Past exam of the mathematics course of the University of Cambridge / 2017 / iii / Paper 329 / 3 / c / Solution

Put the fixed outer centre at the origin, and let the moving inner centre be $\mathbf{d}=(x,z)$. This reverses the sign of the vertical offset used in the previous parts: now $h=\Delta -\mathbf{d}\cdot \mathbf{n}$. At concentricity rotation produces no force and translation has isotropic resistance $R_0=12\pi \mu a^3/{\Delta }^3$. Balancing the external load givesThere is no inertial acceleration in this Stokes flow model; velocity is determined instantaneously by force balance.

For the subsequent motion, the full-film journal bearing assumption matters: pressure is single-valued around the whole annulus, with no specified cavitation boundary. Let $r=|\mathbf{d}|$, $e=r/\Delta$, $\varphi =\arg (x+iz)$, and ${\mathbf{e}}_r,{\mathbf{e}}_{\varphi }$ be its radial and tangential unit vectors. The squeeze resistance of an eccentric journal bearing has radial and tangential coefficientsFor completeness, rotate coordinates until $\mathbf{d}$ points along $x$, so $h=\Delta (1-e\cos \theta )$. Radial translation gives squeezing flux $q=a{v}_r\sin \theta$ and resistance proportional to $\int {\sin }^2\theta /(1-e\cos \theta {)}^3=\pi /(1-e^2{)}^{3/2}$. Tangential translation gives $q=a{v}_t[-\cos \theta +3e/(2+e^2)]$; the constant enforces periodic pressure. Its resistance usesThese expressions supply both translation directions; retaining only the vertical squeezing formula would miss the moving axle's horizontal dynamics.

The rotational fluid force from the earlier calculation points along ${\mathbf{e}}_{\varphi }$, with magnitude $R_t\Omega r/2$. Superposing rotation and translation by Linearity of Stokes flow, and resolving the load, givesFor positive $\Omega$ the axle initially moves down, then to the right because the rotation-generated force is perpendicular to the downward eccentricity. The equilibrium is displaced horizontally to the right, at the same height as the outer axis. Its eccentricity $0<e_{\ast }<1$ is the unique solutionThe rotational force then points upwards and balances the load. For negative $\Omega$ the displacement is to the left, using $|\Omega |$ in the magnitude equation. Uniqueness follows because the logarithmic derivative of $e/[\sqrt{1-e^2}(1+e^2/2)]$ has positive numerator $1-e^2/2+e^4$ after a common positive denominator. If $\Omega =0$ and $F>0$, no separated static equilibrium exists in this ideal model. If $F=0$, concentricity remains a solution.

One should not assert that this full-film model necessarily settles to the equilibrium. It predicts full-film journal-bearing whirl. For $F,\Omega >0$, eliminate time from the two boxed evolution equations and put $K=6\pi \mu \Omega a^3/{\Delta }^2$. The orbit starting at concentricity obeysso, with its zero initial horizontal displacement,The orbit begins on the lower branch, turns at $z=0$ before contact, and returns along the upper branch, forming a closed orbit through concentricity. The ratio of the right-hand side to $e$ increases from zero to infinity, giving a unique turning eccentricity below one. Linearizing at equilibrium likewise gives $\delta \dot{r}=-(F/R_{r,\ast })\delta \varphi$ and $\delta \dot{\varphi }=c_{\ast }\delta r$ with $c_{\ast }>0$: the eigenvalues are imaginary, not damped. Thus the equilibrium position exists, but the initially concentric axle circulates around it in the stated ideal full-film approximation. Cavitation or additional mechanical constraints would change the model; none were specified.

The normalized centre trajectory below illustrates the full-film prediction for $K/F=1$. The dashed circle is the contact locus of the axle centre, not either cylinder wall.