John2020

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1 minute ago, Ghideon said:
Yes, in the rotating frame of reference centripetal force is equal and opposite to the fictitious centrifugal force
No, in the nonrotating frame of reference there is no equilibrium. centripetal force is zero (it does not exist)Clear?
Clear. I just addressed above the case with the rotating frame. Of course in the nonrotating frame we have no centripetal force.
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8 minutes ago, Ghideon said:
There is always a centrifugal force in the rotating frame, and it is constant when angular velocity, mass and radius is constant and greater than zero.
For constant angular velocity there is an equilibrium in your example where the centripetal equals to the centrifugal force. The same conditions apply for my example when the angular velocity is constant.
8 minutes ago, Ghideon said:Notes:
I asked about this your previous examples; why you never mentioned the centrifugal forces when angular velocity w was constantForget about the previous examples, I made some errors there. Let us focus on the example with the rotating mass. Is the drawing clear for you? Do you have any questions?
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2 hours ago, Ghideon said:
Ok, Ask a more precise question and I'll add more explanations.
In your example the system is in equilibrium , thus no acceleration and no centrifugal force is being developed on the rotating frame (assuming the ball is not thrown). That is all OK. My example is different since the system is not in equilibrium because of the change in angular velocity. Consequently, in my example there is an inertial centrifugal at play. Is it clear for you?
2 hours ago, Ghideon said:Once we fully agree on the very basics of rotation and frames of reference.
All clear. We may proceed now.
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1 hour ago, swansont said:
See how r increases? No forces are acting on it once it's released.
I think you misinterpret the drawing. Initially the ball is at r_{1} when the rotor turns with constant angular velocity. r does not increases by its own. We have a fixed length of a rigid rod where mass m is allowed to slide over it but it is held in r_{1 }position_{ }due to friction (static friction should be better to say). On increasing the angular velocity of the rotor, the inertial centrifugal force has to overcome the static friction and push mass m (acceleration) upto position r_{2}. Up until the r_{2} point the angular velocity increases thus the centrifugal too. Assuming being accelerated all the way down to r_{2}.
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1 minute ago, swansont said:
(You showed an arrow on your diagram. No mention of a motor, or that it would have its own mass and angular velocity)
The rotor turns counterclockwise (red arrow) and the stator clockwise (blue arrow).
8 minutes ago, Ghideon said:Right picture: In this coordinate system the ball is not moving or accelerating. Newton says that since acceleration a=0 then m*a=0 so F=0. The force point inwards to the circle's center is still present, otherwise the string would have different tension in left and right image (physically impossible). The solution is to add a fictitious force, pointing outwards. This force, blue in the right picture below, is equal and opposite of the red force. This means that, in this coordinate system, the ball is not accelerating; F=ma=0.
Ok?
It is not so clear to me. Anyway, proceed further by analyzing my example to see if it fits to the above description. I have to go and I will be back a little bit later.
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Just now, swansont said:
You need a motor to keep this thing rotating. You could spin it up and turn the motor off, but the system wouldn't maintain constant w when the mass moved. Also, the motor would be spinning in the opposite direction to conserve angular momentum.
I think this is already shown on the drawing (see the red and blue direction vectors).
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1 minute ago, Ghideon said:
But is it clear why there is no centrifugal fore at play in the nonrotating frame of reference?
I am not commenting your advanced example until agreement is reached on how the basics work.
Yes.
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8 minutes ago, Ghideon said:
Further; fictitious forces, when added to the pictures, will be different in the left and the right picture. Ok?
If not obvious we can investigate further.In my example it is clear there is centrifugal force at play in the rotating frame of reference.
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26 minutes ago, swansont said:
First critique, of course, is that there is no centrifugal force, and no reaction force should appear in a freebody diagram.
The acceleration (i.e. from the net force) is centripetal, and exerted by the rod. a_{c} = v^2/r = w^2 r. (v = wr)
You want to increase r (from, say, r_{1} to r_{2}) while keeping w constant.
Your assertion does not represent the situation above. Maybe I was not clear (I thought it was obvious). In order the mass m to go to r_{2}, presupposes an increase in angular velocity ω_{2 }that implies a centrifugal force will rise pushing mass m to r_{2} point. This is how I present the situation with the drawing.
26 minutes ago, swansont said:Another thing we know is that if there are no external torques, angular momentum will be conserved. But we know the larger mass is rotating at a constant w and also transferring angular momentum and doing work on the small mass — the only way for that to happen is if it's being driven by an external torque, so angular momentum is not conserved.
