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.
Best Answer
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!