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Internal Forces Within Rotating Rigid Body

Posted on 2010-08-24
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When a rod undergoes rotational acceleration, I cannot understand how the tangential internal forces get distributed the way they do.

A uniform density thin rod is pinned at its center. If a perpendicular force, F, is applied to a rod a distance d from the center then
     Torque = d × F = I_cm * a_rot
     (BTW, I_cm = (1/12) M * L² )

The ith atom in the rod has a tangential acceleration, a_i = a_rot × r_i, where r_i is the distance vector from the rod's center for the ith atom.

If the atoms in the rod have mass, µ, then since the ith atom has a tangential acceleration of a_i, then there must be tangential force on that ith atom:
     F_i = µ * a_i
(I know there is also a centripetal acceleration and a corresponding centripetal force.)

So, relating the applied force to the resulting tangential force on individual atoms, I get:
    F_i = µ * a_i =  µ * a_rot × r_i =  µ * (d × F / I_cm) × r_i
    F_i =  (µ/I_cm) * (d × F)  × r_i

What this tells me is that if F is constant, then the tangential force on the individual atoms is proportional to d. I've derived it from basic equations which I saw derivations of. Yet, although the derivations of these equations made sense at the time, I do not understand how the application of a force at d got distributed in such a way so that this last observation is true. I push on a point at a distance d from the CM; and that force get propagated down towards the CM in such a way that as I get closer to the CM, this tangential force gets smaller.

And on the other side of the CM, moving further away from the application point of the force, then those atoms start experiencing a greater tangential force as r_i gets larger.

I suppose one answer is that the internal forces align themselves to satisfy Newton's Laws or Conservation Laws; but I was hoping to get more of an intuitive feeling for how this works internally. In HS, I think I learned how to work out mechanical advantage of tools, yet now I am not sure why I feel less resistance opening a door near the doorknob than pushing the door open near the hinge. (Again, high level concepts like conservation of energy probably simplify the answer, and that is great when solving problems - the high level equations simplify getting an answer to a complex system; but somewhere along the line, I have lost any intuitive notions.)

Granted, even if I apply a force horizontally to a block on a table, internally, each atom must feel the same translational force as every other atom - not clear as to how the laws of nature make that work (although it must because the equations say so), but at least each force in this scenario is the same on each atom (assuming each atom has equal mass, µ).

I'm hoping I'll get an answer that I can understand; I'll up the points as I get more insight.
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good question
With a perfectly rigid rod you would have difficulties. The rod bend slightly giving the required radial force to the ith particle
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The rod possesses stiffness http://en.wikipedia.org/wiki/Stiffness which redistributes forces among its atoms
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>>A uniform density thin rod is pinned at its center

The clue is PINNED at the center. Action and Reaction are equal and opposite. Since the center does not move there is an equal reaction in the direction of the force at the center. This forms a couple, the couple makes the rod turn.

[A body is at rest (or in equilibrium) if the forces acting upon it all meet in a point and their resultant is zero. Here the action and reaction are parallel so a couple results. If the resultant of the forces is a force in a fixed direction, then the center of gravity of the body will more in that direction. If the resultant of the forces produces parallel forces then a couple will result, and an angular acceleration will result.]

Ozo's Stffness is a sort of friction. There is friction between the atoms. This friction is the result of a sheer force and this sheer force will be balanced by reaction force which is perpendicular to it. Assuming 100% stiffness, this is how the Action force gets to the pinned center of the rod.
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>>even if I apply a force horizontally to a block on a table,

The weight of the block lying on a horizontal table causes an equal reaction from the table to support the block [Newton: Action and Reaction are equal and opposite]

On a rough surface friction is proportional to the reaction normal to that surface and is independant of the surface area on which the body sits (to a large degree). The frictional force is always normal to the Reaction, ie along the table, and against any moving force. A body will slip if the force moving the body is greater than the frictional force which maintains it in position (again in the line of action of the frictional force). Once a body moves there is a sliding or rolling frictional force which is not necessarily the same as the stationary force.

