Is Centrifugal Force Real or Fictitious?

[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 postive 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 postive 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 postive 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.

 

[e2m2a]

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

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

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

y'(t) = - a/b cos(wt)
where, a = Rwm_r, b = m_r+m_s
x'(t) = a/b sin(wt), where wt is less than or equal to pi/2
x'(t) = a/b, where wt > pi/2

Thus, at 6 o'clock, the total initial velocity of the center of mass is:

y'(pi/2) = 0
x'(pi/2) = a/b

At 3 o'clock, the total final velocity of the center of mass is:

y'(pi) = a/b
x'(pi) = a/b

Or:
total final speed = (2a²/b²)^1/2

Or:
total final speed = 1.41 a/b
initial speed = a/b

Therefore,
total final speed > initial speed

Thanks. Your first-order approximation of the motion of the center of mass confirms the increase in speed.

 

[brainstorm]

Notice how there isn't a force that's trying to break your arms off? There is instead a force that's trying to keep your arms in place. That's something you might need to sit down and think about. If there are no forces acting between parts of your body, they'll be able to float away on their own. The forces are there trying to keep it all together. And depending on what your body is doing, there might not be enough force to do so.

What force is pulling your arms inward except the tensile strength of what they're made of? If some force was pulling the balls inward, wouldn't you have to push them outward to keep them at arm's length?

The balls feel like they're getting pulled outward because they are trying to continue in a straight line and your arms keep pulling them out of that straight line of motion. There is no centrifugal force because it is just objects in motion tending to stay in motion, in a straight line.

 

[e2m2a]

y'(t) = - a/b cos(wt)
where, a = Rwm_r, b = m_r+m_s
x'(t) = a/b sin(wt), where wt is less than or equal to pi/2
x'(t) = a/b, where wt > pi/2

Thus, at 6 o'clock, the total initial velocity of the center of mass is:

y'(pi/2) = 0
x'(pi/2) = a/b

At 3 o'clock, the total final velocity of the center of mass is:

y'(pi) = a/b
x'(pi) = a/b

Or:
total final speed = (2a²/b²)^1/2

Or:
total final speed = 1.41 a/b
initial speed = a/b

Therefore,
total final speed > initial speed

Thanks. Your first-order approximation of the motion of the center of mass confirms the increase in speed.

Correction. On the last post it originally read: "At 9 o'clock, the total final velocity of the center of mass is:" It should read: "At 3 o'clock, the total final velocity of the center of mass is:"

 

[K^2]

y'(t) = - a/b cos(wt)
where, a = Rwm_r, b = m_r+m_s
x'(t) = a/b sin(wt), where wt is less than or equal to pi/2
x'(t) = a/b, where wt > pi/2

Thus, at 6 o'clock, the total initial velocity of the center of mass is:

y'(pi/2) = 0
x'(pi/2) = a/b

At 3 o'clock, the total final velocity of the center of mass is:

y'(pi) = a/b
x'(pi) = a/b

Or:
total final speed = (2a²/b²)^1/2

Or:
total final speed = 1.41 a/b
initial speed = a/b

Therefore,
total final speed > initial speed

By which you can correctly conclude that there is a force acting on the center of mass.

Lets look at the direction of that force, shall we?

Well, the average acceleration in x-direction is (a/b - a/b)/t = 0.
Average acceleration in y-direction is (a/b-0)/t = a/(bt) > 0.

So the acceleration, and therefore the force, is always in the y direction.

Now why don't you explain to me why your "centrifugal force" acts only in y direction?

Or maybe you'll finally realize that it's the constraint of the rail and centrifugal force isn't involved?

What force is pulling your arms inward except the tensile strength of what they're made of? If some force was pulling the balls inward, wouldn't you have to push them outward to keep them at arm's length?

Your arms move in circles. That's accelerated motion. The pull is required to keep them going in circles. Imagine you cut the tension. Are your arms going to accelerate away? No, they'll continue at uniform speed in whichever direction they were moving. Zero acceleration. No force.

