Classical-Mechanics · Unit 10 · Video 2 · Interactive Practice

When Drag Wins: Terminal Velocity of a Marble in Olive Oil

IKey Formulas

Formula	Name	Key idea
$mg - 6\pi\eta R\,v = m\dfrac{dv}{dt}$	Equation of motion	Gravity vs. Stokes drag
$v(t) = \dfrac{g}{\gamma}\left(1 - e^{-\gamma t}\right),\quad \gamma = \dfrac{6\pi\eta R}{m}$	Velocity	Rises to a ceiling
$v_\infty = \dfrac{mg}{6\pi\eta R} = \dfrac{2\rho R^2 g}{9\eta}$	Terminal velocity	Grows as $R^2$
$y(t) = \dfrac{g}{\gamma}\,t + \dfrac{g}{\gamma^2}\left(e^{-\gamma t} - 1\right)$	Position	Integral of $v(t)$

Key Insight: Terminal velocity is the single speed at which drag exactly cancels gravity: $6\pi\eta R\,v_\infty = mg$ .

IIApproach to Terminal Velocity

Drag grows with speed until it balances gravity, and the marble settles at one constant velocity.

t 0.00 s

IIIBigger Marbles Fall Faster

Terminal speed scales with the square of the radius: $v_\infty = \dfrac{2\rho R^2 g}{9\eta}$ .

Radius R 5.0 mm

💡 The $R^2$ law holds at fixed material density $\rho$ . A denser marble of the same radius also falls faster, since $v_\infty \propto \rho$ .

IVPosition: Glide vs. Free Fall

How far does the marble actually fall, and how much does drag hold it back?

Guess y(2 s) 6.0 m

VQuiz Questions

Problem 1 · Terminal Velocity from the Formula

Given: a steel marble with $m = 4.08\times10^{-3}\ \text{kg}$ , $R = 5.00\times10^{-3}\ \text{m}$ in oil of viscosity $\eta = 8.10\times10^{-2}\ \text{kg}\cdot\text{m}^{-1}\cdot\text{s}^{-1}$ ( $g = 9.8\ \text{m/s}^2$ ) — find the terminal velocity $v_\infty$ .

✅ Correct! At terminal velocity drag balances gravity:

6\pi\eta R\,v_\infty = mg

, so

v_\infty = mg/(6\pi\eta R) = 5.24

m/s.

❌ Close, but check R. That used the diameter. Stokes' law takes the radius, so the denominator is

6\pi\eta R

, not

6\pi\eta(2R)

❌ Not quite. Stokes' drag has coefficient

6\pi\eta R

, not

3\pi\eta R

— you are off by a factor of 2.

❌ Not quite. Use

v_\infty = mg/(6\pi\eta R)

with

mg = 0.0400

N and

6\pi\eta R = 7.63\times10^{-3}

kg/s.

Show solution

Set drag equal to gravity, or take $t\to\infty$ in $v(t)$ ; either way:

v_\infty = \frac{mg}{6\pi\eta R} = \frac{(4.08\times10^{-3})(9.8)}{6\pi(8.10\times10^{-2})(5.00\times10^{-3})} = \frac{0.0400}{7.63\times10^{-3}} = 5.24\ \text{m/s}

Problem 2 · Double the Radius

Given: a second marble of the same material but twice the radius — by what factor does its terminal velocity change?

✅ Correct! With density fixed,

v_\infty = \dfrac{2\rho R^2 g}{9\eta} \propto R^2

, so doubling

R

multiplies the speed by

2^2 = 4

❌ That's linear scaling. The radius appears squared:

v_\infty \propto R^2

, so the factor is

2^2 = 4

, not

2

❌ Not quite. Mass grows as

R^3

, but

v_\infty = mg/(6\pi\eta R)

divides by another

R

, leaving

v_\infty \propto R^2

❌ Not quite. Terminal speed depends on radius as

R^2

— bigger marbles fall faster.

Show solution

Write the mass through the density: $m = \tfrac{4}{3}\pi\rho R^3$ . Substituting into $v_\infty = mg/(6\pi\eta R)$ cancels $\pi$ and one power of $R$ :

v_\infty = \frac{2\rho R^2 g}{9\eta} \propto R^2

Doubling $R$ therefore scales $v_\infty$ by $2^2 = \mathbf{4}$ .

Problem 3 · Measure the Viscosity

Given: the marble from Problem 1 is timed at a steady $v_\infty = 5.24\ \text{m/s}$ — find the oil's viscosity $\eta$ .

✅ Correct! Rearranging

v_\infty = mg/(6\pi\eta R)

gives

\eta = mg/(6\pi R\,v_\infty) = 8.1\times10^{-2}

kg·m⁻¹·s⁻¹.

❌ Close, but check R. That used the diameter

2R

. Use the radius:

\eta = mg/(6\pi R\,v_\infty)

❌ Not quite. The coefficient is

6\pi

, not

3\pi

— halving it doubles your

\eta

❌ Not quite. Solve

v_\infty = mg/(6\pi\eta R)

for

\eta = mg/(6\pi R\,v_\infty)

Show solution

The terminal-velocity equation inverts to give the viscosity directly:

\eta = \frac{mg}{6\pi R\,v_\infty} = \frac{0.0400}{6\pi(5.00\times10^{-3})(5.24)} = \frac{0.0400}{0.494} = 8.1\times10^{-2}

in units of kg·m⁻¹·s⁻¹ — a falling marble becomes a viscometer.

Problem 4 · Distance While Gliding

Given: the marble is already gliding at terminal velocity $v_\infty = 5.24\ \text{m/s}$ — find how far it falls during the next $2.0\ \text{s}$ .

✅ Correct! At terminal velocity the acceleration is zero, so the marble moves at constant speed:

d = v_\infty t = 5.24 \times 2.0 = 10.5

❌ That's free fall.

\tfrac12 g t^2

assumes acceleration

g

, but at terminal velocity

a = 0

; the marble glides at constant

v_\infty

, so

d = v_\infty t

❌ Not quite. Constant-velocity distance is

v_\infty t

, not

v_\infty t^2

— there is no extra power of

t

❌ Not quite. Constant speed for time

t

covers

d = v_\infty t = 5.24 \times 2.0

Show solution

Once drag cancels gravity the net force is zero, so $a = 0$ and the motion is constant-velocity:

d = v_\infty\,t = (5.24)(2.0) = 10.5\ \text{m}

Free fall over the same $2.0$ s would cover $\tfrac12 g t^2 = 19.6$ m — drag more than halves the distance.

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