Use viscosity/shear stress relations to find torque needed to rotate disk

1. Sep 20, 2014

rcummings89

The problem asks for an approximation for the torque needed to rotate a disk that is separated from a stationary boundary by a viscous fluid, given that $\tau = \mu \frac{U}{H}$.

I did it first using like this:
$T = F R$
$F = \tau A = \mu \frac{U}{H} \pi R^2$ where $U = \Omega R$
And thus $T = \mu \frac{\Omega \pi R^4}{H}$

Now making the following assumptions...
Flow is a function of height (z) only
Film is thin enough that the pressure gradient is negligible
...I started with momentum:

$\rho \frac{du}{dt} = -\nabla p + \mu \nabla^2 u$ which reduces to $0 = 0 + \mu \nabla^2 u(z)$
Where $u(z) = \frac{C_1 z}{\mu} + C_2$
Using the boundary conditions
$u(0) = 0$ and $u(H) = U$ yields $C_1 = \mu \frac{U}{H} = \tau$ and $C_2 = 0$ so that
$u(z) = \tau \frac{z}{\mu} = \frac{U z}{H} = \frac{R \Omega z}{H}$

Here is where I have trouble; a classmate told me to square the velocity and multiply it with mass
$T = M u(z)^2 = M \frac{\Omega^2 R^2 z^2}{H^2}$ and let z = R and $\mu = M \frac{\Omega}{H}$ so that the equation becomes
$T = \mu \frac{\Omega R^4}{H}$
Which is the same as I got before, but without the constant $\pi$
I'm note very comfortable with this because it takes the same form as kinetic energy $(\frac{1}{2}Mv^2)$ but it does seem to yield the correct answer. Is this true? Another problem I have is keeping track of the vectors vs. scalars but I'll get into that later on. Any help is greatly appreciated.

2. Sep 21, 2014

Staff: Mentor

This is not done correctly, because the shear stress on the disk is a function of radial location. The calculation needs to involve an integration with respect to r, over annular increments.

Chet