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Old 21st August 2018, 10:24 AM   #241
SDG
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Originally Posted by MRC_Hans View Post
No. You are simply mistaken. Look, even when the train is stopped, Earth is rotating, and orbiting the sun, and moving through the galaxy, all at rather high speeds. Why do you assume that this won't affect your wheel, if the train speed does?

Hans
The rotation of the Earth is in hundreds of meters per second. The cycloid effect is negligible.
Having said that, I am curious, the critical speed of centrifuges.
https://www.beckman.com/search#q=cri...sort=relevancy
Backman Coulter centrifuge manual page 1.

a) The critical speed range is the range of speeds over which the rotor shifts so as to rotate about its center of mass.
Passing through the critical speed range is characterized by some vibration.

Blue - radius, red - \omega


For the cycloid where v=400m/s (an approximate tangential velocity of the Earth at some latitudes), \omega = 350rad/s, r=0.2m - the centrifuge parameters.

Look, how evenly spaced radius changes, nice conditions for the critical speed, vibration.
Is it possible that this is one of the main contributing factors of the critical speed?
SDG
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Old 21st August 2018, 10:40 AM   #242
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Originally Posted by Dave Rogers View Post
This is getting pathetic. You've just admitted that you've been wrong all along, though you don't even realise it, and now you're trying to invoke physical absurdities to make it look like you had a point all along.

I've been telling you all along that your claim violates, not just Einsteinian, but also Galilean, relativity; that you are, in effect, trying to revoke Newtonian mechanics. You're now suggesting that the inertial mass of a specific body is a variable in the limit of low velocities. Newton would have you sitting in the corner wearing a pointed hat.

Dave
Look here:
Originally Posted by SDG View Post
Now I would like to discuss the rotation, angular velocity \omega, acceleration and force on a circle trajectory at different radii.

The lines across the ball represent radius, equipotential line.
Black - radius infinity, \omega = 0rad/s, straight line acceleration: 1m/s^2
Green - radius 100m, \omega = 0.1rad/s, acceleration on a circle: 1m/s^2
Orange - radius 1m, \omega = 1rad/s, acceleration on a circle: 1m/s^2

The balls undergo the same acceleration.
Are the F_cp forces equal in all instances or F_cp_1m > F_cp_100m > F_cp_infinity?
This has to be settled before we continue.
SDG
... and here:

Originally Posted by MRC_Hans View Post
OK, stop right there. This is a game, of course, but not that game. You don't get to demand partial answers in steps. As I said before, YOU have the burden of proof (that is one rule of this game), so YOU do the work. If you want to know the centripetal forces in those instances, go calculate them.

... But I can tell you they won't be equal.

Hans
Hans agrees with me that the forces are going to be different.
So it is two of us who are "invoking physical absurdities".

Please, show us what is wrong with the claim that the center of the balls determined at rest will have the same acceleration on different radii but forces are not going to be equal.
SDG
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Old 21st August 2018, 10:46 AM   #243
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Originally Posted by Myriad View Post
What is causing the mass of the balls to change?

Or when you say "inertial mass" do you actually mean moment of inertia?
I mean energy stored in 1/2 I_ball \omega^2
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Old 21st August 2018, 11:05 AM   #244
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Originally Posted by SDG View Post
Look here:

... and here:



Hans agrees with me that the forces are going to be different.
So it is two of us who are "invoking physical absurdities".
No. Your drawings of three different trajectories do not equate to the movement of the ball in your hypothetical wheel-on-a-train scenario, so all you're seeing is that different trajectories involve different forces.

Originally Posted by SDG View Post
Please, show us what is wrong with the claim that the center of the balls determined at rest will have the same acceleration on different radii but forces are not going to be equal.
SDG
I already have, at least ten times.

Dave
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Old 21st August 2018, 11:12 AM   #245
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Originally Posted by Dave Rogers View Post
This is getting pathetic. You've just admitted that you've been wrong all along, though you don't even realise it, and now you're trying to invoke physical absurdities to make it look like you had a point all along.

I've been telling you all along that your claim violates, not just Einsteinian, but also Galilean, relativity; that you are, in effect, trying to revoke Newtonian mechanics. You're now suggesting that the inertial mass of a specific body is a variable in the limit of low velocities. Newton would have you sitting in the corner wearing a pointed hat.

