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When does Newton's Second Law not hold?

by shinwolf14
Tags: hold, newton
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DeltaČ
#19
Jul25-14, 02:23 PM
P: 450
Apparently there is a catch with stevendaryls example, that equation holds for extended objects like a rocket where c.o.m of rocket changes because they lose fuel mass due to their burning to produce exhaust gases.
Jano L.
#20
Jul25-14, 04:18 PM
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P: 1,168
Quote Quote by stevendaryl View Post
So any equation of motion of the form F = ma would presumably tell you that there is no acceleration in the y-direction. But that might not be true. If by "the location" of the object, you mean the location of the center of mass, then the location is changing with time, because when pieces break off, the center of mass of the remaining matter shifts. The equation of motion governing this shift of center of mass would depend on the details of how the crumbling or burning is propagating, and would have nothing much to do with forces in the y-direction.
Interesting example. I did not think about such case, but I think now even here the equation applies. It is all about the meaning of ##\mathbf v##. The derivation of the equation introduces the quantity ##\mathbf v## as

$$
\mathbf v = \frac{\mathbf p}{m}
$$

where ##\mathbf p## is momentum of the remaining body and ##m## is mass of the remaining body. In your example with the cigarette, this ##\mathbf v## is constant so both sides of the equation of motion vanish.

I did said above that ##\mathbf v## in the equation is velocity of the center of mass of the remaining body. Your example shows that sometimes this meaning is not exactly the same as the one given by the definition above.

Luckily in rocket science, the difference is negligible, so we can use center of mass velocity in the equation without making too great an error :-)
Jano L.
#21
Jul25-14, 04:36 PM
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P: 1,168
Now that I think about it, the equation of motion with variable ##m## can also be written in this funny way:

$$
\mathbf F_{ext} + \mathbf F_{exh} = m\frac{d}{dt}\left(\frac{\mathbf p}{m}\right).
$$
If ##m## is not constant, it cannot be cancelled out:-)
DrStupid
#22
Jul25-14, 06:13 PM
P: 482
Quote Quote by Jano L. View Post
For systems of variable mass, the equation

$$
\mathbf F_{ext}=\frac{d\mathbf p}{dt}
$$
with ##\mathbf p =m\mathbf v## is generally invalid.
Is there a difference between ##F## and ##F_{ext}## or are you still talking about the same equation?
Jano L.
#23
Jul25-14, 06:38 PM
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Quote Quote by DrStupid View Post
Is there a difference between ##F## and ##F_{ext}##
I used ##F## as a wildcard for any of the relevant forces, total or external. No matter which of these is used, the equation
$$
force=dp/dt
$$
is not valid if mass changes. Then I used ##F_{ext}## for external force due to bodies other than the departed parts of the body in question and ##F_{exh}## for total force due to departed parts. Alternatively one may include forces due to these parts into ##F_{ext}## and discard the symbol ##F_{exh}##.
stevendaryl
#24
Jul25-14, 07:16 PM
Sci Advisor
P: 2,210
Quote Quote by Jano L. View Post
Interesting example. I did not think about such case, but I think now even here the equation applies. It is all about the meaning of ##\mathbf v##. The derivation of the equation introduces the quantity ##\mathbf v## as

$$
\mathbf v = \frac{\mathbf p}{m}
$$

where ##\mathbf p## is momentum of the remaining body and ##m## is mass of the remaining body. In your example with the cigarette, this ##\mathbf v## is constant so both sides of the equation of motion vanish.
Yes, it's kind of an odd situation. Normally, in Newtonian mechanics,

[itex]\vec{P} = M \frac{d}{dt} \vec{R}[/itex]

where [itex]\vec{R}[/itex] is the position vector for the center of mass. But in the case of variable mass, the left side can be zero while the right side is non-zero.
DeltaČ
#25
Jul25-14, 11:16 PM
P: 450
Hmmm, strange i cant find where the "catch" is but wikipedia explicitly states that v is the c.o.m velocity of the main body... Is wikipedia wrong in this case?

Update: I still think wikipedia is correct. We just have to distinct between the imaginary c.o.m and the material c.o.m. The material c.o.m has always zero velocity in stevendaryl's example.
Jano L.
#26
Jul26-14, 08:02 AM
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P: 1,168
Quote Quote by DeltaČ View Post
Hmmm, strange i cant find where the "catch" is but wikipedia explicitly states that v is the c.o.m velocity of the main body... Is wikipedia wrong in this case?

