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- Thread starter jbunten
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[tex]\hat{\theta}[/tex]

term, the key is expressing things using the cross product:

[tex]

\frac{d}{dt}\left(\frac{\mathbf r}{\Vert \mathbf r \Vert}\right)

= \left(\hat{\mathbf r} \times \frac{1}{{\Vert \mathbf r \Vert}} \frac{d \mathbf r}{dt}\right) \times \hat{\mathbf r}

[/tex]

Give it a try and see where you get with it. I have a write up I did for myself of such a derivation (since I didn't like the way my mechanics book pulled all this out of a magic hat). I can post that if you want (uses Geometric Algebra though), but it's probably more instructive for you to try yourself with the hint above.

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I'm not sure about this, but I think it's just to do with the development of the rotational equivalents of the translation dynamics equations. Torque is defined as [tex]r\times[/tex][tex]F[/tex]. We need the r here because physically there is a dependence on the distance of the force from the rotation axis. And we want the rotational equivalent of [tex]\sum[/tex] F = [tex]dP/dt[/tex] which is [tex]\sum\tau = dL/dt[/tex] and so we define angular momentuml as [tex]r\times[/tex][tex]p[/tex]

Perhaps somebody more knowledgeable could clarify?

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Good ol' Newton's second law is a vector equation, and that means we can do any of the vector things to it.

In particular, if we cross-product it with another vector, it's still valid.

r is a vector, and the cross-product with r is [itex]\sum\tau = r\times p[/itex] .

erm … why r and not something else?

hmm … not sure …

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The cool thing about defining [itex]\vec{L} = \sum_i \vec{r_i} \times \vec{p_i}[/itex] is that it follows that [itex]d\vec{L}/dt = \vec{\tau}_{\textup{net}}[/itex] because [itex]\vec{v_i}\times \vec{p_i} = 0[/itex]. If you used any other vector other than r, you would have another term in the product rule. And defining [itex]\vec{\tau}_{\textup{net}} = \sum_i \vec{r_i}\times\vec{F_i}[/itex] is intuitive (just use a wrench or play on a see saw) but also can be shown by direct application of the translational form of Newton's Second Law to be proportional to the angular acceleration, which makes it the rotational analog of force. That implies that defining angular momentum as we do yields us a rotational analog of momentum that is conserved when the rotational analog of force vanishes.

The Lagrangian is isotropic, this symmetry is associated with a Noether current in the form of the quantity we would define as angular momentum. It just pops right out of the math that you can find in any introductory classical field theory text.

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In particular, if we cross-product it with another vector, it's still valid.

r is a vector, and the cross-product with r is [itex]\sum\tau = r\times p[/itex] .

note: small typo above (should be angular momentum not torque)

Since jbunten was working out [itex]\mathbf{r}'[/itex] for [itex]\mathbf{r} = r\hat{\mathbf{r}}[/itex] in a different thread, I suggested doing the same thing for the acceleration. I don't know how far he/she got, but the end result, with [itex]

\boldsymbol{\omega} = \frac{\mathbf{r} \times \mathbf{v}}{r^2}

[/itex], is:

you can calculate the velocity and acceleration in terms of radial components:

[tex]

\mathbf{v} = r'\hat{\mathbf{r}} + \boldsymbol{\omega} \times \mathbf{r}

[/tex]

[tex]

\mathbf{a} = r''\hat{\mathbf{r}} - \mathbf{r} {\lvert \boldsymbol{\omega} \rvert

}^2 - \frac{\mathbf{r}}{r^2} \times \frac{d}{dt}(\mathbf{r} \times \mathbf{v})

[/tex]

Thus for constant m you get:

[tex]

F = m\mathbf{a} = mr''\hat{\mathbf{r}} - m\mathbf{r} {\lvert \boldsymbol{\omega} \rvert

}^2 - \frac{\mathbf{r}}{r^2} \times \frac{d}{dt}(\mathbf{r} \times \mathbf{p})

[/tex]

and therefore see directly why the quantity [itex]\mathbf{r} \times \mathbf{p}[/itex] becomes significant (this is given the name angular momentum). It's derivative, which for constant m, (r x p)' = r x F, is given the name torque, and one can see how it is related to the Force on the object.

This is how I liked to view this ... the cross products are just conquences of looking at things from a radial point of view since they can be used to express the difference from the projection onto the radial direction.

