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Two questions regarding the conservation of linear momentum

  1. May 29, 2014 #1
    Lately I have been thinking about the conservation of linear momentum, and there are two cases that I can think of that seem to violate the conservation principle.

    1. Throwing a rubber ball against a wall. Imagine you threw a rubber ball in what we will call the positive x direction, if the ball were to hit a wall we all know from experience what will happen next, the ball will bounce off the wall. However, the ball started off with linear momentum in the plus x direction and now it has its momentum in the negative x direction. How in this case is momentum conserved?

    2. Imagine you had a spinnng wheel which you gently lower down onto the ground, as soon as it touches the ground it will start moving away from you. Once again, before it touches the ground it has only angular momentum and no linear momentum, yet as soon as it touches the ground it will move away from you and therefore must have linear momentum. Where is the momentum coming from?

    Thank you
  2. jcsd
  3. May 29, 2014 #2
    Direction is not considered in momentum calculation. Momentum = Mass * Velocity. This satisfies the first question.

    Secondly, overall momentum is conserved. Changing "types" of momentum does not violate the Law of Conservation of Momentum.
  4. May 29, 2014 #3
    I disagree on both counts, first of all momentum is a vector, direction does matter, and I'm pretty sure linear and angular momentum are both conserved independently.
  5. May 29, 2014 #4
    You can disagree all you like. Is there such thing as 'negative momentum?' If it suits you, consider the equation for momentum to be such: |v| * |m| = momentum.

    Also, let's pretend that they are conserved independently. You're forgetting that it does in fact have linear momentum, even while spinning. v = rw

    Velocity (v) equals the radius (r) of rotating object times the angular speed (omega).
  6. May 29, 2014 #5


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    Huh? This is patently false. Of course momentum depends on direction, it is a vector quantity, just like the velocity! ##\vec{p}=m\vec{v}##

    @ Hideelo:

    1) The wall (if it's attached to the Earth, then the Earth) will move, just very very little because it's very very massive. This effect will assure momentum conservation between the ball+wall system. You have to consider the wall as well in the calculation for momentum conservation.

    2) In a perfectly symmetrical system, the spinning wheel would not move linearly, but just continue to spin. The fact that real tops do move linearly have to do with the fact that in the real world, we cannot make the initial conditions perfectly symmetric. This is essentially an example of spontaneous symmetry breaking! And again, in order to consider the total linear momentum, one would have to consider the ground into this calculation. The ground moves VERY VERY little, but it does move!
  7. May 29, 2014 #6


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    The wall + Earth receive twice the original momentum of the ball in positive direction, so total linear momentum remains constant.

    Here again the Earth receives the opposite linear momentum that the wheel receives, so total linear momentum remains constant.
  8. May 29, 2014 #7


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    Wrong. Total momentum is a vector and is conserved as a vector, including its direction

    Wrong. Linear and angular momenta are conserved individually not "overall". There is no "overall momentum", because linear and angular momenta have different units and cannot be added.

    Its a vector that might have a negative component.

    Wrong. This is not the equation for momentum.

    Wrong. If the center doesn't move the total linear momentum of the wheel is zero.
    Last edited: May 29, 2014
  9. May 29, 2014 #8
    Okay, yes, we've covered this. Then the rebound is not due to it's momentum, but rather to Newton's third law of motion.

    What? This doesn't quite make sense to me. Also, it does have linear momentum even when only spinning.
  10. May 29, 2014 #9


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    Newton's third law is closely related to momentum conservation.

    If the center doesn't move the total linear momentum of the wheel is zero.
  11. May 29, 2014 #10
    Obviously. But, if there is angular momentum, there is linear momentum. Period.
  12. May 29, 2014 #11


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    Wrong. If it spins in place the total angular momentum of the wheel is not zero, but the total linear momentum of the wheel is zero.
  13. May 29, 2014 #12
    Actually, THAT'S wrong. If it has a linear velocity and a mass, then it has a linear momentum.
  14. May 29, 2014 #13


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    It doesn't. It spins in place.
  15. May 29, 2014 #14
    Yes it does. I believe I posted the equation earlier, but I will repeat myself: V=R*w
  16. May 29, 2014 #15
    Linear velocity equals the radius of the rotating object times the angular velocity.
  17. May 29, 2014 #16
    Therefore, if it has a linear velocity and a mass, then it must have a linear momentum.
  18. May 29, 2014 #17


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    The linear velocity of a point on a rotating object is equal to the vector cross product of the radius vector from the center to that point times the angular velocity of the object.

    That linear velocity is different for every point because it points in a different direction. If you add up the momenta of all the points on a suitably symmetric rotating object, the total comes to zero.
  19. May 29, 2014 #18
    Okay, that makes sense.
  20. May 29, 2014 #19


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    Remember that momentum is only conserved for an isolated system where there are no external forces acting on it. Neither the ball nor the wheel is isolated. The ball has a force from the wall and the wheel has a force from the ground.

    You can include the whole earth in the system to get rid of external forces (or at least ignore them) and then you get momentum conservation.
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