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Derivation of the principle of moments

  1. Mar 25, 2006 #1
    Derivation of the principle of moments....

    i know the principle of moments might seem very obvious at the beginnning... but really, how was it derived or proposed? and by what proof?
     
    Last edited: Mar 26, 2006
  2. jcsd
  3. Mar 25, 2006 #2
    am i not clear about something?
     
  4. Mar 25, 2006 #3
    Conservation of energy.
     
  5. Mar 25, 2006 #4

    ZapperZ

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    It is not conservation of energy. It is conservation of angular momentum.

    Your question is rather vague. Notice that you never bother to define what "t" is. I can only assume you mean "T" as in torque based simply on another assumption by what you mean by "moment".

    The conservation of angular momentum L implies that dL/dt = 0. This allows you to define a quantity called torque when an external force to the system acts in changing the angular momentum. This is an identical procedure used in the linear version that results in F=dp/dt.

    Zz.
     
  6. Mar 25, 2006 #5
    Really? Think twice.
    What are you going to say if I can derieve it using conservation of energy and not using conservation of angular momentum? Can you say only one of us is right? According to you ("It is not conservation of energy"), yes.
     
  7. Mar 25, 2006 #6

    ZapperZ

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    Go right ahead.

    Conservation of energy is independent of conservation of momentum and angular momentum. I can show in one example where one is conserved while the other isn't simultaneously.

    Zz.
     
  8. Mar 25, 2006 #7

    robphy

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  9. Mar 25, 2006 #8

    arildno

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    Hmm..I didn't know Varignon's theorem had a name at all.
    Why has this trivial observation upon the distributivity of the cross product been elevated to the status of a theorem? :confused:

    there are many mysteries in the world..
     
  10. Mar 25, 2006 #9
    Feynman Lectures on Physics, Vol I, Chapter 4.

    Your turn.
     
  11. Mar 25, 2006 #10

    ZapperZ

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    Really! Where exactly is the derivation using conservation of energy? In fact, why don't you use conservation of energy to derive F=ma. And in case you forget Noether's theorem, please show me the symmetry principle associated with conservation of linear/angular momentum, and conservation of energy. Would you care to show me how you would actually COMBINE these two?


    When a spinning object reduces its moment of inertia, it spins faster. Angular momentum is conserved, but rotational energy isn't! This is an analogous situation with an inelastic collision where linear momentum is conserved, but kinetic energy isn't.

    So when rotational energy is NOT a conserved quantity, how do you propose to "derive" the dynamics of that system?

    Zz.
     
  12. Mar 25, 2006 #11
    First of all, as you can see from my previous posts, I didn't say conservation of rotational kinetic energy, I said conservation of energy. The energy conserved, in your example, because one must have to do work in order to change the moment of inertial.

    Now this clashes with what you've said before:
    You said you're going to show an example where angular momentum is conserved while energy is not. But you couldn't.

    I'm too lazy to type the entire thing, but here's some places:
    First of all, Feyman assumed "that there is no such thing as perpetual motion", or "energy is conserved". Then this assumption was applied to a reversible machine (fig 4-2). Then Feyman showed the relation between the lengths of lever arms (which are defined by "principle of moments" for systems that are balanced), using the assumptions that: this's a reversible machine & energy is conserved.

    And here's a spacific system that can build a relation between conservation of angular momentum and energy:

    If F is gradient of a potential, [tex]\vec{(\nabla)} \times \vec{F} = 0[/tex], it is a conserved force, ie it cannot cause a perpetual motion. This also means conservation of energy.

    [tex]\vec{r} \times \vec{F} = \vec{r} \times \vec{\nabla} U[/tex]

    Work is defined as the line integral of force on path r, so if we have a path that is parallel to force, rxF should do no work, since [tex]\vec{\nabla} U \times d\vec{r}[/tex] is 0. When we use such path and force, the total work done is 0.

