- #36
ThEmptyTree
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https://ocw.mit.edu/courses/physics...2016/readings/MIT8_01F16_chapter12.1_12.3.pdf
Example 12.2
It is basically what you did.
Example 12.2
It is basically what you did.
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.bob012345 said:I said it is the momentum of cart + remaining sand in the cart plus the sand that just left the cart at that instant.
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.bob012345 said: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)##.
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.haruspex said: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.
I tried to tell you I was using the differential form. I thought it was understood I was deriving the relationship for this case.haruspex said: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.
haruspex said: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.
It is necessary like teaching limits to calculus students.haruspex said:I am surprised the MIT text teaches such unnecessarily verbose methods. Life's too short.
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.bob012345 said:$$\vec P(t ) = (m_c + m_s -bt)\vec v_c(t)$$ and
$$\vec P(t + Δt) = (m_c + m_s - b(t +Δt))(\vec v_c(t) + Δ\vec v_c) + Δm_s(v_c(t) + Δ\vec v_c)$$
I disagree with that assessment . We both did define the system.haruspex said: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.
It does.haruspex said:You later supplied a definition, but it doesn't match the equations.
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.haruspex said: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 $$\vec P(t + Δt) = (m_c + m_s - b(t +Δt))(\vec v_c(t) + Δ\vec v_c)$$
Try it please. Maybe it will help diagnose where your issue really is.haruspex said: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.
Now that is confusing to me. Is that whole thing equal to ##FΔt##? Please carry it through and show what it leads to.haruspex said: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:
$$\vec P(t + Δt)-\vec P(t) = (m_c + m_s - b(t +Δt))(\vec v_c(t) + Δ\vec v_c)- (m_c + m_s -bt)\vec v_c(t)+\vec F.Δt+bΔt\vec v_c(t)$$
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).bob012345 said:It does.
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.bob012345 said:I'm not defining momentum
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).bob012345 said:Try it please. Maybe it will help diagnose where your issue really is.
sorry, typo:bob012345 said:Now that is confusing to me. Is that whole thing equal to FΔt? Please carry it through and show what it leads to.
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.haruspex said: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).
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.haruspex said: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 ##.
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 said: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.
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.hutchphd said: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.hutchphd said: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.
The entire argument is over notation.bob012345 said: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.
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 said: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 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##.hutchphd said: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.
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.hutchphd said:So in the end your system
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?haruspex said: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.
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##.haruspex said: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.
I disagree. Tell me again the definition of your system at any time tbob012345 said: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.
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;hutchphd said:I disagree. Tell me again the definition of your system at any time t
So your working becomesbob012345 said: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##.
I like the conserve momentum approach except for or up to ##\vec FΔt##.haruspex said: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.