Dave
Dave,
this is one of your posts:
Originally Posted by Dave Rogers View Post
Sorry, yes, you're right; the ball has rotational energy because it isn't a point mass. OK, that's established. What's the next step?

Dave
The point of discussion from the beginning is if the balls can pickup the same rotational kinetic energy in both reference frames at the same rate.
SDG
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Old 21st August 2018, 11:32 AM   #246
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Originally Posted by SDG View Post
I mean energy stored in 1/2 I_ball \omega^2
SDG

That's the equation for rotational kinetic energy. When you said "inertial mass" you actually meant "rotational kinetic energy?"

That's bizarre, but okay. I'll substitute that back into the statement of yours I was questioning:

Originally Posted by SDG View Post
Dave,
Yes, the accelerations are identical. I agree with that.
Nevertheless, even though the accelerations are equal the forces does not have to be equal.
F=ma
The rate of change in force does not mean the acceleration will change in the same rate as well. The mass rotational kinetic energy could be changing too.

Sorry, that is incoherent. "F = ma" relates force, inertial mass, and acceleration. A change in rotational kinetic energy might indeed result from force applied to a mass causing acceleration, or the acceleration of a mass causing a force, but that does not alter the equation. If masses and accelerations are equal, forces are equal as well.

Also, as you've been repeatedly told before, kinetic energy (rotational or otherwise) is already known to be reference frame dependent. In the reference frame that's stationary with respect to the train and the wheel's axis of rotation, all the balls on the wheel have the same magnitude of rotational kinetic energy. In the reference frame that's stationary with respect to the track, when the train is moving and the wheel is rotating, balls on different parts of the wheel have different rotational and linear kinetic energies. That does not contradict known physics in the slightest.
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Old 21st August 2018, 12:44 PM   #247
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Originally Posted by Dave Rogers View Post
No. Your drawings of three different trajectories do not equate to the movement of the ball in your hypothetical wheel-on-a-train scenario, so all you're seeing is that different trajectories involve different forces.



I already have, at least ten times.

Dave
Let us forget the cycloid for a moment.

Do you agree with the claim that the center of the balls determined at rest will have the same acceleration on different radii but forces are not going to be equal?
It cannot be denied that the accelerations are the same at the center points.
The only conclusion is that the inertial mass changed. The rest mass is still the same of course.
The 1/2 I_ball \omega^2 is different in those examples.
This is a big deal!!! How come nobody sees it this way?
If F=ma is true then m has to change!!!

Quote:
You're now suggesting that the inertial mass of a specific body is a variable in the limit of low velocities. Newton would have you sitting in the corner wearing a pointed hat.
What would Newton do to you?

SDG
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Old 21st August 2018, 01:01 PM   #248
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no, d(omega)/dt is the same for balls at different radii if they are on the same wheel that gets an angular acceleration
angular acceleration is generated by a torque (units N m) and not by a force (units N)
learn some basic physics
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Old 21st August 2018, 01:08 PM   #249
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Originally Posted by SDG View Post
Look here:

... and here:



Hans agrees with me that the forces are going to be different.
So it is two of us who are "invoking physical absurdities".

Please, show us what is wrong with the claim that the center of the balls determined at rest will have the same acceleration on different radii but forces are not going to be equal.
SDG
OK, next time you mis-use an answer from me, you are going on ignore.

Hans
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Old 21st August 2018, 01:11 PM   #250
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Originally Posted by SDG View Post
Let us forget the cycloid for a moment.

Do you agree with the claim that the center of the balls determined at rest will have the same acceleration on different radii but forces are not going to be equal?
It cannot be denied that the accelerations are the same at the center points.
The only conclusion is that the inertial mass changed. The rest mass is still the same of course.
The 1/2 I_ball \omega^2 is different in those examples.
This is a big deal!!! How come nobody sees it this way?
If F=ma is true then m has to change!!!



What would Newton do to you?

SDG
Stop the nonsense. You are asking unclear questions and you are abusing the answers. You are trolling.

State your claim, present your evidence.

