Doubt about variable mass systems

[Hak]

Speaking of variable mass systems... it seems to me that there is a flaw in Halliday's reasoning when he talks about this subject. But the formula he derives seems to work to me!

He starts with a system of points of total mass $$\ M$$ and cm velocity $$\ v$$. Due to the action of external forces after a certain time t the system has clearly split into two parts. The first has mass $$\ M -\Delta M$$ and cm velocity $$\ v +\Delta v$$; the second has mass $$\Delta M$$ and velocity $$\ u$$.

Applying $$F_{est} = \frac{\Delta P}{\Delta t}$$ finds that for a finite interval of time it holds (approximately)
$$\ F_{est} = M \frac{\Delta v}{\Delta t} + [u-(v+\Delta v)]\frac{\Delta M}{\Delta t}$$.
Moving on to the limit for t tending to 0
replaces $$\frac{\Delta v}{\Delta t}$$ with $$\frac{dv}{dt}$$
replaces $$\frac{\Delta M}{\Delta t}$$ with $$-\frac{dM}{dt}$$
places $$\Delta v = 0$$.

Thus $$\ F_{est} = M \frac{dv}{dt} + v \frac{dM}{dt} -u \frac{dM}{dt}$$.
But shouldn't it have taken into account that $$\ u$$ also goes to zero (or at least changes) for t that tends to zero?

Am I wrong in making such an assumption or am I right? Thanks in advance.

 

[Hill]

Why would $u$ change?

 

[Hak]

Why would $u$ change?

Maybe because $t$ tends to zero, no? I am slightly confused.

 

[Hill]

Seems to me that if no force acts on $\Delta M$, then its velocity would not change and is the same as $\Delta M \to dM$.

 

[erobz]

https://www.physicsforums.com/insights/how-to-apply-newtons-second-law-to-variable-mass-systems/

 

[Hak]

https://www.physicsforums.com/insights/how-to-apply-newtons-second-law-to-variable-mass-systems/

Thank you very much. Is there an answer to my doubt here? Or is it just for information?

 

[erobz]

Thank you very much. Is there an answer to my doubt here? Or is it just for information?

It's been summarized like this for a reason. There has been much back and forth about how to handle this topic. Here is a somewhat recent thread:

https://www.physicsforums.com/threa...ed-about-newtons-2nd-law.1050482/post-6861856

None of what is being said in post #1 seems to be clearly defined. The author is using $-\Delta M$ when the $\Delta$ captures both positive and negative changes, is $u$ lab frame, relative velocity? I don't personally like the derivation.

Here is another thread where the OP call into question a derivation by Taylor that uses $M - \Delta M$:

https://www.physicsforums.com/threads/how-to-develop-the-rocket-equation.961707/page-2

Maybe just read through some of it to see if your particular issues are addressed too.

 

[Hak]

It's been summarized like this for a reason. There has been much back and forth about how to handle this topic. Here is a somewhat recent thread:

https://www.physicsforums.com/threa...ed-about-newtons-2nd-law.1050482/post-6861856

None of what is being said in post #1 seems to be clearly defined. The author is using $-\Delta M$ when the $\Delta$ captures both positive and negative changes, is $u$ lab frame, relative velocity? I don't personally like the derivation.

Here is another thread where the OP call into question a derivation by Taylor that uses $M - \Delta M$:

https://www.physicsforums.com/threads/how-to-develop-the-rocket-equation.961707/page-2

Maybe just read through some of it to see if your particular issues are addressed too.

I read the posts in these two threads. Unfortunately, I understood very little. The notations and justifications that follow each other are conflicting and help very little in terms of clarity. I am still very confused, especially about the second problem you presented. Any further input is appreciated.

 

[erobz]

I read the posts in these two threads. Unfortunately, I understood very little. The notations and justifications that follow each other are conflicting and help very little in terms of clarity. I am still very confused, especially about the second problem you presented. Any further input is appreciated.

One man's trash is another's treasure...

You should probably just stick to the insight, it seems to be handled consistently?

 

[Hak]

One man's trash is another's treasure...

You should probably just stick to the insight, it seems to be handled consistently?

