Newton's Second Law - variable mass case

In summary, the equation F = m(dv/dt) + v(dm/dt) cannot be used if the dust particles are not stationary initially.
  • #1
Nikhil_RG
4
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TL;DR Summary
Can the equation F = m(dv/dt) + v(dm/dt) be used directly in all cases where the mass is variable.
Dear Experts.

In a problem where we need to calculate the acceleration of a satellite in a force free space which sweeps and collects interplanetary dust and a certain rate (dm/dt), I believe that the equation F = m(dv/dt) + v(dm/dt) can be used, by putting F=0 and substituting the function for (dm/dt). Can the same equation be used if the dust particles are not stationary initially?

The doubt arises from the fact that, for a cart carrying sand, moving along a smooth friction less surface, if there is a hole in it through which sand drains out at a constant rate, the velocity of the cart will remain constant. This complies very well with momentum conservation and can be easily verified. But, while trying to use the above equation for the same problem. since (dm/dt) is non zero, (dv/dt) cannot be zero.

I feel that it has to do with the relative velocity of the drained out mass with respect to the cart. But how can we mathematically derive and verify it?
 
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  • #2
[tex]\mathbf{F}=\frac{d\mathbf{p}}{dt}[/tex]
where
[tex]\mathbf{p}=m\mathbf{v}[/tex]
Both m and v can be function of time.
 
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  • #3
The analogy is a boat travelling through a rainstorm. Each raindrop must be accelerated to the velocity of the hull, before it settles in the bilge, increasing the mass of the boat.
 
  • #4
Nikhil_RG said:
TL;DR Summary: Can the equation F = m(dv/dt) + v(dm/dt) be used directly in all cases where the mass is variable.
This equation is not a valid version of Newton's second law. Wikipedia has a good explanation:

https://en.wikipedia.org/wiki/Variable-mass_system

Note the there are many sources online (even some reputable sources) that get this wrong. For example, in the "rocket equation" the second ##v## term is actually the relative velocity of the expellant; and not related to the instantaneous velocity of the rocket.
Nikhil_RG said:
The doubt arises from the fact that, for a cart carrying sand, moving along a smooth friction less surface, if there is a hole in it through which sand drains out at a constant rate, the velocity of the cart will remain constant. This complies very well with momentum conservation and can be easily verified. But, while trying to use the above equation for the same problem. since (dm/dt) is non zero, (dv/dt) cannot be zero.
Correct. This is a good example of why the above equation is not valid.
Nikhil_RG said:
I feel that it has to do with the relative velocity of the drained out mass with respect to the cart. But how can we mathematically derive and verify it?
Correct. See the Wikipedia page. You should forget that equation, and in general keep track of the momentum of all mass entering or leaving a system.

Finally, here is another ongoing thread on the subject:

https://www.physicsforums.com/threa...ss-contribution-to-the-moving-object.1054473/
 
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  • #5
anuttarasammyak said:
[tex]\mathbf{F}=\frac{d\mathbf{p}}{dt}[/tex]
where
[tex]\mathbf{p}=m\mathbf{v}[/tex]
Both m and v can be function of time.
This is wrong!
 
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  • #6
Nikhil_RG said:
TL;DR Summary: Can the equation F = m(dv/dt) + v(dm/dt) be used directly in all cases where the mass is variable.
PS note that equation is frame-dependent (because of the term with ##v##). That means you get a different value for ##F## in different inertial reference frames. It cannot, therefore, be part of standard Newtonian physics, where force, mass and acceleration are independent of the choice of inertial reference frame.
 
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  • #7
Nikhil_RG said:
TL;DR Summary: Can the equation F = m(dv/dt) + v(dm/dt) be used directly in all cases where the mass is variable.
No, only if there's no momentum transported via other stuff, like in the derivation of the rocket equation. The most careful analysis with some illuminating problems is given in A. Sommerfeld, Lectures on Theoretical Physics, vol. 1 (Mechanics).
 
  • #8
Nikhil_RG said:
The doubt arises from the fact that, for a cart carrying sand, moving along a smooth friction less surface, if there is a hole in it through which sand drains out at a constant rate, the velocity of the cart will remain constant. This complies very well with momentum conservation and can be easily verified. But, while trying to use the above equation for the same problem. since (dm/dt) is non zero, (dv/dt) cannot be zero.

