# Variable mass dynamics: Particle and Rigid Body

I'm encountering some issues in the understanding of some basic concepts about the dynamics of variable-mass particles and rigid bodies.

For what I found, for example reading On the use and abuse of Newton's Second Law for Variable Mass Problems (Plastino,Muzzio) and also Lectures On Theoretical Physics: Mechanics (Sommerfield -- p28) the second law of Dynamics is not suitable for a variable mass particle; instead you should use the momentum conservation:

## e.g. Rocket:

Applying the conservation of momentum for an isolated system:

$$p_t=mv$$ $$p_{t+\Delta t}=(m-\text{d}m)(v+\text{d}v) +\text{d}m\,(v-u_e)$$

$$\frac{\text{d}p}{\text{d}t}=\frac{mv + mdv -vdm +v\text{d}m -u_edm - mv}{\text{d}t}=0$$ $$0 =\frac{mdv-u_edm}{\text{d}t}$$ $$dv=-u_e\frac{dm}{m}$$

that is the classical Tsiolkovsky rocket equation, that can be integrated $\int_{t_0}^t$: $$\boxed{\Delta v = u_e\ln\frac{m_0}{m}}$$

Where $u_e$ is the velocity of the gases that are coming out from the nozzle.

In other books the same equation is obtained from Newton's Law: $$m\frac{\text{d}v}{\text{d}t} = \sum_i F^{\text{ext}}_i$$ where the external forces are basically just the thrust (simplest case) that is given by: $$T = \dot{m}u_e + A_e(p_e - p_a) = \dot{m}c$$

where $A_e$ is the area of outflow for the nozzle, $p_e$ the outflow pressure, $p_a$ the ambient pressure (hence $c=u_e + \frac{A_e(p_e - p_a)}{\dot{m}}$ is the equivalent velocity)

substituting: $$m\frac{\text{d}v}{\text{d}t} = \dot{m}c$$ being $\dot{m} = -\frac{\text{d}m}{\text{d}t}$ because of the mass loss we get: $$\text{d}v = -c\frac{\text{d}m}{m}$$

that is basically the same equation but with the equivalent velocity instead of the real convective gas velocity. This is the first confusing passage...

## And what about rigid bodies?

The equation would be: $$\frac{\text{d}\mathbf{Q}}{\text{d}t} =\sum_i \mathbf{F_i^\text{ext}}$$ so should I do: $$\frac{\text{d}\mathbf{Q}}{\text{d}t} = m\mathbf{\dot{v}} + \mathbf{v}\dot{m}=\sum_i \mathbf{F_i^\text{ext}}$$

or not? I'm really confused about that.

• Your question is on the rocket equation for rigid body dynamics, so you probably don't need the discourse on the Tsiolkovsky equation (a link the Wiki page and the key result probably will suffice). Nov 20, 2014 at 1:46
• @KyleKanos I was really interested also on the particle equation because I found different derivations as I explained. Nov 20, 2014 at 9:55

Newton's second law originally assumed that the mass was a constant of nature, at least if you write it as F=dp/dt. It will only work with a changing mass if that mass leaves the body at the same speed than the original object. To understand why, just think you have a composite object moving a constant speed. If you now only watch at one half of the object, the mass will be reduced to half, but the speed will stay constant (we are assuming no internal forces so both halves keep moving at the same speed. Now, if both halves interact so that the one at the "front" pushed the one at the back apart ("a digital one step fluid"), you will have an interaction between the two halves, and the right way to describe it is to use the initial speeds and the interaction. Or using that the total moment is a constant, but always considering the mass of each subpart as constant. If you just use the second law with the derivative of the mass, you will get a different (and incorrect) result. Your last equation for rigid bodies is incorrect (in the sense of non-physical). The correct one is:

$\frac{\text{d}\mathbf{Q}}{\text{d}t} = m\mathbf{\dot{v}} =\sum_i \mathbf{F_i^\text{ext}}$

Because in Newtonian mechanics mass is a constant for a rigid body. I do not have any references beyond my professor telling me that and solving problems (such as the rocket) in both ways and getting different results, with the result using non-variable mass being the correct one. Just think that in nature there is no classical mechanism that allows a rigid body to loose or change mass (unless it is composite and losses some parts). Now, the change of mass due to relativity theory is correct, but Newton laws no longer apply in that case. Just a funny note: Both Plastino's father and Muzzio were professors of mine!

• This might be more useful if you could provide the correct relation OP should use (perhaps even provide a source, if possible). Nov 20, 2014 at 3:19
• So in a rigid body-modeled rocket there's no term about mass variation? I've been looking for this answer for months, It's incredible that nobody examines this problem. All the books I found on missiles consider the mass constant, but this puzzles me because this is a wrong assumption! Nov 20, 2014 at 10:08
• The mass changes, but due to expulsion of mass that carries momentum with it. Second Newton;s law with variable mass do not work for it (I need to do the calculations, but it should work if mass is ejected isotropically). Do you have any online reference? I could tell you if that specific one is wrong or not.
– user65081
Nov 20, 2014 at 14:26
• @julianfernandez I just extracted few pages about that from a book e from a report I'm writing down. The problem is that the book considers the mass constant! And this puzzles me... I higlighted on the report the parts I'm not sure about. Thank you. Nov 23, 2014 at 18:43
• @SolidSnake in the book they assume that the mass loss is minimal, that is, the amount of fuel is negligible, in weight, compared to the rest of the missil. They only consider the effects of the fuel as being an external force that imparts momentum.
– user65081
Nov 25, 2014 at 1:15

Let's address the linear problem first.

