Worked examples — Vorticity — ω = ∇ × v, circulation Γ

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The scenario matrix

Before touching numbers, let us map what kinds of situations vorticity/circulation problems come in. Every worked example below is tagged with the cell it fills.

Cell	Case class	What makes it tricky	Example
A	Solid-body rotation, ${\omega }_z>0$	sign of curl, factor of 2	Ex 1
B	Clockwise rotation, ${\omega }_z<0$	negative vorticity, direction	Ex 2
C	Free vortex, $r\ne 0$	circles but irrotational	Ex 3
D	Free vortex, the singular point $r=0$	degenerate input, delta of vorticity	Ex 3 (part b)
E	Pure shear, straight streamlines	vorticity without visible curving	Ex 4
F	Superposition / cancelling vorticity	$\Gamma =0$ but $\omega \ne 0$ inside	Ex 5
G	Real-world word problem	translate words → field, units	Ex 6 (draining sink)
H	Exam twist — lift from circulation	connect $\Gamma$ to a force	Ex 7
I	Limiting behaviour	what happens as $r\to 0$ or $R\to \infty$	Ex 8

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Notation warm-up (read once)

Every field below is a velocity field $\mathbf{v}=(v_x,v_y,v_z)$: at each point of space it tells you the arrow showing which way the water moves and how fast. We work in 2D (flat flow), so the only spin axis is $z$ (straight up out of the page), and the only vorticity component we need is

${\omega }_z=\frac{\partial v_y}{\partial x}-\frac{\partial v_x}{\partial y}.$

• The symbol $\partial v_y/\partial x$ reads "how fast the up-down velocity $v_y$ changes as you step in the $x$ direction." It is a rate of change — a slope. That is why the derivative is the right tool: rotation is about neighbours moving differently, and a derivative is exactly "how a quantity differs between neighbours."
• ${\omega }_z>0$ means counter-clockwise local spin (the standard positive sense, like a clock run backwards). ${\omega }_z<0$ means clockwise.

Keep the paddle-wheel picture from the parent: drop a tiny cross into the flow; ${\omega }_z$ is (twice) how fast the cross itself pivots.

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Ex 1 — Cell A: solid-body rotation, positive spin

Step 1. Read off the components: $v_x=-3y$, $v_y=3x$. Why this step? The curl formula needs the two partial derivatives; identifying the components cleanly prevents sign slips.

Step 2. Differentiate: $\frac{\partial v_y}{\partial x}=\frac{\partial (3x)}{\partial x}=3,\frac{\partial v_x}{\partial y}=\frac{\partial (-3y)}{\partial y}=-3.$ Why this step? ${\omega }_z$ is precisely the difference of these two "neighbour differences."

Step 3. Combine: ${\omega }_z=3-(-3)=6=2\Omega .$ Why this step? This confirms the parent's headline result $\mathbf{\omega }=2\mathbf{\Omega }$ — the factor 2 is real, not a typo.

Step 4. Circulation via Stokes: ${\omega }_z$ is constant, and by our convention we walk the circle counter-clockwise, so $\Gamma ={\omega }_z\cdot (\text{area})=6\cdot \pi R^2=6\pi (2{)}^2=24\pi \approx 75.40{\text{m}}^2/\text{s}.$ Why this step? When vorticity is uniform, the surface integral is just vorticity times area — no calculus needed.

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Ex 2 — Cell B: clockwise rotation, negative vorticity

Step 1. Components: $v_x=5y$, $v_y=-5x$. Why this step? Same discipline — name the pieces before differentiating.

Step 2. Derivatives: $\frac{\partial v_y}{\partial x}=-5$, $\frac{\partial v_x}{\partial y}=5$. Why this step? These now carry the opposite signs from Ex 1, which is the whole point of the case.

Step 3. ${\omega }_z=-5-5=-10{\text{s}}^{-1}$. Why this step? The negative value encodes clockwise spin — direction lives in the sign, and dropping it would reverse the physics.