I feel that it has to do with the relative velocity of the drained out mass with respect to the cart. But how can we mathematically derive and verify it?
Indeed, you have to take into account that the momentum of the sand in horizontal direction is lost per unit time, i.e., you have
$$\dot{p}=\dot{m} v + m \dot{v}=-\dot{m} v \; \Rightarrow \; \dot{v}=0 \; \Rightarrow \; v=\text{const}.$$
More fundamental is the argument that the total momentum ##P=p+p_{\text{sand}}##, where ##p_{\text{sand}}## is the momentum of the sand still on the cart, is conserved, i.e.,
$$\dot{P} + \dot{p}_{\text{sand}}=m \dot{v} + \dot{m} v - \dot{m} v=0.$$
 
  • #9
PeroK said:
This is wrong!
In familiar rocket case
1690976349829.png


[tex]\frac{d\mathbf{P}}{dt}=M\frac{d\mathbf{V}}{dt}-\frac{dM}{dt}\mathbf{u}=\mathbf{F}_{ext}[/tex]
I assume you say
[tex]\frac{d\mathbf{P}}{dt}\neq M\frac{d\mathbf{V}}{dt}+\frac{dM}{dt}\mathbf{V}\neq\mathbf{F}_{ext}[/tex]
 
  • #10
anuttarasammyak said:
In familiar rocket caseView attachment 329995

[tex]\frac{d\mathbf{P}}{dt}=M\frac{d\mathbf{V}}{dt}-\frac{dM}{dt}\mathbf{u}=\mathbf{F}_{ext}[/tex]
Isn't this along with post #2 ?
There are two different velocities in the rocket equation. The instantaneous velocity of the rocket and the relative velocity of the expellant. That yields a fundamentally different equation, as explained in the Wikipedia page.
 
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  • #11
Thanks.
[tex]p=mv[/tex]
[tex]\frac{dp}{dt}=m\frac{dv}{dt}+v\frac{dm}{dt}[/tex]
is mathematics. I suspect the relation of physics
[tex]F=\frac{dp}{dt}[/tex]
is not universal but has some conditions to stand.
 
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  • #12
anuttarasammyak said:
Thanks.
[tex]p=mv[/tex]
[tex]\frac{dp}{dt}=m\frac{dp}{dt}+v\frac{dm}{dt}[/tex]
is mathematics. I suspect the relation of physics
[tex]F=\frac{dp}{dt}[/tex]
is not universal but has some conditions to stand.
As discussed previously, it's not valid at all because the term in ##v## makes the equation frame dependent.
 
  • #13
anuttarasammyak said:
I suspect the relation of physics
[tex]F=\frac{dp}{dt}[/tex]
is not universal but has some conditions to stand.
It assumes constant mass, if F is to be a classical frame invariant Newtonian force.

Without that assumption, F becomes a more abstract rate of momentum flow, which can be frame dependent.
 
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  • #14
anuttarasammyak said:
Thanks.
[tex]p=mv[/tex]
[tex]\frac{dp}{dt}=m\frac{dp}{dt}+v\frac{dm}{dt}[/tex]
is mathematics. I suspect the relation of physics
[tex]F=\frac{dp}{dt}[/tex]
is not universal but has some conditions to stand.
The assumption that ##F=\mathrm{d} p/\mathrm{d} t## were universally valid also for systems of changing mass. If there are parts of the system moving away from it, then it may carry momentum, and this change in momentum has to be added to the force too.

Take the rocket equation. Let the velocity of the exhaust relative to the rocket (which makes it a frame-indpendent quantity) be ##-u##, then you get
$$\dot{p}=m \dot{v} + \dot{m} v=F_{\text{ext}}+ \dot{m} (v-u),$$
where ##\dot{m}(v-u)## is the momentum taken away from the rocket per unit time.
or the Galilei invariant equation
$$m \dot{v}=F_{\text{ext}}-\dot{m} u.$$
For ##F_{\text{ext}}=0## you get the famouls Tsiolkovsky rocket equation,
$$m \dot{v}=-\dot{m} u \;\Rightarrow \; \frac{\mathrm{d} v}{\mathrm{d} m} = -u/m$$
with the solution
$$v=v_0+u \ln(m_0/m).$$
 
  • #15
A very primitive case to investigate post #2.
A homogenous cylinder is in inertial motion of V in a IFR in the direction along its axis.
We divide its head part and tail part with border moving with time as we like.
[tex]M=M_{head}+M_{tail}[/tex]
[tex]P=MV=M_{head}V+M_{tail}V=P_{head}+P_{tail}[/tex]
[tex]\dot{P}=\dot{MV}=\dot{M}_{head}V+\dot{M}_{tail}V+M\dot{V}[/tex]
Looking at a part, e.g. the head, with no external force the equation of motion expected in #2 is
[tex]\dot{P_{head}}=\dot{M}_{head}V+M_{head}\dot{V}=-\dot{M}_{tail}V-M_{tail}\dot{V} \neq 0[/tex]
so we see even when no external force applied, RHS force applies which shows the fail of the expected equation of motion. With external force applied on head
[tex]\dot{P_{head}}=\dot{M}_{head}V+M_{head}\dot{V}=-\dot{M}_{tail}V-M_{tail}\dot{V}+F_{ext\ on\ head}[/tex]
 
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