Is Centrifugal Force Real or Fictitious?

[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 &gt; 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) &gt; 0

Or simply...

cos(\theta) &gt; 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 &gt; 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) &gt; 0

Or simply...

cos(\theta) &gt; 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.

 

[e2m2a]

Now, what do you have to add to all of this?

I admit I have been wrong about Lagrangian analysis. After seeing your derivations, I have more respect for it and I am beginning to appreciate its beauty and power in analyzing systems. And, if you are absolutely certain your derivation of the angular velocity is constant, then I admit the Lagrangian does yield exact solutions for the equations of motion. I can solve simple Lagrangian problems ( I had to teach myself) concerning pendulums and masses on inclined planes, etc. , but the advanced analysis you have done with Lagrangian multipliers and constraints is beyond my experience right now. I could learn it if I took the time, but I am busy with other things. The reason I doubted an exact parametric equation was possible is based on my experience with this experiment. I have actually been working on it for about 4 years. I have done different versions of it, and always the angular velocity was not constant. I struggled with deciding if this was due to friction or some other fundamental law. If your analysis is correct, then you have answered the question for me and I thank you for that.
I thought I would point out some typos and other errors in your analysis. Under section 4, where you express the Lagrangian with the lambda multipliers, the last term reads lambda33. I think you mean lambda3f3: Under section 6, where you take the first derivative of y2, your first term on the right of the equality is x dot sub y. I think you mean y dot sub 1. In the same section at the end, I think your lambda1 equation is inverted, and I think you should not have an r factor left in the numerator. (I say this because the dimensional analysis of this equation does not make sense.) If you check on the previous page, where you have the two equations without the rigorous proof, I think you’ll see the algebraic error. Since, you have asked me about how I could demonstrate that centrifugal force has anything to do with the increase in speed of the center of mass, and since you have given me the full Lagrangian analysis, I am now going to give you my full explanation of how centrifugal force accelerates the speed of the center of mass. After seeing your Lagrangian analysis I was pleasantly surprised to see something that I think you may find interesting also. I will talk about it at the end.
Bear with me if I seem to go off on tangents with stories, but it is one of the ways I explain or teach a concept. Suppose an individual visits a science museum and after seeing all the standard demonstrations, Foucault’s pendulum, etc., he runs across an exhibit that has a rotating body pivoted at one end inside a glass box that has a near-vacuum inside. The sign on the exhibit states this body has been rotating for 11 years, and has been maintaining a constant angular velocity without the aid of a motor, and rotates solely by its own rotational inertia around high-tech magnetic bearings. The sign further states this exhibit demonstrates Newton’s first law as pertaining to rotation: a body in rotation stays in rotation, unless acted upon by a torque. The individual is intrigued by this. Not having a physics background, the individual wonders how the body could do this without something continuously pushing it. He did read somewhere years ago that the Greek philosopher Aristotle believed that the natural state of bodies was to stay at rest, not to move. In fact Aristotle taught the reason why an arrow continues to fly through the air is because “something” keeps pushing it to maintain its flight. At this time a museum guide notices the individual is perplexed and asks if he has any questions? The individual responds that he does, and the guide starts answering them. The individual asks what force keeps the body rotating? The guide answers its on rotational inertia, there is no outside force necessary to keep it rotating, only an internal force to change its direction. What causes it to change direction, he asks? Centripetal force, the guide answers. It acts perpendicular to the tangential motion of the body, and therefore, only changes the direction of the body, not its speed. Its tangential speed, he continues, is 10 meters per second. Will it always be 10 meters per second, the individual asks? Yes, the guide answers, unless you have relative motion with respect to the body. What do you mean, the individual asks? Wait until the body is at the 6 o’clock position, the guide answers, then quickly step back with continual motion and observe what the speed is at the 3 o’clock position. What has happened to the speed? The individual steps back while still moving and observes the body appears to have gained speed. I guess its going faster when I move back, the individual observes. Right, the guide responds.
