Cylinders filled with liquid; placed on a trolley

[Saitama]

Homework Statement

Two cylinder shaped vessels of cross section A are fixed vertically to a trolley, which was initially at rest. The two vessels are connected with a thin horizontal tube which is equipped with a tap. The distance between the axes of the vessels is L. The vessel in the left hand side is filled with some liquid of density $\rho$; and at this time the total mass of the system at rest is m. After opening the tap what is the speed of the cart at the moment when the speed of the water levels in the vessels is v? (Rolling friction, friction in the bearings and air-drag are negligible.)

(Answer: $vAL\rho /m$)

Homework Equations

The Attempt at a Solution

I think I have to use conservation of linear momentum but I am unsure about how to do that. The motion of fluid in the cylinder confuses me. I need a few hints to get started with.

Thanks!

 

[Simon Bridge]

Oh I see, mass transfers from the LHC to the RHC at a rate that depends on the difference in heights - which makes the cart move the other way ... it's like a continuous version of the problem where you sit at one end of a closed box and throw rotten fruit at the other end where they stick.

 

[voko]

The key to the problem, obviously, is finding the momentum of the liquid in the pipe.

 

[Saitama]

The key to the problem, obviously, is finding the momentum of the liquid in the pipe.

And how should I do that?

 

[voko]

And how should I do that?

Momentum is mass times velocity. Find these.

 

[Saitama]

Momentum is mass times velocity. Find these

The total mass of the system is m. Initially all the water is in the left cylinder. Mass of liquid is $Ah\rho$ where h is the height of the liquid in the left cylinder initially. But I don't think calculating these would help. Also, how do I use $L$ here?

 

[voko]

Again, you need the momentum in the pipe. It is the mass of the liquid in the pipe multiplied by the velocity of the liquid in the pipe.

What is the mass of the liquid in the pipe?

 

[Saitama]

What is the mass of the liquid in the pipe?

I don't have the cross section for the pipe. :(

 

[voko]

I don't have the cross section for the pipe. :(

Assume you do: call it $a$.

 

[Saitama]

Assume you do: call it $a$.

The length of the pipe is $L-2R$ where R is the radius of cylinder.

The mass of liquid in the pipe is $\rho a(L-2R)$. Momentum is $\rho a(L-2R)v$.

 

[voko]

The length of the pipe is $L-2R$ where R is the radius of cylinder.

I believe you are supposed to think that $L$ is much greater than $R$ and ignore the latter. You will have to, anyway, because you do not have enough information to model the behavior of the liquid in the cylinder, and it is not physical to assume it just moves horizontally wall to wall and then goes straight up.

The mass of liquid in the pipe is $\rho a(L-2R)$. Momentum is $\rho a(L-2R)v$.

$v$ is the velocity in the cylinders, not in the pipe.

 

[Saitama]

$v$ is the velocity in the cylinders, not in the pipe.

Sorry about that.

Using the equation of continuity (?),
$$Av=av' \Rightarrow v'=Av/a$$
Hence momentum is $\rho aLv'=\rho ALv$. From conservation of linear momentum, $\rho ALv=mv_f \Rightarrow v_f=\rho ALv/m$. Looks good?

 

[voko]

This agrees with the answer given.

However, I have a reservation. The velocity $v'$ is with respect to the pipe, which is fixed to the trolley. You are required to find the velocity of the trolley with respect to the ground.

 

[Saitama]

This agrees with the answer given.

However, I have a reservation. The velocity $v'$ is with respect to the pipe, which is fixed to the trolley. You are required to find the velocity of the trolley with respect to the ground.

So $v'=v_L-v_P$? where $v_L$ is velocity of liquid and $v_P$ is velocity of pipe with respect to ground. What should I replace $v_P$ with?

 

[voko]

$v_p$ is what you are supposed to find.

 

[Saitama]

$v_p$ is what you are supposed to find.

So what about $v_L$ then?

 

[voko]

Obviously, $v_L = v' + v_P$ :)

The mass of the liquid in the pipe has velocity $v_L$; everything else has velocity $v_P$,

 

[Saitama]

Obviously, $v_L = v' + v_P$ :)

The mass of the liquid in the pipe has velocity $v_L$; everything else has velocity $v_P$,

Conserving linear momentum,
$$\rho aL(v'+v_P)=mv_f$$
Is this what I am supposed to solve?

 

[voko]

$v$ is the vertical velocity in the cylinders; I do not see how it could possibly be in that conservation law. Besides, $m$ is the mass of the entire system, which includes the mass in the pipe, which is traveling at a different velocity, so that cannot be right, either.

 

[Saitama]

$v$ is the vertical velocity in the cylinders; I do not see how it could possibly be in that conservation law.

Woops sorry, I should have used a different symbol.

Besides, $m$ is the mass of the entire system, which includes the mass in the pipe, which is traveling at a different velocity, so that cannot be right, either.

So what am I supposed to do now?

 

[voko]

So what am I supposed to do now?

Correct your mistakes in the equation for conservation of momentum, I would think.

 

[haruspex]

This agrees with the answer given.

However, I have a reservation. The velocity $v'$ is with respect to the pipe, which is fixed to the trolley. You are required to find the velocity of the trolley with respect to the ground.

I don't see the difficulty. Let m' be the mass of the water in the pipe. If it has velocity v' wrt the trolley, and the trolley has velocity u wrt ground, we have m'(v'+u) + (m-m')u = 0, so m'v' + mu = 0.

 

[voko]

I don't see the difficulty. Let m' be the mass of the water in the pipe. If it has velocity v' wrt the trolley, and the trolley has velocity u wrt ground, we have m'(v'+u) + (m-m')u = 0, so m'v' + mu = 0.

The difficulty was that Pranav was having difficulty rolling out this argument, which I hoped he would in the end, but you ruined the party :)

 

[haruspex]

The difficulty was that Pranav was having difficulty rolling out this argument, which I hoped he would in the end, but you ruined the party :)

Sorry, I read it as if you had found a fault in the given answer.

 

[Saitama]

I don't see the difficulty. Let m' be the mass of the water in the pipe. If it has velocity v' wrt the trolley, and the trolley has velocity u wrt ground, we have m'(v'+u) + (m-m')u = 0, so m'v' + mu = 0.

Thank you voko and haruspex!