Emptying a Freight Car

[ThEmptyTree]

https://ocw.mit.edu/courses/physics...2016/readings/MIT8_01F16_chapter12.1_12.3.pdf

Example 12.2

It is basically what you did.

 

[haruspex]

I said it is the momentum of cart + remaining sand in the cart plus the sand that just left the cart at that instant.

What I am trying to get to grips with in your algebra is how you are defining $P(t)$ in general. If it is the momentum of the cart plus remaining sand at time $t$ then necessarily $P(t+\Delta t)$ is the momentum of the cart plus remaining sand at time $t +\Delta t$. You can't arbitrarily define it to be something else.
You probably knew what you meant with your equations but expressed them wrongly.

My method was using the figure in post #10 and writing the momentum before and after the mass $Δm_s$ is released. Then I subtracted those two momentum equations and divided by $Δt$ and took the limit at $Δt →0$ to get a differential equation that can be solved for $v_c(t)$.

Fine, so as I suspected you did not actually use $F=dp/dt$ but instead derived it all from first principles. I see the linked text does the same.

Which still leaves us with the question of whether we can avoid resorting to all this detailed Delta and limit analysis and apply a standard differential equation.
As you saw in post #28 it can be done with F=dp/dt, but really the simplest is to use F=ma, where F is the stated applied force and m is the cart with its remaining sand at time t.
In the case of a rocket or of sand being dropped into a moving cart we can add the actual force (rather than a virtual one to account for momentum brought in or removed by the gained or lost mass) that applies. Much less confusing all round.

I am surprised the MIT text teaches such unnecessarily verbose methods. Life's too short.

 

[bob012345]

What I am trying to get to grips with in your algebra is how you are defining $P(t)$ in general. If it is the momentum of the cart plus remaining sand at time $t$ then necessarily $P(t+\Delta t)$ is the momentum of the cart plus remaining sand at time $t +\Delta t$. You can't arbitrarily define it to be something else.
You probably knew what you meant with your equations but expressed them wrongly.

My equations are exactly what the MIT solution did and I meant the same thing they did. I'll try one more time to explain it by showing you exactly what I said in post #12; I used this diagram from @ThEmptyTree.

Please study this diagram. It is the system at times $t$ and $t +Δt$. In this arbitrary slice of time there is one piece at $t$ and two pieces at $t +Δt$. Nothing is being redefined. Note $Δm_c = -Δm_s$.

Starting with the variables in the diagram of post $10$ we can write the system momentum at times $t$ and $t + Δt$;

$$\vec P(t ) = (m_c + m_s -bt)\vec v_c(t)$$ and

$$\vec P(t + Δt) = (m_c + m_s - b(t +Δt))(\vec v_c(t) + Δ\vec v_c) + Δm_s(v_c(t) + Δ\vec v_c)$$

Then we can write $\vec F = \large \frac{Δ\vec P}{Δt} = \large \frac{\vec P(t + Δt) - \vec P(t )}{Δt}$

Now compare that to what MIT used ;
$$\vec P_{sys} (t) = m_c(t) \vec v_c (t)$$ .
$$\vec P_{sys} (t + Δt) = (Δm_s + m (t) + Δm_c)(\vec v_c(t) + Δ \vec v_c )$$
where $Δm_c = −Δm_s$ , $m_c (t + Δt) = m_c (t) + Δm_c$ and $m_c(t) =m_c + m_s − bt$

substituting their notation and rearranging my equations a bit we get;
$$\vec P(t ) = (m_c + m_s -bt)\vec v_c(t)~~ →~~\vec P_{sys}(t ) = m_c(t)\vec v_c(t)$$ and
$$\vec P(t + Δt) = (m_c + m_s - b(t +Δt))(\vec v_c(t) + Δ\vec v_c) + Δm_s(v_c(t) + Δ\vec v_c)$$
which becomes
$$\vec P_{sys}(t + Δt) = (m_c(t) - bΔt))(\vec v_c(t) + Δ\vec v_c) + Δm_s(v_c(t) + Δ\vec v_c)$$
or
$$\vec P_{sys}(t + Δt) = (Δm_s +m_c(t) +Δm_c))(\vec v_c(t) + Δ\vec v_c)$$
since $-bΔt = -Δm_s = Δm_c$

They are the same and they mean the same.

Fine, so as I suspected you did not actually use $F=dp/dt$ but instead derived it all from first principles. I see the linked text does the same.

I tried to tell you I was using the differential form. I thought it was understood I was deriving the relationship for this case.

Which still leaves us with the question of whether we can avoid resorting to all this detailed Delta and limit analysis and apply a standard differential equation.

Yes. I said that in post #14. Nothing in $\vec F = m \vec a$ says $m$ can't be a function of time.

I am surprised the MIT text teaches such unnecessarily verbose methods. Life's too short.