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What was your question?

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What was your question?

i want to know where this angular momentum or torque equaton comes from and why we use the cross product.In a lot of sites they only gave the definiton of angular momentum and torque and said that theses definitions are rotational equivelants of 2nd law.But how do they get there, i mean from pure 2nd law which doesn't say anytihng about rotation as far as i can see, how can we derive these torque equations?

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In the Newtonian limit it is really F=p' that is fundamental, so this radial acceleration is only valid for constant mass. I'm not sure if I've seen this mentioned in intro texts that describe conservation of angular momentum is discussed, but not where it comes from. If you have mass loss, like in a rocket ship problem, then conservation of angular momentum only applies if you continue to include all the mass loss in your "system".

Note that I didn't derive this above, just posted the end result. There is a hint above on how to actually derive this yourself (start with r=rhat |r| and use rhat' above). It is probably more instructive for you to do that yourself (or at least attempt) than for me to post this derivation directly.

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i want to know where this angular momentum or torque equaton comes from and why we use the cross product.In a lot of sites they only gave the definiton of angular momentum and torque and said that theses definitions are rotational equivelants of 2nd law.But how do they get there, i mean from pure 2nd law which doesn't say anytihng about rotation as far as i can see, how can we derive these torque equations?

It is not possible to derive the conservation of angular momentum from Newton's laws. It a simple thing to postulate a counter example that satisfies the Newton's laws, but violates conservation of angular momentum. So you cannot derive the torque equations either, from Newton's laws alone.

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It is not possible to derive the conservation of angular momentum from Newton's laws. It a simple thing to postulate a counter example that satisfies the Newton's laws, but violates conservation of angular momentum. So you cannot derive the torque equations either, from Newton's laws alone.

violates conservation of angular momentum,what do you mean?if we can not derive them from Newton's eqs, than they didn' completely explain the motion and angular momentum is a whole different thing?

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Peeter could you put the derivation or send it to me?I tried but my maths is weak.

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violates conservation of angular momentum,what do you mean?

I'll use coordinates for two dimension in the example: The particle A is in position (-1,0), the particle B is in position (1,0), the particle A exerts a force F_(AB)=(0,1) to the particle B, and particle B exerts a force F_(BA)=(0,-1) to the particle A. There force vectors satisfy the Newton's third law, because they are equal in magnitude, and in opposite direction. However, the angular momentum is not conserving, because the two particles will start rotating counter clockwise on their own.

This example proves, that if you merely assume that in some system Newton's laws are being satisfied, it is not yet possible to prove that under zero external torque, the angular momentum would be conserving.

if we can not derive them from Newton's eqs, than they didn' completely explain the motion and angular momentum is a whole different thing?

In order to have a complete explanation of the motion, we need to of course specify what kind of forces particles use to interact. This is something different than the Newton's laws. Newton's laws are only a set of rules which the forces must obey, but there can be more rules for the forces too. On possible restriction to the forces is that the force vector must point in the same direction as the spatial separation vector between the particles. In this case the angular momentum will be conserved.

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I'll use coordinates for two dimension in the example: The particle A is in position (-1,0), the particle B is in position (1,0), the particle A exerts a force F_(AB)=(0,1) to the particle B, and particle B exerts a force F_(BA)=(0,-1) to the particle A. There force vectors satisfy the Newton's third law, because they are equal in magnitude, and in opposite direction. However, the angular momentum is not conserving, because the two particles will start rotating counter clockwise on their own.

This example proves, that if you merely assume that in some system Newton's laws are being satisfied, it is not yet possible to prove that under zero external torque, the angular momentum would be conserving.

In order to have a complete explanation of the motion, we need to of course specify what kind of forces particles use to interact. This is something different than the Newton's laws. Newton's laws are only a set of rules which the forces must obey, but there can be more rules for the forces too. On possible restriction to the forces is that the force vector must point in the same direction as the spatial separation vector between the particles. In this case the angular momentum will be conserved.

so you're saying that the static equilibrium conditions: sum of all forces must be zero come from Newtons law and the sum of the moment must be zero comes from somewhere else.Then where is this moment conditon come from?

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it's a cosequence of the seceond rule ,and somewhere there must be a relation between 2nd rule and rotational stuff.

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so you're saying that the static equilibrium conditions: sum of all forces must be zero come from Newtons law and the sum of the moment must be zero comes from somewhere else.Then where is this moment conditon come from?

