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VARIPEND


butovsv

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22 minutes ago, butovsv said:
Does the mass of the “body” change?
On the “case”, does the “force” act?



How will the “body-case” behave if it is allowed to move in its own way?

What will the constant total momentum of this system look like in the absence of external forces?

!

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This thread is for you to explain your idea, therefore you should be answering questions not asking them.

If this continues, this thread will be closed.

 
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will show how the coordinates of the CM of the Varipend system are determined:



body1_body2_e.png



[math]{r_c}(t) = \frac{{{m_{liquid}}(t){r_1}(t) + {m_{case}}(t){r_2}(t)}}{M}[/math]




What is necessary and (or) enough to know in order to calculate the CM of this system at any given time?




a) it is necessary to know the Law of Conservation of Momentum and on the basis of this law to calculate the coordinates and masses of parts of the system at any time?

b) it is necessary to know the Law of Conservation of Energy and based on this law to calculate the coordinates and masses of parts of the system at any time?

c) firmly enough to grammatically “know” how a brick behaves, which lies on the road in anticipation of external forces? Because it should be “obvious” to any dunce familiar with “twice two equals four”?

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1 minute ago, butovsv said:


What is necessary and (or) enough to know in order to calculate the CM of this system at any given time?




a) it is necessary to know the ZSI and on the basis of this law to calculate the coordinates and masses of parts of the system at any time?

b) it is necessary to know the WES and on the basis of this law to calculate the coordinates and masses of parts of the system at any time?

c) firmly enough to grammatically “know” how a brick behaves, which lies on the road in anticipation of external forces? Because it should be “obvious” to any dunce familiar with “twice two equals four”?

!

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34 minutes ago, Ghideon said:

Try answering questions instead of posting new questions. What will happen according to you? Will there be acceleration a>0 even if F=0? 

 

For a brick, the unequivocal answer is NO!

 

[math]{a_{c1}} = \frac{{\left[ {{m_1}{{\ddot r}_1}} \right]}}{{{m_1}}} = {{\ddot r}_1} = 0[/math] if [math]{{\vec F}^e} = 0[/math]


For the CM system of bodies, the answer is YES!
Acceleration of the Center of Mass is not a function of "external force."

[math]{a_c} = \frac{{{d^2}{R_c}}}{{d{t^2}}} = \frac{{\sum {{m_i}{{\ddot r}_i}} }}{{\sum {{m_i}} }} + \frac{{\sum {{{\dot m}_i}{{\dot r}_i}} }}{{\sum {{m_i}} }} = ?[/math] if [math]{{\vec F}^e} = 0[/math]

 

The acceleration of the CM of the system of bodies is the result of both external and internal forces.

 

 

 

But the most interesting is another!
The CM of a system of bodies can change without acceleration at all!

You talk about CM, having in your head the concepts of "force", "acceleration", "velocity"

[math]{{\vec R}_c}(t) = {{\vec R}_c}(0) + {{\vec V}_c}(t)t + \frac{1}{2}\frac{{{{\vec F}^e}}}{M}{t^2}[/math]

 

But the CM "moves" like this:

[math]{{\vec R}_c}(t) = \frac{{\sum {{m_i}(t){r_i}(t)} }}{{\sum {{m_i}} (t)}}[/math]

 

The CM cannot move!

CM is not a material object!
A CM can only CHANGE how the result of a function of 2 (two) variables changes.

 

20 minutes ago, Strange said:

Moderator Note

This thread is for you to explain your idea, therefore you should be answering questions not asking them.

If this continues, this thread will be closed.

I also have the right to ask you questions.

 

varipend1.gif

 

 


Do you see the CM of this system?
Do you see how he moves?

 

 

 

===

The Varipend task is not mathematically complicated and has a very accessible explanation from a physical point of view, which does not go beyond the school textbook.
but this task is extremely difficult in terms of psychology.

Therefore, I must ask you questions so that you LEARN to answer them YOURSELF.

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23 minutes ago, butovsv said:

For a brick, the unequivocal answer is NO!

Ok!

I will simplify for you.

Here is a brick, at rest. Nothing is pushing the brick, sum of external forces are zero.

image.png.e8751c449bf0b6ccd8129ccdd5e80b04.png

The brick is hollow. Hidden inside is a varipend, the device described by you. Will the brick accelerate due to the variepend that is active inside? You seem to claim, but avoid to answer, that you think that F=ma does not apply to your device*
 

 

*) This is not the only physical laws that the description seems to break but one law at a time is enough.

