# Variable-mass systems

1. Dec 26, 2011

### Undoubtedly0

Newton's second law gives that

$$\sum\mathbf{F} = \frac{d\mathbf{p}}{dt} = \frac{d(m\mathbf{v})}{dt}$$

In a system where mass varies with time, m(t), one would simply think that this would lead to

$$\sum\mathbf{F} = m\frac{d\mathbf{v}}{dt} + \mathbf{v}\frac{dm}{dt}$$

Yet everywhere online I see that this is not the case, (especially here, page 228) says that this is not the case. In fact, it is said that

$$\sum\mathbf F + \mathbf{u} \frac{\mathrm{d} m}{\mathrm{d}t} = m {\mathrm{d} \mathbf v \over \mathrm{d}t}$$

where $\mathbf{u}$ is the velocity of dm, the mass entering or leaving the main body. What then is wrong with $\sum\mathbf{F} = m\frac{d\mathbf{v}}{dt} + \mathbf{v}\frac{dm}{dt}$?

Happy holidays!

2. Dec 26, 2011

### VortexLattice

I took a quick look at the book, and my immediate guess is that they have absorbed a negative sign into either the velocity or the dm/dt. In the case of a rocket, for example, dm/dt is negative, because mass is leaving the rocket and thus the mass of the rocket is decreasing. Also be careful with the velocities of the rocket and propellant. Draw out a simple force diagram with a rocket and its propellant.

3. Dec 27, 2011

### D H

Staff Emeritus
You won't see F=dp/dt in Newton's Principia. You won't see F=ma, either. Newton's Principia is pretty much calculus free.

There's a big problem with using ƩF=dp/dt as the definition of force. This makes force a frame dependent quantity for variable mass systems. This problem vanishes if one uses ƩF=ma. This introduces a new problem. Now the connection with the conservation laws is broken. There are two ways around this. One is to rewrite Newton's second law as ƩFext+u*dm/dt=ma. Another is to define thrust (u*dm/dt) as one of the external forces acting on the system, from which one gets ƩF=ma once again.

One last way around this problem is to use an inertial frame that is instantaneously co-moving with the center of mass of the variable mass system. This makes the v*dm/dt term that arises in ƩF=dp/dt vanish. In this frame, ƩF=dp/dt is equivalent to ƩF=ma, but the connection with the conservation laws is clearly maintained if one uses ƩF=dp/dt.

Last edited: Dec 27, 2011
4. Dec 27, 2011

### Undoubtedly0

Thanks DH. Just a couple of questions. First of all, what do you mean by the conservation laws, and how does this relate to the form of Newton's second law?

Also, I noticed that in many derivations of $\sum\mathbf F + \mathbf{u} \frac{\mathrm{d} m}{\mathrm{d}t} = m {\mathrm{d} \mathbf v \over \mathrm{d}t}$ the author suddenly says that the ejected mass has velocity $\mathbf{v} + d\mathbf{v} - \mathbf{u}$ relative to the ground. However, if the ejected mass is launched with a relative velocity $\mathbf{u}$ to the main body, then is not its velocity relative to the ground simply $\mathbf{v} + d\mathbf{v} + \mathbf{u}$? See here for example.

5. Dec 27, 2011

### D H

Staff Emeritus
You'll see people using v-u as opposed to v+u when they reduce the rocket problem to a 1D problem and treat velocity as a scalar as opposed to a vector. Let v is the velocity of the rocket with respect to the ground, positive upward, and u be the speed (positive) at which the exhaust leaves the rocket relative to the rocket. This means the velocity of the exhaust relative to ground is v-u. Treating everything as a vector leads to v+u.

Regarding the v+u versus v+dv+u imbroglio, that's a silly argument based on a typical physicist abuse of mathematics. If you want to be pedantically correct, the momentum of the exhaust cloud is given by the integral
$$\mathbf p_e(t) = \int_{t_0}^t \dot m_e(\tau)(\mathbf v_r(\tau)+\mathbf u(\tau))\,d\tau$$
Differentiating with respect to time via the Leibniz integral rule yields
$$\dot{\mathbf p}_e(t) = \dot m_e(t)(\mathbf v_r(t)+\mathbf u(t))$$

Edit
One last point with regard to the article cited in post #4. Regardless of the author's opinion, there is an ambiguity with regard to the projectile velocity. Someone buying such a system had better read the specs and read how the spec values were tested. Those specs and the test reports are inevitably classified. A university professor without proper clearance knows naught of the topic on which he is writing.

Last edited: Dec 27, 2011
6. Dec 27, 2011

### Undoubtedly0

Interesting! Yet I noticed that the derivation using F = dp/dt only works using v-u. How then does one obtain $\sum\mathbf F + \mathbf{u} \frac{\mathrm{d} m}{\mathrm{d}t} = m {\mathrm{d} \mathbf v \over \mathrm{d}t}$? My thought would be to proceed exactly as was done here, yet it seems that they reached the same (incorrect) result as I did in my first post. Are they wrong in saying that

$$\mathbf{p}_{\mathrm{initial}} = M\mathbf{v}$$

$$\mathbf{p}_{\mathrm{final}} = (M - dm)(\mathbf{v} + d\mathbf{v}) + dm(\mathbf{v} + d\mathbf{v} + \mathbf{u})$$

or is their problem elsewhere? Thanks again DH!