You are right. But for me it is not rubbish, on the contrary. The problem is that I had not thought about issues concerning the sign of the mass (as you have exposed instead), so now I am even more confused than before because you have opened my mind by showing how I accept as true issues that are anything but trivial. I thank you very much for that. I still don't know how to approach the question of negative or positive mass and the derivation made by Taylor. As for the velocity of the centre of mass, I am even more confused. Try having a look at the treatment of variable mass systems in Halliday, Resnick, Krane (you most likely have it), maybe I've misunderstood or missed something I can't understand. Let me know. Thanks again.
The Insight by @kuruman is fine, but it does not cover all the issues in my OP or others we are discussing now.

 

[erobz]

You are right. But for me it is not rubbish, on the contrary. The problem is that I had not thought about issues concerning the sign of the mass (as you have exposed instead), so now I am even more confused than before because you have opened my mind by showing how I accept as true issues that are anything but trivial. I thank you very much for that. I still don't know how to approach the question of negative or positive mass and the derivation made by Taylor. As for the velocity of the centre of mass, I am even more confused. Try having a look at the treatment of variable mass systems in Halliday, Resnick, Krane (you most likely have it), maybe I've misunderstood or missed something I can't understand. Let me know. Thanks again.
The Insight by @kuruman is fine, but it does not cover all the issues in my OP or others we are discussing now.

My personal opinion is multiple systems need to be carved out of $M$. One of them is losing mass, and another gaining mass. Then all the changes are handled by the $\Delta$'s.

Before the impulse there is one system with momentum $Mv$. After the impulse there are two systems one with momentum $( M + dM ) ( v + dv )$ and another with momentum $dm( v - u )$ where $u$ is the velocity of $dm$ w.r.t. $v$, which results in:

$$ \sum F ~dt = ( M + dM ) ( v + dv ) + dm (v-u) - Mv $$

$$ \sum F ~dt = Mv + dM~v + M~dv + dm~v - dm~u - Mv $$

$$ \sum F ~dt = M~dv + [ dM~v + dm~v ] -dm u $$

We know that $dM = -dm$, so that terms in brackets add to zero

$$ \sum F = M~\frac{dv}{dt} -\frac{dm}{dt} u $$

Or in terms of $M$:

$$ \sum F = M~\frac{dv}{dt} +\frac{dM}{dt} u $$I don't own the text for the derivation in the OP.

 

[Hak]

My personal opinion is multiple systems need to be carved out of $M$. One of them is losing mass, and another gaining mass. Then all the changes are handled by the $\Delta$'s.

Before the impulse there is one system with momentum $Mv$. After the impulse there are two systems one with momentum $( M + dM ) ( v + dv )$ and another with momentum $dm( v - u )$ where $u$ is the velocity of $dm$ w.r.t. $v$, which results in:

$$ \sum F ~dt = ( M + dM ) ( v + dv ) + dm (v-u) - Mv $$

$$ \sum F ~dt = Mv + dM~v + M~dv + dm~v - dm~u - Mv $$

$$ \sum F ~dt = M~dv + [ dM~v + dm~v ] -dm u $$

We know that $dM = -dm$, so that terms in brackets add to zero

$$ \sum F = M~\frac{dv}{dt} -\frac{dm}{dt} u $$

Or in terms of $M$:

$$ \sum F = M~\frac{dv}{dt} +\frac{dM}{dt} u $$I don't own the text for the derivation in the OP.

Thank you very much. But, for example, one of the objections in the threads you referred me to is that the mass does not change in such a way that one can make a derivation with respect to time and that, therefore, such a statement is inaccurate.
Anyway, I will attach the part of the text where the derivation in the OP is made.

 

[erobz]

Thank you very much. But, for example, one of the objections in the threads you referred me to is that the mass does not change in such a way that one can make a derivation with respect to time and that, therefore, such a statement is inaccurate.
Anyway, I will attach the part of the text where the derivation in the OP is made.

Can you point me to the objection so I can see it in context?

 

[Hak]

Can you point me to the objection so I can see it in context?

None, your assumptions are correct. It was me who misunderstood post #2 of the first thread you referred me to. My fault.

 

[Hak]

My personal opinion is multiple systems need to be carved out of $M$. One of them is losing mass, and another gaining mass. Then all the changes are handled by the $\Delta$'s.

