Exercises — Vorticity — ω = ∇ × v, circulation Γ

WHAT we do: plug into the spin-counter. WHY: the definition is the fastest test. Both $v_x=3$ and $v_y=-5$ are constants — they do not change with position, so every partial derivative is $0$: ${\omega }_z=\frac{\partial (-5)}{\partial x}-\frac{\partial (3)}{\partial y}=0-0=0.$ Answer: ${\omega }_z=0$. A uniform flow carries a paddle-wheel along but never spins it — makes sense, since there is no difference in speed anywhere to twist it.

WHAT: apply the memorised relation ${\omega }_z=2\Omega$. WHY: the parent note proved curl of this field is exactly twice the angular velocity, so we skip the algebra. ${\omega }_z=2\Omega =2\times 4=8{\text{s}}^{-1}.$ Answer: ${\omega }_z=8{\text{s}}^{-1}$ (constant everywhere — a rigid spin has the same vorticity at every point).

(a) ${\omega }_z=\partial (0)/\partial x-\partial (y)/\partial y=0-1=-1\ne 0$. Rotational. (b) ${\omega }_z=\partial (y)/\partial x-\partial (x)/\partial y=0-0=0$. Irrotational (this is a pure stretch flow — it spreads fluid out, it does not spin it). Answer: only (a) has vorticity.

WHAT: differentiate $v_x=\alpha y^2$ with respect to $y$. WHY: only $v_x$ depends on position, and only through $y$, so that is the only surviving term. ${\omega }_z=\frac{\partial v_y}{\partial x}-\frac{\partial v_x}{\partial y}=0-\frac{\partial (\alpha y^2)}{\partial y}=-2\alpha y.$ At $y=3$: ${\omega }_z=-2(2)(3)=-12{\text{s}}^{-1}$. Answer: ${\omega }_z=-12{\text{s}}^{-1}$. Note the vorticity now varies with $y$ — unlike simple shear, this "curved shear" spins fluid harder the farther out you go.

WHAT: walk a circle of radius $r$; at every point the flow speed $v_{\theta }$ is entirely along the walking direction (it is the swirling part, pointing along $\hat{\theta }$), so $\mathbf{v}\cdot d\mathbf{l}=v_{\theta }(rd\theta )$. WHY we write $d\mathbf{l}=rd\theta$: as introduced above, an arc length on a circle is radius times the angle swept. $\Gamma =\oint v_{\theta }rd\theta ={\int }_0^{2\pi }\frac{k}{r}rd\theta =k{\int }_0^{2\pi }d\theta =2\pi k=2\pi (6)=12\pi \approx 37.70{\text{m}}^2/\text{s}.$ Answer: $\Gamma =12\pi \approx 37.70{\text{m}}^2/\text{s}$, the same for every radius — the hallmark of the free vortex. All swirl sits at the singular centre.

Look at the figure below: the dashed circles are three loops of different radius; the teal (inner) arrows are longer than the orange (outer) arrows because $v_{\theta }=k/r$ falls off with radius. Yet the shorter arrows on the big loop are compensated exactly by the longer walk around it, so the walk-around total $\Gamma =2\pi k$ comes out identical on every dashed circle — that is the picture behind "independent of $r$." The purple star marks the singular centre where all the spin is hidden.

WHAT: integrate the (constant) vorticity over the disc. WHY use Stokes here: vorticity is uniform, so the area integral is just ${\omega }_z$ times the area — far easier than a line integral. $\Gamma ={\iint }_S{\omega }_{z}dA={\omega }_z\cdot \pi R^2=8\cdot \pi (2{)}^2=32\pi \approx 100.53{\text{m}}^2/\text{s}.$ Answer: $\Gamma =32\pi \approx 100.53{\text{m}}^2/\text{s}$.

WHAT (spin): differentiate the field with the spin-counter. WHY we differentiate at all: the whole question is whether a paddle-wheel spins here, and vorticity — the curl — is the only quantity that answers that pointwise, so we must compute $\partial v_y/\partial x-\partial v_x/\partial y$ directly. Let $s=x^2+y^2$. $\frac{\partial v_y}{\partial x}=\frac{\partial }{\partial x}\left(\frac{x}{s}\right)=\frac{s-x(2x)}{s^2}=\frac{y^2-x^2}{s^2},$ $\frac{\partial v_x}{\partial y}=\frac{\partial }{\partial y}\left(\frac{-y}{s}\right)=\frac{-s+y(2y)}{s^2}=\frac{y^2-x^2}{s^2}.$ Subtracting: ${\omega }_z=\frac{y^2-x^2}{s^2}-\frac{y^2-x^2}{s^2}=0$ (for $s\ne 0$). WHY it fails at the origin: there $s=0$, the field blows up, derivatives are undefined — all the vorticity is a spike hidden at that one point. WHAT (circulation): convert to polar form $v_{\theta }=1/r$ and reuse the walk-around integral. WHY we switch to polar here: the loop is a circle, and in polar language the swirling speed $v_{\theta }$ is constant along that circle, turning a messy Cartesian line integral into the one-line result of L2.2. With $k=1$, $\Gamma =2\pi$ for any loop around the origin. The paradox resolved: Stokes still holds — the disc contains the singular point, so the "area" integral is not zero; it picks up the concentrated spike worth exactly $2\pi$.

