Let $z$ increase downward. The <slender viscous thread> is locally in uniaxial extension. The transverse stress equals the ambient pressure, so the <Trouton ratio> gives the excess axial stress $\sigma_{zz}+p_e=3\mu w_z$. <Mass conservation> and axial force balance therefore give
$$
\boxed{A_t+(Aw)_z=0,}
\qquad
\boxed{\rho gA+\partial_z(3\mu A w_z)=0.}
$$

In steady flow $Aw=Q$. Dividing the momentum equation by $A=Q/w$ gives
$$
3\mu w\left(\frac{w_z}{w}\right)_z+\rho g=0.
$$
With $w=\widehat wW$ and $z=\widehat zZ$, choose
$$
\boxed{\widehat z=\left(\frac{3\mu\widehat w}{\rho g}\right)^{1/2}.}
$$
The dimensionless equation becomes
$$
W\left(\frac{W'}W\right)'=-1,
\qquad
WW''-(W')^2=-W.
$$
Treating $(W')^2$ as a function of $W$ and using an <integrating factor> yields
$$
\boxed{\frac12(W')^2=CW^2+W.}
$$
Because $A=Q/(\widehat wW)$, the area sketches are the reciprocals of the functions $W$ found below. The dimensional vertical deviatoric stress is
$$
\sigma_{zz}+p_e=\frac{3\mu\widehat w}{\widehat z}W'.
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 329 / 2 / a / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2025/iii/paper-329/2/a/solution