Fluid Drag Forces, Terminal Velocity, Stokes' Law, and Elasticity

Drag Forces in Fluids

• Fluid Definition: A fluid is defined as any gas or liquid (e.g., air, water, oil). Drag force equations apply universally to motion through both gases and liquids.

• General Drag Force Formula:      $F_D = \frac{1}{2} C \rho A v^2$

• Components of the Drag Force Equation:

  • $C$ (Drag Coefficient): A dimensionless value representing the surface smoothness and aerodynamic properties of an object. Smoother, more aerodynamic objects have smaller $C$ values, resulting in smaller drag forces.
  • $\rho$ (Fluid Density): The mass per unit volume of the fluid medium ($\text{kg/m}^3$). For example, $\rho$ represents the density of air when walking in a room, or the density of water when operating a submarine underwater.
  • $A$ (Facing Cross-Sectional Area): The area directly facing the direction of motion ($\text{m}^2$). It is not the total surface area of the object (e.g., air pushes against the face, not the back of the head, when moving forward).
  • $v$ (Speed): The relative velocity of the object moving through the fluid ($\text{m/s}$), raised to the second power ($v^2$).

Terminal Velocity Mechanics

• Concept of Terminal Velocity ($v_t$): The theoretical maximum velocity an object reaches when falling through a fluid.
• Free Fall vs. Falling in a Fluid:
  • Free Fall: Absence of air resistance leads to constant gravitational acceleration ($-g$) and indefinite speeding up without speed limits.
  • Fluid Fall: Gravity accelerates the object downwards, increasing speed ($v$). Because drag force depends on $v^2$, $F_D$ increases as speed grows.
• Force Balance at Terminal Velocity:
  • Upon release ($v = 0$\text{ m/s}$), drag force is zero ($F_D = 0$\text{ N}$).
  • As speed increases, $F_D$ grows upwards to oppose downward gravitational force (weight, $W = m g$).
  • Eventually, $F_D$ equals the object's weight $m g$ in magnitude.
  • Net vertical force equals zero:

    $\sum F_y = F_D - m g = 0$

• At this equilibrium point, acceleration drops to zero ($a = 0$\text{ m/s}^2$), and the object continues falling at a constant terminal velocity ($v_t).\n- **Derivation of the Terminal Velocity Equation**:\n\n  \frac{1}{2} C \rho A v_t^2 = m g\n\n  v_t = \sqrt{\frac{2 m g}{C \rho A}}\n\n- **Human Skydiving Example**:\n - A human jumping from an airplane spreads their body horizontally to maximize facing cross-sectional area A.\n - Without a parachute, horizontal human skydivers reach a terminal velocity of approximately 90\,\text{mph}.\n - Parachutes drastically increase area A, lowering terminal velocity to safe landing speeds.\n\n# Example Problem: Falling Squirrel Terminal Velocity\n\n- **Problem Statement**: Estimate the terminal velocity of a 515\,\text{g}$squirrel falling from a$4\,\text{m}$tall tree to the ground. Assume the squirrel reaches terminal velocity and can be approximated as a rectangular prism with a cross-sectional area$11.5\,\text{cm}$wide and$23\,\text{cm} long. Use the drag coefficient for a horizontal skydiver.\n- **Identified Parameters and Unit Conversions**:\n - Mass (m$):$515\,\text{g} = 0.15\,\text{kg}$(converted by dividing by$1000\,\text{g/kg} as specified in text context).\n - Drag Coefficient (C$):$C = 1 (looked up from Section 5.2 table for a horizontal skydiver).\n - Density of Air (\rho$):$\rho = 1.21\,\text{kg/m}^3 (looked up from textbook reference tables).\n - Gravitational Acceleration (g$):$g = 9.8\,\text{m/s}^2.\n - Distractor Surface Area: 925\,\text{cm}^2 total surface area is ignored because drag depends solely on cross-sectional area facing the motion.\n - Cross-Sectional Area (A):\n\n    w = 11.5\,\text{cm} = 0.115\,\text{m}\n\n    l = 23\,\text{cm} = 0.23\,\text{m}\n\n    A = 0.115\,\text{m} \times 0.23\,\text{m} = 0.02645\,\text{m}^2\n\n- **Calculation**:\n\n  v_t = \sqrt{\frac{2 \times 0.15\,\text{kg} \times 9.8\,\text{m/s}^2}{1 \times 1.21\,\text{kg/m}^3 \times 0.02645\,\text{m}^2}}\n\n- **Result**:\n - Speed Magnitude: v_t = 17.8\,\text{m/s}.\n - Vector Quantity: v_t = -17.8\,\text{m/s} (negative sign denotes downward direction).\n\n# Stokes' Law and Small Particle Drag Forces\n\n- **Scope and Applicability**: Standard drag equations (F_D = \frac{1}{2}C\rho A v^2) apply to macroscopic objects (e.g., humans, dry erase markers, planets). Very small particles (e.g., single pollen grains, micro dust particles) approximate spheres and require Stokes' Law.\n- **George Gabriel Stokes' Experiment**:\n - Dropping pollen grains in air is ineffective due to horizontal wind gusts disrupting vertical motion and rapid fall times (< 0.5\,\text{s}) being untrackable with stopwatches.\n - Stokes filled a tall graduated cylinder with vegetable oil (a highly viscous, dense fluid) and released pollen grains inside.\n - The high viscosity of vegetable oil eliminated wind interference and slowed vertical fall, allowing precise measurement of terminal velocity and force balance.\n- **Stokes' Law Equation**:\n\n  F_s = 6 \pi r \eta v\n\n- **Stokes' Law Variables**:\n - F_s$: Stokes drag force ($\text{N}).\n - r$: Radius of the spherical particle ($\text{m}).\n - \eta$(eta): Dynamic viscosity of the fluid ($\text{N}\cdot\text{s/m}^2$or$\text{Pa}\cdot\text{s}), measuring fluid thickness and internal friction resistance.\n - v$: Velocity of the spherical particle ($\text{m/s}).\n- **Viscosity Comparisons**:\n - *Air*: Low viscosity (\eta is small); easy to move through, producing low resistive drag force.\n - *Maple Syrup*: High viscosity (\eta is huge); moving through a swimming pool of maple syrup requires immense force due to massive resistive drag.\n\n# Elasticity and Hooke's Law\n\n- **Solid Material Behavior**: All solid objects possess internal elasticity and behave like springs when subjected to external tensile or compressive forces.\n- **Hooke's Law (Force Magnitude)**:\n\n  F_s = k \Delta L\n\n- **Hooke's Law Variables**:\n - F_s$: Restorative spring force ($\text{N}).\n - k$: Spring constant ($\text{N/m}), representing stiffness and the strength of inter-molecular bonds.\n - \Delta L$: Distance of compression or elongation from neutral equilibrium ($\text{m}).\n- **Comparative Spring Constants (k)**:\n - *Automotive Suspension Springs*: Very stiff to support vehicular weight; k \approx 15000\,\text{N/m}.\n - *Mechanical Pencil Springs*: Soft and flexible; k \approx 3\,\text{N/m}.\n - *Play-Doh Springs*: Extremely weak inter-molecular bonds; pliable with an extremely small spring constant, providing negligible restorative force when deformed.\n- **Hooke's Law (Vector Form)**:\n\n  F_s = -k x\n\n- **Restorative Force Sign Convention**:\n - The negative sign indicates force acts opposite to displacement (x).\n - *Stretching*: Displacement in negative direction (-x$) yields positive force ($+F_s) pulling back toward equilibrium.\n - *Compression*: Displacement in positive direction (+x$) yields negative force ($-F_s) pushing back toward equilibrium.