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4 hours ago, butovsv said:

This force is equal to the change in the momentum of the moving substance moving along the trajectory.

The force is not the change in momentum.

 

Quote

CENTER Mass of the body system is associated with the total momentum?

 

If the center of mass is is stationary, the system has no net momentum, and vice-versa.

 

4 hours ago, butovsv said:

How long ago?

 

How long ago what?

 

4 hours ago, butovsv said:

Can you write an equation to calculate the center of mass of a system of bodies?

Can you write an equation that allows you to calculate the SUM MOMENTUM of all the bodies in the system?

Can you write the equation of CHANGE in the total momentum of a system of bodies?

Yes.

What’s more, I can apply these equations, which is what you need to do, rather than making gifs, thinking it proves anything.

 

3 hours ago, butovsv said:

will show how the coordinates of the CM of the Varipend system are determined:



body1_body2_e.png



rc(t)=mliquid(t)r1(t)+mcase(t)r2(t)M




 

Why do you have a variable called mliquid? You described this as a pendulum, and have not said anything about a liquid.

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21 hours ago, Mordred said:

First off the CM cannot change direction without acceleration that is basic physics. Acceleration includes changes in direction so some force must be involved internal or external. 

Why can't it?
It may well!

 

Brick     brick.png - can't!

[math]{{\vec R}_c}(t) = {{\vec R}_c}(0) + {{\vec V}_c}(t)t + \frac{1}{2}\frac{{{{\vec F}^e}}}{M}{t^2}[/math]

 

But the system of bodies    galaktic.jpg- it can even!

 

[math]{{\vec R}_c}(t) = \frac{{\sum {{m_i}(t){r_i}(t)} }}{{\sum {{m_i}} (t)}}[/math]

 

The CM equation for a complex system does not have a unique solution for a future point in time.

You can easily extrapolate the brick moving function - 
[math]{{\vec R}_c}(t) = {{\vec R}_c}(0) + {{\vec V}_c}(t)t + \frac{1}{2}\frac{{{{\vec F}^e}}}{M}{t^2}[/math] - without the “external forces” the brick path or “point” or “straight line”.
 


But for a system of bodies, the result of solving the CM equation is neither a straight line nor a line.
 

The center of Mass of a complex system of bodies is a REGION OF SPACE!

 

[math]{{\vec R}_c}(t) = \frac{{\sum {{m_i}(t){r_i}(t)} }}{{\sum {{m_i}} (t)}} = ?[/math]

 

 

 

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22 hours ago, Ghideon said:

Ok!

I will simplify for you.

Here is a brick, at rest. Nothing is pushing the brick, sum of external forces are zero.

image.png.e8751c449bf0b6ccd8129ccdd5e80b04.png

The brick is hollow. Hidden inside is a varipend, the device described by you. Will the brick accelerate due to the variepend that is active inside? You seem to claim, but avoid to answer, that you think that F=ma does not apply to your device*

If brick 1 (one), then the path of the brick is determined like this:

[math]{{\vec R}_c}(t) = {{\vec R}_c}(0) + {{\vec V}_c}(t)t + \frac{1}{2}\frac{{{{\vec F}^e}}}{M}{t^2}[/math]

 

If inside the brick, “something” “somewhere” moves relative to the brick

For example, like this:

varipend1.gif

Then the Mass Center of such a "brick" is calculated like this:

[math]{r_c}(t) = \frac{{{m_{liquid}}(t){r_1}(t) + {m_{case}}(t){r_2}(t)}}{M}[/math]

 

Yes!
Parts of the system can interact!
Their pulses can change in very large limits!
They can accelerate and slow down!

But at the same time, the total momentum of the system remains unchanged.

 

 

 

20 hours ago, swansont said:
On 8/20/2019 at 2:54 PM, butovsv said:

This force is equal to the change in the momentum of the moving substance moving along the trajectory.

The force is not the change in momentum. 

oh wow .....
You do not confuse Mechanics with "Star Wars": "May the force be with you"?

 

Let's look at the ABC book?

