UPSC 2016 Maths Optional Paper 2 Q5b — Step-by-Step Solution

10 marks · Section B

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

Question

Does a fluid with velocity $\vec q=\left[z-\dfrac{2x}{r},\ 2y-3z-\dfrac{2y}{r},\ x-3y-\dfrac{2z}{r}\right]$ possess vorticity, where $\vec q(u,v,w)$ is the velocity in the Cartesian frame, $\vec r=(x,y,z)$ and $r^2=x^2+y^2+z^2$ ? What is the circulation in the circle $x^2+y^2=9,\ z=0$ ?

Technique

Split into polynomial + radial parts; radial part is $\nabla(2r)$ hence curl-free; compute $\nabla\times\vec q_1$ directly (all components cancel); circulation via Stokes = flux of the (zero) vorticity, cross-checked by a direct line integral.

Solution

Setup. The vorticity is $\vec\omega=\nabla\times\vec q$ . The field splits as a polynomial part plus a radial part:

\vec q=\underbrace{(z,\ 2y-3z,\ x-3y)}_{\vec q_1}\ -\ \underbrace{\frac{2}{r}(x,y,z)}_{=\,2\vec r/r}.

Note $\dfrac{2\vec r}{r}=\nabla(2r)$ since $\nabla r=\vec r/r$ ; any gradient is curl-free. So the radial part contributes nothing to $\vec\omega$ , and we only need the curl of $\vec q_1$ .

Step 1 — Vorticity of the radial part

\frac{2\vec r}{r}=2\,\nabla r=\nabla(2r)\ \Rightarrow\ \nabla\times\Big(\frac{2\vec r}{r}\Big)=\nabla\times\nabla(2r)=\vec 0.

Step 2 — Vorticity of the polynomial part

With $u_1=z,\ v_1=2y-3z,\ w_1=x-3y$ :

\nabla\times\vec q_1=\left(\frac{\partial w_1}{\partial y}-\frac{\partial v_1}{\partial z},\ \frac{\partial u_1}{\partial z}-\frac{\partial w_1}{\partial x},\ \frac{\partial v_1}{\partial x}-\frac{\partial u_1}{\partial y}\right).

\frac{\partial w_1}{\partial y}-\frac{\partial v_1}{\partial z}=(-3)-(-3)=0,

\frac{\partial u_1}{\partial z}-\frac{\partial w_1}{\partial x}=1-1=0,

\frac{\partial v_1}{\partial x}-\frac{\partial u_1}{\partial y}=0-0=0.

So $\nabla\times\vec q_1=\vec 0$ as well.

Step 3 — Conclusion on vorticity

\boxed{\;\vec\omega=\nabla\times\vec q=\vec 0\quad\Rightarrow\quad\text{the fluid has \textbf{no vorticity} (flow is irrotational).}\;}

(The polynomial part is antisymmetric in just the right way — each off-diagonal derivative cancels its partner — and the radial part is a pure gradient.)

Step 4 — Circulation around the circle

By Stokes’ theorem, the circulation around the circle $C:\ x^2+y^2=9,\ z=0$ (a disc $D$ in the plane $z=0$ , normal $\hat k$ ) is

\Gamma=\oint_C\vec q\cdot d\vec r=\iint_D(\nabla\times\vec q)\cdot\hat k\,dS=\iint_D 0\,dS=0.

Answer

\boxed{\;\Gamma=0.\;}