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Linear drag, its terminal speed $mg/b$ and time constant $m/b$, exponential approach and coasting, quadratic drag and the terminal speed $\sqrt{2mg/\rho CA}$.
Paper packet. Every task here also exists on screen, where it is checked automatically; answers written on paper are not assessed by Nydus. When you are back at a device, enter your answers there.
By the end of this lesson you will be able to solve the equation of motion with linear drag and find terminal speeds for linear and quadratic drag.
From earlier in this course you can write Newton's second law as $m\,dv/dt = F(v)$ and solve simple differential equations by separating variables. From Physics 1 you know that drag opposes motion and that falling bodies reach a terminal speed. This lesson turns that qualitative idea into equations you can solve exactly.
| Term | What it means |
|---|---|
| Resistive force | A force from a fluid that opposes motion and grows with speed. |
| Linear drag | $F = -bv$, for slow, small bodies in viscous fluids. |
| Quadratic drag | $F = \tfrac{1}{2}\rho CAv^2$, for fast or large bodies in air or water. |
| Terminal speed | The speed at which drag balances the driving force, so acceleration stops. |
| Time constant | $\tau = m/b$, the time for linear drag to close 63 percent of the gap to terminal speed. |
| Drag coefficient | $C$, a number near $1$ that captures the body's shape. |
| Separation of variables | Rearranging so each side has one variable, then integrating. |
A body falling through a fluid feels its weight down and a drag force up that grows with speed. For slow motion through a thick fluid the drag is linear, $bv$, and the second law is
$$m\frac{dv}{dt} = mg - bv.$$
The acceleration is largest at release, $g$, and falls to zero at the terminal speed $v_T = mg/b$, where drag balances weight. Solving the equation from rest gives
$$v(t) = v_T\left(1 - e^{-t/\tau}\right), \qquad \tau = \frac{m}{b}.$$
The speed rises quickly at first and then creeps toward $v_T$, reaching 63 percent of it after one time constant and 95 percent after three. For fast bodies in air the drag is quadratic, $\tfrac{1}{2}\rho CAv^2$, and the terminal speed is $\sqrt{2mg/\rho CA}$.
Another way: picture
Picture a marble dropped into a tall jar of honey. It lurches downward, then almost at once settles to a steady slow sink. Every bit of extra speed brings extra drag, so the speed is pulled toward the value where drag and weight balance. The speed closes the gap to that value by the same fraction in each time constant, which is what an exponential means.
Another way: steps
Write the equation in terms of the gap to terminal speed: $m\,dv/dt = -b(v - v_T)$. Separating, $\dfrac{dv}{v - v_T} = -\dfrac{dt}{\tau}$, and integrating from $v = 0$ at $t = 0$ gives $\ln\dfrac{v_T - v}{v_T} = -\dfrac{t}{\tau}$. So the gap $v_T - v$ shrinks as $v_Te^{-t/\tau}$, and $v = v_T(1 - e^{-t/\tau})$.
Differentiating checks it: $dv/dt = (v_T/\tau)e^{-t/\tau}$, which equals $g$ at $t = 0$, since $v_T/\tau = g$, and falls to zero. Integrating once more gives the position, $y = v_T t - v_T\tau(1 - e^{-t/\tau})$. At long times the body moves at $v_T$ but lags a body that had started at $v_T$ by a distance $v_T\tau$.
Take away the driving force and keep the drag, as for a boat that cuts its engine. Now $m\,dv/dt = -bv$, and separating gives $v = v_0e^{-t/\tau}$. The speed halves every $\tau\ln 2$, like a radioactive sample, and never quite reaches zero.
The distance, however, is finite: $\int_0^\infty v_0e^{-t/\tau}dt = v_0\tau$. That also follows from impulse. The total impulse from drag is $\int bv\,dt = b \times$ distance, and it must remove all the momentum $mv_0$, so the distance is $mv_0/b$. A heavier boat, or a faster one, glides proportionally further.
