UY1: Resistive Forces

Resistive forces (drag) matter when an object moves through a fluid (air/water) and the fluid exerts a force opposite the motion. This lesson focuses on the standard 1D vertical-fall models and what they predict for v(t) and terminal speed.

At a glance

• Prerequisites: Newton’s 2nd law, vectors (Mathematics for Undergraduate Physics), First-order differential equations (optional refresher)
• You will learn: how to model drag as f ∝ v or f ∝ v²; how terminal speed emerges; what the time constant τ means; how to sanity-check v(t)
• Key results (downward positive): vₜ = mg/k (linear drag), vₜ = square root of (mg/b) (quadratic drag)
• Common trap: forgetting drag always opposes the velocity (sign errors), or swapping the meaning of C_D (dimensionless) with a lumped constant in bv²

---

Setup (model + sign convention)

We model the object as a particle falling vertically through a fluid.

Choose the + y axis downward, so:

• velocity v(t) ≥ 0 means “moving downward”,
• weight is + mg,
• drag is negative (upward) when the object moves downward.

Neglect buoyancy unless stated (we revisit it briefly at the end).

With signed v (downward positive), it is often cleanest to write drag laws in a way that automatically handles the sign:

• Linear drag: F_drag = -kv
• Quadratic drag: F_drag = -b v|v|

(If you know the motion stays downward, v ≥ 0 so v|v| = v² and the quadratic law becomes F_drag = -bv².)

---

1) Linear drag: F_drag = -kv

This model is often used for small objects at low speeds (or as a first approximation).

Governing equation

With signed v (downward positive), Newton’s 2nd law gives: mdv/dt = mg-kv.

Terminal speed

Terminal speed means the acceleration is zero: dv/dt = 0. mg-kvₜ = 0 ⇒ vₜ = mg/k.

Units check: since kv is a force, [k] = N/(m s⁻¹) = kg s⁻¹.

Solution for v(t) and the time constant

The ODE dv/dt = g-(k/m)v has a standard “relaxation” form. Since k/m has units of s⁻¹, define the time constant τ = m/k ⇒ k/m = 1/τ. Then the ODE becomes dv/dt = g-(1/τ)v.

Now use the terminal-speed relation vₜ = mg/k = gτ to rewrite it as: dv/dt = (vₜ-v)/τ. This makes the physics clear: the acceleration is proportional to the “gap” to terminal speed.

The general solution (with v(0) = v₀) is: v(t) = vₜ + (v₀-vₜ)e^(-t/τ), vₜ = mg/k, τ = m/k.

From rest ( v₀ = 0): v(t) = vₜ(1-e^(-t/τ)).

This is an exponential approach to terminal speed:

• At t = 0, v = 0 and the initial slope is .dv/dt|ₜ₌₀ = g (drag is initially zero).
• The “gap to terminal speed” shrinks exponentially: vₜ-v(t) = vₜ e^(-t/τ).

Interpretation: τ sets the approach rate to terminal speed. For a release from rest: v(τ) = vₜ(1-e⁻¹) ≈ 0.63 vₜ.

More generally, from rest the time to reach a fraction p of terminal speed is: v = pvₜ ⇒ t = -τ ln(1-p). (e.g. p = 0.95 gives t ≈ 3.00 τ.)

Rule of thumb (from rest):

• t = τ: v ≈ 0.63 vₜ
• t = 2τ: v ≈ 0.86 vₜ
• t = 3τ: v ≈ 0.95 vₜ
• t = 5τ: v ≈ 0.99 vₜ

How parameters affect the approach:

• Larger m increases both vₜ and τ (heavier objects fall faster, but “relax” more slowly in time).
• Larger k decreases both vₜ and τ (stronger drag lowers the terminal speed and is reached sooner).

Checks (non-optional)

• t → 0: e^(-t/τ) ≈ 1-t/τ so v ≈ vₜ(t/τ) = gt (free fall initially).
• t → ∞: e^(-t/τ) → 0 so v → vₜ (terminal speed).

Scroll diagram horizontally to read all labels.

---

2) Quadratic drag: F_drag = -b v|v|

This model is typical for larger objects moving faster through air (e.g. a skydiver).

Model and parameter meanings

For signed v (downward positive), use: mdv/dt = mg-b v|v|.

If the motion stays downward ( v ≥ 0), this reduces to: mdv/dt = mg-bv².

A common microscopic form is: F_drag = (1/2)ρ C_D A v², so in this notation: b = (1/2)ρ C_D A, where ρ is fluid density, A is cross-sectional area, and C_D is the (dimensionless) drag coefficient.

Units check: since bv² is a force, [b] = N/(m²s⁻²) = kg m⁻¹.

Terminal speed

Set dv/dt = 0: mg-bvₜ² = 0 ⇒ vₜ = square root of (mg/b) = square root of (2mg/(ρ C_D A)) .

Solution for v(t) (from rest)

With v(0) = 0: v(t) = vₜ tanh(gt/vₜ).

Intuition: the net downward force is mg-bv², so as v grows the resistive term grows rapidly, reducing the acceleration until it tends to zero at v = vₜ.

If you differentiate the solution you can see the acceleration decay explicitly: a(t) = dv/dt = g sech² (gt/vₜ).

The dimensionless time variable here is gt/vₜ, so a natural time scale is τ_q = vₜ/g. At t = τ_q: v(τ_q) = vₜ tanh(1) ≈ 0.76 vₜ.

More generally, from rest: v = pvₜ ⇒ p = tanh(gt/vₜ) ⇒ t = (vₜ/g)artanh(p).

Checks

• t → 0: tanh(x) ≈ x so v ≈ gt (free fall initially).
• t → ∞: tanh(x) → 1 so v → vₜ.
• Scaling: if you double A (same m), then vₜ decreases by 1/square root of 2.

