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Conservation of Linear/Angular Momentum?

  1. May 9, 2009 #1

    bfr

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    When exactly are they conserved?

    For example, one question involved a person jumping off of a circular platform (that rotated without friction), and when the student jumped off tangentially, the platform started spinning. The question asked which of the following is conserved - angular momentum, linear momentum, and kinetic energy - and only angular momentum ended up being the correct answer. It is clear that linear momentum wouldn't be conserved using the linear speed of the entire platform, but if the instantaneous linear speed of the point on the platform where the student jumped off were used as the speed of the platform, then would linear momentum be conserved? And kinetic energy wasn't conserved because the person and the platform were stuck together, kind of like an inelastic collision?

    I know that linear momentum is conserved in a system without any external forces, and angular momentum is conserved in a system without any external torques, but I'm still not completely sure for situations like the one I mentioned, and when angular momentum and linear momentum are combined in general.
     
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  3. May 10, 2009 #2

    tiny-tim

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    Hi bfr! :smile:

    Good question …

    Conservation of momentum or angular momentum comes from the RHS of good ol' Newton's second law …

    Force = (rate of) change of momentum

    this is a vector equation, so it works fine one component at a time …

    (or torque = (rate of) change of angular momentum) …

    so you choose a component (or a centre of moments) which makes the LHS zero.

    For the turntable, there is always an external force at the axle, and you don't know what direction that is, so you can't choose a component which makes the LHS zero, but you can choose the axle as the point about which you take moments, which makes the external torque about the axle zero, which gives you zero on the RHS also, which is conservation of angular momentum about the axle. :wink:

    (and KE usually isn't conserved anyway)
     
  4. May 10, 2009 #3
    to bfr,
    I think it is natural to be confused, because I do think that the question is a bit obscuring. I mean if one consider the system to be the platform and the person (probably with the air between them, and around them?), then all three of them should be conserved because there is no external force acting on the system.
    I think the reason you are confused, or the point that you are confused is that you define something other than the platform to be your system. I am not quite certain, but it seems that you define your system to be the point that the person jumped away?
    if so, then it is probably not quite true. Without getting into any math, you can think it this way. For that point A, or the system, the person jumps off from that point and exerts some forces onto that point. So for that point and that point only, there is an external force acting on it, therefore the momentum would not conserve.
    (actually, to think about it, now I am more confused. Because by this theory, non of the three should conserve...So take my words with a grain of salt).
     
  5. May 10, 2009 #4
    That's a really crisp insight, and it actually answers your question.

    What forces can the Earth (which is defined to be outside the system) exert on the system? On the axle, whose position does not change, despite the force the jumper exerts on the platform. So, by your own argument quoted above, linear momentum is not conserved.

    What torques can the Earth exert on the system? None, because the platform rotates without friction. So, angular momentum is conserved.

    By the way, when talking about the linear momentum of a rigid body like the spinning platform, the velocity that matters is that of the center of mass, not an arbitrary point like the edge where the jumper leaves the platform.
     
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