the ship.
We could spend all day talking about various designs, but the fact of the matter is that the laws of physics will not allow them to work. We know before we begin any kind of analysis that the total momentum of the system must be identical at all stages of the machine's working. If you discover at the end of your analysis that the total momentum has changed, then you must have made an error in your analysis somewhere. This is demanded by conservation of momentum. Conservation of momentum, in turn, is demanded by Noether's theorem: https://en.wikipedia.org/wiki/Noether%27s_theorem
Take note how Noether's theorem is a theorem and not a theory. A theory is supported by scientific evidence, whereas a theorem is supported by mathematical proof: https://en.wikipedia.org/wiki/Theorem
Post by: Jaaanosik on 09/04/2020 23:16:35
Please, have a look here: https://en.wikipedia.org/wiki/Momentum#Multiple_dimensions
(https://upload.wikimedia.org/wikipedia/commons/2/2c/Elastischer_sto%C3%9F_2D.gif)
I rotated the figure to follow wiki example.
How big momentum goes to the bottom absorber (on the right now)?
(https://i.imgur.com/APhikFt.png)
Do you have a new theory about momentum?
Jano
Post by: Kryptid on 09/04/2020 23:26:55
How big momentum goes to the bottom absorber (on the right now)?
All of the ball's forward momentum goes into it.
Do you have a new theory about momentum?
No. I have no need for it. The existing one is good enough.
Post by: Jaaanosik on 10/04/2020 01:07:42
How big momentum goes to the bottom absorber (on the right now)?
All of the ball's forward momentum goes into it.Do you have a new theory about momentum?
No. I have no need for it. The existing one is good enough.
Kryptid,
The bold part, you are not serious, are you?
Jano
Post by: Kryptid on 10/04/2020 01:44:43
Kryptid,
The bold part, you are not serious, are you?
Jano
On second thought, perhaps not. I imagine some of the forward momentum could be transmitted to the ship floor through the bottom spring (not by compressing the bottom spring, mind you, but by applying a sideways force through the rod connecting it to the floor). The exact ratio of how much of the forward momentum goes into compressing the right spring and how much gets transmitted through the rod holding the bottom spring, I do not know. It ultimately doesn't matter though, as the total must necessarily be equal and opposite to the amount of the ship's momentum (due to Newton's third law) it cancels out. If the ball comes to a complete stop, then the ship has to as well. That's Newton's third law.
Post by: Jaaanosik on 10/04/2020 02:34:08
Kryptid,Kryptid,
The bold part, you are not serious, are you?
Jano
On second thought, perhaps not. I imagine some of the forward momentum could be transmitted to the ship floor through the bottom spring (not by compressing the bottom spring, mind you, but by applying a sideways force through the rod connecting it to the floor). The exact ratio of how much of the forward momentum goes into compressing the right spring and how much gets transmitted through the rod holding the bottom spring, I do not know. It ultimately doesn't matter though, as the total must necessarily be equal and opposite to the amount of the ship's momentum (due to Newton's third law) it cancels out. If the ball comes to a complete stop, then the ship has to as well. That's Newton's third law.
there is nothing wrong to follow the free body diagram.
1/2 of the initial momentum stays in the ball and 1/2 goes to the tilted plate.
1/2 of the tilted plate momentum (1/4 of the original momentum) goes to the side spring and stays there locked.
The second 1/4 goes to the bottom spring, it can be locked as well.
The cycle continues, the ball will hit another pair of absorbers, then another, ...
We can have an array of this deflectors.
The end result is obvious.
A significant amount of the original ball kinetic energy is locked in the side springs.
Not all energy goes to the bottom springs.
This is obvious now, I hope.
We have a forward net momentum.
The pressure on the sides is not going to slow down the spaceship forward motion,
Jano
Post by: Kryptid on 10/04/2020 05:58:44
We have a forward net momentum.
I think you missed this:
We could spend all day talking about various designs, but the fact of the matter is that the laws of physics will not allow them to work. We know before we begin any kind of analysis that the total momentum of the system must be identical at all stages of the machine's working. If you discover at the end of your analysis that the total momentum has changed, then you must have made an error in your analysis somewhere. This is demanded by conservation of momentum. Conservation of momentum, in turn, is demanded by Noether's theorem: https://en.wikipedia.org/wiki/Noether%27s_theorem
Take note how Noether's theorem is a theorem and not a theory. A theory is supported by scientific evidence, whereas a theorem is supported by mathematical proof: https://en.wikipedia.org/wiki/Theorem
Post by: Jaaanosik on 10/04/2020 06:40:44
Where did I break the conservation of energy or conservation of momentum?We have a forward net momentum.
I think you missed this:We could spend all day talking about various designs, but the fact of the matter is that the laws of physics will not allow them to work. We know before we begin any kind of analysis that the total momentum of the system must be identical at all stages of the machine's working. If you discover at the end of your analysis that the total momentum has changed, then you must have made an error in your analysis somewhere. This is demanded by conservation of momentum. Conservation of momentum, in turn, is demanded by Noether's theorem: https://en.wikipedia.org/wiki/Noether%27s_theorem
Take note how Noether's theorem is a theorem and not a theory. A theory is supported by scientific evidence, whereas a theorem is supported by mathematical proof: https://en.wikipedia.org/wiki/Theorem
What is wrong with changing the linear momentum to angular momentum?
Jano
Post by: Kryptid on 10/04/2020 06:59:52
Where did I break the conservation of energy or conservation of momentum?
When the spacecraft starts off, it is stationary. The total momentum of the system is zero. So if the total momentum ever changes from zero, then conservation of momentum has been violated. You claim that your system creates net momentum and therefore you claim that your system can violate conservation of momentum.
What is wrong with changing the linear momentum to angular momentum?
Linear momentum and angular momentum are both their own, conserved phenomena. They are not interchangeable. If the linear momentum and angular momentum both start off at zero, then both must remain at zero.
Post by: Jaaanosik on 10/04/2020 17:26:18
Kryptid,Where did I break the conservation of energy or conservation of momentum?
When the spacecraft starts off, it is stationary. The total momentum of the system is zero. So if the total momentum ever changes from zero, then conservation of momentum has been violated. You claim that your system creates net momentum and therefore you claim that your system can violate conservation of momentum.What is wrong with changing the linear momentum to angular momentum?
Linear momentum and angular momentum are both their own, conserved phenomena. They are not interchangeable. If the linear momentum and angular momentum both start off at zero, then both must remain at zero.
we can transform kinetic energy to rotational energy but we cannot transform linear momentum to angular momentum?
Please, remember what I said.
A spaceship collects the interstellar hydrogen and helium would be fused with huge amount of excess energy.
The hydrogen potential energy is incoming into the system, no physical objects are ejected out (radiation is different).
So the increase of the internal kinetic energy cannot be transformed to external kinetic energy?
The work energy principle does not apply, F*s (dot product) is not good, correct?
Quote
Linear momentum and angular momentum are both their own, conserved phenomena. They are not interchangeable. If the linear momentum and angular momentum both start off at zero, then both must remain at zero.This is important!
Are you sure the linear momentum cannot be converted to angular momentum?
My original proposal revolves around 'bending' the laws of physics.
The external potential energy transforms into linear kinetic energy at the front of the spaceship.
The balls/cylinders are thrown back and the kinetic energy is transformed into rotational energy.
The balls/cylinders can generate electricity till they stop at the back of the spaceship. The electricity can generate a heat and it can be radiated outside at the back.
Then they can be transferred to the front of the spaceship in a manner that they do not pickup rotational energy.
The front: hydrogen -> potential energy -> electricity -> gun -> linear kinetic energy
The back: linear kinetic energy -> rotational energy -> electricity -> heat -> radiation
Where is the conservation of energy broken here?
The most important step here is the incoming linear kinetic energy = outgoing kinetic energy + rotational energy.
Does energy rule or not?
Jano
Post by: Kryptid on 10/04/2020 17:31:33
we can transform kinetic energy to rotational energy but we cannot transform linear momentum to angular momentum?
No, you can't: https://van.physics.illinois.edu/qa/listing.php?id=24173&t=is-linear-momentum-converted-to-angular-momentum
So the increase of the internal kinetic energy cannot be transformed to external kinetic energy?
It can't produce net momentum, no.
Are you sure the linear momentum cannot be converted to angular momentum?
Yes. See the link above.
Where is the conservation of energy broken here?
It isn't. Conservation of momentum is what is broken.
Post by: Jaaanosik on 10/04/2020 18:08:14
Kryptid,we can transform kinetic energy to rotational energy but we cannot transform linear momentum to angular momentum?
No, you can't: https://van.physics.illinois.edu/qa/listing.php?id=24173&t=is-linear-momentum-converted-to-angular-momentumSo the increase of the internal kinetic energy cannot be transformed to external kinetic energy?
It can't produce net momentum, no.Are you sure the linear momentum cannot be converted to angular momentum?
Yes. See the link above.Where is the conservation of energy broken here?
It isn't. Conservation of momentum is what is broken.
No, there is only one momentum, angular momentum.
The linear momentum is angular momentum!!!!
The linear momentum is angular momentum that has infinite radius, the axis of rotation is infinitely far away,
Jano
Post by: Jaaanosik on 10/04/2020 18:09:48
you do not trust me?
Here: https://www.khanacademy.org/science/ap-physics-1/ap-torque-angular-momentum/conservation-of-angular-momentum-ap/v/ball-hits-rod-angular-momentum-example
Jano
Post by: Jaaanosik on 10/04/2020 18:11:00
Jano
Post by: Kryptid on 10/04/2020 20:28:18
Kryptid,
No, there is only one momentum, angular momentum.
The linear momentum is angular momentum!!!!
The linear momentum is angular momentum that has infinite radius, the axis of rotation is infinitely far away,
Jano
The Department of Physics website I linked stated:
Quote
The two conservation laws- linear and angular momentum- are absolutely separate. Neither one can be converted to the other.
Kryptid,
you do not trust me?
Here: https://www.khanacademy.org/science/ap-physics-1/ap-torque-angular-momentum/conservation-of-angular-momentum-ap/v/ball-hits-rod-angular-momentum-example
Jano
That website does not support your position that linear momentum is angular momentum. All it shows is that a moving ball can have angular momentum even if it isn't rotating. The moving ball transferred both its linear momentum and its angular momentum to the rod.
The link you posted actually demonstrates clearly that linear and angular momentum are different. In the scenario where the ball hits the rod, it causes the rod to rotate and thus shows that the ball possessed angular momentum in that situation. When the ball hits the axis instead, however, there is no rotation. The ball had no angular momentum in that case. In both cases, the ball was moving at the same velocity and therefore had the same linear momentum. So angular momentum can obviously vary independently of linear momentum.
This is not new theory, this is physics!
Jano
Physics states that linear momentum and angular momentum are different. Again, refer back to the Department of Physics website I linked.
Post by: Jaaanosik on 11/04/2020 03:07:20
...Are you sure the linear momentum cannot be converted to angular momentum?
Yes. See the link above.Where is the conservation of energy broken here?
It isn't. Conservation of momentum is what is broken.
Kryptid,
where do we go from here after you watched the linked video?
Do see how the linear momentum can be converted to the angular momentum?
Jano
(https://i.imgur.com/gZVGLHc.png)
Post by: Kryptid on 11/04/2020 05:14:33
Do see how the linear momentum can be converted to the angular momentum?
You can't do that any more than you can convert mass to electric charge.
In that picture you posted, both the angular and linear momentum of the ball are transferred to the rod. It's not a case of linear momentum being turned into angular momentum.
Post by: Jaaanosik on 11/04/2020 15:51:24
Please, look here again:
(https://upload.wikimedia.org/wikipedia/commons/2/2c/Elastischer_sto%C3%9F_2D.gif)
Is there an angular momentum in that figure?
Jano
Post by: Kryptid on 11/04/2020 17:09:24
Is there an angular momentum in that figure?
Yes. Objects don't have to be spinning to have angular momentum. Your link already explained how a non-spinning object can have angular momentum.
Post by: Jaaanosik on 11/04/2020 20:16:56
Great,Is there an angular momentum in that figure?
Yes. Objects don't have to be spinning to have angular momentum. Your link already explained how a non-spinning object can have angular momentum.
From a textbook: MATHEMATICAL METHODS FOR PHYSICISTS, 6th edition by George B. Arfken and Hans J. Weber
(https://i.imgur.com/fjL1BnX.png)
I can show that the equation 5.124 is not correct equation even if we consider the string to be "mass-less".
Do you trust my claim? :)
Jano
Post by: Kryptid on 11/04/2020 20:38:16
I can show that the equation 5.124 is not correct equation even if we consider the string to be "mass-less".
Do you trust my claim?
Jano
I'm not going to trust anyone's claim that modern physics is wrong unless they can provide compelling evidence for it.
Post by: Jaaanosik on 11/04/2020 21:08:46
I can show that the equation 5.124 is not correct equation even if we consider the string to be "mass-less".
Do you trust my claim?
Jano
I'm not going to trust anyone's claim that modern physics is wrong unless they can provide compelling evidence for it.
Kryptid,
what do think about this?
(https://i.imgur.com/AXweaNq.png)
The equation Eq. 23 is correct equation for a pendulum with a bob that does rotate when it falls down. The equation Eq. 23 is more complex than Eq. 22 and it shows that Eq. 23 tangential velocity will be less than Eq. 22 tangential velocity. The smaller tangential velocity leads to smaller tangential and centripetal accelerations.
Jano
Post by: Kryptid on 11/04/2020 21:27:32
Post by: Jaaanosik on 11/04/2020 22:52:35
To be honest, I'm not familiar with either equation. What are they even calculating? I don't see anywhere that it is stated that it is calculating either angular momentum or linear momentum.Kryptid,
I am sure you know
If we know rotational energy/angular velocity of a body we can determine angular momentum.
Are you going to claim you did not know it?
Here goes the comparison to the video.
We change the rod for the pendulum.
If we change the original potential energy in the pendulum textbook example for the kinetic energy of the ball in the video then we have pretty much the same scenario and the analysis holds.
Do you understand what I said or I need to draw it?
The rotational energy has two components, I hope you see that.
It follows that the pendulum bob design plays very important role in determination of the velocity.
Do you agree?
The important question is if the the same original momentum mv can generate two different angular momentums.
Do you see why I am asking this question?
Jano
Post by: Kryptid on 11/04/2020 23:18:30
Are you going to claim you did not know it?
I'm sure I knew that back when I was still in school, but it's been so long since I was exposed to that equation that I forgot about it.
It follows that the pendulum bob design plays very important role in determination of the velocity.
Do you agree?
Yes, the pendulum design does affect the pendulum velocity.
The important question is if the the same original momentum mv can generate two different angular momentums.
Mass times velocity alone can't generate any angular momentum, because it is a measure of linear momentum. Angular momentum requires a radius of rotation. If there is no radius of rotation, then there is no angular momentum.
Do you see why I am asking this question?
Nope.
Not that this matters anyway, as your spaceship design doesn't have any net angular momentum to begin with. The two balls spin in opposite directions, so the net angular momentum is zero. So even if it was possible to convert angular momentum into linear momentum, your ship still could not go anywhere because zero net angular momentum could not produce non-zero net linear momentum.
Post by: Jaaanosik on 15/04/2020 17:09:09
(https://i.imgur.com/qXctQmF.png)
The symmetrical wheel has its axle attached to the spaceship.
When the ball hits the arm of the symmetrical wheel attached to the spaceship it will not take anything from the spaceship forward momentum.
The forward spaceship momentum will be unchanged.
The symmetrical wheel will start to rotate.
Do you agree?
Jano
Post by: Kryptid on 15/04/2020 20:33:28
Do you agree?
Nope. The wheel will start to rotate, but it will also transmit linear momentum from the ball to the ship. If you want to eliminate that, you would need to have another ball coming from the opposite direction hitting the other arm.
Post by: Jaaanosik on 15/04/2020 22:08:14
Do you agree?
Nope. The wheel will start to rotate, but it will also transmit linear momentum from the ball to the ship. If you want to eliminate that, you would need to have another ball coming from the opposite direction hitting the other arm.
Kryptid,
you should have said the rotation of the wheel is going to have a torque and this torque will rotate the spaceship.
I'd say the second wheel on the same axle with the opposite rotation caused by the ball on the other side will cancel the first wheel torque.
If you remember the arm from the video, the inertia of the arm, centrifugal force would be pulling the spaceship back, but here the wheel is symmetrical on purpose.
Every point on the wheel has a symmetrical point across the axis of rotation and the only result is a torque.
When the torques are cancelled out then there is no force pulling the spaceship backwards or rotating it.
Jano
Post by: Kryptid on 15/04/2020 23:50:20
When the torques are cancelled out then there is no force pulling the spaceship backwards or rotating it.
