Introduction
Objects moving through fluids (air, water) experience drag forces opposing motion. Linear (Stokes) and quadratic drag lead to different behaviors and terminal velocities. Understanding resistive forces is essential for modeling realistic motion, from falling raindrops to aircraft flight.
Types of Drag
Drag forces depend on velocity: (1) Viscous/Stokes drag (linear): F_d = -bv, dominant for small, slow objects in viscous fluids, laminar flow; (2) Inertial/Newton drag (quadratic): F_d = -cv2, dominant for larger, faster objects, turbulent flow. The drag coefficient depends on object shape and flow regime. Reynolds number Re = ρvL/μ determines which regime applies (Re < 1: Stokes; Re > 1000: Newtonian).
\nIntroduction
Objects moving through fluids (air, water) experience drag forces opposing motion. Linear (Stokes) and quadratic drag lead to different behaviors and terminal velocities. Understanding resistive forces is essential for modeling realistic motion, from falling raindrops to aircraft flight.
Types of Drag
Drag forces depend on velocity: (1) Viscous/Stokes drag (linear): F_d = -bv, dominant for small, slow objects in viscous fluids, laminar flow; (2) Inertial/Newton drag (quadratic): F_d = -cv2, dominant for larger, faster objects, turbulent flow. The drag coefficient depends on object shape and flow regime. Reynolds number Re = ρvL/μ determines which regime applies (Re < 1: Stokes; Re > 1000: Newtonian).
\nLinear Drag (Stokes' Law)
For a sphere of radius r moving slowly through fluid of viscosity η, Stokes' law gives: F_d = 6πηrv = bv, where b = 6πηr. Characteristic time τ = m/b. Velocity approaches terminal velocity exponentially: v(t) = v_t(1 - e^(-t/τ)). Terminal velocity: v_t = mg/b = mg/(6πηr). Small objects (dust, bacteria, cloud droplets) follow this regime. Position as function of time also follows exponential approach to steady-state motion.
Quadratic Drag
For quadratic drag F_d = -cv2, c = 1/2ρC_dA where ρ is fluid density, C_d is drag coefficient (~0.1-1.0 depending on shape), and A is cross-sectional area. Terminal velocity: v_t = √(mg/c) = √(2mg/(ρC_dA)). This explains why heavy objects fall faster than light objects of same size in air. Objects reach terminal velocity when drag equals weight. For skydiver: v_t ≈ 50-60 m/s (180-200 km/h) belly-to-Earth.
\nTerminal Velocity
Terminal velocity occurs when drag force equals gravitational force (or other applied force), producing zero net force and thus zero acceleration. For linear drag: v_t = mg/b. For quadratic drag: v_t = √(mg/c). Reaching terminal velocity takes infinite time theoretically, but practically achieved after several characteristic times. At terminal velocity, object falls at constant speed, kinetic energy constant, potential energy lost equals work done against drag (dissipated as heat).
Applications
Applications: Parachute design (large area increases drag, reduces v_t for safe landing); Car aerodynamics (reduce C_d for fuel efficiency, typical sedan C_d ≈ 0.3); Sports (shot put vs badminton shuttlecock - different drag, different trajectories); Sediment settling (geology, water treatment); Microorganism swimming (Stokes regime, low Reynolds number, different swimming physics); Ballistics (projectile range reduced by air drag compared to vacuum).
\nSolved Example: Falling with Quadratic Drag
A 75 kg skydiver with cross-sectional area 0.7 m2 falls belly-to-Earth (C_d ≈ 1.0). Air density ρ = 1.2 kg/m3. Find terminal velocity and time to reach 90% of it. Solution: c = 1/2ρC_dA = 0.5 × 1.2 × 1.0 × 0.7 = 0.42 kg/m. Terminal velocity: v_t = √(mg/c) = √(75×9.8/0.42) = √1750 = 41.8 m/s (≈ 150 km/h or 94 mph). Time to reach 90% v_t: For quadratic drag, velocity approaches v_t as tanh function. v/v_t = tanh(gt/v_t) for early times approximately. Setting v = 0.9v_t and solving numerically: t ≈ 2.3v_t/g ≈ 2.3 × 41.8/9.8 ≈ 9.8 s. Actual fall from plane at 3000 m takes about 60 s to reach terminal velocity, then continues at constant speed.
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