Quote

By the definition of momentum,

F→=dp→dt=d(mv→)dt,{\displaystyle {\vec {F}}={\frac {\mathrm {d} {\vec {p}}}{\mathrm {d} t}}={\frac {\mathrm {d} \left(m{\vec {v}}\right)}{\mathrm {d} t}},}[math]\vec F = \frac{{{\rm{d}}\vec p}}{{{\rm{d}}t}} = \frac{{{\rm{d}}\left( {m\vec v} \right)}}{{{\rm{d}}t}}[/math]

where m is the mass and v→{\displaystyle {\vec {v}}}{\displaystyle {\vec {v}}} is the velocity.[3]:9-1, 9-2

In Mechanics - just like that and no other way!
Are we talking about Mechanics here?

 

20 hours ago, swansont said:
22 hours ago, butovsv said:

will show how the coordinates of the CM of the Varipend system are determined:



body1_body2_e.png



rc(t)=mliquid(t)r1(t)+mcase(t)r2(t)M




 

Why do you have a variable called mliquid? You described this as a pendulum, and have not said anything about a liquid. 

 

good question...

 


Material Point System:

Varipend9.gif

 

The total momentum of this point system is defined as follows:

[math]\vec P = \sum {{{\vec p}_i}}  = \sum {{m_i}{v_i}}  = {m_1}{{\vec v}_1} + {m_2}{{\vec v}_2} + {m_3}{{\vec v}_3} + ... + {m_i}{{\vec v}_i}[/math]

If all these points would have the same speed, then this speed could be taken out of the sign of the sum, and sum the mass:

And it would turn out like this:

[math] \ vec P = M {\ vec V_c} [/math]

In order for the speeds of all points of the system to become equal, all these points need to be simply connected!
Connect inelastically!


[math]{m_1}{{\vec v}_1} + {m_2}{{\vec v}_2} + {m_3}{{\vec v}_3} + ... + {m_i}{{\vec v}_i} \to Boom!!! \to ({m_1} + {m_2} + {m_3} + ... + {m_i}){V_c} = M{V_c}[/math]


Actually, what you schoolchildren only know how to do! ....


But here is the misfortune in this "simple" task ....

As the author of this task, I came up with an inelastic connection of system components according to a special law.

Like this:

logo1.gif

 

 

And in the total momentum for this system, the following regularity was determined:

[math]{m_1}{{\vec v}_1} + {m_2}{{\vec v}_2} + {m_3}{{\vec v}_3} + ... + {m_n}{{\vec v}_n} = ({m_{i1}} + {m_{i2}} + {m_{i3}} + ... + {m_{in}}){{\vec V}_{ic}} + ({m_{k1}} + {m_{k2}} + {m_{k3}} + ... + {m_{kn}}){{\vec V}_{kc}}[/math]


And what is very curious  - the masses of particle systems with the same velocity  (i.e. material points, material bodies!) - are constantly changing!


Look:

[math]{m_1}{{\vec v}_1} + {m_2}{{\vec v}_2} + {m_3}{{\vec v}_3} + ... + {m_n}{{\vec v}_n} = ({m_{i1}} + {m_{i2}} + {m_{i3}} + ... + {m_{in}}){{\vec V}_{ic}}({t_1}) + ({m_{k1}} + {m_{k2}} + {m_{k3}} + ... + {m_{kn}}){{\vec V}_{kc}}({t_1})[/math]

[math]{m_1}{{\vec v}_1} + {m_2}{{\vec v}_2} + {m_3}{{\vec v}_3} + ... + {m_n}{{\vec v}_n} = ({m_{i1}} + {m_{i2}} + {m_{i3}} + ... + {m_{in - 1}}){{\vec V}_{ic}}({t_2}) + ({m_{k1}} + {m_{k2}} + {m_{k3}} + ... + {m_{kn + 1}}){{\vec V}_{kc}}({t_2})[/math]

[math]{m_1}{{\vec v}_1} + {m_2}{{\vec v}_2} + {m_3}{{\vec v}_3} + ... + {m_n}{{\vec v}_n} = ({m_{i1}} + {m_{i2}} + {m_{i3}} + ... + {m_{in - 2}}){{\vec V}_{ic}}({t_3}) + ({m_{k1}} + {m_{k2}} + {m_{k3}} + ... + {m_{kn + 2}}){{\vec V}_{kc}}({t_3})[/math]

 

Spoiler

logo1_cm.png

 


Therefore, the total momentum of the entire system as a whole can be written as follows:

[math]{p_{var}}(t) = {m_{liquid}}(t){{\vec v}_1}(t) + {m_{case}}(t){{\vec v}_2}(t)[/math]

 

 

 

 

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28 minutes ago, butovsv said:

oh wow .....
You do not confuse Mechanics with "Star Wars": "May the force be with you"?

