Is Centrifugal Force Real or Fictitious?

[K^2]

Could you please be more specific on what you mean by the constraint force that is acting between the body and the mass?

A constraint force is a force arising due to the constraint added to the Lagrangian, and can be easily found using undetermined multipliers.

If you were actually asking us about some properties of rotating frames, I'd be compelled to explain. Since you are insisting that you are right, and people who actually study physics full time are wrong, I'm just telling you that you do not understand classical mechanics, and suggest that you go and read up on it. Constraint forces are an important part of classical mechanics, and any survivor of a classical mechanics course, whether at an actual university or self-taught, would be familiar with these.

 

[Studiot]

e2m2a

Since you don't like Newton's approach, perhaps you would prefer D'Arlembert's?

http://en.wikipedia.org/wiki/D'Alembert's_principle

 

[e2m2a]

Because there is a constraint force acting between your body and the mass you are holding on to.

Why don't you sit down and write the Lagrangian for that situation? You'll see that
a) There is no external force needed in the Lagrangian for this behavior and that
b) The force of constraint arising from undetermined multiplier does account for it.

If the above is Greek to you, you should go and read a text on classical mechanics.

K^2, I have the greatest respect and admiration for anyone who has completed an undergraduate or graduate-level degree in physics. I studied physics at the university level for a little over a year, and know of the rigors involved in this discipline. I was unable to complete my studies, and most of my knowledge of physics comes from self-study. I am not gifted mathematically, but I believe I do understand physical concepts well. One law I understand is the conservation of momentum. In your response to my question, what is causing the speed of the center of mass to increase, you answered: "Because there is a constraint force acting between your body and the mass you are holding on to..." Well,
if that is your explanation, then this could not possibly have any impact. A force between my body and the mass I am holding on to, actually it would be a force pair, would constitute an internal interaction between two members within a system. Such internal force pair interactions cannot affect the speed of the center of mass of a system. Consider this quote from Wikipedia, under the section, Conservation of Linear Momentum: "The law of conservation of linear momentum is a fundamental law of nature, and it states that the total momentum of a closed system of objects (which has no interactions with external agents) is constant. One of the consequences of this is that the center of mass of any system of objects will always continue with the same velocity unless acted on by a force from outside the system."
The interaction between my body and the mass I am holding would be an internal interaction within the system, and therefore, the interaction would be unable to change the speed of the center of mass of the system, anymore than a drowning man in a lake could save himself by pulling up on his hair with his hands.

 

[e2m2a]

Correct!



Correct !



NOT AT ALL.

If the rod and the chariot on the track are massless, then your rod stops rotating immediately, and your subsystem starts running immediately at the tangent velocity.

You can see this if instead, you would imagine that both masses broke off at the same time. In that case, as you said, the mass on the left would fly off along the direction of the negative y axis, and the mass on the right would fly off along the direction of the positive y-axis.

Now, if you add a *massless* rod and chariot (slider) to this, this doesn't change anything.
The rod and slider have no inertia, and so the y-balance of forces must be essentially 0 in the subsystem.

It's different if you consider the slider to have mass. Then the rod will still rotate (but at a different angular velocity, which is probably not even going to be constant) and there will be a constant exchange of momentum between the slider and the remaining mass.

In the beginning, all of the momentum will be carried by the mass, and none by the slider, and as the axis rotates, the y-component of the momentum of the mass will diminish, and the y-component of the slider will increase by exactly the same amount (the y-component of the action and reaction forces on mass and slider). The mass will be slowest when it has turned over 180 degrees, and then the slider will be fastest. Next, the mass will accelerate, and the slider decelerate. The slider will continuously accelerate and decelerate, like a bad driver making a humping driving with a car.

(all this in the fixed xy frame of reference).

The center of mass of both will however have a steady velocity in the y-direction, which is given by m2 / (m1 + m_slider) x v2

m1 = mass right hand mass
m2 = mass left hand mass (the one that flew away)
m_slider is mass of slider
v2 = speed with which mass 2 is flying off

Note however that the center of gravity of the subsystem is not being at rest along x: it will wobble from left to right and back. That's because there is a binding force on the track. The subsystem is not free in the x-direction and hence there's no conservation of momentum there.

Your reply is getting very close to what I am trying to explain, but there are a few points I need to clarify. I am working on using latex so I can add some mathematical equations Meanwhile, your comment about the angular velocity not being constant is correct, and is demonstrated in the experiment in the video recording. When I am able to put the math on this forum, I can demonstrate that even though the angular velocity decreases, the tangential speed of the center of mass remains constant.

 

[K^2]

e2m2a, CM of the body + weight never shifts along the line in which the chair is free to move.

If you results differ, you made a mistake.

 

[e2m2a]

e2m2a, CM of the body + weight never shifts along the line in which the chair is free to move.

