Mechanics around a rail tank wagon
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.
OK, that is my second tentative to solve this problem. I think I have a solution this time, thank to the discussion of others in that thread. The solution is $v_{\text{final}}=\sqrt{2gh(0)}\frac{ls}{h(0)S}(1-\frac{\pi}{2})$ if $m\gg M$. This corresponds to a few millimetres per second towards the left for a wagon full of water.
Here is how I've derived it :
Notation
In order not to neglect not negligible contributions, I will pose the problem for a cart of a quite arbitrary shape, before restricting it to our cart.
We have :
- $S(z)$ : section of the cart at altitude $z$
- $h(t)$ : height of water at time $t$
- $l(z)$ : abscissa of the centre of mass (CoM) of the slice of water at altitude $z$ - $M$ : Mass of the empty cart
- $m = \int_0^h(0) dz S(z) \rho$ : initial mass of water
- $\mu(t)$ : remaining mass of water at time $t$
- $f(t)=-dµ/dt > 0$ is the mass flow of water
- $v_v(z,t) < 0$ : vertical speed of the water slice at altitude $z$
- $v_h(z,t)$ : horizontal speed of its CoM.
In the case of the cart, we will have :
- $S(z)$ is constant above the nozzle. Let $\delta+\epsilon$ be the nozzle height. We then have $S(z)=S$ for $z>\delta+\epsilon$. For numerical appplications, we'll suppose a $3\times3\times10$ m³ cart, with $S=30$ m².
- The last part of the nozzle is a pipe of height $\delta\ll h(0)$. In this pipe $S(z<\delta)= s\ll S$. If the output has a 10 cm side, $s=1O^{-2}$ m².
- $h(0) = 3$ m
- Above the nozzle, the CoM of the water is fixed at $l(z>\delta+\epsilon)=0$, while in the lower part, $l(z<\delta)=-l$, wher $l=5$ m.
- I'll assume $M=10^4$ kg, but I've no idea whether it's realistic.
- $\rho = 10^3$ kg·m⁻³
- $m=\rho S h(0) =$ 9·10⁴ kg
- $g=10 m·s⁻²$
Vertical movement of water
In the following, we will assume that the horizontal acceleration $a$ of the cart stays $a\ll g$ during the movement. A nonzero acceleration would induce correction terms proportional to $\frac{a^2}{g^2}$, and we will check that this hypothesis is consistent later. This assumption allows us to neglect any motion of the cart when looking at the movement of water in the cart referential, and then compute $f(t)$, $h(t)$ and $\mu(t)$. We will then use the resulting f this computation to find the horizontal movement of the cart.
The incompressibility of water allows us to write
$$ f(t)=-\rho S(z,t) v_v(z,t) =\rho S(h(t)) \frac{dh}{dt} =- \rho s v_v(0,t) \quad(*)$$
Bernoulli, at altitude $h$ and $0$ gives us
\begin{gather} \left(\frac{dh}{dt}\right)^2 + 2gh = (v_v(0,t))^2 \\ 2gh=(dh/dt)² (\frac{S(h)²}{S(0)}² -1) \end{gather}
In our case, except in the nozzle, $\frac{S(h)^2}{S(0)^2}=\frac{S^2}{s^2}\simeq 10^7$. We will therefore neglect the $-1$ in the following.
This equation has the following solution : $$ h(t)=h(0)(1-t/t_m)^2 \text{ for } t\in[0, t_m]$$
and $h(t>t_m)=0$, with $t_m=\frac Ss \sqrt{2h(0)/g}$. Here $t_m=3\cdot 10^3 \sqrt(6/10) \sim 2000$ s.
We have then $\mu(t)=m (1-t/t_m)^2$ and $f(t)=f(0)(1-t/t_m)$ with $f(0)=\rho s \sqrt{2gh(0)}\sim=10^{-2+3}\sqrt{60}\sim80$ kg·s⁻¹.
Conservation of the horizontal momentum
Now comes the interesting part of the problem, the horizontal movement.