 

Let's look at the ABC book?

 

 

F = dp/dt

Force is not change in momentum.  (F ≠ ∆p) The former has units of force, the latter has units of momentum. They cannot be equal

 

28 minutes ago, butovsv said:

In Mechanics - just like that and no other way!
Are we talking about Mechanics here?

Most of us are.

 

I asked you what in your post is liquid, since you have mliquid terms that keep popping up in your equations. Yet your OP said it was a pendulum, and you describe it as a mechanical system. You also never actually explain what's supposed to be going on in your animations. What the heavy grey and black lines represent, for example. 

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39 minutes ago, butovsv said:

 

But here is the misfortune in this "simple" task ....

As the author of this task, I came up with an inelastic connection of system components according to a special law.

Like this:

logo1.gif

 

 

And in the total momentum for this system, the following regularity was determined:

m1v⃗ 1+m2v⃗ 2+m3v⃗ 3+...+mnv⃗ n=(mi1+mi2+mi3+...+min)V⃗ ic+(mk1+mk2+mk3+...+mkn)V⃗ kc

 

 

 

I can identify, at most, two masses in your animation. The ball, and the rim (which ideally could be zero)

You have done nothing to provide sufficient information to clarify what you are talking about. Based on what you have given, this is nonsense.

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17 minutes ago, swansont said:

F = dp/dt

Force is not change in momentum.  (F ≠ ∆p) The former has units of force, the latter has units of momentum. They cannot be equal 

Oh?!

rifle.gif

"Force" = "change in momentum"

[math]\vec F = \frac{{{\rm{d}}\vec p}}{{{\rm{d}}t}} = \frac{{{\rm{d}}\left( {m\vec v} \right)}}{{{\rm{d}}t}}[/math]

 

This is Physics.
The science is ...

 

 

11 minutes ago, swansont said:

this is nonsense. 

Bad answer.
The response of the layman.

I repeat again.
Watch and read carefully.

 

 


Material Point System:

Varipend9.gif

 

The total momentum of this point system is defined as follows:

[math]\vec P = \sum {{{\vec p}_i}}  = \sum {{m_i}{v_i}}  = {m_1}{{\vec v}_1} + {m_2}{{\vec v}_2} + {m_3}{{\vec v}_3} + ... + {m_i}{{\vec v}_i}[/math]

If all these points would have the same speed, then this speed could be taken out of the sign of the sum, and sum the mass:

And it would turn out like this:

[math] \ vec P = M {\ vec V_c} [/math]

In order for the speeds of all points of the system to become equal, all these points need to be simply connected!
Connect inelastically!


[math]{m_1}{{\vec v}_1} + {m_2}{{\vec v}_2} + {m_3}{{\vec v}_3} + ... + {m_i}{{\vec v}_i} \to Boom!!! \to ({m_1} + {m_2} + {m_3} + ... + {m_i}){V_c} = M{V_c}[/math]


Actually, what you schoolchildren only know how to do! ....


But here is the misfortune in this "simple" task ....

As the author of this task, I came up with an inelastic connection of system components according to a special law.

Like this:

logo1.gif

logo1_cm.png

 

 

And in the total momentum for this system, the following regularity was determined:

[math]{m_1}{{\vec v}_1} + {m_2}{{\vec v}_2} + {m_3}{{\vec v}_3} + ... + {m_n}{{\vec v}_n} = ({m_{i1}} + {m_{i2}} + {m_{i3}} + ... + {m_{in}}){{\vec V}_{ic}} + ({m_{k1}} + {m_{k2}} + {m_{k3}} + ... + {m_{kn}}){{\vec V}_{kc}}[/math]


And what is very curious  - the masses of particle systems with the same velocity  (i.e. material points, material bodies!) - are constantly changing!


