Notes on Drag Forces and Terminal Velocity

Review and Introduction

• Prior Topics: This session begins with a review of previous material, specifically Newton's Laws. Students should be prepared for a two-sided quiz covering these concepts.
• Homework: An assigned homework set focuses on in-class material, including a practical problem about slipping on wet leaves on the road.
• Lecture Approach: The instructor often uses videos, especially on Fridays, to provide a different learning style, allow for a mental break, and visually emphasize concepts for a non-English speaking student. The belief is that "pictures are telling you something so much more than equations or tables or even graphs."

James Bond and Terminal Velocity (Video Example)

• Movie Reference: A scene from a James Bond movie (starring Roger Moore), specifically involving Bond fighting villains, including the character Jaws (who has stainless steel teeth), aboard an airplane as they begin to fall.
• Scenario: Bond and his adversaries are in freefall from an airplane.
• Observation: Both Bond and the villain eventually reach terminal velocity. This occurs when the upward drag force perfectly balances the downward force of gravity (weight), resulting in zero net acceleration.
• Key Concept Demonstrated: The video dramatically illustrates the effect of an object's area and drag coefficient on its velocity during freefall.

Quadratic Drag Force

• Application: Drag forces experienced by objects moving at high speeds, such as cars, airplanes, or a person falling from an airplane, are proportional to the square of their velocity.
• Formula: The magnitude of the quadratic drag force (often denoted as $F<em>D$ or $F</em>{drag}$) is given by: $F<em>D = \frac{1}{2} \rho v^2 C</em>D A$ Where:
  • $\rho$ (rho) = density of the fluid (e.g., air).
  • $v$ = velocity of the object relative to the fluid.
  • $C<em>D$ = drag coefficient, a dimensionless quantity that depends on the shape of the object. A flatter shape generally has a larger $C</em>D$.
  • $A$ = cross-sectional area of the object perpendicular to the direction of motion.
• James Bond's Strategy: To increase his speed and catch the villain, James Bond changes his body shape from a "spread eagle" position to a more streamlined, "bullet-like" form. This action significantly decreases both his drag coefficient $(C<em>D)$ and his cross-sectional area $(A)$, thereby reducing the drag force $(F</em>D)$ acting on him and allowing him to accelerate.
• Calculating Terminal Velocity ($v<em>T$): Terminal velocity is reached when the drag force balances the gravitational force (weight). Thus, one can solve for $v</em>T$ by setting $F<em>D = mg$. $\frac{1}{2} \rho v</em>T^2 C_D A = mg$
• Numerical Examples (Skydiving):
  • Spread Eagle Position: For a person in a spread-eagle orientation, the effective area is approximately $A \approx 1 m^2$ and the drag coefficient is about $C<em>D = 1$. If the combined mass (e.g., instructor and student) is assumed to be $m = 75 kg$ (the instructor notes this is an underestimate, closer to $100-130 kg$), the terminal velocity is calculated to be about $v</em>T = 42 m/s$ (approximately $93 miles/hour$).
  • Diving Position: When a person streamlines into a diving position, their drag coefficient $(C<em>D)$ is much smaller, and their effective area $(A)$ is also significantly reduced. This results in a much higher terminal velocity, around $v</em>T = 220 miles/hour$.

Linear Drag Force

• Application: Linear drag forces apply to small objects moving very slowly through fluids (e.g., a tiny sphere, pollen spores). For such objects, drag is directly proportional to velocity.
• Impact on Small Organisms: For insects, drag forces can be deadly. While they might stay on top of a fluid (like water), if submerged, the drag can be too great for them to overcome, making movement impossible.
• Formula: The linear drag force is given by: $F_D = Bv$ Where:
  • $B$ = a constant that depends on the properties of the fluid, the object itself, and its shape. This constant is typically provided in problems.
  • $v$ = velocity of the object.
• Example (Pollen Spore): For a pollen spore, a very small object, its terminal velocity due to linear drag would be extremely low, approximately $v_T \approx 1 mm/s$.

