Program glitch or discovery?

[ABV]

Base on this theory

http://knol.google.com/k/alex-belov/paradox-of-classical-mechanics-2/1xmqm1l0s4ys/9#

Simulator got this glitch.

Is it discovery?

...

Hi Alex,

 

I'm not sure to understand all what you mean and may be I'm wrong on some

points.

So here is my simulation to clarify the problem (a hard work :

http://exvacuo.free.fr/div/Sciences/Experiences/Meca/Momentum/com.wm2d

 

If you have not Workingmodel software (it is free), here is the video (6Mb,

probably long to download):

http://exvacuo.free.fr/div/Sciences/Experiences/Meca/Momentum/com.avi

 

Here are 3 jpg snapshots. N�0 is the initial position. N�1 is the state just

after the pulse. N�2 is some seconds later.

http://exvacuo.free.fr/div/Sciences/Experiences/Meca/Momentum/com0.jpg

http://exvacuo.free.fr/div/Sciences/Experiences/Meca/Momentum/com1.jpg

http://exvacuo.free.fr/div/Sciences/Experiences/Meca/Momentum/com2.jpg

 

The speed vector of each rod is displayed in the animation. The impulse force it

not displayed. It provides 100 N during 0.01s between the rods, at points of

same coordinates (= the position of the center of mass of the green rod).

 

In the red and green small windows, you can monitor the speeds, kinetic energies

and momenta of each rod. The default equations of the program are used.

 

In the small blue windows, I have put the equations to obtain the angular speed,

the moment of inertia and the angular momentum of each rod, calculated from the

common center of mass of the system.

This center of mass is displayed in the simulation (it is calculated by the

program, it's not a fixed point. It is at rest because there is no external

force acting onto the system. It is the mid-point of the line joining the

centers of mass of the two rods).

To simplify the calculi, the positions of the rods are chosen in order the

common center of mass to be at position x,y = 0,0.

 

With this simulation, I have discovered that from the common center of mass C,

each rod possesses a "hidden" angular momentum: each rod flies horizontally away

one another but one is above and the other under the horizontal x axis

containing their common center of mass. Thus the angle delimited by the

horizontal x axis and the line joining the centers of mass of the two rods

(crossing at C), decreases when the distance of the rods from the origin

increases. This variation of the angle is to be considered as an angular

velocity. Then from it, we can calculate the moment of inertia of each rod and

its angular momentum. The result is displayed in the small blue windows.

 

IMPORTANT : We see that the sum of the angular momenta of the two rods,

calculated in the referential frame of their common center of mass, is equal but

with opposite sign, to the angular momentum of the only rotating rod, calculated

in its proper frame.

Thus by adding these two momenta, we find zero. I guess the key of the problem

lies around this, but it is not yet completely clear for me.

 

Fran�ois

[D H]

Moved to pseudoscience.

 

The forces involved are not obeying the strong form of Newton's third law, so of course angular momentum is not conserved.

 

Edit

In addition, you are not considering the angular momentum due to translation.

Edited August 18, 2009 by D H

[J.C.MacSwell]

Moved to pseudoscience.

 

The forces involved are not obeying the strong form of Newton's third law, so of course angular momentum is not conserved.

 

Edit

In addition, you are not considering the angular momentum due to translation.

 

I think this is where he seems to have trouble.

[ABV]

I think this is where he seems to have trouble.

 

You're still thinking in classical mechanics existing laws.

This is the problem.

 

Even simulator got the glitch and you're still argue about my missunderstading.

This is so funny

 

Angular and translation movements are not same thing.

They not correlate to each other.

If body starts rotating then it won't be change body's translation movement.

However, body can start translation and rotation movement from one hit (momentum). Good example - experiment 2.

It's possible to make these equations:

P=m2*v2=m1*v1

L=I*w=m1*v1*R

However, it won't describe experiment 2. It describes 2 movements. One is translation. Another one is rotation. These 2 movements cannot use same momentum m1*v1. It's wrong. These 2 equations describe 2 unrelated movements and they should use 2 independent momentums.

To solve this problem for experiment 2, please look on my site.

