In my fluid mechanics course, I was exposed to some cases where I need to calculate the torque due to the pressure and all solutions manuals or online tutorials take it as a known fact that $d\tau=rdF$ ($r$ is perpendicular to $F$). If I apply the normal differential rules to it, it should be $$d\tau = r dF + F \mathrm{d}r.$$ I'm trying to understand why the second term is cancelled out.

The best answer I thought of was that because the force is due to a pressure so at each point it would have a value of $p\cdot \mathrm{d}\mathbf{A}$ (with $p$ the pressure and $\mathbf{A}$ the area) which is an infinitely small force and thus F is 0. Is that correct?

What if instead the force was defined as a function of r (continuous force distribution), so it does have a value at each point; for example $F = kr^2$?

Here is an example of such a case:
The oil film causes stress proportional to the speed gradient

  • $\begingroup$ You do realize that they are working with shear stress here, not normal stress (pressure), right? And since when is torque the dot product of force and radius? $\endgroup$ – Chet Miller Oct 30 '19 at 14:13
  • $\begingroup$ If you mean in this example yes I do, but I'm asking in general really. This was not meant to be in vector form, someone edited it since I didn't use the correct formatting. Since I only have force and distance in one direction perpendicular to each other, I didn't bother to use vectors and complicate things. However, do correct me if I did something wrong. @ChetMiller $\endgroup$ – Peter Oct 30 '19 at 15:55

I really don't like the way your book explained this, and it is definitely not the way I would have done it. My problem is with the equation $dF=\tau_wdA=\tau_w(2\pi r)dr$. Force is supposed to be a vector quantity (with a single specified direction), and yet the shear stress $\tau_w$ varies with direction around the circumference, so their expression for dF can't represent a force. Here is how I would have done it, which I hope makes more sense:

The differential force acting on a differential area of the disk $dA=rd\theta dr$ is $$\mathbf{dF}=\tau_w rd\theta dr \mathbf{i_{\theta}}$$where $\mathbf{i_{\theta}}$ is the unit vector in the tangential ($\theta$) direction. The differential torque acting on this area of the disk is obtained by taking the cross product of the radius vector ($\mathbf{r}=r\mathbf{i_r}$) with the differential force such that $$d\mathbf{T}=\mathbf{r}x\mathbf{dF}=\tau_wr^2d\theta dr(\mathbf{i_r}x\mathbf{i_{\theta}})=\tau_wr^2d\theta dr\mathbf{i_z}$$where $\mathbf{i_r}$ is the unit vector in the radial direction and $\mathbf{i_z}$ is the unit vector in the axial direction. The total torque on the disk is obtained by integrating this between $\theta = 0$ and $\theta = 2\pi$ and between the inner and outer radii (also taking into account the fact that $\tau_w$ is proportional to r, given by $\tau_w=\frac{r\Delta \omega}{h}$).

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  • $\begingroup$ If you switch the dr and $d\theta$ and integrate over $d\theta$ you get the same expression as in the solution but without the unit vector, which is great but my question persists. Why did you say that dT = r x dF and ignore dr x F, my problem is that torque is not the summation of the change of force multiplied by distance at each point but rather the full force multiplied by the full distance at each point. $\endgroup$ – Peter Oct 30 '19 at 19:01
  • $\begingroup$ It is ignored (and is incorrect to include) because torque is force times moment arm, and not force times differential moment arm. It's just like work is equal to Fds, not d(Fs). $\endgroup$ – Chet Miller Oct 30 '19 at 19:55
  • $\begingroup$ The difference here is work is Force * change in distance and thus in the distance - force graph, the work can be considered to be area under the graph. However the same is not true for Torque which is Force*Distance and can not be represented as area under the graph of a force-distance graph. As you would be multiplying only the change in force by the corresponding distance and not the full value of the force. In other words, Torque is not moment arm time differential force either, it is moment arm by total force. $\endgroup$ – Peter Oct 30 '19 at 20:33
  • $\begingroup$ The total torque is the sum of the forces times moment arms. In this case, the forces occur over differential areas, and are thus differential. That was especially emphasized in my answer because the directions of the differential forces change with theta. You are wrong about torque being total force times moment arm. If there are multiple forces acting on a body, It is the sum of each individual force times its moment arm. $\endgroup$ – Chet Miller Oct 30 '19 at 21:13
  • $\begingroup$ Note that, if the differential forces on each disk are added vectorially (as they must be added), the net (total) force on each disk is zero. So the product of the total force multiplied by any moment arm is zero. However, the sum of the moments of the differential forces acting on the surface of the disk is not zero, and is equal to the total torque. $\endgroup$ – Chet Miller Oct 30 '19 at 22:04

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