# Fluid Mechanics newtons law of viscosity

1. Jul 25, 2013

### wozzers

1. The problem statement, all variables and given/known data
having problems differentiating

2. Relevant equations
Tw ( wall shear stress)= -U(viscosity)*du/dr
been given u as 2V(1-(r/r)^2)

3. The attempt at a solution
i substituted u in and got

d/dr (2V(1-(r/r)^2)
i tried to multiply out the minus sign and 2V and got a very different answer i got -2V+2V (r/r)^2

when the answer is supposed to be -4VR/R^2

please help

2. Jul 26, 2013

### janhaa

What do you mean here:

been given u as
2V(1-(r/r)^2) = 2V(1 - 1) = 0

?

3. Jul 26, 2013

### pasmith

Do you mean
$$u = 2V\left(1 - \left(\frac{r}{r_0}\right)^2\right)$$
where $r_0$ is the radius of what I assume to be a cylindrical pipe?

The given answer is for $du/dr$; to get $T_w$ you have to multiply that by $-\mu$.

So far all you've done is multiply $u$ by -1. Try differentiating with respect to r first.

4. Jul 26, 2013

### wozzers

r=R and its (r/R)^2

5. Jul 26, 2013

### wozzers

okay now i have differentiated it with respect to r i get 2V2r/R^2) or -(4Vr/R^2) thanks again i was then required to use the wall shear stress to determine the frictional force but it is very straight forward from there

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