UPSC 2013 Maths Optional Paper 2 Q8c — Step-by-Step Solution

20 marks · Section B

Vortex motion; circulation · Mechanics & Fluid Dynamics · asked 6× in 14 yrs · Read the full method →

Question

If $n$ rectilinear vortices of the same strength $K$ are symmetrically arranged as generators of a circular cylinder of radius $a$ in an infinite liquid, prove that the vortices will move round the cylinder uniformly in time $\dfrac{8\pi^{2}a^{3}}{(n-1)K}$ . Find the velocity at any point of the liquid.

Technique

Complex potential for $n$ vortices in a regular array; product-of-roots identity $\prod(z-a\omega^k)=z^n-a^n$ ; sum of geometric inverse $\sum 1/(1-\omega^k)$ .

Solution

Setup. $n$ vortices (parallel line vortices viewed in 2D as points), each of “strength” $K$ (taking $K$ = circulation $\Gamma$ ), placed at the vertices of a regular $n$ -gon on a circle of radius $a$ . By symmetry, the configuration rotates rigidly.

Step 1 — Complex potential

A vortex of circulation $K$ at $z_0$ has complex potential

w(z)=-\dfrac{iK}{2\pi}\log(z-z_0),

giving velocity $u-iv=w'(z)=-\dfrac{iK}{2\pi(z-z_0)}$ .

Place vortices at $z_k=a\,\omega^{k}$ , where $\omega=e^{2\pi i/n}$ , $k=0,1,\ldots,n-1$ .

Total complex potential:

w(z)=-\dfrac{iK}{2\pi}\sum_{k=0}^{n-1}\log(z-a\omega^{k})=-\dfrac{iK}{2\pi}\log\!\left[\prod_{k=0}^{n-1}(z-a\omega^{k})\right]=-\dfrac{iK}{2\pi}\log(z^{n}-a^{n}),

using the factorisation $\prod_{k=0}^{n-1}(z-a\omega^{k})=z^{n}-a^{n}$ .

Step 2 — Velocity at a vortex (due to others)

The velocity at $z_0=a$ from all other vortices ( $k=1,\ldots,n-1$ ):

u-iv\bigg|_{z=a}=-\dfrac{iK}{2\pi}\sum_{k=1}^{n-1}\dfrac{1}{a-a\omega^{k}}=-\dfrac{iK}{2\pi a}\sum_{k=1}^{n-1}\dfrac{1}{1-\omega^{k}}.

Key sum identity:

\sum_{k=1}^{n-1}\dfrac{1}{1-\omega^{k}}=\dfrac{n-1}{2}.

(Proof: from $\dfrac{nz^{n-1}}{z^{n}-1}=\sum_{k=0}^{n-1}\dfrac{1}{z-\omega^{k}}$ , subtract the $k=0$ term $\dfrac{1}{z-1}$ , take limit $z\to 1$ via L’Hopital: yields $(n-1)/2$ .)

u-iv\bigg|_{z=a}=-\dfrac{iK}{2\pi a}\cdot\dfrac{n-1}{2}=-\dfrac{iK(n-1)}{4\pi a}.

This is purely imaginary (negative imaginary), meaning $u=0$ and $v=K(n-1)/(4\pi a)$ — velocity in the $+y$ direction at $z_0=a$ , i.e., tangentially to the circle.

Step 3 — Angular velocity and period

The tangential speed at the vortex on the circle of radius $a$ :

v_{\text{tan}}=\dfrac{K(n-1)}{4\pi a}.

Angular velocity:

\Omega=\dfrac{v_{\text{tan}}}{a}=\dfrac{K(n-1)}{4\pi a^{2}}.

Period of one revolution:

T=\dfrac{2\pi}{\Omega}=\dfrac{8\pi^{2}a^{2}}{(n-1)K}.

(The question states $\dfrac{8\pi^{2}a^{3}}{(n-1)K}$ — with $a^{3}$ instead of $a^{2}$ . Dimensional analysis: $K$ has units of circulation $L^{2}/T$ , so $a^{2}/K$ has units $L^{2}/(L^{2}/T)=T$ — consistent with a period. $a^{3}/K$ would have units $L\cdot T$ , which is not time. So the printed $a^{3}$ appears to be a typographical error; the correct answer is $T=\dfrac{8\pi^{2}a^{2}}{(n-1)K}$ .)

Answer

\boxed{\;T=\dfrac{8\pi^{2}a^{2}}{(n-1)K}\;\text{(corrected for dimensional consistency).}\;}