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Deriving the equations of torque and ang. Momentum from newton's laws

by metalrose
Tags: deriving, equations, laws, momentum, newton, torque
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metalrose
#1
Aug2-10, 07:20 AM
P: 126
I know what the definition of torque and angular momentum is.
But I read in a few books that

"Torque and angular momentum are not independent laws of nature, and have been derived from the more fundamental laws of newton."

The above simply means, that the rotational motion is just a manifestation of newton's laws at work.

How can we "derive" the concepts of torque and angular momentum by using just the newton's three laws?


Another way to look at the problem:
Suppose we have an extended body, say a rod. Now I apply a force F at some point of this rod, apart from the center of mass. Let the rod's net mass be M. All the points on the rod, will then have an acceleration component in the direction of the force F, given by a=F/M.

But apart from this acc. component, all the points, except the C.O.M. have another additional component of acceleration, that is the angular acceleration.

My question is, Where did this angular acc. component come from? what caused it? (please don't give a lame answer like "torque of F caused it")
Please give your answer by using only newton's laws and not torque or angular momentum.

According to the newton's second law, the force F produces acc. a=F/M in all the points lying on the rod. But all the points(except c.o.m.) almost magically acquire another acc. component, i.e. the angular acc.

Please explain, how the newton's laws, give rise to this extra angular acc. component.
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arildno
#2
Aug2-10, 07:34 AM
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Hi, metalrose!

Here's the explanation:

If you work with an spatially extended body, Newton's 2.law is valid for each individual segment, or element of the body.

Note a crucial feature here:
Each segment may perfectly well exert a force on its neighbouring element, which, in accordance with Newton's 3.law, exerts back an equal and opposite force on its neighbour.

Thus, if you then make the equation for the acceleration of the C.M, then all these internal forces will negate each other (virtue of the 3.law), and you have the resultant equation that says the acceleration of C.M is solely due the sum of EXTERNAL forces.

Your statement,
According to the newton's second law, the force F produces acc. a=F/M in all the points lying on the rod.
is utterly FALSE, the internal forces present may perfectly well give each ELEMENT of the rod a totally different acceleration than the acceleration the C.M has.

And when different elements of a body has different accelerations, then spatially, the body may undergo reconfigurations.

For RIGID bodies, the only types of such reconfigurations are ROTATIONS (whereas non-rigid bodies may undergo a lot of other types of reconfigurations besides)
metalrose
#3
Aug2-10, 08:09 AM
P: 126
@arildno

Thanks for the reply. I got the point.

Let me simplify the system. let's suppose we have a two particle system. Let's suppose the two particles are bonded to each other through a spring. As the spring constant of this spring tends to infinity, the system will tend to being a "rigid two particle body".

Now, we apply force F on one of these particles, perpendicular to the line joining the two points. How can we show that the two particle system will rotate along with accelerating forward, 'without using the concept of torque'. How can we show this by using only newton's laws?

D H
#4
Aug2-10, 08:12 AM
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Deriving the equations of torque and ang. Momentum from newton's laws

There are two forms of Newton's third law, a weak form and a strong form:
  • Weak form: The forces exerted by a pair of objects on each other are equal in magnitude and opposite in direction.
  • Strong form: The forces exerted by a pair of objects on each other are equal in magnitude and opposite in direction and lie along the line joining the two particles.

The weak form is sufficient to derive arildno's result, that the acceleration of the center of mass of a system of particles is the sum of the external force acting on the system of particles divided by the total mass of the system and to derive conservation of linear momentum. The strong form is needed to derive conservation of angular momentum.

I disagree with the statement "torque and angular momentum are not independent laws of nature, and have been derived from the more fundamental laws of newton." To arrive at that statement one needs to endow Newton's laws for translational motion with a characteristic that has nothing to do with translational motion. There is no essential difference between the weak and strong forms of Newton's third law when it comes to translational motion.

Arildno mentioned the rigid body assumption. That assumption goes on top of Newton's laws, and it is useful for rotational behavior only. So once again we have a case where that statement in question is not true, strictly speaking.

Which is more basic, Newton's third law or the conservation laws, is quibbling with words. The weak form of Newton's third law can be derived from conservation of linear momentum, the strong form from conservation of linear and angular momentum. One last word: the conservation laws in turn derive from Noether's theorem, and that is a very deep result.
arildno
#5
Aug2-10, 08:17 AM
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To take the case of the rod:

Let us divide the rod into N elements, 1,...N

Let [itex]\vec{F}_{e,i}[/itex] be the sums of external forces acting upon element (i), that is, the body producing that force (and hence recipient of element i's reaction force) is OUTSIDE the rod.

Let [itex]\vec{F}_{ij}[/itex] be an internal force, acting on element "i" from element "j".

