Physics in troubles: the real equation of force is F = ma and not F = dp/dt

[martillo]

5 minutes ago, Genady said:

You say that you are considering only momentum of a rocket, but when you write P = mv, you work with the momentum of the entire system, and it does not matter what you say about it.

You can insist on making the same mistake, and it will not lead you anywhere. But if it makes you happy, enjoy!

I can consider anytime separately both the momentum for the total system say P, and the momentum of the rocket alone say p_r. Do you have a problem with that? Why?

Edited February 22, 2023 by martillo

[Genady]

3 minutes ago, martillo said:

I can consider anytime separately both the momentum for the total system say P, and the momentum of the rocket alone say p_r. Do you have a problem with that? Why?

It has been already explained to you in various ways by various contributors in various replies in this thread. Enjoy!

[martillo]

7 minutes ago, Genady said:

It has been already explained to you in various ways by various contributors in various replies in this thread. Enjoy!

I don't think so. I'm considering the momentum on the rocket alone just now. Not before.

You know, you can't escape from the fact that what works for the rockets is the definition of force as F = ma and not F = dp/dt. It is what works in practice...

Edited February 22, 2023 by martillo

[studiot]

10 hours ago, martillo said:

We all already had a lot of discussions in the thread. Please summarize in what you disagree with me.

I don't necessarily disagree with you.

In fact I think the translation barrier has not helped added to the fact we have got several members offering different approaches and/or explanations.

This must be very confusing.

I also note that different texts derive different (not necessarily incompatible) conclusions and results from the same material and starting point.
This fact is why I asked you what you want to calculate.
Since you have told me several times that you do not want to calculate anything I now think perhaps you are actually asking to understand certain terms in one version of 'the rocket equation'.

Another sidetrack is the introduction of 'generalised coordinates' and 'generalised momenta', both of which are coordinates and momenta in name only.
This really is sidetracking a steamroller to crack a nut and quite over the top.
Studying this updating of simpler mechanics is really important to Physicists but takes at least a year of hard study.

You have a track record on not answering the one or two questions I aks in a post.

They are all designed to help us (especially me) help you so please answer this one.

Are you familiar with the dot and double dot notation for derivatives with respect to time ?

[math]\mathop r\limits^{ \bullet  \bullet } [/math]

and

[math]\mathop r\limits^ \bullet  [/math]

 

 

 

[martillo]

18 minutes ago, studiot said:

Are you familiar with the dot and double dot notation for derivatives with respect to time ?

r∙∙

and

r∙

 

Yes.

[studiot]

2 hours ago, martillo said:

Yes.

That's good I can use them then.

But so as I don't spring this idea on you let me explain why I am using r not x for distance.

Consider Newtons second Law  Force  = mass times acceleration.

Easy peasy yes ?

But there is a twist.

This refers to a single body (or system) that remains intact and does not shed (or gain) material.

In fact it refers to the centre of mass (COM) of that body.

And r is a common symbol for the position of that COM in your coordinate system.

So when we quote Newton's 2nd law we really mean that

The product of the mass of a body and the acceleration of its COM equals the force imposed on it.

Swansont refers to this as the net force which is important for a rocket system, because the net force is exactly zero, in the absence of drag, gravity etc.

This means that the acceleration of the COM of the rocket system must be zero since the mass is not zero.

This is both good and bad.

First the bad news:  we cannot use f = ma directly since it only tells us the 0 = 0.

Now the good news:  we can use it to form a suitable momentum equation to work with. ( I will post this next time)

 

I have slyly slipped in the phrase 'rocket system' there.

That is because we normally speak of rigid body dynamics.

The point about a rigid body for the purposes of Newton's 2nd law is that however the COM accelerates, so will any other point of the body.

But we don't have a rigid body.

We have a system of bodies that deforms with time.

 

So your title has some truth in it, but not quite for the reasons you propose.

You are however right to be suspicious of some of the explanations offered online.

Edited February 22, 2023 by studiot

[swansont]

10 hours ago, martillo said:

You know, you can't escape from the fact that what works for the rockets is the definition of force as F = ma

Then go ahead and use F=ma to come up with the rocket equation.

 

studiot’s comments above brings to mind that martillo never actually stated newton’s 2nd law - that the common statement F =ma (and the correct F = dp/dt) refers to the net external force (as studiot states it, the force imposed on the body)  

 

[studiot]

OK then to continue.

