Classical-Mechanics · Unit 10 · Video 1 · Interactive Practice

Fast and Slow: The Two Regimes of Fluid Drag

IKey Formulas

Regime	Drag force (magnitude)	Grows with
Fast — object in air	$F = \tfrac{1}{2} C_D A \rho v^2$	$v^2$ (quadratic); no viscosity
Slow — small sphere in a thick fluid	$F = 6\pi \eta R v$	$v$ (linear); all viscosity
Sedimentation	$\dfrac{\text{drag}}{\text{mass}} \propto \dfrac{1}{R^2}$	falls off as radius grows

Key Insight: Both forces oppose the motion — Stokes' law carries a minus sign, $\vec{F} = -6\pi\eta R\,\vec{v}$ . What changes between regimes is the speed dependence: $v^2$ when fast, $v$ when slow.

IIThe Two Regimes: Linear vs Quadratic Drag

Fast drag climbs as $v^2$ ; slow drag climbs only as $v$ — the same speed, very different forces.

Speed v 1.00

💡 Which law an object obeys is fixed by its speed and size (its Reynolds number): a thrown ball is quadratic, a settling grain linear.

IIIDrag Coefficient — Shape Sets the Drag

How much can shape alone change the drag coefficient $C_D$ in $F = \tfrac{1}{2} C_D A \rho v^2$ ?

💡 This coefficient is the only shape-dependent factor in $F = \tfrac{1}{2} C_D A \rho v^2$ ; the area $A$ and density $\rho$ carry the rest.

IVSedimentation — Why Size Decides the Fall

Why does molecule $B$ , feeling more total drag, still settle faster than the smaller molecule $A$ ?

Step 1

💡 Ultracentrifuges exploit this $1/R^2$ scaling to sort molecules by molecular weight.

VQuiz Questions

Problem 1 · Quadratic Scaling

Given: A ball flies through air, where the high-speed drag is $F = \tfrac{1}{2} C_D A \rho v^2$ . If its speed triples with everything else fixed, by what factor does the drag force grow?

✅ Correct!

F \propto v^2

, so tripling the speed multiplies the drag by

3^2 = 9

❌ Not quite. That is the linear (slow-regime) answer. The high-speed law goes as

v^2

, so the factor is

3^2 = 9

, not

3

❌ Close. The speed enters as

v^2

, not

v^3

— so the factor is

3^2 = 9

, not

3^3 = 27

❌ Not quite. With

F \propto v^2

, tripling

v

scales

F

3^2 = 9

Show solution

Only $v$ changes, so treat the other factors as constant:

F = \tfrac{1}{2} C_D A \rho v^2 \;\Rightarrow\; F \propto v^2

Replacing $v$ with $3v$ multiplies $F$ by $3^2 = 9$ . The drag grows nine-fold — the hallmark of the quadratic regime.

Problem 2 · Which Power of Speed?

Given: A grain of sand settles slowly through water, obeying Stokes' law $F = 6\pi\eta R v$ . If it settled twice as fast, the drag force would ___.

✅ Correct! Stokes' drag is linear in

v

, so doubling the speed doubles the force.

❌ Close. Quadrupling (

v^2

) is the fast-regime rule. Stokes' law

F = 6\pi\eta R v

is linear in

v

, so doubling

v

only doubles

F

❌ Not quite. Speed-independence describes sliding friction between solids; fluid drag depends sharply on speed — here

F \propto v

❌ Not quite.

F = 6\pi\eta R v

is linear in

v

: double the speed, double the force.

Show solution

In Stokes' law the speed appears to the first power:

F = 6\pi\eta R v \;\Rightarrow\; F \propto v

So $v \to 2v$ gives $F \to 2F$ — the force doubles. (In the fast regime it would quadruple, and for solid-on-solid friction it would barely change at all.)

Problem 3 · Compute the Stokes Drag

Given: A sphere of radius $R = 1.0\times10^{-3}\ \mathrm{m}$ settles through glycerol ( $\eta = 1.2\ \mathrm{Pa\cdot s}$ ) at $v = 5.0\times10^{-3}\ \mathrm{m/s}$ .

Which law governs the drag?

Drag force magnitude?

✅ Correct! A millimetre sphere creeping through thick glycerol is the slow regime, and

6\pi\eta R v \approx 1.1\times10^{-4}\ \mathrm{N}

❌ Wrong regime. A tiny sphere creeping through thick glycerol is slow and viscous — use Stokes' law, not the

v^2

law.

❌ Close — you dropped the $\pi$ . Stokes' law is

6\pi\eta R v

, not

6\eta R v

; restoring the

\pi

gives

1.1\times10^{-4}\ \mathrm{N}

❌ Recompute. Evaluate

6\pi\eta R v

carefully, keeping the factor

6\pi

and adding the powers of ten.

Show solution

Step 1 — Pick the regime. The sphere is small ( $R = 1\ \mathrm{mm}$ ) and slow, so viscous (linear) drag governs: use Stokes' law.

Step 2 — Substitute.

F = 6\pi\eta R v = 6\pi\,(1.2)\,(1.0\times10^{-3})\,(5.0\times10^{-3})

= 6\pi\,(6.0\times10^{-6}) \approx 1.13\times10^{-4}\ \mathrm{N}

Dropping the $\pi$ would give $3.6\times10^{-5}\ \mathrm{N}$ ; dropping the $6$ would give $1.9\times10^{-5}\ \mathrm{N}$ . The correct magnitude is $\mathbf{1.1\times10^{-4}\ \mathrm{N}}$ .

Problem 4 · Sedimentation Scaling (Transfer)

Given: Two spherical molecules settle in the same fluid, and molecule $B$ has twice the radius of molecule $A$ . Using drag per unit mass $\propto 1/R^2$ , the drag per unit mass on $B$ compared with $A$ is:

✅ Correct!

\text{drag/mass} \propto 1/R^2

, so doubling

R

gives

(1/2)^2 = \tfrac{1}{4}

— the bigger molecule settles faster.

❌ Close.

1/R

is how the drag force scales; per unit mass it is

1/R^2

, so the factor is

(1/2)^2 = \tfrac{1}{4}

, not

\tfrac{1}{2}

❌ Inverted. Drag per mass falls as

1/R^2

, so the larger sphere has

\tfrac{1}{4}

as much, not

4\times

❌ Not quite. Mass grows as

R^3

but drag only as

R

, so drag per mass scales as

1/R^2

— not equal for the two sizes.

Show solution

Mass scales with volume and Stokes drag with radius:

\text{mass} \propto R^3, \qquad \text{drag} \propto R \;\Rightarrow\; \frac{\text{drag}}{\text{mass}} \propto \frac{R}{R^3} = \frac{1}{R^2}

With $R_B = 2R_A$ :

\frac{(\text{drag/mass})_B}{(\text{drag/mass})_A} = \left(\frac{R_A}{R_B}\right)^2 = \left(\frac{1}{2}\right)^2 = \frac{1}{4}

The larger molecule feels less drag per unit mass, so it accelerates more and settles faster — the principle behind centrifugal separation by molecular weight.

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