Some time ago I came across a problem which might be of interest to the physics.se, I think. The problem sounds like a homework problem, but I think it is not trivial (i am still thinking about it):
Consider a rail tank wagon filled with liquid, say water.
Suppose that at some moment $t=0$, a nozzle is opened at left side of the tank at the bottom. The water jet from the nozzle is directed vertically down. Question:
What is the final velocity of the rail tank wagon after emptying?
Simplifications and assumptions:
Rail tracks lie horizontally, there is no rolling (air) friction, the speed of the water jet from the nozzle is subject to the Torricelli's law, the horizontal cross-section of the tank is a constant, the water surface inside the tank remains horizontal.
Data given:
$M$ (mass of the wagon without water)
$m$ (initial mass of the water)
$S$ (horizontal cross-section of the tank)
$S\gg s$ (cross sectional area of the nozzle)
$\rho$ (density of the water)
$l$ (horizontal distance from the nozzle to the centre of the mass of the wagon with water)
$g$ (gravitational acceleration)
My thinking at the moment is whether dimensional methods can shed light on a way to the solution. One thing is obvious: If $l=0$ then the wagon will not move at all.
Answer
Interesting problem. I think my approach and answer is very close to other posted solutions. I also added a possible scenario. The basic summary is it is the change in the average momentum of the water in the wagon that causes the wagon to move. Requiring the water to distribute it self evenly in the wagon causes this relation:
- average momentum of water in the wagon = $l\times$ mass flow out of wagon
In cases where the wagon has been and forever shall expel water at a constant rate, the wagon stands still. Imagine it being refilled from above its center of mass. You can actually do this same problem with an empty cart being filled from above instead of emptying below. With $l$ being the horizontal point from the wagon's center of mass at which the water falls down.
The wagon does move if there is some fluctuation in the mass flow out of the wagon either by abrupt starts/stops or by running out of water.
Variables
- $t_{c}\to$ time when wagon runs dry
- $l\to$ distance from center of mass of wagon to nozzle, positive $l$ implies nozzle is on the right side of the wagon
- $x(t)\to$ center of mass of wagon
- $x_{cm}(t)\to$ center of mass of everything
- $h(t)\to$ height of water in the container
- $m(t)\to$total mass of the wagon including any water it holds
- $m_{w}\to$ mass of initial water
$m_{c}\to$ mass of the wagon; the c is for the critical point of $m(t)$ when all the water is gone.
Originally c was for container but it makes sense $m(t_{c})=m_c$
Frame of Reference
- $x(0)=0$
- $\dot{x}(0)=0$
Drainage
I'm going to side step the issue of initial conditions for now. I'm going to treat the system as if the nozzle was always open and water has always been running. Only concerned with how a container with a constant cross section, S, would drain.
- Torricelli's Law : Mass Flow =$-\dot{m}(t)$ : Mass of System
$$v(t)=\sqrt{2 g h(t)}$$ $$-\dot{m}(t)=\rho s v(t)$$ $$m(t)=\rho S h(t) + m_{c}$$ Combine to eliminate $m(t)$ and $v(t)$ $$\frac{\dot{h}}{\sqrt{h(t)}}=-\frac{s}{S}\sqrt{2 g}$$
The answer to the differential equation: $$h(t)=h(0){\left(1-t\sqrt{\frac{g {s}^{2}}{2 {S}^{2} h(0)}}\right)}^{2}$$ $$h(t)=h(0){\left(1-\frac{t}{t_{c}}\right)}^{2}$$ where $t_{c}=\sqrt{\frac{2 {S}^{2} h(0)}{g {s}^{2}}}$ and $h(t>t_c)=0$
from there we get $m(t)$: $$m(t)=\rho S h(0) {(1-\frac{t}{t_{c}})}^{2} + m_{c}$$ $$m(t)=m_{w} {(1-\frac{t}{t_{c}})}^{2} + m_{c}$$ and for $m(t>t_{c})$ is simply $m_{c}$, the mass of the wagon
Center of Mass
In order to find the center of mass we will account for all of it. At $t=0$, $x_{cm}(0)=x(0)$=0 since all the mass is in the wagon and we assumed equally distributed.
- The Wagon and its contents $$m(t)x(t)$$
- Water that has left the wagon
If water leaves the the wagon at $t=\tau$, then it will have speed $\dot{x}(\tau)$. Therefore its location is $f(t,\tau)$: $$f(t,\tau) = l+x(\tau)+\dot{x}(\tau)(t-\tau)$$ Then we just integrate to get their contributions. We get their infinitesimal masses from our mass flow: $$\int_0^t f(t,\tau) [-\dot{m}(\tau)]d\tau$$
- Combine $$m(0)x_{cm}(t)=m(t)x(t)-\int_0^t f(t,\tau)\dot{m}(\tau)d\tau$$
Differentiating gives us: $$m(0)\dot{x_{cm}}(t)=\dot{m}(t)x(t)+m(t)\dot{x}(t)-f(t,t)\dot{m}(t)-\int_0^t \frac{df(t,\tau)}{dt}\dot{m}(\tau)d\tau$$
Simplifying: $$f(t,t)=x(t)+ l$$ $$\frac{df(t,\tau)}{dt}=\dot{x}(\tau)$$
Integration by parts: $$\int_0^t\dot{m}(\tau)\dot{x}(\tau)d\tau=m(t)\dot{x}(t)-\int_0^tm(\tau)\ddot{x}(\tau)d\tau$$
Repalce: $$m(0)\dot{x_{cm}}(t)=\dot{m}(t)x(t)+m(t)\dot{x}(t)-\dot{m}(t)(x(t)+ l)-m(t)\dot{x}(t)+\int_0^tm(\tau)\ddot{x}(\tau)d\tau$$
Explanation - In order these terms stand for:
- mass dissapearing from wagon at the center of mass
- momentum of wagon and its contents
- mass appearing outside of wagon at the nozzle
- last two terms account for momentum of water outside of the wagon
Combining the first and third terms gives us the average momentum the water in the wagon must have to maintain its even distribution horizontally in the container. They are not evidence for instantaneous dissapearance from the center and reappearance at the nozzle.