I cannot follow you. The rotor is powered internally being the system in outer space. We start the rotor until it acquires a constant angular velocity and then we leave it. After some secs, there is an internal mechanism that will increase its angular velocity. Moreover, the rod is rigid that means the angular momentum can be transferred to mass m.
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12 minutes ago, Ghideon said:
Ok so far?
If so the next step is to introduce the forces and make sure they have the same physical effect in both frames of reference.
It is not clear to me, however not so important for the analysis. Does the rectangular frame has an opposite direction with that of the rotating circle? If yes then, OK. It is like the drawing depicts a situation while being on the circle rotating counterclockwise, the surrounding space turns clockwise. If this is the case then, I agree, otherwise it doesn't make sense to me. However this would negligibly affect the analysis (just the direction the ball follows in your example).
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1 hour ago, Ghideon said:
Are you following so far? Once you get the difference between the two frames of reference and their coordinate system we may introduce real and fictitious forces and from there move on to:
Yes, we can move on to the analysis.
2 hours ago, Ghideon said:Right frame. We place a coordinate system on the rotating circle, for instance painted on the circle's surface. We also choose to place ourselves on the circle and rotate with it.
The right frame rotates clockwise. Seeing the rotation of the circle on the left, shouldn't the right frame rotate counterclockwise in order to agree with the rotation of the circle on the left?
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13 minutes ago, Ghideon said:
I think we should:
Understand the frame of reference (started above)
Do the math in rotating frame and inertial frame for your example (a) and (b)
Compare the outcome and discussAgree?
Very nice animation that demonstrates the difference between the nonrotating and the rotating frame of reference!
I agree (analysis for my example).
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6 hours ago, swansont said:
Such analyses happen routinely. Compliance with and adherence to Newton’s laws is not a wrong conclusion
Could you make the analysis on the last drawing with the rotating mass and to answer on (a) and (b)?
2 hours ago, The victorious truther said:There is no centrifugal force when seen from your inertial frame of reference.
Please present an analysis as also give your answer for (a) and (b).
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10 minutes ago, swansont said:
No, you can analyze it from an inertial frame. In many cases, it’s probably easier to do it that way.
I think that would lead to wrong conclusions. For e.g. analyzing the rotating mass in outer space (see the drawing I shared a couple of posts above) from an inertial frame of reference, will the motion of the ball (m) be attributed to a real force (and not to inertial centrifugal) that implies a reaction force (upon the rest of the system) appears at the same moment?
I have to go to sleep. See you tomorrow.
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1 minute ago, swansont said:
No, I mean there are no fictitious forces whatsoever in an analysis in an inertial frame.
So, when I need to make an analysis of a rotating object, I have to place the frame of reference rotating along with the rotating object, right? After finding what is going on there, I switch to the inertial reference (my PC screen) and check how the finding in the rotating object may affect the observation from the inertial frame of reference (my PC screen). I think the @The victorious truther mentioned something similar.
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2 minutes ago, swansont said:
A rotating object does not create fictitious forces.
You mean it should have a mass attached on its surface, at least in order the fictitious forces to manifest?
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3 minutes ago, Ghideon said:
This is your speculative thread, what is your prediction? Also please state the frame of reference used.
Frame of reference: I am sitting in front of my PC screen (I don't know how you interpret this), seeing the picture of this rotating mass device (Prediction: accelerating in one direction).
Prediction: Mass (m) will accelerate due to the inertial centrifugal force (no reaction will appear at the rest of the system), creating an accelerating change in CoM that will eventually accelerate the system (M+m) as a whole.
Could someone please do the math for whatever prediction? I am just curious how will this problem be handled.
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1 hour ago, The victorious truther said:
If it doesn't work when seen from an inertial frame of reference then it doesn't work at all.
I have done several mistakes describing the situation with the Fig.1Upper. I should have mentioned while the bolt turns with constant angular velocity the frame of the construction rotates in the opposite direction by simultaneously displacing the mass m_T to the right. Under such conditions, the analysis makes sense to be done as seen from the rotating frame of the construction.
1 hour ago, The victorious truther said:The only force that can be utilized is the one that is actually accelerating you. Fictitious force only arise and are present in noninertial frames of reference because mathematically we attempt to treat an accelerating frame of reference as if it actually is at rest so we have to make up other forces to give rise to the phenomenon we observe while remaining at rest in that noninertial frame of reference. Those forces which do not disappear after we switch frames of reference from say noninertial to inertial (centripetal force for example) are the only real forces that you can do anything with.