For a baseball, a car or a skydiver, the drag comes from pushing air out of the way, and it grows as the square of the speed: $F_D = \tfrac{1}{2}\rho CAv^2$, where $\rho$ is the air's density, $A$ the frontal area, and $C$ a shape factor, about $0.3$ for a car and $1.0$ for a spread-out person.
Setting drag equal to weight gives $v_T = \sqrt{2mg/\rho CA}$. Doubling the mass now raises the terminal speed by only $\sqrt{2}$. The equation $m\,dv/dt = mg - cv^2$ can also be solved by separation; its solution is $v = v_T\tanh(gt/v_T)$, which rises like $gt$ at first and levels off at $v_T$, much as the linear case does.
Whether drag is linear or quadratic depends on the Reynolds number, $\rho vD/\eta$, where $D$ is the body's size and $\eta$ the fluid's viscosity. Below about $1$, viscosity dominates and drag is linear: pollen settling in air, bacteria swimming, a bead sinking in syrup. Above about $1000$, the fluid's inertia dominates and drag is quadratic: nearly everything people throw, drive or fly.
For a sphere in the linear range, Stokes found $b = 6\pi\eta r$. Robert Millikan used exactly this to find the charge of the electron in 1909 at the University of Chicago: timing tiny oil drops falling at terminal speed gave their size, and then the electric field needed to hold them still gave their charge.
Checking an answer. At $t = 0$ the acceleration must be the drag-free value. At long times the speed must approach $v_T$, never pass it. The units of $\tau = m/b$ must be seconds, since $b$ is in kg/s. And setting $b \to 0$ must recover free fall, $v = gt$: expand $1 - e^{-t/\tau} \approx t/\tau$ to check.
Galileo's rule that all bodies fall alike holds only without drag. With drag, the terminal speed $mg/b$ or $\sqrt{2mg/\rho CA}$ depends on mass and size. Two balls of the same size, one of lead and one of wood, fall together at first, when drag is small, but the lead ball pulls ahead as speeds grow.
Size matters too. For bodies of the same shape and material, mass grows as the cube of size and area as the square, so terminal speed grows as the square root of size. A mouse falling down a mine shaft lands at a gentle speed; a horse does not. Raindrops fall at a few meters per second, and the mist in a cloud barely falls at all.
Drag sets the fuel economy of a car at highway speeds. The power to push against quadratic drag is $F_Dv = \tfrac{1}{2}\rho CAv^3$, so driving at $75$ mph instead of $60$ takes nearly twice the power to overcome air resistance. That is why car makers reshape mirrors and underbodies to shave a few percent off $C$.
Drag is also what makes parachutes, air brakes and the heat shields of returning spacecraft work. A capsule re-entering from orbit sheds nearly all of its kinetic energy to drag in a few minutes, and the time constant of that slowing, set by its mass and area, determines how hot the shield becomes.
A few terminal speeds give a sense of scale. A raindrop $2$ mm across falls at about $6.5$ m/s. A skydiver belly-down falls at about $55$ m/s, $120$ mph; head-down, over $80$ m/s. A baseball dropped from a great height would reach about $42$ m/s, and a Ping-Pong ball only $9$ m/s.
The time to approach terminal speed is roughly $v_T/g$: under a second for the raindrop, about $5$ seconds for the Ping-Pong ball's big cousin, and a dozen seconds for a skydiver, who falls about $450$ m before reaching steady speed. These estimates let you judge quickly whether drag matters in a problem at all.
Separating variables treats $dv/dt$ as a ratio, which is legitimate because it is shorthand for the chain rule: integrating $\frac{1}{v - v_T}\frac{dv}{dt}$ with respect to $t$ is the same as integrating $\frac{1}{v - v_T}$ with respect to $v$. The logarithm that appears must be of a positive quantity, which is why the solution is written with $v_T - v$, positive for a body starting below terminal speed.
The starting condition fixes the constant of integration, and there is only one, because the equation is first order: knowing the speed at one instant determines it at all others. That is a real difference from the oscillator of the last lessons, which needed both a position and a velocity. It also means a body launched downward faster than $v_T$ slows down toward $v_T$ along the same kind of exponential, approaching from above instead of below; the same formula covers both cases with the right starting speed.
