Timeline for Can we theoretically balance a perfectly symmetrical pencil on its one-atom tip?
Current License: CC BY-SA 4.0
25 events
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| S Oct 4, 2021 at 23:44 | history | suggested | Xfce4 | CC BY-SA 4.0 |
It was difficult to read the former formula
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| Oct 4, 2021 at 21:03 | review | Suggested edits | |||
| S Oct 4, 2021 at 23:44 | |||||
| Oct 23, 2018 at 17:21 | comment | added | Apoorv Potnis | The same question is asked in David Morin's Introduction to Classical Mechanics. In the solution given in the book itself, he too calculates $t$ approximately to be $3.5~\mathrm{s}$! | |
| Jan 12, 2016 at 16:26 | comment | added | Floris | @Walter - why is the pencil not just a superposition of the wave functions of the individual atoms in a potential well with negative curvature? | |
| Jan 12, 2016 at 16:11 | comment | added | Walter | considering the quantum effects on an unstable classical situation is not a helpful exercise. Certainly, you cannot use HUP to say "you can only get so close". You could say that the unstable classical equilibrium state has no QM equivalent and hence is not a valid description of the system. | |
| Jan 11, 2016 at 22:55 | comment | added | Floris | @DanielSank i would encourage it. Please leave a link here when you're done. | |
| Jan 11, 2016 at 22:48 | comment | added | DanielSank | @Floris I think there's a fundamental misunderstanding of quantum mechanics at work here. I should probably write a self-answered post about it because I doubt I can explain it well in comments. | |
| Jan 11, 2016 at 21:20 | comment | added | Floris | @DanielSank I believe there is no need for a collapse of the wave function. It is sufficient that there is no way to have a pencil which is properly vertical, and with zero momentum. All we need is an infinitesimal displacement from the vertical and the unstable equilibrium will do the rest. We have an ensemble of atoms - are you telling me that QM allows them to be placed with perfect knowledge of position and no lateral momentum? I'm simply using the HUP to say "you can only get so close" - and that provides a starting point for the pencil falling. Even a tiny, tiny displacement suffices. | |
| Jan 11, 2016 at 20:41 | comment | added | DanielSank | I have a problem with these uncertainty relation explanations of a pencil falling over. In the absence of any external degrees of freedom, there's no wave function collapse, so while the pencil's wave function may broaden over time etc. etc. there's no reason to believe it will fall. I think this is actually a really unfortunate abuse of the uncertainty principle which leads to much confusion. All this loose talk about "averages" in the quantum case only makes sense if you have an ensemble of pencils interacting with an environment. The uncertainty principle alone does not suffice. | |
| Jan 11, 2016 at 16:19 | comment | added | Floris | @Walter there are different ways of arriving at the conclusion that a pencil that appears, at one instance, to be perfectly balanced, cannot remain in that state forever. If you use (classical) statistical thermodynamics, you would need to have precise information about position and momentum of every atom. At that scale QM says I can't have that, and the momentum/position of the ensemble is therefore also not perfectly known. But even if it was, a single photon would be sufficient to destroy the equilibrium because of the shape of the potential well. | |
| Jan 11, 2016 at 16:12 | comment | added | Walter | I don't think you should/can use the unsharpness principle (I refuse to use an incorrect English translation) in classical calculations like this. Said principle does not state that one cannot be certain or cannot measure things to full sharpness/accuracy, but that the very concept of such accuracy is unnatural and hence wrong. If you want to use QM in this answer, you better work out the relevant quantum states. The classical equilibrium state on the tip is unstable and has no corresponding quantum state. | |
| Aug 11, 2015 at 0:41 | comment | added | Selene Routley | You might like to check out this question, and the objections I raise to KleinGordon's argument as well as, in all likelihood, those of the book's author since I think the author is simply arguing like KleinGordon. I'm guessing that you are one of the people on this site who might be up to speed with a full analysis to a whirling pendulum. | |
| Jul 25, 2015 at 2:32 | history | bounty ended | HyperLuminal | ||
| Apr 19, 2015 at 0:30 | history | edited | Floris | CC BY-SA 3.0 |
deleted 1 character in body
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| Apr 17, 2015 at 22:12 | history | edited | Floris | CC BY-SA 3.0 |
deleted 5 characters in body
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| Apr 17, 2015 at 21:51 | history | edited | Floris | CC BY-SA 3.0 |
Fixed the calculation of I (I=\frac13 m \ell^2 not 1/12... since pivoting at the end.)
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| Apr 16, 2015 at 16:00 | history | edited | Floris | CC BY-SA 3.0 |
added 2731 characters in body
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| Apr 15, 2015 at 12:44 | history | edited | Floris | CC BY-SA 3.0 |
fixed factor 2 in the conversion from exponential notation to sinh
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| Apr 14, 2015 at 19:09 | comment | added | Floris | @jamesqf valid quibble. Pencil hardness goes from "9H" to "9B" and there will be quite a bit of variability. I found a factor 4x change in Knoop hardness (from 51.5 for 5H to 12.7 for B) in table 1 on page 288 of "Microindentation techniques in Materials Science, issue 889 (P.J.Blau). Converting to MPa is not straightforward... but according to tedpella.com/company_html/hardness.htm the hardest pencil in the above is somewhere around the hardness of silver (Knoop 60) - 250 MPa. That's not too far different from the 170 MPa I used above. | |
| Apr 14, 2015 at 18:35 | comment | added | jamesqf | Just a minor quibble: pencil leads are not made of pure graphite, but of a mixture of graphite and clay, which allows for various hardness levels: en.wikipedia.org/wiki/Pencil Pure graphite probably wouldn be too soft to be usable. | |
| Apr 14, 2015 at 16:10 | history | edited | Floris | CC BY-SA 3.0 |
added information about graphite
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| Apr 14, 2015 at 13:28 | comment | added | Floris | @AndréNeves - the equation of motion stops being "nice" at larger angles, and you would have to integrate numerically. But the additional accuracy is not worth it given the current assumptions. If we used the same equation all the way to 90°, the answer would be 46 seconds. That's the ball park. | |
| Apr 14, 2015 at 13:21 | comment | added | André Chalella | Also, may I lazily ask what would $t$ be for $\theta = 90°$? | |
| Apr 14, 2015 at 12:55 | history | edited | Floris | CC BY-SA 3.0 |
added detailed calculation
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| Apr 14, 2015 at 2:22 | history | answered | Floris | CC BY-SA 3.0 |