Being in the rotating frame of the construction, I was expecting the acceleration of the screw (increasing angular velocity) to induce (over the translation screw) an Euler like force (more accurately an artificially conducted by the translation screw mechanism) that will accelerate the mass m_T to the right. Consequently, since this force wouldn't be created by a real force (like pushing by contact, again according to my expectation), the acceleration of mass m_T would result in the redeployment of the CoM without being able to transfer mass in the opposite direction due to the construction topology, thus the system would accelerate as a whole. This was the initial idea behind the construction in Fig.1Upper.
Regarding inertial and noninertial frame of reference, I have to admit is very confusing for me, as also irrelevant (a rotating object will induce those fictitious forces that has to induce and a nonrotating one, will have just plain Newtonian acceleration (if any)).
1 hour ago, The victorious truther said:Energy was not in fact conserved if this bolt went from not moving to moving.
If the construction could work then, the change in the rotational energy (accelerating screw) would be converted to a change in the kinetic energy of the mass m_T and eventually to the whole system. This transition would force the construction frame to stop rotating (resulting in a nonconservation of angular momentum due to the conversion of rotational to kinetic energy). Again this is my view, which is probably wrong as all the above.
Rotating mass m in outer space.
M: mass of the rotor
m: mass that may move radially
ω: angular velocity
yellow rod: rigid rodFcp: centripetal force
Fra: reaction force
Ffr: friction force
Fcf: inertial centrifugal force while Δω ≠ 0The rotor provides a constant angular velocity while the mass is at distance (r). When the angular velocity increases by Δω within 0.1 radians:
a) How mass (m) will be affected from the change in angular velocity?
b) How the system (M + m) will be affected from the change in angular velocity?Note: Some years ago, I conducted a very simple experiment with a sample of Pb (lead) metal being suspended by a thread hooked under a weight balance (0.001 grams resolution) and applied a DC magnetic field of Nd Magnet that showed something interesting (video available) that depicts the above situation (according to my view).
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13 minutes ago, Ghideon said:
Also true, for regular (real) ropes. I tried to analyse the imagined string introduced by John2020. Since it is stiff like a a rod except for lengthwise one can increase the speed by increasing angular velocity, the push is not radial.
I am preparing a drawing to explain all these and we will make the analysis for three cases: rigid rope, semirigid and nonrigid rope.
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5 hours ago, swansont said:
Dropping to a lower energy doesn’t require energy. Energy is released.
You need an example that’s physically allowed
Just assume for a moment it is possible without releasing the energy over a photon. The nucleus and the electron absorb this energy by converting to radial momentum.
5 hours ago, swansont said:To get a larger angular velocity with no change in tangential velocity, you have to decrease r. a_{c} = r*w^2
i.e. pull in. That means you need to increase the centripetal force.
Why do you think there is a centrifugal force?
I mentioned above we have a rope that is non rigid radially that means the tangential velocity may also change. Thus, at that moment as Ghideon also noted, the Centrifugal will be larger than the centripetal resulting in small mass displacement radially.
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4 minutes ago, Ghideon said:
Then point out the specifics. Trying to describe your fictive scenarios using words opens for mistakes:
Let others answer this if they like. In the meantime when you have a motor suspended by a thread and start it and let us say starts and stop aftet 1/4 of rotor complete cycle what would the motor do?
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14 minutes ago, Ghideon said:
Please get to the point
Your analysis is simply wrong and contrary to observation.
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2 minutes ago, Ghideon said:
*)The man getting tired is one possibility in reality
Correct.
5 minutes ago, Ghideon said:In outer space the analysis need to be different. I was assuming (maybe wrongly) that his was taking place on Earth:
Next time take the hammer you mentioned and try those steps I proposed and tell what you experience.
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7 minutes ago, Ghideon said:
NOTE: again this only applies when analysing the motion from the rotating frame of reference, rotating with the man along the same axis. From a stationary observer beside the man there are no centrifugal forces, they are fictitious. This is important. Failing to realise the importance of this leads to wrong conclusions.
So, what do you predict to happen after all these? The small mass will move for a short time radially and at the same moment what is expected for the man if being in outer space?
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Circumventing Newton's third law through Euler Inertial Forces
in Speculations
Posted
OK!