Look:

[math]{m_1}{{\vec v}_1} + {m_2}{{\vec v}_2} + {m_3}{{\vec v}_3} + ... + {m_n}{{\vec v}_n} = ({m_{i1}} + {m_{i2}} + {m_{i3}} + ... + {m_{in}}){{\vec V}_{ic}}({t_1}) + ({m_{k1}} + {m_{k2}} + {m_{k3}} + ... + {m_{kn}}){{\vec V}_{kc}}({t_1})[/math]

[math]{m_1}{{\vec v}_1} + {m_2}{{\vec v}_2} + {m_3}{{\vec v}_3} + ... + {m_n}{{\vec v}_n} = ({m_{i1}} + {m_{i2}} + {m_{i3}} + ... + {m_{in - 1}}){{\vec V}_{ic}}({t_2}) + ({m_{k1}} + {m_{k2}} + {m_{k3}} + ... + {m_{kn + 1}}){{\vec V}_{kc}}({t_2})[/math]

[math]{m_1}{{\vec v}_1} + {m_2}{{\vec v}_2} + {m_3}{{\vec v}_3} + ... + {m_n}{{\vec v}_n} = ({m_{i1}} + {m_{i2}} + {m_{i3}} + ... + {m_{in - 2}}){{\vec V}_{ic}}({t_3}) + ({m_{k1}} + {m_{k2}} + {m_{k3}} + ... + {m_{kn + 2}}){{\vec V}_{kc}}({t_3})[/math]

 

 

Therefore, the total momentum of the entire system as a whole can be written as follows:

[math]{p_{var}}(t) = {m_{liquid}}(t){{\vec v}_1}(t) + {m_{case}}(t){{\vec v}_2}(t)[/math]

 

 

 

 

10 minutes ago, butovsv said:

Therefore, the total momentum of the entire system as a whole can be written as follows:

[math]{p_{var}}(t) = {m_{liquid}}(t){{\vec v}_1}(t) + {m_{case}}(t){{\vec v}_2}(t)[/math]

And this total momentum cannot change without an external energy source.

This is the Law!
Called: Law of Conservation of Momentum.

 

[math]{p_{var}}(t) = {m_{liquid}}(t){{\vec v}_1}(t) + {m_{case}}(t){{\vec v}_2}(t)=const=0[/math] if [math]{{\vec F}^e} = 0[/math]



Can anyone solve this differential equation?

Or is it not necessary to solve it, because instead of this equation it is possible to solve the "brick equation" easily, even in the mind?

[math]{{\vec R}_c}(t) = {{\vec R}_c}(0) + {{\vec V}_c}(t)t + \frac{1}{2}\frac{{{{\vec F}^e}}}{M}{t^2}[/math]

Just because this equation is much simpler and more understandable to any student?

 

20 hours ago, swansont said:

If the center of mass is is stationary, the system has no net momentum, and vice-versa. 

 

Write this mathematical dependence, "a connoisseur of physics"!

Write, please!

 

 

 

 

 

I really hope that I will be able to explain to you what Real Physics is.

 

 

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1 hour ago, butovsv said:

No!

UgC8AD.jpg

 

Where did you see the dependence of the Center of Mass on "forces" ?!

 

 

 

Look in every physics textbook or for example the barycentre for the Earth Sun system.  The Centre of mass is off centre of the sun due to the gravitational force of attraction of other planets such as Jupiter 

If you wish confirmation study Kepler's laws. The centre of mass is always the vectoral sum of forces. 

From your own link

Choose a reference point R in the volume and compute the resultant forceand torque at this point,

what do you think a resultant force is ? 

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3 hours ago, butovsv said:

Bad answer.
The response of the layman.

I really hope that I will be able to explain to you what Real Physics is.

!

Moderator Note

Dr Swanson is a working atomic physicist, not a "layman". His atomic clocks help regulate our global GPS system. Perhaps you should examine your own failings instead.

Please focus on answering some of the questions you've been asked, and leave the professional qualifications out of it. It's clear you need some help with your concept, and you won't get it by rejecting mainstream physics. Focus on the details of your idea, and stop trying to teach physics.

 
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5 hours ago, butovsv said:

Yes!
Parts of the system can interact!
Their pulses can change in very large limits!
They can accelerate and slow down!

But at the same time, the total momentum of the system remains unchanged.

It seems like I'm asking the wrong questions since the answers are still not very clear. So I'll try another approach. 
I am not yet asking how your device is working. I am asking what you think the device is able to do*.

Your descriptions contains both very basic mechanics but also seems to claim existence of physics far beyond what is currently described by mechanical laws. Laws that have been used and tested for many years. To be able to address your statements it would be good to know if you claim to have evidence that currently known physics is wrong, or you have failed to understand how current physics works. 

