Viscosity — dynamic μ, kinematic ν = μ - ρ; Newtonian vs non-Newtonian: Worked examples

Before we start, three symbols the parent earned that we reuse. If any feels unfamiliar, re-read the parent first.

Recall The three symbols we lean on

$τ$ (Greek "tau") ::: shear stress — force per unit area that neighbouring fluid layers exert on each other, measured in pascals (Pa). $μ$ (Greek "mu") ::: dynamic viscosity, the constant in $τ = μ d u / d y$ , units Pa·s. $\overset{γ}{˙}$ (gamma with a dot) ::: shear rate = velocity gradient $d u / d y$ , "how fast speed changes across the gap", units s⁻¹ (per second).

The dot over $γ$ means "rate" — a change per second — exactly like a dot over a distance would mean speed. So $\overset{γ}{˙}$ is literally the rate at which the layers shear (slide) apart.

The scenario matrix

Every problem this topic can throw at you falls into one of these cells. The rest of the page fills each cell with a worked example.

Cell	What changes	Example
A. Positive gradient	speed rises with height, $d u / d y > 0$	Ex 1
B. Negative gradient	speed falls with height, $d u / d y < 0$ — sign of $τ$ flips	Ex 2
C. Zero gradient (degenerate)	uniform flow, $d u / d y = 0$	Ex 3
D. Thin-gap limit	$h \to 0$ : stress blows up	Ex 4
E. $μ$ vs $ν$ ranking	which fluid diffuses momentum faster	Ex 5
F. Non-Newtonian, $n < 1$	shear-thinning apparent viscosity	Ex 6
G. Non-Newtonian, $n > 1$	shear-thickening apparent viscosity	Ex 7
H. Bingham yield (degenerate flow)	won't move until $τ > τ_{0}$	Ex 8
I. Real-world word problem	falling sphere, momentum-diffusion time	Ex 9
J. Exam twist	two fluids stacked, matching stress	Ex 10

Figure — Viscosity — dynamic μ, kinematic ν = μ - ρ; Newtonian vs non-Newtonian

The figure shows the four gradient signs side by side — look at the amber slope arrows: up-slope (A), flat (C), down-slope (B), and the near-vertical thin-gap case (D). The sign and steepness of that slope is the entire story of Newton's law.

Cell A — Positive gradient

Worked example Ex 1 — Standard Couette drag

Water-like oil, $μ = 0.8$ Pa·s, fills a gap $h = 2 mm = 0.002$ m. The top plate ( $y = h$ ) moves at $U = 0.3$ m/s; bottom fixed. Plate area $A = 0.5 m^{2}$ . Find the shear stress and the force.

Forecast: guess — will the force be closer to 6 N or 600 N? Write your guess down.

Shear rate. $\overset{γ}{˙} = \frac{d u}{d y} = \frac{U - 0}{h - 0} = \frac{0.3}{0.002} = 150 s^{- 1}$ . Why this step? The profile is a straight line from $0$ to $U$ , and the slope of a straight line is (rise)/(run) $= U / h$ . That slope is $d u / d y$ .
Stress. $τ = μ \overset{γ}{˙} = 0.8 \times 150 = 120$ Pa. Why this step? Newton's law: stress = viscosity × gradient.
Force. $F = τ A = 120 \times 0.5 = 60$ N. Why this step? Stress is force per area; multiply back by area to recover force.

Verify: units — $(Pa\cdots) (s^{- 1}) = Pa$ ✓, then $(Pa) (m^{2}) = N$ ✓. The answer 60 N sits near your first guess bracket. Positive $U$ gave positive $τ$ : the fluid pushes back on the moving plate, dragging it, as expected.

Cell B — Negative gradient

Worked example Ex 2 — Speed decreasing with height

Same oil ( $μ = 0.8$ Pa·s), but now the velocity decreases with height: at the bottom $y = 0$ the fluid moves at $u = 0.4$ m/s, and at the top $y = h = 0.002$ m it moves at $u = 0.1$ m/s. Find $τ$ and interpret its sign.

Forecast: will $τ$ come out positive or negative this time?

