core33 min studyLesson 114 in Physics

Choose a drag law by its regime

Compare linear and quadratic resistance through relative fluid velocity, dimensional coefficients, terminal speed, and experimentally testable scaling.

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What you’ll understand

  • Distinguish fluid-relative speed from laboratory speed.
  • Derive terminal speeds for linear and quadratic drag.
  • Use Reynolds number and measured scaling to qualify a drag approximation.

A slow bead settling through oil and a fast ball traveling through air both experience fluid resistance, but they need not obey the same speed dependence. Newton's law supplies the momentum balance; a drag law supplies an additional model of the fluid interaction. Start by naming the surrounding fluid's velocity, the object's shape, and the speed range. Then ask which approximation can be defended by the physical regime and checked against data.

Drag uses velocity relative to the fluid

Let the object's velocity relative to the local undisturbed fluid be u. In an isotropic drag approximation, resistance points opposite u. Wind therefore matters even if a projectile's velocity relative to the ground is unchanged. A stationary object in flowing air can experience drag; a freely drifting object moving with uniform fluid has zero relative speed in this simplified model. Lift, buoyancy, added-mass effects, and history-dependent fluid forces are separate contributions. Omitting them is an assumption to evaluate for the apparatus, not a definition that every fluid interaction must fit the drag arrow.

Linear resistance uses a positive coefficient b with units kilograms per second. Quadratic resistance uses a positive coefficient c with units kilograms per metre, so multiplying it by speed squared produces newtons. The quadratic vector form contains speed times velocity, ensuring that it reverses direction when motion reverses. Writing a one-dimensional force as minus c times velocity squared for all signed velocities is wrong: it points in the same direction on both sides of zero. Express the vector or signed law first, and specialize to a chosen direction only after identifying the relevant branch.

Flin=−bu,Fquad=−c∣u∣u\mathbf F_{\mathrm{lin}}=-b\mathbf u,\qquad \mathbf F_{\mathrm{quad}}=-c|\mathbf u|\mathbf u
Here u is object velocity minus local fluid velocity, b > 0 has units kg/s, and c > 0 has units kg/m. Both models oppose relative motion; neither includes lift or buoyancy automatically.↔ Scroll wide equations horizontally.

A terminal speed is a force balance

Take downward positive for a dense falling object in still fluid and neglect buoyancy. Under linear drag the acceleration decreases as speed approaches the value at which drag balances weight. Solving the first-order velocity equation gives exponential relaxation with time constant mass divided by b. Terminal speed is approached asymptotically in this model; it is not a maximum allowed by Newtonian mechanics. An object released faster than terminal speed slows down toward it. Changing the initial velocity changes the transient while leaving the same equilibrium speed for fixed coefficients and constant gravity.

Quadratic drag produces a different terminal-speed scaling: doubling mass at fixed c increases terminal speed by the square root of two, whereas linear drag increases it by two at fixed b. A geometric change can also change the coefficient, so this comparison requires controlled shape and fluid conditions. If buoyancy is appreciable, replace weight by the net constant downward driving force before finding terminal speed. A nearly neutrally buoyant object cannot be modeled by simply carrying over the dense-object result. The equilibrium tests the complete force balance, including every retained constant force.

vt,lin=mgb,τ=mb,vt,quad=mgcv_{t,\mathrm{lin}}=\frac{mg}{b},\qquad \tau=\frac mb,\qquad v_{t,\mathrm{quad}}=\sqrt{\frac{mg}{c}}
These terminal downward speeds assume still fluid, constant g, constant positive coefficients, and negligible buoyancy. The time constant τ belongs to the linear velocity equation, not generally to quadratic drag.↔ Scroll wide equations horizontally.

Regime checks keep a useful fit honest

For a small sphere in creeping flow, Stokes resistance is linear with b equal to six pi times dynamic viscosity times radius. Its standard assumptions include a Newtonian fluid, no-slip surface, negligible inertial fluid effects, and boundaries sufficiently far away. Reynolds number compares fluid inertia with viscous effects using density, speed, characteristic size, and dynamic viscosity. Very small Reynolds number supports the creeping-flow limit. At larger Reynolds number a quadratic representation may be useful over a range where the drag coefficient is approximately constant, but that coefficient can vary strongly across transitions or with shape.