 

http://knol.google.com/k/alex-belov/paradox-of-classical-mechanics-2/1xmqm1l0s4ys/9#

Edited August 23, 2009 by ABV

[D H]

The only glitch is in your understanding of mechanics. You are ignoring that linear momentum contributes to angular momentum via [math]\mathbf r \times \mathbf p[/math]. You need to account for this or you will get nonsensical results (which is exactly what you have obtained).

[ABV]

The only glitch is in your understanding of mechanics. You are ignoring that linear momentum contributes to angular momentum via [math]\mathbf r \times \mathbf p[/math]. You need to account for this or you will get nonsensical results (which is exactly what you have obtained).

 

Ok. How this linear momentum to angular momentum contribution works?

Just add rotation from nothing?

2 movements. Each of them should have own initial momentum. According to classical mechanics 2 movements cannot start from same momentum. To solve this problem needs to define another standalone rotation with translation movement and describe his law of momentum conservation.

This is what I did.

[D H]

Wrong. Learn your classical mechanics. Applying a force [math]\mathbf F[/math] to a rigid body with the point at which the force is applied located at a displacement [math]\mathbf r[/math] from the body's center of mass makes the body undergo linear and rotational acceleration according to

 

[math]

\aligned

\mathbf F &= m\mathbf a \\

\boldsymbol{\tau} &=

\mathbf r \times \mathbf F =

\boldsymbol I\,\frac{d}{dt} \boldsymbol{\omega} +

\boldsymbol{\omega}\times (\boldsymbol I \boldsymbol{\omega})

\endaligned[/math]

 

Here the angular velocity [math]\boldsymbol{\omega}[/math] is the angular velocity of the body with respect to some inertial frame but represented in some frame fixed with respect to the body.

Edited August 24, 2009 by D H

[ABV]

Wrong. Learn your classical mechanics. Applying a force [math]\mathbf F[/math] to a rigid body with the point at which the force is applied located at a displacement [math]\mathbf r[/math] from the body's center of mass makes the body undergo linear and rotational acceleration according to

 

[math]

\aligned

\mathbf F &= m\mathbf a \\

\boldsymbol{\tau} &=

\mathbf r \times \mathbf F =

\boldsymbol I\,\frac{d}{dt} \boldsymbol{\omega} +

\boldsymbol{\omega}\times (\boldsymbol I \boldsymbol{\omega})

\endaligned[/math]

 

Here the angular velocity [math]\boldsymbol{\omega}[/math] is the angular velocity of the body with respect to some inertial frame but represented in some frame fixed with respect to the body.

 

Really?

2 equations. one for translation [math]\mathbf F[/math]

another for rotation

[math]

\aligned

\mathbf F &= m\mathbf a \\

\boldsymbol{\tau} &=

\mathbf r \times \mathbf F =

\boldsymbol I\,\frac{d}{dt} \boldsymbol{\omega} +

\boldsymbol{\omega}\times (\boldsymbol I \boldsymbol{\omega})

\endaligned[/math]

 

Is this F same force for both equations? Or different?

Could these equations describe experiment 2?

My opinion:

Тhis F is different force for these 2 equations.

This equations may describe experiment 2 only if this force F is different.

One force needs for body's translation movement.

Another force needs for body's rotation movement.

Both will be apply to body.

If describe more details. These 2 forces initially will apply to body with same direction. When body changed position, the force for angular rotation will apply with small angle, which will increase during body’s rotation from 0 to 90 degree. This rotation force will change value following by [math]\cos \alpha [/math].

If use constant net force then translation and angular forces are variable. Their sum should be equal to constant net force.

If use variable net force then only angular force may be variable. The translation force can be constant. Their sum should be equal to the net force.

 

Base on this forces behavior easy to make a model

Anyway it won't bring initial momentum to same value translation momentum plus extra angular momentum from nowhere

Part of initial momentum should going to body's angular momentum part. As it described before.

 

Nothing never come from nowhere

---

Merged post follows:

Consecutive posts merged

Picture with program glitch details:

[insane_alien]

notice how your program also says there is an angular momentum around the center of mass for both scenarios that is the same, going against what you claim.