 

[e2m2a]

Attached is my paper that shows the derivation of the equations describing the motion of the center of mass of the system. I tried to use latex, but the equations wouldn't "translate" on my reply. One attachment, entitled "prephi postphi" was previously posted on this thread. You'll have to go back and find it if you want to refer to it. Sorry, the system would not allow me to attach it twice.

 

[e2m2a]

By which you can correctly conclude that there is a force acting on the center of mass.

Lets look at the direction of that force, shall we?

Well, the average acceleration in x-direction is (a/b - a/b)/t = 0.
Average acceleration in y-direction is (a/b-0)/t = a/(bt) > 0.

So the acceleration, and therefore the force, is always in the y direction.

Now why don't you explain to me why your "centrifugal force" acts only in y direction?

Or maybe you'll finally realize that it's the constraint of the rail and centrifugal force isn't involved?

I am glad you agree there is a force acting on the center of mass. Your derivation implies it by the increase in the speed of the center of mass. We can continue this discussion if we keep it scientific, civilized, and respectful. For the sake of neutrality, maybe we should call this force the x-force because we don't agree on what the force is, only its consequences-- an increase in speed of the center of mass. (I have just posted my paper that derives the equations of motion. In the paper I assert as a postulate at the end that inertia is the cause of the increase in speed. But you can ignore that, and I'll just call it the x-force in our discussions. Incidentally, you will find that your time-domain equations yield the same results as my angle domain equations for some conditions. I admit this surprised me at first because I didn't think a parametric equation was possible. However, I am still not sure if it is exact for all conditions). Continuing, an increse in speed implies an increase in kinetic energy. Now, this should raise a few eyebrows. An increase in kinetic energy? Now, we have to bring in the conservation of energy and maybe the laws of thermodynamics and the work-kinetic energy theorem. If there is an increase in kinetic energy, there has to be an influx of energy into the system, and it appears it must be done by the mechanism of work. Now you talk about the constraint force. I think you mean some kind of reaction force between the rail and slider. Correct? Okay, if this is the case, then we must show that this "rail force" must displace in space relative to our laboratory frame in such a way that it does positve work on the rotator-slider system. How would this be done? Also, if the rail force is responsible for the work, how does this link to an energy reservoir? By the conservation of energy, since the kinetic energy of the center of mass increases, energy somewhere else must be decreasing. But again, what is this energy reservoir and how is it linked to the constraint force? I will work on answering your question about the centrifugal force. It is subtle and difficult to explain.

 

[brainstorm]

Your arms move in circles. That's accelerated motion. The pull is required to keep them going in circles. Imagine you cut the tension. Are your arms going to accelerate away? No, they'll continue at uniform speed in whichever direction they were moving. Zero acceleration. No force.

What is pulling them in, then, except the tensile strength of the bones and joints? As you say, without your arms being attached, the balls would fly off at constant speed. By "force," you must just be referring to the effect of the arms to counteract the balls' linear momentum.

 

[K^2]

e2m2a

First, there is no such thing as kinetic energy of CM, unless you want to call motion of slider a thermodynamic effect. Kinetic energy of a system is sum of kinetic energies of its parts, and only that must be conserved for this system.

Secondly, Φ = π/2. Fact that I have to keep pointing it out is sad. But enough of that siliness.

Here is the proper treatment from very beginning to the very end. m₁ and m₂ are masses of the slider and rotator respectively. Their coordinates will have same quantities. m₁ will be constrained to remain on positive x axis. Dots over variables denote time derivatives. Double dots denote second time derivatives.

1) Kinetic energy of the system is given by:

$$T = \frac{1}{2}m_1(\dot{x}_1^2+\dot{y}_1^2) + \frac{1}{2}m_2(\dot{x}_2^2+\dot{y}_2^2)$$

2) Potential is zero. (Inertial frame, no external forces.)

$$U = 0$$

3) Constraints.

$$f_1 = (x_1 - x_2)^2 + (y_1 - y_2)^2 - R^2 = 0$$
$$f_2 = y_1 = 0$$
$$f_3 = x_1 \geq 0$$

4) Lagrangian.