I am presenting this a Physics exercise, not a formal maths one, although maths is involved.

Physicists are allowed to bend the maths rules a bit.

 

Let us consider a rocket travelling along.

The trick to analysing this is to use the fact that momentum is constant for the whole rocket system (rocket plus exhaust), although it is variable for each part separately.

Thus system momentum = p = a constant.

But we analyse this by considering the parts of the system separately.

At some time t the rocket has mass M and velocity V.

At some time (t +dt) these have changed so the rocket now has mass (m + dm)   and velocity (V + dv).

Note dm is negative .

Furthermore -dm is the mass of the exhaust, which has a velocity v relative to the rocket or velocity (v-V) to a ground observer

Splitting the system mass in this way saves introducing m for the exhaust mass and simplifies the maths.

Here is a diagram showing these essentials.

 

So the system momentum at time t =  p  MV    (first figure).................1

Note this excludes the need for considering the exhaust momentum at this stage

But the system momentum at time (t + dt) is also = p = (M + dM)(V + dV) - dM(V-v) .......................2

Multiplying out the terms and taking care of the signs then either subtracting equation 1 from equation 2 or simply using the fact that the right hand sides are both equal to p we can equate them directly leads to the expression

MdV + vdM  =  0  .........3

Two notes here.

This is a momentum equation, not a force equation.
This equation is the springboard to answer questions about the desired properties of the rocket system.

So for instance we can remember our schoolboy calculus and note it can be rewritten as (after dividing through by M)

dV = -v d(lnM)     where lnM is the natural log of M

 

But you have asked for the force on the rocket also known as the thrust.
To get this we proceed as follows

We divide equation3 through by dt to obtain

[math]M\frac{{dV}}{{dt}} + v\frac{{dM}}{{dt}} = 0[/math]

If we let

[math]\alpha  =  - \frac{{dM}}{{dt}}[/math]

Then

[math]M\frac{{dV}}{{dt}} = \alpha v[/math]

This is the required force on the rocket.

[martillo]

1 hour ago, studiot said:

This is the required force on the rocket.

You see, in the last step you arrive at Mdv/dt = αv and just say "this is the required force on the rocket".  This is to consider that the force F is F = mdv/dt as everybody does. So the definition of the force that works on the rocket is F = mdv/dt = ma and not F = dp/dt. Just note that the mass M of the rocket alone is variable and so the formula F = ma is valid even for variable mass.

The claim on the title of this thread stands.

 

Edited February 23, 2023 by martillo

[Genady]

Imagine a chuck of ice of mass m flying peacefully in space with the speed v. There are no forces, F = 0.

The ice slowly melts. The momentum of the ice P = mv. Thus, dP/dt = mdv/dt + vdm/dt.

So, either the chunk of ice is accelerating, or F = dP/dt is wrong.

Or, I don't know what I'm talking about.

Or, just maybe, I am a troll. 

[martillo]

29 minutes ago, Genady said:

Imagine a chuck of ice of mass m flying peacefully in space with the speed v. There are no forces, F = 0.

The ice slowly melts. The momentum of the ice P = mv. Thus, dP/dt = mdv/dt + vdm/dt.

So, either the chunk of ice is accelerating, or F = dP/dt is wrong.

Or, I don't know what I'm talking about.

Or, just maybe, I am a troll. 

You are right! No problem with this. It's just that as there are no forces and considering force = mdv/dt then the term mdv/dt = 0 and you have dP/dt = vdm/dt.

But here you are also applying F = mdv/dt = ma even for a variable mass.

That's the point F = ma ever. Is the definition of the force that really works.

 

Note:  

p = mv

dp/dt = mdv/dt + vdm/dt

dp/dt = F + dm/dt

and everything runs perfect!

but F ≠ dp/dt and the problem is the consequences of this in Physics... Physics must be reviewed from scratch, from the review of the Newton Second Law...

Don't you find this something good?

I find it exciting, although lot of work... What could be better? 😄 

Edited February 23, 2023 by martillo

[Genady]

37 minutes ago, Genady said:

Imagine a chuck of ice of mass m flying peacefully in space with the speed v. There are no forces, F = 0.

The ice slowly melts. The momentum of the ice P = mv. Thus, dP/dt = mdv/dt + vdm/dt.

So, either the chunk of ice is accelerating, or F = dP/dt is wrong.

Or, I don't know what I'm talking about.