[D H]

Really?

2 equations. one for translation [math]\mathbf F[/math]

another for rotation

[math]

\aligned

\mathbf F &= m\mathbf a \\

\boldsymbol{\tau} &=

\mathbf r \times \mathbf F =

\boldsymbol I\,\frac{d}{dt} \boldsymbol{\omega} +

\boldsymbol{\omega}\times (\boldsymbol I \boldsymbol{\omega})

\endaligned[/math]

 

Is this F same force for both equations? Or different?

Really. Those are the correct equations of motion. One force causes translation and rotation. Read a physics text.

 

Could these equations describe experiment 2?

My opinion:

Тhis F is different force for these 2 equations.

Look at your own example as simulated by Francois. When the two rods are viewed collectively as a system, any forces between the rods are internal to the system: They are internal forces. If these internal forces obey the strong form of Newton's third law, they cannot change the total momentum and angular momentum of the system.

 

The rods are initially at rest. The total linear momentum and angular momentum are trivially zero because neither rod is rotating or translating. After applying the initial forces (one on the green rod and a corresponding equal-but-opposite force on the red rod), the momenta of the rods are

• Linear momentum
  • Green rod: [math]-1.042\,\text{kg-m/s}\;\hat {\mathbf x}[/math]
    
  • Red rod: [math]+1.042\,\text{kg-m/s}\;\hat {\mathbf x}[/math]
    
  • Total linear momentum: 0

  [*]Angular momentum

  • Green rod
    • Angular momentum due to translation: [math]+0.521\,\text{kg-m$^2$/s}\;\hat {\mathbf z}[/math]
      
    • Angular momentum due to rotation: 0

    [*]Red rod

    • Angular momentum due to translation: [math]+0.521\,\text{kg-m$^2$/s}\;\hat {\mathbf z}[/math]
      
    • Angular momentum due to rotation: [math]-1.042\,\text{kg-m$^2$/s}\;\hat {\mathbf z}[/math]

    [*]Total angular momentum: 0

 

There is no glitch. The only "glitch" here is that you do not understand classical physics, and yet you are trying again and again to find paradoxes within it. Read a physics text.

[ABV]

There is no glitch. The only "glitch" here is that you do not understand classical physics, and yet you are trying again and again to find paradoxes within it. Read a physics text.

 

The simulator program got a glitch. The TOTAL ANGUAL MOMENTM is changing during the experiment.

from:

to:

Edited August 25, 2009 by ABV

[D H]

Note well, Alex: That is angular momentum due to translation. That is not a discovery. It most likely results from using single precision arithmetic in the simulation. We do not have adequate insight into Francois' simulation to pinpoint the problem. (I for one I am not going to download a binary).

[ABV]

The rods are initially at rest. The total linear momentum and angular momentum are trivially zero because neither rod is rotating or translating. After applying the initial forces (one on the green rod and a corresponding equal-but-opposite force on the red rod), the momenta of the rods are

• Linear momentum
  • Green rod: [math]-1.042\,\text{kg-m/s}\;\hat {\mathbf x}[/math]
    
  • Red rod: [math]+1.042\,\text{kg-m/s}\;\hat {\mathbf x}[/math]
    
  • Total linear momentum: 0

  [*]Angular momentum

  • Green rod
    • Angular momentum due to translation: [math]+0.521\,\text{kg-m$^2$/s}\;\hat {\mathbf z}[/math]
      
    • Angular momentum due to rotation: 0

    [*]Red rod

    • Angular momentum due to translation: [math]+0.521\,\text{kg-m$^2$/s}\;\hat {\mathbf z}[/math]
      
    • Angular momentum due to rotation: [math]-1.042\,\text{kg-m$^2$/s}\;\hat {\mathbf z}[/math]

    [*]Total angular momentum: 0

 

There is no glitch. The only "glitch" here is that you do not understand classical physics, and yet you are trying again and again to find paradoxes within it. Read a physics text.

 

Fine.

I just looked on equations for angular velocities of this simulation.