Step 4. $\Gamma ={\omega }_z\cdot \pi R^2=-10\pi (1{)}^2=-10\pi \approx -31.42{\text{m}}^2/\text{s}.$ Why this step? By our counter-clockwise convention, a negative $\Gamma$ means the flow pushes you against your walking direction — exactly what a clockwise swirl does.

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Ex 3 — Cells C & D: the free vortex and its singular heart

Step 1. Circulation is the line integral around the circle, walked counter-clockwise. On a circle the tangential velocity is aligned with $d\mathbf{l}$, and $dl=rd\theta$: $\Gamma =\oint v_{\theta }rd\theta ={\int }_0^{2\pi }\frac{k}{r}rd\theta =k{\int }_0^{2\pi }d\theta =2\pi k.$ Why this step? We use the tangential component directly because for a circular loop, walking direction and velocity are parallel — the dot product is just the product of magnitudes.

Step 2. Plug in: $\Gamma =2\pi \cdot 4=8\pi \approx 25.13{\text{m}}^2/\text{s}$, independent of $r$. Why this step? Every loop around the centre gives the same $\Gamma$. That constancy is the fingerprint of the free vortex.

Step 3 (Cell C, $r\ne 0$). Take two nested loops. By Stokes, the vorticity flux between them is ${\Gamma }_{\text{outer}}-{\Gamma }_{\text{inner}}=8\pi -8\pi =0$. Since this holds for any annulus, ${\omega }_z=0$ everywhere off the axis. Why this step? This is the rigorous version of "same circulation ⇒ no swirl in between." The paddle-wheel here revolves but does not spin: the inner edge (fast) and outer edge (slow) torque it in opposite senses and cancel.

Step 4 (Cell D, $r=0$). All the circulation is trapped at the singular centre. Formally the vorticity is a spike (a delta function) at the origin whose total flux equals $8\pi$. The formula $v_{\theta }=k/r$ blows up as $r\to 0$, so this point is a degenerate input — you cannot evaluate ${\omega }_z$ pointwise there. Why this step? Every case-complete analysis must name what happens at the excluded point, not just wave it away.

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Ex 4 — Cell E: pure shear, vorticity with straight streamlines

Step 1. Components: $v_x=\alpha y=7y$, $v_y=0$. Why this step? $v_y=0$ everywhere kills one derivative, isolating the effect of the shear.

Step 2. $\frac{\partial v_y}{\partial x}=0$, $\frac{\partial v_x}{\partial y}=\alpha =7$. Why this step? The nonzero piece comes from velocity changing with height — top of a fluid element moves faster than its bottom.

Step 3. ${\omega }_z=0-7=-7{\text{s}}^{-1}$. Why this step? Nonzero despite straight streamlines — this is the whole lesson of the cell. A tiny cross gets sheared, and its diagonal tilts clockwise, so ${\omega }_z<0$.

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Ex 5 — Cell F: cancelling vorticity, $\Gamma =0$ but $\omega \ne 0$ inside

Step 1. Upper half ($y>0$): $v_x=\beta y$, so $\frac{\partial v_x}{\partial y}=\beta =2$, giving ${\omega }_z=-2$. Why this step? Same shear computation as Ex 4, done piecewise on the region where the formula applies.

Step 2. Lower half ($y<0$): $v_x=-\beta y$, so $\frac{\partial v_x}{\partial y}=-\beta =-2$, giving ${\omega }_z=+2$. Why this step? The sign flips, so the two halves spin oppositely — this is the engineered cancellation we want to test.

