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When does Newton's Second Law not hold? 
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#19
Jul2514, 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.



#20
Jul2514, 04:18 PM

PF Gold
P: 1,168

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


#21
Jul2514, 04:36 PM

PF Gold
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:) 


#22
Jul2514, 06:13 PM

P: 482




#23
Jul2514, 06:38 PM

PF Gold
P: 1,168

$$ force=dp/dt $$ 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}##. 


#24
Jul2514, 07:16 PM

Sci Advisor
P: 2,210

[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 nonzero. 


#25
Jul2514, 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. 


#26
Jul2614, 08:02 AM

PF Gold
P: 1,168




#27
Jul2614, 08:15 AM

Sci Advisor
P: 2,210

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. 


#28
Jul2614, 08:39 AM

P: 482




#29
Jul2614, 09:00 PM

P: 450

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. 


#30
Jul2614, 11:05 PM

PF Gold
P: 1,168

$$ \mathbf r_{COM} = \frac{\sum m_k \mathbf r_k}{\sum_k m_k}. $$ It is just an abstract point. There is no "material center of mass". 


#31
Jul2714, 11:28 PM

P: 450

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. 


#32
Jul2814, 12:49 AM

PF Gold
P: 1,168




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