Yes there is. For the reason I just stated in my last post. Conservation of linear momentum won't allow it to disappear into nothingness (nor can it be converted into angular momentum. Remember this? https://van.physics.illinois.edu/qa/listing.php?id=24173&t=is-linear-momentum-converted-to-angular-momentum).
Post by: Jaaanosik on 16/04/2020 01:57:19
When the torques are cancelled out then there is no force pulling the spaceship backwards or rotating it.
Yes there is. For the reason I just stated in my last post. Conservation of linear momentum won't allow it to disappear into nothingness (nor can it be converted into angular momentum. Remember this? https://van.physics.illinois.edu/qa/listing.php?id=24173&t=is-linear-momentum-converted-to-angular-momentum).
Kryptid,
Please, go ahead, do an experiment as shown in the linked video.
Are you saying that the water mills should not work? They are miracles?
Jano
Post by: Kryptid on 16/04/2020 06:24:52
Kryptid,
Please, go ahead, do an experiment as shown in the linked video.
Are you saying that the water mills should not work? They are miracles?
Jano
I don't see a link.
Post by: Jaaanosik on 16/04/2020 15:09:49
Kryptid,
Please, go ahead, do an experiment as shown in the linked video.
Are you saying that the water mills should not work? They are miracles?
Jano
I don't see a link.
Kryptid,
post #80
https://www.khanacademy.org/science/ap-physics-1/ap-torque-angular-momentum/conservation-of-angular-momentum-ap/v/ball-hits-rod-angular-momentum-example
The explanation to your confusion is around the 3rd minute.
Do you really think the water mills are miracles?
Jano
Post by: Kryptid on 16/04/2020 16:34:32
The explanation to your confusion is around the 3rd minute.
I don't have any confusion.
Do you really think the water mills are miracles?
No. Water mills are attached to buildings which, in turn, are attached to the ground. The linear momentum they receive from the falling water is transmitted to the entire planet. Since the Earth is so incredibly massive, you don't notice its change in momentum.
Post by: Jaaanosik on 16/04/2020 17:56:25
The explanation to your confusion is around the 3rd minute.
I don't have any confusion.Do you really think the water mills are miracles?
No. Water mills are attached to buildings which, in turn, are attached to the ground. The linear momentum they receive from the falling water is transmitted to the entire planet. Since the Earth is so incredibly massive, you don't notice its change in momentum.
Kryptid,
what happens when two torques of the opposite rotations of two wheels cancel out?
Jano
Edit:
to help you: https://en.wikipedia.org/wiki/Control_moment_gyroscope
Post by: Kryptid on 16/04/2020 20:07:53
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?
Post by: Jaaanosik on 16/04/2020 20:31:55
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?
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
Post by: Kryptid on 16/04/2020 21:59:46
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.
Quote
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
Quote
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.
Post by: Jaaanosik on 16/04/2020 22:23:57
we can play this game for a long time:
https://ccrma.stanford.edu/~jos/pasp/Relation_Angular_Linear_Momentum.html
(https://i.imgur.com/QeR5osQ.png)
(https://i.imgur.com/oQBfoFQ.png)
Jano
Post by: Kryptid on 16/04/2020 22:45:28
Post by: Jaaanosik on 16/04/2020 23:01:24
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
Post by: Jaaanosik on 17/04/2020 19:14:38
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
(https://i.imgur.com/HDTzHZl.png)
Post by: Kryptid on 17/04/2020 20:57:04
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.
Post by: Jaaanosik on 17/04/2020 21:49:00
...
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
(https://i.imgur.com/c5Q33VW.png)
Post by: Kryptid on 17/04/2020 21:51:24
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.
Post by: Jaaanosik on 17/04/2020 22:06:25
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.
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
Post by: Kryptid on 17/04/2020 22:11:21
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.
Post by: Jaaanosik on 17/04/2020 22:22:37
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
Post by: Kryptid on 17/04/2020 22:25:18
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.
Post by: Jaaanosik on 17/04/2020 22:29:45
So the gain is 10m forward.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.
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
Post by: Kryptid on 17/04/2020 22:34:33
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?
Post by: Jaaanosik on 17/04/2020 22:51:16
We need to remember conservation of energy as well.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?
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
Post by: Kryptid on 17/04/2020 23:26:15
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?
Quote
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?
Post by: Jaaanosik on 18/04/2020 17:06:08
I got a hand on Blender sw to create a video.
Here is the script what's going to be in it.
(https://i.imgur.com/zDnpUHs.png)
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
Post by: Kryptid on 18/04/2020 17:14:17
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?
Post by: Jaaanosik on 18/04/2020 17:20:26
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?
Right,
0am + 0lm = 0am + 0lm
... however...
7am + 3lm = 3am + ?lm
Back to you,
Jano
Post by: Kryptid on 18/04/2020 17:26:03
0am + 0lm = 0am + 0lm
So then you agree that a complete lack of net momentum at the beginning of a stroke means that there must also be a complete lack of net momentum at the end of a stroke as well.
7am + 3lm = 3am + ?lm
That would be 7. But, of course, your ship starts out with zero total momentum before the engine turns on. So those numbers should all still be zero.
Now about your newest design: I'm not sure whether the round things are balls or whether they are reels. I'm going to assume they are reels given your description. In that case, the ship will gain forward momentum as the CMGs are reeled in and the CMGs will gain momentum equal in magnitude and opposite in direction so that the total momentum remains zero.
Post by: Jaaanosik on 18/04/2020 17:28:02
What reference frame?0am + 0lm = 0am + 0lm
So then you agree that a complete lack of net momentum at the beginning of a stroke means that there must also be a complete lack of net momentum at the end of a stroke as well.7am + 3lm = 3am + ?lm
That would be 7. But, of course, your ship starts out with zero total momentum before the engine turns on. So those numbers should all still be zero.
Now about your newest design: I'm not sure whether the round things are balls or whether they are reels. I'm going to assume they are reels given your description. In that case, the ship will gain forward momentum as the CMGs are reeled in and the CMGs will gain momentum equal in magnitude and opposite in direction so that the total momentum remains zero.
Jano
Post by: Kryptid on 18/04/2020 17:30:10
What reference frame?
The barycenter's.
Post by: Jaaanosik on 18/04/2020 19:53:19
The barycenter moves in the background frame.What reference frame?
The barycenter's.
This not the barycenter of plain linear momentum when barycenter is stationary in the background frame,
Jano
Post by: Bobolink on 18/04/2020 20:15:37
Post by: Jaaanosik on 18/04/2020 21:04:17
Sorry Jano, but there some laws you just can't break. Your attempt is doomed for failure.
Bobolink,
here is the problem that Kryptid and it appears you too do not understand.
The system is a system of particles.
Writing 0am + 0lm = 0am + 0lm is wrong.
This is not a description of the system.
The systems has many particles so writing
0am_1 + 0lm_1 + 0am_2 + 0lm_2 + ... + 0am_n + 0lm_n = 0am_1 + 0lm_1 + 0am_2 + 0lm_2 + ... + 0am_n + 0lm_n
is wrong as well.
1. What is the left side of the equation?
- it is the system, let us take all the particles in the universe
2. What is the right side of the equation?
- it is the same system but after a delta time.
If we put zeros for all the particles on the left side then we do not have motion, we do not have particles.
We have nothing.
Ooops, ... we do not have nothing, we do not exist!!!!!
This is nothingness!!!!!!!!!
I hope you see the big problem!
We have to write some values on the left side ... and we have to have some values on the right side!!!
7am_1 + 3lm_1 + 5am_2 + 5lm_2 + ... + 2am_n + 8lm_n = 3am_1 + 7lm_1 + 5am_2 + 5lm_2 + ... + 8am_n + 2lm_n
I hope this helps,
Jano
Post by: Kryptid on 18/04/2020 22:23:10
The barycenter moves in the background frame.
This not the barycenter of plain linear momentum when barycenter is stationary in the background frame,
Jano
So the background frame is that of the CMGs? In that case, what you are doing is abandoning an inertial frame and entering an accelerating frame (for a moment), then entering a different inertial frame once the acceleration ceases. In the new inertial frame, the ship will have net forward momentum from the point of view of the CMGs. However, there can't be a contradiction between different frames: a device cannot look like reactionless propulsion in one frame but not another. So analyzing the system from the point of view of the barycenter is equally valid as analyzing it from the point of view of the CMGs.
I hope you see the big problem!
We have to write some values on the left side ... and we have to have some values on the right side!!!
7am_1 + 3lm_1 + 5am_2 + 5lm_2 + ... + 2am_n + 8lm_n = 3am_1 + 7lm_1 + 5am_2 + 5lm_2 + ... + 8am_n + 2lm_n
You're ignoring the fact that the sum of the momentum of all of the constituent particles is zero for an object that isn't moving. So although it's perfectly true that individual molecules will have momentum, we know that the total momentum of all the particles must add up to zero because the stationary object has no net momentum.
Before the engine turns on, the net linear and angular momentum of all of the particles in the system is zero because every piece of machinery in the ship is motionless relative to every other piece of machinery. And because of conservation of momentum, the net momentum must remain zero for every step of the engine's action. So it isn't possible to gain net linear momentum without violating conservation of momentum.
Post by: alancalverd on 18/04/2020 23:25:56
But there's very little difference between elementary physics and pendantry.
Post by: Kryptid on 19/04/2020 00:21:28
At the risk of being labelled a pedant, may I remind Jano that the dimensions of momentum are MLT-1 and angular momentum ML2T-1. They are not the same thing, and not being dimensionally equivalent, cannot be added.
I'm sure he'll find some excuse to find this "wrong"...
Post by: Jaaanosik on 19/04/2020 00:23:46
the CMG is group of rotating wheels, gyros, there is a center of axis of rotations.
The intersect of axes of rotation is like a potential well because any rotational disturbance to the CMG will bring the intersect back to the center.
It is like ECI (Earth Center Inertial) reference frame.
This point is pretty stable in space, just an oscillation. It can be extrapolated one point that is 'fixed' to the background.
This point does not change during the startup or shutdown of the rotation.
Having said that, when the CMG does not rotate then the potential well does not exist.
The collection of wheels will behave as a body of parts.
If you look here:
(https://i.imgur.com/HDTzHZl.png)
You are saying that v_L - v_R = 0
Right.
So go ahead, grab the device with the trapped balls inside, point to a bottle of beer and open one end.
There is no linear momentum, so nothing happens, correct?
I am not going to do that, I'd rather enjoy my beer. :)
The momentum can 'disappears' for days and then it can appear out of nothing?
That is some serious 'magic'. :)
Not only that but when one ball is shot out it will have higher velocity than it came in.
True miracle! Increase of speed out of nothing!
If v_L - v_R = 0 then v_L = v_R is true as well. Where did 0 go?
Jano
Post by: Kryptid on 19/04/2020 00:37:55
There is no linear momentum, so nothing happens, correct?
Just because the net momentum is zero doesn't mean that nothing happens. The ball will be shot out by the spring and the rest of the device will recoil in the opposite direction. The total momentum, however, is still zero.
Not only that but when one ball is shot out it will have higher velocity than it came in.
True miracle! Increase of speed out of nothing!
There is no miracle. All that has happened is that potential energy was converted into kinetic energy. If you calculate the momentum, you will see that it is conserved. The momentum of the recoil is equal to the momentum of the ball.
If v_L - v_R = 0 then v_L = v_R is true as well. Where did 0 go?
It's still there.
Post by: Jaaanosik on 19/04/2020 00:47:50
Is the increased tension in spring, atoms moving faster caused by something?There is no linear momentum, so nothing happens, correct?
Just because the net momentum is zero doesn't mean that nothing happens. The ball will be shot out by the spring and the rest of the device will recoil in the opposite direction. The total momentum, however, is still zero.Not only that but when one ball is shot out it will have higher velocity than it came in.
True miracle! Increase of speed out of nothing!
There is no miracle. All that has happened is that potential energy was converted into kinetic energy. If you calculate the momentum, you will see that it is conserved. The momentum of the recoil is equal to the momentum of the ball.If v_L - v_R = 0 then v_L = v_R is true as well. Where did 0 go?
It's still there.
Where did atoms momentum come from?
Jano
Post by: Jaaanosik on 19/04/2020 00:58:14
Force, m*dv/dt and impulse are essential for momentum, energy transfer between systems.
If you say the momentum is 0 then can we have dp/dt?
p - momentum
Jano
Post by: Kryptid on 19/04/2020 01:23:30
Is the increased tension in spring, atoms moving faster
If the atoms move any faster, it would be due to the conversion of the ball's kinetic energy into heat energy. But that's not where the spring's tension is coming from, because you can let the spring cool back down to room temperature and the potential energy will still be there.
Quote
caused by something?
It's caused by the conversion of kinetic energy into potential energy.
Where did atoms momentum come from?
They don't receive any net increase in momentum.
The problem is delta time. We can talk all day about energy/momentum, where they are.
Force, m*dv/dt and impulse are essential for momentum, energy transfer between systems.
If you say the momentum is 0 then can we have dp/dt?
p - momentum
Jano
Conservation of momentum will not be broken no matter how long you wait.
Post by: Jaaanosik on 19/04/2020 03:28:25
...
It's caused by the conversion of kinetic energy into potential energy.
...
Kryptid,
I have no idea why it is so difficult to admit that linear momentum is like kinetic energy and angular momentum is like potential energy.
If we convert kinetic energy to potential then we have to convert linear momentum to angular momentum.
If a magnet flies in space around a rotating magnetic field generator.
When the generator is off then nothing happens the magnet flies straight ahead.
If we turn on the magnetic field then we trap the magnet into a 'circular orbit', potential energy, the magnet has angular momentum.
The magnet can orbit the generator multiple times around their barycenter.
When the generator is turned off the magnet flies away, the potential energy is transformed into kinetic energy.
The angular momentum into the linear momentum.
Is flying out magnet going to have the same linear momentum?
Not necessarily, there are flyby accelerations and decelerations.
How come you admit kinetic energy to potential energy conversion but you deny linear momentum to angular momentum conversion?
You keep talking that the system HAS TO BE DEFINED in a way that linear momentum is 0.
The system definition can be changed. We can change the boundaries - space and time.
Initial and end conditions...
So let say:
generator mass m=10
magnet mass m=1
IN
lm - linear momentum
magnet_lm = 10
generator_lm = 0
OUT
magnet_lm = 20
generator_lm = 2
Can we get the numbers above?
Sure we can, it all depends on the generator design, how it will increase the angular momentum.
Jano
Post by: Kryptid on 19/04/2020 03:41:03
I have no idea why it is so difficult to admit that linear momentum is like kinetic energy and angular momentum is like potential energy.
That is incorrect. An object with either linear momentum or angular momentum will have kinetic energy. Potential energy does not need anything to be moving. None of the components of a spring need to be moving for it to have potential energy. You can suppress its temperature down to absolute zero and it will still have potential energy.
f we convert kinetic energy to potential then we have to convert linear momentum to angular momentum.
No.
If a magnet flies in space around a rotating magnetic field generator.
When the generator is off then nothing happens the magnet flies straight ahead.
If we turn on the magnetic field then we trap the magnet into a 'circular orbit', potential energy, the magnet has angular momentum.
The magnet can orbit the generator multiple times around their barycenter.
When the generator is turned off the magnet flies away, the potential energy is transformed into kinetic energy.
Orbital kinetic energy is a form of kinetic energy, not potential energy.
The angular momentum into the linear momentum.
No. The angular momentum was there before the magnetic field was ever turned on.
Is flying out magnet going to have the same linear momentum?
The system as a whole will have the same linear momentum, but, as you say, the individual components of the system can have different momenta. It just all has to add up to the original momentum.
How come you admit kinetic energy to potential energy conversion but you deny linear momentum to angular momentum conversion?
Maybe if you actually went back and read that link I posted where it was stated that linear momentum and angular momentum had separate conservation laws, you'd know. But as I said before, converting angular momentum into linear momentum won't help you because both linear and angular momentum start at zero before the engine is turned on. Then there is the fact that your machine is symmetrical: any angular momentum possessed by the left gyroscope will be cancelled out by that of the right gyroscope. So starting out by assuming that the gyroscopes are rotating won't help either.
You keep talking that the system HAS TO BE DEFINED in a way that linear momentum is 0.
The system definition can be changed. We can change the boundaries - space and time.
Initial and end conditions...
Before the engine is turned on, all of the internal components are stationary. You realize that means zero linear and zero angular momentum, don't you? Both linear and angular momentum require motion.
So let say:
generator mass m=10
magnet mass m=1
IN
lm - linear momentum
magnet_lm = 10
generator_lm = 0
OUT
magnet_lm = 20
generator_lm = 2
Can we get the numbers above?