Let's put a varipend in vacuum in space. No forces are pushing the varipend, it is at rest in our frame of reference, and it is not yet running**. Are you of the opinion that the varipend will be able to propel itself through space once the varipend is started***?
Does the varipend move at constant speed or stop after one cycle?
If the varipend are run through multiple cycles (your animations shows only one cycle) does it speed up for each cycle?

 

 

*) It does not yet matter if I agree or not. It is as always possible that I misunderstand the proposed device, hence I need to ask more about it. 
**) That is, the internal parts of the varipend is not yet in movement. Started means that the cycle, displayed in your animations, begins.
***) Math and animations may not be important at this time since it is not clear what kind of movement you are trying to describe. 

 

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3 hours ago, butovsv said:

Oh?!

rifle.gif

"Force" = "change in momentum"

F⃗ =dp⃗ dt=d(mv⃗ )dt

 

This is Physics.
The science is ...

 

Did you notice that pesky little "dt" in those equations?

What are the fundamental units of force? Of momentum? Are they equal?

 

3 hours ago, butovsv said:

Bad answer.
The response of the layman.

 

On the contrary. My professional opinion.

 

3 hours ago, butovsv said:

I repeat again.
Watch and read carefully.

 

 


Material Point System:

Varipend9.gif

 

The total momentum of this point system is defined as follows:

P⃗ =p⃗ i=mivi=m1v⃗ 1+m2v⃗ 2+m3v⃗ 3+...+miv⃗ i

If all these points would have the same speed, then this speed could be taken out of the sign of the sum, and sum the mass:

And it would turn out like this:

 vecP=M vecVc

In order for the speeds of all points of the system to become equal, all these points need to be simply connected!
Connect inelastically!


m1v⃗ 1+m2v⃗ 2+m3v⃗ 3+...+miv⃗ iBoom!!!(m1+m2+m3+...+mi)Vc=MVc


Actually, what you schoolchildren only know how to do! ....


But here is the misfortune in this "simple" task ....

As the author of this task, I came up with an inelastic connection of system components according to a special law.

Like this:

logo1.gif

logo1_cm.png

 

And, once again, you have failed to explain what is going on in your diagrams, that people might have a chance to make some sense out of them.

 

 

3 hours ago, butovsv said:

And in the total momentum for this system, the following regularity was determined:

m1v⃗ 1+m2v⃗ 2+m3v⃗ 3+...+mnv⃗ n=(mi1+mi2+mi3+...+min)V⃗ ic+(mk1+mk2+mk3+...+mkn)V⃗ kc


And what is very curious  - the masses of particle systems with the same velocity  (i.e. material points, material bodies!) - are constantly changing!

None of those black dots has the same velocity as any of the others, and therefore do not have the same momentum. They do have the same speed, however.

If they are evenly space, one can make the argument that for any particle moving at velocity v, there is one moving at -v, so if the masses are all equal, the net momentum is zero. That's consistent with the center of mass not changing, and being located at the center.

 

3 hours ago, butovsv said:

 


Look:

m1v⃗ 1+m2v⃗ 2+m3v⃗ 3+...+mnv⃗ n=(mi1+mi2+mi3+...+min)V⃗ ic(t1)+(mk1+mk2+mk3+...+mkn)V⃗ kc(t1)

m1v⃗ 1+m2v⃗ 2+m3v⃗ 3+...+mnv⃗ n=(mi1+mi2+mi3+...+min1)V⃗ ic(t2)+(mk1+mk2+mk3+...+mkn+1)V⃗ kc(t2)

m1v⃗ 1+m2v⃗ 2+m3v⃗ 3+...+mnv⃗ n=(mi1+mi2+mi3+...+min2)V⃗ ic(t3)+(mk1+mk2+mk3+...+mkn+2)V⃗ kc(t3)

 

 

Therefore, the total momentum of the entire system as a whole can be written as follows:

pvar(t)=mliquid(t)v⃗ 1(t)+mcase(t)v⃗ 2(t)

 

 

 

I will ask yet again: what is liquid in your diagram?

 

3 hours ago, butovsv said:

And this total momentum cannot change without an external energy source.

Momentum can indeed change without an energy source. A ball moving in a circle at constant speed does not change its energy, but does not have constant momentum.

 

 

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20 hours ago, Mordred said:
22 hours ago, butovsv said:

No!

UgC8AD.png

 

Where did you see the dependence of the Center of Mass on "forces" ?!

