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

[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 steady
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.

 

[Chestermiller]

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