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## Homework Statement

Show that the force resisting change of the minimum distance

*h*between the surfaces of two rigid spheres of radii

*a*and

*b*which are nearly touching is:

$$6\pi\frac{\mu}{h({a^{-1} + b^{-1}})^2}\frac{dh}{dt}$$

provided

$$\frac{\rho h}{\mu}\frac{dh}{dt}$$

## Homework Equations

Reynolds lubrication equation:

$$(h^3p_x)_x +(h^3p_y)_y = 6\mu(h(U_x+V_y) - (Uh_x+Vh_y)+2W)$$

where

*U(x,y,t), V(x,y,t)*and

*W(x,y,t)*describe the motion of the surface

## The Attempt at a Solution

I suppose we can say that $$W=\dot{h}$$ and $$U=V=0$$ Since

*h*is uniform in

*x*and

*y*, the Reynolds lubrication equation may be simplified to:

$$h^3{\nabla}^2p = 12\mu \dot{h}$$

Assuming symmetry around

*θ:*

$$p_{rr}+\frac{1}{r}p_r = 12\mu \frac{dh}{dt}\frac{1}{h^3}$$

$$ p = \frac{3\mu \dot{h}r^2}{h^3} + A $$

Assume on

*r = a, p = p_a:*

$$ p = p_a + \frac{3\mu \dot{h}}{h^3}(r^2 - a^2)$$

The total upward force is:

$$2\pi\int{p-p_a}rdr$$

But in here I am struggling to put up the limits for the integral as because we have two spheres nearly touching,

*r*would vary similarly as it would vary between two parabolas.