Why does it speed up? Has the rotation of the body changed in any way? The individual thinks about it for a moment. No, you said the body by its own inertia continues to rotate on its own at a constant speed, and I don’t see how any force could act on it just by my stepping back. Right, the guide answers. The reason it speeds up is because, relative to you, the rotator at its 3 o’clock position has gained an additional speed, pointing to the right because you are moving to the left. In fact, if you recorded the motion of the body with a high-speed video camera while you moved to the left, and then played it back at slow motion, you would see the body inscribe an arc of a circle from 9 o’clock to 6 o’clock, but the shape of the path would “stretch” out from 6 o’clock to 3 o’clock for the same interval of time. Then measuring the two different paths, you would find the post-6 o’clock path is longer than the pre-6 o’clock path, demonstrating the body has gained in total speed. The individual thinks about it. First he observes the motion of the body without moving back and notices the speed at 3 o’clock does not change. He then moves back beginning at 6 o’clock, and notes the total speed of the body speeds up. He does it over and over again. No movement back, speed is constant. Move back, speed increases. He concludes correctly, its only due to relative motion of the body to the right at the 3 o’clock position that the body has sped up. Now, I think you get the point I’m making. The reason why the center of mass of the rotator-slider system increases is because of the gain in speed of the slider to the right. Period. By you own derivation, since the rotational velocity is constant, there is no other factor necessary to explain the increase in speed of the center of mass. The center of mass has simply acquired an additional velocity to the right when it is at its 3 o’clock position due to the speed of the slider to the right. Furthermore, there is no need to find an external force, such as a rail constraint force, acting on the center of mass of the system to account for the increase in speed of the center of mass because it is only the increase in speed of the system to the right that accounts for the overall gain in speed of the center of mass as demonstrated in my “story”. Surely, this constraint force from the rail, could not act to the right on the center of mass of the system, since this normal force can only act perpendicular to the motion of the slider. By the way, concerning constraint forces, I looked it up, and everything that I read states that the constraint forces used in Lagrangian analysis cannot do virtual work on a body because there is no real or virtual displacement associated with them. More on this later. So here is the crux of the matter concerning the gain in speed of the center of mass. It is due to and only due to the increase in speed of the slider to the right, which speed is added to the speed of the center of mass at the 3 o’clock position. But now, we do have to find a force that is responsible for the acceleration of the slider to the right in the positive x-direction. What causes the slider’s speed to increase to the right? The x-component of the centrifugal force. This is where the inertial force steps in. It never acts in the positive y-direction from 6 o’clock to 3 o’clock. It doesn’t have to. It only has to act in such a way to increase the speed of the slider to the right, and there is an x-component of the inertial centrifugal force that acts in the positive x-direction as the rotator rotates from 6 o’clock to 3 o’clock. Now I am going to show a mathematical derivation how this happens and how this exactly matches your Lagrangian analysis.
To do this, I am going to use my hanging-weight, swinging-weight explanation. Let's begin with a simple high-school experiment. Imagine two bodies. Both are connected by a wire. One body hangs over the edge of a table. We call this body m₂. The other body initially rests on a rail on top of the table. We call this body m₁. At some point in time we release the hanging body, and the whole two-body system begins to accelerate as the hanging mass accelerates downward. Simultaneously, the speed of the center of mass of the system increases. Obviously, all of this is due to the weight acting on m2, the overhanging mass. Now every physicist on this planet has no problem in understanding the mechanism of this motion and accepting weight as the “real” force that causes it. But now watch what happens next. Imagine this two-body system is a Dali surreal painting. We “morph” the overhanging mass so that it transforms into the rotator, and instead of hanging down, its “lifted” up, so that it can rotate in a plane parallel to the table. As it rotates, an inertial force manifests in the center of mass of the rotator. This inertial force arises by the same mechanism that the weight manifests in the overhanging system – the forces arises from a departure of the bodies from their natural path in spacetime. That is, weight manifests when a body departs from its locally-curved geodesic in curved spacetime; Inertial force manifests when a body departs from its locally-straight line geodesic in un-curved spacetime. Einstein made a big deal about this equivalence between inertial force and gravitational weight, which led to his strong and weak principle of equivalence and to his theory of general relativity. For Einstein, there really was no difference between gravitational weight and inertial force. Yet, it fell on deaf ears then, and it still falls on deaf ears today. But now I will apply this equivalence literally, and show how it predicts the motion of the rotator-slider system. Instead of thinking of centrifugal inertial force acting on the rotating rotator, just replace this with a “weight force”, acting in an outward radial direction on the rotator. Every dynamic and kinematics motion of the system can be predicted by using this substitution.
First, we take your equation from your Lagrangian analysis. Now, at this point I am going to make a change in coordinates to simplify for me the use of the trigonometric functions. I am going to rotate the coordinate system so that the rotator-slider system moves in the positive y-direction. This change will not affect you Lagrangian analysis since it is not bound to any coordinate system or any orientation of a coordinate system. (One of the usefulness of the Lagrangian). Recall after taking the first derivative of your derived equation, we have this for the total constant speed of the center of mass of the system in the positive y-direction, for wt > 0 radians:

y`(t) = (rwm2)/(m1 + m2) equation (1).

