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I need help to understanding this concept: We all know that Torque is $T=F\times d$ , but what I want to know is the differential form of it, and its meaning. $$ (a) dT=F\times dr$$ or $$ (b) dT=dF\times r$$

I think the option (a) makes more sense, because when I integrate $dT$, I will get the whole Torque on a disk resulting of the Force F applied on several points dr

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Since is a cross product the ordering is important. If $\vec{T}=\vec{r}\times \vec{F}$ then $d\vec{T}=\vec{dr}\times \vec{F}+\vec{r}\times \vec{dF}$.

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Interesting question. As Julio correctly pointed out, $d\vec{T}=d\vec{r}\times \vec{F} + \vec{r}\times d\vec{F}$. (Note that the original question had the order of $\vec{r}$ and $\vec{F}$ reversed.)

How do we interpret these two parts? The first term $(1)=d\vec{r}\times \vec{F}$ contains the change in torque due to a change in the moment arm $\vec{r}$, while the second term $(2)=\vec{r}\times d\vec{F}$ would correspond the change in torque due to a change in a force $\vec{F}$.

If you have forces distributed over a volume $\mathcal{V}$, and would like to calculate net torque, then the relevant term is the second one. E.g. if we want to calculate torque to the weight of an object, the infinitesimal force $d\vec{F}=-\rho(\vec{r})g\hat{z}dV$ \begin{equation} \vec{T}=\int_{\mathcal{V}}\vec{r}\times \left(-\rho(\vec{r})g\hat{z}\right)dV \end{equation} We can of course move the constant stuff outside of the integral \begin{aligned} \vec{T}&=-g\left[\int_{\mathcal{V}}\vec{r}\rho(\vec{r})dV\right]\times\hat{z} \\ &= - gM\vec{r}_{CM} \times \hat{z} \\ &= \vec{r}_{CM} \times \left(-Mg\hat{z}\right) \end{aligned} And this last expression is the standard result that when you calculate the torque due to the weight of an object, you can treat it as if the weight was applied at the center of mass.

Anyhow this is an example where the second term is the relevant one. But I'm sure you can cook up an example where the first term is relevant. E.g. if you are applying a constant force at a point that is moving relative the the origin, at what rate is the Torque changing?

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If $T=Fr$ then $dT = Fdr + rdF$
That's calculus 101.
You can put anything you want in the denominator.

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  • $\begingroup$ In an exercise of a book of fluid mechanics, they do dT= r dF. I can show it to you. $\endgroup$ – VitorAguiar68 Oct 1 '17 at 22:40
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    $\begingroup$ If the moment arm is not changing then $dr=0$. Also be sure to understand that the cross product will need to expanded into scalar products before you apply the chain rule in general. $\endgroup$ – Bobak Hashemi Oct 1 '17 at 22:57
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Torque is defined as $$\vec{T} = \int \vec{r} \times d \vec{F},$$ and, therefore, the correct differential is $d \vec{T} = \vec{r} \times d \vec{F}$. There is nothing particularly clever about it, it's just the definition. It is analogous to the definition of work $$W = \int \vec{F} \cdot d\vec{r}$$ which has the differential $dW = \vec{F} \cdot d\vec{r}$.

Note that torque is not defined as $$\vec{T} = \vec{r} \times \vec{F},$$ and, consequently, does not have the differential of the form $d \vec{T} = d \vec{r} \times \vec{F} + \vec{r} \times d \vec{F}$.

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  • $\begingroup$ Where are you getting this information? I cannot find a single definition of torque that shows the integral, and multiple that show without. $\endgroup$ – JMac Oct 3 '17 at 21:27
  • $\begingroup$ Have you ever studied continuum mechanics? Take a look at Euler's laws of motion: en.wikipedia.org/wiki/Euler%27s_laws_of_motion. In particular, look at the definition of the torque. $\endgroup$ – Fizikus Oct 3 '17 at 21:34
  • $\begingroup$ The definitions that have integrals are not integrals of $\vec r \times d \vec F$ on that page. $\endgroup$ – JMac Oct 3 '17 at 21:53
  • $\begingroup$ Yes, they are exactly that! Look at how the force is defined as an integral over dF (it's just that they use a different notation to separate the volume and surface forces), and then you will see that the torque is define exactly as I've written it down. $\endgroup$ – Fizikus Oct 3 '17 at 21:59
  • $\begingroup$ You're taking the integral of $dF$ over what though? In those cases, it was a surface and volume integral you took. $\endgroup$ – JMac Oct 3 '17 at 22:08

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