(Body[1].v.x*Body[1].p.y + Body[1].v.y*Body[1].p.x) / (sqr(Body[1].p.x)+sqr(Body[1].p.y))

i.e.

[math]\omega=\frac{v_xS_y+v_yS_x}{\sqrt{S_x}+\sqrt{S_y}}[/math]

where Vx, Vy rod's translation velocities from center mass

Sx,Sy rod's distance from center mass.

Units is not accurate for this equations.

The angular velocity should equal to [math]\omega=\frac{V}{R}[/math]

If calculate velocities sum vector dividing to the distance then it should be:

[math]\omega=\frac{\sqrt{V_x^2+V_y^2}}{\sqrt{S_x^2+S_y^2}}[/math]

Parameters for this simulation gives zero to total angular if reverse one of velocity value.

However, these rods displacement from center mass base on these equations gives total angular momentum more or less zero.

If use another equation like

[math]\omega=\frac{\sqrt{V_x^2+V_y^2}}{\sqrt{S_x^2+S_y^2}}[/math] then total angular mometum is striving to zero.

 

This equations are not adequate for angulal momentum calculation.

====

As I see on simulator, just one of rod has angular momentum.

Another rod has nothing.

The question "how total kinetic energy for rotating rod growth from nothing?" is still open.

Edited August 25, 2009 by ABV
Consecutive posts merged.

[D H]

I have given you the correct equations. The only glitch here is your understanding and Francois' calculation of angular momentum due to translation. The correct equation is [math]\mathbf L = m\mathbf r\times \mathbf v[/math].

 

BTW, sqr(x) calculates x squared, not the square root of x.

[ABV]

I have given you the correct equations. The only glitch here is your understanding and Francois' calculation of angular momentum due to translation. The correct equation is [math]\mathbf L = m\mathbf r\times \mathbf v[/math].

 

BTW, sqr(x) calculates x squared, not the square root of x.

 

Let's simulate these experiments.

 

Using

Workingmodel software

(Thank you to François Guillet guillet.francois@wanadoo.fr for good idea)

The experiment 1.

 

This simulation shows all adeqate to theory results.

Before start.

 

Durings the experiment

 

The experiment 2.

 

This simulation shows all extra rotation without additional torque.

Before start

Durings the experiment

Let's simulate same result using experiment 1 plus additional torque.

 

Before start

During the experiment

 

As it show on pervious experiments print screens, experiment 2 and experiment with additional torque have same results. However, these experiments have different initial conditions.

 

The red rod for experiment 2 following these equations:

 

 

 

note: P - rod's translation momentum, mr - rod's mass, vr -rod's translation velocity, Fi - initial pulse force, ti - initial pulse force time

L - rod's angular momentum, I - rod's moment of inertia, w - rod's angular velocity, R - rod's unit radius.

 

 

The red rod for experiment with additional torque following these equations:

 

 

 

note: P - rod's translation momentum, mr - rod's mass, vr -rod's translation velocity, Fi - initial pulse force, ti1 - initial pulse force, ti2 - initial torque time

L - rod's angular momentum, I - rod's moment of inertia, w - rod's angular velocity, tau - torque.

 

These experiments use different parameters for angular momentum equations. For this equation the simulator for experiment 2 uses same initial pulse force. However, for experiment with additional torque simulator uses another parameter (torque).

 

experiment 1 + additional torque = experiment 2

 

Thse experiment 2 shows how simulator generate additional torque for Isolated System from initial pulse force displacement.

The isolated systems for these experiments are not the same. The experiment with additional torque should not have same results as experiment 2. The experiment 2 simulation is not adequate to real world. The new rotation with translation movement should be added to classical mechanics.

[D H]

Where do you think the torque comes from? (Hint: Think force.)

[ABV]

Where do you think the torque comes from? (Hint: Think force.)

 

Both rods have same mass and moment of inertia.

However, on experiment 2 the rotating rod has extra torque, which easy to simulate. Please look on experiment 3.

This means the simulator apply extra translation force for rotation rod. Which is should not be on real world.