Before the impulse there is one system with momentum $Mv$. After the impulse there are two systems one with momentum $( M + dM ) ( v + dv )$ and another with momentum $dm( v - u )$ where $u$ is the velocity of $dm$ w.r.t. $v$, which results in:

$$ \sum F ~dt = ( M + dM ) ( v + dv ) + dm (v-u) - Mv $$

$$ \sum F ~dt = Mv + dM~v + M~dv + dm~v - dm~u - Mv $$

$$ \sum F ~dt = M~dv + [ dM~v + dm~v ] -dm u $$

We know that $dM = -dm$, so that terms in brackets add to zero

$$ \sum F = M~\frac{dv}{dt} -\frac{dm}{dt} u $$

Or in terms of $M$:

$$ \sum F = M~\frac{dv}{dt} +\frac{dM}{dt} u $$I don't own the text for the derivation in the OP.

However, I fail to understand why the form presented by $\Delta$ is contradictory and unhelpful: so much evidence and justification has been put forward by the OP of the second thread, and much has been rejected or objected to by experts, and finally you tell the OP that you understand his frustration with this (contradictions of $\Delta$). This has added to my confusion, which is why I cannot shed light on these doubts. Could you explain to me why it is controversial and forced, gives rise to misunderstandings, etc.?
Why is it not advisable to use $M - \Delta M$? Thanks.

 

[erobz]

However, I fail to understand why the form presented by $\Delta$ is contradictory and unhelpful: so much evidence and justification has been put forward by the OP of the second thread, and much has been rejected or objected to by experts, and finally you tell the OP that you understand his frustration with this (contradictions of $\Delta$). This has added to my confusion, which is why I cannot shed light on these doubts. Could you explain to me why it is controversial and forced, gives rise to misunderstandings, etc.?
Why is it not advisable to use $M - \Delta M$? Thanks.

It just seems to make a mess of things. After the ejection the mass $M$ has lost mass, $\Delta M < 0$ ( i.e. it's negative). The equation $M - \Delta M$ would lead you to believe that paradoxically after losing mass, the mass $M$ has in fact gained mass. So when its being said the momentum afterward is $( M - \Delta M ) ( v+ \Delta v )$, what's really being said?

 

[Hak]

It just seems to make a mess of things. After the ejection the mass $M$ has lost mass, $\Delta M < 0$ ( i.e. it's negative). The equation $M - \Delta M$ would lead you to believe that paradoxically after losing mass, the mass $M$ has in fact gained mass. So when its being said the momentum afterward is $( M - \Delta M ) ( v+ \Delta v )$, what's really being said?

Thank you very much. I'm attaching the derivation by Halliday: looking at it again, it looks slightly different from what I wrote, I don't know why. Some time ago I must have taken notes from the Halliday textbook and reworked it according to my version. I don't know what to say. What could you tell me? Do you have any advice or suggestions for correcting mistakes and thinning out my doubts? Thank you very much.

 

[erobz]

Thank you very much. I'm attaching the derivation by Halliday: looking at it again, it looks slightly different from what I wrote, I don't know why. Some time ago I must have taken notes from the Halliday textbook and reworked it according to my version. I don't know what to say. What could you tell me? Do you have any advice or suggestions for correcting mistakes and thinning out my doubts? Thank you very much.

I can't read it.

 

[Hak]

I can't read it.

Thanks for the feedback. How about now?

 

[erobz]

Thanks for the feedback. How about now? View attachment 334604View attachment 334605

Good, none of this derivation is bothering in my opinion, it’s almost exactly what I did just without the presence of external forces. What is bothering you about it?

 

[Hak]

Good, none of this derivation is bothering in my opinion, it’s almost exactly what I did just without the presence of external forces. What is bothering you about it?

1) The role of $u$. i.e. the question in post #1.
2) Where did $M - \Delta M$ come from in the OP? Quite a mystery, this is the document I had taken the demonstration from, who knows what I had in mind then. This is the other problem.
Thank you.

 

[erobz]

1) The role of $u$. i.e. the question in post #1.
2) Where did $M - \Delta M$ come from in the OP? Quite a mystery, this is the document I had taken the demonstration from, who knows what I had in mind then. This is the other problem.
Thank you.

The derivation in post 1 is not the derivation you said it was. Just forget it. What don’t you understand about the correct derivation?

 

[Hak]

The derivation in 1 is not the derivation you said it was.

Wait, can you rephrase that a little better?

 

[erobz]

Wait, can you rephrase that a little better?

Post 1 is not the derivation in the textbook you quote. You didn’t copy it properly. Why are you trying to understand the butchered derivation in post 1?

