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Newton's 2nd law for open mass systems

  1. Mar 31, 2014 #1
    Sorry, but this will be a long post.

    This isn't homework, I'm reviewing physics after many years of neglect.

    Halliday & Resnick, 4th Ed., section on variable mass and rockets, refers the interested reader to an article, "Force, Momentum Change, and Motion," Martin S. Tiersten, Am. J. Phys., Jan. 1969. I'm trying to apply Tiersten's methodology (which seems to me to be more straightforward than other methods) to 2 problems.

    The usual representations of Newton's 2nd law for open mass systems:

    [tex]M\frac{d\vec{v}}{dt} = \sum\vec{F}_{ext.} + (\vec{u}-\vec{v})\frac{dM}{dt} = \sum\vec{F}_{ext.} + \vec{v}_{rel.}\dot{M}, (1)[/tex]
    [tex]\frac{d}{dt}(M\vec{v}) = \sum\vec{F}_{ext.} + \vec{u}\frac{dM}{dt} = \sum\vec{F}_{ext.} + \vec{\phi}_{\text{net mom.in}}, (2)[/tex]
    where M = system mass, [itex]\vec{v}[/itex] = system velocity, [itex]\vec{u}[/itex] = velocity of incoming/outgoing mass at time t.
    Please note:
    1. Tiersten helpfully points out: (let the sums be taken over all of the particles in the system at time t)
    [tex]\vec{v} = \frac{\vec{p}}{M} = \frac{\sum m_i\vec{v}_i}{M}\mbox{. M not constant implies, in general, }\vec{v}\neq\vec{v}_{cm} = \frac{d}{dt}\vec{x}_{cm} = \frac{d}{dt}(\frac{\sum m_i\vec{x}_i}{M})[/tex]
    [tex]\vec{a} = \dot{\vec{v}} = \frac{d}{dt}(\frac{\vec{p}}{M})\mbox{. In general, }\vec{a}\neq\frac{\sum m_i\vec{a}_i}{M} = \frac{\sum\vec{F}_{ext.}}{M}\mbox{ and }\vec{a}\neq\vec{a}_{cm} = \frac{d^2}{dt^2}\vec{x}_{cm}[/tex]
    2. For rockets, [itex]\vec{v}_{rel.}\dot{M}[/itex] is usually called momentum thrust. (1) is Galilean invariant simply because [itex]\vec{u}-\vec{v}[/itex] is a relative velocity and therefore clearly invariant in any inertial frame (M, [itex]\dot{\vec{v}}, \sum\vec{F}_{ext.}[/itex] are clearly invar.).
    3. (2) is frame dependent (clearly [itex]\dot{\vec{p}},\phi[/itex] are frame dependent), though [itex]\dot{\vec{p}} - \vec{\phi}[/itex] isn't.

    The examples given in the article are explained well, but there are 2 from other sources which are confusing me.