If part of applied force spent for rotation movement (torque) then this portion of force should be removed from translation part of applied force. The sum of these rotation and translation portions of applied force should be equal to applied force for non-rotating rod. (3-rd Newton’s law)

The translation velocities of these rods should be different. However, they are equal on simulator. Because simulator knows nothing about rotation with translation movement and two independent rotation and translation movements cannot describe this experiment 2 correctly.

The classical mechanics should include new standalone translation with rotation movement to describe natural phenomenon correctly.

 

This is how should be happen on real world

 

This a movie capture from experiment 3 (simulates velocities difference)

 

This a movie capture from experiment 2 with accurate rods alignments.

 

 

Please check my site.

http://knol.google.com/k/alex-belov/paradox-of-classical-mechanics-2/1xmqm1l0s4ys/9#

I added:

Steps How to build these experiments on working model demo version.

[D H]

Alex, you have not discovered a paradox or a new feature in classical mechanics. You have made several mistakes in the google.docs document cited in this thread, and the simulation done by your friend also appears to be buggy.

 

Alex, you have a tendency to create overly complicated systems. To make things worse, you use some poorly defined concepts in those systems. This makes it a bit tougher to find where your reasoning has gone awry. So, let's simplify and clarify this system.

 

Simplification #1: In both experiments a pair of equal-but-opposite forces are applied to the two rods. The force is applied at the center of mass of rod #1 in both experiments.

 

A question arises: Why bother with a rod for rod #1? There is zero difference in the resulting physics if you use a point mass, so you might as well use a point mass.

 

Clarification: What is the nature of the force? If you make the energy source that applies the force internal to the system then the total linear momentum and total angular momentum of the system comprising the two objects and the energy source will be conserved. If you make that force a conservative force then energy will be conserved as well. For example, a spring will do the trick. All of the conserved quantities are conserved: energy, linear momentum, and angular momentum.

 

You can still make the mass+rod+spring system rather complex. For example, the finite amount of time it takes for the spring to fully propel the point mass away from the rod means that the rod rotates while it is ejecting the mass. One way around this:

 

Simplification #2: Make the spring extremely stiff. Making the spring very stiff makes it so one can ignore this rotation issue. Whether you do this or not, Alex, the bottom line is that energy, linear momentum, and angular momentum will be conserved if you use a conservative force such as a spring.

 

 

So what is this new contraption? It's a spaceship. Aerospace engineers have known for a long time that firing a single rocket engine whose thrust vector is not directed along the line from the vehicle's center of mass to the thruster will cause the vehicle to accelerate in translation and in rotation.

 

Thrusters can be used in pairs to help eliminate this cross-coupling between translational and rotational acceleration. This is not always possible. For example, consider the Mars Climate Orbiter. From http://www.jamesoberg.com/mars/loss.html

The use of jet thrusters for attitude control raises further operational issues. In a world of perfect symmetry and unlimited payload size and budget, a spacecraft could rotate cleanly about its center of mass (often carelessly called its center of gravity) if opposing ends were equipped with jets and if those jets pointed in opposite directions and were set at right angles to the axis to be turned. In the real world, rotational jets may not be arranged in such a theoretically perfect alignment. ... On the Mars Climate Orbiter, four separate clusters of jets were located around the vehicle's waist. However, because of the large solar array extending from one side, the craft's center of mass did not coincide with the center point of the waist. Thus there was a significant imbalance each time these small thrusters fired.

 

The Mars Climate Orbiter needed to use its attitude jets once or twice a day to dump angular momentum from its momentum wheels. The use of these attitude jets also made the vehicle accelerate translationally. Failing to account for this properly was the ultimate cause of the demise of the Mars Climate Orbiter. The mission controllers did try to account for this undesired acceleration, but there was a units mixup. The vehicle reported the undesired acceleration as a number in English units. The controllers interpreted that number to be in metric units.

According to a JPL spokesman, every maneuver intended to dump momentum added a velocity error of about 0.001 meter per second, on a probe that was traveling at a rate of tens of kilometers per second. These deflections themselves were not the problem, but their incorrect modeling was, when the computer was told the spacecraft had received a force of four or five times as great as it really had.