Where, m1 is the mass of the slider and m2 is the mass of the rotator.
Now, there are two components that make up this total velocity of the center of mass of the system in the positive y-direction: a component due to the instantaneous speed of the slider in the positive y-direction and the y-component of the tangential velocity of the center of mass of the system as it rotates from 3 o’clock to 12 o’clock. Now, the y-component of the tangential velocity of the center of mass as the rotator rotates from 3 o’clock to 12 o’clock is:

vcom y = [rwm2cos(wt)]/(m1 + m2) equation (2).

Now, I am ready to use the centrifugal or weight force to find the speed of the slider in the positive y-direction. Now this weight force has a postive x-component. This force “pulls” the slider against the rails in the positive x-direction. There is a reaction, normal constraint force to this from the rail that acts on the slider in the negative x-direction. Can this force effect the system in anyway? Possibly, if you assume centrifugal force is a fictitious force and not real. Absolutely not, if you realize the centrifugal “weight” force is as real as a weight force. If we do a free-body diagram of the slider, we find that the x-component of the centrifugal weight force in the postive x-direction cancels out the normal reaction rail force acting on the slider in the negative x-direction. Since it is axiomatic that the net effect on any part of a multi-body system affects the center of mass of the system, this cancellation of forces has no effect on the center of mass. This is no different than the case where we have a body sliding down an inclined plane. There is a normal constraining force that keeps the body at an angle as it slides down. However, this constraining force is always normal to the motion of the body, so it can never do virtual work on the body. And, since we can find a component of the weight acting on the sliding body that cancels this normal force, there is no net force acting in the direction of the normal force that can do virtual work in this direction. Now, we return to the hanging weight system and find the force acting on m1, the mass of the object on the table. Without derivation, we find it is equal to:

f weight m1 = m1m2g/(m1 + m2) equation (3).

Now I “lift” this expression and apply it to the rotator system as follows:

fcenm1y = [m1m2 wsq r sin(wt)]/(m1 + m2) equation (4).

This is the y-component of the centrifugal or weight force acting on the slider in the positive y-direction. Note, I have only substituted wsq r sin(wt) for g. Now I seek the change in velocity of the slider due to this force. I can find it if I first find the change in momentum of the slider by the action of this force. This is easy if I integrate 4 with respect to time. Taking the constants out of the integral, I end up with this:
integral sin(wt) dt. equation 5
Now it turns out the indefinite integral of equation 5 is:

-1/w cos(wt) + C.

Now, taking the definite integral of equation 5 from 0/w seconds to wt/w seconds, I end up with:

-cos(wt) + cos(0 radians). Factoring in the constants taken out of the integral, I get:

py = rwm1m2[ 1 – cos wt]/(m1 + m2) equation (6).

Or:

py = rwm1m2/(m1 + m2) - rwm1m2 cos wt/(m1 + m2) equation (7)

Or, the momentum of the slider in the positive y-direction.
Finally, dividing out by m1, the mass of the slider, we get the velocity of the slider in the positive y-direction, or:

y`(t) = rwm2/(m1+m2) - rwm2 cos wt/(m1+m2) equation (8).

But alas, the second term in the above equation is just the negative of equation 2, the tangential velocity of the center of mass in the positive y-direction. As stated before, the total velocity of the center of mass of the system in the positive y-direction is the sum of the velocity of the slider and the tangential velocity of the center of mass in the positive y-direction. When we add equation 2 to equation 8, it reduces to:

y`(t) = rwm2/(m1+m2) equation (9).

Which is in exact agreement with your derived equation, equation 1, the constant speed of the center of mass of the system in the positive y-direction! All this from assuming a component of the centrifugal force acts on the slider in the positive y-direction!
Now, finally, I am ready to show you the key equation that matches your Lagrangian analysis. Look at my equation 4, the y-component of the centrifugal force acting in the positive y-direction:

fcenm1y = [m1m2 r wsq sin(wt)]/(m1 + m2) equation (4).

I am going to rotate the coordinate system back to its original orientation, changing the above to:

fcenm1x = [m1m2 r wsq cos(wt)]/(m1 + m2) equation (4).