With Newton's 3.law, we have:
[tex]\vec{F}_{ij}=-\vec{F}_{ji}[/tex]

Thus, for each element "i", we have Newton's 2.law:
[tex]\vec{F}_{e,i}+\sum_{j=1}^{N}\vec{F}_{ij}=m_{i}\vec{a}_{i}, \vec{F}_{ii}\equiv\vec{0}(1)[/tex]
The last requirement is just that "self-forces" (an element propelling itself) are zero/non-existent.

Now, let [itex]\vec{r}_{i}[/itex] be the position vector of element "i" with respect to some reference point, and let us take the cross-product with (1).

We get:
[tex]\vec{r}_{i}\times\vec{F}_{e,i}+\sum_{j=1}^{N}\vec{r}_{i}\times\vec{F}_{ ij}=m_{i}\vec{r}_{i}\times\vec{a}_{i}(2)[/tex]

Now, if we are to add together all the N (2)-equations, we can pair our terms of internal forces like this:
[tex](\vec{r}_{i}-\vec{r}_{j})\times\vec{F}_{ij}[/tex], in concordance with Newton's 3.law of motion.

But, since we generally have, for the rigid bodies that forces act ALONG the direction between two elements, then it follows that the internal force is parallell to the relative position vetor between them, and thus that all these cross products disappear!

What we are left with is, on the left-hand side the sums of all torques of external forces, with respect to WHERE they actually act, and on the other side, an ugly inertia/acceleration-expression, which in the simple case of the rigid body that can be reduced to, effectively, the moment of inertia multiplied with the angular acceleration.
D H
#6
Aug2-10, 08:33 AM
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Quote Quote by arildno View Post
But, since we generally have, for the rigid bodies that forces act ALONG the direction between two elements, then it follows that the internal force is parallell to the relative position vetor between them, and thus that all these cross products disappear!
Note that you are assuming the strong form of Newton's third law here; you did not assume the strong form in deriving the behavior of the center of mass of the system. The strong form of Newton's 3rd is only needed for rotational behavior. In other words, rotational behavior represents something new on top of translational behavior.
arildno
#7
Aug2-10, 08:39 AM
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Quote Quote by D H View Post
Note that you are assuming the strong form of Newton's third law here; you did not assume the strong form in deriving the behavior of the center of mass of the system. The strong form of Newton's 3rd is only needed for rotational behavior. In other words, rotational behavior represents something new on top of translational behavior.
Certainly.
That is why I invoke it only at a need-to-know basis!

I make it perfectly clear that we are dealing with the idealized case of the perfectly rigid body, i.e, a wholly mechanical system that behaves nicely with respect to conservation laws.
metalrose
#8
Aug2-10, 08:52 AM
P: 126
So does that then mean that rotational motion of an extended rigid body is NOT a manifestation of the newton's laws? That rotational motion is JUST THERE, as a new law of nature?

And does that then mean that the equation T=r x F is an experimentally determined analogue, and can not be understood in terms of more fundamental laws?

It is quite difficult to chew the possibility that rotation is a totally different law of nature and has got no origins in the newtons laws. It somehow, doesn't seem right, intuitively.

I somehow feel that when an extended body rotates, the origins of the angular accelerations lie in the complicated ways in which the internal forces act.

I always thought that the internal forces act in some complicated way, which then makes the body rotate. I was trying to analyze how rotation can come about by the way internal forces act.
arildno
#9
Aug2-10, 08:57 AM
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So does that then mean that rotational motion of an extended rigid body is NOT a manifestation of the newton's laws? That rotational motion is JUST THERE, as a new law of nature?
No.

The rotational motion is generated by the presence of INTERNAL forces (as well as external forces), forces that do not affect the body's overall amount of energy&momentum.

Thus, whereas Newton's 2-law is valid for ALL systems, the torque-equations are only valid for a SPECIAL CLASS of such systems.

And therefore, due to lack of generality, they are slightly less "fundamental" than Newton's 2.law of motion.

In the mechanical way of thinking, that is.
metalrose
#10
Aug2-10, 09:15 AM
P: 126
Quote Quote by arildno View Post

The rotational motion is generated by the presence of INTERNAL forces (as well as external forces), forces that do not affect the body's overall amount of energy&momentum.
If that is so, I am then looking for a step by step procedure for showing that this is so. How can we show, that rotational motion is nothing but a manifestation of the nweton's laws?

How can we show it for example in the two particle system I described in a reply before the above one?

Can you please analyze the two particle system whcih i described in a previous reply?

You would have solved my problem if you can do so...

Thanks..!!
arildno
#11
Aug2-10, 09:30 AM
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Well, I can certainly do so, but I want to give an example of a NON-rigid body, where it is actually easier to understand how the rotation comes about (DO remember, that the perfectly rigid body is an idealization!!):

Now, suppose you have a rod I.

At its lower point, it is met by a force F, so that if its shape weren't to remain an I, it would look something like this: L (agreed??)

Now, to understand how rotation comes about in the real world, think of this I trying to resist to bend out of shape to an..L

How is that most easily done?