If you results differ, you made a mistake.

k^2, maybe if you visually saw my experiment, you would see things differently. Verbal descriptions too often don't convey as well as a picture or video, something is always lost in the translation. I may get in trouble for this, but if you would like me to send a short, about 1.5 mb, avi video of the experiment,send me a message in my account.

 

[K^2]

Sure. Sent you an e-mail. You might want to edit your address out now, before addvertising bots find it. By the way, PM system works pretty if you want to send someone your e-mail address privately.

 

[K^2]

Received video.

Clever machine, but there is still a problem with your reasoning. You seem to be under impression that center of mass is at rest, and then starts moving. It's not. The extended mass is moving to the right initially, while the slider is at rest. The CM is therefore moving to the right. It IS accelerating, but because the slider is pushing off the wall. As soon as extended mass passes the "lowest" point (lowest in reference to the screen), the relative velocity of extended mass and the slider begin to decrease. That means, to keep CM speed constant, the slider must accelerate to the right.

If you really want me to, I can go through this video frame by frame and show exact CM speed for every frame.

Part which I did not catch from your explanation is that the slider rests against the wall initially. But once it separates, there is no difference. CM continues at constant speed if there is no friction.

 

[e2m2a]

Received video.

Clever machine, but there is still a problem with your reasoning. You seem to be under impression that center of mass is at rest, and then starts moving. It's not. The extended mass is moving to the right initially, while the slider is at rest. The CM is therefore moving to the right. It IS accelerating, but because the slider is pushing off the wall. As soon as extended mass passes the "lowest" point (lowest in reference to the screen), the relative velocity of extended mass and the slider begin to decrease. That means, to keep CM speed constant, the slider must accelerate to the right.

If you really want me to, I can go through this video frame by frame and show exact CM speed for every frame.

Part which I did not catch from your explanation is that the slider rests against the wall initially. But once it separates, there is no difference. CM continues at constant speed if there is no friction.

Glad you saw the video. Some definitions for clarity. The part you see rotating that has the red circle is called the rotator. The red circle is the center of mass of the rotator-slider system. The rectangular part you see moving to the right is called the slider. No, I am not saying the com is intially at rest. Obviously, you can see in the video that the com is moving before it reaches phi. (I hope you read my power point presentation, so that you know what I mean by phi.) I am saying the com speed increases as soon as it passes phi. You stated the com continues at constant speed if there is no friction. In the power point presentation it shows that this is not happening. How do I know? Well, by the very definition of speed-- distance over time. After phi, the distance the com travels with respect to our laboratory frame (the camera was shot from a ladder attached to the earth) increases for a given increment of time. This clearly means the speed is increasing. You mentioned the slider pushes off the wall. No sir, that does not happen. The reason why it seperates from the wall is because the inertial force, acting on the center of mass of the rotator pulls it away from the left wall or bumper. The slider does not push away from the wall. This would be impossible. If you would like to go over the video, frame by frame, you are welcome to. Anyone else reading this reply who would like to see the video and the short power presentation that explains how the measurement of speed was made, send me a message in my private account.

 

[e2m2a]

k^2, about you remark the slider is pushing off the wall. I think you are assuming there is some kind of elastic collision going on here with the left bumper. The left bumper gets a little compressed as the inertial force pulls the rotator-slider system to the left when the rotator is at the 9 o'clock position, and then this elastic potential energy is released, pushing the slider to the right. However, if you did a careful analysis of the amount of compression of the bumper, it is made of steel, and knowing the Young's modulus for steel, which is very high, the amount of elastic potential energy stored in the momentarily compressed bumper would be so tiny, that the amount of kinetic energy it would impart to the slider when it expands back out would be inconsequential.

 

[K^2]

Lets denote angle by clock-face. Once the rotator passes 6 o'clock, the red circle moves to the right at uniform speed. That can be easily verified.

If you would like, I can put together a video that both demonstrates that and actually shows all the forces at every instant of time.

And you COMPLETELY mis-understand dynamics at the left bumper. Yes, I guess I'll have to put together a video.

 

[e2m2a]

Lets denote angle by clock-face. Once the rotator passes 6 o'clock, the red circle moves to the right at uniform speed. That can be easily verified.

If you would like, I can put together a video that both demonstrates that and actually shows all the forces at every instant of time.

And you COMPLETELY mis-understand dynamics at the left bumper. Yes, I guess I'll have to put together a video.