$$L = T - U - \sum_i \lambda_i f_i$$

$$L = \frac{1}{2}m_1(\dot{x}_1^2+\dot{y}_1^2) + \frac{1}{2}m_2(\dot{x}_2^2+\dot{y}_2^2) - \lambda_1 f_1 - \lambda_2 f_2 - \lambda_3 \f_3$$

5) Equations of motion.

$$\frac{d}{dt}\frac{\partial L}{\partial \dot{x}_1} - \frac{\partial L}{\partial x_1} = m_1\ddot{x}_1 - 2 \lambda_1 (x_1 - x_2) - \lambda_3 = 0$$

$$\frac{d}{dt}\frac{\partial L}{\partial \dot{x}_2} - \frac{\partial L}{\partial x_2} = m_2\ddot{x}_2 + 2 \lambda_1 (x_1 - x_2) = 0$$

$$\frac{d}{dt}\frac{\partial L}{\partial \dot{y}_1} - \frac{\partial L}{\partial y_1} = m_1\ddot{y}_1 - 2 \lambda_1 (y_1 - y_2) - \lambda_2 = 0$$

$$\frac{d}{dt}\frac{\partial L}{\partial \dot{y}_2} - \frac{\partial L}{\partial y_2} = m_2\ddot{y}_2 + 2 \lambda_1 (y_1 - y_2) = 0$$

6) Performing variable substitution and accounting for rotation.

$$x_2 = x_1 + r cos(\theta)$$
$$y_2 = y_1 + r sin(\theta)$$

This leads to:

$$f_1 = (x_1 - x_2)^2 + (y_1 - y_2)^2 - R^2 = r^2(cos^2(\theta) + sin^2(\theta)) - R^2 = r^2 - R^2 = 0$$

Now that r is a constant, we can easily differentiate to find:

$$\dot{x_2} = \dot{x_1} - r\dot{\theta}sin(\theta)$$
$$\dot{y_2} = \dot{x_y} + r\dot{\theta}cos(\theta)$$

And again...

$$\ddot{x_2} = \ddot{x_1} - r\ddot{\theta}sin(\theta) - r\dot{\theta}^2cos(\theta)$$
$$\ddot{y_2} = \ddot{y_1} + r\ddot{\theta}cos(\theta) - r\dot{\theta}^2sin(\theta)$$

EQM for x₂ now takes this form.

$$m_2(\ddot{x_2} - r\ddot{\theta}sin(\theta) - r\dot{\theta}^2cos(\theta)) + 2 \lambda_1 (-rcos(\theta)) = 0$$

And rearranging a few terms in EQM for x₁ we substitute it into the above...

$$\frac{m_2}{m_1}(2 \lambda_1 (-rcos(\theta)) + \lambda_3) - m_2(r\dot{\theta}^2cos(\theta) + r\ddot{\theta}sin(\theta)) + 2 \lambda_1 (-rcos(\theta)) = 0$$

Multiplying throuhg by m₁ and rearranging some terms...

$$m_1 m_2 (r\dot{\theta}^2cos(\theta) + r\ddot{\theta}sin(\theta)) + (m_1 + m_2) \lambda_1 rcos(\theta) = 0$$

Separating the above into terms with sin(θ) and cos(θ) we get two equations. (If you want rigorous proof that I can do that, you need to consider equations for y. I'm a bit lazy to type it all out, but feel free to check it out. This is the only way that EQMs for x and y can hold at the same time.

$$m_1 m_2 r\dot{\theta}^2cos(\theta) + (m_1 + m_2) \lambda_1 rcos(\theta) = 0$$
$$m_1 m_2 r\ddot{\theta}sin(\theta) = 0$$

Second equation is trivial.

$$\ddot{\theta} = 0$$

In other words, angular velocity does not change. (So it is friction after all.)
First equation is more useful. It gives us one of the lambdas.

$$\lambda_1 = - \frac{m_1 + m_2}{m_1 m_2} r\dot{\theta}^2 = \left(\frac{1}{m_1} + \frac{1}{m_2}\right) r\omega^2$$

Hope it looks familiar, except for the 1/m terms. Since both r and ω have been shown to be constant, so is λ₁.

7) Accounting for rail.