Momenta will be computed in the cart referential ($P^{CR}$) and in the rail referential ($P^RR$).If you look at the water inside the cart, its momentum will be
$$P^{CR}_{\text{water}}=\rho\int_{0}^{h(t)}dz S(z) v_h(z,t)$$
with $v_h(z,t)= dl/dz v_v(z,t)$. From that and the expression $(*)$, we have
$$P^{CR}_{\text{water}}=- f(t) \int_{0}{h(t)}dz dl/dz= f(t) (l(0)-l(h(t))).$$
Going back to the more physical rail-refrential, we have then
$$P^{RR}_{\text{water}}=µ(t)v(t) + f(t) (l(0)-l(h(t)))$$
We also have, for the cart,
$$ P^{RR}_{\text{cart}}=M v(t)$$
As stated in other answers (but not my previous one :-(), one should not forget the momentum of the water which has left the cart in previous time :
$$P^{RR}_{\text{leaked water}}=\int_0^t d\tau f(\tau) v(\tau)$$
Summing these term, together with the momentum conservation, we have :
$$ 0=P^{RR}_{\text{total}}=(M+\mu(t))v(t) + f(t) (l(0)-l(h(t))) + \int_0^t d\tau f(\tau) v(\tau) $$
For example when the cart is empty, $f(t)=0$, $\mu(t)=0$ and the above equations becomes : $$ 0=P^{RR}_{\text{total}}=Mv_{\text{final}} + \int_0^t d\tau f(\tau) v(\tau) $$ The cart can have a final nonzero speed, if its momentum is compensated by the net momentum of the water having left the cart.
Differentiating the momentum conservation relatively to $t$, we obtain,
$$ 0=(M+µ(t))\frac{dv}{dt} - f(t) v(t) + \frac{df}{dt}(l(0)-l(h(t))) - f(t) \frac{dh}{dt} \frac{dl}{dz} + f(t) v(t)$$
This equation can be simplified into
$$ \frac{dv}{dt}=\frac{1}{M+\mu(t)}\left[\frac{df}{dt}[l(h(t))-l(0)] - \frac{dl}{dz}\frac{f(t)^2}{\rho S(h(t))}\right] $$
Knowing $f(t)$ as per the previous section allows us to integrate this equation, at least numerically, for any cart. In the following, we solve the equation for our cart geometry, distinguishing three steps.
Step 1: opening the nozzle
When the nozzle is quickly opened at $t=0$, the cart is full and $\mu=m$ is constant. the equation we have to solve is then $$\frac{dv}{dt}=\frac{1}{M+m}\frac{df}{dt}l-0 $$ from which we easily deduce $$\Delta v = \frac{l\Delta f}{M+m}=\frac{lf(0)}{M+m}.$$ With the numerical values above, this corresponds to a speed of 4 mm·s⁻¹. This movement of the cart compensate the internal acceleration of the water inside the cart towards the nozzle.
As wee will see later, this abrupt speed change is the biggest acceleration taken by the cart. If the nozzle is opened in one second, which is still quickly enough to keep $\mu=m$ approximation valid, the horizontal acceleration $a$ is still small : $\frac{a}{g}=4\cdot10^{-4}$.
Step 2: Emptying the cart above the nozzle
Above the nozzle, we have a constant $l(h)=0$ and the differential equation is $$\frac{dv}{dt}=\frac{l}{M+\mu(t)}\frac{df}{dt}$$.
If the cart is emptied with a constant $f(t)$, it does not accelerate nor slow down, until the f(t) is cut. In that moment the back action is the same in a reverse direction, but with a lower mass. (M instead of M+m). We end therefore with a net speed towards the left of value $lf(1/(M+m)-1/M)$
In the more general case where f slowly decrease to 0, $df/dt <0$, implying a slow down, and indeed a reversal of the speed, since the total mass $M+µ(t)$ decreases.
If we plug into the above equation the values we have for $f(t)$ and $\mu(t)$, we have
$$\frac{dv}{dt}=\frac{lf(0)}{t_m(M+m(1-t/t_m)^2)}=-g\frac{ls^2m}{h(0)S^2M}\frac1{1+\frac mM(1-t/t_m)^2}$$ which can be analytically integrated using $\int dt/(1+t^2)= \arctan t$. We have then $$v(t)-v(0)=-\frac{ls}{h(0)S}\sqrt{2gh(0)}\left[\arctan\sqrt{\frac mM} - \arctan\left(\frac{t_m-t}{t_m}\sqrt{\frac mM}\right)\right]$$.