Solving the Differential Equation for Linear Drag

• Significance: This type of differential equation, which describes how the speed of an object changes over time as it approaches terminal velocity, is also encountered in other fields like electricity and magnetism.
• Goal: Determine the velocity of an object as a function of time, $v(t)$, under linear drag.
• Newton's Second Law with Drag: Assuming downward motion is positive, the net force on a falling object with linear drag is the gravitational force minus the drag force:
  $F<em>{net} = mg - Bv$ According to Newton's Second Law, $F</em>{net} = ma = m \frac{dv}{dt}$.
  Therefore, the differential equation is:
  $m \frac{dv}{dt} = mg - Bv$
• Separation of Variables: To solve for $v(t)$, we separate the variables $v$ and $t$:
  $\frac{dv}{dt} = g - \frac{B}{m} v$
  $\frac{dv}{g - \frac{B}{m} v} = dt$
• Integration: Integrate both sides:
  $\int \frac{dv}{g - \frac{B}{m} v} = \int dt$
  Let $u = g - \frac{B}{m} v$. Then $du = -\frac{B}{m} dv$, so $dv = -\frac{m}{B} du$.
  Substituting this into the integral:
  $\int \frac{1}{u} \left(-\frac{m}{B}\right) du = \int dt$
  $-\frac{m}{B} \int \frac{1}{u} du = \int dt$
  $-\frac{m}{B} \ln|u| = t + C<em>1$ Substitute $u$ back: $-\frac{m}{B} \ln\left|g - \frac{B}{m} v\right| = t + C</em>1$
• Solving for $v(t)$:
  $\ln\left|g - \frac{B}{m} v\right| = -\frac{B}{m} (t + C<em>1)$$g - \frac{B}{m} v = e^{-\frac{B}{m} (t + C</em>1)} = e^{-\frac{B}{m} t} e^{-\frac{B}{m} C<em>1}$ Let $A = e^{-\frac{B}{m} C</em>1}$ (This is a positive constant).
  $g - \frac{B}{m} v = A e^{-\frac{B}{m} t}$
  $\frac{B}{m} v = g - A e^{-\frac{B}{m} t}$
  $v(t) = \frac{mg}{B} - \frac{mA}{B} e^{-\frac{B}{m} t}$
• Applying Initial Conditions: Assume the object is dropped from rest, meaning $v(0) = 0$ at time $t=0$.
  $0 = \frac{mg}{B} - \frac{mA}{B} e^0 = \frac{mg}{B} - \frac{mA}{B}$
  From this, we find that $A = \frac{mg}{m} = g$. (Wait, this is wrong. $A = \frac{mg}{B} \cdot \frac{B}{m} = g$ if $m/B$ is canceled. Let's re-evaluate. $0 = \frac{mg}{B} - \frac{mA}{B} \Rightarrow \frac{mA}{B} = \frac{mg}{B} \Rightarrow mA = mg \Rightarrow A = g$). No, the constant multiplies the exponential. Let's check with the final form. The lecturer mentions $v_T = mg/B$.
  Let's re-evaluate the constant more carefully: $v(t) = \frac{mg}{B} - \frac{C'}{B} e^{-\frac{B}{m} t}$, where $C' = mA$. So if $v(0) = 0$, then $0 = \frac{mg}{B} - \frac{C'}{B}$, which means $C' = mg$. This makes sense.
• Final Solution for $v(t)$:
  $v(t) = \frac{mg}{B} - \frac{mg}{B} e^{-\frac{B}{m} t}$
  $v(t) = \frac{mg}{B} \left(1 - e^{-\frac{B}{m} t}\right)$
• Terminal Velocity ($v<em>T$): As time $t \to \infty$, the exponential term $e^{-\frac{B}{m} t}$ approaches $0$. Therefore, the terminal velocity is:
  $v</em>T = \lim_{t \to \infty} v(t) = \frac{mg}{B}$
  This shows that the factor $rac{mg}{B}$ in the solution is indeed the terminal velocity.