    Ex. 1. A force lifts a length L mass m uniform chain at constant speed, v, vertically from a pile. Let N = normal force, F = lifting force, x = length of chain already lifted at time t.
    1. Let S = the system = the entire chain = closed. [itex]\sum F_{ext} = F+N-Mg = \dot{p} - \phi = \frac{d}{dt}((m/L)xv) - 0 = (m/L)v^2\Rightarrow F+N = (m/L)v^2+Mg = constant[/itex]. It seems clear that (a) system closed [itex]\Rightarrow\sum F_{ext} = \dot{p}[/itex] in all frames and therefore [itex]\dot{p}[/itex] is frame invariant and (b) system closed [itex]\Rightarrow\phi = 0[/itex] in all frames.
    2. Now let S = just the pile = open system. [itex]N-(m/L)(L-x)g = \dot{p} - \phi = 0 - 0[/itex] gives, when combined with the above result, the expected [itex]F = (M/L)(v^2+xg)[/itex]. Here's where the confusion begins. [itex]\dot{p} = 0[/itex] because we are observing the system from the pile's restframe and therefore [itex]v_{pile} = 0 \ and \ \dot{v}_{pile} = 0. \ \phi = 0[/itex] because I intuitively know that the rising chain applies no thrust to the pile, i.e., [itex]\phi \ (= u\dot{M})[/itex], when combined with [itex]-v\dot{M}[/itex] (= 0), should give zero thrust. But this seems mathematically insatisfactory because mass leaves the pile at speed u = v, suggesting that [itex]\phi = u\dot{M}\neq0[/itex]. I suppose we could reason this away by observing that the mass doesn't actually leave the pile at any speed, rather it simply disappears, and therefore the speed of the disappearing mass will always be zero relative to the pile. But let us now keep the same open system = just the pile, but observe it from another inertial frame, say the restframe of the rising chain. Now [itex]\dot{p}[/itex] seems to be [itex]\neq0[/itex] since [itex]v_{pile}\neq0[/itex], i.e., [itex]\dot{p} = -v\dot{M}[/itex]. And if we remain consistent, insisting that the speed of the disappearing mass is zero relative to the pile in all inertial frames, then [itex]\phi[/itex] still = 0, giving N-(m/L)(L-x)g = [itex]-v\dot{M}[/itex], which is incorrect.

    We can't have it both ways. Either the speed of the disappearing mass relative to the pile is zero in all frames ([itex]\Rightarrow\phi = 0[/itex] in all frames [itex]\Rightarrow\sum F_{ext} = N-(m/L)(L-x)g = \dot{p}\neq0[/itex] in all frames other than the pile's restframe) or the speed of the disappearing mass relative to the pile is not zero in all frames ([itex]\Rightarrow u_{\text{disappearing mass}}\neq0[/itex] in the pile's restframe [itex]\Rightarrow\phi\neq0[/itex] in the pile's restframe [itex]\Rightarrow\sum F_{ext} = N-(m/L)(L-x)g\neq0[/itex] in the pile's restframe). Either way implies [itex]\sum F_{ext}\neq0[/itex] in some inertial frame.

    Ex. 2. Halliday & Resnick, 4th ed., ch.9, problem 55. "To keep a conveyor belt moving when it transports luggage requires a greater driving force than for an empty belt. What additional driving force is needed if the belt moves at a constant speed of 1.5 m/s and the rate at which luggage is placed on one end of the belt and removed at the other end is 20 kg/s? Assume that the luggage is dropped vertically onto the belt; persons removing luggage grab hold of it and bring it to rest relative to themselves before lifting it off the belt."

    The intuitive and correct answer is [itex]2v\dot{M}[/itex] since if no mass ever left the belt, the driving force would = [itex]v\dot{M}[/itex]. Halliday & Resnick evidently agree, since the answer provided in the back of the book is 60N. But I can't see the exact steps in deriving this solution.

    And forcing this example to conform to the constraints of [itex]\sum F_{ext} = \dot{p} - \phi[/itex] is giving me the same trouble as Ex. 1. I prefer to think of this problem as sand instead of luggage, since finer particles make it easier to define the system. Therefore assume sand falls vertically onto the belt, moves a certain distance at constant speed, then hits a vertical wall, which stops the sand while it's still on the conveyor belt. But the sand doesn't accumulate in a pile, rather it falls vertically through a hole.