Adding EQMs for y yields:

$$m_1 \ddot{y}_1 + m_2 \ddot{y}_2 - \lambda_2 = 0$$

Substitution...

$$(m_1 + m_2) \ddot{y}_1 - m_2 r\dot{\theta}^2sin(\theta) - \lambda_2 = 0$$

Using constraint f₂...

$$\ddot{y_1} = \ddot{f_2} = \ddot{0} = 0$$

Ergo:

$$m_2 r\dot{\theta}^2sin(\theta) + \lambda_2 = 0$$

Or to make θ dependence a little more clear:

$$\lambda_2 = - m_2 r \omega^2 sin(\theta)$$

Last bit before we are done.

8) Accounting for bumpre.

Same trick as above, but we are working with EQMs for x.

$$(m_1 + m_2) \ddot{x}_1 - m_2 r\dot{\theta}^2cos(\theta) - \lambda_3 = 0$$

Unfortunatley, f₃ is not as helpful as f₁.

$$x_1 \geq 0$$

Gives us

$$\ddot{x_1} = 0$$

ONLY if x₁=0. It is unrestricted otherwise. However, for x₁>0, λ₃=0. So we have two cases.

I) x₁=0

$$\lambda_3 = -m_2 r\dot{\theta}^2cos(\theta)$$

II) λ₃=0

$$(m_1 + m_2) \ddot{x}_1 = m_2 r\dot{\theta}^2cos(\theta)$$

This gives us a very interesting insight. For slider to transition from I) to II) the following condition must be met at x<sub>1[/sub=0.

$$\ddot{x}_1 > 0$$

Lets use equation from II) to expand on that.

$$\ddot{x}_1 = \frac{m_2}{m_1 + m_2} r\dot{\theta}^2cos(\theta) > 0$$

Or simply...

$$cos(\theta) > 0$$

This condition is met for the following range

$$\theta \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$$

Naturally, the transition must be continuous. So let's see what happens to λ3 in I) at the end points.

$$\lambda_3 = -m_2 r\dot{\theta}^2cos\left(\pm\frac{\pi}{2}\right) = 0$$

Which means that these end points are the transition angle we've been looking for.

$$\Phi = \pm \frac{\pi}{2}$$

Which is exactly 12 o'clock and exactly 6 o'clock as advertised.

9) Integrating EQMs.

Starting with θ.

$$\ddot{\theta} = 0$$

$$\dot{\theta} = \int \ddot{\theta} d\theta = \omega$$

$$\theta = \int \dot{\theta} d\theta = \omega t + \theta_0$$

Where θ0 is angle of rotator at t=0.

Coordinates for case I)

$$x_1 = 0$$

$$x_2 = x_1 + rcos(\theta) = Rcos(\omega t + \theta_0)$$

$$y_1 = 0$$

$$y_2 = y_1 + rsin(\theta) = Rsin(\omega t + \theta_0)$$

Coordinates for case II)

$$\ddot{x}_1 = \frac{m_2}{m_1 + m_2} r\dot{\theta}^2cos(\theta)$$

$$\ddot{x}_1 = \frac{m_2}{m_1 + m_2} R\omega^2cos(\omega t + \theta_0)$$

$$\dot{x}_1 = \int \ddot{x} dt = \frac{m_2}{m_1 + m_2} R\omega sin(\omega t + \theta_0) + v_0$$

$$x_1 = \int \dot{x} dt = x_0 + v_0t - \frac{m_2}{m_1 + m_2} Rcos(\omega t + \theta_0)$$

$$x_2 = x_1 + rcos(\theta) = x_0 + v_0t \left(1 - \frac{m_2}{m_1 + m_2}\right) Rcos(\omega t + \theta_0)$$

$$y_1 = 0$$

$$y_2 = y_1 + rsin(\theta) = Rsin(\omega t + \theta_0)$$

Where x0 and v0 are position and velocity of slider at time t=0 respectively.

And that is all. If you bother to find CM coordinates from coordinates above and impose the same initial conditions I used earlier, you'll get the same parametric equations as I posted.

These are exact. These equations were derrived from Lagrangian for this problem. That means it accounts for all possible mechanical interactions, for all effects of rotational motion. The only thing this doesn't account for is friction, which is the ONLY cause for your deceleration of the rotator, and I have been able to account for it quite easily.

