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Mechanics
Stokes' Law and Terminal Velocity
Stokes' law describes the viscous drag force experienced by a small spherical body moving through a viscous fluid at low Reynolds numbers. It forms the theoretical basis for determining the coefficient of viscosity and calculating the terminal velocity of falling droplets.
UP TGT ScienceUP PGT PhysicsStudy Notes + Practice
Scope and prerequisites
TGT core
Definition of Stokes' force, dimensional formula of viscosity, dependence of viscous force on radius, velocity, and coefficient of viscosity, and simple terminal velocity problems.
PGT extension
Detailed dimensional derivation of Stokes' law, analysis of Reynolds number limitations, motion of bodies in rotating or accelerated frames involving viscous drag, and transient approach to terminal velocity.
Prerequisites
Coefficient of viscosity and Newton's law of viscosity
Dimensional analysis and Buckingham Pi theorem
Buoyancy and Archimedes' principle
Viscous Drag Force on Spheres
When a solid sphere of radius r moves with a uniform velocity v through an infinite, incompressible fluid of viscosity eta, it experiences a retarding force known as viscous drag.
The magnitude of this force is directly proportional to the radius of the sphere, its velocity, and the dynamic viscosity of the medium.
The direction of the Stokes' force is always opposite to the instantaneous relative velocity vector between the sphere and the fluid.
Terminal Velocity
As a spherical body falls freely under gravity through a viscous fluid, its downward acceleration decreases due to the simultaneous increase in upward viscous drag and buoyant force.
When the magnitude of the gravitational force equals the sum of the upward buoyant force and the Stokes' drag force, the net force becomes zero.
At this stage, the body continues to fall with a constant maximum velocity known as the terminal velocity (v_t).
Dependence and Scaling Factors
Terminal velocity varies directly as the square of the radius of the spherical body (v_t proportional to r2).
It is directly proportional to the density difference between the sphere (rho) and the fluid (sigma), and inversely proportional to the coefficient of viscosity (eta).
If the radius is doubled, the terminal velocity increases by a factor of four, assuming laminar flow conditions are maintained.
Reynolds Number Validity Criterion
Stokes' law is strictly valid only when the flow pattern around the sphere is purely laminar and creeping, characterized by a very low Reynolds number (Re less than 0.1).
At higher Reynolds numbers, inertial forces dominate over viscous forces, leading to the formation of eddies and wake separation, causing deviations from Stokes' law.
Corrections (such as the Oseen correction) are applied when the Reynolds number approaches unity.
Derivations
Dimensional Derivation of Stokes' Law
Conditions: Assuming the viscous force F depends only on sphere radius r, velocity v, and fluid viscosity eta.
Express the dependence in functional form: $F = k * (r^{a}) * (v^{b}) * (\eta^{c})$, where k is a dimensionless constant.
Substitute the dimensional formulas for each quantity: [F] = MLT-2, [r] = L, [v] = LT-1, and [eta] = ML-1T-1.
Equate the powers of fundamental dimensions M, L, and T on both sides: For M: c = 1; For T: -b - $c = -2$, yielding b = 1; For L: a + b - c = 1, yielding a = 1.
Substitute the exponents back into the initial expression to obtain $F = k * \eta * r * v$.
Result: $F = 6 * \pi * \eta * r * v (where hydrodynamic integration yields k = 6 * \pi)$
Derivation of Terminal Velocity
Conditions: A solid sphere falling freely under gravity in a viscous fluid under laminar flow conditions.
Identify the forces acting on the falling sphere of volume $V = (4/3)*\pi*r^{3}$ and density rho: Downward gravitational force $W = V * \rho * g$.
Identify upward buoyant force $B = V * \sigma * g$, where sigma is fluid density.
Identify upward viscous drag force $F_s = 6 * \pi * \eta * r * v_t$.
Apply equilibrium condition at terminal velocity: $W = B + F_s$, which gives V * (rho - sigma) * $g = 6 * \pi * \eta * r * v_t$.
Substitute $V = (4/3)*\pi*r^{3}$ and solve for v_t.