Not unless you violate conservation of momentum. 10 + 0 does not equal 20 + 2. If the total momentum of the system starts out at 10, then it can never be anything other than 10 (unless you add momentum in from an outside source).
Post by: Bobolink on 19/04/2020 13:11:37
Bobolink,I think we both understand that you want your conjecture to be correct so you are ignoring all attempts to help you understand the reality of the situation. It is clear that you won't listen, because that would destroy your fantasy that you have figured out something that scientists say is not possible. I don't think you are a troll, you are just someone who has a conjecture based on a limited and flawed understanding and are excited that you think you have made a great discovery. So, I guess all I can say is have fun living this fantasy. [Shrug]
here is the problem that Kryptid and it appears you too do not understand.
Post by: Jaaanosik on 19/04/2020 13:59:22
...
Not unless you violate conservation of momentum. 10 + 0 does not equal 20 + 2. If the total momentum of the system starts out at 10, then it can never be anything other than 10 (unless you add momentum in from an outside source).
Kryptid,
exactly my point!
We are adding only angular momentum from the generator!!!
We are not adding linear momentum!!!
Jano
Post by: Jaaanosik on 19/04/2020 14:01:10
Bobolink,Bobolink,I think we both understand that you want your conjecture to be correct so you are ignoring all attempts to help you understand the reality of the situation. It is clear that you won't listen, because that would destroy your fantasy that you have figured out something that scientists say is not possible. I don't think you are a troll, you are just someone who has a conjecture based on a limited and flawed understanding and are excited that you think you have made a great discovery. So, I guess all I can say is have fun living this fantasy. [Shrug]
here is the problem that Kryptid and it appears you too do not understand.
Sailboats can sail 3x faster than wind.
Please, explain, how is that possible,
Jano
Post by: Kryptid on 19/04/2020 15:08:04
We are adding only angular momentum from the generator!!!
We are not adding linear momentum!!!
Adding extra exclamation points does not make your argument stronger.
If linear momentum is added to the magnet in orbit, then an equal and opposite amount of momentum must be added to the generator.
Post by: Jaaanosik on 19/04/2020 15:17:50
How is it possible that magnet comes to the generator proximity slower than it leaves.We are adding only angular momentum from the generator!!!
We are not adding linear momentum!!!
Adding extra exclamation points does not make your argument stronger.
If linear momentum is added to the magnet in orbit, then an equal and opposite amount of momentum must be added to the generator.
Where is the extra linear momentum coming from?
Yes, we added the linear momentum to both magnet and the generator.
Did we add it out of nothing?
Jano
Post by: Kryptid on 19/04/2020 15:22:27
Where is the extra linear momentum coming from?
The magnetic field in itself will cause the magnet to accelerate as it approaches the generator, which adds kinetic energy. It's a conversion of magnetic potential energy into kinetic energy. That net momentum is still zero.
Post by: Jaaanosik on 19/04/2020 17:24:52
Kryptid,Where is the extra linear momentum coming from?
The magnetic field in itself will cause the magnet to accelerate as it approaches the generator, which adds kinetic energy. It's a conversion of magnetic potential energy into kinetic energy. That net momentum is still zero.
is magnetic field linear or it rotates in our example,
Jano
Post by: Jaaanosik on 19/04/2020 19:50:08
Bobolink,Bobolink,I think we both understand that you want your conjecture to be correct so you are ignoring all attempts to help you understand the reality of the situation. It is clear that you won't listen, because that would destroy your fantasy that you have figured out something that scientists say is not possible. I don't think you are a troll, you are just someone who has a conjecture based on a limited and flawed understanding and are excited that you think you have made a great discovery. So, I guess all I can say is have fun living this fantasy. [Shrug]
here is the problem that Kryptid and it appears you too do not understand.
Sailboats can sail 3x faster than wind.
Please, explain, how is that possible,
Jano
For those that don't know how to google youtube, the solution,
Jano
Post by: Jaaanosik on 19/04/2020 20:31:08
Different classes of iceboat can achieve speeds, as follow.
International DN: 55–68 miles per hour (89–109 km/h).[7][10]
Classic iceboats and Skeeters: 100–150 miles per hour (160–240 km/h). Record: Das Boot, 155.9 miles per hour (250.9 km/h).[2] Record: classic iceboat, Debutaunte, 143 miles per hour (230 km/h).[7][11]
Post by: Kryptid on 19/04/2020 23:27:26
Kryptid,
is magnetic field linear or it rotates in our example,
Jano
I've never heard of a "linear" magnetic field before. Did you mean an unchanging field? And I presume, by rotating, you mean a field of this type: https://en.wikipedia.org/wiki/Rotating_magnetic_field
In either case, it doesn't matter. Linear momentum will be conserved because the generator must necessarily experience the exact same force that the magnet does, only in the opposite direction. Newton's third law guarantees it.
Post by: Jaaanosik on 20/04/2020 00:18:34
Kryptid,Kryptid,
is magnetic field linear or it rotates in our example,
Jano
I've never heard of a "linear" magnetic field before. Did you mean an unchanging field? And I presume, by rotating, you mean a field of this type: https://en.wikipedia.org/wiki/Rotating_magnetic_field
In either case, it doesn't matter. Linear momentum will be conserved because the generator must necessarily experience the exact same force that the magnet does, only in the opposite direction. Newton's third law guarantees it.
you are like 45 years late, linear magnetic field, magnetic river:
Go to 8:30 for the linear magnetic motor.
Did ROTATING magnetic field increased linear KINETIC energy?
Did increase of angular momentum increased the linear momentum?
Jano
Post by: Jaaanosik on 20/04/2020 00:35:09
Bobolink,I think we both understand that you want your conjecture to be correct so you are ignoring all attempts to help you understand the reality of the situation. It is clear that you won't listen, because that would destroy your fantasy that you have figured out something that scientists say is not possible. I don't think you are a troll, you are just someone who has a conjecture based on a limited and flawed understanding and are excited that you think you have made a great discovery. So, I guess all I can say is have fun living this fantasy. [Shrug]
here is the problem that Kryptid and it appears you too do not understand.
Bobolink,
This thread was moved to new theories because somebody does not understand that linear momentum can be converted to angular momentum and angular back to linear.
Please, tell me.
Did the increase of the angular momentum increased the linear momentum in the ON/OFF rotating magnetic field, orbiting manget example?
Jano
Post by: Kryptid on 20/04/2020 01:53:51
Did ROTATING magnetic field increased linear KINETIC energy?
Yes, but the linear momentum remains unchanged because any increase in linear momentum of the piece of aluminium being propelled is perfectly offset by the increase in linear momentum in the motor in the opposite direction.
Did increase of angular momentum increased the linear momentum?
The total angular momentum didn't increase.
This thread was moved to new theories because somebody does not understand that linear momentum can be converted to angular momentum and angular back to linear.
No, it was moved to New Theories because a reactionless drive would violate Newton's third law.
Post by: Jaaanosik on 20/04/2020 03:55:46
...
No, it was moved to New Theories because a reactionless drive would violate Newton's third law.
Kryptid,
Is the International Space Station violating the Newton's third law?
The ISS is pure inertial reference frame, no force acting on it, just falling.
Why does the ISS keep its attitude?
Why the ISS is not moving around the Earth like this pendulum?
(https://i.imgur.com/AXweaNq.png)
Why it is not upside-down on one side of the Earth and downside-up on the other?
Thrusters are not used to do the attitude management.
Where does the rotation of the ISS come from?
If there was one wheel inside the ISS trying to rotate it then it would fail because of the Newton's third law.
A wheel not used inside a big gimbal, but a wheel just to rotate the ISS.
So ISS is not suppose to rotate, to keep its attitude, agreed?
Jan
Post by: Kryptid on 20/04/2020 04:30:38
Is the International Space Station violating the Newton's third law?
No, but it isn't a reactionless drive either.
The ISS is pure inertial reference frame
No it isn't. It's in an accelerating frame because it's traveling in a (roughly) circular orbit.
no force acting on it, just falling.
No force acting on it? I don't think you know what the word "falling" means...
Why does the ISS keep its attitude?
Due you mean altitude? Because the centrifugal force of its orbital motion counteracts the gravitational force pulling it towards the Earth.
Why it is not upside-down on one side of the Earth and downside-up on the other?
I don't know because I've never looked into it. It could either be due to some kind of gyroscopes keeping it in the desired orientation, or due to tidal locking (admittedly, I don't know if tidal locking works for an object that small on such small time scales).
Thrusters are not used to do the attitude management.
Nor are they needed.
Where does the rotation of the ISS come from?
See my previous statement about gyroscopes and tidal locking.
If there was one wheel inside the ISS trying to rotate it then it would fail because of the Newton's third law.
You have just demonstrated that you don't know how Newton's third law works. If a wheel is spun aboard a spacecraft and it is not counterbalanced by another wheel spinning in the opposite direction, then the spacecraft itself will spin in the opposite direction of the wheel. That is precisely what Newton's third law means: the force causing the wheel to rotate results in an equal and opposite force that forces the spacecraft to spin in the opposite direction.
So ISS is not suppose to rotate, to keep its attitude, agreed?
I'm not completely sure what you are asking here, but absolutely none of this has anything to do with a reactionless drive. A reactionless drive is a machine that can produce net momentum in a closed system. The ISS does no such thing.
So let me ask you this: if there is a spacecraft floating in free space, do you agree that the total momentum off all of its constituent particles must add up to zero if the spacecraft itself has no net movement?
Post by: Colin2B on 20/04/2020 08:22:46
,Newton’s First law. Once you set it rotating at one rev per orbit it just keeps doing it, no correction required.
Why does the ISS keep its attitude?
Why the ISS is not moving around the Earth like this pendulum?
Why it is not upside-down on one side of the Earth and downside-up on the other?
Thrusters are not used to do the attitude management.
Where does the rotation of the ISS come from?
So ISS is not suppose to rotate, to keep its attitude, agreed?
Jan
Clever folks those flyers
Post by: Jaaanosik on 20/04/2020 17:26:45
Colin,,Newton’s First law. Once you set it rotating at one rev per orbit it just keeps doing it, no correction required.
Why does the ISS keep its attitude?
Why the ISS is not moving around the Earth like this pendulum?
Why it is not upside-down on one side of the Earth and downside-up on the other?
Thrusters are not used to do the attitude management.
Where does the rotation of the ISS come from?
So ISS is not suppose to rotate, to keep its attitude, agreed?
Jan
Clever folks those flyers
Yes, very good point, agreed.
Still, this is digging deeper:
https://en.wikipedia.org/wiki/Control_moment_gyroscope#International_Space_Station
Quote
In addition, it seeks a Torque Equilibrium Attitude (TEA), in which the combined torque contribution of gravity gradient, atmospheric drag, solar pressure, and geomagnetic interactions are minimized.
The ISS is influenced by these disturbances.
If ISS did not have CMGs to compensate then it would be in a chaotic motion like astreroids in this arcade game.
(https://i.imgur.com/QZ4vBFm.png)
The disturbances would got the ISS into a chaotic rotations and nothing would help if we gave up.
CMGs give us a 'firm/fixed' point to push against and to balance the ISS.
How is that possible? What is that 'fixed point'?
If this 'magical point' did not exist then how we can get any torque for the Torque Equilibrium Attitude (TEA)?
I call this point a potential well, because the effect really appears to be similar to a potential well,
Jano
Post by: Jaaanosik on 20/04/2020 17:33:12
Post by: Kryptid on 20/04/2020 19:40:34
Newton’s First law. Once you set it rotating at one rev per orbit it just keeps doing it, no correction required.
Clever folks those flyers
Now I feel kind of stupid for not realizing that.
How is that possible?
Gyroscopes work through conservation of angular momentum. It has nothing to do with a reactionless drive.
Post by: Jaaanosik on 20/04/2020 21:32:03
Kryptid,Newton’s First law. Once you set it rotating at one rev per orbit it just keeps doing it, no correction required.
Clever folks those flyers
Now I feel kind of stupid for not realizing that.How is that possible?
Gyroscopes work through conservation of angular momentum. It has nothing to do with a reactionless drive.
if you watch the above video and you compare 45s when the CD players are OFF.
The CD players rotate and the center of mass moves away almost in a 'straight line'.
He gave the players linear impulse, linear momentum, he did not give them any angular momentum.
Why are the CD players rotating around the center of mass?
Jano
Post by: Kryptid on 20/04/2020 21:35:49
He gave the players linear impulse, linear momentum, he did not give them any angular momentum.
Why are the CD players rotating around the center of mass?
Jano
I'll answer that once you've answered the question that I posed to you earlier:
So let me ask you this: if there is a spacecraft floating in free space, do you agree that the total momentum off all of its constituent particles must add up to zero if the spacecraft itself has no net movement?
Post by: Jaaanosik on 20/04/2020 21:56:33
Yes!He gave the players linear impulse, linear momentum, he did not give them any angular momentum.
Why are the CD players rotating around the center of mass?
Jano
I'll answer that once you've answered the question that I posed to you earlier:So let me ask you this: if there is a spacecraft floating in free space, do you agree that the total momentum off all of its constituent particles must add up to zero if the spacecraft itself has no net movement?
Do you agree that if you drop a cat backwards (with cat's back facing the ground) then the cat should not rotate and fall on her legs?
Do cats generate angular momentum out of nothing?
Jano
Post by: Kryptid on 20/04/2020 22:03:55
Yes!
All right, good. Then here is the next question: do you agree that, before the ship's engine is turned on, the total linear and angular momentum of a stationary ship is also zero?
In regards to your question about the CD players: he did give them angular momentum. Pushing an object off-center results in a transfer of angular momentum.
In regards to your question about the cat, the total angular momentum is unchanged. The total angular momentum starts at zero and ends at zero. The cat did not generate any net angular momentum. If you actually paid attention to the very video you posted, you would realize that.
Post by: Jaaanosik on 21/04/2020 15:23:03
Kryptid,Yes!
All right, good. Then here is the next question: do you agree that, before the ship's engine is turned on, the total linear and angular momentum of a stationary ship is also zero?
In regards to your question about the CD players: he did give them angular momentum. Pushing an object off-center results in a transfer of angular momentum.
In regards to your question about the cat, the total angular momentum is unchanged. The total angular momentum starts at zero and ends at zero. The cat did not generate any net angular momentum. If you actually paid attention to the very video you posted, you would realize that.
I am paying attention to the video I posted. This is digging deeper.
It is very import, we have to settle this.
Cat started with zero angular momentum, generated half a turn and stopped the rotation, ended up with zero angular momentum again.
Where did the NET half turn come from?
Jano
Post by: Jaaanosik on 21/04/2020 15:56:18
...Kryptid,
In regards to your question about the CD players: he did give them angular momentum. Pushing an object off-center results in a transfer of angular momentum.
...
what transfer you are talking about?
I thought you were saying a linear momentum is a linear momentum and it does not have an angular momentum.
I thought you were saying the linear momentum cannot be converted to angular momentum.
Jano
Post by: Kryptid on 21/04/2020 16:20:46
Then here is the next question: do you agree that, before the ship's engine is turned on, the total linear and angular momentum of a stationary ship is also zero?
Then I will answer yours.
Post by: Jaaanosik on 21/04/2020 16:29:39
It would be nice if you would be respectful enough to stop skipping over my questions. This is a give and take. You can't expect me to keep answering your questions if you won't answer mine in return. If you answer this question:Then here is the next question: do you agree that, before the ship's engine is turned on, the total linear and angular momentum of a stationary ship is also zero?
Then I will answer yours.
..Yes, agreed,
...
All right, good. Then here is the next question: do you agree that, before the ship's engine is turned on, the total linear and angular momentum of a stationary ship is also zero?
...
Jano
Post by: Kryptid on 21/04/2020 16:38:30
Yes, agreed,
Okay then, thank you. Since we are in agreement that the ship starts out with zero linear momentum and zero angular momentum, we know that the total momentum of the system starts out at zero. If this is the case, then do you also agree that, in accordance with the conservation of momentum, the total momentum of the ship at the end of stroke one must also be zero?
Kryptid,
I am paying attention to the video I posted. This is digging deeper.
It is very import, we have to settle this.
Cat started with zero angular momentum, generated half a turn and stopped the rotation, ended up with zero angular momentum again.
Where did the NET half turn come from?
Jano
Net angular momentum is not required in order to perform a turn. The solution to the problem is explained at about minute four in the video you linked. For more information, Wikipedia has an article explaining it as well: https://en.wikipedia.org/wiki/Falling_cat_problem
Kryptid,
what transfer you are talking about?
I thought you were saying a linear momentum is a linear momentum and it does not have an angular momentum.
I thought you were saying the linear momentum cannot be converted to angular momentum.