Shear rate. $\overset{γ}{˙} = \frac{u _{top} - u _{bottom}}{h} = \frac{0.1 - 0.4}{0.002} = \frac{- 0.3}{0.002} = - 150 s^{- 1}$ . Why this step? The gradient is (value at top − value at bottom)/gap. Because speed drops going up, the numerator is negative.
Stress. $τ = μ \overset{γ}{˙} = 0.8 \times (- 150) = - 120$ Pa. Why this step? Newton's law carries the sign straight through.
Interpret. The minus sign says the stress on the upper layer points in the negative flow direction — the slow top fluid is dragged forward while it drags the fast bottom fluid back. The magnitude, 120 Pa, is identical to Ex 1. Why this step? Sign encodes direction of momentum transfer; magnitude encodes how much.

Verify: flipping which plate is faster flips the sign but not the size — a good symmetry check. $∣ τ ∣ = 120$ Pa matches Ex 1's magnitude exactly. ✓

Cell C — Zero gradient (the trap)

Worked example Ex 3 — Uniform stream

A whole river of the same oil moves as a rigid slab: every layer travels at $u = 10$ m/s, top and bottom alike. Find the internal viscous stress.

Forecast: 10 m/s is fast — surely there's big stress? Guess before reading.

Shear rate. $\overset{γ}{˙} = \frac{u _{top} - u _{bottom}}{h} = \frac{10 - 10}{h} = 0 s^{- 1}$ . Why this step? Viscosity cares about the difference between layers, not their common speed. No difference ⇒ zero gradient, whatever $h$ is.
Stress. $τ = μ \times 0 = 0$ Pa. Why this step? Newton's law: no gradient, no stress — the whole point of writing $τ = μ d u / d y$ rather than $τ = μu$ .

Verify: dimension-free sanity — a uniformly moving fluid can be viewed from a frame moving with it, where everything is at rest and obviously stress-free. Physics can't depend on that frame choice, so the stress must be $0$ in every frame. ✓ This is the parent's #1 misconception, defused.

Cell D — Thin-gap limit (degenerate)

Worked example Ex 4 — Squeezing the gap to zero

Take Ex 1 ( $μ = 0.8$ Pa·s, $U = 0.3$ m/s fixed) and shrink the gap: $h = 0.2$ mm $= 2 \times 1 0^{- 4}$ m. What happens to $τ$ ? Then reason about $h \to 0$ .

Forecast: ten times thinner gap — stress goes up by what factor?

New shear rate. $\overset{γ}{˙} = \frac{0.3}{2 \times 1 0 ^{- 4}} = 1500 s^{- 1}$ . Why this step? Same speed spread over a tenth of the distance ⇒ ten times steeper slope.
New stress. $τ = 0.8 \times 1500 = 1200$ Pa — exactly $10 \times$ the Ex 1 value. Why this step? $τ \propto 1/ h$ at fixed $U$ ; cut $h$ by 10, multiply $τ$ by 10.
Limit. As $h \to 0$ , $\overset{γ}{˙} = U / h \to \infty$ , so $τ \to \infty$ . Why this step? This is why real machines keep a finite oil film — an infinitely thin film would demand infinite force (and shred the fluid). This is the seed of the Boundary Layer idea: near a wall the gradient is steepest.

Verify: ratio check $τ_{new} / τ_{old} = 1200/120 = 10 = h_{old} / h_{new}$ ✓ — inverse proportionality confirmed.

Cell E — μ vs ν ranking

Worked example Ex 5 — Which spreads motion faster, air or water?

Air: $μ_{a} = 1.8 \times 1 0^{- 5}$ Pa·s, $ρ_{a} = 1.2 kg/m^{3}$ . Water: $μ_{w} = 1.0 \times 1 0^{- 3}$ Pa·s, $ρ_{w} = 1000 kg/m^{3}$ . Compute both $ν$ and rank them.

Forecast: water is obviously "thicker" than air — so water's $ν$ is bigger, right?

