I am facing problems in concept of torque and its applications.
\begin{align} \vec\tau & = \vec r \times \vec F \quad\checkmark\\ \vec\tau & = \vec F \times \vec r \quad ✗ \end{align} Why it happens, please explain in detail.
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1$\begingroup$ 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... $\endgroup$– FlorisCommented Oct 4, 2017 at 13:22
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3$\begingroup$ 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. $\endgroup$– Sayan MandalCommented Oct 4, 2017 at 13:22
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$\begingroup$ Dear Floris & Sayan Mandal.... I know this vectorial realtion..but problems is that why we write T= r×F instead of F×r. $\endgroup$– Krishan MalikCommented Oct 4, 2017 at 13:35
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5$\begingroup$ 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 $\frac{d\vec L'}{dt}=\vec\tau'$. $\endgroup$– DiracologyCommented Oct 4, 2017 at 13:41
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$\begingroup$ @KrishanMalik: The answer by Diracology is perfect. $\endgroup$– Sayan MandalCommented Oct 4, 2017 at 13:45
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This comes entirely down to convention. You could swap all the definitions that include cross products around, and you'd get exactly the same physical theory as far as predictions of observable physical phenomena are conserved. Or, more practically, you could do exactly the same by the definition of cross products and swapping all mentions of "left hand" with "right hand", which would be exactly equivalent as the swap in the order of factors you propose.
That means that we have two equivalent but different ways to define pseudovector quantities (including the torque, but any cross product is subject to the same); they're both fine, but we do need to choose one of the two to be able to work. It so happens that, historically, people defined torque as $\vec\tau = \vec r\times\vec F$, and it's too late to switch conventions now.
answered Oct 4, 2017 at 13:36