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
Frame of Reference
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.
It was already discussed in the comments that a water rocket needs to push "something" out. It is instructive to do the calculation in a little more detail to see where the "energy" goes. For this I will consider the relative share of energy going to the rocket and the "expelled matter" (gas, or water) as a function of the expelled mass. To simplify things, we will assume that all matter is expelled as a single entity with a certain velocity; in reality you might need to integrate, but any inequality that holds for a small amount of expelled matter will hold for the integral over many such amounts.
I will use upper case symbols for quantities relating to the "rest of the" rocket (mass M, velocity V, momentum P - without the expelled mass) and lower case for the expelled matter(m, v, p). From conservation of momentum, $P = -p$ so $M\cdot V = - m\cdot v$. The energy of the rocket $E_r$ and expelled mass $E_m$ will be respectively:
$$E_{r} = \frac12 M V^2 = \frac{P^2}{2M}\\
E_m = \frac12 m v^2 = \frac{p^2}{2m} = \frac{P^2}{2m}$$
It follows that the ratio of (energy in rocket)/(energy in expelled matter) is
$$\frac{E_r}{E_m} = \frac{m}{M}$$
In other words - the lower the mass of the expelled matter, the greater the relative amount of energy it contains. In the limit of "no water", the little bit of air mass contains virtually all the energy.
Best Answer
Great problem! Simple to state, incredibly complex to solve.
I have three suggestions:
I'm not sure that it's a good idea to consider an infinite cylinder. It will have an infinite mass an might just sit there.
Your question is not formulated very precisely. To study the time evolution of a dynamical system you need to fully specify it's initial conditions. In your case you should specify the speed (and position) of the cylinder as well as the state of the water. The second specification might be quite tricky. You need to specify the velocity field of the water as well as the shape of its interface with the air. Moreover you need to make sure that the velocity field is incompressible $\boldsymbol{\nabla}\cdot \boldsymbol{v}=0$, and that the physical boundary conditions are enforced at the interior edge of the cylinder. If you have a non-vanishing viscosity the velocity field must match the velocity of the cylinder when the water touches it. In conclusion, it gets really complicated.
It might be simpler to place you cylinder on an inclined plane. Then your cylinder speeds up at first, but reaches a constant speed once the viscosity of the water is able to dissipate the gravitational potential energy as it is converted to kinetic energy. You can then look for a steady state without having to worry about initial conditions. I guess that depending on the inclination of the plane, there will be different regimes. For small inclinations the speed of the cylinder might saturate and you might find a steady state with the water surface being smooth. Then if your plane is steeper you might (???) find another steady state with turbulent water in the cylinder. Finally for a very steep plane, you will find no steady state. The water will just stick to the edge of the cylinder and roll with it as it accelerates down.
I would start with the third quasi-stationary regime (accelerating cylinder with water turning with it). It looks like it's the simplest. You could try to find out the minimum speed above which it is realised.