UPSC 2024 Maths Optional Paper 2 Q8c — Step-by-Step Solution

20 marks · Section B

Vortex motion; circulation · Mechanics & Fluid Dynamics · asked 6× in 13 yrs · Read the full method →

Question

An infinite liquid contains two parallel, equal and opposite rectilinear vortices at distance $2a$ . Show that streamlines relative to the vortex are

\log\frac{x^2+(y-a)^2}{x^2+(y+a)^2}+\frac{y}{a}=C.

Technique

Stream function of the pair; compute the translation velocity $U=\kappa/(4\pi a)$ ; subtract $Uy$ to get the relative stream function.

Solution

$Comoving streamlines of two equal-opposite rectilinear vortices on the y-axis: +\kappa at (0,a) and -\kappa at (0,-a), separation 2a. The dividing streamline \psi_{\mathrm{rel}}=0 is a Kelvin oval passing through the stagnation points (\pm\sqrt3\,a,0) on the x-axis and reaching apexes (0,\pm2.07a); it encloses two recirculation cells, one looping each vortex. Ambient fluid streams past the oval, and the pair drifts rigidly with velocity U=\kappa/(4\pi a) in the +x direction.$

Setup. Vortex $V_1$ at $(0,a)$ with circulation $+\kappa$ ; vortex $V_2$ at $(0,-a)$ with $-\kappa$ .

Step 1 — Stream function in the lab frame.

For a single vortex of strength $\Gamma$ at $(x_0,y_0)$ : $\psi=-\dfrac{\Gamma}{2\pi}\ln r$ where $r=\sqrt{(x-x_0)^2+(y-y_0)^2}$ .

Summing the two vortices:

\psi=\frac{\kappa}{4\pi}\ln\frac{x^2+(y+a)^2}{x^2+(y-a)^2}.

Step 2 — Translation velocity of the pair.

$V_2$ (strength $-\kappa$ ) induces velocity $U=\kappa/(2\pi\cdot 2a)=\kappa/(4\pi a)$ at $V_1$ , directed in the $+x$ direction (perpendicular to the join). The pair translates rigidly with velocity $(U,0)$ .

Step 3 — Relative stream function.

In the frame moving with the vortices, the stream function is

\psi_{\text{rel}}=\psi-Uy=\frac{\kappa}{4\pi}\ln\frac{x^2+(y+a)^2}{x^2+(y-a)^2}-\frac{\kappa}{4\pi a}y.

Step 4 — Streamlines.

Setting $\psi_{\text{rel}}=$ const and multiplying by $4\pi/\kappa$ :

\ln\frac{x^2+(y+a)^2}{x^2+(y-a)^2}-\frac{y}{a}=K'.

Negating (absorbed into the arbitrary constant):

\ln\frac{x^2+(y-a)^2}{x^2+(y+a)^2}+\frac{y}{a}=C.

Answer

\boxed{\;\log\frac{x^2+(y-a)^2}{x^2+(y+a)^2}+\frac{y}{a}=C.\;}