Expert: Daniel Mazur - 7/23/2010

Question
QUESTION: Hi Daniel,

I know it might be bothering to answer a typical coursework problem, which is what I have asked in this question, but as I am now studying physics on my own, I have no other source of help. I looked up the formal solution to the problem below and it didn't match with what I had worked out. And that is where my problem lies.

1. The problem statement, all variables and given/known data

A freight car of mass M contains a mass of sand m'. At t = 0 a
constant horizontal force F is applied in the direction of rolling and at
the same time a port in the bottom is opened to let the sand flow out at
constant rate dm/dt. Find the speed of the freight car when all the sand
is gone. Assume the freight car is at rest at t = o.



2. The attempt at a solution

I applied the following equation to the system of "freight car".

Which is,

F = d(P)/dt, where P is the momentum of the freight car at any instant of time.

Now P = mv, where m and v are the total mass and velocity of the freight car at any instant. Now, d(P)/dt should be

v(dm/dt) + m(dv/dt) .................. (right?)

But I looked up the solution, and over there it is instead given,

d(P)/dt = d(mv)/dt = m(t) x (dv/dt)

where m(t) is the mass of the freight car at time 't'.

So essentially, in the solution, the first term on the right side of my equation has been dropped. My question is, why is it so?

What is wrong with my differential equation, d(P)/dt = d(mv)/dt = v(dm/dt) + m(dv/dt) ?


Thanks,
Shikhin

ANSWER: Shikhin,

I am late with my response and I think my answer will not satisfy you much either. One half of the answer to your question is that
F = dP/dt = m(t) dv/dt
completely describes reality completely already and no further term is needed. Note that it is nothing other than writing
a(t) = dv/dt = F/m(t)
which is something none would dare question. Or would you? Instantaneous acceleration a(t) is the ratio of a constant force F and an instantaneous mass m(t).

The second half, i.e. why is your equation F = d(mv)/dt = v(dm/dt) + m(dv/dt) wrong is more opaque. One thing is "It doesn't describe reality", but this is not enough I dare say. Mathematically, your eq is completely correct, so how can it be "outperformed" by something different?!

Personally, I don't have an answer. It may lie in Classical Mechanics, which shows the conservation laws... But it may also lie just in rehearsing the definitions of derivatives and their operations (like this rule: (fg)' = f'g + fg' ). I am really at a loss, sorry.

Good luck with looking for the answer some more.
Daniel


---------- FOLLOW-UP ----------

QUESTION: Hi Daniel,

Thanks a lot for replying.

I had asked the same question on physics forums, and I recieved a response, that, to me seams to be quite possibly the correct answer to this question. I'll paste the link below. Could you please tell me wether this guy who replied to my question on physics forums right or not? I couldn't find much of a fault in his arguments.

The link:

Please scroll down and you'll see a reply by a user named "hikaru1221". Please look at his very first reply. This reply is the fourth last reply from the bottom. here's the link:

http://physicsforums.com/showthread.php?t=416757

It would be great if you could read the above, and tell me wether what the guy said is correct or not.


Thanks a lot,
Shikhin

ANSWER: Hi Shikhin,

hikaru1221 may be correct, but I the answer does not satisfy me. He spends most of the post with answering a different question (at least not the one I understood you were asking), I threw the "term relative to system" model away from the start. But in the part "The problem with the differentiation method (d(mv) = mdv+vdm)..." he is at least heading the same direction with me.

Indeed, dripping of the sand separates the system into two parts, one of which we are describing and the other we're not interested in. I do not agree that pouring of sand into the car would use both terms of your equation. The problem can still be described fully by dv/dt = F/m(t), can it not? What do you think?

Let me suggest something: In an ODR we never work with 3 differentials, always with 2. Equation F = m dv/dt + v dm/dt has 3 and before integration you need to either substitute for dv or for dm (dt should stay, it is the only thing in the denominators). Say I want to keep the dv and dt, so I do the following
F = m dv/dt + v dm/dt = (m + v [dm/dv]) dv/dt
where the [dm/dv] is something like the Jacobi matrix element. As mass has been completely parametrized using variable t and not variable v, m=m(t) independent of v and so dm/dv = 0. And as an immediate consequence... Q.E.D. Now, how do you like that? ;-)

Have fun and let me know!
Daniel



---------- FOLLOW-UP ----------

QUESTION: Hi Daniel,

I dodn't fully understand your method and that is because I lack some technical knowledge as of now. First of I don't know what you used "ODR" for. Second, I don't know anything about the Jacobian matrix.

But what I understood is, that you eliminated dm, and then , you showed that mass m becomes a func. of time and thus dm/dv=0.

But I didn't understand how mass has become a func of time.

I guess that's because of my lack of knowledge about certain things like the jacobian. I'll try to study about the jacobian and try to get back at this from time to time. But if it is possible to reduce the argument down even further, please do that.

If not, no problems. I'll understand it in sometime for sure.

Thanks,
Shikhin

Answer
Hi Shikhin,

ODR is short for the ordinary differential equation. Your problem stated that "port in the bottom is opened to let the sand flow out at constant rate dm/dt" and I took it to mean that dm/dt has no explicit dependence on the speed of the car v. Would you agree?

[q]But I didn't understand how mass has become a func of time.[/q]
If dm/dt = const, then m = m0 - (dm/dt)t = m(t). This is the most basic integration you can have.

Jacobian matrix is just what I showed you: the transform from one set of integration parameters to another. It is most commonly used to transform integration between Cartesian, spherical and cylindrical coordinates. In our case it is used to transform to a common integration variable and thus solve our mystery. This is not something to be explained here, you need matrix calculus and integral calculus theory for that.

Sorry I couldn't be of more help.
Daniel