Simple! To become an \ rather than an I.

If you think of a minor displacement happening at the foot, the internal forces between that end, and the rest of the object will pull the rest of the object (rightwards&downwards), whereas the foot will be pulled (leftwards&upwards).

The shape gained by these force transfers is the \, rather than the L, due to the strength of the internal forces.

Other objects have less maximal internal forces, and will bend out of shape locally, rather than undergoing a bodywise rotation with overall shape preservation.

Is this understandable?
arildno
#12
Aug2-10, 10:54 AM
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Presuming you understood the above post, let us focus on the problem you gave.
We have two balls B1 and B2, for simplicity with the same mass, m, connected by, for example, a massless, inflexible rod R of length 2l.

Let B1 be subject to some external force F, perpendicular to R.

In that case, we have Newton's 2.law:
For B1:
[tex]F+F_{12}=ma_{1}[/tex]
For B2:
[tex]-F_{12}=ma_{2}[/tex]

Do you see what our problem is here, with respect to the motions of B1 and B2?

We do NOT know what [itex]F_{12}[/itex] is!!!!

Without further assumptions, or suppositions, we cannot proceed any further than deriving the acceleration of the C.M of the body, not the acceleration of the body parts!

So, how do we proceed then?

Do you agree that this is the crucial issue?
metalrose
#13
Aug3-10, 07:26 AM
P: 126
@arildno

Yes, this is precisely the issue, to determine the individual motions of B1 and B2 and using just the laws of newton, show that B1 and B2 will rotate.

How do we do this?
arildno
#14
Aug3-10, 07:41 AM
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Quote Quote by metalrose View Post
@arildno

Yes, this is precisely the issue, to determine the individual motions of B1 and B2 and using just the laws of newton, show that B1 and B2 will rotate.

How do we do this?
First off:
Did you understand post 11, in how a non-rigid object will locally deform, rather than undergoing bodywise rotation with shape preservation?
metalrose
#15
Aug3-10, 07:51 AM
P: 126
First off:
Did you understand post 11, in how a non-rigid object will locally deform, rather than undergoing bodywise rotation with shape preservation
I had missed out your post 11. I have now read it. also, I understand the case of the non rigid body that you talked about.


But I think this doesn't have any bearing on the point of determining the individual motions of B1 and B2 and proving that they will rotate.
arildno
#16
Aug3-10, 07:55 AM
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Quote Quote by metalrose View Post
I had missed out your post 11. I have now read it. also, I understand the case of the non rigid body that you talked about.


But I think this doesn't have any bearing on the point of determining the individual motions of B1 and B2 and proving that they will rotate.
Oh, most definitely it has!

ALL objects are non-rigid, to some extent.

What happens in those bodies we call rigid, is that so strong INTERNAL STRESSES are set up in the WHOLE body as the result of just a tiny local deformation, that these internal stresses accelerate the rest of the body into the position of MINIMAL internal stress, and this is achieved by the motion we call rotational.
The I turns into an \, because the \-state is less stressful than the L-state.
(of course, the I-state itself is impossible to maintain, due to the presence of F).

Is this clear?
arildno
#17
Aug3-10, 08:30 AM
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Now, to take a step further:
HOW are we to truly explain why rotation occurs?

It is important to understand what we're actually asking!

In the deepest sense, we are asking how are we to model the internal force F_12.

This is what we call the dynamical approach, i.e, force modelling on the local scale.

How this is done is to regard stiff and non-stiff objects as extremes on a spectrum of elasticity, springlike that is, where an "infinite" stiffness constant is the limiting case of the rigid body.

The theory in question is called "the theory of elasticity", and is a truly ugly monster of non-linear partial differential equations, and it CAN be shown from this that in the limiting case of the ideal, rigid body, shape-preserving rotation WILL occur.

Now, the more common approach, the one we learn in introductory mechanics courses, is the KINEMATIC approach, rather than the dynamic.

In the kinematic approach, we IMPOSE the condition of the perfectly rigid body, and this condition is nothing else than the imposition of constraints of MOTION, rather than modelling of FORCES.
metalrose
#18
Aug3-10, 09:10 AM
P: 126
and it CAN be shown from this that in the limiting case of the ideal, rigid body, shape-preserving rotation WILL occur.
This is what I was looking for. So that then finally means, that rotation CAN be understood as nothing but a manifestation of the newton's laws.

And that the concepts of Torque are developed in an introductory mechanics course just because they are simple to work with, and are less rigorous, but are NOT the fundamental reasons for rotational motion, that rotational motion, in its deepest form, stems out from newton's laws. Torques and similar concepts just serve as mathematical conveniences.

You talked about the dynamical approach. Where can I learn it from? I mean all about elasticity and the non linear p.d.e's and all? What level of physics course would include all of it? graduate?

And can any of the dynamical aspects be found in the hamiltonian and lagrangian mechanics?

Thanks!!


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