Jano
They are separate quantities. I'm actually rather amazed that you didn't pick up the solution from the very video you linked earlier from Khan Academy. It explained that the ball hitting the paddle has angular momentum because it is hitting off of the axis' center. It stated that there would be no angular momentum if the ball hit exactly on the axis, despite the fact that the ball has identical linear momentum in both cases. It's exactly the same scenario with pushing the CD player. A force pushing off-center applies angular momentum.
Post by: Jaaanosik on 21/04/2020 17:12:37
Yes, agreed,
Okay then, thank you. Since we are in agreement that the ship starts out with zero linear momentum and zero angular momentum, we know that the total momentum of the system starts out at zero. If this is the case, then do you also agree that, in accordance with the conservation of momentum, the total momentum of the ship at the end of stroke one must also be zero?
...
No, because release of potential energy changes boundaries/dynamics of the system.
You hold a kilo in front of you in your hand - momentum zero, you let it go and the momentum is not zero.
A release of potential energy can change arrangement of inertial forces that can lead to conversion/transfer between linear and angular momentum.
Jano
Post by: Jaaanosik on 21/04/2020 17:33:00
...Kryptid,Kryptid,
I am paying attention to the video I posted. This is digging deeper.
It is very import, we have to settle this.
Cat started with zero angular momentum, generated half a turn and stopped the rotation, ended up with zero angular momentum again.
Where did the NET half turn come from?
Jano
Net angular momentum is not required in order to perform a turn. The solution to the problem is explained at about minute four in the video you linked. For more information, Wikipedia has an article explaining it as well: https://en.wikipedia.org/wiki/Falling_cat_problem
...
can cat turn around without releasing a potential energy?
Where is the change of 0 angular momentum to some angular momentum to 0 angular momentum coming from?
The same as the above, reconfiguration of the inertial forces. This costs energy.
potential energy - angular momentum
kinetic energy - linear momentum
Post by: Jaaanosik on 21/04/2020 17:47:13
...Kryptid,
They are separate quantities. I'm actually rather amazed that you didn't pick up the solution from the very video you linked earlier from Khan Academy. It explained that the ball hitting the paddle has angular momentum because it is hitting off of the axis' center. It stated that there would be no angular momentum if the ball hit exactly on the axis, despite the fact that the ball has identical linear momentum in both cases. It's exactly the same scenario with pushing the CD player. A force pushing off-center applies angular momentum.
Yes, but the effect might be different.
See the CD players ON and OFF.
When the CD players are OFF then the center of mass moves away at constant velocity.
It has a constant linear momentum.
CD players have also constant angular momentum.
When the CD players are ON then the center of mass moves on a chaotic trajectory, forward, sideways, up, down, backwards, ...
The rotating CDs generate torque, all the torque force vectors are directed to an intersect - the potential well.
Pushing the CD players will cause the wobble.
A part of the original linear momentum/kinetic energy will be spent on climbing from the potential well so not all momentum/energy generates the forward and rotational motion,
Jano
Post by: Jaaanosik on 21/04/2020 17:52:48
...Kryptid,
They are separate quantities. I'm actually rather amazed that you didn't pick up the solution from the very video you linked earlier from Khan Academy. It explained that the ball hitting the paddle has angular momentum because it is hitting off of the axis' center. It stated that there would be no angular momentum if the ball hit exactly on the axis, despite the fact that the ball has identical linear momentum in both cases. It's exactly the same scenario with pushing the CD player. A force pushing off-center applies angular momentum.
When the CD players are OFF.
Can we say the original linear momentum is split into the new linear momentum plus the new angular momentum?
lm_orig = lm_new + am_new
Jano
Post by: Kryptid on 21/04/2020 21:39:19
No, because release of potential energy changes boundaries/dynamics of the system.
No it doesn't. That would violate conservation of momentum. A closed system cannot experience a change in net momentum. Period. That's not debatable.
You hold a kilo in front of you in your hand - momentum zero, you let it go and the momentum is not zero.
The momentum is still zero because the Earth is pulled up towards the kilogram with just as much force as the kilogram is pulled down towards the Earth. The momentum of the Earth becomes equal and opposite to the momentum of the kilogram. So the total momentum is unchanged.
A release of potential energy can change arrangement of inertial forces that can lead to conversion/transfer between linear and angular momentum.
First of all, no, it can't. Noether's theorem makes it clear that linear momentum and angular momentum are their own separate conservation laws. Arguing against Noether's theorem is like arguing against the Pythagorean theorem.
Second of all, your ship starts off with neither linear nor angular momentum. So the ability to convert one into the other would not make the engine work either. You can't get any linear momentum from angular momentum because there isn't any angular momentum present to work with.
can cat turn around without releasing a potential energy?
The potential energy comes from the chemical energy in the cat's metabolism.
Where is the change of 0 angular momentum to some angular momentum to 0 angular momentum coming from?
It isn't. The angular momentum remains zero throughout the entire process of the cat turning.
potential energy - angular momentum
kinetic energy - linear momentum
Restating this nonsense will not make it true.
When the CD players are ON then the center of mass moves on a chaotic trajectory, forward, sideways, up, down, backwards, ...
If you knew what "center of mass" meant, you would know that isn't true. The center of mass will move along an identical trajectory whether the CD players are turned on or turned off. The reason for this is because the linear momentum received is identical in both scenarios.
A part of the original linear momentum/kinetic energy will be spent on climbing from the potential well so not all momentum/energy generates the forward and rotational motion,
That would violate conservation of momentum.
Can we say the original linear momentum is split into the new linear momentum plus the new angular momentum?
No, we can't. That would go against Noether's theorem. Linear momentum remains linear momentum and angular momentum remains angular momentum. No amount of arguing will change that.
Post by: Jaaanosik on 21/04/2020 22:55:58
Kryptid,No, because release of potential energy changes boundaries/dynamics of the system.
No it doesn't. That would violate conservation of momentum. A closed system cannot experience a change in net momentum. Period. That's not debatable.You hold a kilo in front of you in your hand - momentum zero, you let it go and the momentum is not zero.
The momentum is still zero because the Earth is pulled up towards the kilogram with just as much force as the kilogram is pulled down towards the Earth. The momentum of the Earth becomes equal and opposite to the momentum of the kilogram. So the total momentum is unchanged.
...
You and kilo in your hand. Are you a closed system with zero momentum?
You are saying no because the Earth got moved by the kilo.
We cannot stop, the Sun got moved by the Earth.
The galaxy got moved by the Sun.
The other galaxies got moved by our galaxy.
The Universe is unchanged only the internal conversions between its particles.
As mentioned before, all the particles in the Universe:
7am_1 + 3lm_1 + 5am_2 + 5lm_2 + ... + 2am_n + 8lm_n = 3am_1 + 7lm_1 + 5am_2 + 5lm_2 + ... + 8am_n + 2lm_n
Please, tell me where are you going to draw the boundaries of a closed system?
Jano
Post by: Kryptid on 21/04/2020 22:58:47
Please, tell me where are you going to draw the boundaries of a closed system?
You can technically say that the entire visible Universe represents the closed system in question. But that is, of course, because gravity has an infinite range. Such is not the case if all you are doing is moving balls around inside of a spaceship. The ship itself represents the closed system then.
Post by: Jaaanosik on 21/04/2020 23:12:04
...Kryptid,
Second of all, your ship starts off with neither linear nor angular momentum. So the ability to convert one into the other would not make the engine work either. You can't get any linear momentum from angular momentum because there isn't any angular momentum present to work with.
...
I do not follow what you are saying.
A loaded spring, potential energy, can start a rotation of a wheel, angular momentum.
Are you saying this cannot happen??
Jano
Post by: Kryptid on 22/04/2020 00:35:06
A loaded spring, potential energy, can start a rotation of a wheel, angular momentum
The recoil of the spring will send it in the opposite direction, creating an equal and opposite angular momentum to the wheel. The net angular momentum of the system remains zero.
Are you saying this cannot happen??
It can happen. The problem is that you are considering various parts of the system in isolation and not taking into account that momentum (angular or otherwise) is not a one-way street. Newton's third law will cause any donation of momentum to be counteracted by an equal amount of momentum in the opposite direction.
Just to make sure that you actually understand what conservation of momentum means (because it's far from clear that you do), take a look at this quote from "the Physics Classroom" website: https://www.physicsclassroom.com/class/momentum/Lesson-2/Momentum-Conservation-Principle
Quote
One of the most powerful laws in physics is the law of momentum conservation. The law of momentum conservation can be stated as follows.
For a collision occurring between object 1 and object 2 in an isolated system, the total momentum of the two objects before the collision is equal to the total momentum of the two objects after the collision. That is, the momentum lost by object 1 is equal to the momentum gained by object 2.
The above statement tells us that the total momentum of a collection of objects (a system) is conserved - that is, the total amount of momentum is a constant or unchanging value.
I bolded the important bit. Based on this, do you understand that conservation of momentum means that the total momentum of a closed system cannot change?
Post by: Jaaanosik on 22/04/2020 00:47:43
...What is causing the wobble of the CD players when they are ON and when they receive gentle external impulse?
It can happen. The problem is that you are considering various parts of the system in isolation and not taking into account that momentum (angular or otherwise) is not a one-way street. Newton's third law will cause any donation of momentum to be counteracted by an equal amount of momentum in the opposite direction.
...
Where is the action and counter-reaction coming from?
Jano
Post by: Kryptid on 22/04/2020 00:49:56
What is causing the wobble of the CD players when they are ON and when they receive gentle external impulse?
Where is the action and counter-reaction coming from?
Jano
And, once again, I will answer your question once you've answered mine...
Post by: Jaaanosik on 22/04/2020 01:28:24
From https://en.wikipedia.org/wiki/Closed_system
Quote
A closed system is a physical system that does not allow certain types of transfers (such as transfer of mass in or out of the system), though the transfer of energy is allowed.
The cat burning potential energy is losing atoms, mass is getting out of the system.
The rotating cat is not a closed system.
Jano
Post by: Jaaanosik on 22/04/2020 05:20:57
(https://i.imgur.com/DXoArdR.png)
(https://i.imgur.com/sVEiex9.png)
Quote
Thus, in the absence of an external impulse, the linear momentum of a system remains unchanged.
(https://i.imgur.com/qXctQmF.png)
The black ball is a closed system.
The dark blue wheel is a closed system.
The momentum is conserved till there is an external impulse.
The black ball hitting the wheel system is the 'external' impulse to the wheel system.
Yes, blue wheel and the black ball are subsystems of the bigger system.
Still this is the proper way to analyze the situation presented in the diagram.
The linear impulse from the black ball can act as an external angular impact of the blue wheel.
The conclusion is that the internal subsystems can change their values, conversion from one momentum to another.
Even a conversion from linear to angular.
I hope this settles the problem we were talking about for the last week or so,
Jano
Post by: Kryptid on 22/04/2020 05:24:46
What is a closed system?
From https://en.wikipedia.org/wiki/Closed_systemQuoteA closed system is a physical system that does not allow certain types of transfers (such as transfer of mass in or out of the system), though the transfer of energy is allowed.
The cat burning potential energy is losing atoms, mass is getting out of the system.
The rotating cat is not a closed system.
Jano
That didn't answer my question, now did it?
Post by: Jaaanosik on 22/04/2020 05:30:51
I am not sure what is your question,What is a closed system?
From https://en.wikipedia.org/wiki/Closed_systemQuoteA closed system is a physical system that does not allow certain types of transfers (such as transfer of mass in or out of the system), though the transfer of energy is allowed.
The cat burning potential energy is losing atoms, mass is getting out of the system.
The rotating cat is not a closed system.
Jano
That didn't answer my question, now did it?
Jano
Post by: Kryptid on 22/04/2020 05:34:06
I am not sure what is your question,
It's here. The sentence that is underlined:
Just to make sure that you actually understand what conservation of momentum means (because it's far from clear that you do), take a look at this quote from "the Physics Classroom" website: https://www.physicsclassroom.com/class/momentum/Lesson-2/Momentum-Conservation-PrincipleQuoteOne of the most powerful laws in physics is the law of momentum conservation. The law of momentum conservation can be stated as follows.
For a collision occurring between object 1 and object 2 in an isolated system, the total momentum of the two objects before the collision is equal to the total momentum of the two objects after the collision. That is, the momentum lost by object 1 is equal to the momentum gained by object 2.
The above statement tells us that the total momentum of a collection of objects (a system) is conserved - that is, the total amount of momentum is a constant or unchanging value.
I bolded the important bit. Based on this, do you understand that conservation of momentum means that the total momentum of a closed system cannot change?
Post by: Jaaanosik on 22/04/2020 05:39:34
Yes, please, see the post #179 for the explanation,I am not sure what is your question,
It's here. The sentence that is underlined:Just to make sure that you actually understand what conservation of momentum means (because it's far from clear that you do), take a look at this quote from "the Physics Classroom" website: https://www.physicsclassroom.com/class/momentum/Lesson-2/Momentum-Conservation-PrincipleQuoteOne of the most powerful laws in physics is the law of momentum conservation. The law of momentum conservation can be stated as follows.
For a collision occurring between object 1 and object 2 in an isolated system, the total momentum of the two objects before the collision is equal to the total momentum of the two objects after the collision. That is, the momentum lost by object 1 is equal to the momentum gained by object 2.
The above statement tells us that the total momentum of a collection of objects (a system) is conserved - that is, the total amount of momentum is a constant or unchanging value.
I bolded the important bit. Based on this, do you understand that conservation of momentum means that the total momentum of a closed system cannot change?
Jano
Post by: Kryptid on 22/04/2020 05:49:36
Yes
Good. Now, the following things are true:
(1) The ship starts out with no momentum (angular or linear).
(2) The ship is a closed system (nothing is entering or leaving it).
Given that you have just agreed that a closed system must conserve momentum, we must conclude the total momentum of the ship at the end of stroke one must be equal to the total momentum of the ship before stroke one. Do you agree with this?
The cat burning potential energy is losing atoms, mass is getting out of the system.
The rotating cat is not a closed system.
Jano
Although this is true on a technical level, the cat is not receiving any angular momentum from the environment so it's a moot point.
What is causing the wobble of the CD players when they are ON and when they receive gentle external impulse?
Where is the action and counter-reaction coming from?
This is called gyroscopic precession. This video explains the physics behind it:
Nothing about it violates conservation of angular momentum.
The black ball is a closed system.
The dark blue wheel is a closed system.
The very moment that those two objects interact with each other, they cease to be closed systems. A closed system is one that does not interact with other systems.
The conclusion is that the internal subsystems can change their values, conversion from one momentum to another.
Even a conversion from linear to angular.
Not if Noether's theorem has anything to say about it.
Post by: Jaaanosik on 22/04/2020 06:08:22
The black ball is a closed system.
The dark blue wheel is a closed system.
The very moment that those two objects interact with each other, they cease to be closed systems. A closed system is one that does not interact with other systems.
...
The text book is clear, a close system does not change its momentum till there is an external impulse.
Where do you get your definition that "A closed system is one that does not interact with other systems"?
This is a little nuance or playing on words but very important.
It appears the only system that fits your definition is the whole Universe, if we assume there is nothing outside the Universe.
As mentioned above, the black ball is a closed system but it can change the momentum when it interacts with the wheel, the external impulse as per the textbook.
The wheel is a closed system, it is not moving, till there is an external impulse that is going to cause the rotation, as per the textbook,
Jano
Post by: Kryptid on 22/04/2020 06:15:48
Where do you get your definition that "A closed system is one that does not interact with other systems"?
I have to admit, I had not encountered the definition of closed system that you posted before. An isolated system is more what I was thinking.
In any case, the spaceship is not only a closed system, but also an isolated system (it doesn't utilize any external mechanisms. It tries to move using only its internal mechanisms). That being said, do you agree that its momentum must remain unchanged since it is an isolated system?
Post by: Jaaanosik on 22/04/2020 06:16:13
...If this was a simple gyro precession then the CD players body would not wobble.
This is called gyroscopic precession. This video explains the physics behind it:
Nothing about it violates conservation of angular momentum.
...
It would keep some kind of rotation.
There are 3 gyros. That's the difference. Therefore there is a wobble, the potential well,
Jano
Post by: Jaaanosik on 22/04/2020 06:18:47
We need to know where you got your definition.Where do you get your definition that "A closed system is one that does not interact with other systems"?
I have to admit, I had not encountered the definition of closed system that you posted before. An isolated system is more what I was thinking.
In any case, the spaceship is not only a closed system, but also an isolated system (it doesn't utilize any external mechanisms. It tries to move using only its internal mechanisms). That being said, do you agree that its momentum must remain unchanged since it is an isolated system?
As mentioned, the only system is the whole Universe based on your definition.
... and even that one is conditional that there is nothing outside the Universe,
Jano
Post by: Kryptid on 22/04/2020 06:21:34
If this was a simple gyro precession then the CD players body would not wobble.