---

Worked examples

Example: linear drag time constant

Problem

• Given: m = 0.020 kg, k = 0.080 kg s⁻¹, g = 9.80 m s⁻², released from rest
• Find: τ, vₜ, and v(0.50 s)

Approach

• Use the linear-drag model v(t) = vₜ(1-e^(-t/τ)) with τ = m/k and vₜ = mg/k.

Working

1. Time constant: τ = m/k = 0.020/0.080 = 0.25 s.
2. Terminal speed: vₜ = mg/k = 0.020(9.80)/0.080 = 2.45 m s⁻¹.
3. Speed at t = 0.50 s: v(0.50) = vₜ(1-e^(-t/τ)) = 2.45(1-e^(-0.50/0.25)) = 2.12 m s⁻¹.

Answer + checks

• τ = 0.25 s, vₜ = 2.45 m s⁻¹, v(0.50 s) = 2.12 m s⁻¹
• v(0.50) < vₜ and increases toward vₜ as expected

Example: quadratic-drag terminal speed estimate

Problem

• Given: m = 80 kg, ρ = 1.2 kg m⁻³, C_D = 1.0, A = 0.70 m², g = 9.80 m s⁻²
• Find: terminal speed vₜ

Approach

• Use vₜ = square root of (2mg/(ρ C_D A)) from the quadratic-drag model.

Working vₜ = square root of (2(80)(9.80)/(1.2)(1.0)(0.70)) = square root of 1867 = 43.2 m s⁻¹.

Answer + checks

• vₜ ≈ 43 m s⁻¹ ( ≈ 155 km h⁻¹)
• Units: inside the square root is m²s⁻² so vₜ is m s⁻¹

---

Practice set (with hints + answers)

1. Linear drag: show that v(t) = vₜ(1-e^(-t/τ)) implies a(t) = g e^(-t/τ).
2. Linear drag: an object reaches 0.90 vₜ after time t. Express t in terms of τ.
3. Linear drag: starting from rest, derive the displacement y(t) = y₀ + vₜ[t-τ(1-e^(-t/τ))] and show the small-t limit is y-y₀ ≈ (1/2)gt².
4. Quadratic drag: verify that v(t) = vₜ tanh(gt/vₜ) satisfies m dv/dt = mg-bv².
5. Quadratic drag: show that from rest y(t) = y₀ + vₜ²/g ln cosh(gt/vₜ).
6. Sign matters: explain why the 1D quadratic drag law should be F_drag = -b v|v| rather than -bv² if the object can move upward ( v < 0).
7. Buoyancy: a falling object experiences buoyancy B upward (constant). Write the modified terminal speed for (i) linear drag, (ii) quadratic drag.
8. Data/estimation (linear drag): suppose a best-fit curve gives vₜ = 3.00 m s⁻¹ and τ = 0.306 s. Use the linear-drag model to estimate g and comment.

Hints

2. Use 0.90 = 1-e^(-t/τ).
3. Differentiate tanh and use sech² = 1- tanh².
4. Use ∫ tanh(ax) dx = (1/a) ln cosh(ax).
5. Use “drag opposes motion”: if v < 0 then F_drag > 0 in the downward-positive convention.
6. Replace mg by the net constant downward force mg-B (assume mg > B).
7. Use vₜ/τ = g for linear drag.

Answers

1. a = dv/dt = (vₜ/τ)e^(-t/τ) = ge^(-t/τ) since vₜ/τ = (mg/k)/(m/k) = g.
2. e^(-t/τ) = 0.10 ⇒ t = τ ln 10 ≈ 2.30 τ.
3. y(t) = y₀ + ∫₀^t vₜ(1-e^(-t'/τ)) dt' = y₀ + vₜ[t-τ(1-e^(-t/τ))]. Small t: e^(-t/τ) ≈ 1-t/τ + (1/2)(t/τ)² gives y-y₀ ≈ (1/2)(vₜ/τ)t² = (1/2)gt².
4. dv/dt = g sech² (gt/vₜ), 1- tanh² = sech² so mg-bv² = mg(1-v²/vₜ²) = mg sech²(gt/vₜ) = mdv/dt.
5. y(t) = y₀ + ∫₀^t vₜ tanh(gt'/vₜ) dt' = y₀ + vₜ²/g ln cosh(gt/vₜ).
6. If v < 0 (object moving upward), drag must act downward, i.e. F_drag > 0 in a downward-positive convention. But -bv² is always negative, so it cannot represent drag for upward motion. Using -b v|v| fixes the sign automatically.
7. Linear: vₜ = (mg-B)/k. Quadratic: vₜ = square root of ((mg-B)/b) = square root of ((2(mg-B))/(ρ C_D A)) .
8. g = vₜ/τ = 3.00/0.306 = 9.80 m s⁻² (consistent with near-Earth gravity).

---

Quiz (MCQ)

Quiz — Resistive Forces (UY1)

Notes and extensions (optional)

Buoyancy (constant upthrust)

If buoyancy B is not negligible, include it as an upward force. With downward-positive convention, the constant-force part becomes mg-B. For example, linear drag becomes: mdv/dt = (mg-B)-kv, so the terminal speed is vₜ = (mg-B)/k.

For quadratic drag with downward-only motion: mdv/dt = (mg-B)-bv², vₜ = square root of ((mg-B)/b) .

Summary + next steps

• Drag always opposes velocity; keep signs explicit to avoid errors.
• Linear drag gives exponential approach to vₜ = mg/k with time constant τ = m/k.
• Quadratic drag gives hyperbolic-tangent approach to vₜ = square root of (mg/b).
• Terminal speed is where net force (and acceleration) becomes zero.

Next: Uniform Circular Motion & Non-uniform Circular Motion Back To Mechanics (UY1)