I am pleased that progress is being made in our discussion. We have agreement. First some preliminaries. The rotating body you see in the video I denote as the rotator. The rectangular body you see, I denote as the slider. The rotator and slider together constitute the rotator-slider system, or just simply the system. And the red circle on the rotator is where the center of mass of the rotator-slider system is. The video was edited or clipped so I could send the video to you. Some servers do not take large files. The full, un-clipped video showed the rotator initially at about 11 o'clock. An impulse was given to the rotator, and where you see the video start, the rotator was already in motion. Let's define an x-y coordinate system, such that the motion of the rotator-slider system to the right is along the x-axis in the positive direction. Up is the positive y-direction. Before the slider moves to the right, the origin of the x-y coordinate system is at the center of the axis of rotation. One other thing, you see the video stop when the rotator is at 3 o'clock. In actuality the full-video shows the system colliding with the right bumper when the rotator is at about 2 o'clock. In a real-world application, the rotator would never get passed the 3 o'clock point before a collision of the rotator-slider system would occur. Collisions could occur before this, such at 4 o'clock, but never after 3 o'clock. At collision, the kinetic energy of the system is converted to another form of energy, such as heat. Then, the rotator would be given a boost up to a critical velocity and the whole process would repeat in the opposite direction. Thus, there would be 2 collisions per cycle of rotation of the rotator. One other thing, if you look carefully at the video, (and I recommend you use quicktime, you can use the left and right arrows to advance the video frame by frame), notice when the slider begins its movement. You can see a crack form between the slider and the left bumper. The angle of the rotator at this point with respect to the y-axis, I denote as the angle phi. This angle is important in deriving equations and making quantitative predictions about the system because at this angle, we can determine the initial momentum of the center of mass of the system with respect to the x-axis.
Ok, now that I have that out of the way, I agree with you that the red circle moves to the right at uniform speed. This uniform motion begins at phi. The red circle must move at uniform speed in order for conservation of linear momentum to be conserved with respect to the x-axis. But now here is a critical point. Think about this. At the point where the rotator has rotated to its 3 o'clock position, the system is still moving to the right at its previous uniform speed, but now we need to include the tangential velocity of the red circle which is pointing in the positive y-direction at this point. When you do a vectorial addition of this velocity and the velocity of the red circle in the positive x-direction, you will find that the magnitude of the total vector is greater than the initial tangential velocity of the red circle at phi! This means, the total magnitude of the speed of the red circle, the center of mass, has increased with respect to our laboratory frame! This is why the video analysis with an AUTOCAD program of the motion of the red circle shows that its speed is increasing. The measurement was accomplished by advancing the video frame by frame. Since the video shot at 210 frames per second, this represented about 4.7 milliseconds per frame. At each frame a dot was carefully placed at the center of the red circle on a flat computer monitor. Then an AUTOCAD program was used to connect these dots to determine the total distance. Equal frame increments were made prior to phi and after phi, such as 90 pre and post-phi measurements. The AUTOCAD program then "connected the dots" to determine the length of each path. The pre-phi path was simple to compute. It was just the locus of a section of a circle. The post-phi path used a spline method to connect the dots. In every case, the post-phi length was always greater then the pre-phi length for the same time increment for both pre and post-phi measurements. We even used a non-spline method on a post-phi measurement, where we used tiny line segments to connect the dots. Since these line segments were made on the concave side of the curve, the total length of these segments understated the true length of the post-phi path. Yet, the post-phi path was shown to still be greater than the pre-phi path. The differences in paths ranged from about 4 to 6 per cent. In these test runs we checked to see that the rotator-slider system was level to the earth, so that gravity would not bias the results of the experiment. What caused this increase in speed of the red circle? Inertia. Thus, inertia is not a "fictitious force", it may not be a true force in a Newtonian, physical-contact sense, but its consequences are real. Its consequences are as real as the action of weight on a system. For example, an over-hanging body, tied by a string to a second body on a table, which rides on a rail, will cause the speed of the center of mass of the two-body system to increase by the action of the weight on the over-hanging mass as the over-hanging mass accelerates to the earth.

 

[K^2]

Ok, now that I have that out of the way, I agree with you that the red circle moves to the right at uniform speed. This uniform motion begins at phi. The red circle must move at uniform speed in order for conservation of linear momentum to be conserved with respect to the x-axis. But now here is a critical point. Think about this. At the point where the rotator has rotated to its 3 o'clock position, the system is still moving to the right at its previous uniform speed, but now we need to include the tangential velocity of the red circle which is pointing in the positive y-direction at this point. When you do a vectorial addition of this velocity and the velocity of the red circle in the positive x-direction, you will find that the magnitude of the total vector is greater than the initial tangential velocity of the red circle at phi!

Dude. There is a force acting in the y direction. The force that prevents the slider from moving towards the rotator in the y direction? One that keeps the slider on the rail? That force? Yeah, that's what is accelerating the CM (red circle) along y.

ONLY in the x direction is your system free to move, so ONLY in the x direction do you need to watch for conservation of momentum. And momentum in the x direction IS unchanging. That means there is NO NET FORCE in x direction. And that means no centrifugal force.

 

[e2m2a]

Dude. There is a force acting in the y direction. The force that prevents the slider from moving towards the rotator in the y direction? One that keeps the slider on the rail? That force? Yeah, that's what is accelerating the CM (red circle) along y.