At drop zones like Perris, California, jumpers leave a plane at about $13{,}000$ feet and fall for about a minute before opening their parachutes. Within about $12$ seconds they reach terminal speed, near $55$ m/s belly-down, and then fall steadily while the air pushes up as hard as gravity pulls down.
Skydivers steer their speed by changing their area: arching flat to slow down, tucking head-down to go faster when catching up with a partner. The parachute raises $CA$ by a factor of about $40$, cutting the terminal speed to about $5$ m/s, which is about the speed of jumping off a two-meter wall.
At the University of Chicago in 1909, Robert Millikan and his student Harvey Fletcher sprayed tiny oil drops between two metal plates and watched them through a microscope. A drop falls at terminal speed almost at once, since its time constant is microseconds, and from that speed and Stokes's linear drag law they found its radius and mass.
Switching on an electric field between the plates, they could hold a charged drop still or drive it upward, and the field needed gave the drop's charge. Every charge came out as a whole-number multiple of $1.6 \times 10^{-19}$ C, the charge of one electron. Millikan received the Nobel Prize in 1923; the measurement rests entirely on linear drag.
Without air, a falling body speeds up without limit; with air, drag grows until it balances the weight and the body falls steadily. It is also tempting to think a heavy body reaches its terminal speed sooner. With linear drag the time constant is $m/b$, so a heavier body of the same size takes longer, because it is heading for a higher terminal speed.
A second error is to put terminal speed at the moment the parachute opens. Opening the chute raises the drag suddenly, and the jumper slows toward a new, lower terminal speed, again exponentially.
A $2.0$ g bead in oil feels linear drag with $b = 0.050$ kg/s. Find its terminal speed.
$v_T = \dfrac{mg}{b} = \dfrac{0.0020 \times 9.8}{0.050} = 0.392\ \text{m/s}$
Drag balances weight.
Find the time constant.
$\tau = \dfrac{0.0020}{0.050} = 0.040\ \text{s}$
Very short: light body, thick fluid.
Write its speed after release.
$v = 0.392(1 - e^{-t/0.040})\ \text{m/s}$
From rest.
Find its speed after $0.080$ s.
$v = 0.392(1 - e^{-2}) = 0.392 \times 0.865 = 0.339\ \text{m/s}$
Two time constants in.
Check the starting acceleration.
$\dfrac{v_T}{\tau} = \dfrac{0.392}{0.040} = 9.8\ \text{m/s}^2$
Free fall at the first instant.
A $300$ kg boat at $6.0$ m/s cuts its engine. Water drag is $bv$ with $b = 60$ kg/s. Find the time constant.
$\tau = \dfrac{300}{60} = 5.0\ \text{s}$
Mass over drag coefficient.
Write the speed.
$v = 6.0e^{-t/5.0}\ \text{m/s}$
Exponential decay.
Find the speed after $10$ s.
$v = 6.0e^{-2} = 0.81\ \text{m/s}$
Two time constants.
Find the time to halve its speed.
$t = 5.0\ln 2 = 3.5\ \text{s}$
Set $e^{-t/\tau} = 1/2$.
Find how far it coasts in all.
$x = v_0\tau = 6.0 \times 5.0 = 30\ \text{m}$
The integral of the speed.
Check with impulse.
$bx = 60 \times 30 = 1800 = mv_0 = 300 \times 6.0$
Drag's impulse removes all the momentum.
A $75$ kg skydiver falls belly-down with $C = 1.0$ and $A = 0.70$ m², in air of density $1.2$ kg/m³. Write the drag.
$F_D = \tfrac{1}{2} \times 1.2 \times 1.0 \times 0.70 \times v^2 = 0.42v^2\ \text{N}$
Quadratic drag.
Balance drag against weight.
$0.42v_T^2 = 75 \times 9.8 = 735$
Terminal speed.
Solve for the terminal speed.
$v_T = \sqrt{\dfrac{735}{0.42}} = \sqrt{1750} = 41.8\ \text{m/s}$
About 94 mph.
