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jbunten
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Is there a particularly good reason that angular momentum was defined as the cross product of position and linear momentum vectors?
jbunten said:Is there a particularly good reason that angular momentum was defined as the cross product of position and linear momentum vectors?
tiny-tim said: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] .
Peeter said:What was your question?
teodorakis said: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?
jostpuur said: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.
teodorakis said: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?
jostpuur said: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.
teodorakis said: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?
dear jostpuur you're saying that it's a postulate and deal with it:), but i still believe that
it's a cosequence of the seceond rule ,and somewhere there must be a relation between 2nd rule and rotational stuff.
jostpuur said:It is not possible to derive the conservation of angular momentum from Newton's laws.
Peeter said:Jostpuur, Can you point me to a reference (preferably online) that discusses this in more detail.jostpuur said:It is not possible to derive the conservation of angular momentum from Newton's laws.
teodorakis said:Peeter could you put the derivation or send it to me?I tried but my maths is weak.
jostpuur said: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
tiny-tim said:Yes … obviously, we could define forces in a way that violates Newton's second and third law, which your example does.
jostpuur said:The Newton's laws were not being violated in my example.
tiny-tim said:But your FAB and FBA are not opposite … they're offset.
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).jostpuur said: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.
teodorakis said:if i could undersand the peeter's derivation
Weak form: Forces are equal in magnitude and opposite in direction.jostpuur said:I had never heard of strong or weak form of the Newton's third law.
jostpuur said: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.
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.
Peeter said: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]
teodorakis said: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.
Angular momentum is a physical quantity that measures the amount of rotational motion of an object. It is important in science because it helps us understand and predict the behavior of rotating objects, such as planets, atoms, and galaxies.
Angular momentum was first defined by the physicist Isaac Newton in his famous laws of motion. He defined it as the product of an object's moment of inertia and its angular velocity.
This definition of angular momentum was chosen because it accurately describes the relationship between an object's rotational motion and its mass distribution. It also allows for the conservation of angular momentum, which is a fundamental principle in physics.
Yes, there is a difference between angular momentum and linear momentum. Angular momentum is a measure of rotational motion, while linear momentum is a measure of an object's straight-line motion. They are both conserved quantities, but they apply to different types of motion.
Angular momentum is used in many real-world applications, including engineering, astronomy, and physics. It is used to design and control rotating machinery, to study the motion of planets and galaxies, and to understand the behavior of particles at the atomic level.