ONLY in the x direction is your system free to move, so ONLY in the x direction do you need to watch for conservation of momentum. And momentum in the x direction IS unchanging. That means there is NO NET FORCE in x direction. And that means no centrifugal force.

Dude? I didn't know I was a dude. K^2, here is something to think about while I try to master latex equations. I could just easily do this experiment with a "dual" rotator system. The rotators could be synchronized by gears so that each rotator rotates in the opposite sense at the same rotational speed, and in such a way that the y-components of the centrifugal reactive forces acting on each axis of each rotator cancel out along the y-axis. That is, there are no inertial forces pulling the slider in the negative or positive y-direction against the rails because all y-forces cancel out within the slider. Yet, we will still see the constant speed in the positive x-direction, beginning at phi, and at a higher speed than with just one rotator, and each rotator will have a tangential velocity pointing along the y-axis at the 3 o'clock position. There would be no forces possible from the rails, acting on the rotators, to cause these y-component tangential velocities. The reason why the rotator in the video has a tangential velocity in the y-direction at the 3 o'clock position is because of rotational inertia. The instantaneous centripetal force, acting on the tangential velocity of the center of mass of the rotator, causes the direction of the rotator to change, but not its tangential magnitude.

 

[K^2]

That's the whole point. CONSTANT SPEED. Means NO NET FORCE.

Oh, and it's the same speed as right before phi. The red dot only accelerates while the slider is in contact with the bumper. Why? Because bumper applies force on the slider in the positive x direction. Why? because if red dot starts out at rest when rotator is at 9 o'clock (we are looking at x-direction only), in order for it to remain in place, slider would have to slide to the left. Normal force from the bumper prevents that. This normal force acting on the slider, is the net force acting on slider-rotator system, providing acceleration to the red dot.

Want proof? Set up a force sensor at the bumper. The integral of the force from 9 oclock to 6 o'clock will be EXACTLY equal to the momentum of the slider-rotator system past phi.

Don't have a force sensor? I can set it up. I still have a key card to the intro labs, and they have rails, carts, force sensors, and sonars. I can measure the force the bumper applies to the slider, the acceleration of the slider at every moment, and get all the other relevant data.

Do you need me to do all that, or can you see why bumper applies force to slider?

 

[e2m2a]

That's the whole point. CONSTANT SPEED. Means NO NET FORCE.

Oh, and it's the same speed as right before phi. The red dot only accelerates while the slider is in contact with the bumper. Why? Because bumper applies force on the slider in the positive x direction. Why? because if red dot starts out at rest when rotator is at 9 o'clock (we are looking at x-direction only), in order for it to remain in place, slider would have to slide to the left. Normal force from the bumper prevents that. This normal force acting on the slider, is the net force acting on slider-rotator system, providing acceleration to the red dot.

Want proof? Set up a force sensor at the bumper. The integral of the force from 9 oclock to 6 o'clock will be EXACTLY equal to the momentum of the slider-rotator system past phi.

Don't have a force sensor? I can set it up. I still have a key card to the intro labs, and they have rails, carts, force sensors, and sonars. I can measure the force the bumper applies to the slider, the acceleration of the slider at every moment, and get all the other relevant data.

Do you need me to do all that, or can you see why bumper applies force to slider?

Let me think about what you just said and I'll get back to you. Meanwhile I want to complete the mathematical derivations. Some on this forum have stated I need more equations and less verbosity.

 

[e2m2a]

That's the whole point. CONSTANT SPEED. Means NO NET FORCE.

Oh, and it's the same speed as right before phi. The red dot only accelerates while the slider is in contact with the bumper. Why? Because bumper applies force on the slider in the positive x direction. Why? because if red dot starts out at rest when rotator is at 9 o'clock (we are looking at x-direction only), in order for it to remain in place, slider would have to slide to the left. Normal force from the bumper prevents that. This normal force acting on the slider, is the net force acting on slider-rotator system, providing acceleration to the red dot.

Want proof? Set up a force sensor at the bumper. The integral of the force from 9 oclock to 6 o'clock will be EXACTLY equal to the momentum of the slider-rotator system past phi.

Don't have a force sensor? I can set it up. I still have a key card to the intro labs, and they have rails, carts, force sensors, and sonars. I can measure the force the bumper applies to the slider, the acceleration of the slider at every moment, and get all the other relevant data.

Do you need me to do all that, or can you see why bumper applies force to slider?