Hans
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Old 21st August 2018, 01:12 PM   #251
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Originally Posted by SDG View Post
The rotation of the Earth is in hundreds of meters per second. The cycloid effect is negligible.
Having said that, I am curious, the critical speed of centrifuges.
https://www.beckman.com/search#q=cri...sort=relevancy
Backman Coulter centrifuge manual page 1.
https://i.imgur.com/RuBSNXb.png
a) The critical speed range is the range of speeds over which the rotor shifts so as to rotate about its center of mass.
Passing through the critical speed range is characterized by some vibration.

Blue - radius, red - \omega
https://i.imgur.com/cLhTOAh.png?1

For the cycloid where v=400m/s (an approximate tangential velocity of the Earth at some latitudes), \omega = 350rad/s, r=0.2m - the centrifuge parameters.

Look, how evenly spaced radius changes, nice conditions for the critical speed, vibration.
Is it possible that this is one of the main contributing factors of the critical speed?
SDG
Gibberish. Look, I'm about done with you.

Hans
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Old 21st August 2018, 02:02 PM   #253
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Originally Posted by MRC_Hans View Post
Stop the nonsense. You are asking unclear questions and you are abusing the answers. You are trolling.

State your claim, present your evidence.

Hans
Here is the 'side' claim:
Originally Posted by SDG View Post
Now I would like to discuss the rotation, angular velocity \omega, acceleration and force on a circle trajectory at different radii.

The lines across the ball represent radius, equipotential line.
Black - radius infinity, \omega = 0rad/s, straight line acceleration: 1m/s^2
Green - radius 100m, \omega = 0.1rad/s, acceleration on a circle: 1m/s^2
Orange - radius 1m, \omega = 1rad/s, acceleration on a circle: 1m/s^2

The balls undergo the same acceleration.
Are the F_cp forces equal in all instances or F_cp_1m > F_cp_100m > F_cp_infinity?
This has to be settled before we continue.
SDG
F_cp_1m > F_cp_100m > F_cp_infinity
even though
a_cp_1m = a_cp_100m = a_cp_infinity
the acceleration at the center of the balls where the center is determined at the rest (no rotation).
Let's call the acceleration only a_cp because they are all equal.

If F=ma then
m_1m a_cp > m_100m a_cp > m_infinity a_cp
this leads to
m_1m > m_100m > m_infinity

Inertial mass (m) represents the Newtonian response of mass to forces.

Where is an error in this?
SDG
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Old 21st August 2018, 02:11 PM   #254
Reality Check
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Thumbs down A "This is how the F_cp depends on the train speed" ignorance

Originally Posted by SDG View Post
This is how the F_cp depends on the train speed.
SDG
22 August 2018 SDG: "This is how the F_cp depends on the train speed" ignorance because centripetal force depends on rotational velocity.
Centripetal force

F = ma. For someone on the train there are no forces due to the train speed. For someone on a station platform there are no forces due to the train speed. Thus a pendulum on a train travelling at a constant speed has the same dynamics as a pendulum on a station platform.
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Old 21st August 2018, 02:13 PM   #255
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Originally Posted by SDG View Post
Where is an error in this?
The error is repeated ignorance of physics: Centripetal force for a pendulum does not depend on the size of the bob (your ball).
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Old 21st August 2018, 02:24 PM   #256
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Originally Posted by tusenfem View Post
no, d(omega)/dt is the same for balls at different radii if they are on the same wheel that gets an angular acceleration
angular acceleration is generated by a torque (units N m) and not by a force (units N)
learn some basic physics
This is from a few pages back:
Originally Posted by SDG View Post
...
We can even seat the balls on a 'friction less' bearing at the top of the elastic rod so the rod will not torque the ball when the rod rotates beneath the ball.
The only forces applied on the ball is the rod pushing the ball and the string attaching the ball to the axle.
...
and
Originally Posted by SDG View Post
The string! Exactly like in the 'simple pendulum' example.
The torque will come from the point where the string is attached to the ball.
That's the question, how fast will string apply the rotational kinetic energy to the balls at different points on the cycloid.
SDG
Loading the elastic rods with energy and releasing the energy in combination with the string is more interesting scenario than the ball attached to the wheel rigidly.
When the rod is released then the balls are moving freely without any torque till the string is going to catch them. That's why the analysis is not that straightforward as it seems at the first look.
SDG
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Old 21st August 2018, 03:02 PM   #257
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It's indeed pointless. It's not the balls what goes around the circle. It's this discussion :-D
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Old 21st August 2018, 03:52 PM   #258
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Originally Posted by Dr.Sid View Post
It's indeed pointless. It's not the balls what goes around the circle. It's this discussion :-D
Heh! You beat me to it. Back a few pages, I was wondering why the experiment was with balls. Why not use tiny wheels? That way we could name the thread "Wheels Within Wheels" which would seem much more fitting.
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Old 21st August 2018, 04:00 PM   #259
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Originally Posted by SDG View Post
Here is the 'side' claim:

F_cp_1m > F_cp_100m > F_cp_infinity
even though
a_cp_1m = a_cp_100m = a_cp_infinity
the acceleration at the center of the balls where the center is determined at the rest (no rotation).
Let's call the acceleration only a_cp because they are all equal.

If F=ma then
m_1m a_cp > m_100m a_cp > m_infinity a_cp
this leads to
m_1m > m_100m > m_infinity

Inertial mass (m) represents the Newtonian response of mass to forces.

Where is an error in this?
SDG

Let's start with: F_cp_1m > F_cp_100m > F_cp_infinity

What is your rationale for this claim?

F_cp is basically defined as being the unbalanced force required to produce the centripetal acceleration a_cp that keeps a mass (in this case a ball) moving in a circular path. You acknowledge that those accelerations are equal for the green and orange balls. (In the black ball case the path is a straight line so there's no centripetal acceleration or centripetal force, but you've specified an equivalent "straight line acceleration" instead, so the accelerations are equal for all three balls.)

The balls' masses are equal as well, because no mass is being added or removed from them. (Calling rotational kinetic energy "mass" doesn't make it so.)

There are equal centripetal forces, equal masses, and equal centripetal accelerations, for balls at all parts of the wheel in all reference frames. The cycloid is merely the result of vector-adding the train's constant lateral linear velocity to the rotational motion of a ball on the wheel, which changes the magnitudes of the forces on and accelerations of the ball not at all.
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Old 21st August 2018, 04:00 PM   #260
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Originally Posted by Reality Check View Post
The error is repeated ignorance of physics: Centripetal force for a pendulum does not depend on the size of the bob (your ball).
You clearly do not understand the 'simple' pendulum:




When the bob 'I' changes then the equation changes.
The centripetal force does depend on the size of the bob.
SDG
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Old 21st August 2018, 04:35 PM   #261
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Originally Posted by Myriad View Post
Let's start with: F_cp_1m > F_cp_100m > F_cp_infinity

What is your rationale for this claim?


F_cp is basically defined as being the unbalanced force required to produce the centripetal acceleration a_cp that keeps a mass (in this case a ball) moving in a circular path. You acknowledge that those accelerations are equal for the green and orange balls. (In the black ball case the path is a straight line so there's no centripetal acceleration or centripetal force, but you've specified an equivalent "straight line acceleration" instead, so the accelerations are equal for all three balls.)

The balls' masses are equal as well, because no mass is being added or removed from them. (Calling rotational kinetic energy "mass" doesn't make it so.)

There are equal centripetal forces, equal masses, and equal centripetal accelerations, for balls at all parts of the wheel in all reference frames. The cycloid is merely the result of vector-adding the train's constant lateral linear velocity to the rotational motion of a ball on the wheel, which changes the magnitudes of the forces on and accelerations of the ball not at all.


Do you see how the 1m orange radius cuts the ball?
Compare that to how the 100m green radius cuts the ball.

Every atom below the radius line has higher acceleration than above the radius line and it requires bigger centripetal force than atoms above the line in order to keep the center of the ball at the same radius.
When we sum all forces for every ball atom then we will get bigger force for the 1m radius compared to 100m radius.