Line integral (counter-clockwise loop, the standard positive direction). Parametrise $x=\cos \theta ,y=\sin \theta$, $d\mathbf{l}=(-\sin \theta ,\cos \theta )d\theta$. Then $\mathbf{v}=(\sin \theta ,-\cos \theta )$, so $\mathbf{v}\cdot d\mathbf{l}=(\sin \theta )(-\sin \theta )+(-\cos \theta )(\cos \theta )=-{\sin }^2\theta -{\cos }^2\theta =-1.$ $\Gamma ={\int }_0^{2\pi }(-1)d\theta =-2\pi .$ Stokes. ${\omega }_z=\partial (-x)/\partial x-\partial (y)/\partial y=-1-1=-2$. Over area $\pi (1{)}^2$: $\Gamma =-2\cdot \pi =-2\pi$. ✓ Answer: $\Gamma =-2\pi$. The minus sign is honest: the flow spins clockwise while we chose to walk counter-clockwise, so the flow opposes our walk — negative circulation.

WHAT: add the two flux contributions. Each half has area $1\times 2=2{\text{m}}^2$. $\Gamma =(+5)(2)+(-5)(2)=10-10=0.$ Answer: $\Gamma =0$, yet the flow is not irrotational — vorticity is $\pm 5$ almost everywhere. The positives and negatives merely cancel in the sum. A paddle-wheel on the left spins one way, on the right the other; the global loop cannot see this.

WHAT: solve the lift law for $\Gamma$. WHY this law: the Kutta–Joukowski theorem says lift per unit span is directly the fluid density times flight speed times the circulation the wing "binds" around itself — circulation is the wing's secret swirl. $\Gamma =\frac{L}{\rho U}=\frac{900}{1.2\times 50}=\frac{900}{60}=15{\text{m}}^2/\text{s}.$ Answer: $\Gamma =15{\text{m}}^2/\text{s}$. A wing with more bound circulation lifts harder at the same speed.

WHAT: conservation of circulation. WHY: Kelvin's theorem says for an ideal fluid the circulation round a loop that moves with the fluid is frozen — it cannot change from its initial value $0$. If the loop now encloses both the wing's bound vortex ($+15$) and the shed starting vortex (${\Gamma }_s$): ${\Gamma }_b+{\Gamma }_s=0\Rightarrow {\Gamma }_s=-{\Gamma }_b=-15{\text{m}}^2/\text{s}.$ Answer: ${\Gamma }_s=-15{\text{m}}^2/\text{s}$ — an equal and opposite vortex is shed off the trailing edge, which is exactly what wind-tunnel smoke shows.

(a) Vorticity. For a purely swirling flow (only $v_{\theta }$, with $v_r=0$) the axial vorticity is ${\omega }_z=\frac{1}{r}\frac{d}{dr}\left(r{v}_{\theta }\right).$ Where this comes from (WHY, not magic): take the walk-around total $\Gamma$ over a small circle of radius $r$; since $v_{\theta }$ is constant along it, $\Gamma (r)=v_{\theta }\cdot 2\pi r=2\pi (r{v}_{\theta })$. Stokes says $\Gamma$ is the vorticity flux through the disc, so the flux out to radius $r$ is $\Phi (r)=2\pi (r{v}_{\theta })$. Vorticity is that flux density: the extra flux in the thin ring between $r$ and $r+dr$ is ${\Phi }^{\prime }(r)dr$, and the ring's area is $2\pi rdr$, so ${\omega }_z=\frac{{\Phi }^{\prime }(r)dr}{2\pi rdr}=\frac{1}{2\pi r}\frac{d}{dr}\left(2\pi r{v}_{\theta }\right)=\frac{1}{r}\frac{d}{dr}\left(r{v}_{\theta }\right).$ That is the same 2D curl of the toolbox, just written in circular coordinates. Now apply it:

• Inside ($r\le a$): $r{v}_{\theta }=\Omega r^2$, so $\frac{d}{dr}(\Omega r^2)=2\Omega r$, giving ${\omega }_z=\frac{1}{r}(2\Omega r)=2\Omega$. A uniform core spin — solid-body, exactly ${\omega }_z=2\Omega =6{\text{s}}^{-1}$.
• Outside ($r>a$): $r{v}_{\theta }=\Omega a^2$ (constant), so its derivative is $0$, giving ${\omega }_z=0$. Irrotational halo — like a free vortex.

(b) Circulation.

• Inside: $\Gamma (r)=\oint v_{\theta }rd\theta =2\pi r\cdot \Omega r=2\pi \Omega r^2$. (Equivalently ${\omega }_z\cdot \pi r^2=2\Omega \pi r^2$. ✓)
• Outside: $\Gamma (r)=2\pi r\cdot \frac{\Omega a^2}{r}=2\pi \Omega a^2$ — the $r$ cancels.

(c) Constant outside. $\Gamma (r>a)=2\pi \Omega a^2=2\pi (3)(2{)}^2=24\pi \approx 75.40{\text{m}}^2/\text{s}$, independent of $r$. Check continuity at the boundary $r=a$: the inside formula gives $\Gamma (a)=2\pi \Omega a^2$, which matches the outside value exactly — the two pieces join smoothly. Conclusion: all the vorticity is bottled inside the core (${\omega }_z=2\Omega$ there, $0$ outside), so beyond $r=a$ you still feel the full accumulated swirl $\Gamma =24\pi {\text{m}}^2/\text{s}$ but no local spin — a paddle-wheel out there merely revolves, it does not turn on its pin.

Numbers at $r=4\text{m}$ (outside): ${\omega }_z=0$, $\Gamma =24\pi \approx 75.40{\text{m}}^2/\text{s}$.

Can you, without notes: (1) compute ${\omega }_z$ from a 2D field, (2) do a circulation line integral including the $rd\theta$ arc length, (3) explain why $\Gamma =0$ ≠ irrotational, (4) relate $\Gamma$ to lift, and (5) split a Rankine vortex into rotational core + irrotational halo? If any is shaky, revisit that level.