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Action without equal and opposite reaction

  1. Sep 10, 2005 #1
    Hi All,

    I just read an interesting website talking about Newtons laws and how they apply in one respect but not in another (not referring to the quantum area). I personally have no opinion one way or another but thought it would be interesting to see what you all think :-).


    - Jason O
  2. jcsd
  3. Sep 10, 2005 #2


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    Well, I think that the author is very wrong.

  4. Sep 10, 2005 #3


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    Yeah, he's wrong. Total momentum is what matters - having part of the momentum angular and part linear is not a contradiction of Newton's first.

    And the device is an application of that error and doesn't work. If the writer chooses to do the math on that device, he'll find that it will not move. Better yet, he could build it and compare its motion to his prediction.
  5. Sep 10, 2005 #4
    The author seems to forget that angular motion has linear motion as a part of it and vice versa. For example, look at the first diagram on the page, the one where a mass of mass M approaches a barbell system where two masses of mass M/2 are separated by a rod. You might make the argument that angular momentum is not conserved in the collision because initially, there's only linear momentum and the final product involves angular momentum. However, there actually is angular momentum involved before the collision. Picture a vector drawn from the center of mass of the system to the approaching mass M. The angle of this vector changes with time, so there is angular momentum (I can't remember the exact equation, but it involves a cross product). It should be reasonable obvious that linear momentum is conserved because when mass M hits the barbell, the barbell will not only spin, it will translate to the right. Finally, you could make the argument that energy that energy that goes into linear momentum doesn't go into angular momentum, but that ignores the fact that linear momentum and angular momentum are two different things that can't be added like translational and rotational energy.
  6. Sep 10, 2005 #5


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    From montalk.net:
    (Note that this is for the first mass being M and the masses at the ends of the barbell each being M/2.)

    This is entirely wrong! The only way for the first mass to lose all of its speed is for the masses at the barbell ends to have mass M. If they have mass M/2, then [itex]v_1'=v_1/3,\ v_2'=(4/3)v_1[/itex], and the speed of the barbell c.of.m is [itex]v_2'/2[/itex]. The fact that the mass that is hit is attached to another by a rod is irrelevant to the initial evaluation of the problem, since as is usual for these idealized cases, the interaction is assumed to be instantaneous.

    Similarly, the cart that can begin moving just by internal interactions inside it, is wrong. The brick that kicks off the left end and then transfers motion to the internal wheel will bring the cart entirely to rest when it does so. Linear momentum is conserved with no qualifications.
  7. Sep 11, 2005 #6


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    A good homework problem is to demonstrate that the contents of the website are pure nonsense. Further, this problem is worked out in many text books; and, the normal way of solving the problem actually agrees with experiment. In fact, spending any more time on such nonsense is a waste of time.

    That the conservation laws work for isolated systems has constantly been verified by experiment, no matter what the components of the system are. They, for example, were crucial in the establishment of the reality of neutrinos.

    Beware of one-trick horses.

    Reilly Atkinson
  8. Sep 11, 2005 #7


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    So tell me, out of curiosity, how in the world did you find such a silly website?

  9. Sep 12, 2005 #8


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  10. Sep 12, 2005 #9
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