The end result: The Mars Climate Orbiter entered Mars atmosphere, something it was never intended to do. The vehicle most likely blew up 70 kilometers or so above Mars.

[ABV]

Whether you do this or not, Alex, the bottom line is that energy, linear momentum, and angular momentum will be conserved if you use a conservative force such as a spring.

They are.

The question only HOW.

If use new translation with rotation movement the momentum is conserve by this law.

This movement has translation and rotation parts of momentum.

If use both movement instead one. It will be P=const , L=const

The difference is these two movements use same momentum. However, new movement is using just one momentum for both his movement parts.

If one force applied to the body it should for one of the movement. By classical mechanics law, the movement should be chosen first. If translation then all value of force should be used for translation movement. If rotation then all value of force should be used for rotation movement. But how about if this force is starting both movement together? If it one force then it should be spit by parts. One part should be for rotation movement. Another part - for translation movement. Sum of these parts of force will be equal to full value of force. This is not covering by classical mechanics. Using regular calculations, for both movements is using same initial value of force. Sum of these parts of force is doubling initial value. This is not right.

Your example with spaceship shows how engineers implement miscalculation, base on wrong classical mechanics model. It should not be 2 movements. It should be one new movement. My equation may be so empirical and need more studies about it. However, this new movement gives more understanding about nature.

For example. The Dark matter. This natural phenomenon does not exist. It just miscalculation, because our physics deny situation where object can move by itself (without external forces). Correct. The object cannot move by itself. However the set of objects can do this. The law of momentum conservation of new rotation with translation movement is allowing this.

[D H]

If one force applied to the body it should for one of the movement. By classical mechanics law, the movement should be chosen first. If translation then all value of force should be used for translation movement. If rotation then all value of force should be used for rotation movement. But how about if this force is starting both movement together? If it one force then it should be spit by parts. One part should be for rotation movement. Another part - for translation movement.

 

That is not how physics works. A force, and all forces, cause both translational and rotational acceleration. You really need to learn how classical physics is rather than making things up yourself. Your stubbornness to learn does not do you any service.

 

Sum of these parts of force will be equal to full value of force. This is not covering by classical mechanics.

That is not covered by classical mechanics because that is not how classical mechanics works.

 

Read a book or take a class.

[ABV]

That is not how physics works. A force, and all forces, cause both translational and rotational acceleration. You really need to learn how classical physics is rather than making things up yourself. Your stubbornness to learn does not do you any service.

 

I don't see logic. Each part of force must do own job

2 Movements. Each movements has own acceleration and force. Net of these forces equal to applied to body force.

 

 

For example, if apply these force separately then part of force need for translation movement and another part for rotation movement. If apply same value of initial forces for both movements then rod will have const translation velocity plus depended on applied radius angular velocity. But sum of forces values is doubling initial value. This is normal simulator behavior on wrong classical mechanics model.

This is from real world:

According to a JPL spokesman, every maneuver intended to dump momentum added a velocity error of about 0.001 meter per second, on a probe that was traveling at a rate of tens of kilometers per second. These deflections themselves were not the problem, but their incorrect modeling was, when the computer was told the spacecraft had received a force of four or five times as great as it really had.

[mooeypoo]

If F(net) is the red arrow, then your image is incorrect. It should be:

 

[math]F_{net}=F_{2}cos(\theta) + F_{1}cos(\alpha) [/math]

 

Where [math]\theta[/math] is the angle between F2 and F(net) and [math]\alpha[/math] is the angle between F1 and F(net).

[D H]

For example, if apply these force separately then part of force need for translation movement and another part for rotation movement.

No. All of the force makes a body undergo translational acceleration and all of the force makes the body undergo angular acceleration; see post #7. It might seem paradoxical, but it is not. That is the only way that the conserved quantities, energy, linear momentum, and angular momentum, can be conserved simultaneously.

[ABV]

No. All of the force makes a body undergo translational acceleration and all of the force makes the body undergo angular acceleration; see post #7. It might seem paradoxical, but it is not. That is the only way that the conserved quantities, energy, linear momentum, and angular momentum, can be conserved simultaneously.

