Drag and terminal velocity

AS · 12 min

Real objects do not fall at a constant $9.81\ \text{m s}^{-2}$ for long. Air and liquids push back on anything moving through them, and that push grows with speed until it balances the driving force. This note covers friction and drag qualitatively, motion in a gravitational field with air resistance, and terminal velocity. The syllabus needs no formulae for friction or viscosity, but you must describe and explain the motion using forces and Newton's laws, and interpret the graphs. These explanations are a regular six-mark-style question in Paper 2.

Frictional forces

Friction is the force that opposes the relative motion (or attempted motion) of two surfaces in contact. It acts parallel to the surfaces, in the direction opposite to the sliding.

If you push a heavy box gently and it does not move, static friction matches your push exactly, so the resultant force is zero.
Once the box slides, friction acts against the direction of motion. On a level floor it is roughly constant regardless of speed.
Friction converts kinetic energy into internal energy (thermal energy) of the surfaces.

No treatment of the coefficient of friction is required.

Drag (viscous) forces

A drag force (or viscous force) acts on an object moving through a fluid (a liquid or a gas). Air resistance is the drag force in air. Drag:

always acts in the direction opposite to the velocity of the object relative to the fluid;
is zero when the object is at rest relative to the fluid;
increases as the speed increases;
depends on the shape and cross-sectional area of the object (streamlining reduces it), and on the fluid (drag in water is much bigger than in air at the same speed).

Key result

The simple model you need: drag increases with speed. A useful version for calculations, when a question gives it, is

D = kv^2

where $k$ is a constant for a given object and fluid. (For slow, small objects in viscous liquids drag is closer to proportional to $v$ .) You will always be told which model to use.

Falling with air resistance

Think about a skydiver jumping from a hovering helicopter (so the initial velocity is zero). Two forces act: weight $W = mg$ (constant, downwards) and air resistance $D$ (upwards, growing with speed).

Method

Explaining motion with drag, stage by stage

Name the forces and their directions.
Say how the drag changes as the speed changes.
Compare the forces to find the resultant, and use $F = ma$ to say what happens to the acceleration.
Conclude what the velocity does.

Stage 1: just after jumping. The speed is zero, so the drag is zero. The only force is weight, so the resultant force is $mg$ and the acceleration is $g = 9.81\ \text{m s}^{-2}$ .

Stage 2: speeding up. As the speed increases, drag increases. The resultant force $mg - D$ decreases, so the acceleration decreases. The skydiver still speeds up, but more and more slowly.

Stage 3: terminal velocity. Eventually the drag equals the weight. The resultant force is zero, so the acceleration is zero (Newton's first law) and the skydiver falls at a constant speed, the terminal velocity.

Definition

Terminal velocity is the constant (maximum) velocity reached by an object moving through a fluid when the resistive force on it equals the driving force (for a falling object, when drag plus any upthrust equals its weight), so that the resultant force and the acceleration are zero.

Stage 4: parachute opens. The area increases enormously, so at the same speed the drag becomes much larger than the weight. The resultant force is now upwards, so the skydiver decelerates. As the speed falls, the drag falls, until it again equals the weight at a new, much lower terminal velocity.

y = 50 tanh(0.1962 x) (1 - sign(x - 20))/2 + (5 + 45 exp(-(x - 20)/1.2)) (1 + sign(x - 20))/2

The velocity–time graph of the skydiver (downwards positive). The gradient starts at $9.81\ \text{m s}^{-2}$ and falls to zero as the first terminal velocity ( $50\ \text{m s}^{-1}$ here) is approached. When the parachute opens at $20\ \text{s}$ the velocity falls steeply (a large negative gradient), then levels off at the second terminal velocity ( $5\ \text{m s}^{-1}$ ).

Watch out

When the parachute opens, the skydiver does not start moving upwards. The velocity is still downwards; it is the acceleration (and the resultant force) that is upwards. Many answers say "the skydiver goes up" and lose the mark.

Also, at terminal velocity there is still a weight and still a drag: they are equal, not absent. "There are no forces acting" is wrong.

The acceleration–time graph for stages 1 to 3 starts at $9.81\ \text{m s}^{-2}$ and decreases (more and more slowly) towards zero. The displacement–time graph starts as a curve getting steeper and becomes a straight line once terminal velocity is reached.

Projectiles with air resistance

For an object thrown at an angle, drag acts opposite to the velocity, so it has a horizontal component (slowing the horizontal motion) and a vertical component (opposing whichever way the object is moving vertically). The result is a lower, shorter, asymmetric path (see Projectile motion). On the way up, weight and drag both act downwards, so the vertical deceleration is greater than $g$ ; on the way down, drag acts upwards, so the acceleration is less than $g$ . The ball therefore takes longer to fall than to rise.

Terminal velocity in other situations

The same reasoning applies to anything with a constant driving force and a resistive force that grows with speed.

A car on a level road with a constant driving force accelerates until air resistance and friction add up to the driving force; that speed is its top speed. (When the limit is the engine's power rather than a constant force, see Power.)
A ball bearing dropped into oil reaches terminal velocity quickly because drag in a viscous liquid is large. In a liquid the upthrust matters too: at terminal velocity, weight $=$ upthrust $+$ drag.
A cyclist freewheeling down a hill reaches a steady speed when the component of weight down the slope equals the resistive forces.

Drag proportional to speed squared

A skydiver of total mass $80\ \text{kg}$ has a terminal velocity of $50\ \text{m s}^{-1}$ . Assume the drag force is $D = kv^2$ . (a) Find $k$ . (b) Find the acceleration of the skydiver when the speed is $30\ \text{m s}^{-1}$ .