It is necessary like teaching limits to calculus students.

 

[haruspex]

$$\vec P(t ) = (m_c + m_s -bt)\vec v_c(t)$$ and

$$\vec P(t + Δt) = (m_c + m_s - b(t +Δt))(\vec v_c(t) + Δ\vec v_c) + Δm_s(v_c(t) + Δ\vec v_c)$$

Although the definitions of $m_c$ etc. need a little translation, I agree that your equations match those at the link. And they share the sin of not defining what the 'system' is.
You later supplied a definition, but it doesn't match the equations. If you define p(t) as the momentum of the cart and remaining sand at time t then it is just a matter of algebra that $$\vec P(t + Δt) = (m_c + m_s - b(t +Δt))(\vec v_c(t) + Δ\vec v_c)$$

To get your equation and the one at the link, you have to define P(t) in a very awkward way; so awkward I have failed to find a form of words to describe it.

The whole matter can be resolved by accepting that $\vec P(t + Δt)$ is as I have written it above and then observing that the change in the momentum of this system is due not only to the applied force F but also because we are no longer including a bit of momentum that has gone off with the lost sand:
$$\vec P(t + Δt)-\vec P(t) = (m_c + m_s - b(t +Δt))(\vec v_c(t) + Δ\vec v_c)- (m_c + m_s -bt)\vec v_c(t)+\vec F.Δt+bΔt\vec v_c(t)$$

Edit: should be
$$\vec F.Δt-bΔt\vec v_c(t)=\vec P(t + Δt)-\vec P(t) = (m_c + m_s - b(t +Δt))(\vec v_c(t) + Δ\vec v_c)- (m_c + m_s -bt)\vec v_c(t)$$
The hard part is getting the sign right.

Really it is rather a sloppy and confusing text at that link. I'll try to contact the page owner.
As to the use of limit methods in all the exercises, if it is just to practise such methods it seems like overkill. Do it once then show the workaday method.

 

[bob012345]

Although the definitions of $m_c$ etc. need a little translation, I agree that your equations match those at the link. And they share the sin of not defining what the 'system' is.

I disagree with that assessment . We both did define the system.

You later supplied a definition, but it doesn't match the equations.

It does.

If you define p(t) as the momentum of the cart and remaining sand at time t then it is just a matter of algebra that $$\vec P(t + Δt) = (m_c + m_s - b(t +Δt))(\vec v_c(t) + Δ\vec v_c)$$

That ignores the sand that just left the cart. Why do you want to do that? It is part of the problem. Besides, I'm not defining momentum, I am describing it and how it changes for the whole system as I've said before.

To get your equation and the one at the link, you have to define P(t) in a very awkward way; so awkward I have failed to find a form of words to describe it.

Try it please. Maybe it will help diagnose where your issue really is.

The whole matter can be resolved by accepting that $\vec P(t + Δt)$ is as I have written it above and then observing that the change in the momentum of this system is due not only to the applied force F but also because we are no longer including a bit of momentum that has gone off with the lost sand:
$$\vec P(t + Δt)-\vec P(t) = (m_c + m_s - b(t +Δt))(\vec v_c(t) + Δ\vec v_c)- (m_c + m_s -bt)\vec v_c(t)+\vec F.Δt+bΔt\vec v_c(t)$$

Now that is confusing to me. Is that whole thing equal to $FΔt$? Please carry it through and show what it leads to.

 

[haruspex]

It does.

No, you added a term in your expression for $\vec P(t + Δt)$ to represent the sand lost during (t, t + Δt). That is no longer in the cart. If P(t) is defined generally as "the momentum of cart+remaining sand at time t" then by that definition $\vec P(t + Δt)$ is "the momentum of cart+remaining sand at time t+ Δt". It does not include sand lost during (t, t + Δt).

I'm not defining momentum

I didn't say you were defining momentum. What you need is a clear definition of the function p(t) and equations that match that.

Try it please. Maybe it will help diagnose where your issue really is.

I don't think there's a way. You'd need another parameter, something like p(t,t') = momentum at time t' of cart plus the sand that was in it at time t. What you write as p(t+Δt) is then p(t,t+Δt).

Now that is confusing to me. Is that whole thing equal to FΔt? Please carry it through and show what it leads to.

sorry, typo:
$F.Δt-bΔt\vec v_c(t)=\vec P(t + Δt)-\vec P(t) = (m_c + m_s - b(t +Δt))(\vec v_c(t) + Δ\vec v_c)- (m_c + m_s -bt)\vec v_c(t)$
$F.Δt= (m_c + m_s - bt )Δ\vec v_c$.

