Expert: Daniel Mazur - 11/16/2009

Question
QUESTION: Hi Daniel,

I once again want to go back to the problem of what forces and masses are. Though we have discussed it before, i'll just give the model that I have in my head once again and I want to know wether it's correct or not. And then I want to discuss the distribution of forces. And then if my model is correct I would like to apply it further to a situation, in a follow up perhaps.

Let's consider a Universe made up of just one kind of particles to simplify the model.
Also let there be just one kind of force operating in the universe.
Let the universe, for the moment have just one of these particles present in it and let us position ourselves on an inertial frame and let's also assume that our body doesn't interact with these particles, i.e. we are mere observers.
Now when we observe this particle from the inertial frame S, it's not accelerating. Let that be the initial condition.
Now when we insert another of these particles in the universe, we obsereve that both these particles are now accelerating w.r.t. S.

And now if take out the 1st particle from the universe, the second particle attains a constant velocity.

So this fact of nature can be put down as:
Whenever two of these particles are present, they accelerate w.r.t. an inertial frame.
Or it's just by the virtue of prescence of each other that they start accelerating.
But to put this idea on a quantitative basis, we make up a new thing called "force" and we can then chose 1 unit of this thing called force as follows:
We can keep two of these particles in the universe and can then measure there acceleration w.r.t. S(assuming that they have a constant acc. independent of the seperation b/w them).
When we have measured the acceleration in this simple two particle case, let's say 'a' m/s^2 we can call this "something" b/w them that we called a force as 1 unit force.

And we also experiment to see that this quantity that we have called a force is directly propotional to the acceleration of the particles.
Since we have called the force 1 unit when the acceleration was a, we can say that for every 'a' unit of acceleration, there is a force of 1 unit.
And this ratio is fixed in nature, once we have decided on our unit of force.
If suppose in the above situation, we had called the force as 2 units, i.e. when the particles acc. with 'a', then this ratio of force per acc. would also get doubled.
But once we have fixed our unit of force, this ratio becomes a fixed quantity called 'mass'.

I want to know wether this is the case or not.
Am i right when I say that it is basically just by virtue of presence of these particles that they acc.?? and that our saying that there actually is "something" that goes on b/w them or there's an "interaction" b/w them , is just our way of looking at it and giving this "interaction" (which might or might not be there) a numerical value to put it on a quantitativee basis?

about the distribution of a force on a rigid finite body:
Let's consider a big rigid block.
Let this block consist of 'n' number of particles each having an equal inertial mass 'm'. Let the total mass of the block be M=nm.
Now let us apply a force somewhere on the block.
Now using the newton's second law, we get F=Ma
And a=F/M
now since all particles move with this very acceleration and since all of them have equal masses we should expect all of them to have been acted upon by same force.
And then we can write the above eq. as
a = F/n*m
a = f/m     where f=F/n

So this shows that the force F that we apply gets equally distributed among n particles.
Am i right in the above.
if i am , then will the force get distributed the same way, in equal amount, irresepective of what particular point of the body i apply the force on???

Thanks
Shikhin  




ANSWER: Hi Shikhin,

it depends on the nature of the interaction that you impose on your particles. But as you have chosen a radial force with magnitude independent of the particles' separation, it is indeed only the the presence of the particles together in one universe that "turns on" the interaction and, therefore, their acceleration. That part should work, I see no problem with it now.

About the rigid block:
[Q]Now let us apply a force somewhere on the block.[/Q]
I notice that you don't mention the Center Of Mass here, which is important afterward.
[Q]So this shows that the force F that we apply gets equally distributed among n particles.
Am i right in the above.
if i am , then will the force get distributed the same way, in equal amount, irresepective of what particular point of the body i apply the force on???[/Q]
No. The force will be distributed so that it causes both linear and angular (rotational) acceleration of the body, which means that the trajectories of each of the n particles will be different. Only in the special case of COM force the rotational effect of the force will be zero.

Personally, I think that we have covered this topic entirely before - there has been nothing in this "rigid body" question that we haven't talked about. Have you reviewed the communication?

All Best!
Daniel


---------- FOLLOW-UP ----------

QUESTION: Hi daniel,

thank you for the reply.
I know it might be bothersome for you to answer the same thing again and again.
But I just can't get it.

[Q]
No. The force will be distributed so that it causes both linear and angular (rotational) acceleration of the body, which means that the trajectories of each of the n particles will be different. Only in the special case of COM force the rotational effect of the force will be zero.
[/Q]
The question then is, why does it distribute itself so as to cause a rotational motion.
Also since the entire body consisting of n particles of equal mass m moves forwards (along with rotation) with an acc. a= F/nm.
This implies that even when the particles go about the axis, they always have an acc. given by the above eq. in the forward direc.
This means each particle experienced a force f=ma in the forward direc. where f is F/n.
Now the different particles also have a changing acc. i.e. the angular acc. "over and above" the linear eq.
There must have been some extra force that produced this extra acc. a=r_i*@
@ being the angular acc??