The ball centripetal acceleration is v^2/r or \omega^2*r for the point mass at the center of the ball when the ball center is determined on the stationary ball (no rotation).
It is 1m/s^2 for both instances when \omega_1m = 1rad/s and \omega_100m = 0.1rad/s.
SDG
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Old 21st August 2018, 04:42 PM   #262
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Originally Posted by Dr.Sid View Post
It's indeed pointless. It's not the balls what goes around the circle. It's this discussion :-D
It will continue like that till the issue presented in the post #253 is resolved.
SDG
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Old 21st August 2018, 05:25 PM   #263
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Thumbs down Clear ignorance about the 'simple pendulum' (m is a point mass)

Originally Posted by SDG View Post
You clearly do not understand the 'simple' pendulum:
22 August 2018 SDG: Clear ignorance about the 'simple pendulum' (m is a point mass).
For a simple gravity pendulum, the mass m is a point mass.
If there is a spherical bob then I suspect that nothing changes so long as L is the distance between the pivot and the center of the bob. But you are the one with the assertions about pendulums so it up to you to support them.

Last edited by Reality Check; 21st August 2018 at 05:28 PM.
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Old 21st August 2018, 06:01 PM   #264
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Originally Posted by SDG View Post
https://i.imgur.com/lYtMT9w.png

Do you see how the 1m orange radius cuts the ball?
Compare that to how the 100m green radius cuts the ball.

Every atom below the radius line has higher acceleration than above the radius line and it requires bigger centripetal force than atoms above the line in order to keep the center of the ball at the same radius.
When we sum all forces for every ball atom then we will get bigger force for the 1m radius compared to 100m radius.


The ball centripetal acceleration is v^2/r or \omega^2*r for the point mass at the center of the ball when the ball center is determined on the stationary ball (no rotation).
It is 1m/s^2 for both instances when \omega_1m = 1rad/s and \omega_100m = 0.1rad/s.
SDG
Iím quoting this one for posterity with respect to the highlighted part.
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Old 21st August 2018, 07:28 PM   #265
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Originally Posted by SDG View Post
It will continue like that till the issue presented in the post #253 is resolved.
SDG
Call me skeptical (and a slow learner) but somehow, I suspect this might be a sticking point.

I ran into this whilst researching falling bodies

https://arxiv.org/ftp/physics/papers/0702/0702155.pdf

Sounds a bit like your theory.
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Old 21st August 2018, 11:35 PM   #266
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Originally Posted by SDG View Post
Here is the 'side' claim:


F_cp_1m > F_cp_100m > F_cp_infinity
even though
a_cp_1m = a_cp_100m = a_cp_infinity

Where is an error in this?
SDG
The error is that you are using a angular acceleration d(omega)/dt and NOT a
I already told you that above.
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Old 21st August 2018, 11:50 PM   #267
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Originally Posted by SDG View Post
This is from a few pages back:

and

Loading the elastic rods with energy and releasing the energy in combination with the string is more interesting scenario than the ball attached to the wheel rigidly.
When the rod is released then the balls are moving freely without any torque till the string is going to catch them. That's why the analysis is not that straightforward as it seems at the first look.
SDG
And completely NOT addressing what I said. I did not say that the rods "torque" I told you that to accelerate the angular frequency of the wheel YOU NEED A TORQUE and not a FORCE.

The balls on the same wheel at different radii, iike in your black/green/blue "equipotential" (whatever that means) lines, all have the same angular frequency. When you then "accelerate" the wheel, they will all have the same angular acceleration, and not all the same a in your F = m a. because the "a" will of course be dependent on the radius they are away from the centre of the wheel.

It is really very simple. Like I said, first learn some basic physics.
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Old 22nd August 2018, 12:55 AM   #268
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Originally Posted by SDG View Post
The point of discussion from the beginning is if the balls can pickup the same rotational kinetic energy in both reference frames at the same rate.
SDG
You disposed of that one in one of your more recent goalpost moving events by putting the balls on frictionless bearings.

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Old 22nd August 2018, 12:56 AM   #269
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Originally Posted by SDG View Post
Do you agree with the claim that the center of the balls determined at rest will have the same acceleration on different radii but forces are not going to be equal?
No. The very suggestion violates F=ma.

