Turbulent Air Drag: Horizontal Motion

From: Classical Physics / Chapter 2 - Projectile Motion and Air Drag

When an object moves through a fluid at higher speeds or for larger objects, the drag force is often better modeled as proportional to the square of the velocity rather than the velocity itself. This is called quadratic air drag or turbulent drag, and it occurs when the flow around the object becomes turbulent rather than laminar.

This page will solve the equations of motion for a particle experiencing turbulent (quadratic) air drag and no other forces (i.e. no gravity).

System Definition: A particle with quadratic air drag:
  • Mass: $m$
  • Initial velocity: $v_0$
  • Drag coefficient: $c$ (proportionality constant for quadratic drag)
  • Drag force magnitude: $F_d = cv^2$ (opposes motion direction)
  • For motion in $+x$ direction: $F_{d,x} = -cv^2$ (negative sign indicates opposition)
  • For motion in $+y$ direction: $F_{d,y} = -cv^2$ (negative sign indicates opposition)
  • Initial position: $(x_0, y_0) = (0, 0)$

The Equation of Motion

To simplify the equations, we assume the object is moving in the $+x$ direction, i.e. $v>0$. With this assumption, the quadratic drag force is $F_d = -cv^2$.

We start with Newton's $2^{nd}$ Law:

$$F_{\text{net}} = ma.$$

Since the drag is the only force acting on the object, the net force equals the drag force, $F_{\text{net}} = -cv^2$. We also write the acceleration as $a = \frac{dv}{dt}$ to give:

$$-cv^2 = m\frac{dv}{dt}.$$

Rearranging, we obtain the equation of motiomn:

Equation of Motion for quadratic drag without gravity:
$$\frac{dv}{dt} = -\frac{c}{m}v^2.$$

This is a first-order nonlinear ordinary differential equation.

Analytical Solution for Velocity

To solve the differential equation $\frac{dv}{dt} = -\frac{c}{m}v^2$, we use separation of variables:

$$\frac{dv}{v^2} = -\frac{c}{m}dt.$$

Integrating both sides with the initial condition $v(0) = v_0$:

$$\int_{v_0}^{v(t)} \frac{dv'}{(v')^2} = -\frac{c}{m}\int_0^t dt'.$$

The left and right sides integrate to:

$$\left[-\frac{1}{v'}\right]_{v_0}^{v(t)} = -\frac{1}{v(t)} + \frac{1}{v_0} = -\frac{c}{m}t.$$

Rearranging:

$$\frac{1}{v(t)} = \frac{1}{v_0} + \frac{c}{m}t$$.

Taking the reciprocal gives

$$v(t) = \frac{v_0}{1 + \frac{cv_0}{m}t}.$$

Following the approach used in the linear drag case, we can define the characteristic time as $\tau = m/cv_0$. (Note: this $\tau$ is different from the characteristic time in the linear drag equations.) After substituting, we have

Velocity Solution For Quadratic Drag (Without Gravity):
$$v(t) = \frac{v_0}{1 + t/\tau}.$$

This shows that the velocity decays as $1/t$ for large times, much slower than the exponential decay of linear drag. The velocity never reaches zero in finite time, but approaches zero asymptotically.

Interactive Plot: Velocity Comparison

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Comparison of velocity decay for linear drag ($F_d = -bv$) and quadratic drag ($F_d = -cv^2$)

Reynolds number at crossing: calculating...

Analytical Solution for Position

To find the position as a function of time, we integrate the velocity:

$$x(t) = \int_0^t v(t') dt' = \int_0^t \frac{v_0}{1 + t'/\tau} dt'.$$

Using the substitution $u = 1 + t'/\tau$, we have $du = dt'/\tau$, so $dt' = \tau du$:

$$x(t) = v_0\tau\int_1^{1 + t/\tau} \frac{du}{u} = v_0\tau\ln\left(1 + \frac{t}{\tau}\right).$$
Position Solution (Horizontal Motion):
$$x(t) = v_0\tau\ln\left(1 + \frac{t}{\tau}\right)$$

Unlike linear drag, where position approaches a finite maximum, with quadratic drag the position grows logarithmically with time. This means the object continues to move (albeit very slowly) for all time, never reaching a maximum distance.

Part II: Vertical Motion With Gravity

The Equation of Motion

For vertical motion with gravity (taking $+y$ downward), we have two forces:

$$F_{\text{net}} = mg - cv^2 = m\frac{dv}{dt}$$

Rearranging:

$$\frac{dv}{dt} = g - \frac{c}{m}v^2$$

Terminal Velocity

Before solving the full equation, we can find the terminal velocity by setting the acceleration to zero:

$$g - \frac{c}{m}v_{\text{term}}^2 = 0$$
Terminal Velocity (Quadratic Drag):
$$v_{\text{term}} = \sqrt{\frac{mg}{c}}$$

Notice that terminal velocity depends on the square root of mass, unlike linear drag where it was proportional to mass. This is a key difference between the two drag models.