The Newton's approach is that we set a postulate, that for all forces there are corresponding counter forces. In similar spirit, we can simply set a postulate, that for all torques there are corresponding counter torques.

The Lagrange's and Noether's approach reveals something else. If the Lagrange's function is invariant under translations, then it implies that the total momentum is conserved, which is equivalent with the existence of counter forces. If the Lagrange's function is invariant under rotations, then it implies that the total angular momentum is conserved, which is equivalent with the existence of the counter torques.

Notice, translation invariance does not imply rotational invariance.

it's a cosequence of the seceond rule ,and somewhere there must be a relation between 2nd rule and rotational stuff.

Good luck on your search

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It is not possible to derive the conservation of angular momentum from Newton's laws.

Jostpuur, Can you point me to a reference (preferably online) that discusses this in more detail. My mechanics texts do not really discuss this in a way that I find satisfactory. My assumption was that Newton's laws were more fundamental (in the v << c limit) than angular momentum conservation, and that angular momentum was only conserved under specific circumstances (like constant and non-relativistic mass).

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Jostpuur, Can you point me to a reference (preferably online) that discusses this in more detail.It is not possible to derive the conservation of angular momentum from Newton's laws.

Not really... I don't remember very well where I have learned this myself. I don't think I've ever read that claim from anywhere. I've only thought about it myself.

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Peeter could you put the derivation or send it to me?I tried but my maths is weak.

I looked, but can't find a derivation in my notes that does not use Geometric Algebra (I'm sure I did one, but it must be on paper somewhere). I've attached what I do have typed up... you can probably get an idea how to directly approach the same radial acceleration decomposition using traditional cross products. Like I said, the key is exploiting a cross product expansion of \rcap', and taking derivitives of r = rcap |r|. If you can't derive the rcap' expression yourself just use what I posted above (my old Salus/Hille Calculus text did that rcap' derivation but it was a bit involved and indirect).

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With the cross product, we see that two vectors interact within a plan to induce another vector in a direction perpendicular to the plane. Also take notice that you could take the cross product of any two interacting vectors...although the procedure may or may not hold much meaning to it.

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I'll use coordinates for two dimension in the example: The particle A is in position (-1,0), the particle B is in position (1,0), the particle A exerts a force F_(AB)=(0,1) to the particle B, and particle B exerts a force F_(BA)=(0,-1) to the particle A. There force vectors satisfy the Newton's third law, because they are equal in magnitude, and in opposite direction. However, the angular momentum is not conserving, because the two particles will start rotating counter clockwise on their own.

This example proves, that if you merely assume that in some system Newton's laws are being satisfied, it is not yet possible to prove that under zero external torque, the angular momentum would be conserving.

Hi jostpuur!

Newtonian mechanics essentially does not recognise force at a distance, except as allowed by Newton's third law (action and reaction).

When equal but opposite forces are applied perpendicular to opposite ends of a rigid rod, you might say that Newton's second law only requires that the centre of the rod remains stationary, and says nothing about the rotation.

But that fails to treat the ends of the rod as separate bodies. Each end feels a tension force along the rod (it must do, or the rod wouldn't be rigid), and Newton's second law then makes the end of the rod go in a circle.

Although the applied forces are not equal and opposite (and so do not obey Newton's third law), the tension forces are.

All central forces (such as gravity and electric charge) comply with Newton's third law, and they can easily be proved to conserve angular momentum.

By contrast, a uniform magnetic field (which is perpendicular-to-central), for example, sends charged bodies in circles, and that obviously does

In order to have a complete explanation of the motion, we need to of course specify what kind of forces particles use to interact. This is something different than the Newton's laws. Newton's laws are only a set of rules which the forces must obey, but there can be more rules for the forces too. On possible restriction to the forces is that the force vector must point in the same direction as the spatial separation vector between the particles. In this case the angular momentum will be conserved

Yes … obviously, we could define forces in a way that violates Newton's second and third law, which your example does.

Forces which *don't* meet at a point must be augmented by connecting forces obeying Newton's third law.

Thus all forces can be grouped according to the point at which they meet.

And, at each such point, forces which meet at a point can be proved to conserve angular momentum about any axis:

Angular momentum can only fail to be conserved in Newtonian mechanics if there is a field which does not obey Newton's third law.

