# Force Balance Question

Hello, I've been asked to show that: $$d\Gamma=\tau 2\pi r^2 dr$$

Where $\Gamma$ is the torque applied to the shaft and $\tau$ is the shear stress of the sample fluid at radius $dr$. So, I know that $d\Gamma= Force\cdot dr$ and that $Force = \tau \cdot Area$ and this is where I start encountering issues. If we're looking at some radius $dr$, isn't the area $2\pi dr$ or just the circumference of the local circle? So I end up with $d\Gamma = \tau 2 \pi dr\cdot dr$ which isn't right, can someone help?

The problem here is that $dr$ is not the radius, $r$ is! Your expression for the infinitesimal area should be $2\pi r dr$: this is for a thin loop of radius $r$. (Excuse the crude diagram):

So $r$ is a coordinate which is the radius of the loop, and $dr$ is a infinitesimal increment of $r$, i.e. the thickness of the loop. Since the loop is infinitely thin , the loop is like a rectangle curved around, so it's area will be the circumference of the circle, $2\pi r$, times it's thickness, $dr$.

Hope this helps :)

• Thanks, that's what I thought it was, the 'dr' is the radius threw me off Apr 22, 2016 at 20:03

Your Average MechEng's answer is fantastic and it is often how I explained the area of a thin circular shell to my students. But the mathematical derivation of such formula might be of interest to some. Using the above answer's diagram:

we are interested in computing the area between the two dashed lines. The inner circle has area $\pi r^2$ while the outer circle has area $\pi (r+\Delta r)^2$. The area between the two is obtained by subtraction: \begin{align} \text{Area} &= \pi (r+\Delta r)^2 - \pi r^2 \\ &= \pi (r^2 + 2r\Delta r + \Delta r^2 - r^2) \\ &= \pi (2r\Delta r + \Delta r^2) \end{align} Taking the limit as $\Delta r$ goes to zero yields the desired answer.

As an aside, it does not seem like any of these two answers resolve your issue, although they get you closer.