Timeline for A More Intuitive Solution for the Kinematics of the Pulley-Rope System
Current License: CC BY-SA 3.0
13 events
| when toggle format | what | by | license | comment | |
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| May 13, 2017 at 18:12 | comment | added | Shashaank | @shivams I meant that KenG said that "........one would need to include both the projected components" ( in his last comment when he talks about breaking about OA) .So I wanted to know which all components will we take then ? | |
| May 13, 2017 at 17:14 | comment | added | shivams | @Shashaank: Your question is not clear. Could you try to rephrase it better? | |
| May 12, 2017 at 20:25 | comment | added | Shashaank | @shivams will you like to have a look at the above comment of mine since I think you have understood the question | |
| May 11, 2017 at 11:10 | comment | added | Shashaank | Year back I had this precise doubt ! I need a bit more clarification. You say that it doesn't work to keep projecting like that , one would need to include both the projected components. Could you please show how or which both components should be added to get the correct answer if we use the triangle OAB or u as the velocity and break it into "which all components " to get the correct answer. It would be really helpful if you could add this bit in the answer too. Thanks | |
| Dec 8, 2016 at 18:29 | comment | added | shivams | @KenG: Your comment clarifies a lot. My doubt was precisely this that why are we taking the component of v along u and not u along v. Your argument that "the actual velocity is along AB, and hence that velocity should be projected along OA and not the other way round" is clarifying. Thanks for that. +1 | |
| Dec 8, 2016 at 18:13 | history | edited | Ken G | CC BY-SA 3.0 |
added 50 characters in body
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| S Dec 8, 2016 at 18:11 | history | suggested | user135951 | CC BY-SA 3.0 |
added LaTeX
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| Dec 8, 2016 at 18:09 | review | Suggested edits | |||
| S Dec 8, 2016 at 18:11 | |||||
| Dec 8, 2016 at 18:09 | history | edited | Ken G | CC BY-SA 3.0 |
added 233 characters in body
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| Dec 8, 2016 at 18:06 | comment | added | user135951 | I think you should add your comment as an edit to the answer :). Also, please use LaTeX while writing the math symbols. | |
| Dec 8, 2016 at 18:04 | comment | added | Ken G | Good point, that's the easiest solution. I misspoke, the issue is not the rate of change of theta, as the projection is instantaneous. The student who used the ABC triangle is correct because the actual velocity is along AB and can be projected along OA. The student who used the OAB triangle is wrong because that solution treats the actual velocity as if it is along OA and can be projected along AB. It doesn't work to keep projecting like that, one would need to include both the projected components. | |
| Dec 8, 2016 at 17:57 | comment | added | user135951 | Actually, at any instant we can directly say $u=v\cos(\theta)$. $\theta$ is the instantaneous angle, and that first method isn't wrong. $u$ is simply the component of $v$ along the string. The length of string must remain constant and so the constraint relation holds. | |
| Dec 8, 2016 at 17:40 | history | answered | Ken G | CC BY-SA 3.0 |