I have two responses to your last post. First response:
Imagine we take the rotator-slider system and the base—the object that the slider slides on—and we put the whole thing in a cart. The cart itself is on a linear track and can slide left or right with minimal friction along the x-axis. We rigidly attach the base to the cart with bolts and we glue the slider to the base so that the slider cannot move within the base. We set the rotator initially at 9 o’clock. We then give it a quick impulse in the negative y-direction. We measure the displacement of the above system, when the rotator has rotated to 6 o’clock, and find that the cart, base, and slider have shifted to the left by, let's say, 1 cm. This is necessary so that the center of mass of the whole system remains rock-solid at that point it was initially because the center of mass of the rotator has shifted to the right with respect to the x-axis when it moved from its 9 o’clock to 6 o’clock position. To repeat, the slider has shifted to the left by 1 cm, meaning a net force to the left had to act on the slider to do this. Now, we repeat the above procedure exactly as before, except the slider is not glued to the base. It is free to slide in the base as in the video. Yet everything as before will happen. When the rotator is at 9 o’clock, the slider still will have shifted 1 cm to the left, meaning even though there was a reaction force from the left bumper, acting on the slider to the right, the inertial force of the rotator acting to the left on the slider was greater than this reaction force from the bumper, so that the slider still had a net force to the left, shifting the slider 1 cm to the left. Now, let us repeat the above procedure, but increase the mass of the cart this time. Everything will happen as before, except the slider will have shifted to the left by a less amount, let's say .5 cm. We repeat the procedure again, increasing the mass of the cart even more, and the slider shifts to the left only by .0l cm. Finally, we keep doing this experiment over and over again, while increasing the mass of the cart, until the mass of the cart equals the mass of the earth. Which way will the slider shift now with respect to an inertial frame outside of the earth, even though there is a reactive force from the left bumper acting to the right on the slider? Will the net force on the slider be to the right or left? Has anything changed just because we have increased the mass of the cart to the mass of the earth?
Second response: The increase in speed of the center of mass of the system shown in the video begins right after phi, when there is no physical contact between the left bumper and the slider. If the left bumper is the cause of the acceleration, it has to be acting on the center of mass of the system simultaneously as the speed of the center of mass increases. Of course, this is impossible if the left bumper no longer has any physical contact with the slider. It is impossible in Newtonian mechanics for acceleration to occur if there is no force acting simultaneously during the acceleration.

 

[K^2]

You are forgetting many things. Mass of the cart, for example.

This is why I earlier suggested you read a text on classical mechanics. You are making a lot of mistakes by overlooking simple things that you would learn in a mechanics course. And if you write down a Lagrangian for this problem with relevant constraints in places, you can see where all the forces come from, what they act on, and that the net force remains zero.

At no point in any of these setups does the center of mass shift in direction where all components are free to move. If you add the cart, you have to also consider cart's mass when you look for CM. It will no longer be at the red dot. But the CM will still move uniformly, because there are no external forces.

If you think it will help, I'm absolutely serious about setting up the same setup as in the video, but with sensors on board to show forces and accelerations.

 

[e2m2a]

You are forgetting many things. Mass of the cart, for example.

This is why I earlier suggested you read a text on classical mechanics. You are making a lot of mistakes by overlooking simple things that you would learn in a mechanics course. And if you write down a Lagrangian for this problem with relevant constraints in places, you can see where all the forces come from, what they act on, and that the net force remains zero.

At no point in any of these setups does the center of mass shift in direction where all components are free to move. If you add the cart, you have to also consider cart's mass when you look for CM. It will no longer be at the red dot. But the CM will still move uniformly, because there are no external forces.

If you think it will help, I'm absolutely serious about setting up the same setup as in the video, but with sensors on board to show forces and accelerations.

First of all, to have the same set-up you would have to invest some money to build the rotator-slider system. I can give you the dimensions, masses, etc., but I don't think you would want to spend the money. We are talking thousands of dollars in design and machining costs. Second, your emphasis on measuring the forces is not necessary in my opinion. The kinematics of the system tells everything that is needed. The video recording clearly shows the speed of the center of mass increases, something which you seem to downplay. When I say the speed increases, I am not talking about the speed of the center of mass with respect to the x-axis. I agree with you that it remains constant. I am talking about the total speed of the center of mass when you include the y-component of the velocity. Would you use a high-speed video camera to record the motion of the center of mass of the system or would you think it is redundant and unnecessary? If you so, we differ in how to make a meaningful measurment of the experiment. Your emphasis is on the dynamics of the forces acting. My emphasis is on the kinematics. Also, my understanding of the Lagrangian is that Lagrangian formalism is meaningful only when applied to a system where the total energy-- kinetic and potential-- is conserved. I maintain the kinetic energy of the center of mass increases, so the energy of the system is not conserved. And in this experiment there is no clearly defined potential-energy function, so I don't see how the Lagrangian can be used because gravitational potential energy is not involved in this experiment. If you did this experiment on your own, I would like others of your colleagues to interpret the results to see if they concur with you. But again, the video camera is the essential measuring tool. Would you use a high-speed video camera, which in of itself can cost some money? Again, I assert measuring the speed of the center of mass of the system tells all. It should be the basis of measurement of the experiment. Everything else is just fluff.