Update: I still think wikipedia is correct. We just have to distinct between the imaginary c.o.m and the material c.o.m. The material c.o.m has always zero velocity in stevendaryl's example.
Center of mass is usually defined as a geometrical point. If some parts are removed from one side of the body, its center of mass has to move. It appears that here the concept of center of mass is not that useful as for constant mass motion.
stevendaryl
#27
Jul26-14, 08:15 AM
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P: 2,210
Quote Quote by DeltaČ View Post
Hmmm, strange i cant find where the "catch" is but wikipedia explicitly states that v is the c.o.m velocity of the main body... Is wikipedia wrong in this case?

Update: I still think wikipedia is correct. We just have to distinct between the imaginary c.o.m and the material c.o.m. The material c.o.m has always zero velocity in stevendaryl's example.
Well, the definition of what mass constitutes the "object" is subjective. If I have a cigarette whose end is burning, it's my choice whether to consider the smoke and ash part of the cigarette, or not. If I include the smoke and ash, then the burning process doesn't change the center of mass. If I don't include the smoke and ash, then the burning process does change the center of mass.

This really shows why there is something a little fishy about variable mass problems. An object can only have a variable mass (classically, anyway) because we have drawn a completely subjective boundary between what counts as part of the object, and what doesn't. Since that boundary was the creation of our minds, there is really no reason for the boundary to obey any particular physical law for its propagation.
DrStupid
#28
Jul26-14, 08:39 AM
P: 482
Quote Quote by Jano L. View Post
Then I used ##F_{ext}## for external force due to bodies other than the departed parts of the body in question and ##F_{exh}## for total force due to departed parts.
Does that mean you changed your original statement in order to limit it to external forces? That would be acceptable.
DeltaČ
#29
Jul26-14, 09:00 PM
P: 450
Quote Quote by Jano L. View Post
Center of mass is usually defined as a geometrical point. If some parts are removed from one side of the body, its center of mass has to move. It appears that here the concept of center of mass is not that useful as for constant mass motion.
In the case of mass variable systems, the geometric c.o.m does not always correspond (as time passes) to the same unique material point. The geometric c.o.m gains an extra velocity term due to the asymmetry in the way that new matter is lost (or gained). The material c.o.m (that is the point of matter that lies at the spatial coordinates defined by the geometric c.o.m) isnt aware of this asymmetry, its velocity is only due to F_ext+F_exh.

If we have constant mass or there is symmetry (like saying the stick is burning the exact same way from both edges) then the geometric c.o.m corresponds to a unique material point and its velocity equals the velocity of the material c.o.m.
Jano L.
#30
Jul26-14, 11:05 PM
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P: 1,168
Quote Quote by DeltaČ View Post
In the case of mass variable systems, the geometric c.o.m does not always correspond (as time passes) to the same unique material point.
Yes, but this may happen already for systems which do not lose parts. Center of mass is a geometrical point. It does not matter which material point it coincides with. For some systems like a wedding ring the center of mass does not coincide with a material point at all.

If we have constant mass or there is symmetry (like saying the stick is burning the exact same way from both edges) then the geometric c.o.m corresponds to a unique material point and its velocity equals the velocity of the material c.o.m.
For given set of particles, there is only one commonly used concept of center of mass - average of constituents positions weighted by their mass:
$$
\mathbf r_{COM} = \frac{\sum m_k \mathbf r_k}{\sum_k m_k}.
$$
It is just an abstract point. There is no "material center of mass".
DeltaČ
#31
Jul27-14, 11:28 PM
P: 450
Quote Quote by Jano L. View Post
It is just an abstract point. There is no "material center of mass".
And how do you refer to the point of matter that lies in the spatial coordinates of the geometric c.o.m? (there might be the case that there is no matter there but the usual case is that there is).

I ve to say, though i cannot prove it, my intuition tells me that the velocity of the material c.o.m (in the usual case it exists) and the geometric c.o.m is the same in all cases except in the case there is asymmetry in the way that mass is gained or lost.
Jano L.
#32
Jul28-14, 12:49 AM
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P: 1,168
Quote Quote by DeltaČ View Post
And how do you refer to the point of matter that lies in the spatial coordinates of the geometric c.o.m? (there might be the case that there is no matter there but the usual case is that there is).
I do not know any special name for it. The material point is not important.

I ve to say, though i cannot prove it, my intuition tells me that the velocity of the material c.o.m (in the usual case it exists) and the geometric c.o.m is the same in all cases except in the case there is asymmetry in the way that mass is gained or lost.
That is true for rigid bodies. If the parts move with respect to each other, the center of mass of the body may move as well and is not attached to any particular mass point.


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