Or, just maybe, I am a troll. 

BTW, the law of momentum conservation is wrong, too. You see, as the mass of the chunk of ice goes down, its momentum goes down as well. So, the momentum is not conserved.

[martillo]

3 minutes ago, Genady said:

BTW, the law of momentum conservation is wrong, too. You see, as the mass of the chunk of ice goes down, its momentum goes down as well. So, the momentum is not conserved.

I was editing the post. May be you didn't see the last part now. You are right here that there is a problem with the current law of conservation of momentum. Consider my edition in the post:

Note:  

p = mv

dp/dt = mdv/dt + vdm/dt

dp/dt = F + dm/dt

and everything runs perfect!

but F ≠ dp/dt and the problem is the consequences of this in Physics... Physics must be reviewed from scratch, from the review of the Newton Second Law...

Don't you find this something good?

I find it exciting, although lot of work... What could be better? 😄 

 

Edited February 23, 2023 by martillo

[Lorentz Jr]

5 minutes ago, martillo said:

What could be better? 😄 

Some real physics? 🤔

[martillo]

Just now, Lorentz Jr said:

Some real physics? 🤔

Right!

 

[Genady]

6 minutes ago, Genady said:

BTW, the law of momentum conservation is wrong, too. You see, as the mass of the chunk of ice goes down, its momentum goes down as well. So, the momentum is not conserved.

Moreover, the kinetic energy

E = mv²/2 = mvv/2 = Pv/2

Since P is not conserved, E is not conserved.

Tha law of energy conservation is wrong.

The entire Physics must be rewritten now. How exciting.

[studiot]

1 hour ago, martillo said:

You see, in the last step you arrive at Mdv/dt = αv and just say "this is the required force on the rocket".  This is to consider that the force F is F = mdv/dt as everybody does. So the definition of the force that works on the rocket is F = mdv/dt = ma and not F = dp/dt. Just note that the mass M of the rocket alone is variable and so the formula F = ma is valid even for variable mass.

The claim on the title of this thread stands.

I'm sorry you are so quick to challenge, rather than ask for further detail as to why I said what I said.

No it is not exactly the same as the work of the others, although there are commonalities.

If you were not so hasty you would see that you even misquoted my statement.

 

I did not say Mdv/dt any more than I used m or F = dp/dt anywhere.

I hoped you would follow my extended commentary instead of reaction with a Putin style knee-jerk.

 

The whole point is to distinguish between M and m as well as  V and v.

 

If you did not understand it ASK.

 

Edited February 23, 2023 by studiot

[martillo]

44 minutes ago, Genady said:

Moreover, the kinetic energy

E = mv²/2 = mvv/2 = Pv/2

Since P is not conserved, E is not conserved.

Tha law of energy conservation is wrong.

The entire Physics must be rewritten now. How exciting.

The Momentum and the Energy MUST be conserved ALWAYS!

Let me try in your case. Just help me if needed please.

You have

p = mv

dp/dt = F + vdm/dt where F= mdv/dt

Conservation of momentum: 

p = constant if dp/dt = 0

Follows that the Principle of Conservation of Momentum is: Momentum p is conserved  IF external forces F = 0 AND there's no variation in the mass: dm/dt =0.

dp/dt = F + vdm/dt = if F = 0 and dm/dt = 0

Although the case F = -vdm/dt also verify it but I cannot imagine now an example for this...

May be the law of conservation of energy must be rewritten too. I said, lot of problems to solve now. From scratch...

 

Edited February 23, 2023 by martillo

[swansont]

1 hour ago, martillo said:

Note:  

p = mv

dp/dt = mdv/dt + vdm/dt

dp/dt = F + dm/dt

and everything runs perfect!

 

You mean dp/dt = F + vdm/dt

An object at constant speed v, with no external force on it, shoots equal mass out of opposite ends at a rate of dm/dt, at the same velocity, so the ejected mass has equal momentum in each direction.

Your equation predicts that momentum is not conserved. F = 0 so there’s no acceleration, so the object remains at speed v.

Where does the momentum change come from?

[martillo]

10 minutes ago, swansont said:

 

You mean dp/dt = F + vdm/dt

An object at constant speed v, with no external force on it, shoots equal mass out of opposite ends at a rate of dm/dt, at the same velocity, so the ejected mass has equal momentum in each direction.

Your equation predicts that momentum is not conserved. F = 0 so there’s no acceleration, so the object remains at speed v.