In a perfect body all the atoms have a frictional force such that it is immaterial at which point the force is applied. In practise this is not so.

With this information you ought to be able to solve "Park the Car in Harvard yard".
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>> Since the center does not move there is an equal reaction in the direction of the force at the center.
>> This forms a couple, the couple makes the rod turn.
But at the center, the forces are ~0, so how is there a couple?

>> With a perfectly rigid rod you would have difficulties.
So, I take it that Newton's laws work because in reality that there is no such thing as 100% stiffness (just like there are no such things as point mass), but the laws still work on a simpler model of reality.

>> Assuming 100% stiffness
But, now I'm contradicting myself with the previous assumption that 100% stiffness results in difficulties.
(Having trouble resolving these two.)

BTW - I'm not sure it matters, but let's assume that the rod is very heavy and is on a frictionless ice pond.

So, if I push the rod with a force half way between the center and the end, I think I feel a greater resistance to my pushing than if I push with the same force at the end of the rod (i.e., I think it is easier to move the rod when I push closer to the end). It's as if the rod has more inertia (rotational inertia, I guess) the further from the end that I push. Any idea what is going on internally to resist my push more when not at the end of the rod?
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"So, if I push the rod with a force half way between the center and the end, I think I feel a greater resistance to my pushing than if I push with the same force at the end of the rod (i.e., I think it is easier to move the rod when I push closer to the end). It's as if the rod has more inertia (rotational inertia, I guess) the further from the end that I push. Any idea what is going on internally to resist my push more when not at the end of the rod?"
actually I do not think this is what happens. If the rod if on a perfectly frictionless surface I do not think it rotates if you push on the end.
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in the center, the force from the pin is opposes the  force, F, applied to a rod a distance d from the center

If infinite stiffness is a problem for you, you can take the limit as stiffness approaches infinity,
(stiffness is bounded in relativistic or quantum physics, but newtonian physics should not object)

on a frictionless surface, with the center unpined,
If you push near the center, the point that you push moves less with the same force (which is what I think you mean by feel a greater resistance)
than if you push near the middle, however the center moves more and the rod rotates less
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>>But at the center, the forces are ~0, so how is there a couple?

You are forgetting Newton's Laws.

A body will move in the direction of a force and will continue to do so until acted upon by another force.

The rod is smoothly pinned at the center. A force at one end should, by Newton's first Law, make the body move in the direction of that force. It does not, it rotates. Why?

A body, if acted upon by TWO forces, will move according to Newton's first law in the direction of (and acceleration of the magnitude with) the resolute of the two forces. What if the forces are parallel? If the two forces are equal and opposite and act in the same line of direction, the body will remain stationary. If the two forces are in the same direction then they will simply sum. But if the two forces are parallel and opposite then a couple is produced which will cause the body to rotate.

Now since the rod rotates under the action of ONE force, there must be an equal and opposite force at some point which causes the body to rotate. This is the force at the pin. Since this force is equal and opposite, it is, by Newton's Third law, a reaction - since action and reaction are equal and opposite.
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>> I do not think it rotates if you push on the end
Why not? I have the notion of degrees of resistance to a push, but I can't see how the rod could resist completely the push (unless the push were at the pin location).

>>Since the center does not move there is an equal reaction in the direction of the force at the center.
So, at the pin, there must be a force equal and opposite direction to the force applied at the end. Makes sense since otherwise the entire rod would have translational acceleration. Thanks.

>> This forms a couple, the couple makes the rod turn.
I see the couple as noted above, but since one of the two forces is on the pin, it seems that a rotational couple does not exist. But, now I ask is a couple even necessary to make the rod rotate? If it were stationary in outerspace (not anchored - free floating) and I pushed on one end (perpendicular to the rod), then it would experience a rotational acceleration (and for lack of a couple, it would also experience a translational acceleration - at least so say the formula).