Result: $$m(0)\dot{x_{cm}}(t)=-\dot{m}(t) l+\int_0^tm(\tau)\ddot{x}(\tau)d\tau$$
where: $$m(t)=m_{w} {(1-\frac{t}{t_{c}})}^{2} + m_{c}$$
Wagon w/ Brakes
In this scenario, the wagon has been losing water before $t=0$. However the force of the brakes keeps $\dot{x}(t)=0$. At $t=0$ the brakes are released and it is allowed to move. This avoids any instantaneous jump in velocity by the wagon. It also allows $x_{cm}$ to be a non-zero constant after $t=0$.
Setting $t=0$: $$m(0)\dot{x_{cm}}(0)=-\dot{m}(0) l+\int_0^0m(\tau)\ddot{x}(\tau)d\tau$$ $$m(0)\dot{x_{cm}}(0)=-\dot{m}(0) l$$ $$\dot{x_{cm}}(0)=-\frac{\dot{m}(0)}{m(0)} l$$ $$\dot{x_{cm}}(0)=\frac{2 l m_w}{t_c m(0)}$$
For $t>0$ there is no force from the brakes: $$\ddot{x_{cm}}(t\ge0)=0$$ $$\dot{x_{cm}}(t\ge0)=\frac{2 l m_w}{t_c m(0)}$$
In other words in this situation at $t=0$ the momentum of the whole system matches that of the water in side the wagon. The only question now is as time evolves how is that momentum transfered to the wagon and water leaving the moving wagon.
Differentiate the system's momentum: $$m(0)\ddot{x_{cm}}(t)=-\ddot{m}(t) l+\frac{d}{d t}\int_0^tm(\tau)\ddot{x}(\tau)d\tau$$ $$0=-\ddot{m}(t)l+m(t)\ddot{x}(t)$$ $$\ddot{x}(t)=\frac{\ddot{m}(t)l}{m(t)}$$
Physical Considerations
Therefore we have a simple system as long as $\ddot{m}(t)$ is continuous. The physical explanation is that if we abruptly closed the nozzle the water in the wagon does not come to an immediate stop relative to the wagon. It sloshes around and after a certain relaxation time redistributes its momentum to the system as a whole. Similarly with the quick turn on, the water in the container can't just gain an average momentum to match $-\dot{m}(t)l$. Again there must be some relaxation time for the water to hit that equilibrium where it can evenly distribute itself in the wagon. It is not that these situations are impossible but that my equations would not take into account these relaxation times.
My situation just avoids that. The water in the wagon has already hit some equilibrium before $t=0$. Also having the water move under its own weight provides a slow turn off.
Velocity of Wagon
Combining the results from previous sections: $$\ddot{x}(t)=\frac{2\frac{m_w}{{t_c}^2}l}{m_{w} {(1-\frac{t}{t_{c}})}^{2} + m_{c}}$$ $$\ddot{x}(t)=\frac{2 l m_w}{{t_c}^2 m_c}{\left[\frac{m_w}{m_c}{(1-\frac{t}{t_c})}^{2}+1\right]}^{-1}$$
$$\int\frac{du}{1+u^2}=\arctan(u)$$ $$u=\sqrt{\frac{m_w}{m_c}}(1-\frac{t}{t_c})$$ $$\dot{x}(t)=-\frac{2 l}{t_c}\sqrt{\frac{m_w}{m_c}}\int\frac{du}{1+u^2}$$
$$\dot{x}(t)=\frac{2 l}{t_c}\sqrt{\frac{m_w}{m_c}}\left[\arctan\sqrt{\frac{m_w}{m_c}}-arctan\sqrt{\frac{m_w}{m_c}}\left(1-\frac{t}{t_c}\right)\right]$$
Extremely Heavy Wagon: $\sqrt{\frac{m_w}{m_c}}\ll1$ $$\arctan(x)\to x-\frac{1}{3}x^3$$ $$\dot{x}(t_c)=\frac{2 l m_w}{t_c m_c}$$ $$\dot{x_{cm}}(t\ge0)=\frac{2 l m_w}{t_c m(0)}$$
This makes physical sense. The wagon's final momentum is just about equal to our initial momentum. The higher order terms would account for the momentum that the dispensed water has.
Regular Wagon: $\sqrt{\frac{m_w}{m_c}}\gg1$ $$\arctan(x)\to \frac{\pi}{2}$$ $$\dot{x}(t_c)=\frac{\pi l}{t_c}\sqrt{\frac{m_w}{m_c}}$$ $$\dot{x_{cm}}(t\ge0)=\frac{2 l m_w}{t_c m(0)}$$ $$p_{cm}(t\ge0)=\frac{2}{\pi}\sqrt{\frac{m_w}{m_c}}p(t)$$
This case has the wagon with a significantly smaller portion of the systems momentum.
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