Gyroscopic precession does cause a wobble. It's what causes a spinning top to wobble.
So how about answering my question?
We need to know where you got your definition.
As mentioned, the only system is the whole Universe based on your definition,
Jano
This is one such example: https://www.dictionary.com/browse/closed-system
Post by: Jaaanosik on 22/04/2020 06:31:20
Really???If this was a simple gyro precession then the CD players body would not wobble.
Gyroscopic precession does cause a wobble. It's what causes a spinning top to wobble.
So how about answering my question?We need to know where you got your definition.
As mentioned, the only system is the whole Universe based on your definition,
Jano
This is one such example: https://www.dictionary.com/browse/closed-system
A dictionary definition???
We need a better one.
What is your question?
Jano
Post by: Kryptid on 22/04/2020 06:33:01
Really???
A dictionary definition???
And what is your problem with dictionaries?
What is your question?
This:
In any case, the spaceship is not only a closed system, but also an isolated system (it doesn't utilize any external mechanisms. It tries to move using only its internal mechanisms). That being said, do you agree that its momentum must remain unchanged since it is an isolated system?
Remember, no momentum is entering or leaving the ship.
Post by: Jaaanosik on 22/04/2020 06:37:40
If this was a simple gyro precession then the CD players body would not wobble.
Gyroscopic precession does cause a wobble. It's what causes a spinning top to wobble.
So how about answering my question?We need to know where you got your definition.
As mentioned, the only system is the whole Universe based on your definition,
Jano
This is one such example: https://www.dictionary.com/browse/closed-system
How is the gyro precession causing a wobble?
Is it in the youtube videos? Where the gravity is involved?
Do you have a single wheel precession in micro-gravity?
I don't think so: Jano
Post by: Kryptid on 22/04/2020 06:41:14
How is the gyro precession causing a wobble?
It's explained by Wikipedia here: https://en.wikipedia.org/wiki/Precession#Classical_(Newtonian)
Is it in the youtube videos? Where the gravity is involved?
Do you have a single wheel precession in micro-gravity?
Gravity is not involved.
Now answer my question.
Post by: Jaaanosik on 22/04/2020 06:42:58
Really???
A dictionary definition???
And what is your problem with dictionaries?What is your question?
This:In any case, the spaceship is not only a closed system, but also an isolated system (it doesn't utilize any external mechanisms. It tries to move using only its internal mechanisms). That being said, do you agree that its momentum must remain unchanged since it is an isolated system?
Remember, no momentum is entering or leaving the ship.
No momentum is entering or leaving the ship?
I was saying the ship is collecting hydrogen from the space.
Still this is not relevant because the question is whether a system can generate something that resembles external momentum from within the system.
What momentum entered or left the cat when it rotated half a turn?
Jano
Post by: Jaaanosik on 22/04/2020 06:45:06
There is no wobble from a single gyro in micro-gravity, check the video.How is the gyro precession causing a wobble?
It's explained by Wikipedia here: https://en.wikipedia.org/wiki/Precession#Classical_(Newtonian)Is it in the youtube videos? Where the gravity is involved?
Do you have a single wheel precession in micro-gravity?
Gravity is not involved.
Now answer my question.
Please, what is your question?
There were many posts, I do not know what question you are asking,
Jano
Edit: this one?
That being said, do you agree that its momentum must remain unchanged since it is an isolated system?
There is no answer till we settle the closed system definition.
Edit2:
The answer is in pointing out how the cat rotated without external momentum.
Post by: Kryptid on 22/04/2020 06:49:29
There is no wobble from a single gyro in micro-gravity, check the video.
I can see it wobble. The wobble is just very small.
Please, what is your question?
You already asked me this before. My question is in reply #191.
Post by: Jaaanosik on 22/04/2020 06:53:53
Dictionary is not a textbook, and even textbooks can have mistakes,There is no wobble from a single gyro in micro-gravity, check the video.
I can see it wobble. The wobble is just very small.Please, what is your question?
You already asked me this before. My question is in reply #191.
Jano
Post by: Kryptid on 22/04/2020 06:55:37
Dictionary is not a textbook
Your point being what?
There is no answer till we settle the closed system definition.
The spaceship is an isolated system. That is clearly defined. No matter or energy is entering or leaving the system. No momentum is entering or leaving the system either.
So if a system does not have momentum entering or leaving it, you understand that the momentum must remain unchanged, yes?
Post by: Jaaanosik on 22/04/2020 07:00:14
Yes, there is very tiny wobble on the gyro ring, but check the axle, no wobble at all.There is no wobble from a single gyro in micro-gravity, check the video.
I can see it wobble. The wobble is just very small.Please, what is your question?
You already asked me this before. My question is in reply #191.
The wheel could have a wobble due to bearings the axle tells the story though.
It appears the axle might be slightly bent on the bottom part, the top part seems straight, without any wobble,
Jano
Post by: Jaaanosik on 22/04/2020 07:03:41
Check the cat.Dictionary is not a textbook
Your point being what?There is no answer till we settle the closed system definition.
The spaceship is an isolated system. That is clearly defined. No matter or energy is entering or leaving the system. No momentum is entering or leaving the system either.
So if a system does not have momentum entering or leaving it, you understand that the momentum must remain unchanged, yes?
Forget that the cat stopped the rotation.
What is more important is to realize that the cat was able to start the rotation without external momentum.
Do you understand what it means?
The cat is isolated system, correct?
How the cat could start the rotation without an external impulse?
Jano
Post by: Kryptid on 22/04/2020 07:08:16
Do you understand what it means?
Yes, it means that executing a rotation does not require net momentum (unlike your engine, which does require net momentum).
The cat is isolated system, correct?
For the sake of discussion, yes.
How the cat could start the rotation without an external impulse?
Because net momentum isn't needed for a rotation (but net momentum is needed for your spaceship to go forward). I've already said that before.
The thing about the gyroscope is ultimately irrelevant. Even if I was utterly unable to explain anything about gyroscopes, I can still point to Noether's theorem as proof that you can't interconvert angular and linear momentum.
Post by: Jaaanosik on 22/04/2020 07:21:40
It's getting late here, 2:20am, I am going to bed.Do you understand what it means?
Yes, it means that executing a rotation does not require net momentum (unlike your engine, which does require net momentum).The cat is isolated system, correct?
For the sake of discussion, yes.How the cat could start the rotation without an external impulse?
Because net momentum isn't needed for a rotation (but net momentum is needed for your spaceship to go forward). I've already said that before.
The thing about the gyroscope is ultimately irrelevant. Even if I was utterly unable to explain anything about gyroscopes, I can still point to Noether's theorem as proof that you can't interconvert angular and linear momentum.
Still, I think we made a good progress.
I'll just say an isolated system is not a closed system.
The isolated system can generate momentum, the closed system requires an external momentum.
You are mistaken about the wobble.
Check how big wobble is for the CD players.
Here is a quote from wiki about CMGs: 'large CMGs have produced thousands of newton meters of torque.'
What is this torque? How is this torque used on ISS?
Jano
Post by: Kryptid on 22/04/2020 07:32:18
I'll just say an isolated system is not a closed system
According to the definitions that you have provided in your Wikipedia link, an isolated system is even more cut off from the outside world than a closed system is. A closed system allows for the transfer of energy but not matter, whereas an isolated system does not allow for the transfer ot either energy or matter.
The isolated system can generate momentum
How is it going to generate momentum when (1) it can't receive momentum from an outside source, and (2) momentum is conserved?
You are mistaken about the wobble.
Check how big wobble is for the CD players.
Even if I am, so what? It won't help your engine work.
What is this torque?
https://en.wikipedia.org/wiki/Torque
How is this torque used on ISS?
I presume it is used to reorient the ISS and make it stable against changes in orientation. But that doesn't change the net angular momentum of the ISS nor does it help your engine work.
You have agreed with me that your ship starts with zero angular momentum and zero linear momentum. You have agreed that conservation of momentum is true. The ship is an isolated system. Putting these three things together, the ship cannot ever have a net momentum that isn't zero. There is no angular momentum present, so you can't argue that it can be changed into linear momentum because zero can't become a positive number. No momentum is entering the ship from an outside source nor is momentum leaving the ship to an outside source. So tell me already: how can zero net momentum in an isolated system turn into positive net momentum without violating conservation of momentum?
Post by: Jaaanosik on 22/04/2020 18:53:14
I'll just say an isolated system is not a closed system
According to the definitions that you have provided in your Wikipedia link, an isolated system is even more cut off from the outside world than a closed system is. A closed system allows for the transfer of energy but not matter, whereas an isolated system does not allow for the transfer ot either energy or matter.The isolated system can generate momentum
How is it going to generate momentum when (1) it can't receive momentum from an outside source, and (2) momentum is conserved?
The cat is a system.
We can call it whatever - closed/isolated, I do not care.
What is important is that the system can turn 180 degrees by itself and it costs energy.
Are we sure the cat generated no net angular momentum?
(https://i.imgur.com/CEqFij3.png)
The orientation of the body seems to be different.
Quote
You are mistaken about the wobble.
Check how big wobble is for the CD players.
Even if I am, so what? It won't help your engine work.What is this torque?
https://en.wikipedia.org/wiki/TorqueHow is this torque used on ISS?
I presume it is used to reorient the ISS and make it stable against changes in orientation. But that doesn't change the net angular momentum of the ISS nor does it help your engine work.
You have agreed with me that your ship starts with zero angular momentum and zero linear momentum. You have agreed that conservation of momentum is true. The ship is an isolated system. Putting these three things together, the ship cannot ever have a net momentum that isn't zero. There is no angular momentum present, so you can't argue that it can be changed into linear momentum because zero can't become a positive number. No momentum is entering the ship from an outside source nor is momentum leaving the ship to an outside source. So tell me already: how can zero net momentum in an isolated system turn into positive net momentum without violating conservation of momentum?
Please, have a look here:
(https://i.imgur.com/9wOAjyi.png)
Just one side. The CMG will not rotate, the spaceship will rotate.
There is an accelerometer attached to the CMG body - 'no acceleration'.
There is an accelerometer attached to the spaceship - there is an acceleration during the reeling in.
'No acceleration' - meaning if the motion is done right with the VSCMGs then there is no acceleration measured.
https://en.wikipedia.org/wiki/Control_moment_gyroscope#Variable-speed
The VSCMGs can eliminate the wobble.
We can do the reeling in - time interleaved left, right short impulses - the same result, no acceleration on both VSCMG systems but the spaceship accelerates.
We stop the CMG rotations before they hit the spaceship back.
What would be the result? A forward net momentum,
Jano
Edit: Stopping the CMGs is important to get rid of the big torque force vectors.
The torque vectors will not cause any havoc when they are not present.
Post by: Jaaanosik on 22/04/2020 19:52:59
Jano
Post by: Kryptid on 22/04/2020 21:05:15
Are we sure the cat generated no net angular momentum?
Yes, because Noether's theorem says so. Do you even know what a conservation law is?
The orientation of the body seems to be different.
Of course it's different. Changing orientation doesn't require net momentum.
We stop the CMG rotations before they hit the spaceship back.
What force is stopping the CMG?
What would be the result? A forward net momentum
I don't really need to explain to you why zero angular momentum plus zero linear momentum equals positive linear momentum is bad math, do I?
As long as you keep arguing that 0 + 0 = positive number I am going to stand against it because it is wrong.
Post by: Jaaanosik on 22/04/2020 23:26:59
Kryptid,Are we sure the cat generated no net angular momentum?
Yes, because Noether's theorem says so. Do you even know what a conservation law is?The orientation of the body seems to be different.
Of course it's different. Changing orientation doesn't require net momentum.We stop the CMG rotations before they hit the spaceship back.
What force is stopping the CMG?What would be the result? A forward net momentum
I don't really need to explain to you why zero angular momentum plus zero linear momentum equals positive linear momentum is bad math, do I?
As long as you keep arguing that 0 + 0 = positive number I am going to stand against it because it is wrong.
did you check the last video?
Anything to say?
Jano
Post by: Kryptid on 23/04/2020 00:01:57
Anything to say?
No. The Cubli doesn't violate conservation of angular momentum (nothing ever does), so there is nothing to say.
Post by: Jaaanosik on 23/04/2020 01:09:54
Is Cubli a closed system?Anything to say?
No. The Cubli doesn't violate conservation of angular momentum (nothing ever does), so there is nothing to say.
If yes then it needs an external force to change the momentum.
Do you see any external force that causes the jump up?
Jano
Post by: Kryptid on 23/04/2020 01:24:32
Is Cubli a closed system?
Depends on which definition of "closed system" you are going for.
If yes then it needs an external force to change the momentum.
The momentum doesn't change...
Do you see any external force that causes the jump up?
No, because no such force is needed. I already told you once that changing orientation does not require a change in net angular momentum.
Momentum is conserved. Period. If you think you've found a case where it isn't, it's only because your analysis of it was either flawed or incomplete. We know that beforehand even without having to get into the details of how the system works. Noether's theorem guarantees it.
Post by: Jaaanosik on 23/04/2020 03:18:54
The textbook definition for now.Is Cubli a closed system?
Depends on which definition of "closed system" you are going for.
Cubli is not a closed system. Agreed?
Quote
Cubli is not a closed system.If yes then it needs an external force to change the momentum.
The momentum doesn't change...Do you see any external force that causes the jump up?
No, because no such force is needed. I already told you once that changing orientation does not require a change in net angular momentum.
Momentum is conserved. Period. If you think you've found a case where it isn't, it's only because your analysis of it was either flawed or incomplete. We know that beforehand even without having to get into the details of how the system works. Noether's theorem guarantees it.
CMGs are not a closed system.
There is no point in talking that a momentum is conserved if the system is not closed.
Jano
Post by: Kryptid on 23/04/2020 05:58:19
The textbook definition for now.
The one that says a closed system doesn't allow matter to be transferred but energy can?
Cubli is not a closed system. Agreed?
If it was an open system, then matter would be transferred in and out of it. What matter is being transferred in and out of the Cubli?
CMGs are not a closed system.
Of course not. They are interacting with the other components of the ship. But the other components of the ship don't have any momentum that they can give to the CMGs. The CMGs don't have any momentum to give to the other components of the ship either (remember, you agreed that the system has zero momentum before the engine is cut on).
There is no point in talking that a momentum is conserved if the system is not closed.
You know why that is, don't you? It's because an open system can change its momentum by interacting with outside systems. Your ship is not interacting with other systems because it is an isolated system. No mechanism exists by which it can receive momentum from somewhere else. And if you did rely on receiving momentum from an outside source, then that would pretty much mean that you have abandoned your claim that you can give the ship net momentum just by manipulating its internal components.
To drive this home, consider a universe that is completely empty of all energy and matter with the exception of your spaceship and an astronaut that is floating right behind it. The spaceship is turned off and the astronaut sees the ship as stationary in their reference frame. Since the astronaut obviously sees themself as being stationary as well, then they know that the total momentum in their universe is zero (the ship's momentum plus the astronaut's momentum equals zero).
Now you turn the engine on. According to you, the ship will begin to move forward and leave the astronaut behind. The astronaut goes about measuring the total momentum of the universe a second time. This time, the moving ship does have momentum from the astronaut's reference frame. They take the measurements, do the calculations and find that the ship's momentum plus the astronaut's momentum is now a positive number. This means that a closed system (the entire universe) has experienced an increase in net momentum.
We know, of course, that such a thing is impossible because momentum is a conserved quantity. A closed system cannot go from zero momentum to positive momentum. This is how we know that the scenario I just described cannot happen in the real world. It isn't possible for the ship to move because it will increase the total momentum of a closed system.
Post by: Jaaanosik on 23/04/2020 17:44:27
The textbook definition for now.
The one that says a closed system doesn't allow matter to be transferred but energy can?Cubli is not a closed system. Agreed?
If it was an open system, then matter would be transferred in and out of it. What matter is being transferred in and out of the Cubli?
This is from the textbook:
Quote
Thus, if there is no angular impulse about a fixed point (or about the mass center), the angular momentum of the system about the fixed point (or about the mass center) remains unchanged.
This is the definition of the closed system for the angular momentum.
No angular impulse - conservation of the angular momentum.
It is the angular impulse that makes the Cubli jump up, the definition for the conservation of the angular momentum does not hold anymore.
If the definition of the closed system has to include the conservation of the angular momentum then Cubli is an open system.
Quote
CMGs are not a closed system.
Of course not. They are interacting with the other components of the ship. But the other components of the ship don't have any momentum that they can give to the CMGs. The CMGs don't have any momentum to give to the other components of the ship either (remember, you agreed that the system has zero momentum before the engine is cut on).
So when CMGs try to rotate the ISS then the ISS is not resisting the rotation? The angular momentum in opposite direction?
The CMGs balance the disturbances and CMGs make sure the attitude is steady in the ECI frame.