Measure a drag curve rather than deciding the law solely from the phrase low speed or high speed. Plot force against speed and against speed squared, inspect residuals, and check that the inferred coefficients remain consistent across repeated trials. A mixture of effects may require a combined or more detailed model. Even a good empirical curve does not automatically predict unsteady motion, rapid rotation, or flow near a wall. State the interval over which the fit was tested and propagate coefficient uncertainty into the predicted terminal speed. Modeling precision should not exceed the reliability of the chosen fluid approximation.

Two laws with different terminal-speed scaling

A 2 kg2\,\mathrm{kg} object falls in still fluid with g=10 m s−2g=10\,\mathrm{m\,s^{-2}}. Compare a linear model with b=4 kg s−1b=4\,\mathrm{kg\,s^{-1}} and a separate quadratic model with c=0.8 kg m−1c=0.8\,\mathrm{kg\,m^{-1}}. Neglect buoyancy.

  1. Weight is mg=20 Nmg=20\,\mathrm N. The linear balance bvt=20bv_t=20 gives vt=5 m s−1v_t=5\,\mathrm{m\,s^{-1}}, and its relaxation time is m/b=0.5 sm/b=0.5\,\mathrm s.

  2. The quadratic balance cvt2=20cv_t^2=20 gives vt=20/0.8=5 m s−1v_t=\sqrt{20/0.8}=5\,\mathrm{m\,s^{-1}} as well. Matching one terminal-speed measurement therefore does not distinguish these laws.

  3. At a downward speed of 2 m s−12\,\mathrm{m\,s^{-1}}, linear drag is 8 N and acceleration is (20−8)/2=6 m s−2(20-8)/2=6\,\mathrm{m\,s^{-2}}. Quadratic drag is 3.2 N and acceleration is 8.4 m s−28.4\,\mathrm{m\,s^{-2}}. A transient measurement can separate them.

The result

The two fitted laws agree at one terminal speed yet disagree away from it. Matching an equilibrium is weaker evidence than matching a resolved velocity history.

Explore the model

Interactive lab

Interactive lab · function

Inspect quadratic resistance scaling

Set coefficient a to 1 and position x to 1, then 2: y = ax² changes from 1 to 4. Interpret positive a as normalized c and nonnegative x as speed, so the curve represents a quadratic drag magnitude. Negative a or x is outside this magnitude interpretation; this lab is not a falling-body solver.

Inspect quadratic resistance scalingThis lab uses only y = ax². Playback traces the curve within the fixed integration interval from 0 to x; it does not estimate numerical error or change the chosen point. Current readings: y at chosen x: 0.608; Tangent slope · 2ax: 1.35; Signed area · ax³/3: 0.182. All parameters are adjustable with the labeled controls below.−2.4−1.201.22.4−9−4.504.59xy(x, ax²)
y = ax²TangentSigned integral
y at chosen x
0.608
Tangent slope · 2ax
1.35
Signed area · ax³/3
0.182
t = 0 s

This lab uses only y = ax². Playback traces the curve within the fixed integration interval from 0 to x; it does not estimate numerical error or change the chosen point. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forA speed-squared law makes testable scaling predictions beyond one fitted point.

Check your understanding

For a falling object whose shape and drag coefficient c stay fixed, how does the quadratic-drag terminal speed change if mass becomes four times larger?

Try it yourself

A sphere of radius 1 mm1\,\mathrm{mm} moves at 1 mm s−11\,\mathrm{mm\,s^{-1}} through a fluid with density 1000 kg m−31000\,\mathrm{kg\,m^{-3}} and dynamic viscosity 1 Pa s1\,\mathrm{Pa\,s}. Use the diameter for Reynolds number. Estimate Re and the Stokes drag magnitude.

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

Use Re = density × speed × diameter / viscosity and b = 6π times viscosity times radius.

Compare with the solution

The diameter is 0.002 m0.002\,\mathrm m, giving Re=1000(0.001)(0.002)/1=0.002\mathrm{Re}=1000(0.001)(0.002)/1=0.002, consistent with creeping flow if the other assumptions hold. Stokes drag is 6π(1)(0.001)(0.001)=6π×10−6 N≈1.88×10−5 N6\pi(1)(0.001)(0.001)=6\pi\times10^{-6}\,\mathrm N\approx1.88\times10^{-5}\,\mathrm N. A nearby container wall could still require a correction despite the small Reynolds number.

Take it with you

  • The drag direction follows fluid-relative velocity, not an arbitrary laboratory axis.
  • Terminal-speed and transient measurements test different aspects of a resistance law.
  • Reynolds number, geometry, and coefficient stability qualify the regime of validity.

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