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UPSC 2022 Maths Optional Paper 2 Q5e — Step-by-Step Solution

10 marks · Section B

Potential flow · Mechanics & Fluid Dynamics · asked 10× in 13 yrs · Read the full method →

Question

Velocity components in spherical polar $(r,\theta,\psi)$: $v_r=2Mr^{-3}\cos\theta$, $v_\theta=Mr^{-2}\sin\theta$, $v_\psi=0$. Show velocity is of potential kind. Find velocity potential and streamlines.

Technique

Standard 3D doublet flow; verify irrotationality, recover potential by $\nabla\phi=\vec v$, recover stream function by $v_r=(1/(r^2\sin\theta))\partial_\theta\psi$.

Solution

Wait — re-read: the velocity is $(2Mr^{-3}\cos\theta, Mr^{-2}\sin\theta, 0)$. The $r^{-3}$ and $r^{-2}$ factors suggest dipole-like flow.

Step 1 — Test for potential flow

Velocity is of potential kind iff $\nabla\times\vec v=0$ (irrotational).

In spherical: $\nabla\times\vec v$ has $\psi$-component $\dfrac{1}{r}\!\left[\dfrac{\partial(rv_\theta)}{\partial r}-\dfrac{\partial v_r}{\partial\theta}\right]$.

$rv_\theta=r\cdot Mr^{-2}\sin\theta=Mr^{-1}\sin\theta$.

$\partial(rv_\theta)/\partial r=-Mr^{-2}\sin\theta$.

$\partial v_r/\partial\theta=2Mr^{-3}(-\sin\theta)=-2Mr^{-3}\sin\theta$.

Hmm wait — the formula for curl in spherical (assuming axisymmetric, no $\psi$ dependence):

$(\nabla\times\vec v)_\psi=\dfrac{1}{r}\!\left[\dfrac{\partial(rv_\theta)}{\partial r}-\dfrac{\partial v_r}{\partial\theta}\right]$

Hmm — let me look this up. The full curl in spherical $(r,\theta,\phi)$:

$(\nabla\times\vec v)_\phi=\dfrac{1}{r}\!\left[\dfrac{\partial(rv_\theta)}{\partial r}-\dfrac{\partial v_r}{\partial\theta}\right]$.

Substitute: $\dfrac{\partial(rv_\theta)}{\partial r}=\dfrac{\partial(Mr^{-1}\sin\theta)}{\partial r}=-Mr^{-2}\sin\theta$.

$\dfrac{\partial v_r}{\partial\theta}=\dfrac{\partial(2Mr^{-3}\cos\theta)}{\partial\theta}=-2Mr^{-3}\sin\theta$.

$(\nabla\times\vec v)_\phi=\dfrac{1}{r}[-Mr^{-2}\sin\theta-(-2Mr^{-3}\sin\theta)]=\dfrac{1}{r}\!\left[-Mr^{-2}+2Mr^{-3}\right]\sin\theta$.

For irrotational: need this $=0$. But $-Mr^{-2}+2Mr^{-3}$ does not vanish identically; it depends on $r$.

Hmm — likely a typo or misreading. Let me re-examine the problem.

Actually, re-reading the velocity components: $v_r=2Mr^{-3}\cos\theta$, $v_\theta=Mr^{-3}\sin\theta$ (note $r^{-3}$ not $r^{-2}$). Let me retry with $v_\theta=Mr^{-3}\sin\theta$.

Hmm but the question states $v_\theta=Mr^{-2}\sin\theta$. Let me just check if this is the dipole formula.

Standard dipole flow. A 3D doublet of strength $\mu$ aligned with $z$-axis has velocity potential $\phi=-\mu\cos\theta/r^2$. Then $v_r=\partial\phi/\partial r=2\mu\cos\theta/r^3$ and $v_\theta=(1/r)\partial\phi/\partial\theta=(1/r)\cdot(\mu\sin\theta/r^2)=\mu\sin\theta/r^3$.

So the standard dipole has $v_r\sim r^{-3}$ and $v_\theta\sim r^{-3}$, both with the same factor.

The question’s $v_\theta=Mr^{-2}\sin\theta$ has a different exponent. Likely a typo; assuming $v_\theta=Mr^{-3}\sin\theta$ (matching dipole):

$\partial(rv_\theta)/\partial r=\partial(Mr^{-2}\sin\theta)/\partial r=-2Mr^{-3}\sin\theta$.

$\partial v_r/\partial\theta=-2Mr^{-3}\sin\theta$.

$(\nabla\times\vec v)_\phi=\dfrac{1}{r}\!\left[-2Mr^{-3}\sin\theta-(-2Mr^{-3}\sin\theta)\right]=0$ ✓.

Assuming the corrected formula $v_\theta=Mr^{-3}\sin\theta$, flow is irrotational.

Step 2 — Find velocity potential

$v_r=\partial\phi/\partial r=2M\cos\theta/r^3$.

Integrate w.r.t. $r$: $\phi=-M\cos\theta/r^2+g(\theta)$.

$v_\theta=(1/r)\partial\phi/\partial\theta=(1/r)\cdot[M\sin\theta/r^2+g'(\theta)]=M\sin\theta/r^3+g'(\theta)/r$.

Match $v_\theta=M\sin\theta/r^3$: $g'(\theta)/r=0$, so $g'(\theta)=0$, $g=$ const.

This is the classical 3D doublet (dipole) potential of strength $M$ aligned with the $z$-axis.

Step 3 — Streamlines

For axisymmetric flow, the stream function $\psi$ satisfies $v_r=\dfrac{1}{r^2\sin\theta}\partial_\theta\psi$, $v_\theta=-\dfrac{1}{r\sin\theta}\partial_r\psi$.

From $v_r=2M\cos\theta/r^3$: $\partial_\theta\psi=v_r\cdot r^2\sin\theta=2M\cos\theta\sin\theta/r$.

Integrate w.r.t. $\theta$: $\psi=-M\cos^2\theta/r+h(r)$? Let me use $2\sin\theta\cos\theta=\sin 2\theta$ — then $\int 2\sin\theta\cos\theta\,d\theta=-\cos 2\theta/...$

Actually $\int\sin\theta\cos\theta\,d\theta=\sin^2\theta/2$ (since $d(\sin^2\theta)/d\theta=2\sin\theta\cos\theta$).

So $\psi=\dfrac{M\sin^2\theta}{r}+h(r)$.

Wait: $\partial_\theta\psi=2M\cos\theta\sin\theta/r=(M/r)\sin 2\theta=(M/r)\cdot 2\sin\theta\cos\theta$. Integrating w.r.t. $\theta$: $(M/r)\cdot\sin^2\theta+h(r)$ (since $d(\sin^2\theta)/d\theta=2\sin\theta\cos\theta$).

So $\psi=M\sin^2\theta/r+h(r)$.

Check $v_\theta=-(1/(r\sin\theta))\partial_r\psi=-(1/(r\sin\theta))[-M\sin^2\theta/r^2+h'(r)]$.

$=\dfrac{M\sin\theta}{r^3}-\dfrac{h'(r)}{r\sin\theta}$.

Match $v_\theta=M\sin\theta/r^3$: $h'(r)/(r\sin\theta)=0$, so $h'(r)=0$, $h=$ const.

Answer