 

[K^2]

I can build a rotator out of Legos and mount it on a cart. I don't need any money invested into it at all, because I can re-scale all the masses.

Oh, and you insist that there is a net force on the system. I'm prepared to show that the system only accelerates when it's pushing off the bumper, and that the force of the bumper accounts for acceleration of CM. No centrifugal force necessary.

The speed of the CM is irrelevant. It's the velocity that's important. It's a vector quantity. The x-velocity is CONSTANT as soon as slider separates from bumper. Therefore, NO NET FORCE. Done. What else is there?

Yes, the y-velocity changes, but you have rail providing normal force to the slider in y-direction. So that's irrelevant. Only the x-direction matters.

 

[e2m2a]

i can build a rotator out of legos and mount it on a cart. I don't need any money invested into it at all, because i can re-scale all the masses.

Oh, and you insist that there is a net force on the system. I'm prepared to show that the system only accelerates when it's pushing off the bumper, and that the force of the bumper accounts for acceleration of cm. No centrifugal force necessary.

The speed of the cm is irrelevant. It's the velocity that's important. It's a vector quantity. The x-velocity is constant as soon as slider separates from bumper. Therefore, no net force. Done. What else is there?

Yes, the y-velocity changes, but you have rail providing normal force to the slider in y-direction. So that's irrelevant. Only the x-direction matters.

I have attached a picture of the pre-phi and post-phi path of the center of mass of the system. This image was made by the AUTOCAD program. Note that the post-phi path is greater in length than the pre-phi length by about 4 per cent. The pre-phi path is just the locus of the center of mass of the system following the path of a circle. The tangential speed is constant during the pre-phi path, except for a tiny decrease in speed due to friction. The centripetal force acting during this pre-phi path cannot increase or decrease the speed of the center of mass. Now look at the point where phi is. Here, the total speed of the center of mass is increasing, demonstrated by the observable fact that the post-phi length is greater than the pre-phi length for same time increment. The total speed of the center of mass is relevant because I am only concerned with the energy consequences of this experiment. Energy is a scalar. The direction is irrelevant. I am interested in a scalar measurement. It doesn't matter if a Tesla electric car is heading north at 100 mph or south at 100 mph, etc. its the speed that determines how much linear kinetic energy it has, not its direction. I am baffled that you don't see it. I repeat, I am only interested in a scalar measurement-- speed, kinetic energy. Clearly, the image shows the speed increased. Isn't seeing believing? This effect is literally a fleeting effect, lasting only
about .4 seconds. This is something you cannot see with your naked eye. It requires a high-speed video camera to capture the increase in total speed of the center of mass. You know, Michelson, who did the famous experiment that measured the speed of light in the so-called ether of space, who tried to prove the existence of this ether by measuring a change in the velocity of light? Michelson was a very skilled optical experimenter and an intelligent person. He took infinite pains to make this experiment, but the results contradicted what he believed should happen. Even seeing no change in the fringe patterns of the light beam did not convince him that the speed of light was constant. This was a case of a man's thinking getting in the way of what he saw. He went to his grave not believing the results of his own experiment.

 

[K^2]

That angle is exactly 6 o'clock, unless there is significant friction between slider and the rail, which would invalidate your test.

The exact path of center of mass is quarter-circle from 9 o'clock to 6 o'clock, and a sine curve from there on after.

Radius of the rotor: R.
Angular frequency of the rotor: ω
Mass of the rotor: m_r
Mass of the slider: m_s

The exact coordinates of the CM from t=0 at 9 o'clock.

y(t) = -\frac{Rm_r}{m_r+m_s}sin(\omega t)

x(t) = \left\{ \begin{array}{cc}-\frac{Rm_r}{m_r+m_s}cos(\omega t),&\omega t\leq \frac{\pi}{2} \\ \frac{R\omega m_r}{m_r+m_s}\left( t - \frac{\pi}{2\omega}\right), & \omega t>\frac{\pi}{2}\end{array}\right

I suppose, I might as well try to plot this and superimpose it with the video. I'll see what I can do.

 

[e2m2a]

The exact path of center of mass is quarter-circle from 9 o'clock to 6 o'clock.

Actually, the path of the center of mass is a section of a circle from 9 o'clock to phi. Remember, the slider doesn't move unitl phi occurs.

 

[e2m2a]

The exact coordinates of the CM from t=0 at 9 o'clock.

y(t) = -\frac{Rm_r}{m_r+m_s}sin(\omega t)

x(t) = \left\{ \begin{array}{cc}-\frac{Rm_r}{m_r+m_s}cos(\omega t),&\omega t\leq \frac{\pi}{2} \\ \frac{R\omega m_r}{m_r+m_s}\left( t - \frac{\pi}{2\omega}\right), & \omega t>\frac{\pi}{2}\end{array}\right

I suppose, I might as well try to plot this and superimpose it with the video. I'll see what I can do.