Air's $ν$ . $ν_{a} = \frac{μ _{a}}{ρ _{a}} = \frac{1.8 \times 1 0 ^{- 5}}{1.2} = 1.5 \times 1 0^{- 5} m^{2} / s$ . Why this step? Definition $ν = μ / ρ$ ; dividing by density compares the viscous force to the inertia it must move.
Water's $ν$ . $ν_{w} = \frac{1.0 \times 1 0 ^{- 3}}{1000} = 1.0 \times 1 0^{- 6} m^{2} / s$ .
Rank. $ν_{a} / ν_{w} = \frac{1.5 \times 1 0 ^{- 5}}{1.0 \times 1 0 ^{- 6}} = 15$ . Air diffuses momentum 15× faster per unit mass than water. Why this step? Air's $μ$ is ~55× smaller, but its $ρ$ is ~830× smaller — the density factor wins, so $ν_{a} > ν_{w}$ . This is Momentum Diffusion in numbers.

Verify: units $\frac{Pa\cdots}{kg/m ^{3}} = \frac{kg/(m\cdots)}{kg/m ^{3}} = m^{2} / s$ ✓. Answer contradicts the "thicker = higher ν" intuition — exactly the parent's counter-intuitive fact, made quantitative.

Cell F — Non-Newtonian, shear-thinning (n < 1)

Worked example Ex 6 — Paint on a brush

Power-law paint $τ = K \overset{γ}{˙}^{n}$ with $K = 2$ , $n = 0.5$ . Compare the apparent viscosity $μ_{app} = τ / \overset{γ}{˙}$ at a gentle wall shear $\overset{γ}{˙} = 4 s^{- 1}$ versus a brisk brush shear $\overset{γ}{˙} = 100 s^{- 1}$ .

Forecast: does the paint get thicker or thinner when you brush it faster?

Apparent-viscosity formula. $μ_{app} = \frac{τ}{γ ˙} = \frac{K γ ˙ ^{n}}{γ ˙} = K \overset{γ}{˙}^{n - 1} = 2 \overset{γ}{˙}^{- 0.5}$ . Why this step? "Apparent viscosity" is the slope-to-origin you'd read as if it were Newtonian at that shear rate. We use it because $μ$ is no longer constant.
Slow shear. $μ_{app} = 2 \times 4^{- 0.5} = 2/ 4 = 2/2 = 1.0$ Pa·s.
Fast shear. $μ_{app} = 2 \times 10 0^{- 0.5} = 2/ 100 = 2/10 = 0.2$ Pa·s. Why this step? $n - 1 = - 0.5 < 0$ , so raising $\overset{γ}{˙}$ lowers $μ_{app}$ — the definition of shear-thinning.

Verify: ratio $1.0/0.2 = 5$ ; also $(100/4)^{0.5} = (25)^{0.5} = 5$ ✓. On the brush (thin, spreads) then on the wall (thick, doesn't drip) — the paint's whole selling point.

Cell G — Non-Newtonian, shear-thickening (n > 1)

Cell H — Bingham yield (degenerate flow start)

Worked example Ex 8 — Toothpaste won't budge

A Bingham plastic obeys $τ = τ_{0} + μ_{p} \overset{γ}{˙}$ with yield stress $τ_{0} = 60$ Pa and plastic viscosity $μ_{p} = 0.4$ Pa·s. (a) Applied stress $τ = 40$ Pa — does it flow? (b) Applied $τ = 100$ Pa — find the shear rate.

Forecast: at 40 Pa, does anything move at all?

Check the yield. Flow occurs only if $τ > τ_{0}$ . At (a) $40 < 60$ , so $\overset{γ}{˙} = 0$ — it stays put (a solid-like plug). Why this step? Below yield the structure holds like a soft solid; there is no gradient because there is no motion.
Above yield (b). Rearrange: $\overset{γ}{˙} = \frac{τ - τ _{0}}{μ _{p}} = \frac{100 - 60}{0.4} = \frac{40}{0.4} = 100 s^{- 1}$ . Why this step? Only the excess stress beyond $τ_{0}$ drives flow; divide that excess by $μ_{p}$ .