Next, I am going to divide out the mass of the slider on both sides, yielding:

Acceleration x-direction = [m2 r wsq cos(wt)]/(m1 + m2) equation (4).

Now look at your first equation after the “Coordinates for case II” in your analysis, the equation that shows the acceleration of the slider in the positive x-direction. What do we find? They exactly match!
Do you think this is just a coincidence? Or do you think the convergence of your brilliant Lagrangian analysis with my a priori approach to the reality of centrifugal force, which yields the same exact equation for the acceleration of the slider, tells us we have found a mathematical expression that expresses the effect of a subtle but real force of nature?

 

[K^2]

You are making fundamental mistakes.

Centrifugal force is NOT acting on the slider. Slider is not rotating. (ω=0 -> F_C=ω²R=0) That means, whatever's pulling it, is not centrifugal force.

The force acting on the slider is the tension in the arm connecting slider to rotator. The tension on the rotator side of the arm is the centripetal force that keeps rotator moving in circles. Rotator is accelerating towards the slider, not away. That means the force acting on it is towards the slider. And centrifugal force would have to push rotator away from slider. So centrifugal force is not acting on rotator either.

If you go into coordinate system attached to the slider, there is centrifugal force acting on the rotator, but you are obviously in an accelerated frame of reference. Motion of slider is not uniform, so you expect Fictitious forces.

The way you check if there is a net external force is you look at acceleration of center of mass, and it's clear from all of the above, and even your own analysis, that acceleration of CM in the x-direction is zero. That means, no net force in x-direction. That means, centrifugal force is not involved.

If you look in the y-direction, there is net acceleration and net force. But I account for it by showing that this force comes from the rail. Again, centrifugal force not involved.

 

[e2m2a]

You are making fundamental mistakes.

Centrifugal force is NOT acting on the slider. Slider is not rotating. (ω=0 -> F_C=ω²R=0) That means, whatever's pulling it, is not centrifugal force.

The force acting on the slider is the tension in the arm connecting slider to rotator. The tension on the rotator side of the arm is the centripetal force that keeps rotator moving in circles. Rotator is accelerating towards the slider, not away. That means the force acting on it is towards the slider. And centrifugal force would have to push rotator away from slider. So centrifugal force is not acting on rotator either.

If you go into coordinate system attached to the slider, there is centrifugal force acting on the rotator, but you are obviously in an accelerated frame of reference. Motion of slider is not uniform, so you expect Fictitious forces.

The way you check if there is a net external force is you look at acceleration of center of mass, and it's clear from all of the above, and even your own analysis, that acceleration of CM in the x-direction is zero. That means, no net force in x-direction. That means, centrifugal force is not involved.

If you look in the y-direction, there is net acceleration and net force. But I account for it by showing that this force comes from the rail. Again, centrifugal force not involved.

Of course the slider is not rotating. The force acting on the slider is the tension force as you have pointed out, formally known as the centrifugal reactive force. The x-component of this reactive force must be the force responsible for the acceleration of the slider to the right. We can see this if we do a free-body diagram of the slider. We know it can't be the constraint forces on the slider from the rail that accelerates the slider to the right because they are always perpendicular to the travel of the slider and do no virtual work on the slider, and we know it can't be a force from the left bumper because the left bumper is no longer in contact with the slider as it accelerates to the right. This leaves us only with the centrifugal reactive force. The normal explanation for the centrifugal reactive force is that it arises as an equal and opposite force to the centripetal force per Newton's third law. Period.
But now let's look at the system when the rotator is at 3 o'clock. At that precise point in time the slider is still accelerating to the right. (Look at your equation for the second derivative of x in your case 2 derivation, it still shows the slider is accelerating). But the slider and rotator are rigid bodies with a holonomic constraint, such that the distance from the axis of the slider and the center of mass of the rotator is constant, therefore, at that exact point in time at the 3 o'clock position, both the instantaneous velocity and acceleration of the rotator and body must be equal. Now, if you do a free-body diagram of the rotator at that point, you have an impossible situation. If there is only centripetal force acting on the rotator to the left, how can the rotator be instantaneously accelerating to the right?

 

[Dead Boss]

The normal explanation for the centrifugal reactive force is that it arises as an equal and opposite force to the centripetal force per Newton's third law. Period.