F = Viscous drag force, eta = Coefficient of viscosity, r = Radius of sphere, v = Velocity F in N, eta in Pa·s or kg/(m·s), r in m, v in m/s
Laminar flow, spherical body, infinite medium, Re < 0.1
Terminal Velocity Formula
$$v_t = \frac{2r^2(\rho - \sigma)g}{9\eta}$$
$v_t = Terminal velocity$, r = Radius, rho = Density of sphere, sigma = Density of fluid, g = Acceleration due to gravity, eta = Viscosity v_t in m/s, densities in kg/m3, g in m/s2
Object released from rest, constant fluid density and viscosity
Reynolds Number for Sphere
$$Re = \frac{\rho v d}{\eta} = \frac{2 \rho v r}{\eta}$$
Re = Reynolds number, rho = Fluid density, v = Velocity, d = Diameter, eta = Viscosity Dimensionless
Used in experimental determination of viscosity Pa·s
Measured at constant terminal velocity
Drag Acceleration
$$a = g \left(1 - \frac{\sigma}{\rho}\right) - \frac{6 \pi \eta r v}{m}$$
Equation of motion for falling sphere m/s2
General vertical motion
Graphs and experiments
Velocity vs. Time Graph for a Falling Sphere
Draw: A curve starting from origin (v = 0 at t = 0) showing an exponentially decreasing slope, asymptotically approaching a horizontal line representing terminal velocity v_t.
Interpret: Illustrates that acceleration decreases continuously as viscous drag increases with velocity, eventually reaching zero when $v = v_t$.
Exam use: Frequently asked in conceptual multiple-choice questions regarding the shape of the velocity-time curve in viscous media.
Draw: A tall glass jar filled with a transparent viscous liquid (like castor oil), marked with two reference rings separated by a fixed distance, with a small steel ball dropped inside.
Interpret: By measuring the time taken to fall between the reference marks, terminal velocity is computed, allowing the calculation of the liquid's viscosity.
Exam use: Tested via questions on experimental precautions, such as ensuring the jar is wide enough to avoid wall effects.
Logarithmic Plot of Terminal Velocity vs. Radius Squared
Draw: A straight line passing through the origin on a v_t versus r2 plot.
Interpret: Confirms the linear relationship between terminal velocity and the square of the radius, validating Stokes' law experimentally.
Exam use: Used in PGT-level data interpretation and slope-based numerical problems.
Solved numericals
TGT core
A small metal ball of radius 2 mm and density 8000 kg/m3 falls through a liquid of density 1000 kg/m3 and viscosity 0.8 Pa·s. Calculate its terminal velocity. (Take $g = 10 m/s^{2})$
Identify given values: $r = 2 mm = 2 * 10^{-3} m$, rh$o = 8000 kg/m^{3}$, sigm$a = 1000 kg/m^{3}$, eta = 0.8 Pa·s, $g = 10 m/s^{2}$.
Two small identical spherical raindrops reach a terminal velocity of 2 m/s when falling separately through air. If coalesced into a single larger spherical drop, find the new terminal velocity.
Let the radius of each small drop be r and terminal velocity be $v_1 = 2 m/s$.
Volume conservation upon coalescence: \frac{4}{3}\pi $R^3 = 2 * \frac{4}{3}\pi r^3$, which gives $R = 2^{1/3} * r$.
Terminal velocity is proportional to the square of the radius: v_t \propto r^2.
Find ratio of new terminal velocity (v_2) to old (v_1): \frac{v_2}{$v_1 = \left(\frac{R}{r}\right)^2 = (2^{1/3})^2 = 2^{2/3}$.
A spherical ball of mass m and radius r is projected vertically upwards with initial velocity v_0 in a viscous medium obeying Stokes' law. Find the time taken to reach the highest point.
Write down the equation of motion during upward motion: m*(dv/dt) = -mg - 6*pi*eta*r*v.
Rewrite as dv/d$t = -g - (6*\pi*\eta*r / m)*v$. Let $k = 6*\pi*\eta*r / m$. Then dv/d$t = -(g + k*v)$.
Separate variables and integrate from $t = 0 (v = v_0)$ to $t = T (v = 0): \int_{v_0}^{0} \frac{dv}{g + kv} = - \int_{0}^{T} dt$.
Evaluate integral: \frac{1}{k} \ln\left(\frac{g + kv_0}{g}\right) = T.
Substitute $k = 6*\pi*\eta*r / m$ to get final expression for time T.
Answer: $T = \frac{m}{6\pi\eta r} \ln\left(1 + \frac{6\pi\eta r v_0}{mg}\right)$
Assumptions and limitations
The fluid must be unbounded (infinite in extent) so that wall effects do not alter the hydrodynamic drag.
The falling sphere must be perfectly rigid, smooth, and free from internal circulation or surface contamination.
The flow velocity must be sufficiently low to maintain laminar conditions without turbulence or wake separation (Re < 0.1).
