Question 1.10: A oil film of viscosity μ and thickness h«R lies between a s...

• System sketch: Figure E1.10 shows a side view (a) and a top view (b) of the system.
• Assumptions: Linear velocity profile, laminar flow, no-slip, local shear stress given by
Eq. (1.23).
\tau = \mu \frac{d\theta }{dt} =\mu \frac{du}{dy}
• Approach: Estimate the shear stress on a circular strip of width dr and area dA = 2πr dr ,in Fig. E1.10b, then find the moment dM about the origin caused by this shear stress. Integrate over the entire disk to find the total moment M.
• Property values: Constant oil viscosity μ . In this steady flow, oil density is not relevant.
• Solution steps: At radius r, the velocity in the oil is tangential, varying from zero at the
fixed wall (no-slip) to u = Ωr at the disk surface (also no-slip). The shear stress at this
position is thus
\tau =\mu \frac{du}{dy}\approx \mu \frac{\Omega r}{h}
This shear stress is everywhere perpendicular to the radius from the origin (see Fig. E1.10b). Then the total moment about the disk origin, caused by shearing this circular strip, can be found and integrated:
dM=(\tau )(dA)r=(\frac{\mu \Omega r}{h} )(2\pi rdr)r , M=\int{dM=\frac{2\pi \mu \Omega }{h} }\int_{0}^{R}{r^3dr}=\frac{\pi \mu \Omega R^4}{2h} Ans.
• Comments: This is a simplified engineering analysis, which neglects possible edge
effects, air drag on the top of the disk, and the turbulence that might ensue if the disk
rotates too fast.