This doesn't seem right to me. The action and reaction act on different bodies. So if there is, for example, gravitational force on a planet, the reaction is force on the star. The centrifugal force arises as a purely inertial effect.

 

[e2m2a]

This doesn't seem right to me. The action and reaction act on different bodies. So if there is, for example, gravitational force on a planet, the reaction is force on the star. The centrifugal force arises as a purely inertial effect.

The "correct explanation" of centripetal force and centrifugal reactive force is as follows:
A real centripetal force is acting on the center of mass of the rotator, pointing inward toward the axis of rotation. By Newton's third law, there must be an equal and opposite real force on a seperate body. In this case it is the centrifugal reactive force acting on the physical axis of the slider pointing in an outward radial direction. So everything is explained, everything is squared away. Except...when you observe the rotation at 3 o'clock as explained in my previous post. At this point, the rotator must be instantaneously accelerating to the right, but there is only the centripetal force acting on the rotator, pointing to the left as given by the typical explanation. But something else must be going here. You're right, there must be a net inertial force, pointing to the right to account for the instantaneous acceleration of the rotator to the right. To assert otherwise is to assert a non-Newtion dynamic where a body is instantaneously accelerating to the right, but it only has a real force acting on it pointing to the left.

 

[e2m2a]

You are making fundamental mistakes.

Centrifugal force is NOT acting on the slider. Slider is not rotating. (ω=0 -> F_C=ω²R=0) That means, whatever's pulling it, is not centrifugal force.

The force acting on the slider is the tension in the arm connecting slider to rotator. The tension on the rotator side of the arm is the centripetal force that keeps rotator moving in circles. Rotator is accelerating towards the slider, not away. That means the force acting on it is towards the slider. And centrifugal force would have to push rotator away from slider. So centrifugal force is not acting on rotator either.

If you go into coordinate system attached to the slider, there is centrifugal force acting on the rotator, but you are obviously in an accelerated frame of reference. Motion of slider is not uniform, so you expect Fictitious forces.

The way you check if there is a net external force is you look at acceleration of center of mass, and it's clear from all of the above, and even your own analysis, that acceleration of CM in the x-direction is zero. That means, no net force in x-direction. That means, centrifugal force is not involved.

If you look in the y-direction, there is net acceleration and net force. But I account for it by showing that this force comes from the rail. Again, centrifugal force not involved.

k-2, I have gone over your Langrangian derivation and I am beginning to have doubts about the constancy of w. I found a lot of algebraic errors, starting with the eqm's, at the very top. There are a lot minuses for pluses and vice-versa and these ripple down in your other derivations. Plus, I don't see lamba3 in your substitution of the eqm for x1 into the eqm for x2. Also, I'm not sure how you separated the equations, I would like to see the rigorous proof. And did this prove the constancy of w before phi at pi/2 or did this prove it for all angles of the rotator? For pre-phi, it is trivial to prove it is constant, you don't even need Langrangian. The tough proof would be after pi/2. If you could prove the constancy of w for all angles, after correcting your errors, I would appreciate it very much. Once you prove the constancy you can stop there in the analysis. I attempted to prove it myself, but again there are a lot of errors, so I did not proceed.

 

[stevenb]

This doesn't seem right to me. The action and reaction act on different bodies. So if there is, for example, gravitational force on a planet, the reaction is force on the star. The centrifugal force arises as a purely inertial effect.

I agree. As an electrical engineer, I've found that if I want to pick a fight with a physicist, in the fastest way possible, I don't need to slap him in the face with a glove. Just mention the term "reaction centrifugal force", and the fight is on.

If you have a weight on a string and you whirl it around your head, the string places centripetal force on the weight and a centripetal reaction force on your hand. Now, if you look at an element of the string itself, there is a force on each side of the length element. The tension in the string is created by centripetal force and reaction centripetal force. In other fields, and I was even taught this in high school, the reaction force is sometimes called "centrifugal", and it is a "real" force, whatever you call it. This is a completely different animal from the "ficticious" centrifugal force like Coriolis etc.

The thing is that physicists have drawn a line in the sand and have said, "DO NOT CALL REACTION TO CENTRIPETAL FORCE A CENTRIFUGAL FORCE". I guess they feel this is better to avoid confusion. I see no reason to fight with them about this. It's just terminology. If we accept this, then centrifugal force is always "ficticious", as much as I hate the word ficticious in this context.

I've never had any significant issues doing a Newtonian mechanics problem, whether in an inertial or non-inertial reference frame, and I never sweat about the terminology, but I have to say that this thread has confused me more than any real-world classical mechanics problem ever could.