 

[bob012345]

No, you added a term in your expression for $\vec P(t + Δt)$ to represent the sand lost during (t, t + Δt). That is no longer in the cart. If P(t) is defined generally as "the momentum of cart+remaining sand at time t" then by that definition $\vec P(t + Δt)$ is "the momentum of cart+remaining sand at time t+ Δt". It does not include sand lost during (t, t + Δt).

First, I am happy to continue this dialog as long as this is a dialog and not a one sided teaching moment from your point of view. You keep saying If P(t) is defined generally as "the momentum of cart+remaining sand at time t" ... but I did not define it as such so that is irrelevant to my argument. I used the definition in my text that if you have a system of particles and an external force acting on the system, you need to keep account of all the particles not just some.

sorry, typo:
$F.Δt-bΔt\vec v_c(t)=\vec P(t + Δt)-\vec P(t) = (m_c + m_s - b(t +Δt))(\vec v_c(t) + Δ\vec v_c)- (m_c + m_s -bt)\vec v_c(t)$
$F.Δt= (m_c + m_s - bt )Δ\vec v_c$.

This is just quibbling. You are looking at the cart only and just adding a fictitious force to cancel out the $bΔt \vec v_c(t)$ term from $\vec P(t + Δt) - \vec P(t)$. Assuming no friction and level ground, if $F=0$ your equation suggests there would be a force acting on the cart to maintain the same speed because of the momentum loss from the leaving sand or as the sand falls otherwise the cart would slow down. There is no such force on the cart or the sand because it already possesses momentum and is moving at speed $v_c(t)$. The cart would continue to roll at the same speed as the sand fell out assuming no friction. We get to the same answer so at the very least you should admit both views are equivalent ways to look at the problem rather than continuing to say I and MIT are somehow wrong.

 

[hutchphd]

First, I am happy to continue this dialog as long as this is a dialog and not a one sided teaching moment from your point of view.

Think of it this way. Consider the mass leaving the system: if it is shoved out (like a rocket) there will be a reaction force (on the LHS) from it on the rocket that is proportional to exhaust velocity. Here the exhaust velocity is zero. The system is the remaining total mass. That works for me.

 

[bob012345]

Think of it this way. Consider the mass leaving the system: if it is shoved out (like a rocket) there will be a reaction force (on the LHS) from it on the rocket that is proportional to exhaust velocity. Here the exhaust velocity is zero. The system is the remaining total mass. That works for me.

In most problems either the applied force is causing all the momentum change or the momentum change causes all the force (rocket). Here, Interestingly or perhaps frustratingly, the change in momentum by falling sand does not cause a real force and the applied force does not cause the change due to the sand falling out so it's muddled. Fortunately most problems are clearly one or the other.

 

[jbriggs444]

Think of it this way. Consider the mass leaving the system: if it is shoved out (like a rocket) there will be a reaction force (on the LHS) from it on the rocket that is proportional to exhaust velocity. Here the exhaust velocity is zero. The system is the remaining total mass. That works for me.

One should draw the system boundaries carefully.

Does one put the system boundary between the rocket and the exhaust stream at the inlet into the combustion chamber? If so, mass transfer takes place at [near] zero relative velocity and the pressure of the exhaust gasses on the nozzle bell is an external force.

Does one put the system boundary at the mouth of the exhaust bell? If so, mass transfer takes place at a high relative velocity and the pressure of the exhaust gasses on the nozzle bell is an internal force.

You get the same result either way, of course. But if one fails to decide then a muddle can result.

 

[haruspex]

You keep saying If P(t) is defined generally as "the momentum of cart+remaining sand at time t" ... but I did not define it as such so that is irrelevant to my argument. I used the definition in my text that if you have a system of particles and an external force acting on the system, you need to keep account of all the particles not just some.

The entire argument is over notation.

You can draw a diagram of a system of particles in two different states and define p1 as its momentum in one state and p2 as its momentum in the other.
You can then write $F.\Delta t=p_2-p_1$, and if the expressions for p1 and p2 involve such as $v$ and $v+\Delta v$ then you may well obtain a valid differential equation.

The confusion arises because you denoted the momenta of these states generically as a function of time, p(t). That implies a definition that can be interpreted at any given time t, but I cannot see a way to do that with your p1 and p2.
In any algebraic development, if you write $f(t)=x(t)y(t)$ then it follows that $f(t+\Delta t)=x(t+\Delta t)y(t+\Delta t)$, but your equations do not satisfy that.

 

[hutchphd]

I used the definition in my text that if you have a system of particles and an external force acting on the system, you need to keep account of all the particles not just some.

So in the end your system consists of $m_s$ spewed along the roadway and $m_c$ trundling down the road. The center of mass of your system is now moving backwards relative to the cart in an unpleasant way and your system has received unknown forces from the ground to bring $m_s$ to a stop. If figured correctly you will be able to work out the result. This doesn't make it the best solution, which indeed it is not.