Common misconceptions
Common error: Terminal velocity depends on the total distance the body falls through the fluid.
Correct: Terminal velocity is independent of distance fallen once steady-state is achieved.
It is determined purely by the balance between gravitational, buoyant, and viscous forces.
Common error: Stokes' law applies to objects of any shape moving through a viscous fluid.
Correct: Stokes' law is strictly formulated and mathematically derived only for spheres.
Common error: Viscous drag force increases continuously as the object falls indefinitely.
Correct: Viscous drag force increases only until velocity reaches terminal velocity, after which it remains constant.
Once velocity becomes constant, the drag force stops increasing and remains equal to the effective weight.
Quick revision
Stokes' viscous drag is given by $F = 6*\pi*\eta*r*v$.
Terminal velocity varies directly as the square of the radius (v_t proportional to r2).
Stokes' law is valid only for creeping flow with Reynolds number less than 0.1.
The time constant for reaching terminal velocity is ta$u = (2*\rho*r^{2}) / (9*\eta)$.
Must remember
Stokes' Law states that the viscous drag force on a spherical body is given by $F = 6 \pi \eta r v$.
Terminal velocity $v_t$ is directly proportional to the square of the radius ($r^2$) of the spherical body and the density difference $(\rho - \sigma)$.
Stokes' law is strictly valid only for purely laminar and creeping flow where the Reynolds number $Re < 0.1$.
The viscous drag force always acts in a direction opposite to the instantaneous relative velocity vector between the sphere and the fluid.
Terminal velocity is independent of the distance fallen and is achieved when the net force on the falling body becomes zero.
Glossary
Dynamic Viscosity ($\eta$)A fluid's internal resistance to flow, measured in Pa·s or kg/(m·s).
Terminal Velocity ($v_t$)The constant maximum velocity reached by a body falling through a viscous medium when gravitational force balances buoyant and viscous forces.
Reynolds Number ($Re$)A dimensionless number used to predict flow patterns in different fluid flow situations ($Re = \frac{2 \rho v r}{\eta}$). Heightened values signify turbulent flow.
Viscous Drag Force ($F$)The retarding force experienced by an object moving through a fluid medium due to shearing stresses.
Stokes' Flow vs Turbulent Flow
Feature
Stokes' Flow (Creeping Flow)
Turbulent / High Re Flow
Reynolds Number
$Re < 0.1$
$Re > 1000$ (or moderate to high)
Drag Force Dependence
Directly proportional to velocity ($F \propto v$)
Proportional to square of velocity ($F \propto v^2$)
Flow Characteristics
Laminar, smooth streamlines, symmetric
Eddies, wake separation, irregular
Practice Quiz
Choose an option to reveal the explanation.
1. What is the dimensional formula for the coefficient of dynamic viscosity $\eta$ derived from Stokes' law $F = 6 \pi \eta r v$?
Explanation: Rearranging Stokes' law for viscosity gives $\eta = \frac{F}{6 \pi r v}$. Substituting dimensions: $[M L T^{-2}] / ([L][L T^{-1}]) = [M L^{-1} T^{-1}].$
2. If the radius of a spherical rain droplet is doubled, how does its terminal velocity change under laminar conditions?
Explanation: Terminal velocity is proportional to the square of the radius ($v_t \propto r^2$). Doubling the radius increases the terminal velocity by $2^2 = 4$ times.
3. Under what condition is Stokes' law strictly valid for a sphere moving through a fluid?
Explanation: Stokes' law is derived assuming purely creeping and laminar flow, which strictly requires a very low Reynolds number, specifically $Re < 0.1$.
4. The viscous drag force on a sphere of radius $r$ moving with velocity $v$ is proportional to:
Explanation: According to Stokes' law, $F = 6 \pi \eta r v$, which shows direct proportionality to both radius $r$ and velocity $v$ ($F \propto r v$).
5. What direction does the Stokes' viscous drag force act relative to the motion of a sphere?
Explanation: Viscous drag is a retarding force and its vector $\vec{F} = -6 \pi \eta r \vec{v}$ always points opposite to the instantaneous relative velocity.
6. What happens to the acceleration of a small steel ball dropped from rest into a tall jar of glycerine as time passes?
Explanation: As the ball falls, speed increases, increasing the upward viscous drag. Net downward force and hence acceleration decrease continuously until it becomes zero at terminal velocity.
7. Terminal velocity $v_t$ of a sphere in a viscous fluid depends on the density of the sphere $\rho$ and fluid $\sigma$ as:
Explanation: The terminal velocity formula is $v_t = \frac{2r^2(\rho - \sigma)g}{9\eta}$, showing direct dependence on the density difference $(\rho - \sigma)$.