Both the approaches you have written here are correct: they are essentially the same approach to slightly different situations. The first is the clearer one because it is thinking in terms of momentum conservation of the whole system, which is unquestionably the clearest way to think about these kinds of problems.

In the first case, you're simply throwing a stream of mass out of the hinder end of a rocket and deriving Tsiolkovsky equation from thinking of the rocket at time $t$, comparing it to what it has become at time $t+\mathrm{d}t$, namely, the diminished rocket plus the separately ejected mass and equating the momentum of the two. In particular, in the first approach, the ejected mass is assumed to have no interaction with the any masses that have been ejected at times before time $t$.

The second case is slightly different. Here the rocket system (rocket and the mass it throws in time $\mathrm{d}t$) is considered together with a fluid that the rocket is steeped in: partially at least (in deep space there is only fluid behind the rocket, namely, formerly ejected exhaust gasses). See below:

The exhaust gasses ejected before time $t$ are still interacting with the rocket + about-to-be-ejected-mass system: the latter exert pressure $p_e$ on the system. So we get the free body diagram shown. There is a nett force $(p_e-p_0)\,A$ on the system, acting to the right of my drawing. So now we take exactly the same approach as in the first problem, but we note that now momentum of the rocket $m$ + about-to-be-ejected $\mathrm{d}m$ is not conserved: it must change by $(p_e-p_0)\,A\,\mathrm{d}t$. Therefore, as before, but now with the nett impulse acting:

$$(m-\mathrm{d}m)(v+\mathrm{d}v) -\mathrm{d}m\,u_e- m\,v = (p_e-p_0)\,A\,\mathrm{d}t$$

which yields your second equation $m\,\dot{v} = \dot{m}\,v + (p_e-p_0)\,A$.

If you bring drag and gravity into the picture, then you add the impulse of these too. So you'd wind up with $m\,\dot{v} = \dot{m}\,v + (p_e-p_0)\,A - \frac{1}{2}\,\rho_a\,C_D\,v^2 - m\,g$ in that case for a rocket flying straight upwards.

I hope you can see that your first approach applies to something like a stream of sand thrown out of the back of something; the second applies to a punctured can flying through the air.

I think, from memory, the $(p_e-p_0)\,A$ term tends to be quite small in comparison with the $\dot{m}\,v$ term, but hopefully a rocket expert can clarify this point (and there is one writing one of your answers).

So now, you should have the confidence to study rigid body problems. If, for example, you have one of those little thruster thingies that look like Maltese crosses on the bow of your spaceship, you're going to draw a free body diagram like the one below

and write down two equations analogous to the above to express the change in the linear and angular momentum during time $\mathrm{d}t$ of the system comprising:

1. The rocket with mass $m-\mathrm{d}m$ and moment of inertia $I-L^2\,\mathrm{d}m$ about the centre of mass and
2. The ejected mass $\mathrm{d}m$ with momentum $\mathrm{d}m\,u_e$ upwards and angular momentum $L\,\mathrm{d}m\,u_e$ anticlockwise about the centre of mass.

noting that now there is a nett change in momentum $(p_e-p_0)\,A\,\mathrm{d}t$ downwards and nett change in angular momentum of $L\,(p_e-p_0)\,A\,\mathrm{d}t$ clockwise about the rocket's centre of mass.

Disclaimer: the following applied to yours truly, the writer of this answer. Whereas one of your other answers to this question is written by a real rocket scientist.

• The term $A_e(p_e-p_a)$ is usually zero for first stages (you optimize the nozzle to eject gases at pressure equal to the ambient). For upper stages this could lead to huge nozzles, too heavy to bring (infinite if $p_a = 0$, in Space). Anyway I don't get how to conciliate this with rigid-body equations (the most general ones $\frac{\text{d}\mathbf{Q}}{\text{d}t}= ...$) Nov 20, 2014 at 12:08
• @SolidSnake: Write down conservation of both momentum and ang. momentum exactly analogously to the first approach and see what you get. Then you can add the moments of the nett impulse $A\,\Delta p$ and its moment $A\,\Delta p$ to the momentum conservation and angular momentum conservation equations, just as I have sketched in the last paragraph. Nov 20, 2014 at 12:25
• I used to think that "nett" was just a misspelling, but I found out today that it's just an outdated version of "net." Nov 20, 2014 at 15:09
• @KyleKanos Having been born in 1964, I guess I'm a bit outdated too! As you know, there are also slight spelling differences between different dialects of English: you see nett quite often here in the media still. Nov 20, 2014 at 21:48
• @KyleKanos Also, when I was very little, I had a great deal of trouble telling the difference between different homophones and homonymity was a nightmare for me so I tended seize on any reason to break two words' homomnymity. Nov 21, 2014 at 1:08