You'll find that these equations also perfectly describe motion of the slider. If you are having difficulties believing that, I can post another video where I do show that motion of slider follows the above equations perfectly.

And yes, it is absolutely trivial to demonstrate that the speed of CM changes. It's also trivial to use the above to find the source of that change. It is λ2 from above. Setting it to zero will produce equations similar to equations for x1 in II), and that will result in CM velocity remaining perfectly constant.

What does that tell you? That the cause of CM acceleration is the force applied by the rail on the slider. All of the effects of circular motion of the rotator are contained in λ1, and these effects on slider and rotator cancel each other out, resulting in no acceleration to CM.

Now, what do you have to add to all of this?</sub>

 

[e2m2a]

e2m2a

First, there is no such thing as kinetic energy of CM, unless you want to call motion of slider a thermodynamic effect. Kinetic energy of a system is sum of kinetic energies of its parts, and only that must be conserved for this system.

Secondly, Φ = π/2. Fact that I have to keep pointing it out is sad. But enough of that siliness.

Here is the proper treatment from very beginning to the very end. m₁ and m₂ are masses of the slider and rotator respectively. Their coordinates will have same quantities. m₁ will be constrained to remain on positive x axis. Dots over variables denote time derivatives. Double dots denote second time derivatives.

1) Kinetic energy of the system is given by:

$$T = \frac{1}{2}m_1(\dot{x}_1^2+\dot{y}_1^2) + \frac{1}{2}m_2(\dot{x}_2^2+\dot{y}_2^2)$$

2) Potential is zero. (Inertial frame, no external forces.)

$$U = 0$$

3) Constraints.

$$f_1 = (x_1 - x_2)^2 + (y_1 - y_2)^2 - R^2 = 0$$
$$f_2 = y_1 = 0$$
$$f_3 = x_1 \geq 0$$

4) Lagrangian.

$$L = T - U - \sum_i \lambda_i f_i$$

$$L = \frac{1}{2}m_1(\dot{x}_1^2+\dot{y}_1^2) + \frac{1}{2}m_2(\dot{x}_2^2+\dot{y}_2^2) - \lambda_1 f_1 - \lambda_2 f_2 - \lambda_3 \f_3$$

5) Equations of motion.

$$\frac{d}{dt}\frac{\partial L}{\partial \dot{x}_1} - \frac{\partial L}{\partial x_1} = m_1\ddot{x}_1 - 2 \lambda_1 (x_1 - x_2) - \lambda_3 = 0$$

$$\frac{d}{dt}\frac{\partial L}{\partial \dot{x}_2} - \frac{\partial L}{\partial x_2} = m_2\ddot{x}_2 + 2 \lambda_1 (x_1 - x_2) = 0$$

$$\frac{d}{dt}\frac{\partial L}{\partial \dot{y}_1} - \frac{\partial L}{\partial y_1} = m_1\ddot{y}_1 - 2 \lambda_1 (y_1 - y_2) - \lambda_2 = 0$$

$$\frac{d}{dt}\frac{\partial L}{\partial \dot{y}_2} - \frac{\partial L}{\partial y_2} = m_2\ddot{y}_2 + 2 \lambda_1 (y_1 - y_2) = 0$$

6) Performing variable substitution and accounting for rotation.

$$x_2 = x_1 + r cos(\theta)$$
$$y_2 = y_1 + r sin(\theta)$$

This leads to:

$$f_1 = (x_1 - x_2)^2 + (y_1 - y_2)^2 - R^2 = r^2(cos^2(\theta) + sin^2(\theta)) - R^2 = r^2 - R^2 = 0$$

Now that r is a constant, we can easily differentiate to find:

$$\dot{x_2} = \dot{x_1} - r\dot{\theta}sin(\theta)$$
$$\dot{y_2} = \dot{x_y} + r\dot{\theta}cos(\theta)$$

And again...

$$\ddot{x_2} = \ddot{x_1} - r\ddot{\theta}sin(\theta) - r\dot{\theta}^2cos(\theta)$$
$$\ddot{y_2} = \ddot{y_1} + r\ddot{\theta}cos(\theta) - r\dot{\theta}^2sin(\theta)$$