8. Which of the following corrections is applied when the Reynolds number approaches unity in fluid flow around a sphere?
Explanation: The Oseen correction modifies Stokes' equation to account for inertial terms when the Reynolds number approaches unity ($Re \approx 1$).
9. If a sphere falls through a fluid and reaches terminal velocity, what is the net force acting on the sphere?
Explanation: At terminal velocity, the downward gravitational force equals the sum of upward buoyant and viscous forces, resulting in a net force of zero.
10. What is the SI unit of dynamic viscosity $\eta$?
Explanation: Dynamic viscosity $\eta$ has SI units of Pascal-seconds $(\text{Pa}\cdot\text{s})$ or $\text{kg}/(\text{m}\cdot\text{s})$, which is equivalent to $\text{N}\cdot\text{s}/\text{m}^2$.
11. How does the terminal velocity of an air bubble rising in water compare to that of a solid metal ball of the same size falling in water?
Explanation: For a rising air bubble, $\rho_{\text{air}} < \sigma_{\text{water}}$, so the effective density term $(\rho - \sigma)$ is negative, dictating upward motion.
12. The relaxation time (time constant $\tau$) for a spherical body to approach terminal velocity depends on mass $m$ and Stokes' drag coefficient $b = 6\pi\eta r$ as:
Explanation: The transient motion analysis gives the time constant as $\tau = \frac{m}{6\pi\eta r} = \frac{m}{b}$.
13. Which of the following factors does NOT affect the terminal velocity of a spherical body in an infinite fluid?
Explanation: Terminal velocity is a steady-state velocity and is completely independent of the total distance fallen once steady-state is achieved.
14. If the coefficient of viscosity of a medium is halved, what will happen to the terminal velocity of a spherical body falling through it?
Explanation: Since terminal velocity is inversely proportional to viscosity ($v_t \propto 1/\eta$), halving $\eta$ will double the terminal velocity.
15. What is the expression for the effective downward force (effective weight) driving a sphere of radius $r$, density $\rho$, in a fluid of density $\sigma$?
Explanation: Effective weight $W_{\text{eff}}$ is real weight minus buoyant force: $W_{\text{eff}} = \frac{4}{3}\pi r^3 \rho g - \frac{4}{3}\pi r^3 \sigma g = \frac{4}{3}\pi r^3 (\rho - \sigma) g$.
PYQ Patterns
These are pattern-based practice questions. Authentic year labels are only used when verified from a source.
1. Two small spherical drops of water of equal radius $r$ coalesce to form a single spherical drop of radius $R$. What is the ratio of the terminal velocity of the new drop to that of each original drop?
Explanation: Volume is conserved: $\frac{4}{3}\pi R^3 = 2 \times \frac{4}{3}\pi r^3 \implies R = 2^{1/3} r$. Terminal velocity $v_t \propto r^2$. Thus, $v_{t,R} / v_{t,r} = R^2 / r^2 = (2^{1/3})^2 = 2^{2/3}$ ... Wait, volume conservation: $2 \times (4/3)\pi r^3 = (4/3)\pi R^3 \implies R^3 = 2r^3 \implies R = r \cdot 2^{1/3}$. Terminal velocity $v_t \propto r^2$, so $v_{t,R}/v_{t,r} = (R/r)^2 = (2^{1/3})^2 = 2^{2/3}$. Let's re-verify standard options. If volume doubles, radius becomes $2^{1/3}r$, so velocity ratio is $2^{2/3}$. Let's adjust option B to $2^{2/3}:1$ for mathematical precision.
2. A small metal ball of radius $r$ and density $\rho$ falls through a fluid of density $\sigma$ and viscosity $\eta$. If the Reynolds number for this motion is $0.05$, the drag force is best calculated using:
Explanation: Since the Reynolds number $0.05 < 0.1$, the flow is strictly laminar/creeping, and Stokes' law is applicable.
3. A sphere of radius $r$ is dropped in a viscous fluid. The time taken to reach $50\%$ of its terminal velocity depends on the relaxation time $\tau$. What is the expression for $\tau$?
Explanation: The relaxation time constant is given by $\tau = \frac{m}{6\pi\eta r} = \frac{\frac{4}{3}\pi r^3 \rho}{6\pi\eta r} = \frac{2\rho r^2}{9\eta}$.