    Now, for the system described above, [tex]torque = \frac{\vec{dL}}{dt} = \vec{r} \times {\vec{\nabla} U}[/tex]
    But by conservation of energy, we have [tex]\vec{r} \times \vec{F} = \vec{r} \times \vec{\nabla} U = 0[/tex] so [tex]\frac{\vec{dL}}{dt} = 0[/tex]. In other words, L is constant with time.

    In this specific example, angular momentum and conservation of energy is related, for instance.
     
    Last edited: Mar 25, 2006
  13. Mar 25, 2006 #12

    ZapperZ

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    Oh, I didn't know you are going to nitpick over such a thing.

    If you notice, if you simply stick by the conservation of energy laws, you have ZERO ability to predict the new value of an angular momentum upon the change of the moment of inertia. Try it! When I spin on my skates, and when I pull my arms in, you have no way to derive my final angular velocity if you try using conservation of energy. Why? Because the rotational energy isn't conserved AND the fact that you have no ability to predict find how much internal energy my body has expanded simply from looking at the mechanics alone.

    You could do the same with an inelastic collision. The energy of the system has been taken out of the mechanics and transfered elsewhere. And last time I checked, this was a mechanics problem and not a solid state to account for the phonon vibrations in the colliding matter, nor a biology problem to account for the energy expanded in my arm muscles.

    There's something strange here. r x F isn't work. In fact, in the example I gave, work IS done on the system. It isn't zero, because I am displacing the object inwards and I am applying a force radially to pull it in! So dr dot F is NOT zero. It isn't just an angular displacement! This is EXACTLY the reason why the energy IN THE MECHANICS isn't conserved, and that an external energy source (the chemical energy in the body) is being added to the mechanics.

    What you have just described is that there is no net TORQUE onto the system, which is not a contention of debate. If there is a net torque, the system no longer is conservative and the curl is no longer zero.

    However, the very important relation here is the ability to define F as the gradient of the potential field. Again, if we apply Noether's theorem, there is an explicit requirement of how one is able to detect a "net force" of a test object in a field. You require that in a flat space where there is a translational symmetry, dp/dt is conserved unless there is a field gradient. If there is no translational symmetry, dp/dt can be none zero without any field gradient, and the relationship between F and the potential field is no longer valid!

    Every conservation laws reflects an underlying symmetry principle. The conservation laws for momentum and angular momentum are the consequences of different symmetry principle than the conservation laws for energy (or mass+energy). In classical mechanics, these two do not combine.

    Zz.
     
  14. Mar 25, 2006 #13
    rxF is work.
    It's the work done by F, on a path that is perpendicular to r; in this case 0, which also means there's no torque, which is a fact I exploited to show angular momentum remains constant in time.

    Whoops, it seems you're referring to this
    It was clearly a typo, I intended to mean "the energy is not conserved", which is a natural result of "one must have to do work", sorry. (I should've also said inertia, not intertial).

    Since I don't know anything about Noether's theorem (noone has show us anything about that yet - I'm a 2nd year undergrad), I've read some text on it. It involves manifold, QFT, etc which I don't know, so that's my limit. I take field gradient as a fundamental theorem for energy conservation, but if there's something more fundamental, then I can't say a thing.

    However, I do know some SR, and for v/c << 1 (one can say, c goes to infinity or v is too small compared to c to make this condition hold), for this limited case, they should combine in CM too, if they combine in general.
     
  15. Mar 25, 2006 #14

    ZapperZ

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    Where did you learn this? From the Feynman text? What happened with work being defined as

    [tex]W = \int{\vec{F} \cdot \vec{dr}}[/tex]?

    http://arxiv.org/abs/physics/9807044

    There are MANY things you accept as part of the "definition" of various things being used in classical mechanics (Noether theorem is NOT just applied to QFT, etc) that depends on such symmetry principles. I used to teach undergraduate intro physics, and I told my students on the very first day that all they are going to be learning are conservation of momentum and conservation of energy, applied in various disguises. These two are intractable in classical physics.

    Zz.
     