I can tell you now the above equations will not work. You cannot parameterize the motion of the center of mass with respect to time. The main reason is you are assuming the angular velocity is constant. It is not. If you examine the video carefully, and if you measure the angular displacement of the rotator with respect to the y-axis, frame by frame, you will see the angular velocity is decreasing. You would first have to determine the angular velocity as a function of time. Good luck. The reason why the angular velocity is not constant has to do with the curvature k. Recall from mechanics that k is the measure of curvature or the tendency of a particle to turn as it travels along a curvilinear path, where k = theta/s. The k of the center of mass of the system is constant during the pre-phi path travel of the rotator, but it begins to decrease at phi and reaches a minimum value at 3 o’clock. You would have to find k as a function of theta, where theta is the post-phi angle of the rotator with respect to the y-axis. This would be no easy task. If you can do this, I would like to see it. And to find k as a function of time, you would have to know theta as a function of time, which means you would have to know the angular velocity. So, you would be back to where you started, trying to deterimine the angular velocity as a function of time. There is a lot of inter-coupling going on here. The bottom line is you cannot parameterize the x and y coordinates of the center of mass because the angular velocity is not constant, which is another reason you cannot use the Lagrangian. Time is one of the independent variables used in the Lagrangian. More precisely, using this formalism, you want to end up with a partial differential equation as a function of time with an exact solution. It cannot be done. You would have to use an approximation technique to find a parameterized solution. Fortunately, there is a much easier way to find the motion of the center of mass. You have to forget time as the independent variable and use theta. Using the conservation of linear momentum, you can find the total linear velocity of the center of mass as a function of theta. One reason you can do this, is because the magnitude of the tangential velocity of the center of mass is constant. Recall the expression, v = r w, where v is the tangential velocity, r is the radial distance from an axis of rotation to the center of mass of a body, and w is the angular velocity. It turns out the r is equal to the absolute value of the inverse of k. So, the above can be expressed as: v = (ds/d theta) (d theta/ dt). Notice that as the angular velocity decreases, so does the curvature k decrease in inverse proportion, such that the d-theta's cancel out, and you are left with v = ds/dt. Thus, the tangential velocity remains constant. In the equations I am deriving, they are all based on the independent variable theta. All I have to do is invoke the conservation of linear momentum to find the total tangential velocity of the center of mass as a function of theta. Actually, I have already derived these equations years ago, it's just a matter of using latex to get them posted on the forum, which is taking a lot of time for me.

 

[K^2]

If anything, ω should increase past 6 o'clock. The electric motor might react to that in a weird way that causes an overall slowdown. I'll take a look at it. And in either case, CM x-velocity is a constant from that point on. So even if the ω(t) changes in time, it's only going to affect the y motion. (Take ω in second part of x(t) to be some effective ω₀)

But I highly doubt ω changes a whole lot. There are some symptoms of that that would be plainly obvious. I'm going to see if I can make a video with the curve above plotted over the frame.

 

[K^2]

So yeah, got it all worked out. You're right, rotator is decelerating, but at apparently uniform rate, so it doesn't make much of a difference. I've made following adjustments to the above equations.

1) Assumed ω is constant from 9 o'clock to 6 o'clock.
2) Used that omega for x on the entire interval.
3) For y past 6 o'clock I assumed deceleration is constant and made relevant adjustments.

And here is the video.

Notice that at the very, very end, the red dot falls short of the green line. Most likely, that's the effect of friction on the slider. But look! The red dot is MOST CERTAINLY not accelerating to the right, because then it would end up moving to the RIGHT of the green curve. It never does that. That means, centrifugal force from the rotator DOES NOT accelerate center of mass. QED.

Enjoy.

Edit: Yes, just added a little bit of deceleration to account for friction, and it works perfectly now. I even have a little box painted where CM is. If you need, I can make that into a video as well.

 

[e2m2a]

If anything, ω should increase past 6 o'clock. The electric motor might react to that in a weird way that causes an overall slowdown. I'll take a look at it. And in either case, CM x-velocity is a constant from that point on. So even if the ω(t) changes in time, it's only going to affect the y motion. (Take ω in second part of x(t) to be some effective ω₀)

But I highly doubt ω changes a whole lot. There are some symptoms of that that would be plainly obvious. I'm going to see if I can make a video with the curve above plotted over the frame.

