A single asperity sliding contact model for molecularly thin lubricant

Abstract

Molecularly thin lubricants are important in protecting the recording head and the rotating disk in a magnetic storage hard disk drive from mechanical damage induced by contact. Direct contact is more likely to occur at lower head-media spacing, which is the distance between the rotating magnetic disk and the head that reads/writes the data, and is highly desirable to be extremely low (in the nanometer range) for Terabit/in² areal densities. Solid contact mechanics of the head-disk interface has been well addressed through extensive experiments and modeling, by neglecting the effect of the molecularly thin lubricant. Considering the important role of the lubricant, it is necessary to investigate the contact mechanics that include the lubricant. The present study develops a mechanics-based model to account for the nanoscale phenomena including slippage, nonlinear viscosity and bonded lubricant fraction, thus providing a measure to bridge the hydrodynamic lubrication and solid contact regimes at the nanoscale. Results show that the friction coefficient increases with the bonded ratio, which correlates with experimental observations.

Appendix

Fig. 9 shows simplified nonmenclature for lubrication with a slippery wall. Derivation of Eq. (2) begins from the Navier–Stokes equation:

$$\rho \left( {\frac{\partial u}{\partial t} + u\frac{\partial u}{\partial x} + v\frac{\partial u}{\partial y} + w\frac{\partial u}{\partial z}} \right) = - \frac{\partial p}{{\partial x^{{}} }} + \mu \left( {\frac{{\partial^{2} u}}{{\partial x^{2} }} + \frac{{\partial^{2} u}}{{\partial y^{2} }} + \frac{{\partial^{2} u}}{{\partial z^{2} }}} \right)$$

(4)

Assuming steady-state conditions and neglecting the velocity gradients along the z and y directions, i.e., $\partial u/\partial y = \partial u/\partial z = 0$, the Naiver-Stokes equation simplifies to:

$$\rho \left( {u\frac{\partial u}{\partial x}} \right) = - \frac{\partial p}{{\partial x^{{}} }} + \mu \left( {\frac{{\partial^{2} u}}{{\partial z^{2} }}} \right)$$

(5)

The velocity along x keeps constant, thus $\partial u/\partial x = 0$, then we have:

$$- \frac{\partial p}{{\partial x^{{}} }} + \mu \left( {\frac{{\partial^{2} u}}{{\partial z^{2} }}} \right) = 0$$

(6)

Integration with respect to z to obtain:

$$u = \frac{1}{\mu }\frac{\partial p}{\partial x}\left( {\frac{{z^{2} }}{2} + Az + B} \right)$$

(7)

Using the boundary conditions for a slip case: $u(z = 0) = U_{W} ;\;\;u(z = h) = U$, then we have the velocity profile:

$$u(z) = \frac{1}{2\mu }\frac{\partial p}{\partial x}z(z - h) + \frac{{(h - z)U_{W} }}{h} + \frac{z}{h}U$$

(8)

Similarly, we have velocity expression along y by using boundary conditions:

$$v = \frac{1}{\mu }\frac{\partial p}{\partial y}\left( {\frac{{z^{2} }}{2}} \right)$$

(9)

The mass conservation requires that net flow in a finite control volume is zero:

$$\frac{{\partial q_{x} }}{\partial x} - \frac{{\partial q_{y} }}{\partial y} = 0$$

(10)

where,

$$\begin{aligned} q_{x} = \int_{0}^{h} {\left[ {\frac{1}{2\mu }\frac{\partial p}{\partial x}z(z - h) + \frac{{(h - z)U_{W} }}{h} + \frac{z}{h}U} \right]} dz \hfill \\ = \left. {\left[ {\frac{1}{2\mu }\frac{\partial p}{\partial x}\left( {\frac{{z^{3} }}{3} - \frac{{hz^{2} }}{2}} \right) + \left( {z - \frac{{z^{2} }}{2h}} \right)U_{W} + \frac{{z^{2} }}{2h}U} \right]} \right|_{0}^{h} \hfill \\ = - \frac{{h^{3} }}{12\mu }\frac{\partial p}{\partial x} + \frac{h}{2}(U + U_{W} ) \hfill \\ \end{aligned}$$

(11)

$$q_{y} = - \frac{{h^{3} }}{12\mu }\frac{\partial p}{\partial y}$$

(12)

Substitute q _x and q _y into the mass conservation relation:

$$\frac{\partial }{\partial x}\left( {\frac{{h^{3} }}{12\mu }\frac{\partial p}{\partial x}} \right) + \frac{\partial }{\partial y}\left( {\frac{{h^{3} }}{12\mu }\frac{\partial p}{\partial y}} \right) = \frac{\partial }{\partial x}\left( {\frac{{U + U_{W} }}{2}h} \right)$$

(13)

By definition of the slip factor f*:

$$f* = (U - U_{W} )/U = h/\left( {b + h} \right).$$

(14)

The governing equation for pressure becomes:

$$\frac{\partial }{\partial x}\left( {\frac{{h^{3} }}{12\mu }\frac{\partial p}{\partial x}} \right) + \frac{\partial }{\partial x}\left( {\frac{{h^{3} }}{12\mu }\frac{\partial p}{\partial y}} \right) = \frac{\partial }{\partial x}\left[ {\left( {1 - \frac{{f^{*} }}{2}} \right)h} \right]U$$

(15)