 

Correct. This is classical mechanic says so.

D H. I'm not a beginner of physics science. I've got some.

 

I understand you are the one smart of the physics scientist. Because we are still talking about this idea:)

 

Let's think logically.

If an object takes acceleration on translation trajectory then this means a force applied to the object for translation movement. Correct?

If an object takes acceleration on rotation trajectory then this means a force applied to the object for rotation movement. Correct?

The net of these forces is resultant force which applied to the object. It's logic.

The classical mechanics says - No. Force applied to the object with same values for both translation and rotation movements. However, the net of these forces is doubling resultant force. There is no logic. Why? The logic is sacrificing to law of momentum conservation for translation movement, which must work. However, the object starts 2 movements. If look back on experiment 2 then it shows how experimental rods conduct different movements. If base on classical mechanics models compare their kinetic energies and momentums then just one translation momentums have same values. This is not right. No symmetrical action. However, base on 3rd Newton's law symmetrical action must present on experiment.

 

====

Back to your good example about spaceships. You would probably have more information about their trajectories miscalculations relatively to center mass problem. If you would try to create a simple spaceship model which takes rotation with translation movement then the simulator is giving a wrong translation velocity result. If engine has a thrust with fixed value then no matter how spaceship rotates the translation velocity will be constant. Wrong. Practice says the spaceship with this miscalculation toward to cosmic garbage. The engineers usually build spaceships which avoid this center mass problem.

This is not very friendly effects for rocket scientists. However, this effect brings new era to our civilization.

[D H]

D H. I'm not a beginner of physics science. I've got some.

Not near enough. You are a rank beginner. Your continued attempts to find paradoxes in classical mechanics are proof of this.

 

I understand you are the one smart of the physics scientist. Because we are still talking about this idea:)

There is no idea. Just bad physics here.

 

Let's think logically.

That would be a good idea. An even better idea is to think in terms of physics and mathematics, rather than Alex's fantasy physics.

 

If an object takes acceleration on translation trajectory then this means a force applied to the object for translation movement. Correct?

If an object takes acceleration on rotation trajectory then this means a force applied to the object for rotation movement. Correct?

The net of these forces is resultant force which applied to the object. It's logic.

No. One more time. A force causes both translational and rotational motion.

 

Think in term of a single point mass. A point mass has zero moment of inertia. Yet it still has angular momentum with respect to some origin because angular momentum is defined as [math]\mathbf L = \mathbf r \times \mathbf p = m\,\mathbf r \times \mathbf v[/math]. That is the definition of angular momentum in classical mechanics. Now suppose some force [math]\mathbf F[/math] acts on the point mass. By the definition of force, [math]\dot{\mathbf p} = \mathbf F[/math].

 

Differentiate the definition of angular momentum with respect to time: [math]\dot{\mathbf L} = \mathbf r \times \dot{\mathbf p} = \mathbf r \times \mathbf F[/math]. A single force causes to point mass to undergo change in linear and angular momentum via [math]\dot{\mathbf p} = \mathbf F[/math] and [math]\dot{\mathbf L} = \mathbf r \times \mathbf F[/math].

 

Now consider a collection of particles. Some of the forces acting on an individual particle are a result of interactions between the particle and other particles in the system. These are called internal forces. Other forces come from outside the system. So long as the internal forces obey the strong form of Newton's third law, the changes in the total linear and angular momentum of the collection of particles are given by

 

[math]\aligned

\dot{\mathbf p}_{\text{tot}} &= \sum_i \mathbf F_{\text{ext},i} \\

\dot{\mathbf L}_{\text{tot}} &= \sum_i \mathbf r_i \times \mathbf F_{\text{ext},i}

\endaligned[/math]

 

The external forces are the only things that change the collection's linear and angular momentum, and each external force contributes to the change in both linear and angular momentum.

 

Finally, consider a solid body. A solid body is just a special kind of collection of particles. I've given the equations of motion many times. You can find these in previous posts. You can find the derivations in any college level classical mechanics text.

 

Read a physics text. Take a physics class. You do not know physics, yet, and until you do all you are accomplishing is making a fool out of yourself.