There is a problem. In the experiment we did not use a motor. The rotator was "free spinning". We gave it an initial impulse around 11 o'clock, and by its own rotational inertia it rotated counter-clockwise. If you use the image in the attachment I sent to compare with a video of your motorized experiment, it is highly unlikey that it will match. Besides, in my opinion, using a motor just adds an additional complication to the motion of the center of mass. In my experiment I took a minimalist approach, eliminating all unnecessary internal inter-actions. Consider this about the angular velocity. If there was a frame instaneously co-moving with the slider, an observer in this frame would experience a gravitational field because the slider is being accelerated. Albeit the acceleration is not constant, but a complex almost, but not exactly, sinusoidal function of time, nevertheless, there is a gravitational field. Thus, an observer in this non-Minkowski, non-Eucledean accelerated frame would expect "gravity" to slow down the angular velocity of the rotator. He or she would predict it would take more time for the rotator to reach the 3 o'clock position. And it does. We would observe the same effect. Here is why: Since the motion of the rotator-slider system is at non-relativistic velocities, time is essentially invariant across the accelerated frame and our laboratory frame, therefore, we too would measure more time for the rotator to reach the 3 o'clock position-- the same amount of time the observer in the accelerated frame measures.

 

[K^2]

You see the video I posted? See the green curve? That curve assumes NO ACCELERATION of the center of mass.

See how it matches anyways?

See the problem with your reasoning?

Yes, the SLIDER accelerates. Center of mass does not. Where is your centrifugal force?

 

[e2m2a]

So yeah, got it all worked out. You're right, rotator is decelerating, but at apparently uniform rate, so it doesn't make much of a difference. I've made following adjustments to the above equations.

1) Assumed ω is constant from 9 o'clock to 6 o'clock.
2) Used that omega for x on the entire interval.
3) For y past 6 o'clock I assumed deceleration is constant and made relevant adjustments.

And here is the video.

Notice that at the very, very end, the red dot falls short of the green line. Most likely, that's the effect of friction on the slider. But look! The red dot is MOST CERTAINLY not accelerating to the right, because then it would end up moving to the RIGHT of the green curve. It never does that. That means, centrifugal force from the rotator DOES NOT accelerate center of mass. QED.

Enjoy.

Edit: Yes, just added a little bit of deceleration to account for friction, and it works perfectly now. I even have a little box painted where CM is. If you need, I can make that into a video as well.

Okay, I'm glad its posted on youtube, now others can see the experiment. Here is my response, point by point. You assume w is constant from 9 o'clock to 6 o'clock, but w is constant from 9 o'clock to phi. Look at the video, the slider does not begin moving at 6 o'clock. You stated, "For y past 6 o'clock I assumed deceleration is constant." I'm not quite sure what you mean by that because the y-component of the center of mass is actually increasing with respect to our laboratory frame. There is a sin(theta) going on here, where theta is the angle of the rotator with respect to the y-axis. When theta gets bigger, its sine must be getting bigger, approaching the value of 1. The y-component of the tangential velocity of the center of mass is determined by this sine factor. I AGREE with you that the total velocity of the center of mass in the x-positve direction must remain constant in velocity after phi. You don't have to try to prove that, I agree with you. But now, consider this subtlity. Why does the speed remain constant? The x-component of the tangential velocity of the center of mass is determined by cos (theta). What happens to the cosine when theta goes to 90 degrees? It goes to zero. Therefore, as the rotator continues to rotate, the x-component begins to "shrink". But, we know the total velocity in the x-direction remains constant. Something else must be added to this shrinking x-component of the com tangential velocity in order to keep the total x-velocity constant. What is that something else? The velocity of the slider must be added. The x-component of the tangential velocity of the com decreases, the x-velocity of the slider increases in such a way that the total velocity of the com is constant to the right. Remember, the total velocity of the com is determined by the vectorial sum of the tangential velocity of the center of mass and the linear velocity of the slider. What causes the acceleration of the slider to the right?
Yes, I would like to see the new video and I would like to see your "adjusted" equations.
One final point, one key point. You still need to measure the pre-phi path of the com and compare it with the post-phi path for the same time increment. This is the crux of the
experiment. In the experiment we did not try to predict the structure of the path. We just measured it. This path actually is an "invariant" structure. You don't even need a reference system to measure it. All you need is an arbitrarily defined unit measure, then you measure the pre-phi and most phi distance for the same time increment (you still need time), and then compare the two. Its the whole point of the experiment.

 

[K^2]

Wow, you are one confused person.

Velocity in the x-direction does not have to "shrink". The ONLY way that the x-velocity can change is if there is an x-Force applied to the CM. There is no such force here (other than friction). That is entirely consistent with the behavior.

You took a simple system of slider/rotator rotating around CM (red dot), and added a constraint in the y-direction (rail). Then you tried to explain what happens, and you explained it completely wrong, because you don't understand constraints. No surprise that in your wrong model you need an extra force to make it work. That doesn't mean it actually exists.

And yes, slider begins to separate from bumper at exactly 6 o'clock, unless you did a crappy job building a rail, and it gets stuck. The motion of the slider is easy enough to describe as well.

Would you like me to also plot slider's position over that video to prove that my model works EXACTLY to describe the motion of slider, rotator, and the CM without having to invent any new forces?

I already posted to you proof that the model I derived is correct. Your "proof" that there must be a force is that your incorrect curve doesn't match video. All that proves is that you screwed up.