CMGs keep the rotation of the ISS steady in the ECI frame.
The disturbances are external impulses to the ISS and CMGs work against them.
What makes you say the ISS has no momentum to give to CMGs?
What are those disturbances?
Jano
Post by: Jaaanosik on 23/04/2020 18:21:01
...Kryptid,There is no point in talking that a momentum is conserved if the system is not closed.
You know why that is, don't you? It's because an open system can change its momentum by interacting with outside systems. Your ship is not interacting with other systems because it is an isolated system. No mechanism exists by which it can receive momentum from somewhere else. And if you did rely on receiving momentum from an outside source, then that would pretty much mean that you have abandoned your claim that you can give the ship net momentum just by manipulating its internal components.
To drive this home, consider a universe that is completely empty of all energy and matter with the exception of your spaceship and an astronaut that is floating right behind it. The spaceship is turned off and the astronaut sees the ship as stationary in their reference frame. Since the astronaut obviously sees themself as being stationary as well, then they know that the total momentum in their universe is zero (the ship's momentum plus the astronaut's momentum equals zero).
Now you turn the engine on. According to you, the ship will begin to move forward and leave the astronaut behind. The astronaut goes about measuring the total momentum of the universe a second time. This time, the moving ship does have momentum from the astronaut's reference frame. They take the measurements, do the calculations and find that the ship's momentum plus the astronaut's momentum is now a positive number. This means that a closed system (the entire universe) has experienced an increase in net momentum.
We know, of course, that such a thing is impossible because momentum is a conserved quantity. A closed system cannot go from zero momentum to positive momentum. This is how we know that the scenario I just described cannot happen in the real world. It isn't possible for the ship to move because it will increase the total momentum of a closed system.
There is an impulse that propels the spaceship away.
Is it an internal or an external impulse? Is it important?
The system is not closed anymore.
The momentum is frame dependent.
Dynamics/boundaries changed, the astronaut sees dv/dt change of the spaceship in his frame.
If the Universe, the spaceship and the astronaut are one frame in the beginning.
Then the spaceship accelerates.
What is the accelerometer going to say that is attached to the spaceship, the Universe accelerometer and the astronaut accelerometer?
I think you should rethink and rewrite what you wrote, specifically think about the frame dependency of the momentum,
Jano
Post by: Bobolink on 23/04/2020 20:38:07
Do you see any external force that causes the jump up?I saw where it used gravity and ground to push against. Of course the conservation of momentum wasn't broken. You realize that would not work in a zero-g environment with no wall to work against. You know why? That's right the conservation of momentum.
You are tilting at windmills.
If you spent half this effort learning physics you would be very knowledgeable guy...
Post by: Kryptid on 23/04/2020 20:45:18
You might as well be arguing that you can charge up a dead battery using another dead battery, because if you believe that momentum can be created, then you might as well believe that energy can be created too. Conservation laws have no meaning to you.
Post by: Jaaanosik on 23/04/2020 21:24:05
This is a completely worthless discussion. You are just as bad at physics as Dave Lev is. You don't understand what conservation laws are. You don't understand what a closed system is. And nothing I can do will be able to alleviate that ignorance. So I'm ending my attempt to educate you.Kryptid,
You might as well be arguing that you can charge up a dead battery using another dead battery, because if you believe that momentum can be created, then you might as well believe that energy can be created too. Conservation laws have no meaning to you.
You missed that momentum is frame dependent?
When your spaceship accelerates then gas is flying out on the other side.
Momentum is conserved for those 'two entities'.
Both entities had zero momentum in the astronaut frame before but now both entities have momentum left/right in the astronaut frame.
This is you: "They take the measurements, do the calculations and find that the ship's momentum plus the astronaut's momentum is now a positive number."
It has to be positive number because the gas and astronaut is negative number, isn't it?
The spaceship is positive number, the astronaut is unchanged (0) and the gas is negative number in the Universe frame.
You screwed up royally my friend. The spaceship and the astronaut is open system, not closed, you left out the engine gas.
The spaceship has a positive net momentum in the astronaut frame.
Are you going to admit that you made a mistake?
Jano
Post by: Kryptid on 23/04/2020 21:26:10
you left out the engine gas.
Your design doesn't have engine gas.
Post by: Jaaanosik on 23/04/2020 21:31:28
This is a completely worthless discussion. You are just as bad at physics as Dave Lev is. You don't understand what conservation laws are. You don't understand what a closed system is. And nothing I can do will be able to alleviate that ignorance. So I'm ending my attempt to educate you.Bobolink,
You might as well be arguing that you can charge up a dead battery using another dead battery, because if you believe that momentum can be created, then you might as well believe that energy can be created too. Conservation laws have no meaning to you.
What happens when the Cubli rotates in zero-g?
Will the outside frame rotate 45 degrees as in the video if the frame is not attached to anything?
Yes, no?
Jano
Post by: Jaaanosik on 23/04/2020 21:49:25
I missed that you had my design of the spaceship in mind.you left out the engine gas.
Your design doesn't have engine gas.
Then the question is if the CMGs center of mass would have unchanged position in the astronauts frame during/after reeling in the spaceship.
Imagine Cubli doing 45 degree impulse (jump up) as in the video.
The jump up 45 degrees and then it stops.
I am sure you are going to question the stopping part.
That's where the wobble example shows it is possible,
Jano
Edit: The 6:03min into the video is very interesting part.
Post by: Kryptid on 23/04/2020 21:50:58
I am sure you are going to question the stopping part.
Nope. I'm not questioning anything else. I'm going to leave you to your ignorance now.
Post by: Bobolink on 23/04/2020 21:54:35
Bobolink,https://www.google.com/url?sa=t&source=web&rct=j&url=https://www.ethz.ch/content/dam/ethz/special-interest/mavt/dynamic-systems-n-control/idsc-dam/Research_DAndrea/Cubli/RevisedManuscript.pdf&ved=2ahUKEwiLjYjutP_oAhVFZM0KHcH5DccQFjADegQIBBAB&usg=AOvVaw3Q75p6hREw01NhghV2_P53 (https://www.google.com/url?sa=t&source=web&rct=j&url=https://www.ethz.ch/content/dam/ethz/special-interest/mavt/dynamic-systems-n-control/idsc-dam/Research_DAndrea/Cubli/RevisedManuscript.pdf&ved=2ahUKEwiLjYjutP_oAhVFZM0KHcH5DccQFjADegQIBBAB&usg=AOvVaw3Q75p6hREw01NhghV2_P53)
What happens when the Cubli rotates in zero-g?
Will the outside frame rotate 45 degrees as in the video if the frame is not attached to anything?
Yes, no?
Jano
Here is the paper written by the inventors of the cubli. Sad, to say but the conservation of momentum is kind of the basis for the cubli. But what the heck, why let facts and science get in the way of your fantasy.
Post by: Jaaanosik on 24/04/2020 01:42:21
All blue arrows without red and green are CMGs at a stable state.
The blue arrows are gyro angular velocities pointing away from each other.
The green arrow is in the direction of the string towards the spaceship reel motor.
Reeling in starts and at the same time top gyro and bottom left gyro will increase the angular acceleration to generate the red counter torque to the green torque disturbance.
The CMGs center of mass will not move anywhere, CMGs will not turn anywhere if the load and counter torque are balanced properly, yet the spaceship will be pulled.
There is nothing wrong with this free body diagram,
Jano
Edit: fixing wrong description of the vectors.
Post by: Jaaanosik on 24/04/2020 16:46:35
Bobolink,Bobolink,https://www.google.com/url?sa=t&source=web&rct=j&url=https://www.ethz.ch/content/dam/ethz/special-interest/mavt/dynamic-systems-n-control/idsc-dam/Research_DAndrea/Cubli/RevisedManuscript.pdf&ved=2ahUKEwiLjYjutP_oAhVFZM0KHcH5DccQFjADegQIBBAB&usg=AOvVaw3Q75p6hREw01NhghV2_P53 (https://www.google.com/url?sa=t&source=web&rct=j&url=https://www.ethz.ch/content/dam/ethz/special-interest/mavt/dynamic-systems-n-control/idsc-dam/Research_DAndrea/Cubli/RevisedManuscript.pdf&ved=2ahUKEwiLjYjutP_oAhVFZM0KHcH5DccQFjADegQIBBAB&usg=AOvVaw3Q75p6hREw01NhghV2_P53)
What happens when the Cubli rotates in zero-g?
Will the outside frame rotate 45 degrees as in the video if the frame is not attached to anything?
Yes, no?
Jano
Here is the paper written by the inventors of the cubli. Sad, to say but the conservation of momentum is kind of the basis for the cubli. But what the heck, why let facts and science get in the way of your fantasy.
What happens to Cubli if it is in zero-g and it is not attached to anything and the jump-up impulse is generated.
Will Cubli move? Yes? No?
It could be argued that the inverted pendulum is more difficult to control then rotations in micro-gravity when Cubli is not restricted by gravity/external forces.
Jano
Post by: Jaaanosik on 24/04/2020 23:43:46
If I did know it then I would propose this right at the beginning.
This works because it is proven by Cubli that it can generate the angular acceleration also known as torque by breaking the spinning wheel. The breaking generates bigger torque than spin up. Please, don't get fooled by the fact that Cubli is an inverted pendulum in the video.
(https://i.imgur.com/27oqHzz.png)
The setup is in space, 'zero-g'.
1. A simple gyro start up generates 0 net angular momentum.
Let us assume there is a ring within a ring (stator/rotor - an electric motor) when the angular accelerations/torques start they cancel each other in opposite directions. The rings can have high angular velocities even though the velocities are in the opposite directions.
2. The 3 gyros are configured in a triangle layout and the brakes are applied just to one ring (stator) on all gyros.
The brakes are attached to the outside structure (similar to Cubli), they would not be stopping the second ring.
This would be a jump/torque impulse in the up direction. The gyro assembly starts to move up.
3. A steady torque is applied to turn 3 gyros axles 180 degrees.
The brakes are applied to the second ring (rotor) on all gyros.
This would be the second jump/torque impulse in the up direction. The gyro assembly continues to move up even faster.
The cycle can repeat,
Jano
Post by: Colin2B on 25/04/2020 09:11:59
....it can generate the angular acceleration also known as torqueAngular acceleration is NOT known as torque
Please, don't get fooled by the fact that Cubli is an inverted pendulum in the video.No one is getting confused. This device uses conservation of momentum to do what it does. Don’t confuse rotation, inertia, kinetic energy, potential energy, and angular momentum or you will never understand how this and the falling cat work.
Post by: Jaaanosik on 25/04/2020 16:12:18
Colin,....it can generate the angular acceleration also known as torqueAngular acceleration is NOT known as torquePlease, don't get fooled by the fact that Cubli is an inverted pendulum in the video.No one is getting confused. This device uses conservation of momentum to do what it does. Don’t confuse rotation, inertia, kinetic energy, potential energy, and angular momentum or you will never understand how this and the falling cat work.
thank you, I am glad you are paying attention.
I used the term in a 'loose way' to see who is going to pick it up.
Torque is rotational inertia multiplied by the angular acceleration. The relationship is the same as F=ma for the linear systems.
(https://i.imgur.com/msroRjg.png)
This is how it works.
When the gyro is built as a motor and it will start to rotate by the EM field than mechanical parts will rotate in opposite directions. This angular acceleration combine with the mass of the gyro rings that have rotational inertia gives us the torque.
The torque, angular acceleration and the end result, the angular velocity, are vectors in the same direction along the axle because that is where the center of mass is located for a well centered gyro rings.
Let us assume there is uneven distribution of the rotational inertia, the stator - outside part is a bigger ring compared to rotor - the inner part.
Two things can happen. We can compensate by mass or if we do not compensate with mass then the angular velocity will be different so the net angular momentum is conserved, in opposite directions, it will give us 0.
Either way the result is 0 net angular momentum.
The torques are equal in opposite directions but acceleration and angular velocities might be different based on the solution.
The gyro rings center of mass will not change the position during the acceleration startup.
The gyro is attached to a housing/cage/spaceship and this outside structure will not change its position as well.
When we are in steady state, there is no more angular acceleration, no more torque, even though there are angular velocities of they gyro rings.
Now we decide to brake one gyro ring.
The deceleration will cause the torque impulse, the gyro system moves in the direction of the torque.
If this is a spaceship in the interstellar space then it will move in regards to the stars.
Now we rotate the gyro 180 degrees in regards to the stars.
Is it possible to rotate gyro 180 degrees. Yes, if we assume the symmetrical numbers then gyro will rotate 180 degrees and the spaceship will rotate 180 degrees because of 0 net angular momentum in regards to the stars.
Now we brake the second gyro ring.
The deceleration will cause the torque impulse of the gyro system in the direction of the torque.
This is the same direction as the first impulse in regards to the stars even though it appears the spaceship is going backwards in the spaceship reference frame.
It is time to pause, ... to think, ... to think more, ...
Jano
Post by: Colin2B on 25/04/2020 17:58:17
I used the term in a 'loose way' to see who is going to pick it up.Yeah, yeah, of course you did ;)
By the way, I'm not paying attention, I just skim sample posts to check spamming etc
Torque is rotational inertia multiplied by the angular acceleration. The relationship is the same as F=ma for the linear systems.yes, we all know these relationships, but I noticed further back you get very confused about quite a few of them.
We also know how the device works
Post by: Jaaanosik on 25/04/2020 18:02:20
Colin,I used the term in a 'loose way' to see who is going to pick it up.Yeah, yeah, of course you did ;)
By the way, I'm not paying attention, I just skim sample posts to check spamming etcTorque is rotational inertia multiplied by the angular acceleration. The relationship is the same as F=ma for the linear systems.yes, we all know these relationships, but I noticed further back you get very confused about quite a few of them.
We also know how the device works
whatever suits you,
Jano
Post by: Jaaanosik on 29/04/2020 21:48:41
can I ask the following question at the physics forum?
Let us assume an astronaut pushes to the right inside the ISS (the green force vector) and the ISS CMG's compensate with the opposite torque (red arrow).
Is the ISS center for mass going to be disturbed (moved out of position) in the ECI? (Erath Center Inertial reference frame)
Jano
(https://i.imgur.com/H40oNiF.png)
Post by: Colin2B on 29/04/2020 22:54:43
Kryptid,No, keep them here.
can I ask the following question at the physics forum?
Post by: Jaaanosik on 30/04/2020 02:40:11
Colin,Kryptid,No, keep them here.
can I ask the following question at the physics forum?
is there anything wrong with the question?
Jano
Post by: Halc on 30/04/2020 03:11:18
is there anything wrong with the question?The problem is more with the asker, not as much with the question. Yes, keep it here.
The question also has problems. Your description suggests the addition of a force vector to a torque vector, which is adding different things.
The answer to the question is trivial: The ISS is moving at over 7 km/s relative to 'ECI', so of course it is going to change its position in this frame regardless of what minor pushes the astronauts and gyros decide to perform. If it were stationary in that frame, it would immediately fall to the ground like any other dropped rock.
Secondly, Earth Center (like any real object I can think of) is always accelerating, and thus does not define an inertial frame. For thought experiments that take away other objects like the sun, there is an inertial frame defined by the center of gravity of the full system being considered, but you didn't reference that.
Post by: Jaaanosik on 30/04/2020 16:07:53
Halc,is there anything wrong with the question?The problem is more with the asker, not as much with the question. Yes, keep it here.
The question also has problems. Your description suggests the addition of a force vector to a torque vector, which is adding different things.
The answer to the question is trivial: The ISS is moving at over 7 km/s relative to 'ECI', so of course it is going to change its position in this frame regardless of what minor pushes the astronauts and gyros decide to perform. If it were stationary in that frame, it would immediately fall to the ground like any other dropped rock.
Secondly, Earth Center (like any real object I can think of) is always accelerating, and thus does not define an inertial frame. For thought experiments that take away other objects like the sun, there is an inertial frame defined by the center of gravity of the full system being considered, but you didn't reference that.
good points. I wanted to have the question real, I did not describe it in the best way.
Let us make it unrealistic. The ISS is in the intergalactic space, the ISS center of mass is stationary in the reference frame of the surrounding galaxies. The flat space-time as we can get.
The interstellar space-time has too much curvature compared to the intergalactic one.
Let us assume an astronaut pushes to the right inside the ISS (the green force vector) and the ISS CMG's compensate with the opposite torque (red arrow).
Is the ISS center for mass going to be disturbed (moved out of position) in in the reference frame of the surrounding galaxies?
Jano
Post by: Bobolink on 30/04/2020 16:22:28
It is time to pause, ... to think, ... to think more, ...If only you could take your own advice...
By the way, the answer to the OP is still no.
Post by: Halc on 30/04/2020 16:33:02
I wanted to have the question real, I did not describe it in the best way.Short story: We're defining a stable environment to keep outside influences from messing with the answer. That's good.