Step 3. Circulation around the full rectangle, walked counter-clockwise. The top and bottom edges (length 2) carry $v_x$. On the top ($y=1$): $v_x=2$, walking $-x$ contributes $-2\cdot 2=-4$. On the bottom ($y=-1$): $v_x=-\beta (-1)=+2$, walking $+x$ contributes $+2\cdot 2=+4$. Side edges have $v_x\perp d\mathbf{l}$ so contribute $0$. $\Gamma =-4+4=0.$ Why this step? The equal-and-opposite vorticities cancel in the total flux.

Step 4. Conclude: $\Gamma =0$ across the whole loop, yet ${\omega }_z=\pm 2\ne 0$ inside. Why this step? This is exactly the parent's warning: net flux zero does not mean irrotational. Positive and negative vorticity balanced out.

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Ex 6 — Cell G: real-world word problem (draining bathtub)

Step 1. Model: $v_{\theta }=k/r$. Solve for $k$ using the measurement: $k=v_{\theta }r=0.5\times 0.10=0.05{\text{m}}^2/\text{s}$. Why this step? One data point fixes the single unknown of the model.

Step 2. Circulation: $\Gamma =2\pi k=2\pi (0.05)=0.1\pi \approx 0.3142{\text{m}}^2/\text{s}$. Why this step? Same free-vortex formula from Ex 3, now with a physical $k$.

Step 3. Speed at $r=0.02$: $v_{\theta }=k/r=0.05/0.02=2.5\text{m/s}$. Why this step? The $1/r$ law means smaller radius ⇒ faster flow — the water accelerates into the drain, matching the visible fast whirl at the centre.

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Ex 7 — Cell H: exam twist, lift from circulation

Step 1. Write the relation from Lift and the Kutta–Joukowski Theorem: lift per unit span $L=\rho U\Gamma$. Why this step? This is the bridge from circulation (a purely kinematic swirl) to a real force — the reason engineers care about $\Gamma$ at all.

Step 2. Substitute: $L=1.2\times 60\times 30=2160\text{N/m}$. Why this step? Direct plug-in once the formula is chosen.

Step 3. Answer the forecast: $L\propto \Gamma$ linearly, so doubling $\Gamma$ doubles the lift. Why this step? Recognising proportionality is the "twist" examiners test — not the arithmetic.

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Ex 8 — Cell I: limiting behaviour

Step 1 (a, $r\to \infty$). Free vortex: $v_{\theta }=k/r\to 0$. Far away the swirl dies out — the flow is calm. Why this step? Limits reveal the "background" state; a free vortex is a localized disturbance that fades with distance.

Step 2 (a, $r\to 0$). Free vortex: $v_{\theta }=k/r\to \infty$ (the singular core of Cell D). Physically, real fluids cap this with a viscous core, but the ideal model diverges. Why this step? Naming the blow-up completes the degenerate-input picture — you must state what the model does at the point it cannot handle, and how reality tames it.

Step 3 (b, $r\to 0$). Solid body: $v_{\theta }=\Omega r\to 0$. The centre of a rigidly rotating body is at rest — the exact opposite behaviour to the vortex, which blows up there. Why this step? Contrasting the two limits cements that "rotation" is not one single flow: the free vortex is fastest at the centre, the solid body is fastest far out.

Step 4 (c, matching speeds). Set the two speeds equal and solve for the crossover radius: $\Omega r=\frac{k}{r}\Rightarrow r^2=\frac{k}{\Omega }=\frac{4}{3}\Rightarrow r=\sqrt{\frac{4}{3}}=\frac{2}{\sqrt{3}}\approx 1.155\text{m}.$ Why this step? This crossover radius is exactly the Rankine vortex boundary — solid-body core inside, free vortex outside — the two-piece model used for real tornadoes and drain vortices.

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Quick self-test

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Connections

• Parent: Vorticity & Circulation
• Curl and Divergence (vector calculus)
• Stokes' Theorem
• Irrotational Flow and Velocity Potential
• Kelvin's Circulation Theorem
• Bernoulli's Equation
• Lift and the Kutta–Joukowski Theorem
• Angular Velocity of Rigid Bodies