4. In an experiment to determine the viscosity of an oil using Stokes' law, a steel ball of radius $1.5\text{ mm}$ and density $7.8 \times 10^3 \text{ kg/m}^3$ is dropped. Which major experimental precaution must be ensured?
Explanation: Stokes' law assumes an infinite medium; hence, the container must be wide enough to neglect wall effects on the falling sphere.
5. Which of the following graphs best represents the variation of velocity $v$ of a spherical body falling from rest in a viscous fluid with time $t$?
Explanation: Velocity starts at zero and increases exponentially, approaching terminal velocity asymptotically according to $v(t) = v_t(1 - e^{-t/\tau})$.
6. If the terminal velocity of a spherical body inside a given liquid is $v_t$, what will be the terminal velocity if the radius of the body is halved and all other parameters remain constant?
Explanation: Terminal velocity varies directly as the square of the radius ($v_t \propto r^2$). Halving the radius ($r/2$) results in $(1/2)^2 = 1/4$ of the original terminal velocity.
Timed Topic Test
Attempt these questions under a ten-minute limit using a separate timer.
1. Stokes' law fails when the Reynolds number exceeds $0.1$ primarily because:
Explanation: At higher Reynolds numbers, inertial forces dominate over viscous forces, leading to turbulence, eddies, and deviations from Stokes' law.
2. A spherical ball of mass $m$ and radius $r$ is falling in a viscous fluid. When its velocity is $v$, the drag acceleration is given by:
Explanation: The equation of motion accounts for gravity, buoyancy, and viscous drag: $a = g(1 - \sigma/\rho) - \frac{6\pi\eta r v}{m}$.
3. Calculate the terminal velocity of a spherical droplet of radius $r = 1\text{ mm}$, density $\rho = 10^3 \text{ kg/m}^3$ in air where density $\sigma$ is negligible and viscosity $\eta = 1.8 \times 10^{-5} \text{ Pa}\cdot\text{s}$ (taking $g = 9.8 \text{ m/s}^2$). Which order of magnitude is correct?
Explanation: Using $v_t = \frac{2 r^2 \rho g}{9 \eta} = \frac{2 \times (10^{-3})^2 \times 10^3 \times 9.8}{9 \times 1.8 \times 10^{-5}} = \frac{2 \times 10^{-3} \times 9.8}{1.62 \times 10^{-4}} \approx 120 \text{ m/s}$ (Note: exceeds Stokes' limit $Re < 0.1$, but numerical order is around $10^2 \text{ m/s}$).
4. Why does a parachute allow a person to land safely by limiting terminal velocity?
Explanation: A parachute drastically increases the drag force (effective radius/area), which lowers the terminal velocity to a safe value.
5. If the vector form of Stokes' drag force is $\vec{F} = -6 \pi \eta r \vec{v}$, what does the negative sign signify?
Explanation: The negative sign indicates that the viscous drag force always opposes the direction of the relative velocity vector between the sphere and fluid.
6. In terms of dimensions, which of the following quantities has the same dimensions as dynamic viscosity $\eta$ multiplied by radius $r$?
Explanation: From Stokes' law $F = 6 \pi \eta r v$, the combination $\eta r$ has dimensions of $F / v$ (since $F \sim \eta r v$). Surface tension has dimensions of force per unit length ($F/L$). Thus $\eta r$ has dimensions of $F/v$.
7. Consider a tiny oil drop in Millikan's oil drop experiment. When the electric field is switched off, the drop falls with terminal velocity $v_1$. If the droplet evaporates slightly such that its radius becomes half, what happens to its terminal velocity falling under gravity?
Explanation: Terminal velocity is proportional to the square of the radius ($v_t \propto r^2$). When radius becomes $r/2$, the new terminal velocity is $(1/2)^2 v_1 = v_1 / 4$.
8. The Reynolds number $Re$ for a sphere of diameter $d$ moving with velocity $v$ in a fluid of density $\rho$ and viscosity $\eta$ is expressed as:
Explanation: By definition, Reynolds number for a sphere is $Re = \frac{\rho v d}{\eta} = \frac{2 \rho v r}{\eta}$.
9. Which of the following statements about terminal velocity is FALSE?
Explanation: Terminal velocity is independent of the height or distance fallen; statement C is false.
10. What is the physical significance of the ratio $\frac{2 \rho r^2}{9 \eta}$ in the context of a falling sphere?
Explanation: The expression $\frac{2 \rho r^2}{9 \eta}$ is the relaxation time $\tau$ governing the exponential transition of velocity to its terminal value.