 

[kuruman]

The Insight by @kuruman is fine, but it does not cover all the issues in my OP or others we are discussing now.

The equation I derived, $$m~\frac{dv}{dt}=(u-v)\frac{dm}{dt}+F_{\text{ext}}.$$ is a general form of Newton's second law for any variable mass system, not just a rocket in free space.

In it

• $m$ is the variable mass of the system of interest. This is a system that is either ejecting or accruing mass.
• $v$ is the velocity of said system relative to an inertial frame, i.e. the lab frame.
• $\dfrac{dm}{dt}$ is the rate of change of the system's mass. It is a negative quantity if mass is ejected from the system and a positive quantity if mass is accumulated by the system.
• $u$ is the velocity relative to the lab frame of the mass $dm$ after is ejected from the system or before it incorporated into the system.

For a rocket in free space, we set the external force equal to zero and get $$m~\frac{dv}{dt}=(u-v)\frac{dm}{dt}$$. In this form, without using $v_{rel}$, I think it is easier to see what's going on in different cases.

Case I: Rocket is moving in the positive direction and the mass is ejected in the negative direction (usual case).
Then $\dfrac{dm}{dt}=-R,~~~(R>0)$ and $u=-|u|$. Thus, $$m~\frac{dv}{dt}=(-|u|-v)(-R)=R(|u|+v)>0.$$ This says that the speed of the rocket increases with time because $v$ and $\frac{dv}{dt}$ are both positive.

Case II: Rocket is moving in the positive direction and the mass is ejected also in the positive direction Then $$m~\frac{dv}{dt}=(u-v)(-R)=R(v-u)>0.$$ This says that

1. if $v > u$, the speed of the rocket will be increasing
2. if $v < u$, the speed of the rocket will be decreasing
3. if $v=u$, the speed of the rocket will be constant.

These results, especially item 1, may not be as transparent if one uses the "first rocket equation" 9.9.6 and has to sort out the sign of $v_{rel}$.

 

[Hak]

Post 1 is not the derivation in the textbook you quote. You didn’t copy it properly. Why are you trying to understand the butchered derivation in post 1?

My confusion and misrepresentation is entirely similar (almost corresponding) to that of the OP of the following thread:
https://www.physicsforums.com/threads/a-system-with-varying-mass-a-rocket.61358/
I checked, the demonstration that is made in that edition of Halliday's textbook is the same as the one I sent in the two pictures (it has remained unchanged since the first edition, up to the 12th), yet the OP expresses the same perplexities as me and makes the same calculations that are different from those in the book ($M - \Delta M$, etc...). This is truly a mystery. Try to have a look: I thank you in advance, hoping that you will be able to unravel it completely.

 

[Hak]

The equation I derived, $$m~\frac{dv}{dt}=(u-v)\frac{dm}{dt}+F_{\text{ext}}.$$ is a general form of Newton's second law for any variable mass system, not just a rocket in free space.

In it

• $m$ is the variable mass of the system of interest. This is a system that is either ejecting or accruing mass.
• $v$ is the velocity of said system relative to an inertial frame, i.e. the lab frame.
• $\dfrac{dm}{dt}$ is the rate of change of the system's mass. It is a negative quantity if mass is ejected from the system and a positive quantity if mass is accumulated by the system.
• $u$ is the velocity relative to the lab frame of the mass $dm$ after is ejected from the system or before it incorporated into the system.

For a rocket in free space, we set the external force equal to zero and get $$m~\frac{dv}{dt}=(u-v)\frac{dm}{dt}$$. In this form, without using $v_{rel}$, I think it is easier to see what's going on in different cases.

Case I: Rocket is moving in the positive direction and the mass is ejected in the negative direction (usual case).
Then $\dfrac{dm}{dt}=-R,~~~(R>0)$ and $u=-|u|$. Thus, $$m~\frac{dv}{dt}=(-|u|-v)(-R)=R(|u|+v)>0.$$ This says that the speed of the rocket increases with time because $v$ and $\frac{dv}{dt}$ are both positive.

Case II: Rocket is moving in the positive direction and the mass is ejected also in the positive direction Then $$m~\frac{dv}{dt}=(u-v)(-R)=R(v-u)>0.$$ This says that

1. if $v > u$, the speed of the rocket will be increasing
2. if $v < u$, the speed of the rocket will be decreasing
3. if $v=u$, the speed of the rocket will be constant.

