Timeline for Balancing a pencil
Current License: CC BY-SA 3.0
12 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Mar 1, 2015 at 5:57 | vote | accept | Dhruva Patil | ||
| Feb 28, 2015 at 21:07 | history | edited | hft | CC BY-SA 3.0 |
edited body
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| Feb 28, 2015 at 21:07 | comment | added | hft | Yes, I=m(l/2)^2 for a bob on string of length (l/2). Thanks for the correction. | |
| Feb 28, 2015 at 20:24 | vote | accept | Dhruva Patil | ||
| Mar 1, 2015 at 5:57 | |||||
| Feb 28, 2015 at 20:21 | comment | added | Dhruva Patil | Alright, that makes sense. Thank you. And also, the last edit you made, shouldn't it be divide by 4 instead of 2? | |
| Feb 28, 2015 at 20:17 | comment | added | Dhruva Patil | So, the L is really the length of the hypothetical pendulum you are replacing the pencil with? | |
| Feb 28, 2015 at 20:12 | comment | added | hft | In those links "l" is the length of the pendulum. | |
| Feb 28, 2015 at 20:09 | comment | added | Dhruva Patil | Shouldn't it be $\frac{ml^2}{4}$? | |
| Feb 28, 2015 at 20:08 | comment | added | hft | The first link does exactly what I said at the bottom of my post. It models the pencil as an "inverted pendulum". I.e., a bob at the center of mass, which is ignoring the fact that it is actually a solid rod. | |
| Feb 28, 2015 at 20:06 | comment | added | Dhruva Patil | The answer, all over the internet, seems to be the one in the question. Links: thatsmaths.com/2014/06/26/balancing-a-pencil, youtube.com/watch?v=U3vAoJhIWms#t=29, arxiv.org/pdf/1406.1125v1.pdf | |
| Feb 28, 2015 at 20:04 | history | edited | hft | CC BY-SA 3.0 |
added 192 characters in body
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| Feb 28, 2015 at 19:58 | history | answered | hft | CC BY-SA 3.0 |