\n\n# Deformation Regions of Materials\n\n- **Elastic Region**:\n - Initial region follows linear Hooke's Law (\text{Slope} = k).\n - Second region deviates slightly from strict linearity, but material fully returns to original length and un-warped state upon removing applied force. Inter-molecular bonds distort but do not permanently break.\n- **Permanent Deformation Region**:\n - Exceeding the elastic limit pulls molecular bonds beyond recovery thresholds.\n - Material permanently warps and fails to return to initial shape upon release.\n - *Slinky Example*: Over-stretching a metal slinky permanently warps individual coils, preventing clean coiling.\n - *Laboratory Safety Warning*: Over-stretching physics lab springs manually uncoils them into straight wire in seconds, permanently destroying equipment.\n- **Fracture Point**:\n - Applying force beyond permanent deformation uncoils material into straight wire until ultimate material failure occurs.\n - The wire snaps or fractures, resulting in infinite length change (\Delta L \to \infty).\n\n# Young's Modulus, Stress, and Strain\n\n- **Generic Elasticity Formula (Tension and Compression)**:\n\n  \Delta L = \frac{1}{Y} \left(\frac{F}{A}\right) L_0\n\n- **Variables**:\n - \Delta L$: Change in length ($\text{m}).\n - Y$: Young's Modulus ($\text{N/m}^2$or$\text{Pa}), an inherent material property measuring rigidity.\n - F$: Applied external force ($\text{N}).\n - A$: Cross-sectional area ($\text{m}^2).\n - L_0$: Initial original length ($\text{m}).\n- **Young's Modulus Comparisons**:\n - *Play-Doh*: Very small Y$; small applied forces yield large length changes ($\Delta L).\n - *Stainless Steel*: Huge Y$(approx.$10^6$times larger than Play-Doh); requires industrial machine forces to produce visible$\Delta L\n- **Engineering Formulation (Stress and Strain)**:\n\n  \frac{F}{A} = Y \left(\frac{\Delta L}{L_0}\right)\n\n- **Stress**:\n - Defined as force per unit cross-sectional area (\frac{F}{A}$) applied to solid-on-solid systems ($\text{N/m}^2$or$\text{Pa}).\n - Note: Force per unit area exerted by solids on fluids/liquids is termed *pressure*; solid on solid is termed *stress*.\n- **Strain**:\n - Defined as fractional deformation (\frac{\Delta L}{L_0}); a unitless ratio of length change relative to initial length.\n- **Relationship**: \text{Stress} = Y \times \text{Strain}. High stress creates high strain on structural materials.\n\n# Questions & Discussion\n\n- **Question**: Is Young's modulus used in structural engineering, such as evaluating support beams?\n- **Answer**: Yes. Structural engineers utilize Young's modulus equations to determine load capacities, maximum stress limits, and safety flexure thresholds for support pillars before structural failure. Distributing heavy weight across multiple pillars reduces individual pillar force and strain.\n- **Bed of Nails Analogy**: Lying on a single nail causes extreme pressure/stress because entire weight is concentrated at one point. Lying on a bed of nails distributes total downward weight across hundreds of contact points, reducing stress at any single point to safe levels.\n\n# Example Problem: Upper Arm Bone Compression in a Wrestler\n\n- **Problem Statement**: A 140\,\text{kg}$wrestler briefly performs a one-handed stand during a maneuver designed to perplex his moribund adversary. Calculate the shortening in length (in meters) of his upper arm bone. The bone is represented by a uniform rod$39\,\text{cm}$in length with a radius of$1.7\,\text{cm}.\n- **Identified Parameters and Unit Conversions**:\n - Mass (m$):$140\,\text{kg}.\n - Initial Length (L_0$):$39\,\text{cm} = 0.39\,\text{m}.\n - Radius (r$):$1.7\,\text{cm} = 0.017\,\text{m}.\n - Gravitational Acceleration (g$):$9.8\,\text{m/s}^2.\n- **Force Calculation**:\n - Weight of Wrestler: W = -m g = -140\,\text{kg} \times 9.8\,\text{m/s}^2 = -1372\,\text{N}.\n - Compressive Normal Force from Ground (F$): By Newton's Third Law,$F = +1372\,\text{N}.\n- **Cross-Sectional Area Calculation**:\n\n  A = \pi r^2 = \pi (0.017\,\text{m})^2 = 9.0792 \times 10^{-4}\,\text{m}^2\n\n- **Compression Calculation**:\n - Young's Modulus for Bone (Y) obtained from textbook reference table.\n  \n  \Delta L = \frac{1}{Y} \left(\frac{1372\,\text{N}}{9.0792 \times 10^{-4}\,\text{m}^2}\right) (0.39\,\text{m})\n\n- **Result**:\n - Bone Shortening: \Delta L = 6.55 \times 10^{-5}\,\text{m}$$.
• Physical Meaning: The wrestler's arm bone compresses by a tiny, imperceptible distance invisible to the human eye.

Administrative and Course Logistics

• WebAssign Glitch: Homework system malfunction shifted all assignment deadlines by 3 weeks into the post-semester period. System dates have been manually corrected.
• Homework Deadline: Current WebAssign homework submission deadline extended to Friday.
• Lecture & Quiz Schedule: Chapter 5 slides will conclude on Wednesday, followed by the start of Chapter 6/11 material. Chapter 5 Quiz will take place on Friday.