Thus all forces can be grouped according to the point at which they meet.

And, at each such point, forces which meet at a point can be proved to conserve angular momentum about any axis:

∑m**v**´ = 0 implies ∑m**r**x**v**´ = 0

Angular momentum can only fail to be conserved in Newtonian mechanics if there is a field which does not obey Newton's third law.

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Yes … obviously, we could define forces in a way that violates Newton's second and third law, which your example does.

The Newton's laws were not being violated in my example.

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The Newton's laws were not being violated in my example.

But your F

- #28

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But your F_{AB}and F_{BA}are not opposite … they're_{offset}.

We got into definition of the word "opposite" then. Actually I don't know what Newton meant himself, but I insist that what I said was correct with the modern terminology. We are always taught that the vectors don't have position, only the direction and magnitude. Forces are vectors, so now the forces F_(AB) and F_(BA) are opposite, because the vectors (0,1) and (0,-1) are opposite.

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Yes it is, if you assume the strong form of Newton's third law (equal but opposite central forces). Introductory classical mechanics texts (junior year in college) do just that. Your counterexample used the weak form of Newton's third law (non-central forces).It is not possible to derive the conservation of angular momentum from Newton's laws. It a simple thing to postulate a counter example that satisfies the Newton's laws, but violates conservation of angular momentum. So you cannot derive the torque equations either, from Newton's laws alone.

Newton postulated conservation of linear momentum. Conservation of angular momentum can be derived from Newton's laws if one assumes the strong form of Newton's third law. Deriving conservation of energy from Newton's laws requires a yet another assumption, that forces are conservative.

However, the conservation laws (linear momentum, angular momentum, and energy) are in a sense deeper than Newton's laws. They derive from Noether's theorem and apply in regimes where Newton's laws fail or are inapplicable.

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I had never heard of strong or weak form of the Newton's third law.

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if i could undersand the peeter's derivation

Okay, how about this for the angular velocity part (first derivatives). Does this make sense (if it doesn't then it doesn't make much sense to do the next step which is taking second derivatives for acceleration).

Starting point is taking the derivative of:

[tex]

\mathbf{r} = r \mathbf{\hat{r}}

[/tex]

[tex]

\mathbf{v} = \mathbf{r}' = r' \mathbf{\hat{r}} + r \mathbf{\hat{r}}'

[/tex]

It can be shown (see for example, Salus and Hille, "Calculus") that the unit vector derivative can be expressed using the cross product:

[tex]

\mathbf{\hat{r}}' = \frac{1}{r} \left(\mathbf{\hat{r}} \times \frac{d\mathbf{r}}{dt}\right) \times \mathbf{\hat{r}}.

[/tex]

Now, one can express [itex]r'[/itex] in terms of [itex]\mathbf{r}[/itex] as well as follows:

[tex]

\left(\mathbf{r} \cdot \mathbf{r}\right)' = 2 \mathbf{v} \cdot \mathbf{r} = 2 r r'.

[/tex]

Thus the derivative of the vector magnitude is part of a projective term:

[tex]

r' = \mathbf{\hat{r}} \cdot \mathbf{v}.

[/tex]

Putting this together one has velocity in terms of projective and rejective

components along a radial direction:

[tex]

\mathbf{v} = \left(\mathbf{\hat{r}} \cdot \mathbf{v}\right) \mathbf{\hat{r}} + \left(\mathbf{\hat{r}} \times \mathbf{v}\right) \times \mathbf{\hat{r}}.

[/tex]

Now [itex]\boldsymbol{\omega} = \frac{\mathbf{r} \times \mathbf{v}}{r^2}[/itex] term is what we call the angular velocity. The magnitude of this

is the rate of change of the angle between the radial arm and the direction of rotation. The direction of this

cross product is normal to the plane of rotation and encodes both the rotational plane and the direction of the

rotation. Putting these together one has the total velocity expressed radially:

[tex]

\mathbf{v} = \left(\mathbf{\hat{r}} \cdot \mathbf{v}\right) \mathbf{\hat{r}} + \boldsymbol{\omega} \times \mathbf{r}.

[/tex]

- #32

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Weak form: Forces are equal in magnitude and opposite in direction.I had never heard of strong or weak form of the Newton's third law.