    1. Let S = a closed system = a fixed # of sand grains for a time interval [itex]\Delta t[/itex] during the belt's steady state operation. Assume the system boundaries extend a negligible hight above and below the sand on the belt. Therefore the force exerted by the belt-sand on the still-falling input sand, which accelerates the falling sand from 0 to v, and the reaction force of the falling sand on the belt-sand comprise an internal action/reaction force pair. Similarly for the force of friction between belt and belt-sand. Therefore the external forces on the system are the driving force which moves the belt (call it [itex]F_d[/itex]) and the stopping force of the vertical wall ([itex]F_s[/itex]). Assume the fixed # of sand grains comprise a small amount of still-falling sand, call this [itex]\Delta[/itex]m', + the sand on the belt, m, + a small amount of falling output sand, [itex]\Delta[/itex]m". At time t, [itex]\Delta[/itex]m' is still falling and therefore has no horizontal momentum, m+[itex]\Delta[/itex]m" is on the belt moving with speed v. At time t+[itex]\Delta[/itex]t, [itex]\Delta[/itex]m' has already fallen onto the belt, so that m+[itex]\Delta[/itex]m' is moving on the belt at speed v, and [itex]\Delta[/itex]m" has now fallen through the hole at the end and is falling vertically with no horizontal momentum. Steady state [itex]\Rightarrow\Delta[/itex]m'=[itex]\Delta[/itex]m"[itex]\Rightarrow p(t)=p(t+\Delta t)\Rightarrow\dot{p}=0[/itex]. System is closed [itex]\Rightarrow\sum F_{ext} = F_d - F_s = \dot{P} = 0.[/itex]
    2. Now let S= sand on belt + negligible height above and below = open system with sand flowing in and out. [itex]\dot{M}=0 \ and \ \dot{v}=0\Rightarrow\dot{p}=0[/itex] in all frames. But [itex]\phi = \dot{m}'_{in}v_{x,in}-\dot{m}'_{out}v_{x,out}\neq0[/itex] in any frame other than the ground frame.

    So I need some helpful suggestions on how to apply [itex]\sum F_{ext}=\dot{P}-\phi[/itex] consistently in every inertial frame for these 2 examples, and a well as something more than hand-waving to justify getting from [itex]\sum F_{ext}=\dot{P}-\phi \ to \ F_d=2v\dot{M}[/itex] in ex.2.

    This is nonrelativistic, non-Lagrangian, simple Newtonian physics. Please do not overcomplicate anything.
     
  2. jcsd
  3. Mar 31, 2014 #2

    Andrew Mason

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    Famous last words. Your question needs to be less complicated. You seem to be referring to different systems eg. rocket and conveyor belt. The physics of a rocket is different than a conveyor belt even though they are both variable mass systems.

    If you can reduce your question to a simple sentence it would be most helpful. You have given an awful lot of detail but you have not succinctly stated the problem. If you cannot state the problem we cannot give you an answer.

    AM
     
  4. Apr 4, 2014 #3
    I've talked to a few people and made progress.

    Ignore Ex. 1. The rocket eqn. (1) is derived for systems in which the exhaust and the rocket (the total closed system, if you will) are affected by exactly the same external forces. But in Ex. 1, whether one considers the pile to be the rocket and the rising chain to be the exhaust or vice versa, there exists a force, namely the driving force that is raising the chain, which is exerted only on the rising chain, and not on the pile. Therefore eqns. (1) and (2) don't apply. Therefore the issues I was having with frame dependence were meaningless.

    For Ex.2, there exists a force, that of the luggage handler or the wall (if thinking of the problem in terms of sand) which is exerted only on the "rocket,' i.e., only on the mass on the belt, not on the 'exhausts' (the input and output streams). So, again, eqns. (1) & (2) don't apply directly but must be re-derived from conservation of momentum.

    So that clears up the confusion I was having with frame dependence vs. frame invariance, which was the hard part.

    The only problem I'm still having is showing that the force needed to drive the conveyor belt at constant speed in Ex. 2 is exactly [itex]2\dot{m}v[/itex] (btw, someone asked about the difference between M and m. M is the mass of the system [whether open or closed] at time t, [itex]\dot{m}[/itex] is the mass flowrate into and out of the system, which are equal when the system is at steady state.). Everyone I've talked to seems to agree that the total driving force on the belt is [itex]2\dot{m}v[/itex]. No one seems to be able to show it exactly, tho'.