>> with the center unpined
In this Q the center is pinned (but I'll bookmark this response in dealing with another question I asked).

>> the point that you push moves less with the same force (which is what I think you mean by feel a greater resistance)
This is a good way to describe this. The distance in this case would be the arc of the point pushed around the center. So, if I push perpendicular to this (initially stationary) heavy rod in two separate tests, (1) one near the end, and (2) one half way to the end, then when I apply a small force, the rod atoms in (1) travel further (i.e., a larger rotation) than in test 2. Why don't the atoms near the point of application resist the force the same way whether at the end or half way to the end of the rod?
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>>  This friction is the result of a sheer force and this sheer force will be balanced by reaction force
So, I gather that upon contact the atoms touched are compressed perpendicular (along the force dircection), and that these atoms have a "sheer force" on adjacent atoms causing them to compress, and so on up and down the rod. It seems like there is a wave of energy that slides around the pin and all the way to the other end of the rod.

Somehow, if the compressed atoms are half way towards the end from the center, these atoms do not move as far as the corresponding atoms at the end of the rod for a corresponding force.

BTW - re: "friction" analogy. Perhaps there is another term since friction, I thought, always connotes heat emanation with consequent energy loss. I believe we can talk about a totally elastic case where there is no energy loss.

So, instead of a friction analogy, maybe a spring analogy works? At least at the macroscopic level, a frictionless "perpetual motion" spring is a little more intuitive. Is there a wave propagation of the force that somehow causes the "springs" in the rod to have greater resistance to the force (i.e., less displacement) if the point of application is half way towards the end than if the force is applied at the end?

Since this topic is not discussed in basic college physics courses, is this a topic that is studied in any physics course? I imagine that civil engineers may need something like this to analyze forces in the structures they build to make sure they do not break under various force application scenarios, but is this kind of discussion a part of classical physics courses?
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"If it were stationary in outerspace (not anchored - free floating) and I pushed on one end (perpendicular to the rod), then it would experience a rotational acceleration (and for lack of a couple, it would also experience a translational acceleration - at least so say the formula)."
-
But that is the point. It would NOT experience a rotational acceleration BECAUSE of a lack of a couple.
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In my rod questions, the term "couple" has not been introduced until this thread. But reading through this thread, I believe I am seeing different conflicting comments as to whether there is or is not a rotation for two different scenarios - with a pin, and without a pin.
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An off center force will cause rotation, with or without a pin.
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Then is there a couple formed by the force propagation up/down the rod that causes the rotation?
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>>Since this topic is not discussed in basic college physics courses, is this a topic that is studied in any physics course?

I have no idea what one gets taught in basic physics courses these days, I did mine forty-five years ago. In that we had Applied Maths (as one might call it) which was basic Statics and Dynamics. The core feature of these courses was Newton's Laws and they were applied vigoriously. The first thing from these Laws lead to Lami's Theorem (http://en.wikipedia.org/wiki/Lami's_theorem - please learn), Moments (the moment of a force about a point), which leads to parallel forces and couples (or torques if you like that term), coplanar forces (reduction of many forces to a force and a couple - needed in bouancy problems) bending moments, Hookes Law and centers of gravity.

It seems to me phoffric that you need to take a book on statics and go through resolution of forces and moments. That will arm you for a lot of statics problems, since friction and the like are relatively straightforward once one has mastered these. Question in dynamics are often of the type of what happens when a static situation is disturbed, so having a good foundation in Statics will help.
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Never heard of Lami.

>> couples (or torques if you like that term)
I'll go with any term that has a definition. From this discussion and web verification, a couple requires at least 2 forces. I didn't realize that a torque required two forces. So, in the example of the rod with or without the pin, are
you saying that there is a couple that makes the rod rotate?