These results, especially item 1, may not be as transparent if one uses the "first rocket equation" 9.9.6 and has to sort out the sign of $v_{rel}$.

Thanks for your contribution. It seems really good to me.

 

[erobz]

My confusion and misrepresentation is entirely similar (almost corresponding) to that of the OP of the following thread:
https://www.physicsforums.com/threads/a-system-with-varying-mass-a-rocket.61358/
I checked, the demonstration that is made in that edition of Halliday's textbook is the same as the one I sent in the two pictures (it has remained unchanged since the first edition, up to the 12th), yet the OP expresses the same perplexities as me and makes the same calculations that are different from those in the book ($M - \Delta M$, etc...). This is truly a mystery. Try to have a look: I thank you in advance, hoping that you will be able to unravel it completely.

With $u = 0$ ( lab frame ), what is your specific gripe about it? I'm standing on a rocket powered by me throwing baseballs out the back. Its currently moving at 10 m/s to the left, I throw a ball out at 10 m/s relative to the rocket. In the lab frame the velocity of the ball $u$ is zero. Regardless of $u = 0$ I applied in impulse to the rocket, and it accelerated. If I was moving a 20 m/s and I threw a ball out the back at 10 m/s, in the lab frame $u$ is 10 m/s to the left. I still applied an impulse, and the rocket accelerated even though $v$ ( the velocity of the rocket) was initially greater than $v_{rel}$. The velocity $u$ ( lab frame ) being zero is just not special.

 

[Hak]

With $u = 0$ ( lab frame ), what is your specific gripe about it? I'm standing on a rocket powered by me throwing baseballs out the back. Its currently moving at 10 m/s to the left, I throw a ball out at 10 m/s relative to the rocket. In the lab frame the velocity of the ball $u$ is zero. Regardless of $u = 0$ I applied in impulse to the rocket, and it accelerated. If I was moving a 20 m/s and I threw a ball out the back at 10 m/s, in the lab frame $u$ is 10 m/s to the left. I still applied an impulse, and the rocket accelerated even though $v$ ( the velocity of the rocket) was initially greater than $v_{rel}$. The velocity $u$ ( lab frame ) being zero is just not special.

OK, thanks. Were you able to see the posts in the thread I referred you to in post #26?

 

[erobz]

OK, thanks. Were you able to see the posts in the thread I referred you to in post #26?

Yeah, I see them, but we already covered that mistake I thought. The OP thought ($M+dM$ ) was wrong because it looks like you are adding "very small" mass $dM$, but in fact what we are adding is a "very small" negative change in mass in that expression. Then in the system that is gaining mass, we force that it gets the opposite of that change.

 

[kuruman]

OK, thanks. Were you able to see the posts in the thread I referred you to in post #26?

The OP in that thread is guilty of setting up the reader for confusion.
First the OP says "The rocket now has velocity v+dv and mass M+dM, where the change in mass dM is a negative quantity.". That's fine.
Then the OP says "If I'm allowed to change M+dM with M-dM and -dM with dM then Eq. 9-38 becomes
M * v = dM * U + (M - dM) * (v + dv)"
Equation 9.38 is
M * v = -dM * U + (M + dM) * (v + dv) ... (9-38)

Well, having defined dM as a negative quantity, one is not allowed to change dM to -dM but to -|dM| to avoid conflict with the definition plus confusion about the variables. Then Eq. 9-38 should be
M * v = |dM| * U + (M - |dM|) * (v + dv) ... (9-38a)
OP writes the equation
M * v = dM * U + (M - dM) * (v + dv) ... (9.38b)

Do you see the difference between equations (9.38a) and (9.38b)? They are the same but only if dM in (9.39b) is defined as a positive quantity which is in direct contradiction to its original definition. Is that what confused you?

The equation $$m~\frac{dv}{dt}=(u-v)\frac{dm}{dt}$$ gets around the issue whether $dm$ is positive or negative. The mass transfer to or from the system of interest 1 (e.g the rocket) and the secondary system 2 (e.g. the fuel) is such that $dm_1+dm_2=0.$ Study the derivation in the insight. Which one is positive and which one is negative is irrelevant. In the equation $m$ and $dm$ are $m_1$ and $dm_1$ with the subscripts dropped as explained in the derivation.