Let us make it unrealistic. The ISS is in the intergalactic space, the ISS center of mass is stationary in the reference frame of the surrounding galaxies. The flat space-time as we can get.
The interstellar space-time has too much curvature compared to the intergalactic one.
Quote
Let us assume an astronaut pushes to the right inside the ISS (the green force vector) and the ISS CMG's compensate with the opposite torque (red arrow).Conservation of momentum says it cannot be disturbed, so that's the easy answer. You've been told this in countless posts but you continue to ignore it all. Hence the discussion staying here in lighter-side.
Is the ISS center for mass going to be disturbed (moved out of position) in in the reference frame of the surrounding galaxies?
BTW, you're still making the mistake of comparing two vectors of different units. Force and torque cannot compensate for each other.
Force/momentum: Astronaut pushes with F for a second, sending astronaut one way and the rest of the ISS the other way. Those balance, both in force and momentum. Astronaut hits the far wall and both stop, and the center of mass has moved not a bit.The gyros have no effect on this.
Torque: Astronaut applies torque T for 1 second to the ISS structure and the gyros (which can be anywhere) compensate. Assuming the thing has zero angular momentum before (you didn't specify that), then the angular momentum of the astronaut relative to the system is exactly compensated by the opposite angular momentum now acquired by the gyros, until he hits the far wall and the gyros are forced to give it back. At no point was there a nonzero total angular momentum.
Post by: Jaaanosik on 30/04/2020 19:18:26
I wanted to have the question real, I did not describe it in the best way.Short story: We're defining a stable environment to keep outside influences from messing with the answer. That's good.
Let us make it unrealistic. The ISS is in the intergalactic space, the ISS center of mass is stationary in the reference frame of the surrounding galaxies. The flat space-time as we can get.
The interstellar space-time has too much curvature compared to the intergalactic one.QuoteLet us assume an astronaut pushes to the right inside the ISS (the green force vector) and the ISS CMG's compensate with the opposite torque (red arrow).Conservation of momentum says it cannot be disturbed, so that's the easy answer. You've been told this in countless posts but you continue to ignore it all. Hence the discussion staying here in lighter-side.
Is the ISS center for mass going to be disturbed (moved out of position) in in the reference frame of the surrounding galaxies?
BTW, you're still making the mistake of comparing two vectors of different units. Force and torque cannot compensate for each other.
Force/momentum: Astronaut pushes with F for a second, sending astronaut one way and the rest of the ISS the other way. Those balance, both in force and momentum. Astronaut hits the far wall and both stop, and the center of mass has moved not a bit.The gyros have no effect on this.
Torque: Astronaut applies torque T for 1 second to the ISS structure and the gyros (which can be anywhere) compensate. Assuming the thing has zero angular momentum before (you didn't specify that), then the angular momentum of the astronaut relative to the system is exactly compensated by the opposite angular momentum now acquired by the gyros, until he hits the far wall and the gyros are forced to give it back. At no point was there a nonzero total angular momentum.
Halc,
there are two scenarios that need to be analyzed.
1. With the CMGs NOT RUNNING/ACTUATED
2. With the CMGs RUNNING/ACTUATED
Here is the video:
1. is shown at 50s of the video
2. is shown at 56s of the video
1. The center of CD players (ISS) mass moves on a straight line and the CD players rotate when CD players are OFF (CDs inside do not rotate).
2. The center of CD players (ISS) mass does not move on a straight line and the CD players do not rotate, they just wobble when CD players are ON (CDs inside rotate).
Do you see the difference?
This is what is at the core of the problem I am trying to point out,
Jano
Post by: Jaaanosik on 30/04/2020 19:32:59
...Halc, is torque a force vector?
BTW, you're still making the mistake of comparing two vectors of different units. Force and torque cannot compensate for each other.
...
This is a little bit tricky question, because of the well balanced accelerating wheel.
What is a torque for the accelerating wheel?
Jano
Post by: Halc on 30/04/2020 22:33:22
OK, but 3 CD players taped together is not a CMG. Nothing in there attempts to stop its own motion. The object has entirely passive behavior.
Halc,
there are two scenarios that need to be analyzed.
1. With the CMGs NOT RUNNING/ACTUATED
2. With the CMGs RUNNING/ACTUATED
Quote
1. The center of CD players (ISS) mass moves on a straight line and the CD players rotate when CD players are OFF (CDs inside do not rotate).Fine, I see the difference. The guy applies a little torque to the running object and that changes its angular momentum, causing it to wobble. Same thing with a toy top on its side, which precesses due to torque applied to it. The top orientation never rotates in the direction the torque is applied.
2. The center of CD players (ISS) mass does not move on a straight line and the CD players do not rotate, they just wobble when CD players are ON (CDs inside rotate).
Do you see the difference?
This is what is at the core of the problem I am trying to point out,
Halc, is torque a force vector?No. Force is measured in Newtons, torque in Newton-meters. They're different animals.
Your red torque vector should A) come from the center of mass, not the point of force application, and B) point towards the point of view, not to the left. Torque vectors are along the axis of rotation.
Quote
This is a little bit tricky question, because of the well balanced accelerating wheel.It can be expressed as tangential force component multiplied by perpendicular offset from the center of rotation.
What is a torque for the accelerating wheel?
Post by: Jaaanosik on 01/05/2020 04:13:30
here is a wheel, top view and the side view:
(https://i.imgur.com/jI0a70R.png)
The wheel and the brakes calipers have friction less bearings.
The wheel and the calipers can move up along the axles without any obstruction.
The setup is here on the Earth, the axles point up, the wheel is horizontal to the ground.
The wheel is rotating clock wise, omega points down.
Let us apply an instantaneous breaking to the wheel.
The angular acceleration (actually deceleration) is in upward direction.
Is the wheel going to jump up due to the angular acceleration/deceleration, assuming angular acceleration is bigger than gravitational acceleration?
Jano
Post by: Halc on 01/05/2020 12:30:02
Is the wheel going to jump up due to the angular acceleration/deceleration, assuming angular acceleration is bigger than gravitational acceleration?You know it will not.
The fact that you got the α and ω vectors correct suggests you're not completely uneducated, and yet you ask a question like that. Now why is that?
For the third time, you are adding vectors of different units (bold above), meaning you're ignoring everything I say and intend to continue to do so. Hence I see no reason to continue this troll discussion.
Post by: Jaaanosik on 01/05/2020 15:18:15
Halc,Is the wheel going to jump up due to the angular acceleration/deceleration, assuming angular acceleration is bigger than gravitational acceleration?You know it will not.
The fact that you got the α and ω vectors correct suggests you're not completely uneducated, and yet you ask a question like that. Now why is that?
For the third time, you are adding vectors of different units (bold above), meaning you're ignoring everything I say and intend to continue to do so. Hence I see no reason to continue this troll discussion.
I assure you, this is not trolling.
Please, have a look:
(https://i.imgur.com/9YpDTfu.png)
https://en.wikipedia.org/wiki/Torque
The figure shows the setup from my question.
If we take a realistic time, let's say dt = 0.1s for the braking.
When Torque - T, Omega - w, Inertia - I, Angular acceleration - a
T = I (dw/dt) + (dI/dt) w
We know that dI/dt = 0 during the breaking, so the torque
T = I (dw/dt)
or
T = I a
Please, tell me a good reason why the wheel is not going to jump up if the angular acceleration a is bigger than gravitational g?
Jano
Post by: Bobolink on 01/05/2020 16:40:40
Please, tell me a good reason why the wheel is not going to jump up if the angular acceleration a is bigger than gravitational g?
I assure you, this is not trolling.It is trolling, because regardless of ANY information you're given you will continue to maintain that the conservation of momentum does not hold.
Post by: Jaaanosik on 01/05/2020 17:30:26
Bobolink,Please, tell me a good reason why the wheel is not going to jump up if the angular acceleration a is bigger than gravitational g?I assure you, this is not trolling.It is trolling, because regardless of ANY information you're given you will continue to maintain that the conservation of momentum does not hold.
on the contrary, if you say that the wheel is not going to jump up then you are defending position that the conservation of the momentum does not hold,
Jano
Post by: Kryptid on 01/05/2020 17:33:52
Bobolink,
on the contrary, if you say that the wheel is not going to jump up then you are defending position that the conservation of the momentum does not hold,
LOL!
Post by: Jaaanosik on 01/05/2020 19:54:57
Kryptid,Bobolink,
on the contrary, if you say that the wheel is not going to jump up then you are defending position that the conservation of the momentum does not hold,
LOL!
:)
2:50min into the video, how is it possible that the cube jumps?
The axles in my setup keep the reaction wheel on a straight up jump.
As I said the wheel and the breaks have bearings in my example and they allow the motion along the axles.
The cubes in the video jump 'randomly' but they use inertial actuators - reaction wheels, the same principle,
Jano
Post by: Jaaanosik on 01/05/2020 20:07:34
15 seconds into the video and 2:00min,
Jano
Post by: Kryptid on 01/05/2020 20:23:04
Post by: Jaaanosik on 01/05/2020 20:47:58
It amazes me that even after so much explanation, you still just don't understand angular momentum.Really?
Please, explain how the cube can fly above the table?
What caused the jump?
How is it going to be different from my explanation?
Jano
Post by: Kryptid on 01/05/2020 21:08:12
Really?
Please, explain how the cube can fly above the table?
What caused the jump?
How is it going to be different from my explanation?
Jano
If you haven't understood it by now, nothing I can say will fix that problem.
Post by: Bobolink on 01/05/2020 22:37:28
2:50min into the video, how is it possible that the cube jumps?Hint: It wouldn't jump anywhere in a freefall reference frame.
Post by: Jaaanosik on 01/05/2020 23:11:52
Bobolink,2:50min into the video, how is it possible that the cube jumps?Hint: It wouldn't jump anywhere in a freefall reference frame.
here is the spaceship, the axles are attached to the spaceship frame.
(https://i.imgur.com/zBLryB9.png)
Do you see how easy it is to generate counter torque by the CMGs?
The wheel will move in the direction of the yellow angular acceleration and the spaceship has no reason to move.
The sum of all the CMGs torques is going to compensate for the disturbance,
Jano
Post by: Jaaanosik on 02/05/2020 18:18:03
Halc,Halc,Is the wheel going to jump up due to the angular acceleration/deceleration, assuming angular acceleration is bigger than gravitational acceleration?You know it will not.
The fact that you got the α and ω vectors correct suggests you're not completely uneducated, and yet you ask a question like that. Now why is that?
For the third time, you are adding vectors of different units (bold above), meaning you're ignoring everything I say and intend to continue to do so. Hence I see no reason to continue this troll discussion.
I assure you, this is not trolling.
Please, have a look:
(https://i.imgur.com/9YpDTfu.png)
https://en.wikipedia.org/wiki/Torque
The figure shows the setup from my question.
If we take a realistic time, let's say dt = 0.1s for the braking.
When Torque - T, Omega - w, Inertia - I, Angular acceleration - a
T = I (dw/dt) + (dI/dt) w
We know that dI/dt = 0 during the breaking, so the torque
T = I (dw/dt)
or
T = I a
Please, tell me a good reason why the wheel is not going to jump up if the angular acceleration a is bigger than gravitational g?
Jano
my friend, look above and here:
(https://i.imgur.com/jI0a70R.png)
The dark blue force F in wiki diagram - the braking force is not torque.
If two calipers/pads on the wheel are 1m from the axis and the tangential braking force is 1N then we have 2Nm of torque.
If two calipers/pads on the wheel are 0.5m from the axis then we need 2N of the tangential braking force in order to have 2Nm of the torque.
It is important to establish 'the flow', the braking force is the cause the torque is the effect.
The simple F=ma, right, it is simple.
Well, there could be lots of stuff happening to generate force F, the cause, and how it affects the acceleration, the effect.
There is going to be difference in acceleration a if I push a wheelbarrow not lifted and lifted.
The same force will give us different acceleration.
The same goes for the torque. If the brakes on the wheel can supply only 1N then the torque is different based on the position of the brakes, 0.5m gives us smaller torque compared to 1m for the same braking force 1N.
Jano
Post by: pzkpfw on 02/05/2020 23:17:16
The right hand rule as applied to rotation is essentially a mathematical artifact of a cross product.
That the pseudovector L appears to have a "direction" does not imply a force in that direction, in the way you're thinking.
Why would L be a force up or down, due to the disk spinning?
The pseudovector L (being a vector) is more about resistance to a change in the orientation of that vector. e.g. like a gyro wanting to keep spinning in its plane.
When the brakes are applied to your spinning disk, it won't move up (or down).
(Edit: removed section too easily misconstrued.)
Post by: Jaaanosik on 03/05/2020 05:17:54
It seems to me you've misunderstood L (angular momentum).Hi,
The right hand rule as applied to rotation is essentially a mathematical artifact of a cross product.
That the pseudovector L appears to have a "direction" does not imply a force in that direction, in the way you're thinking.
Why would L be a force up or down, due to the disk spinning?
The pseudovector L (being a vector) is more about resistance to a change in the orientation of that vector. e.g. like a gyro wanting to keep spinning in its plane.
When the brakes are applied to your spinning disk, it won't move up (or down).
(Edit: removed section too easily misconstrued.)
No, to the pseudovector (pseudo force?).
The torque is a real force during the braking of the wheel otherwise we would not have the gyro precession.
Here are some analysis:
(https://i.imgur.com/4dJgKGW.png)
https: //www.youtube.com/watch?v=bYF0PGsF92k
(https://i.imgur.com/CQ0ZeUd.png)
https: //www.youtube.com/watch?v=qS_dcNqs3d4
Please, notice the different position of the torque vectors in those two images.
(https://i.imgur.com/6ubvKhG.png)
The top view of the wheel. The darker gray balls represent two centers of mass of two halves of the wheel.
The orange vectors are torques positioned at the same position as in the second video.
It does not matter because they can be summed up and put at the total center of mass, the yellow vector, when angular acceleration is multiplied by the rotational inertia,
Jano
Post by: Kryptid on 03/05/2020 06:36:25
Post by: pzkpfw on 04/05/2020 00:40:48
Post by: Halc on 04/05/2020 00:59:33
A bicycle and a spinning office chair would be enough to put aside this silly idea of a braked wheel experiencing a force in the direction of L.He hasn't explicitly mentioned that since post 252.
Still trolling however. Most recent post had at least four errors, one of which was the lack of direct relevance of the videos to the situation being discussed.
I find it best to just not feed the trolls. He's not here to learn anything, only to push our collective buttons and get his jollies from those who continue to respond.
I try to respond to intelligent questions, but the last several post have neither asked questions nor displayed any deliberate intelligence.
Post by: Jaaanosik on 04/05/2020 14:55:03
A bicycle and a spinning office chair would be enough to put aside this silly idea of a braked wheel experiencing a force in the direction of L.Hi,
Please, provide a diagram and math to support your claim,
Jano
Post by: Jaaanosik on 04/05/2020 14:58:35
Halc,A bicycle and a spinning office chair would be enough to put aside this silly idea of a braked wheel experiencing a force in the direction of L.He hasn't explicitly mentioned that since post 252.
Still trolling however. Most recent post had at least four errors, one of which was the lack of direct relevance of the videos to the situation being discussed.
I find it best to just not feed the trolls. He's not here to learn anything, only to push our collective buttons and get his jollies from those who continue to respond.
I try to respond to intelligent questions, but the last several post have neither asked questions nor displayed any deliberate intelligence.
Please, point the errors or otherwise your posts are trolling,
Jano
Post by: Halc on 04/05/2020 15:50:05
Please, point the errors or otherwise your posts are trolling,Error 6: Failure to respond to a troll is not in itself trolling. It is doing the troll and the rest of the world a service, similar to abstinence of feeding wild geese.
If you wish a response, show some intelligence in your posts, not goose droppings.
The torque is a real force1: Continued equivocation of force and torque, despite numerous posts informing you of this error.
Quote
during the braking of the wheel otherwise we would not have the gyro precession.2: Braking of the wheel does not cause gyro precession.
Quote
https: //www.youtube.com/watch?v=qS_dcNqs3d43: Link to a video about gyro precession, when the topic at hand concerned the braking of a spinning wheel. A) Your example does not involve precession, and B) The video lacks any mention of the wheel being braked or otherwise slowed.
Physics - Mechanics: The Gyroscope (3 of 5) The Torque of a Spinning Gyroscope
Quote
The top view of the wheel. The darker gray balls represent two centers of mass of two halves of the wheel.4: The second video does not depict any masses as two separate centers of mass. These two separate masses play no significant role in the situation at hand. In particular, the angular moment of an object is not a function of the mass and separation of those two centers of mass.
The orange vectors are torques positioned at the same position as in the second video.