EQM for x₂ now takes this form.

$$m_2(\ddot{x_2} - r\ddot{\theta}sin(\theta) - r\dot{\theta}^2cos(\theta)) + 2 \lambda_1 (-rcos(\theta)) = 0$$

And rearranging a few terms in EQM for x₁ we substitute it into the above...

$$\frac{m_2}{m_1}(2 \lambda_1 (-rcos(\theta)) + \lambda_3) - m_2(r\dot{\theta}^2cos(\theta) + r\ddot{\theta}sin(\theta)) + 2 \lambda_1 (-rcos(\theta)) = 0$$

Multiplying throuhg by m₁ and rearranging some terms...

$$m_1 m_2 (r\dot{\theta}^2cos(\theta) + r\ddot{\theta}sin(\theta)) + (m_1 + m_2) \lambda_1 rcos(\theta) = 0$$

Separating the above into terms with sin(θ) and cos(θ) we get two equations. (If you want rigorous proof that I can do that, you need to consider equations for y. I'm a bit lazy to type it all out, but feel free to check it out. This is the only way that EQMs for x and y can hold at the same time.

$$m_1 m_2 r\dot{\theta}^2cos(\theta) + (m_1 + m_2) \lambda_1 rcos(\theta) = 0$$
$$m_1 m_2 r\ddot{\theta}sin(\theta) = 0$$

Second equation is trivial.

$$\ddot{\theta} = 0$$

In other words, angular velocity does not change. (So it is friction after all.)
First equation is more useful. It gives us one of the lambdas.

$$\lambda_1 = - \frac{m_1 + m_2}{m_1 m_2} r\dot{\theta}^2 = \left(\frac{1}{m_1} + \frac{1}{m_2}\right) r\omega^2$$

Hope it looks familiar, except for the 1/m terms. Since both r and ω have been shown to be constant, so is λ₁.

7) Accounting for rail.

Adding EQMs for y yields:

$$m_1 \ddot{y}_1 + m_2 \ddot{y}_2 - \lambda_2 = 0$$

Substitution...

$$(m_1 + m_2) \ddot{y}_1 - m_2 r\dot{\theta}^2sin(\theta) - \lambda_2 = 0$$

Using constraint f₂...

$$\ddot{y_1} = \ddot{f_2} = \ddot{0} = 0$$

Ergo:

$$m_2 r\dot{\theta}^2sin(\theta) + \lambda_2 = 0$$

Or to make θ dependence a little more clear:

$$\lambda_2 = - m_2 r \omega^2 sin(\theta)$$

Last bit before we are done.

8) Accounting for bumpre.

Same trick as above, but we are working with EQMs for x.

$$(m_1 + m_2) \ddot{x}_1 - m_2 r\dot{\theta}^2cos(\theta) - \lambda_3 = 0$$

Unfortunatley, f₃ is not as helpful as f₁.

$$x_1 \geq 0$$

Gives us

$$\ddot{x_1} = 0$$

ONLY if x₁=0. It is unrestricted otherwise. However, for x₁>0, λ₃=0. So we have two cases.

I) x₁=0

$$\lambda_3 = -m_2 r\dot{\theta}^2cos(\theta)$$

II) λ₃=0

$$(m_1 + m_2) \ddot{x}_1 = m_2 r\dot{\theta}^2cos(\theta)$$

This gives us a very interesting insight. For slider to transition from I) to II) the following condition must be met at x<sub>1[/sub=0.

$$\ddot{x}_1 > 0$$

Lets use equation from II) to expand on that.

$$\ddot{x}_1 = \frac{m_2}{m_1 + m_2} r\dot{\theta}^2cos(\theta) > 0$$

Or simply...

$$cos(\theta) > 0$$

This condition is met for the following range

$$\theta \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$$

Naturally, the transition must be continuous. So let's see what happens to λ3 in I) at the end points.

$$\lambda_3 = -m_2 r\dot{\theta}^2cos\left(\pm\frac{\pi}{2}\right) = 0$$

Which means that these end points are the transition angle we've been looking for.

$$\Phi = \pm \frac{\pi}{2}$$

Which is exactly 12 o'clock and exactly 6 o'clock as advertised.