Where does the momentum change come from?

I think what we begun discussing with Genady also applies here::

32 minutes ago, martillo said:

The Momentum and the Energy MUST be conserved ALWAYS!

Let me try in your case. Just help me if needed please.

You have

p = mv

dp/dt = F + vdm/dt where F= mdv/dt

Conservation of momentum: 

p = constant if dp/dt = 0

Follows that the Principle of Conservation of Momentum is: Momentum p is conserved  IF external forces F = 0 AND there's no variation in the mass: dm/dt =0.

dp/dt = F + vdm/dt = if F = 0 and dm/dt = 0

Although the case F = -vdm/dt also verify it but I cannot imagine now an example for this...

May be the law of conservation of energy must be rewritten too. I said, lot of problems to solve now. From scratch...

 

 

10 minutes ago, swansont said:

Where does the momentum change come from?

dp/dt = F +vdm/dt

If F = 0 then dp/dt = vdm/dt

The change in momentum comes from the variation in mass.

This would apply in Genady's example.

41 minutes ago, studiot said:

I'm sorry you are so quick to challenge, rather than ask for further detail as to why I said what I said.

No it is not exactly the same as the work of the others, although there are commonalities.

If you were not so hasty you would see that you even misquoted my statement.

 

I did not say Mdv/dt any more than I used m or F = dp/dt anywhere.

I hoped you would follow my extended commentary instead of reaction with a Putin style knee-jerk.

 

The whole point is to distinguish between M and m as well as  V and v.

 

If you did not understand it ASK.

 

No I don't get it.

But you know, i think you are lagging behind. I'm going one step further with Genady, Swansont, may be Lorentz Jr. Please, give some time for this. If I go wrong in this we can return to your current point of view. Please, it's not easy to discuss with several ones at the same time. Please give me some time...

Edited February 23, 2023 by martillo

[studiot]

17 minutes ago, martillo said:

No I don't get it.

No you don't.

Can you honestly not see the difference between what I wrote and your copy of it ?

BTW

The centre of mass of the rocket changes.

So the rocket at time t and time (t+dt) is effectively a different object.

so you cannot use general formulae.

MdV/dt is the force on the rocket at one instant in time only the force at time t.

This is what is meant by saying the mass must be constant or how we get around that restriction.

 

Suggestion

Instead of all this constant criticism of theory, find yourself a few examples with numbers in or exercises with answers.

Can you obtain the correct numerical answers, using your methods. whatever they may be ?

 

In other words,

Does your way actually work in practice. ?

[martillo]

2 minutes ago, studiot said:

No you don't.

Can you honestly not see the difference between what I wrote and your copy of it ?

BTW

The centre of mass of the rocket changes.

So the rocket at time t and time (t+dt) is effectively a different object.

so you cannot use general formulae.

MdV/dt is the force on the rocket at one instant in time only the force at time t.

This is what is meant by saying the mass must be constant or how we get around that restriction.

 

Suggestion

Instead of all this constant criticism of theory, find yourself a few examples with numbers in or exercises with answers.

Can you obtain the correct numerical answers, using your methods. whatever they may be ?

 

In other words,

Does your way actually work in practice. ?

I asked you to give me some time... 

[Genady]

42 minutes ago, martillo said:

we begun discussing with Genady

 

42 minutes ago, martillo said:

it's not easy to discuss with several ones at the same time.

You have misunderstood. I am not in a discussion with you. I just contributed couple of comments to the thread, but I did not refer to you in those comments, as you can check. You are free to ignore them. I am not looking forward to your reply.

[martillo]

9 minutes ago, Genady said:

 

You have misunderstood. I am not in a discussion with you. I just contributed couple of comments to the thread, but I did not refer to you in those comments, as you can check. You are free to ignore them. I am not looking forward to your reply.

No, no. Your comments are being very important to me. As you say they are being a  contribution, an excellent contribution. It`s a misspelling, by discussing I meant interacting, interchanging ideas, with good criticism of course, just that. Which would be a better word for this?

Edited February 23, 2023 by martillo

[studiot]

I see that hyperphysics does it my way, but with much better pictures.

http://hyperphysics.phy-astr.gsu.edu/hbase/limn2.html#ln21

http://hyperphysics.phy-astr.gsu.edu/hbase/rocket.html#c2

 

Edited February 23, 2023 by studiot