 

[bob012345]

So in the end your system consists of $m_s$ spewed along the roadway and $m_c$ trundling down the road. The center of mass of your system is now moving backwards relative to the cart in an unpleasant way and your system has received unknown forces from the ground to bring $m_s$ to a stop. If figured correctly you will be able to work out the result. This doesn't make it the best solution, which indeed it is not.

The problem at hand consists only of everything at time $t$ and everything at time $t + Δt$. I kept track of that interval only not all $Δt's$.

 

[haruspex]

So in the end your system

In @bob012345's notation, p(t) does not represent the momentum as a function of time of any definable 'system'. That's why we're all confused. See post #46.

 

[bob012345]

In @bob012345's notation, p(t) does not represent the momentum as a function of time of any definable 'system'. That's why we're all confused. See post #46.

You had an issue with my second equation but now you say my notation is wrong? What notation is that? I merely wrote down the momentum before and after based on the figure. Is the figure wrong?

 

[bob012345]

The entire argument is over notation.

You can draw a diagram of a system of particles in two different states and define p1 as its momentum in one state and p2 as its momentum in the other.
You can then write $F.\Delta t=p_2-p_1$, and if the expressions for p1 and p2 involve such as $v$ and $v+\Delta v$ then you may well obtain a valid differential equation.

The confusion arises because you denoted the momenta of these states generically as a function of time, p(t). That implies a definition that can be interpreted at any given time t, but I cannot see a way to do that with your p1 and p2.
In any algebraic development, if you write $f(t)=x(t)y(t)$ then it follows that $f(t+\Delta t)=x(t+\Delta t)y(t+\Delta t)$, but your equations do not satisfy that.

Fine. Pick some time between 0 and $\frac{m_s}{b}$, call it $t'$. Then call the first equation $p_1$ and the second equation at a $Δt$ later $p_2$.

 

[hutchphd]

The problem at hand consists only of everything at time t and everything at time t+Δt. I kept track of that interval only not all Δt′s.

I disagree. Tell me again the definition of your system at any time t

 

[bob012345]

I disagree. Tell me again the definition of your system at any time t

At time $t$ the system is the left hand side of this figure. It consists of the mass of cart plus however much sand is in the cart at that moment traveling with velocity $v_c$. Here it is in all it's glory;

 

[haruspex]

Fine. Pick some time between 0 and $\frac{m_s}{b}$, call it $t'$. Then call the first equation $p_1$ and the second equation at a $Δt$ later $p_2$.

So your working becomes

Let p(t) be the momentum of the system consisting of the cart and its remaining sand at time t:
$\vec P(t ) = (m_c + m_s -bt)\vec v_c(t)$
Thus $\vec P(t + Δt) = (m_c + m_s - b(t +Δt))(\vec v_c(t) + Δ\vec v_c))$.
But the momentum that had been in $\vec P(t )$ is now shared between $\vec P(t + Δt)$ and the sand that was lost in interval $(t, t + Δt)$. That lost momentum is $-Δm_s(v_c(t) + Δ\vec v_c)$. ($Δm_s$ being negative).
[We can simplify to $Δm_sv_c(t)$ since the other part is a second order small quantity.]
Were there no external force, conservation of momentum gives
$\vec P(t ) = \vec P(t + Δt)- Δm_s(v_c(t) + Δ\vec v_c)$, or $\Delta \vec P(t ) = Δm_sv_c(t)$.
But in the same interval, a further $\vec F.\Delta t$ has been injected into the system, so
$\Delta \vec P(t ) = Δm_sv_c(t) +\vec F.\Delta t$.

Looks ok to me.

 

[bob012345]

So your working becomes

Let p(t) be the momentum of the system consisting of the cart and its remaining sand at time t:
$\vec P(t ) = (m_c + m_s -bt)\vec v_c(t)$
Thus $\vec P(t + Δt) = (m_c + m_s - b(t +Δt))(\vec v_c(t) + Δ\vec v_c))$.
But the momentum that had been in $\vec P(t )$ is now shared between $\vec P(t + Δt)$ and the sand that was lost in interval $(t, t + Δt)$. That lost momentum is $-Δm_s(v_c(t) + Δ\vec v_c)$. ($Δm_s$ being negative).
[We can simplify to $Δm_sv_c(t)$ since the other part is a second order small quantity.]
Were there no external force, conservation of momentum gives
$\vec P(t ) = \vec P(t + Δt)- Δm_s(v_c(t) + Δ\vec v_c)$, or $\Delta \vec P(t ) = Δm_sv_c(t)$.
But in the same interval, a further $\vec F.\Delta t$ has been injected into the system, so
$\Delta \vec P(t ) = Δm_sv_c(t) +\vec F.\Delta t$.

Looks ok to me.

I like the conserve momentum approach except for or up to $\vec FΔt$.