9) Integrating EQMs.

Starting with θ.

$$\ddot{\theta} = 0$$

$$\dot{\theta} = \int \ddot{\theta} d\theta = \omega$$

$$\theta = \int \dot{\theta} d\theta = \omega t + \theta_0$$

Where θ0 is angle of rotator at t=0.

Coordinates for case I)

$$x_1 = 0$$

$$x_2 = x_1 + rcos(\theta) = Rcos(\omega t + \theta_0)$$

$$y_1 = 0$$

$$y_2 = y_1 + rsin(\theta) = Rsin(\omega t + \theta_0)$$

Coordinates for case II)

$$\ddot{x}_1 = \frac{m_2}{m_1 + m_2} r\dot{\theta}^2cos(\theta)$$

$$\ddot{x}_1 = \frac{m_2}{m_1 + m_2} R\omega^2cos(\omega t + \theta_0)$$

$$\dot{x}_1 = \int \ddot{x} dt = \frac{m_2}{m_1 + m_2} R\omega sin(\omega t + \theta_0) + v_0$$

$$x_1 = \int \dot{x} dt = x_0 + v_0t - \frac{m_2}{m_1 + m_2} Rcos(\omega t + \theta_0)$$

$$x_2 = x_1 + rcos(\theta) = x_0 + v_0t \left(1 - \frac{m_2}{m_1 + m_2}\right) Rcos(\omega t + \theta_0)$$

$$y_1 = 0$$

$$y_2 = y_1 + rsin(\theta) = Rsin(\omega t + \theta_0)$$

Where x0 and v0 are position and velocity of slider at time t=0 respectively.

And that is all. If you bother to find CM coordinates from coordinates above and impose the same initial conditions I used earlier, you'll get the same parametric equations as I posted.

These are exact. These equations were derrived from Lagrangian for this problem. That means it accounts for all possible mechanical interactions, for all effects of rotational motion. The only thing this doesn't account for is friction, which is the ONLY cause for your deceleration of the rotator, and I have been able to account for it quite easily.

You'll find that these equations also perfectly describe motion of the slider. If you are having difficulties believing that, I can post another video where I do show that motion of slider follows the above equations perfectly.

And yes, it is absolutely trivial to demonstrate that the speed of CM changes. It's also trivial to use the above to find the source of that change. It is λ2 from above. Setting it to zero will produce equations similar to equations for x1 in II), and that will result in CM velocity remaining perfectly constant.

What does that tell you? That the cause of CM acceleration is the force applied by the rail on the slider. All of the effects of circular motion of the rotator are contained in λ1, and these effects on slider and rotator cancel each other out, resulting in no acceleration to CM.

Now, what do you have to add to all of this?</sub>

_{It'll take me some time to study this. I'll attempt it if you will read my paper. I work 12 hour shifts on Fridays and Saturdays, so I won't be able to respond until next week sometime.}

 

[K^2]

I did read it. Scanned through most of the equations. You use the right principles in deriving them, so if you actually use Φ=-π/2 and θ=ωt+θ₀ you should get the right answers. I did not check to make sure that is true. There might be errors, but that in itself is not significant, so I did not check for it. (I know that on the video ω is not constant, but that really is just friction.)

The problem is with your conclusion. Yes, speed of CM changes, but the velocity change, which is more important, is entirely in the y-direction. That immediately tells you that the force responsible is not centrifugal force. Furthermore, the acceleration is up, so the force causing it is pointing in the wrong direction for centrifugal force.

In the complete analysis above, that force is identified. λ₂ is that force. If you analyze the results above, it completely accounts for the speed change you are deriving and observing. Each λ is a generalized force associated with a particular constraint. Constraint y₁=0 in this case. Which means that this force is that of interaction between the rail and the slider.

And so the two solutions are in agreement. You show in your derivation that speed of the CM changes, and I show with a more complete analysis that the cause of the change is force applied by the rail.

And centrifugal force, as promised, is not present.