    Here's what I have:

    Let the system be just the unput flow (falling vertically down, therefore no horiz. mom.) + the mass on the belt at steady state. Therefore no stopping force or output flow. It is straightforward (eqns. (1) & (2) apply) to see that the additional force needed to drive the belt with with mass flowing onto it (as compared to the force needed to drive an empty belt) = [itex]\dot{m}v[/itex]. Call this driving force [itex]F_{in}[/itex]. Therefore [itex]F_{in}=\dot{m}v[/itex].

    Now let the system be just the mass on the belt + the output stream at steady state. Remember that the mass, whether luggage or sand, is stopped, while still on the belt, by an external force, before it leaves with no horiz. momentum. Following a fixed number of particles as they move along the belt, i.e. a closed system, at time t we have m+[itex]\Delta[/itex]m moving on the belt at speed v. At time t+[itex]\Delta[/itex]t we have m still moving on the belt at speed v but [itex]\Delta[/itex]m has been stopped by the external force, call it [itex]F_{wall}[/itex], and has left the belt with no horiz. momentum. Call the force, here, driving the belt, [itex]F_{out}[/itex]. Therefore [itex]\sum F_{ext}=F_{out}-F_{wall}=\dot{p}=\frac{d}{dt}[mv-(m+\Delta m)v]=-\dot{m}v<0[/itex]. Therefore [itex]F_{wall}=F_{out}+\dot{m}v \ and \ F_{wall}>F_{out}[/itex].

    Now let the system be the whole system with input and output streams at steady state. At time t mass m+[itex]\Delta m_{out}[/itex] is on the belt moving at speed v, [itex]\Delta m_{in}[/itex] is entering vertically with no horiz. mom., and at time t+[itex]\Delta[/itex]t, m+[itex]\Delta m_{in}[/itex] is moving on the belt at speed v and [itex]\Delta m_{out}[/itex] is falling with no horiz. momentum. Steady state [itex]\Rightarrow\Delta m_{in}=\Delta m_{out}=\Delta m[/itex]. Therefore [itex]\sum F_{ext}=F_{in}+F_{out}-F_{wall}=\dot{m}v+F_{out}-(F_{out}+\dot{m}v)=0=\dot{p}=\frac{d}{dt}[(m+\Delta m)v-(m+\Delta m)v]=0[/itex]. But we can't solve for [itex]F_{out}[/itex] since it cancels.

    Now let the system be the whole system but with no conveyor belt and with everything floating out in space in an inertial frame far from any external forces (except for [itex]F_{wall}[/itex], clearly). It should be clear that the only driving force necessary here is that needed to accelerate the incoming particles from 0 to v, which then move at constant v to the wall, which stops them. Therefore clearly [itex]F_{\text{wall in space}}=\dot{m}v<F_{wall}[/itex] since [itex]F_{wall}[/itex] must also overcome the frictional force between the belt and the mass.

    The best hand-waving argument I can come up with is that [itex]F_{\text{total driving force}}=F_{in}+F_{out}=\dot{m}v+F_{out}=2\dot{m}v[/itex] since that would mean [itex]F_{wall}=2\dot{m}v[/itex] which exceeds [itex]F_{\text{wall in space}} \ by \ \dot{m}v[/itex] which seems to make sense since the difference between [itex]F_{wall} \ and \ F_{\text{wall in space}}[/itex] is that [itex]F_{wall}[/itex] must overcome the frictional force between the belt and the mass.

    So all I need is a direct mathematical argument showing that [itex]F_{out}=\dot{m}v[/itex].
     
  5. Apr 4, 2014 #4

    Andrew Mason

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    If the rate at which mass is added, [itex]\dot{m}[/itex], to the conveyor belt moving at velocity v is constant then the change in momentum per unit time is [itex]\dot{\vec{p}} = \dot{m}\vec{v}[/itex] is constant.

    Since [itex]\vec{F} = \dot{\vec{p}}[/itex] then [itex]F = \dot{m}\vec{v}[/itex].

