Why was angular momentum defined as it was?

[Peeter]

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:

<br /> \mathbf{r} = r \mathbf{\hat{r}}<br />

<br /> \mathbf{v} = \mathbf{r}&#039; = r&#039; \mathbf{\hat{r}} + r \mathbf{\hat{r}}&#039;<br />

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

<br /> \mathbf{\hat{r}}&#039; = \frac{1}{r} \left(\mathbf{\hat{r}} \times \frac{d\mathbf{r}}{dt}\right) \times \mathbf{\hat{r}}.<br />

Now, one can express r&#039; in terms of \mathbf{r} as well as follows:

<br /> \left(\mathbf{r} \cdot \mathbf{r}\right)&#039; = 2 \mathbf{v} \cdot \mathbf{r} = 2 r r&#039;.<br />

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

<br /> r&#039; = \mathbf{\hat{r}} \cdot \mathbf{v}.<br />

Putting this together one has velocity in terms of projective and rejective
components along a radial direction:

<br /> \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}}.<br />

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

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

 

[D H]

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

Weak form: Forces are equal in magnitude and opposite in direction.
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" .

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

 

[tiny-tim]

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 just vectors … a force has a line of action also.

I agree with you that Newton's second law does treat forces as vectors.

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 words and us, the question is who is to be the master!

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 D H!

That's a really good summary (post #29) … please pretend this is a respect smilie … :respect: !

WE NEED MORE SMILIES!​

 

[teodorakis]

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:

<br /> \mathbf{r} = r \mathbf{\hat{r}}<br />

<br /> \mathbf{v} = \mathbf{r}&#039; = r&#039; \mathbf{\hat{r}} + r \mathbf{\hat{r}}&#039;<br />

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

<br /> \mathbf{\hat{r}}&#039; = \frac{1}{r} \left(\mathbf{\hat{r}} \times \frac{d\mathbf{r}}{dt}\right) \times \mathbf{\hat{r}}.<br />

Now, one can express r&#039; in terms of \mathbf{r} as well as follows:

<br /> \left(\mathbf{r} \cdot \mathbf{r}\right)&#039; = 2 \mathbf{v} \cdot \mathbf{r} = 2 r r&#039;.<br />

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

<br /> r&#039; = \mathbf{\hat{r}} \cdot \mathbf{v}.<br />

Putting this together one has velocity in terms of projective and rejective
components along a radial direction:

<br /> \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}}.<br />

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

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

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.

 

[Peeter]

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:

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

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

<br /> Rej_x(r) = r - Proj_x(r)<br />

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).