As an engineering student, I was taught by my physics teachers that there are 4 real fundamental forces and my engineering teachers promised I would need to deal with only two of those in real engineering problems, - so far so good as I hit the halfway point in my career. So, we can ask, is reaction centripetal force real? Well, yes it is real because it is electromagnetic in the case of a weight on a string, or gravametric in the case of two astronomical bodies orbiting each other. It seems to me that the inertial force due to a noninertial reference frame can not be electromagnetic, nor can it be gravametric, and I don't think anyone is going to claim it is a nuclear force.

A fallback position is to cite Einstein's gravity and note that he established laws valid in any reference frame. In effect, he banished the "inertial frame" and sent it into the wastelands. So, now with inertia and gravity equated, does centrifugal force gain the status of real force by proxy? I say no, since it's the other way around. Gravity now roams the badlands too.

The last fallback position would be to ask, "What will a unified field theory say about inertia, and could it allow us to call inertial force real?" I don't know how to answer that, nor do I know if it is even a meaningful question to ask.

 

[e2m2a]

A fallback position is to cite Einstein's gravity and note that he established laws valid in any reference frame. In effect, he banished the "inertial frame" and sent it into the wastelands. So, now with inertia and gravity equated, does centrifugal force gain the status of real force by proxy? I say no, since it's the other way around. Gravity now roams the badlands too.

The last fallback position would be to ask, "What will a unified field theory say about inertia, and could it allow us to call inertial force real?" I don't know how to answer that, nor do I know if it is even a meaningful question to ask.

Good thoughts. I say we can call inertial force real if we can demonstrate that it can do everything else that a "real" force can do. One reason inertial forces was classified as fictitious was because there was no other physical body one could relate it to. In Newtonian mechanics it is axiomatic that contact forces come in pairs. So, if you are in a rotating frame, and you have a body that is on a radial track, if the body is unrestrained, it will start accelerating away from you. (Assume you are at the axis of the rotating frame.) Now relative to your frame "something" is causing the body to accelerate away from you. Einstein pointed out this something could be viewed as a gravitational field. Now, if you reached out and grabbed the body and restrained it from moving away from you, you would feel a force. Conventional terminology would call this a fictitious force because you cannot associate another body with it, as required by Newton's third law, that is experiencing an equal and opposite force. Time out. This conventional terminology is ignoring the advances of modern physics. Forces can also manifest when a body is accelerating relative to a field. There is a lot of literature out there that relates the manifestation of inertia to the presence of a field in space. Now in the case of general relativity, the field is space itself being curved. (One interpretation.) Another interpretation says its the metric of space that determines gravity-inertia. You have the Machian inertia interpretation of general relativity. You have the formalism of Dennis Sciama, you have gravitomagnetic models, you have the scalar field of Brans-Dicke, you have models that relate inertia to the vacuum energy, and you have an interesting effect tested recently by the Gravity Probe B experiment-- the frame-dragging effect predicted by Lense and Thirring. Some would argue this has a Machian interpretation of the cause of inertia. All of these theories suggest that inertia arises out of a coupling of objects in a ubiquitous field in space. I think the frame-dragging effect is worth considering. If you want to see the latest data and more info about frame-dragging, go to the Gravity Probe B website. They have some interesting videos on the topic. Now, back to the example of the rotating frame. The recent test results of Gravity Probe B suggests that relative to the cosmic mass of the universe which could be viewed as a "hollow shell", the relative rotation of the frame "inside" and with respect to this hollow shell, generates a local frame-dragging effect within the rotating frame which causes the manifestation of both a Coriolis and centrifugal effect. When you restrain the body from moving, some theorists would argue its the frame-dragging effect that is causing the force-- not a fictitious force. But as I said, one way to judge if inertia is real is to test its consequences. I have done this in an experiment. If you want to see a video of it, you'll have to look back through this thread. The bottom line is the experiment showed that the speed of the center of mass of a rotator-slider system increased with respect to a laboratory frame. The only "force" possible that could cause this increase in speed was an inertial force. There is a well-established law in mechanics, known as Euler's first law. It states that only an external force acting on a system can change the velocity of the center of mass of the system. The conservation of linear momentum also requires this. Thus inertial force is real in the sense that it can qualify as an "external" force to a system and impact the momentum of the center of mass of the system. Also, the experiment isn't just demonstrating a local effect. It is indirectly confirming that a global field must exist in space that accounts for the test results. Time will tell which theorist is correct. Incidentally, I lean toward the vacuum energy explanation in combination with a general relativistic metric interpretation of inertia.