  16. Mar 25, 2006 #15
    It's still alive.
    I still say the work is defined that way. And this's how I could say rxF is 0.
    You may ask, why do I call it work. It's because it would be the work if we consider the work done in perpendicular component, which is 0. Put it another way, I intended to mention the line integral of F on a path that it's perpendicular to, and it's 0.

    Thanks for the URL, I've had a quick look at it, and I guess it's not at undergraduate level. I'll try, however.
     
  17. Mar 25, 2006 #16

    ZapperZ

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    So you made one up?

    And note, if you define it as rxF, then "work done in perpendicular component" is NOT zero, because rxF is not zero if r and F are perpendicular. You will also have a lot of explaining to do when I put a moving charge particle in a magnetic field.

    Zz.
     
  18. Mar 25, 2006 #17
    I did not define rxF is zero, it came out of
    [tex]dW = {\vec{F} \cdot \vec{dr}}[/tex]

    Since you're too pedantic, here I derieve it for you:

    [tex]dW = {F dr'} cos(\theta)[/tex]
    For the case where F and dr' are always perpendicular (ie, theta is constant), this equation says
    [tex]dW = {F dr'} cos(\pi/2)[/tex] (I note that this's always zero)
    Now, [tex]cos(\theta) = sin(\pi/2 - \theta)[/tex], so it can also be written as
    [tex]dW = {F dr'} sin(0)[/tex]
    which is of course
    [tex]dW = {\vec{dr} \times \vec{F} }[/tex]
    where dr is a vector perpendicular to dr' (or parallel to F).
    Since I'm writing the same thing in a different form, it should also be 0.
    So I'll simply say magnetic force does no work; because it's perpendicular (to velocity, therefore instantenous path) component counts only, so drxF is always 0.

    Of course, if F is constant with path, I could write rxF as well.
     
    Last edited: Mar 25, 2006
  19. Mar 25, 2006 #18

    ZapperZ

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    Hang on. Don't start accusing me of being pedantic when you were the one who earlier picking apart the difference between "energy" and "rotational energy" while ignoring the CONTEXT of what was meant.

    And of course, you SWITCHED basis vector while using a standard "r" or "dr" from before. Why didn't you defined this earlier? You could have easily used [tex]r d\theta[/tex] for this component. It would be strange, but it wouldn't have cause this much confusion. And why go through all of this gymnastics when the standard definition of work would have done the same thing if that's all you want to prove?

    Zz.
     
  20. Mar 25, 2006 #19
    Sorry, I didn't intend to say it a bad sense...

    Well, I switched it because, in the original statement, dr and F are paralel. So I made up a vector dr' that is perpendicular to dr, put it into line integral (which is a general rule, so it should be entirely valid), and derieved a result in terms of the original vector dr.
    I didn't define it earlier, because I didn't do the line integral before. I'm sorry if this caused a confusion, I just thought my notation was just fine since I've written the meaning of every -well, almost every, except for the obvious ones- variable when I wrote an equation.

    It seemed to tell the situation better for the case of paths perpendicular to forces since it involved a cross product. For instance, it's easy to see it for the case of work done by magnetic force, because it involes a cross product:
    [tex]dW = d\vec{r} \cdot (q\vec{v} \times \vec{B})[/tex]
    [tex]dW = d\vec{r} \cdot (q \frac{d\vec{r}}{dt} \times \vec{B}) = 0[/tex]
    which lets me to see the result is always is zero at the first glance. It was just an attempt of making an analogy.
     
    Last edited: Mar 25, 2006
  21. Mar 25, 2006 #20

    ZapperZ

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    But when you switch coordinate symbol to mean something else, you didn't think that would cause utter confusion? I mean, defining work as rxF where your "r" is NOT the same r that is defined in a plane polar or spherical polar coordinate system is just annoying. What about using what has already been defined in THAT basis, which is the [tex]rd\theta[/tex]? If not, why is that component even there?

    And no one needs to do any of these gymnastics to deal with the Lorentz forces just to prove that the field does no work on the free charges. Just simply from the fact that the F is perpendicular to both v and B is enough.

    Zz.
     
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