Past exam of the mathematics course of the University of Cambridge / 2017 / iii / Paper 329 / 2

Take $z$ positive downwards. The leading axial velocity of a slender viscous thread is independent of radius. Incompressible flow then gives $u_r=-r{w}^{\prime }/2$, while steady mass conservation gives $(a^{2}w{)}^{\prime }=0$. The Newtonian fluid stress tensor has ${\sigma }_{rr}=-p-\mu w^{\prime }$ and ${\sigma }_{zz}=-p+2\mu w^{\prime }$. The Young–Laplace equation, neglecting axial curvature at slender order, gives ${\sigma }_{rr}=-p_e-\gamma /a$ and hence $p=p_e+\gamma /a-\mu w^{\prime }$.

The axial cut transmits both the integrated Newtonian fluid stress tensor and the circumferential surface tension. After subtracting the ambient pressure, its tensile force isThe coefficient three is the Trouton ratio for axial extension. The last term is essential: the capillary tensile force of a slender thread is $\pi \gamma a$, not $2\pi \gamma a$, because the capillary pressure subtracts half the interfacial line contribution. Axial force balance is $T^{\prime }+\pi \rho g{a}^2=0$. Together the governing equations areDefine the dimensionless variablesThen $a=a_0{W}^{-1/2}$ and the axial equation becomesFor the increasing power-law branch, all three terms have the same dependence when $\alpha =2$. Substitution leaves $1-6k^2-\Gamma k=0$. ThusThis remains valid at $\Gamma =0$, when $k=1/\sqrt{6}$. The other root is negative and describes a decreasing local branch up to its singular endpoint, not a thread accelerating downwards indefinitely. Positive surface tension decreases $k$, so it slows the fall at a given height on this branch. Thinning reduces the capillary tensile force $\pi \gamma a$. Its negative axial derivative supplies an upward capillary force on a thread segment, balancing part of the weight. The viscous tensile force therefore supplies a smaller remaining balance; on the quadratic family this reduces the stretching rate.

The condition $W(0)=1$ alone does not select a unique solution of this second-order ordinary differential equation. The boxed power law is the requested verified branch, with tensile force tending to zero at large height; an additional nozzle tension or downstream condition is required to select it physically. The slender, negligible-inertia approximation must remain valid over the height range considered.

Write $D=R_e-R_i$, $S=R_e+R_i$, $A=R_e^2-R_i^2=DS$, and $\Delta p=p_i-p_e$. Incompressible flow with axial strain rate $E$ givesIts radial Stokes equation gives $p_r=0$, because $u_{r,rr}+u_{r,r}/r-u_r/r^2=0$. The radial Newtonian fluid stress tensor component is ${\sigma }_{rr}=-p-\mu E-2\mu C/r^2$. The outer and inner Young–Laplace equation conditions have opposite curvature signs:Subtracting and eliminating $C$ givesIn particular, the boxed pressure is equivalent to the printed identity for $(-p-\mu E)A$. For equal gas pressures, $C=-\gamma R_i{R}_e/(2\mu D)$, so surface tension draws both interfaces inward in addition to the imposed extension.

The kinematic boundary condition is ${\dot{R}}_j=-E{R}_j/2+C/R_j$ at either interface. Taking the difference, and then the difference of squared radii, givesThe cancellation of $C$ in the area equation is mass conservation: axial stretching reduces the liquid area while radial redistribution changes the hole size. For $E=0$, equal pressures, and $R_i(0)=b$, $R_e(0)=2b$, one has $A=3b^2$ and $D=b+\gamma t/(2\mu )$. Since $S=A/D$,Closure occurs at $D=\sqrt{A}=\sqrt{3}b$, soThis capillary collapse of an annular viscous cylinder has finite closure time for $\gamma >0$ and $b>0$. For $\gamma =0$ it does not close in this case. The annular formula is used only up to closure; the singular cylindrical inner curvature is not a boundary condition on the subsequent solid cylinder.

Use $A=R_e^2-R_i^2$, $S=R_e+R_i$, $D=R_e-R_i$, and $Q=Aw=A(0)w_0$. Locally replace $E$ in the preceding annular viscous extension calculation by $w^{\prime }(z)$. Its leading pressure isThe axial liquid stress relative to ambient pressure, together with the two circumferential surface tension forces, transmits a cut forceIt would be incorrect simply to differentiate this and retain an extra gas-pressure term. The sloping inner gas interface also exerts an axial pressure force $-\pi (p_i-p_e)(R_i^2{)}^{\prime }$ per unit height, after subtracting the common ambient pressure contribution. For constant gas pressures these terms cancel. The resulting capillary tensile force of a hollow slender thread therefore givesThus the printed axial equation holds even for unequal constant gas pressures; their influence remains in the radial evolution.

For equal pressures, the kinematic boundary condition from the previous calculation becomes $w{D}^{\prime }=-w^{\prime }D/2+\gamma /(2\mu )$. Multiplying by the appropriate integrating factor givesAt closure $R_i=0$, so $D=\sqrt{A}$ and $D\sqrt{w}=\sqrt{Q}$. HenceThe right side is positive for a genuine annulus. On the gravity-driven quadratic family from the previous part, $w^{1/2}$ grows linearly with height and this integral diverges logarithmically. It must then reach the finite positive threshold at a finite height. This explains the expected capillary closure of a falling hollow thread on that freely falling branch without requiring its full coupled profile.

There is a real boundary-data limitation to that expectation. Neither the printed nozzle speed nor the local equations alone guarantee finite closure for every imposed nozzle tension. To see this within the same leading model, define $T=3\mu Q{w}^{\prime }/w+\gamma S$, so $T^{\prime }=-\rho gQ/w$. For equal pressures and $w^{\prime }>0$, both radii decrease and $S\le S_0$. Ifthen a bootstrap gives $T>T_0/2$ and $w\ge w_0{e}^{cz}$: the maximum accumulated weight is smaller than the assumed tension margin. Consequently the closure integral, including its prefactor, is at most $\gamma /(\mu c{w}_0)$. If this is smaller than $\sqrt{A_0}-D_0$, closure never occurs at finite height. For example, in consistent units choose $\mu =\rho g=w_0=1$, $R_i(0)=\lambda$, $R_e(0)=2\lambda$, $\gamma =\lambda$, $T_0=100{\lambda }^2$. Then $c=47/9$, the weight bound is $27{\lambda }^2/47<50{\lambda }^2$, and $9\lambda /47<(\sqrt{3}-1)\lambda$. Small $\lambda$ makes the initial geometry slender. Thus finite closure is the expected freely falling branch behaviour, not a theorem following from nozzle speed alone. If $\gamma =0$ there is no capillary closure by this mechanism.