Newton's second law clarification - derivative, constant mass


by nitai108
Tags: clarification, constant, derivative, mass, newton
nitai108
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#1
Jun8-11, 12:37 PM
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Newton's second law states F = d(mv)/dt as this law is valid only for constant mass systems it is also written as F = md(v)/dt.
But let's suppose the mass was not a constant, then the derivative of the law would become F = m'v + mv' (where by ' I mean derivative, m' = dm/dt), would that be right?
If the law is valid only for constant mass why are we writing it inside the derivative and only after that we take it out as a constant?
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piyushkumar
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#2
Jun8-11, 01:12 PM
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I doubt your statement that Newton's second law holds good only for constant mass system. I have successfully used it in past for variable mass systems like sand falling on/from a moving vehicle, and he rocket propulsion problems. I think that is what troubled you.
AlephZero
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Jun8-11, 02:06 PM
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Quote Quote by nitai108 View Post
Newton's second law states F = d(mv)/dt as this law is valid only for constant mass systems it is also written as F = md(v)/dt.
But let's suppose the mass was not a constant, then the derivative of the law would become F = m'v + mv' (where by ' I mean derivative, m' = dm/dt), would that be right?
If the law is valid only for constant mass why are we writing it inside the derivative and only after that we take it out as a constant?
Newtons' second law IS valid for variable mass systems. Whatever (or whoever) gave you the idea that it is not, was just wrong.

You are right that for a variable mass F = m'v + mv'.

There are many practical examples of the second law applied to variable mass systems. For example, a rocket burning fuel, or a conveyor belt being loaded by material falling onto it.

nitai108
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#4
Jun8-11, 02:41 PM
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Newton's second law clarification - derivative, constant mass


Thanks for the replies!
I was reading Feynman Lectures and I searched Newton second law to see if it could be used in a variable mass system, of course wikipedia came first and there are a few notes which state that the law is valid only for constant mass systems (note 20 21 22 http://en.wikipedia.org/wiki/Newton%...te-plastino-19 )
Could you please elaborate why did they write this?
arildno
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Jun8-11, 03:09 PM
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It depends if you call the introduced momentum flux as a force, or if you regard it as..just flux of momentum.

If you regard momentum flux as a force, then Newton's 2 law holds good in variable mass problems.

However, in classical mechanics proper, we generally regard momentum flux as just that, bevcause it is, really simpler to model lots of problems from this perspective.

I wrote this thread some years ago, distinguishing between the "material system" and the "geometrical system".

I advise you to read through it:
http://www.physicsforums.com/showthread.php?t=72176
madness
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#6
Jun8-11, 03:13 PM
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It's only valid for closed (constant mass) systems, otherwise you need to introduce an external force to account for the loss of momentum. Imagine a moving particle which loses mass with time - using Newton's 2nd law it will accelerate to conserve momentum. Now analyse the same problem in the frame where the particle is initially at rest - you will find it stays at rest. This breaks Galilean relativity. The rocket equation is a simple modification which generalises the law to variable mass systems.
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Jun8-11, 03:58 PM
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F = d(mv)/dt

When m = constant, F = m dv/dt because the derivative of a constant * variable = constant * derivative of the variable

When m is not constant, F = m dv/dt + v dm/dt by the Chain Rule of derivatives.

This second equation is fundamentally rocket science.
arildno
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Jun8-11, 04:15 PM
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Quote Quote by SteamKing View Post
F = d(mv)/dt

When m = constant, F = m dv/dt because the derivative of a constant * variable = constant * derivative of the variable

When m is not constant, F = m dv/dt + v dm/dt by the Chain Rule of derivatives.

This second equation is fundamentally rocket science.
And is the "v" you differentiate here the same as the "v" multiplied with dm/dt?
And, is your F=0?
And if not, how do you determine it?
nitai108
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#9
Jun11-11, 10:01 AM
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Quote Quote by arildno View Post
It depends if you call the introduced momentum flux as a force, or if you regard it as..just flux of momentum.

I advise you to read through it:
http://www.physicsforums.com/showthread.php?t=72176
Thanks!


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