Analytical Solution for Velocity

To solve $\frac{dv}{dt} = g - \frac{c}{m}v^2$, we rearrange:

$$\frac{dv}{dt} = \frac{c}{m}\left(\frac{mg}{c} - v^2\right) = \frac{c}{m}\left(v_{\text{term}}^2 - v^2\right)$$

Separating variables:

$$\frac{dv}{v_{\text{term}}^2 - v^2} = \frac{c}{m}dt$$

Using partial fractions, we can write:

$$\frac{1}{v_{\text{term}}^2 - v^2} = \frac{1}{2v_{\text{term}}}\left(\frac{1}{v_{\text{term}} - v} + \frac{1}{v_{\text{term}} + v}\right)$$

Integrating both sides with the initial condition $v(0) = v_0$:

$$\int_{v_0}^{v(t)} \frac{dv'}{v_{\text{term}}^2 - (v')^2} = \frac{c}{m}\int_0^t dt'$$

The integral on the left evaluates to:

$$\frac{1}{2v_{\text{term}}}\ln\left|\frac{v_{\text{term}} + v(t)}{v_{\text{term}} - v(t)}\right| - \frac{1}{2v_{\text{term}}}\ln\left|\frac{v_{\text{term}} + v_0}{v_{\text{term}} - v_0}\right| = \frac{c}{m}t$$

Simplifying and solving for $v(t)$:

Velocity Solution (Vertical Motion):
$$v(t) = v_{\text{term}}\frac{(v_{\text{term}} + v_0)e^{2gt/v_{\text{term}}} - (v_{\text{term}} - v_0)}{(v_{\text{term}} + v_0)e^{2gt/v_{\text{term}}} + (v_{\text{term}} - v_0)}$$

For an object starting from rest ($v_0 = 0$), this simplifies to:

$$v(t) = v_{\text{term}}\frac{e^{2gt/v_{\text{term}}} - 1}{e^{2gt/v_{\text{term}}} + 1} = v_{\text{term}}\tanh\left(\frac{gt}{v_{\text{term}}}\right)$$

where $\tanh$ is the hyperbolic tangent function. This shows that the velocity approaches terminal velocity as a hyperbolic tangent, which is different from the exponential approach in linear drag.

Analytical Solution for Position

For the position, we integrate the velocity. For the case starting from rest ($v_0 = 0$):

$$y(t) = \int_0^t v_{\text{term}}\tanh\left(\frac{gt'}{v_{\text{term}}}\right) dt'$$

Using the substitution $u = \frac{gt'}{v_{\text{term}}}$:

$$y(t) = \frac{v_{\text{term}}^2}{g}\int_0^{gt/v_{\text{term}}} \tanh(u) du = \frac{v_{\text{term}}^2}{g}\ln\left(\cosh\left(\frac{gt}{v_{\text{term}}}\right)\right)$$
Position Solution (Vertical Motion, from rest):
$$y(t) = \frac{v_{\text{term}}^2}{g}\ln\left(\cosh\left(\frac{gt}{v_{\text{term}}}\right)\right)$$

For large times, $\cosh(x) \approx \frac{1}{2}e^x$, so the position grows approximately linearly with time at terminal velocity: $y(t) \approx v_{\text{term}} t + \text{constant}$.

Key Observations:
  • Horizontal motion: Velocity decays as $1/t$, position grows logarithmically
  • Vertical motion: Velocity approaches terminal velocity as $\tanh$, position grows as $\ln(\cosh)$
  • Terminal velocity: $v_{\text{term}} = \sqrt{mg/c}$ (depends on square root of mass)
  • No finite stopping distance: Unlike linear drag, quadratic drag allows motion to continue indefinitely

Interactive Plot: Horizontal Motion

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Analytical solutions for position and velocity with quadratic air drag (horizontal motion, no gravity)

Interactive Plot: Vertical Motion

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Analytical solutions for position and velocity with quadratic air drag and gravity

Terminal Velocity: 9.90 m/s

Comparison with Linear Air Drag

Key Differences

Aspect Linear Drag ($F_d = -bv$) Quadratic Drag ($F_d = -cv^2$)
Velocity Decay (Horizontal) Exponential: $v(t) = v_0e^{-t/\tau}$ Inverse: $v(t) = \frac{v_0}{1 + t/\tau}$
Position (Horizontal) Finite maximum: $x(t) = v_0\tau(1 - e^{-t/\tau})$ Logarithmic growth: $x(t) = v_0\tau\ln(1 + t/\tau)$
Terminal Velocity $v_{\text{term}} = \frac{mg}{b} = g\tau$ (proportional to $m$) $v_{\text{term}} = \sqrt{\frac{mg}{c}}$ (proportional to $\sqrt{m}$)
Velocity Approach (Vertical, from rest) Exponential: $v(t) = v_{\text{term}}(1 - e^{-t/\tau})$ Hyperbolic tangent: $v(t) = v_{\text{term}}\tanh(\frac{gt}{v_{\text{term}}})$
Characteristic Time $\tau = \frac{m}{b}$ (natural time scale) $\frac{v_{\text{term}}}{g}$ (time to reach terminal velocity)
Physical Regime Low Reynolds number, laminar flow, small objects, low speeds High Reynolds number, turbulent flow, large objects, high speeds

Physical Interpretation

The quadratic drag model is most appropriate for:

  • High Reynolds numbers: When the flow around the object is turbulent
  • Larger objects: Such as baseballs, raindrops, or skydivers
  • Higher speeds: When inertial effects dominate over viscous effects

The transition from linear to quadratic drag depends on the Reynolds number, which characterizes the ratio of inertial to viscous forces in the fluid flow.

Mathematical Summary

Horizontal Motion (No Gravity):
$$v(t) = \frac{v_0}{1 + t/\tau}$$
$$x(t) = v_0\tau\ln\left(1 + \frac{t}{\tau}\right)$$
Vertical Motion (With Gravity, from rest):
$$v(t) = v_{\text{term}}\tanh\left(\frac{gt}{v_{\text{term}}}\right)$$
$$y(t) = \frac{v_{\text{term}}^2}{g}\ln\left(\cosh\left(\frac{gt}{v_{\text{term}}}\right)\right)$$
$$v_{\text{term}} = \sqrt{\frac{mg}{c}}$$