Estimate the time to approach it.
$t \approx \dfrac{v_T}{g} = \dfrac{41.8}{9.8} = 4.3\ \text{s}$
The natural time scale of the fall.
Find the speed at that time.
$v = v_T\tanh(1) = 0.76 \times 41.8 = 31.8\ \text{m/s}$
The quadratic solution.
Tuck into a head-down dive with a quarter of the area.
$v_T' = 41.8 \times \sqrt{4} = 83.7\ \text{m/s}$
Terminal speed goes as $1/\sqrt{A}$.
Open the parachute, raising $CA$ to $30$ m².
$v_T = \sqrt{\dfrac{2 \times 735}{1.2 \times 30}} = 6.4\ \text{m/s}$
A safe landing speed.
Balance drag against weight.
$v_T = \dfrac{0.50 \times 9.8}{0.25} = 19.6\ \text{m/s}$
Drag equals weight.
Divide the mass by the drag coefficient.
$\tau = \dfrac{0.50}{0.25} = 2.0\ \text{s}$
The time constant.
Find its speed after one time constant.
A small bead falling through oil feels linear drag and reaches a terminal speed of $4$ cm/s. A bead of the same size and twice the mass is dropped. What is its terminal speed?
Complete the worked solution: a $300$ kg canoe gliding at $7$ m/s feels water drag $bv$ with $b = 50$ kg/s. Find its time constant in s, its speed after a time $\tau\ln 2$ in m/s, and how far it coasts in all, in m.
Divide the mass by the drag coefficient.
$\tau = \dfrac{m}{b} =$ t
The time constant of the decay.
Evaluate the speed after a time $\tau\ln 2$.
$v = v_0e^{-\ln 2} = \tfrac{1}{2}v_0 =$ h
One halving time.
Integrate the speed over all time.
$x = \displaystyle\int_0^\infty v_0e^{-t/\tau}\,dt = v_0\tau =$ d
The speed never reaches zero, but the distance stays finite.
Check the distance another way.
$\text{momentum } mv_0 = \text{total impulse } bx$
Drag's impulse is $b$ times the distance.
Match each drag quantity to its expression.
| $mg/b$ | $m/b$ | $\sqrt{2mg/\rho CA}$ | $v_T(1 - e^{-t/\tau})$ | |
|---|---|---|---|---|
| linear terminal speed | ||||
| time constant | ||||
| quadratic terminal speed | ||||
| speed after release |
A $5$ kg probe is dropped from rest into a fluid that exerts linear drag $bv$ with $b = 5$ kg/s. With $g = 10$ m/s², fill in its terminal speed in m/s, its time constant in s, and its speed one time constant after release in m/s.
| value | |
|---|---|
| terminal speed (m/s) | |
| time constant (s) | |
| speed at one time constant (m/s) |
A motorboat moving at $4$ m/s cuts its engine at $t = 0$. The water's drag is $-bv$, and $m/b = 7$ s. Write its speed $v$, in m/s, as a formula in $t$ (s).
Answer:
A $4$ kg package is released from rest in a fluid with linear drag, $b = 5$ kg/s. How long does it take to reach half its terminal speed, in seconds?
Answer: s to reach half terminal speed
A skydiver over Perris, California, has a mass of $60$ kg with gear and falls belly-down with drag coefficient $C = 1.0$ and frontal area $0.70$ m². With air density $1.2$ kg/m³ and $g = 9.8$ m/s², what is the terminal speed, in m/s?
Answer: m/s terminal speed
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
A motorboat moving at $3$ m/s cuts its engine at $t = 0$. The water's drag is $-bv$, and $m/b = 8$ s. Write its speed $v$, in m/s, as a formula in $t$ (s).
Answer:
You can solve drag problems. Explain to someone why a skydiver stops speeding up even though gravity keeps pulling.
20. Your turn: a $0.50$ kg ball falls in a fluid with linear drag, $b = 0.25$ kg/s. Find its terminal speed and time constant., step 3
$v = 0.632 \times 19.6 = 12.4\ \text{m/s}$
63 percent of the way.