 

 

 

Look in every physics textbook or for example the barycentre for the Earth Sun system.  The Centre of mass is off centre of the sun due to the gravitational force of attraction of other planets such as Jupiter 

If you wish confirmation study Kepler's laws. The centre of mass is always the vectoral sum of forces. 

From your own link

Choose a reference point R in the volume and compute the resultant forceand torque at this point,

what do you think a resultant force is ? 

"Resulting force" is the change in the total momentum of a mechanical system.

 

Write the dependence of the "center of mass of the system of bodies" on the "resulting force".

 

 

 

19 hours ago, Mordred said:

In all your equations where is your direction components for your vectors ? 

All equations in this branch are given in vector form.
I cannot give a calculation in the projections of vectors on the axis of the corresponding coordinates here. Very little space.
I can give a link to these calculations!

18 hours ago, Phi for All said:
22 hours ago, butovsv said:

Bad answer.
The response of the layman.

I really hope that I will be able to explain to you what Real Physics is.

!

Moderator Note

Dr Swanson is a working atomic physicist, not a "layman". His atomic clocks help regulate our global GPS system. Perhaps you should examine your own failings instead.

Please focus on answering some of the questions you've been asked, and leave the professional qualifications out of it. It's clear you need some help with your concept, and you won't get it by rejecting mainstream physics. Focus on the details of your idea, and stop trying to teach physics. 

A very serious argument!
And I am a specialist in powerful technological lasers.

But we will not talk about lasers here!

I hope that after Dr Swanson remembers the basic concepts of Mechanics, his "atomic clock" will work even better!

 

"Dr Swanson" wrote:

22 hours ago, butovsv said:
  On 8/20/2019 at 7:08 PM, swansont said:

If the center of mass is is stationary, the system has no net momentum, and vice-versa. 

Such a phrase is permissible for a student.
But VERY embarrassing for an "atomic clock specialist"!

 

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31 minutes ago, butovsv said:

"Resulting force" is the change in the total momentum of a mechanical system.

 

Incorrect force is not momentum. You should really learn the definitions of each term. For example the unit of force is a Newton. Which is the amount of energy required to move a 1 kg mass one metre in one second.

 It should not take long to include your vector directions with your post.  

The SI unit for force is N for newton while the SI unit for momentum is kg*m/s. Momentum has dimensionality MLT^-1

Edited by Mordred
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17 hours ago, Ghideon said:

It seems like I'm asking the wrong questions since the answers are still not very clear. So I'll try another approach. 
I am not yet asking how your device is working. I am asking what you think the device is able to do*.

Nothing complicated!
The device shows you the Law of Conservation of Momentum.

varipend1.gif

17 hours ago, Ghideon said:

Your descriptions contains both very basic mechanics but also seems to claim existence of physics far beyond what is currently described by mechanical laws. Laws that have been used and tested for many years. To be able to address your statements it would be good to know if you claim to have evidence that currently known physics is wrong, or you have failed to understand how current physics works. 

In your head is the "Mechanics of the material body."

[math]{{\vec R}_c}(t) = {{\vec R}_c}(0) + {{\vec V}_c}(t)t + \frac{1}{2}\frac{{{{\vec F}^e}}}{M}{t^2}[/math]
And nothing but this!

But I explain to you the Mechanics of the system of bodies.
I remind you that the Mass Center of the body system is not a material object.

[math]{{\vec R}_c}(t) = \frac{{\sum {{m_i}(t){r_i}(t)} }}{{\sum {{m_i}} (t)}}[/math]
The result of the CM function is not subject to the laws of conservation, because it does not have mass.

 

17 hours ago, Ghideon said:

Let's put a varipend in vacuum in space. No forces are pushing the varipend, it is at rest in our frame of reference, and it is not yet running**. Are you of the opinion that the varipend will be able to propel itself through space once the varipend is started***?
Does the varipend move at constant speed or stop after one cycle?
If the varipend are run through multiple cycles (your animations shows only one cycle) does it speed up for each cycle? 

Let's!

The graph of the total momentum, that is, the momentum of the entire system, coincides with the abscissa of the graph.

If at any moment in time the relative motion of the system of moving elements ceases, the entire mechanical system will have an initial velocity.

If at the initial moment of time the whole mechanical system had zero velocity in the system ... then after stopping the system of moving elements the velocity of the entire mechanical system will also be zero!

image009.gif

Graph of the overall momentum of the system.

 

 

varipend1.gif

animation6.gif

 

Moving the Varipend system can be called a "linear precession".