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Tony Szamboti: That is right
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Old 22nd August 2018, 01:02 AM   #270
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Originally Posted by SDG View Post
Here is the 'side' claim:


F_cp_1m > F_cp_100m > F_cp_infinity
even though
a_cp_1m = a_cp_100m = a_cp_infinity
the acceleration at the center of the balls where the center is determined at the rest (no rotation).
Let's call the acceleration only a_cp because they are all equal.

If F=ma then
m_1m a_cp > m_100m a_cp > m_infinity a_cp
this leads to
m_1m > m_100m > m_infinity

Inertial mass (m) represents the Newtonian response of mass to forces.

Where is an error in this?
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How come that every time we get close to what might be a conclusion, you change the set-up? Are you afraid the conclusion might go against you?

A few posts up, we suddenly also had a vertical wheel (where the conditions will be quite different due to gravity), and here we suddenly have different radii. Stick to your set-up with the 1m radius horizontal wheel with the balls on sticks, and let us finish with that.

Hans
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Old 22nd August 2018, 06:33 AM   #271
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Originally Posted by tusenfem View Post
The error is that you are using a angular acceleration d(omega)/dt and NOT a
I already told you that above.
I am very careful about not mixing reference frames.
This is ground reference frame:

What effect has the same force on different radii when the a_cp is the same but centripetal forces are different?
Each radius is going to see different d\omega\dt in the ground frame.
The resulting forces are not the same as well.
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Old 22nd August 2018, 06:46 AM   #272
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Originally Posted by tusenfem View Post
And completely NOT addressing what I said. I did not say that the rods "torque" I told you that to accelerate the angular frequency of the wheel YOU NEED A TORQUE and not a FORCE.

The balls on the same wheel at different radii, iike in your black/green/blue "equipotential" (whatever that means) lines, all have the same angular frequency. When you then "accelerate" the wheel, they will all have the same angular acceleration, and not all the same a in your F = m a. because the "a" will of course be dependent on the radius they are away from the centre of the wheel.

It is really very simple. Like I said, first learn some basic physics.
Well, just check the image above. F_l is the same for all balls coming from the angular acceleration of the wheel. Does this effect yield the same forces acting on the balls in the ground frame?
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Old 22nd August 2018, 07:00 AM   #273
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Originally Posted by Dave Rogers View Post
You disposed of that one in one of your more recent goalpost moving events by putting the balls on frictionless bearings.

Dave
The bearings are part of the setup for a week. You just might have missed that.
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Old 22nd August 2018, 07:11 AM   #274
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Originally Posted by Dave Rogers View Post
No. The very suggestion violates F=ma.

Dave
You know my take on it.
Would you have a different solution to issue presented by the post # 253 ?
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Old 22nd August 2018, 07:27 AM   #275
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Originally Posted by MRC_Hans View Post
How come that every time we get close to what might be a conclusion, you change the set-up? Are you afraid the conclusion might go against you?

A few posts up, we suddenly also had a vertical wheel (where the conditions will be quite different due to gravity), and here we suddenly have different radii. Stick to your set-up with the 1m radius horizontal wheel with the balls on sticks, and let us finish with that.

Hans
The setup is horizontal from the beginning.
You asked what is the top and the bottom of the cycloid. I used the vertical wheel to explain the top and the bottom, I never changed the experimental setup.
I stick to 1m radius wheel.
Having said that when the train moves at constant velocity in the straight line then balls follow cycloid trajectories that have changing radius in the ground frame.
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Old 22nd August 2018, 07:40 AM   #276
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Originally Posted by Elagabalus View Post
Call me skeptical (and a slow learner) but somehow, I suspect this might be a sticking point.

I ran into this whilst researching falling bodies

https://arxiv.org/ftp/physics/papers/0702/0702155.pdf

Sounds a bit like your theory.
I am with you on this one. The post # 253 is a sticking point.
I have many more questions about this setup.
Our heads are going to be spinning looking for answers. That is one big can of worms.
Resolution of this is essential to the cycloid issue.
SDG
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Old 22nd August 2018, 07:44 AM   #277
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Originally Posted by SDG View Post
The bearings are part of the setup for a week. You just might have missed that.
And the thread's been running for a month, at the begining of which the crux of your argument was the angular energy due to rotation of the balls. Based on how badly you seem to misunderstand your own arguments, you just might have missed that.