    In your example it appears that the mass that falls off the conveyor belt does not apply a horizontal force to the conveyor belt: the mass that falls off takes its horizontal momentum with it. So the mass that falls off does not impart momentum to the mass on the conveyor belt or the conveyor belt itself. Fout = 0. So all you have to do is worry about the mass that enters the conveyor.


    [itex]\vec{F}_{total} = \vec{F}_{in} = \dot{m}\vec{v}[/itex]

    This assumes that no force is required to maintain the conveyor system at constant speed (frictionless).

    AM
     
  6. Apr 5, 2014 #5
    No. The mass that falls off has no horiz. mom. It is stopped by the wall.
    No. At the end of the belt, the wall pushes on the mass, stopping it. As the mass stops, the mass pushes with friction on the belt and the belt pushes back on the mass. This is [itex]F_{out}[/itex] and [itex]F_{out}\neq0[/itex]. Once the mass has stopped, it falls horizontally from the belt.

    This is equivalent to the luggage handler bringing the luggage to rest relative to themselves, as stated in the problem, before lifting it off the belt.

    Note: friction exists between the belt and the mass else the in-falling mass would not initially accelerate.
    The only forces of interest are:
    1. The in-falling mass is pulled by the belt, and the in-falling mass pulls back on the belt.
    2. The wall pushes on the mass at the end of the belt (and the mass pushes pack on the wall, but this force is not useful here).
    3. As the wall slows the mass, the belt pushes with friction on the mass and the mass pushes back on the belt.
     
  7. Apr 5, 2014 #6

    Andrew Mason

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    ??You will have to tell us what the force of friction is. That is Fout. If there is no friction, Fout=0.

    Why did you make up this problem?

    AM
     
  8. Apr 6, 2014 #7
    Please refer to the OP. I didn't make up this problem.

    From the OP:
    Ex. 2. Halliday & Resnick, 4th ed., ch.9, problem 55. "To keep a conveyor belt moving when it transports luggage requires a greater driving force than for an empty belt. What additional driving force is needed if the belt moves at a constant speed of 1.5 m/s and the rate at which luggage is placed on one end of the belt and removed at the other end is 20 kg/s? Assume that the luggage is dropped vertically onto the belt; persons removing luggage grab hold of it and bring it to rest relative to themselves before lifting it off the belt."

    You may find that a careful reading of the thread will clear up any problems you're having about this problem.
     
  9. Apr 7, 2014 #8

    Andrew Mason

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    If the momentum of each infinitesimal unit of mass dm is reduced to zero before it is removed from the conveyor, the conveyor belt receives an impulse [itex]dp = \dot{m}v dt = dm v[/itex].
    So the force opposing the conveyor belt on removal of the mass is [itex]F_{out} = dp/dt = \dot{m}v[/itex].

    If you add this to the force accelerating the mass the total force is: [itex]\vec{F}_{total} = \vec{F}_{in} + \vec{F}_{out} = 2\dot m\vec{v}[/itex]

    AM
     
  10. Apr 7, 2014 #9

    jbriggs444

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    I disagree. The problem is poorly specified.

    The bags are not fixed to the conveyer. If they were, the force required to decellerate the bags to zero velocity would involve decellerating the entire conveyer (including all the other bags) to zero velocity.

    During the decelleration process, the bag will be sliding backwards on the conveyer. The resulting force might be approximated as a function of the weight of the bag, the coefficient of dynamic friction and the time taken to decellerate the bag. The resulting impulse is not determinable from the givens of the problem.

    For a strong baggage handler and a slippery belt, the impulse delivered to the belt during bag removal will be negligible. Accordingly, the problem should have specified that the bags are brought to rest after being removed from the conveyer, not before.

    If you count the force required to accelerate a bag at the one end (supplied by the conveyer) and also count the force required to decellerate it at the other end (supplied by the baggage handler) then you not doing a consistent accounting. You are adding apples and oranges.
     
    Last edited: Apr 7, 2014
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