This movement is inertialess.

The total momentum of the system and the kinetic energy of the entire system, reduced to the CM of the system, are zero.
At every moment in time, the system is formally motionless.
The system does not have inertia to continue moving.

In order for the system to move rectilinearly and evenly - it is necessary to expend energy.

 

 

 

 

2 hours ago, swansont said:
22 hours ago, butovsv said:

F⃗ =dp⃗ dt=d(mv⃗ )dt

 

This is Physics.
The science is ...

 

Did you notice that pesky little "dt" in those equations?

What are the fundamental units of force? Of momentum? Are they equal? 

Of course!

Quote

https://en.wikipedia.org/wiki/Newton_(unit)

n more formal terms, Newton's second law of motion states that the force exerted by an object is directly proportional to the acceleration of that object, namely:[1]

F=ma{\displaystyle F=ma}[math]F = ma[/math]

where the proportionality constant, m{\displaystyle m}m, represents the mass of the object undergoing an acceleration, a{\displaystyle a}a. As a result, the newton may be defined in terms of kilograms (kg{\displaystyle {\text{kg}}}{\displaystyle {\text{kg}}}), metres (m{\displaystyle {\text{m}}}{\displaystyle {\text{m}}}), and seconds (s{\displaystyle {\text{s}}}{\displaystyle {\text{s}}}) by

[math]1{\rm{N}} = 1\frac{{{\rm{kg}} \cdot {\rm{m}}}}{{{{\rm{s}}^2}}}[/math]

 

2 hours ago, swansont said:

rifle.gif

"Force" = "change in momentum"

F⃗ =dp⃗ dt=d(mv⃗ )dt

Add!

[math]{{\vec F}_1} = {{\vec F}_2}[/math]

[math]\frac{{{\rm{d}}{{\vec p}_1}}}{{{\rm{d}}t}} = \frac{{{\rm{d}}{{\vec p}_2}}}{{{\rm{d}}t}}[/math]

 

This is Physics.
The science is ...

 

 

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35 minutes ago, Bufofrog said:

Bustvov, why do you refuse to answer questions asked of you?

Do you not want to reveal too much about your diagrams?  But then what would be the point of the thread?

Maybe you don't know the answers? 

I beg your pardon if I did not answer any question.
Ready to answer any question!
There are no secrets!

22 minutes ago, Mordred said:

Acceleration is not the same as momentum. Why don't you study basic physics terminology 

True?!

[math]\vec F = \frac{{{\rm{d}}\vec p}}{{{\rm{d}}t}} = \frac{{d(m\vec v)}}{{dt}} = m\frac{{d\vec v}}{{dt}} = m\vec a[/math]

 

Something is wrong?
Butov did not study well at school ?!

 

In general, the change in momentum looks like this:

[math]\vec F = \frac{{{\rm{d}}\vec p}}{{{\rm{d}}t}} = \frac{{d(m\vec v)}}{{dt}} = m\frac{{d\vec v}}{{dt}} + {{\vec v}_r}\frac{{dm}}{{dt}} = m\vec a + {{\vec F}^{jet}}[/math]

 

 

 

2 hours ago, swansont said:
Quote

logo1.gif

logo1_cm.png

 

And, once again, you have failed to explain what is going on in your diagrams, that people might have a chance to make some sense out of them.

 

I will show how the Mass Center of the Varipend system is calculated:

body1_body2_e.png

 

[math]{r_c}(t) = \frac{{{m_{liquid}}(t){r_1}(t) + {m_{case}}(t){r_2}(t)}}{M}[/math]

 

In order to calculate the coordinates of the CM system at any time, it is necessary to know the masses and coordinates of the parts of the system.
And the masses and coordinates of the parts of the system change over time.
These CHANGES are subject to conservation laws: the Law of Conservation of Momentum and the Law of Conservation of Energy.

In order to calculate the coordinates of the system at any time , it is enough to solve the differential equation of the momentum balance of the parts of the system:

[math]{p_{var}}(t) = {m_{liquid}}(t){{\vec v}_1}(t) + {m_{case}}(t){{\vec v}_2}(t)=0[/math]

 

Oh!
Or maybe you want to solve the differential equation of a "brick" of the same mass as the Varipend system?

[math]{m_{brik}}{{\vec v}_{brik}} = 0[/math]

Because there is a very respectable competent reason for this - is this equation a little easier?

 

Edited by butovsv
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