Dave
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Old 22nd August 2018, 07:49 AM   #278
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I can kind of understand some of SDG's confusion here, because cycloidal motion isn't intuitively obvious. But it's fairly clear, once you understand it, how the only force on an object moving in a cycloid is identical to the centripetal force it experiences in a frame of reference where the centre of rotation is stationary.

First, a couple of definitions. In what follows, I'll use "stationary frame of reference" to mean the one for a person standing by the track, watching the train go by; and "train frame of reference" to mean the one for a person riding on the train, who sees the wheel's centre as stationary. I've extracted a few frames from Wikipedia's animated GIF for illustration:



The only force acting on the ball is represented by the line to the centre, and it's directed towards the centre of the wheel. The first thing to notice is that, unlike motion in a circle, the force isn't always at right angles to the direction of motion. In fact, it's only perpendicular at one point:



At this point the object (which we can regard as equivalent to the ball on the edge of the wheel in SDG's example) is neither speeding up nor slowing down; it's at its maximum speed of twice that of the centre of rotation, so the only effect of the force at this instant is to cause its path to be curved. Strangely, though, while the component of the force at right angles to the direction of motion is at its greatest here, the radius of curvature is also at its greatest (and therefore the rate of change of direction is at its smallest); this is because the centripetal force is mv^2/R and v is at its greatest here, so the v^2 term dominates.

Just before reaching this point, the force is angled forward of the direction of motion:



We can resolve the force into two components, perpendicular and parallel to the motion. The perpendicular component causes the path to curve, and the parallel component accelerates the ball along the path, so at this point it's speeding up. Further on in the motion:



it's slowing down, because the force is angled backwards. And that speeding and slowing means that the kinetic energy varies in the stationary frame of reference, although it's constant in the train frame of reference. That's not a problem for relativity, which simply says that the laws of physics have to be the same in the two frames, but not that they have to give identical results.

Near the cusp of the cycloid, things also get interesting. Just before it gets there, the force is almost entirely parallel to the direction of motion:



So it's nearly all directed towards slowing the ball, and in fact at the cusp itself:



it brings it to a dead stop. Again, it's counter-intuitive that at this point the radius is at its smallest (in fact it's dropped all the way to zero) yet the force causing the path to curve has also dropped to zero. But agin, going back to mv^2/R, v has also dropped to zero, so this expression is indeterminate. Calculus tells us that the force at this point is given by the value the force is approaching as the radius and velociuty approach zero, and as the force is always constant in magnitude, it remains the same through this point. And just afterwards,



it's directed towards accelerating the ball back up the next segment of the cycloid.

And what all this tells you - or should tell you - is that SDG's over-simplistic belief that the lateral force depends only on the curvature could hardly be further from the truth; in fact, it's dominated much more by the tangential velocity - the rate at which the ball moves along the cycloidal path - so that in fact the curvature is least when the lateral component of force is at its greatest.

Another interesting thing about cycloidal motion is that it doesn't require the constraint imposed by the wheel; all it requires is a moving object with a force of constant magnitude and steadily rotating direction. A rocket in space which is first set moving, spinning end over end, and then has its motor fired up, will follow a cycloid, and for a particular relationship between force and speed it will follow the one shown above, stopping dead at one point in its motion only to reverse direction and follow another segment. And that would be an easier situation in which to make a measurement, because the thrust of the rocket motor is controllable. But, of course, there would be no difference between different inertial frames of reference in this example either.

Dave
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Tony Szamboti: That is right

Last edited by Dave Rogers; 22nd August 2018 at 07:51 AM.
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Old 22nd August 2018, 10:10 AM   #279
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Originally Posted by SDG View Post
Well, just check the image above. F_l is the same for all balls coming from the angular acceleration of the wheel. Does this effect yield the same forces acting on the balls in the ground frame?
F_l CANNOT be the same for all radii when you apply a d(omega)/dt to the wheel.
and no reference frame dan help you with this nonsense
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Old 22nd August 2018, 10:57 AM   #280
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Originally Posted by tusenfem View Post
F_l CANNOT be the same for all radii when you apply a d(omega)/dt to the wheel.
and no reference frame dan help you with this nonsense
Many a soul had been lost in the wonderland of calculus
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