Last week, the problem of the conveyor belt accumulating some material at a steady rate $$r$$ was examined. The motor driving the belt kept it moving at a constant velocity. The power expended by the motor was double the power needed to accelerate the material when it hits the belt. The natural question is, why is it exactly double.
This question provoked a lot of discussion in ‘The Physics Teacher’, starting with an article by Wu, which tried to give a straightforward answer to why the value is exactly $$2$$ and also where did the other half of the energy go. For the most part, the argument by Wu, while physically intuitive, is lacking in logical rigor.
The best argument was given by Marcelo Alonso who starts with the varying-mass form of Newton’s law
\[ m \dot {\vec v} + {\dot m} (\vec v – \vec u) = \vec F_{ext} \; . \]
In this equation, $$m$$ is the system mass, which, in the case of the conveyor problem, is the mass of the belt and the material on it at any given time. The vector $$\vec v$$ is the velocity of the system. The quantities $$\dot m = r$$ and $$\vec u$$ are the rate of change of the system mass (more on this later) and the velocity of the mass that either enters of leaves the system, respectively. Justification for this equation will be given below.
The power imparted by the external force $$\vec F_{ext}$$ is
\[ {\mathcal P} = \vec F_{ext} \cdot \vec v = m \, \dot {\vec v} \cdot \vec v + \dot m (\vec v \cdot \vec v – \vec u \cdot \vec v) \; .\]
For the conveyor belt problem, the material enters with a very small velocity $$\vec u$$ that is perpendicular to the system velocity $$\vec v$$ so the last term vanishes. Since the motor keeps $$\vec v$$ constant it can be moved into and out of any total time derivative. The motor power is now
\[ {\mathcal P} = \dot m (\vec v \cdot \vec v) = \frac{d}{dt} \left( m \, \vec v \cdot \vec v \right) = \frac{d}{dt} ( 2 KE ) \; .\]
Viewed in this light, the factor of two is not so much mysterious as accidental. It is a consequence of the special circumstances that $$\vec u \cdot \vec v $$ is zero and $$\dot {\vec v} = 0$$. If the material were introduced to the belt in some other fashion, for example, thrown horizontally on the belt, the ratio of the power provided to the time rate of change of the materials kinetic energy would have been different and its value wouldn’t attract special attention.
All told this is a mostly satisfactory explanation. The only problem is that the terse response Alonso provided in the form of a letter to the editor didn’t give enough details on the varying-mass form of Newton’s law was derived.
Surprisingly, these details are difficult to find in most textbooks. Modern presentations don’t seem to care too much about varying mass situations although they are important in many applications. Systems in motion that suffer a mass change are almost everywhere in modern life. Examples range from cars that consume gasoline as they move, to ice accumulation on or water evaporation from industrial surfaces, to rocket motion and jet propulsion.
So in the interest of completeness, I derive the varying-mass form of Newton’s equations here.
There are two cases to consider: the system mass increases as material is accreted and the system mass decreases as material is ejected. The specific mechanisms of accretion or ejection are not important for what follows.
I will start with the accreting case just before a bit of matter $$\delta m$$ is introduced into the system mass. The initial momentum of the system and lump is
\[ \vec p_i = m \vec v + \delta m \, \vec u \; .\]
Just after this small lump has combined with the system, the final momentum is
\[ \vec p_f = ( m + \delta m)(\vec v + \delta \vec v) \; .\]
The change in momentum is the difference between these two quantities
\[ \Delta \vec p = \vec p_f – \vec p_i = ( m + \delta m)(\vec v + \delta \vec v) – m \vec v – \delta m \, \vec u \; ,\]
which simplifies to
\[ \Delta \vec p = m \, \delta \vec v + \delta m \, \vec v – \delta m \, \vec u \; .\]
This change is momentum is caused by the impulse due to the external force $$\vec F_{ext}$$
\[ \Delta \vec p = \vec F_{ext} \, \delta t \]
acting over the short time $$\delta t$$.
Equating the terms gives
\[ m \, \delta \vec v + \delta m \, \vec v – \delta m \, \vec u = \vec F_{ext} \, \delta t \; , \]
which becomes
\[ m \frac{\delta \vec v}{\delta t} + \frac{\delta m}{\delta t} ( \vec v – \vec u) = \vec F_{ext} \]
when dividing by the small time $$\delta t$$. Before this equation is taken to the limit and derivatives materialize out of deltas, let’s look at the ejection case.
For ejected mass, the system starts with the momentum
\[ \vec p_i = m \vec v \; .\]
After the lump with mass $$\delta m$$ is ejected with velocity $$\vec u$$, the final momentum takes the form
\[ \vec p_f = ( m – \delta m)(\vec v + \delta \vec v) + \delta m \, \vec u \;. \]
Following the same process above, the analogous equation
\[ m \frac {\delta \vec v}{\delta t} – \frac{\delta m}{\delta t} ( \vec v – \vec u) = \vec F_{ext} \]
results.
The two equations look similar with only a small change in sign in front of the $$\frac{\delta m}{\delta t}$$ to hint at the physical difference of mass inflowing in the first and outflowing in the second.
Sign differences can always be problematic and more so when it comes to ‘in’ versus ‘out’. Confusion can easily insert itself in such cases. As an example where careful track of sign conventions is important, consider the presentation of mass-varying systems in Halliday and Resnick. They present the mass ejection case as a preliminary to a discussion of rocket propulsion but they arrive at an equation with the opposite sign from the one derived here. They get this difference by explicitly setting the sign of $$dm/dt$$ opposite to that of $$\delta m/\delta t$$.
A small change in one’s frame of mind and a careful attention to the difference between $$\delta m/\delta t$$ and $$dm/dt$$ is all that is needed to reconcile these differences. To go from the simple ratios of deltas to actual derivatives, note that by the initial construction $$\delta m$$ was a positive quantity, a small bit of matter. The time increment $$\delta t$$ is also positive. So the ratio of these two quantities is also positive. But the rate of change of the system mass $$m$$ can be either positive or negative. Despite the notational similarity between $$\delta m/\delta t$$ and $$dm/dt$$, the mass $$m$$ being addressed is not the same. The mass in the derivative expression is the system mass which can gain or lose mass and so $$dm$$ can be positive or negative.
For the accreted mass case, $$dm/dt = r > 0$$ and the appropriate limit is
\[ m \frac{d \vec v}{d t} + \frac{d m}{dt} ( \vec v – \vec u ) = \vec F_{ext} \; ,\]
or, with an eye towards combining the two cases,
\[ m \frac{d \vec v}{d t} + \left| \frac{d m}{dt} \right| ( \vec v – \vec u ) = \vec F_{ext} \; .\]
For the ejected mass case, $$dm/dt = r < 0$$ and the appropriate limit is \[ m \frac{d \vec v}{d t} + \left| \frac{d m}{dt} \right| ( \vec v - \vec u ) = \vec F_{ext} \; ,\] with the sign on the second term now changing due to the inclusion of the absolute value.