 

[D H]

For crying out loud, e2m2a! There are these nice little things called paragraphs that help people read and understand what you write. Please do try to use them.

 

[e2m2a]

For crying out loud, e2m2a! There are these nice little things called paragraphs that help people read and understand what you write. Please do try to use them.

Ok. I'll use spaces between paragraphs. Thanks for pointing it out.

 

[brainstorm]

For crying out loud, e2m2a! There are these nice little things called paragraphs that help people read and understand what you write. Please do try to use them.

You're lucky you found one that understands the concept, "paragraph." I usually just ask people to "hit the 'enter' key once in a while," with the hope they'll reflexively do it at logical points in their stream of consciousness.

 

[D H]

Ok. I'll use spaces between paragraphs. Thanks for pointing it out.

Use line returns, two of them, between paragraphs. Continue to use only spaces and this thread is locked. Continue to write in the style you have been using and this thread is locked.

One of the rules of this forum is no posting of personal theories. This thread looks a lot like a personal theory, but I can't really tell because I can't read/parse what you wrote.

 

[e2m2a]

You are making fundamental mistakes.

Centrifugal force is NOT acting on the slider. Slider is not rotating. (ω=0 -> F_C=ω²R=0) That means, whatever's pulling it, is not centrifugal force.

The force acting on the slider is the tension in the arm connecting slider to rotator. The tension on the rotator side of the arm is the centripetal force that keeps rotator moving in circles. Rotator is accelerating towards the slider, not away. That means the force acting on it is towards the slider. And centrifugal force would have to push rotator away from slider. So centrifugal force is not acting on rotator either.

If you go into coordinate system attached to the slider, there is centrifugal force acting on the rotator, but you are obviously in an accelerated frame of reference. Motion of slider is not uniform, so you expect Fictitious forces.

The way you check if there is a net external force is you look at acceleration of center of mass, and it's clear from all of the above, and even your own analysis, that acceleration of CM in the x-direction is zero. That means, no net force in x-direction. That means, centrifugal force is not involved.

If you look in the y-direction, there is net acceleration and net force. But I account for it by showing that this force comes from the rail. Again, centrifugal force not involved.

Actually, a typical analysis would show its the y-compoment of the centripetal force in the positive y-direction that accounts for increase in velocity of the center of mass of the rotator in the positive y-direction. However, the centripetal force is one force of a pair of forces as mandated by Newton's third law. The other force is the centrifugal reactive force acting on the axis attached to the slider.

Since this force pair is internal to the system, by the conservation of linear momentum and Euler's first law, it is impossible for these forces to impact the center of mass of the system in any way.

Also, the constraint forces of the rails can be completely removed, and yet an increase in the speed of the center of mass of the system would be observed. This could be done by having a dual-rotator system in space-- one rotator rotates counter-clockwise, the other clockwise. Initially, the left end of the slider could be up against an object (space shuttle). The two rotators could initially be at 9 o'clock. At some point in time each rotator could be given an impulse, one in the positive y-direction, the other in the negative y-direction.

Since there is no friction. the dual-rotator-slider system would began to move to the right immediately when one rotator is at 12 o'clock and the other rotator is at 6 o'clock. Essentially, everything analyzed for the single rotator system would apply. The speed of the center of mass of the system would increase, but there would be no rail constraint forces to account for it.

 

[e2m2a]

One last afterthought. In a previous post, I doubted the constancy of the angular velocity of the rotator. But I have found a simple, straightforward way to prove it, invoking the conservation of linear momentum.

The speed of the center of mass of the system in the positive x-direction at 6 o'clock is expressed as:

(m_rrw)/(m_r+m_s) = constant (1).

where, m_r is the mass of the rotator, m_s is the mass of the slider, r is the distance from the axis of rotation to the center of mass of the rotator, and w is the angular velocity of the rotator.

By the conservation of linear momentum the speed in the positive x-direction must remain constant as the system begins to move to the right, beginning at 6 o'clock. (We assume no friction.) Now, m_r,m_s, and r is constant. (The r term is a holonomic constraint of the system and is a constant.) Thus, in order for the speed to stay constant to comply with the conservation of momentum, w must always be constant. q.e.d.