1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Force, acceleration, momentum question

  1. Sep 17, 2012 #1
    I was reading through Lewis Epstein's book "Thinking Physics" the other day, and I came across the problem called "Rolling Drain." (This is not a homework problem! It is a question about the definitions of force, momentum, and acceleration.)

    The problem is simple: a cart full of water is rolling horizontally along on a frictionless track, and it is dropping water out a hole in the bottom as it goes. The questions are what happens to the cart's speed, momentum, and kinetic energy. For each quantity, does it a) increase, b) decrease, or c) stay the same.

    Now, the speed is not going to change, because there are no (horizontal) forces acting on the system from outside that would change the speed. However, the mass is decreasing, since the system is losing water. These together mean the momentum and kinetic energy must decrease, since mass decreases while speed stays the same. Right?

    My trouble comes when I try to think of this problem in terms of forces. On the one hand, it seems the system cannot accelerate, because no outside forces act on it. On the other hand, the system's momentum is changing, and according to NL2, force is defined as a change in momentum! So there MUST be a force here, right?

    Typically we first learn NL2 as F = m(dv/dt), since for most problems mass is constant. But for this problem, velocity is constant while mass changes: F = v(dm/dt). (Or at least I think; maybe I am wrong and this is my problem!) This is just as much a force as the first one, right? But I cannot think how to represent this force to myself in the context of the problem. Is it true that force as 'change in momentum' and force as 'cause of acceleration' are distinct? Should I not think of the quantity v(dm/dt) as a force at all?

    Can anyone help to clarify all this for me? (I am quite familiar with the Lagrangian and Hamiltonian formulations of mechanical principles, so I would also welcome any light that could be shed from that perspective, also. But I doubt that will be necessary!)
     
  2. jcsd
  3. Sep 17, 2012 #2
    What are you referring to when you say "the system's momentum is changing"? It seems to me you're defining the system to be the cart plus the remaining water within the cart. Using this definition you are deducing that there is a momentum change (in particular, a horizontal momentum change), and that there must therefore be a horizontal force.

    The error in this logic is that your definition of the system changes as water leaks out of the bottom of the cart. The momentum of your system is changing not because of a force, but because you're continuously changing what you mean by the "system".


    If a train is rolling along a track at 10mph and someone disconnects the rearmost carriage, does the momentum of the train decrease? No, because the "train" is still moving at 10mph, whether it has been separated or not.
     
  4. Sep 17, 2012 #3
    Okay, I think you've spotted my confusion. But just to make sure I understand, if I take the car/water system and keep this definition consistently, then once the water falls, the system's center of mass moves to the right, which makes sense, because there is a right-ward directed force on the water from the Earth once the water hits ground. So the force I'm looking for is the force of the ground on the water.

    And as for velocity, although the velocity of the cart-plus-remaining-water never changes, the velocity of the whole system does change, because the system has to include the lost water, whose velocity of course did change in response to the force of the Earth.

    Is this correct?
     
  5. Sep 17, 2012 #4

    Ken G

    User Avatar
    Gold Member

    Yes that's correct. The problem is that even if m is conserved overall (so forget about nuclear reactions and don't consider any relativity), we still can't say F=mdv/dt, not because "m is changing", but simply because the same v won't always apply at any given moment to every part of m. So what you really have is F = d/dt (sum over i of m_i times v_i), subject to sum over i of m_i = a constant (and you can make these continuous integrals if you need to of course). The point is, you only need F when some m changes v, not just when we change the meaning of "the system," as MikeyW said. So your question exposes the fact that F is really about changing v, not about changing m, but when some of the m is changing to v=0, you can kind of pretend that the F is appearing because m is changing-- when in fact, F is appearing because v is changing for some of the m! So I do think you have a nice question there, because it really exposes the flaw in statements like "F can be needed when either m changes or v changes"-- that's just baloney, F is always about changing v, because m is conserved.
     
    Last edited: Sep 17, 2012
  6. Sep 20, 2012 #5

    Philip Wood

    User Avatar
    Gold Member

    Well said, Ken G.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook