Drag Force and Terminal Speed Study Guide

Definition of Drag: Drag (air resistance) is a resistive mechanical force exerted on an object moving through a fluid medium (liquid or gas).
Causal Mechanism: Drag is caused by the physical interaction and collisions between the object and the fluid medium through which it moves.
Directionality: Drag always acts in the direction strictly opposite to the instantaneous relative motion of the object through the fluid.

Quadratic Drag Model (Standard Macroscopic Regime):
- For macroscopic bodies moving at standard or high speeds through air, the magnitude of the drag force $D$ is proportional to the square of the speed: $D = b v^2$
- $D$ represents the drag force.
- $v$ represents the instantaneous speed of the object.
- $b$ represents a composite drag parameter.
Linear Drag Model (Low-Speed / Small Particle Regime):
- For extremely small particles (such as microscopic dust particles) or for objects moving at very low speeds (laminar flow/low Reynolds number regime), the drag force is modeled as linearly proportional to speed: $D = b v$
Factors Determining the Drag Parameter ( $b$ ):
- Shape: The aerodynamic geometry and streamlining of the object.
- Properties of the Fluid: The density ( $\rho$ ) and viscosity of the surrounding fluid medium.
- Cross-Sectional Area: The frontal projected area ( $A$ ) perpendicular to the direction of travel.
- Drag Coefficient ( $C_D$ ): A dimensionless index characterizing the aerodynamic shape and flow separation properties of the body.

Concept of Terminal Speed ( $v_t$ ):
- When an object falls through a fluid under the influence of gravity, the upward resistive drag force increases as the falling speed increases.
- When the upward drag force equals the downward force of gravity ( $F_g = m g$ ), the net external force on the object reaches zero.
- According to Newton's Second Law, zero net force produces zero acceleration ( $a = 0$ ), causing the object to cease accelerating and descend at a constant, maximum terminal speed ( $v_t$ ).
Mathematical Derivation:
- Setting up the equation of motion along the vertical axis (taking downward as positive): $\Sigma F_y = m g - D = m a$
- Substituting the quadratic drag model $D = b v^2$ : $\Sigma F_y = m g - b v^2 = m a$
- At terminal speed ( $v = v_t$ ), acceleration ceases ( $a = 0$ ): $\Sigma F_y = m g - b v_t^2 = 0$
- Equating the gravitational force and the drag force: $m g = b v_t^2$
- Solving for terminal speed ( $v_t$ ): $v_t^2 = \frac{m g}{b}$ $v_t = \sqrt{\frac{m g}{b}}$

State 1: Initial Release ( $t = 0$ )
- Velocity: $v = 0$
- Drag Force: $D = b (0)^2 = 0$ (no upward vector)
- Downward Force: $F_g = m g$
- Net Force: $\Sigma F_y = m g$
- Instantaneous Acceleration: $a = 9.8\,m/s^2$ downward (or $-9.8\,m/s^2$ if defined coordinate system takes upward as positive)
State 2: Early Descent
- Velocity: $v > 0$ (speed is increasing)
- Drag Force: A non-zero upward drag force $b v^2$ develops opposing motion.
- Downward Force: Gravity remains constant at $m g$ .
- Net Force: $\Sigma F_y = m g - b v^2 < m g$
- Instantaneous Acceleration: Net downward acceleration decreases to $a \approx 7\,m/s^2$
State 3: Intermediate Descent
- Velocity: Speed continues to increase, further elevating the drag force.
- Drag Force: Upward vector $b v^2$ increases in magnitude.
- Downward Force: Gravity remains constant at $m g$ .
- Net Force: The net downward force decreases significantly.
- Instantaneous Acceleration: Net downward acceleration diminishes further to $a \approx 3\,m/s^2$ to $4\,m/s^2$
State 4: Terminal Equilibrium
- Velocity: Speed reaches terminal velocity ( $v = v_t$ ).
- Drag Force: Upward vector $b v_t^2$ is exactly equal in magnitude to downward weight vector $m g$ .
- Downward Force: Gravity remains constant at $m g$ .
- Net Force: $\Sigma F_y = m g - b v_t^2 = 0$
- Instantaneous Acceleration: $a = 0\,m/s^2$

Position vs. Time ( $x(t)$ or $y(t)$ ):
- The curve begins parabolic at $t = 0$ , reflecting early downward acceleration.
- As the resistive drag increases, the position curve straightens out.
- At long times ( $t \gg 0$ ), the slope becomes constant and linear, where: $\text{slope} = v_t$
Velocity vs. Time ( $v(t)$ ):
- The velocity starts at $v(0) = 0$ .
- The initial slope of the curve at $t = 0$ is equal to the gravitational acceleration ( $g = 9.8\,m/s^2$ ).
- The slope progressively flattens out, asymptotically approaching the horizontal terminal velocity line $v = v_t$ .
Acceleration vs. Time ( $a(t)$ ):
- The acceleration curve begins at its maximum magnitude of $9.8\,m/s^2$ .
- Over time, as speed grows and upward drag increases, acceleration decreases continuously.
- Asymptotically decays toward zero ( $a \rightarrow 0$ ) as terminal speed is reached.

Table 6-1: Terminal Speeds and 95% Fall Distances in Air:
- Data adapted from Peter J. Brancazio, Sport Science, Simon & Schuster, New York, 1984.
- Note on 95% Distance: This indicates the vertical distance through which an object must fall from rest to achieve 95% of its terminal speed.

Physical Insights from Empirical Data:
- Heavy, dense, and compact objects (such as the 16-lb shot) require substantial distances ( $2500\,m$ ) to approach terminal velocity and achieve high terminal speeds ( $145\,m/s$ ).
- Lightweight, high-surface-area objects (such as a typical parachutist) encounter dominant drag forces rapidly, reaching terminal velocity ( $5\,m/s$ ) within a short fall distance of only $3\,m$ .