Strong form: Forces are equal in magnitude, opposite in direction, and act along a common line (i.e., central forces).

http://core.ecu.edu/phys/flurchickk/Classes/PHYS4226/Section1/Lecture1-1.xml" [Broken].

Recently there was a very brief discussion at PF on the two forms in https://www.physicsforums.com/showthread.php?t=215360".

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- #33

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We are always taught that the vectors don't have position, only the direction and magnitude. Forces are vectors, so now the forces F_(AB) and F_(BA) are opposite, because the vectors (0,1) and (0,-1) are opposite.

Hi jostpuur!

Yes, I entirely agree that vectors don't have position … we have to be able to shift them about and tack them onto each other!

But forces aren't

I agree with you that Newton's second law

But Newton's third law doesn't … it requires the full monty.

We got into definition of the word "opposite" then. Actually I don't know what Newton meant himself, but I insist that what I said was correct with the modern terminology.

Well, maths is all about definitions.

And, as Humpty Dumpty said, as between

Sometimes the trick is to be able to tell the difference between a problem and a solution …

what would you prefer, a definition which adds one more to life's problems, or a definition which provides a solution?

Hi

That's a

- #34

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Okay, how about this for the angular velocity part (first derivatives). Does this make sense (if it doesn't then it doesn't make much sense to do the next step which is taking second derivatives for acceleration).

Starting point is taking the derivative of:

[tex]

\mathbf{r} = r \mathbf{\hat{r}}

[/tex]

[tex]

\mathbf{v} = \mathbf{r}' = r' \mathbf{\hat{r}} + r \mathbf{\hat{r}}'

[/tex]

It can be shown (see for example, Salus and Hille, "Calculus") that the unit vector derivative can be expressed using the cross product:

[tex]

\mathbf{\hat{r}}' = \frac{1}{r} \left(\mathbf{\hat{r}} \times \frac{d\mathbf{r}}{dt}\right) \times \mathbf{\hat{r}}.

[/tex]

Now, one can express [itex]r'[/itex] in terms of [itex]\mathbf{r}[/itex] as well as follows:

[tex]

\left(\mathbf{r} \cdot \mathbf{r}\right)' = 2 \mathbf{v} \cdot \mathbf{r} = 2 r r'.

[/tex]

Thus the derivative of the vector magnitude is part of a projective term:

[tex]

r' = \mathbf{\hat{r}} \cdot \mathbf{v}.

[/tex]

Putting this together one has velocity in terms of projective and rejective

components along a radial direction:

[tex]

\mathbf{v} = \left(\mathbf{\hat{r}} \cdot \mathbf{v}\right) \mathbf{\hat{r}} + \left(\mathbf{\hat{r}} \times \mathbf{v}\right) \times \mathbf{\hat{r}}.

[/tex]

Now [itex]\boldsymbol{\omega} = \frac{\mathbf{r} \times \mathbf{v}}{r^2}[/itex] term is what we call the angular velocity. The magnitude of this

is the rate of change of the angle between the radial arm and the direction of rotation. The direction of this

cross product is normal to the plane of rotation and encodes both the rotational plane and the direction of the

rotation. Putting these together one has the total velocity expressed radially:

[tex]

\mathbf{v} = \left(\mathbf{\hat{r}} \cdot \mathbf{v}\right) \mathbf{\hat{r}} + \boldsymbol{\omega} \times \mathbf{r}.

[/tex]

i will work on your derivaiton but these projection and rejection terms are so unfamiliar for me, i will study some maths and work o your derivaiton.

- #35

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i will work on your derivaiton but these projection and rejection terms are so unfamiliar for me, i will study some maths and work o your derivaiton.

Projection is the part of a vector in the direction of a unit vector. This is usually expressed in terms of the dot product:

[tex]

Proj_x(r) = \left(r \cdot \frac{x}{\lvert x \rvert} \right) \frac{x}{\lvert x \rvert}.

[/tex]

Rejection is not a standard term but convienient to give a name to non-projective part of an orthogonal decomposition:

[tex]

Rej_x(r) = r - Proj_x(r)

[/tex]

This last term can be expressed in a few ways. One of which is using the cross product as indicated in the previous post.

If you have trouble with the vector relations, a simpler way to derive the radial acceleration relations is to restrict the motion to a plane. Then use complex numbers (polar form) to represent the vectors. I'd suggest that as a alternative approach. You would probably find that more straightforward (presuming you know how to work with complex numbers).

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