Verify: plug back — $τ = 60 + 0.4 \times 100 = 60 + 40 = 100$ Pa ✓. Case (a) correctly gives the degenerate no-flow answer, case (b) a clean 100 s⁻¹.

Cell I — Real-world word problem

Worked example Ex 9 — How long for a jolt to reach across a mug of water

A spoon jolts one wall of a mug. Using momentum diffusion, estimate how long for that disturbance to spread a distance $L = 3$ cm $= 0.03$ m through still water, $ν_{w} = 1.0 \times 1 0^{- 6} m^{2} / s$ .

Forecast: seconds, minutes, or hours?

Diffusion length rule. From the parent: in time $t$ momentum spreads $L \sim ν t$ . Why this step? $ν$ has units m²/s (a diffusion coefficient), so the only length its combination with time can make is $ν t$ — see Momentum Diffusion.
Solve for $t$ . Square both sides: $L^{2} = ν t \Rightarrow t = \frac{L ^{2}}{ν} = \frac{( 0.03 ) ^{2}}{1.0 \times 1 0 ^{- 6}} = \frac{9 \times 1 0 ^{- 4}}{1 0 ^{- 6}} = 900$ s. Why this step? Rearranging the scaling law gives the timescale for viscosity alone to carry the motion.

Verify: units $\frac{m ^{2}}{m ^{2} / s} = s$ ✓. $900$ s $= 15$ minutes — startlingly slow! This is why real stirring works by bulk swirling (convection), not by viscosity; it also explains why viscous effects hide in a thin Boundary Layer and why Reynolds Number compares inertia to viscosity.

Cell J — Exam twist

Worked example Ex 10 — Two stacked fluids, one plate

Two Newtonian layers fill a gap between a fixed floor and a moving ceiling at $U = 0.5$ m/s. Bottom layer: $μ_{1} = 0.2$ Pa·s, thickness $h_{1} = 1 mm$ . Top layer: $μ_{2} = 0.6$ Pa·s, thickness $h_{2} = 1 mm$ . The stress is the same through both (steady, no acceleration). Find the interface speed $u_{i}$ and the common stress $τ$ .

Forecast: is the interface speed above or below the midpoint value $0.25$ m/s?

Stress continuity. In steady shear with no body force, $τ$ is uniform, so $μ_{1} \frac{u _{i} - 0}{h _{1}} = μ_{2} \frac{U - u _{i}}{h _{2}}$ . Why this step? If the two stresses differed, the thin interface layer would have a net force and accelerate — impossible in steady state.
Insert numbers ( $h_{1} = h_{2} = 0.001$ m cancel): $\frac{0.2 u _{i}}{0.001} = \frac{0.6 ( 0.5 - u _{i} )}{0.001} \Rightarrow 0.2 u_{i} = 0.6 (0.5 - u_{i})$ .
Solve. $0.2 u_{i} = 0.3 - 0.6 u_{i} \Rightarrow 0.8 u_{i} = 0.3 \Rightarrow u_{i} = 0.375$ m/s. Why this step? Algebra. The interface sits above the midpoint because the thinner-resisting bottom layer (smaller $μ_{1}$ ) shears more easily, so most of the speed is used up there.
Common stress. $τ = μ_{1} \frac{u _{i}}{h _{1}} = 0.2 \times \frac{0.375}{0.001} = 0.2 \times 375 = 75$ Pa. Why this step? Now that $u_{i}$ is known, either layer gives $τ$ .

Verify: top layer must give the same stress: $μ_{2} \frac{U - u _{i}}{h _{2}} = 0.6 \times \frac{0.5 - 0.375}{0.001} = 0.6 \times 125 = 75$ Pa ✓. Both agree, and $u_{i} = 0.375 > 0.25$ as forecast — the less-viscous layer takes more of the velocity drop.

Recall Which cell was hardest?

The zero-gradient trap (Cell C) ::: stress is $0$ even at 10 m/s — viscosity reads gradients, not speeds. The μ-vs-ν ranking (Cell E) ::: air beats water on $ν$ despite feeling thinner, because dividing by the tiny density dominates. The stacked-fluid twist (Cell J) ::: enforce equal stress across the interface, not equal shear rate.