The force F that I apply does get equally distributed among the n particles, even if I apply F away from COM because as I showed above, every particle moves forward independent of its rotational motion with an acc a = F/n*m or a= f/m.
But its over and above this linear acc. that it has another acc. (that continously changes in direc.). Where or how does this arise???

In a previous answer you said that it's not that torques are less fundamental than forces are.
Does that then mean, that it's just another fact of our universe that on application of forces, rigid bodies undergo rotation as given by the equations of rotational mechanics??? And they just get another acc. component other than the linear one from "somewhere" called angular acceleration???

Is it that we don't know why that happens just as we don't know why particles show acceleration in prescence of each other???


Thanks,
Shikhin



ANSWER: Well Shikhin,

the reason for [Q]The question then is, why does it distribute itself so as to cause a rotational motion.[/Q] is the existence of 1) the inertia and 2) rotational degree of freedom in our Universe. In PURE THEORY, a Universe can be constructed without either or both, we don't know and cannot know the reason, why our Universe has them. In consequence, a question "Why does the rigid body (plank, grid of points) have to rotate?" cannot be answered any better than "because the Universe works this way.

The existence of inertia is formulated by Newton's first law: "All objects in inertial systems of coordinates (SC) tend to conserve their state of motion." Rest is merely a state of motion with zero velocity. The rotational degree of freedom is an axiom, it's not even a law: "Objects of finite dimensions are allowed to rotate unless an external systemic constraint prevents it."

[Q]Also since the entire body consisting of n particles of equal mass m moves forwards (along with rotation) with an acc. a= F/nm.
This implies that even when the particles go about the axis, they always have an acc. given by the above eq. in the forward direc.[/Q]
One has to choose one of the two points of view, which are equivalent, but not only part of one or mixture of both:
View ONE: Describe the each point in the rigid body by a separate equation of motion, which of course includes the constraints of having the the distances fixed between the nearest neighbors. The constraints are what makes the bunch of points a rigid body, of course.
View TWO: Describe the set of points "as a body", i.e. the position of its COM and the body's ORIENTATION (three angles) with respect to an ISC and calculate the total mass and the moment of inertia. Then write only two vector equations of motion that together describe the motion ofthe body entirely: 1. the eq. of motion of COM and 2. the rotational eq. of motion of the whole body about COM. Equation of motion of each point can be then EXACTLY calculated at any time from them, and that is done by "reverse-engineering" of the COM position and body orientation angles.

Your saying that "they always have an acc. given by the above eq. in the forward direc" is using the View ONE and (!) a half of View TWO. No wonder that you keep being stuck. Mathematically, each point's acceleration can be written as a_i = a_COM + a_something. In your above sentence you are trying to tell me that a_i = a_COM, which is not true owing to the presence of the rotational degree of freedom. You keep thinking along the lines: Why does it rotate, when there is no external force-pair that would cause the rotation? The answer is: Rotation is a natural response in our Universe, you see it whenever there is nothing PREVENTING the rotation.

Your last block of questions shows, that you already began realizing all this. I hope my writing has put the matter in another useful perspective.

Cheers,
Daniel

---------- FOLLOW-UP ----------

QUESTION: Hi Daniel,

Thanks for the answer.
I got it, I think.
I just want to confirm it....please tell me if I still haven't got something.

First of all,
I didn't mean  a_i=a_COM
Even I meant to say that a_i=a_COM + a_something.
That is what I meant when I said that every particle has another component of acc. apart from the linear acc. given by a_COM.
And this other component of acc. is because of the rotation of the object.

As observed from an inertial frame, every particle of the rigid body undergoes a translational motion with acc. a_COM which is equal to F/nm  +  it has another part of acc. due to its rotation.

The first part i.e. a_COM = F/nm is due to this force F getting equally distributed.
And the second part, as i get it from your reply, "is just there".
There's nothing preventing it from rotating, and it's just another fact of our universe, that whenever a force is applied anywhere on the rigid body apart from the axis of COM and having some component in the direc. perpendicular to the surface of the body, the body just rotates.
Along with the linear acc. i.e. a_COM, every particle also gets another component of acc. which can be seen as rotation.
And we don't know why that happens.....but it just rotates...

Have I got it correctly or not??

Thanks a lot,
Shikhin

Answer
Yes Shikhin,

this is correct. To go a bit of polishing I would remark that instead of
[Q]The first part i.e. a_COM = F/nm is due to this force F getting equally distributed.
And the second part, as i get it from your reply, "is just there". [/Q]
I would write that F is inherently distributed inequality, but we can write the effect as a sum of a_COM + a_something. And this a_something is - amazingly - the "orbital acceleration" of the point about the COM.

The second and the last remark for now is that only force on the COM point causes linear motion without rotation. Applying force on a COM-axis (of an already rotating object) will cause a combined rotational motion: rotation before force application plus the rotary acceleration of the AXIS of the first motion about the COM. The case, where a rotational axis at rest exists, is a special case....

Good work, take care!

Daniel