If so, where is the other force that composes the couple? (The one force is, of course, the point of application; but the other?)
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The reaction, otherwise the force on the rod would make it move in the direction of that force. Newtons SECOND law.
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If I push the rod, the reaction force (I thought) was just the rod pushing back on me (Newton's third law). I don't see how that reaction force is the couple necessary to rotate the rod. I guess you are talking about a different reaction force. With the pin version, you were referring, I think, to a reaction at the pin; with the unpinned version, I still don't see what 2nd force you are speaking of to form the couple.
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For this couple (that is elusive to me), what is the magnitude, direction, and point(s) of application of the two forces that makes the rod rotate?
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>>With the pin version, you were referring, I think, to a reaction at the pin; with the unpinned version, I still don't see what 2nd force you are speaking of to form the couple.

I'm getting terribly confused as to which senario you are refering. The original question refered to a pinned rod. I'm leaving the office in an hour or so, so I'll see you again in three weeks time with a puzzle about a stairway. For that you need to know how to take moments!
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OK, have a good vacation! 3 weeks - great!!

I'll be happy with just the pinned version for this question. I'll keep this question open if you would like so that it will be fresh on your mind in three weeks. :)
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I just started reading about couples.

"A couple consists of two parallel forces that are equal in magnitude, opposite in sense and do not share a line of action. It does not produce any translation, only rotation. The resultant force of a couple is zero. BUT, the resultant of a couple is not zero; it is a pure moment."

    http://www.uoregon.edu/~struct/courseware/461/461_lectures/461_lecture12/461_lecture12.html

If that is also your definition, then since ozo said that the unpinned version has translation, then maybe a couple does not apply for that case.

Reading more about couples on rigid bodies that are pinned, I found something about a single force being replaced by an equivalent couple. It was brief, and I'm not sure they are saying there really is a couple, or just making an equivalency statement.

I'm still wondering about the force propagation down the rod (if that is the correct terminology), and why the atoms seem to resist a force differently depending upon where the force is applied.
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There were a number of useful comments above, and in further research on the web, I found one derivation of torque that did not start out with a definition of torque, but rather came up with the concept when considering work and resultant increased energy. Now, I think I am beginning to get a more intuitive idea of what is happening at the micro level.

On one topic mentioned in one post, I think I do see a couple around the pin (although the way I am thinking of it, I wouldn't call it a reaction force).

On the original topic, I am inclined to believe that a larger force away from the rod's end is required to move an atom a small delta x distance than a force at the end, because the atoms away from the rod's end have to drag the other atoms in the rod a larger distance, and those atom's inertia pulls harder against the atoms where the external force is applied (i.e., the reaction force).

First, I have to define what I consider intuitive: It is simply that if an external force is applied to a solid block, then the block's translational motion is determined by F_external = ma.

To test out my ideas, I was thinking of modeling the rod in small cubes (one layer high, and three layers wide), and then saying that there were four springs, one per side, that connects to the adjacent cubes. The pin would replace one cube at the rod's center.

If this is a fair idea, then one issue (maybe more for another question, if I actually program this) - I am not sure yet how to determine the propagation speed of the effect of an external force applied to one cube (perpendicular to the rod).
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The propagation speed will be proportional to the spring length times the square root of the spring constant divided by the mass
http://en.wikipedia.org/wiki/Wave_equation#Derivation_of_the_wave_equation
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Where I am getting confused is the discussion of reaction force:
>> Since the center does not move there is an equal reaction in the direction of the force at the center. This forms a couple, the couple makes the rod turn.
   I do not see how the force of the rod's material on the pin and the resultant reaction force by the pin on the rod results in a couple that makes the rod turn.

======

   Also, are very rigid springs connecting point masses a reasonable approach in simulating rod rotation? If so, I am a bit confused about whether the propagation speed will come out of the simulation rather than my computing it in advance and then using it somehow.

    I am not sure how to create incremental time steps that results in the correct propagation speed. If I move one point mass a tiny delta-x, then on the next iteration, I could see that only those adjacent point masses are affected by the spring force, and that the end of the rod is not affected (yet). But using this approach, if there are N point masses between the point of force application and the end of the rod, then after N iterations, the point mass at the close end of the rod begins to move. But this is independent of the propagation speed.

