Timeline for Why is torque defined as $\vec\tau = \vec r \times \vec F$ instead of $\vec\tau = \vec F \times \vec r$?
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Oct 4, 2017 at 13:45 | comment | added | Sayan Mandal | @KrishanMalik: The answer by Diracology is perfect. | |
Oct 4, 2017 at 13:41 | comment | added | Diracology | At the end of the day that is a convention. You could have define the new quantity "ntorque" (negative torque) $\vec\tau '$ and the dynamics would be simply given by $\frac{d\vec L}{dt}=-\vec\tau'$ and in particular $I\alpha=-\tau'$. Everything would work just fine. You could also change conventions for angular momentum and define $\vec L'=\vec p\times\vec r$ and then $\frac{d\vec L'}{dt}=\vec\tau'$. | |
Oct 4, 2017 at 13:41 | history | edited | Qmechanic♦ |
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Oct 4, 2017 at 13:36 | answer | added | Emilio Pisanty | timeline score: 5 | |
Oct 4, 2017 at 13:35 | comment | added | Krishan Malik | Dear Floris & Sayan Mandal.... I know this vectorial realtion..but problems is that why we write T= r×F instead of F×r. | |
Oct 4, 2017 at 13:28 | history | edited | Emilio Pisanty | CC BY-SA 3.0 |
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Oct 4, 2017 at 13:22 | comment | added | Sayan Mandal | Please upload the picture in the correct way! It is upside down. Also, you surely know that $\mathbf{r}\times\mathbf{F}=-\mathbf{F}\times\mathbf{r}$. So, switching the order switches the direction of torque, and we all agree on a particular sign convention for torque with respect to the right hand rule. | |
Oct 4, 2017 at 13:22 | comment | added | Floris | The vector product has a direction - if you change the order of the vectors, you change the direction. But a particular torque can only point one way - so you have to have the order of $\mathbf{r}$ and $\mathbf{F}$ correct. Incidentally you might want to turn your picture the right way up, and explain in more detail what you have a problem with. Right now your question is awfully broad... | |
Oct 4, 2017 at 13:19 | review | First posts | |||
Oct 4, 2017 at 13:26 | |||||
Oct 4, 2017 at 13:18 | history | asked | Krishan Malik | CC BY-SA 3.0 |