Timeline for Normal force of ball sliding on concave surface
Current License: CC BY-SA 3.0
14 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Nov 6, 2013 at 23:55 | comment | added | John Alexiou | I was hoping my reply was of use. So you are welcome for accepting it. | |
| Nov 6, 2013 at 22:14 | vote | accept | 0xFE | ||
| Nov 6, 2013 at 22:14 | comment | added | 0xFE | Wow this is fantastic! As soon as I saw the $v^2/\rho$ term, I connected it with centripetal motion and it all made so much more sense. Thanks!! To clarify for anyone else, gravity is in the $\left(\begin{array}{c} 0\\ -1 \end{array}\right)$ direction as you would expect. | |
| Nov 6, 2013 at 16:16 | comment | added | John Alexiou | let us continue this discussion in chat | |
| Nov 6, 2013 at 15:13 | comment | added | TROLLHUNTER | @0xFE Warning: If he thinks my right answer is wrong, then his answer is probably wrong. | |
| Nov 6, 2013 at 14:44 | comment | added | TROLLHUNTER | Read my answer to see what you can learn. | |
| Nov 6, 2013 at 14:42 | comment | added | John Alexiou | @user1708 No the simplest answer would be to state that the normal force is equal to $N = m v^2/\rho$. But what can one learn from just simple answers? | |
| Nov 6, 2013 at 14:35 | comment | added | TROLLHUNTER | I downvoted your answer for beeing needlessly complicated and hiding all the simple concepts needed. Also you post it when there is already a simple and correct answer, and you didnt upvote it. | |
| Nov 6, 2013 at 14:27 | comment | added | John Alexiou | @user1708 in my defense, I am presenting a complete answer with all the bits and pieces so as to answer all the questions the OP might have. Oh, and it is correct too. | |
| Nov 6, 2013 at 14:25 | comment | added | John Alexiou | @user1708 I understand you being proud of your answer, but it is bad manners in Phys.SE to go around and put down all other answers because you do not like them. It is up to the OP to decide which is the best suited answer. | |
| Nov 6, 2013 at 12:21 | comment | added | TROLLHUNTER | -1. My answer is much better and simpler. | |
| Nov 6, 2013 at 4:00 | comment | added | 0xFE | I'm going to try it to make sure I understand it and that it works properly, but this looks like exactly what I was looking for. | |
| Nov 6, 2013 at 2:09 | history | edited | John Alexiou | CC BY-SA 3.0 |
added 109 characters in body
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| Nov 6, 2013 at 2:02 | history | answered | John Alexiou | CC BY-SA 3.0 |