   I may try to do this if I can figure out how to get started. I would hope that I would then be able to even remove the pin and see translation and rotation from the simulation.
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If your delta-t is small compared to the propagation speed of the spring coupled masses, it should be a good approximation.
You should adjust the masses and spring constants so that the propagation speed equals the speed of sound in the rod material.
You may also want a little bit if damping.
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>> If your delta-t is small
Ok, thanks. I'll give that a try. It'll probably take awhile.

Re: couple and torque wiki links..
  I had looked at them, but I didn't see anything that talks about a reaction force causing the couple (i.e., reaction as in Newton's Third Law of Motion).

What I think causes the couple is that if the force is to the left, then
 -- the atoms close to the pin will soon (at the speed of sound?) feel a force also to the left due to sheer forces
 -- the atoms to the right of the pin will feel a force upwards and to the left and move accordingly in that direction.
 -- This stretches the "string" causing the atoms below the pin to feel a force upwards and to the right (and move in that direction).
 -- And these atoms below the pin will pull the atoms to the left of the pin downwards and to the right.
    -- And apparently, this latter force is stronger than the sheer force of the atoms located at the top left part of the pin moving to the left pulling the atoms to the left of the pin upwards and to the left.
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>> You may also want a little bit if damping
   Before you made this comment, I was wondering about whether the springs would cause an infinite oscillation after the force abated. Is this the reason for the damping suggestion - to avoid this oscillation; or perhaps to add stability given small numerical error buildup?

   Earlier it was mentioned that the shear force is sort of like friction. I think of damping also like a frictional force. So, if the rod is on a frictionless ice pond, then if there is damping, won't that result in the rod losing all its energy?

BTW - If I do run into any questions on an implementation of this simulation, it will be addressed in a different question. It'll be awhile, not just to work out the simulation, but probably longer to work out the graphic visualization and an interactive interface.
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Without damping, there would be an infinite oscillation after the force abated,
But if the force was applied slowly compared to the propagation speed, the oscillations should be small,
and may not cause a problem.  Numerical errors may provide sufficient damping, you wouldn't need or want much,
as long as you always round towards lower energy.

Damping of the coupled spring system should not affect conservation of momentum,
(if it does, you have too much numerical error)
It does lose the energy that went into oscillations, but the oscillation energy had already come out of the
rotational and translational kinetic energy when you decided to use a rod with finite rather than infinite stiffness.

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Thanks for the advice.
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Also, it sounds like your applied force is imparting a specific amount of momentum,
not a specific amount of energy, so more vibrations could just absorb the extra energy by
making the force work through a longer distance than it would with no vibrations.
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In previous questions, there was some disagreement about my talking about applying an Impulse rather than a force. So, what I'm planning on doing to avoid (temporarily) this issue, was to make the rod heavy and apply a force for a specific delta-x. In this way I impart a fixed amount of energy to the rod. Good point about removing the oscillations - I didn't want the energy to go there.

Also, the Impulse issue seemed to be related to the short duration. But, it is my understanding that an impulse need not be short. So, for a very heavy rod, if I take a small constant force over a long time (say, 1 second), then I have an impulse that may bring up the issues I'm asking about without the issue of the impulse having an unknown, or unpredictable force (over a short time interval). At least I think that was the issue raised in previous questions.

>> if the force was applied slowly compared to the propagation speed, the oscillations should be small
    Are you referring just to a simulation issue, or a physics issue. That is, if the force were applied quickly, as in an impulse of short duration, would oscillations occur? This may be a subject for a new question.
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Thank you for all the advice and guidance in helping me think about these types of problems differently. I think I have a better sense of what is going on at a lower level.

In reading about couples, I learned that a single force can be replaced by a couple; and "Conversely, a couple and a force in the plane of the couple can be replaced by a single force, appropriately located."
    http://en.wikipedia.org/wiki/Couple_%28mechanics%29
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