We have then $$v(t_m)=v(0)-\frac{ls}{hS}\sqrt{2gh(0)}\arctan\sqrt{\frac mM}$$ In the limit $m\gg M$, where the mass of water is larger than the cart mass, $\arctan\sqrt{m/M}\simeq\pi/2$ and $v(0)=\sqrt{2gh(0)}\frac{ls}{h(0)S}$, so that : $$v(t_m)=\sqrt{2gh(0)}\frac{ls}{h(0)S}(1-\frac{\pi}{2})$$
step 3: Showing that the nozzle has no influence, so long at it is small
The problem of the nozzle is the zone where $\frac{dl}{dz}$ is not small. let say that this zone is of height $\epsilon$, above a vertical pipe of height $\delta$, with $\epsilon\ll\delta\ll h(0)$. I have the intuition that the problem is not so dangerous, since the $\propto l/\epsilon$ derivative will be only relevant for a time proportional to $\epsilon$, and the small amount of water involved should keep the corrective term small. But I have nothing more rigorous yet :-(
Qualitative Answer
I think the cart exhibits an extremely surprising behavior. The cart begins by sitting still on the track. The hole is to the left of the center. When the nozzle is opened, water in the cart begins a net flow to the left. The cart, conserving momentum, picks up a velocity to the right. In a steady state, the flow of water would be constant and the cart would move at constant velocity. However, as the flow rate begins to decrease, the velocity of the cart decreases. Eventually, the cart comes to a standstill, then actually reverses directions, moving to the left before the last water falls out. When the last of the water is gone, the cart is coasting to the left. The center of mass of the system never moves, because as the center of mass of the cart moves, the center of mass of the water moves oppositely. Momentum is also conserved, because as the cart picks up momentum, the water picks up opposite momentum. If the water also slides after hitting the track, by the end of the process the water will have a net motion somewhat to the right to compensate the motion to the left of the cart.
Quantitative Answer
Let the cart move at a speed $v$ to the right, and the water move at an average speed $w$ to the right. In general, $v \neq w$ because the water's center of mass is moving relative to the cart. The hole is at $l$. If the hole is on the left then $l$ is negative.
The velocity of the water relative to the cart is $w-v$. This velocity comes from the fact that the water, if it were to continue as it is now, would all move from the center of the cart to the hole, a distance $l$, in a time $m/f$, with $f$ the mass flow rate. Thus the kinematic relation
$$w-v = \frac{lf}{m}$$
Next, we want to conserve momentum. This gives
$$\frac{d}{dt}(Mv + mw) = 0$$
Taking this derivative, we have to keep in mind that $M$ and $m$ are changing because water is flowing out of the cart. $m$ is decreasing at the rate $f$, and $M$ is increasing at the rate $f$ when we think of $M$ as the total mass moving at speed $v$ rather than the mass of the cart.
$$M\dot{v} + m\dot{w} + f(v-w) = 0$$
Physically, the first two terms represent the force on the cart and the force on the water in the cart. The last term represents the force on the water entering the nozzle. Water entering the nozzle goes from $w$ to $v$, thus experiencing acceleration. We have an earlier expression for $v-w$, so plug it in.
$$M\dot{v}+m\dot{w} = \frac{lf^2}{m}$$
I would like to solve for $\dot{v}$. To do this, take the time derivative of the kinematic equation for $w-v$
$$\dot{w} - \dot{v} = \frac{l\dot{f}}{m} + \frac{lf^2}{m^2}$$
These last two equations simplify to
$$\dot{v} = \frac{-l\dot{f}}{M+m}$$
When the flow rate is constant, there is no acceleration. This is plausible because we can imagine watching in a center-of-mass frame where the cart moves to the right and the water moves to the left. The water entering the nozzle feels an acceleration, but the water in the cart is also accelerating, and in the opposite direction. (The water in the cart is accelerating because there is less and less of it, so on average it must move faster to deliver the correct flow rate from the center of the cart to the nozzle.)
Right when we release the nozzle, the flow rate very quickly jumps up, and so the cart quickly picks up speed, too. $m$ is essentially constant over the course of this acceleration, so the cart jumps up to a speed
$$v = -\frac{lf}{M+m}$$
If $m$ were to remain constant, we would find that this relation continues to hold, so that when the water stops flowing, the cart also stops. However, $m$ is not constant; it decreases. When the flow slows to a stop, the acceleration of the cart is now larger because $m$ is smaller. Hence, by the time all the water has left the cart, it is actually moving to the left. This is surprising but necessary - the water is mostly moving to the right because the cart initially moved to the right. The cart must wind up moving left when all is said and done to compensate.
If we suppose the flow rate is constant the entire time, except abruptly beginning and ending (an assumption not in the original problem, which is qualitatively similar but more work to calculate), the final velocity of the cart is
$$v_f = \frac{lfm}{M(M+m)}$$
The water is all flowing at the speed the cart originally jumped to,
$$w_f = -\frac{lf}{M+m}$$
so we see that momentum is conserved.