5: Your orange vectors are shown parallel to the disk axis, but in the video the one vector shown is perpendicular to the axis of rotation of the disk. This is not the 'same position', evidence of the irrelevancy of your choice of videos to the topic you're trying to discuss.
OK, more than 4.
Post by: Jaaanosik on 04/05/2020 17:24:13
Halc,Please, point the errors or otherwise your posts are trolling,Error 6: Failure to respond to a troll is not in itself trolling. It is doing the troll and the rest of the world a service, similar to abstinence of feeding wild geese.
If you wish a response, show some intelligence in your posts, not goose droppings.The torque is a real force1: Continued equivocation of force and torque, despite numerous posts informing you of this error.Quoteduring the braking of the wheel otherwise we would not have the gyro precession.2: Braking of the wheel does not cause gyro precession.
...
I'd like to answer 1 and 2 first.
1.
If torque is not a real force then what is preventing the guy to rotate the wheel against the torque (precession direction) at 2:50 min in this video?
2.
(https://i.imgur.com/CQ0ZeUd.png)
The arm (r) of the assembly is like a spoke of a bigger 'wheel'.
The rotation of this 'virtual wheel' is in the F=mg direction, down.
The gyro wheel angular momentum L is 'breaking', preventing the free fall, this generates the torque.
If we increase the mass at the end of the gyro wheel then the angular momentum 'brakes' more, bigger torque, faster precession,
Jano
Post by: Halc on 04/05/2020 20:43:23
I'd like to answer 1 and 2 first.At 2:50 the guy is just hefting the thing around to show that it's heavy. I'm guessing the object to weigh about 150 Newtons. Were he to stand still for a moment holding it like that, he seems to be applying about 300 N upward with his right arm, 150 N downward with his left, and gravity supplying the remaining 150 N of weight downward. That's a net force of zero which means it doesn't accelerate either towards the ground or the sky.Quote from: HalcContinued equivocation of force and torqueIf torque is not a real force then what is preventing the guy to rotate the wheel against the torque (precession direction) at 2:50 min in this video?
On the torque front, the torces applied at the various points along its length add up to zero torque, which is why the thing doesn't begin to rotate while he's holding it.
At 3.04 it show him holding the spinning object to the right. It has zero momentum, zero angular momentum in the direction of the camera (the one we're using, not the other camera in the picture), and positive angular momentum to the left. I'll consider the 5 seconds from that moment, after which the spinning object points at the camera.
Force: For the duration, the guy supplies 150N upward and gravity 150N downward, hence the thing doesn't acquire significant momentum upward or downward. He varies it a bit as he lifts the thing over his head, but the net momentum remains zero. Were he to stop applying this force, there would a net downward force and the object would accelerate to the ground.
Torque: He (and gravity) apply about 150 N-m of torque to the object, initially in the direction away from the camera. Sure enough, the object has something like 600 Nm²/sec angular momentum in the direction away from the camera after those 5 seconds, but has not moved away from the camera since torque is not force. Similarly, as we approach time 3:09, he (and gravity) are applying about 150 N-m of torque to the right, which by 3:09 has entirely cancelled the angular momentum to the left it initially had.
The two (force and torque) are totally separate things, and have totally separate effects. Your continued equivocation of them is a serious error.
Quote
Again you equivocate force and torque. Yes, the forces being applied result in torque on the assembly, but torque doesn't make something move down, net force does, and there is no net downward force in the scenario depicted. It has no reason to move downward.Quote from: Halc2: Braking of the wheel does not cause gyro precession.The arm (r) of the assembly is like a spoke of a bigger 'wheel'.
The rotation of this 'virtual wheel' is in the F=mg direction, down.
Quote
The gyro wheel angular momentum L is 'breaking', preventing the free fall, this generates the torque.Braking requires motion and friction resisting (slowing) that motion. There is no downward motion to resist in this situation, and the assembly is assumed to be free of friction. Yes, toy gyros slow over time and eventually fall. The ones in space stations run in a vacuum and do so for years. There's nothing being slowed down, hence the video does not depict your 'braking' scenario where energy is being dissipated by friction. It is an irrelevant choice of videos to make the (incorrect) point you were attempting.
Of course, posting an irrelevant video obfuscates your point, which serves the purpose of being a troll, so it all depends on what your goals are in the making of that selection.
Post by: Jaaanosik on 04/05/2020 21:05:30
...Halc,QuoteThe gyro wheel angular momentum L is 'breaking', preventing the free fall, this generates the torque.Braking requires motion and friction resisting (slowing) that motion. There is no downward motion to resist in this situation, and the assembly is assumed to be free of friction. Yes, toy gyros slow over time and eventually fall. The ones in space stations run in a vacuum and do so for years. There's nothing being slowed down, hence the video does not depict your 'braking' scenario where energy is being dissipated by friction. It is an irrelevant choice of videos to make the (incorrect) point you were attempting.
...
The torque is a rotational inertia multiplied by an angular acceleration.
There is a torque therefore there has to be an angular acceleration.
Please, show us, where is the angular acceleration so we can have the torque?
The gravitational acceleration g is not the angular acceleration because it is in a different direction.
Where is the angular acceleration?
Jano
Post by: Jaaanosik on 05/05/2020 15:15:32
Halc,The torque is a rotational inertia multiplied by an angular acceleration.Read my 4th paragraph in the previous reply, which answers exactly this. It shows the resulting angular acceleration.
There is a torque therefore there has to be an angular acceleration.
Please, show us, where is the angular acceleration so we can have the torque?
the guy in the video cannot turn counter clockwise at 2:50 min and but he can turn clockwise at 2:55 min without any problem.
There is NO change in the setup, just his intention where to turn.
What is the delta?
Is he wrestling with a force at 2:50? Inertial force to be accurate?
This is very important to settle.
Also going back, where is that angular acceleration? This does not explain it.
Jano
Post by: Jaaanosik on 06/05/2020 04:15:31
I understand your argument about the units of measure for the torque - Nm.
One way to look at it - a force applied/generated at a distance from the axis.
It is an inertial force.
If torque is not a force at a distance from the axis then F=mg is not a force as well.
Right, it is just what people use to actually calculate motion but it is not real force, gravity is a curvature of space-time.
The centripetal force is real but the centrifugal force is not real.
Going on a centripetal/centrifugal ride in a park and having 10 kilo brick in one's lap to check the centrifugal force would not be fun,
Jano
Post by: Jaaanosik on 07/05/2020 03:16:41
when a gyro wheel spins, nothing happens.
The wheel has an angular momentum and it is a stable vector.
Minute 48 in this video:
Do we need a force, some work done to rotate the angular momentum, to rotate the gyro wheel axle?
Jano
Post by: Jaaanosik on 08/05/2020 15:59:57
Minute 29 of the same "8.01x - Lect 24 - Rolling Motion, Gyroscopes, VERY NON-INTUITIVE video."
Professor Lewin says: "There is no net force on that wheel, but there is a net torque."
What is a net torque? Out of nothing?
We can see, that if the torque is not there then there is no precession, minute 48.
The wheel is stable, no net force without the additional weight on the axle.
Also the wheel is stable no net force with the additional weight on the axle with the precession.
That is peculiar.
What are the causes of the precession effect?
The external torque on the angular momentum of a rotating wheel.
Both are required, if one is missing then there is no precession.
Jano
Post by: Halc on 08/05/2020 18:43:15
You're still assuming that force is torque, which is going to give you wrong answers every time.
Professor Lewin says: "There is no net force on that wheel, but there is a net torque."Net torque means that the sum of the torque vectors acting on the wheel is nonzero. No, those torques must be transferred to the wheel. Conservation of angular momentum does not allow torque 'out of nothing'.
What is a net torque? Out of nothing?
He also says net force is zero, meaning the sum of the force vectors is zero. The force vectors are obviously not the torque vectors, since (the part you never remember) force is not torque.
Quote
We can see, that if the torque is not there then there is no precession, minute 48.No, there is still no net force, as evidenced by the fact that the setup doesn't go up, down, left, right, or whatever. f=ma, so absent any linear acceleration, there can be no net force acting on the wheel.
The wheel is stable, no net force without the additional weight on the axle.
Also the wheel is stable no net force with the additional weight on the axle with the precession.
Quote
What are the causes of the precession effect?If it is precessing, then there must be a net torque on it, which necessarily changes (accelerates) the angular velocity vector ω.
Quote
The external torque on the angular momentum of a rotating wheel.The torque does it. It is not 'external' torque, since that word conveys no additional meaning. Angular momentum does not cause precession, as evidenced by the lack of precession when no torque is applied.
Quote
Both are required, if one is missing then there is no precession.Reasonable statement. If angular momentum was zero, there would be no ω vector to precess. Torque would still cause the exact same angular acceleration, but it wouldn't be considered precession in that scenario.
Post by: Jaaanosik on 09/05/2020 16:42:21
I didn't respond to several posts due to obvious lack of reading prior responses.Halc,
You're still assuming that force is torque, which is going to give you wrong answers every time.Professor Lewin says: "There is no net force on that wheel, but there is a net torque."Net torque means that the sum of the torque vectors acting on the wheel is nonzero. No, those torques must be transferred to the wheel. Conservation of angular momentum does not allow torque 'out of nothing'.
What is a net torque? Out of nothing?
He also says net force is zero, meaning the sum of the force vectors is zero. The force vectors are obviously not the torque vectors, since (the part you never remember) force is not torque.
...
If an object is not moving, it is stationary, then it does not have a kinetic energy.
The kinetic energy comes from a force in mechanical systems.
Minute 48 of the video, the weight and the axle have NO kinetic energy.
When professor releases the weight from his hand the weight and the axle start to rotate, they have the rotational kinetic energy.
Where did the rotational kinetic energy come from?
Jano
Post by: Jaaanosik on 09/05/2020 21:21:35
Halc,The kinetic energy comes from a force in mechanical systems.Forces are involved to change kinetic energy, but net forces are not necessarily involved. Torque is not force, and yet X torque applied for time T to a system can (doesn't necessarily) result in some amount of kinetic energy change to the system regardless of the magnitude of the forces used to achive said X torque.
Anyway, in context of your post here, I know what you mean.QuoteWhen professor releases the weight from his hand the weight and the axle start to rotate, they have the rotational kinetic energy.The precessing disk applied a momentary torque (not any net force) to the axle. No continued torque is required once the axle gets its steady state angular momentum (a small vector pointing down)
Where did the rotational kinetic energy come from?
There is only rotating disk/wheel in the beginning.
Professor even says that the rotating wheel is suspended in such a way that there is no gravitational torque on it, 47:14 min.
The rotating disk cannot provide/cause a torque. It is steady by itself, just a constant angular velocity.
It does not have units of measure for it, does it?
You are right, small vector pointing down.
There had to be I*a (rotational inertia * angular acceleration); what is the cause?
The only cause is a net force. An object cannot start rotating without a net force.
A torque is a net force (90 degrees component) applied at a distance from the axis of rotation, Nm units of measure.
The w_precession = T/L torque over the rotational inertia of the wheel.
If one of them is missing then there is no precession.
Both have to be present.
Therefore: "The precessing disk applied a momentary torque (not any net force) to the axle. No continued torque is required once the axle gets its steady state angular momentum (a small vector pointing down)"
is wrong. It is not a momentary torque. It is a constant torque.
You can attach the weight to the rotating wheel axle on the ISS and it is not going to precess,
Jano
Post by: Halc on 09/05/2020 23:46:12
There is only rotating disk/wheel in the beginning.Which means that at no time does the weight of the disk contribute to the torque acting on the disk, quite unline the other video with the guy supporting the thing well away from its center of gravity.
Professor even says that the rotating wheel is suspended in such a way that there is no gravitational torque on it, 47:14 min.
There's no way to work out the weight of the spinning disk from what he's doing, even knowing the weight of the objects he adds, and the distances, etc.
Quote
The rotating disk cannot provide/cause a torque.It can, but while undisturbed, it doesn't. This is no different than saying a mass in space cannot provide/cause a force, but if a force X is applied to said mass with my hand, the mass will provide a equal an opposite force -X to my hand, all per Newton's laws of motion. Likewise, the rotating disk is capable of exerting opposite torque to any object exerting torque on the disk.
Quote
It is steady by itself, just a constant angular velocity.Angular velocity is measured typically in radians/sec. The disk has quite a bit of angular velocity. The axle has none before the precession starts.
It does not have units of measure for it, does it?
Quote
You are right, small vector pointing down.Wrong again. Force is not torque. You are incapable of learning this.
There had to be I*a (rotational inertia * angular acceleration); what is the cause?
The only cause is a net force.
There is no net force on the axle ever, as evidenced by the fact that its center of gravity never moves.
Quote
An object cannot start rotating without a net force.Same mistake, again and again. Sad really.
Quote
The w_precession = T/L torque over the rotational inertia of the wheel.That's wrong as well, but at least you're using torque now instead of force. L is angular momentum, not rotational inertia.
Quote
Therefore: "The precessing disk applied a momentary torque (not any net force) to the axle. No continued torque is required once the axle gets its steady state angular momentum (a small vector pointing down)"The angular velocity of the axle is constant, changing only when the weight is added or removed. That means the torque is momentary. No additional torque is needed to rotate something that's already rotating. Instead of asserting all these things, why don't you consider the implications of your assertions, which run into contradictions. If there is continuous torque on the axle, why does its angular velocity not change after that moment when the weight is added. That's would be a contradiction.
is wrong. It is not a momentary torque. It is a constant torque.
Quote
You can attach the weight to the rotating wheel axle on the ISS and it is not going to precessThere is no weight on the ISS.
Please reply with some intelligent thought put into your remarks. I see little point in responding to what are obviously either troll assertions or massive ignorance. Retake your middle-school physics classes, or at least get a refund from your education system that so grossly failed you.
Post by: Jaaanosik on 11/05/2020 15:40:20
...]The angular velocity of the axle is constant, changing only when the weight is added or removed. That means the torque is momentary. No additional torque is needed to rotate something that's already rotating. Instead of asserting all these things, why don't you consider the implications of your assertions, which run into contradictions. If there is continuous torque on the axle, why does its angular velocity not change after that moment when the weight is added. That's would be a contradiction.Halc,QuoteYou can attach the weight to the rotating wheel axle on the ISS and it is not going to precessThere is no weight on the ISS.
...
this is our major disagreement.
The wheel axle rotation, the precession, is not going to happen if there is no torque from the weight.
The torque and the angular momentum have to be present to cause the precession.
The precession, the rotation, is not going to happen when the weight is taken away of the axle.
The torque has to be constant, continuous.
Yes, I should have said mass of the brass cylinder weight added to the axle of the rotating wheel on the ISS.
I had this physical weight object in mind when I said that, not weight as a force.
This is the proof that the precession does not happen when there is no torque but there is an angular momentum of the wheel,
Jano
Post by: Jaaanosik on 11/05/2020 15:48:10
...]The angular velocity of the axle is constant, changing only when the weight is added or removed. That means the torque is momentary. No additional torque is needed to rotate something that's already rotating. Instead of asserting all these things, why don't you consider the implications of your assertions, which run into contradictions. If there is continuous torque on the axle, why does its angular velocity not change after that moment when the weight is added. That's would be a contradiction.Halc,QuoteYou can attach the weight to the rotating wheel axle on the ISS and it is not going to precessThere is no weight on the ISS.
...
Why a boat does not continue moving at a constant velocity on water once it is moving and boat engines are turned off?
The rotational inertia of the wheel does not want to rotate in the precession motion by itself.
The rotational inertia resists the precession, it is being forced into the precession by the torque,
Jano
Post by: Jaaanosik on 15/05/2020 01:49:24
...]The angular velocity of the axle is constant, changing only when the weight is added or removed. That means the torque is momentary. No additional torque is needed to rotate something that's already rotating. Instead of asserting all these things, why don't you consider the implications of your assertions, which run into contradictions. If there is continuous torque on the axle, why does its angular velocity not change after that moment when the weight is added. That's would be a contradiction.Halc,QuoteYou can attach the weight to the rotating wheel axle on the ISS and it is not going to precessThere is no weight on the ISS.
...
The bold statement is true for a rigid body, a mass, ...
The bold statement is not true for the rotational inertia.
This is very similar to the following scenarios:
- holding a weight in a stretched hand, no potential energy change for the weight but it costs energy to hold the weight.
- an astronaut going in a circle around the ISS with the jets on his back costs energy though the kinetic energy does not change for the astronaut in the ISS frame.
- hanging the weight on a string, no potential energy change for the weight and it does not cost any energy.
- the astronaut going in a circle around the ISS attached with a string, no kinetic energy change and it does not cost any energy.
... the same results as the above but no energy cost.
The torque rotating the angular momentum is doing work even though it appears there is no potential energy change.
Jano