Solution

(a) At terminal velocity, drag $=$ weight:

k(50)^2 = 80 \times 9.81 = 784.8\ \text{N} \quad\Rightarrow\quad k = \frac{784.8}{2500} = 0.314\ \text{kg m}^{-1}

(The unit of $k$ is $\text{N}/(\text{m s}^{-1})^2 = \text{kg m}^{-1}$ .)

(b) At $30\ \text{m s}^{-1}$ : $D = 0.314 \times 30^2 = 283\ \text{N}$ .

a = \frac{mg - D}{m} = \frac{784.8 - 282.5}{80} = 6.3\ \text{m s}^{-2} \text{ downwards}

Opening the parachute

The same skydiver opens the parachute while falling at $50\ \text{m s}^{-1}$ . With the parachute open, $k$ becomes $31.4\ \text{kg m}^{-1}$ . Calculate (a) the new terminal velocity and (b) the acceleration immediately after the parachute opens.

Solution

(a) $kv^2 = mg$ :

v = \sqrt{\frac{784.8}{31.4}} = \sqrt{25.0} = 5.0\ \text{m s}^{-1}

(b) At $50\ \text{m s}^{-1}$ with the new $k$ : $D = 31.4 \times 2500 = 7.85 \times 10^{4}\ \text{N}$ .

a = \frac{mg - D}{m} = \frac{784.8 - 78\,500}{80} = -971\ \text{m s}^{-2}

A deceleration of about $970\ \text{m s}^{-2}$ , nearly $100g$ . Real parachutes open over a second or two precisely to avoid this: the model assumes the full area appears instantly.

Explaining a velocity–time graph

A small steel ball is released from rest at the surface of a tall cylinder of oil. Describe and explain how its velocity changes as it falls.

Solution

Forces on the ball: weight downwards (constant), upthrust upwards (constant, since the volume of oil displaced does not change), and viscous drag upwards (increasing with speed).

At release the speed is zero, so drag is zero; the resultant force (weight minus upthrust) is downwards and the ball accelerates.

As the speed increases, the drag increases, so the resultant force decreases and the acceleration decreases.

When drag $+$ upthrust $=$ weight, the resultant force is zero, the acceleration is zero, and the ball moves at a constant terminal velocity.

The initial acceleration is less than $g$ , because the upthrust already opposes the weight at the start.

Measuring terminal velocity in a liquid

Apparatus. A tall, wide measuring cylinder (or tube) of a viscous liquid such as glycerol or oil; small steel ball bearings of different diameters; rubber bands around the cylinder as markers; a metre rule; a stop-watch; a magnet to retrieve the balls; a micrometer to measure the diameters.

Method.

Place three or more rubber bands around the tube at equal spacings (for example $10.0\ \text{cm}$ ), the top one well below the surface so the ball has reached terminal velocity before it.
Drop a ball from just above the surface, at the centre of the tube.
Time the ball between successive bands. If the times for equal distances are equal, the ball is at terminal velocity: $v = \text{distance}/\text{time}$ .
Repeat for each ball several times and average; vary the diameter to investigate how terminal velocity depends on size.

Sources of uncertainty and improvements.

Parallax when judging when the ball passes a band: view at eye level, or film the fall with a scale behind the tube and step through frame by frame.
Reaction time over short intervals: use longer distances between bands, or light gates.
The ball may not have reached terminal velocity at the first band: check that consecutive intervals give the same time.
The walls of the tube slow the ball: use a wide tube and drop the ball down the centre.
Viscosity depends strongly on temperature: keep the temperature constant and record it.

Exam tip

"Describe and explain" questions are marked point by point: forces named with directions; drag increases with speed; resultant force decreases; acceleration decreases; drag equals weight so resultant force is zero; constant (terminal) velocity. Use the words "resultant force" and "acceleration" explicitly.
On a sketched $v$ – $t$ graph, the line must start with a gradient of $g$ (if released from rest in air), curve smoothly and become horizontal; it must never overshoot and come back down unless the drag suddenly increases (parachute).
When a constant upthrust is present, the starting acceleration is less than $g$ . Say so.
If a question gives a model such as $D = kv^2$ , write the terminal condition $kv_t^2 = mg$ as the first line: that equation usually earns the first mark.

Summary

Friction opposes sliding between surfaces; drag opposes motion through a fluid and increases with speed.
Falling from rest in air: acceleration starts at $g$ and decreases as drag grows.
Terminal velocity is reached when the resistive forces equal the driving force (drag $+$ upthrust $=$ weight): resultant force zero, acceleration zero.
Opening a parachute increases drag, giving an upward resultant force and a deceleration to a lower terminal velocity; the velocity stays downwards.
In liquids include upthrust; the initial acceleration is less than $g$ .
The same reasoning gives the top speed of a car or cyclist.

Practice

Question

State two factors that affect the drag force on an object moving through air.
Explain why a skydiver's acceleration is $9.81\ \text{m s}^{-2}$ at the moment of jumping from a stationary balloon, but decreases afterwards.
Sketch the acceleration–time graph for an object released from rest that falls through air and reaches terminal velocity.
A sphere falls through oil. Its weight is $0.050\ \text{N}$ and the upthrust on it is $0.012\ \text{N}$ . Calculate the drag force at terminal velocity.
A car of mass $1200\ \text{kg}$ has a constant driving force of $2400\ \text{N}$ . The total resistive force is $D = 0.80v^2$ (in newtons, with $v$ in $\text{m s}^{-1}$ ). Calculate (a) its top speed and (b) its acceleration at $25\ \text{m s}^{-1}$ .
Two balls of the same size, one steel and one plastic, are dropped from a high tower. Explain which reaches the ground first.
A ball is thrown vertically upwards in air. Explain why the time it takes to fall back is longer than the time it takes to rise.
In a terminal-velocity experiment, a ball takes $1.62\ \text{s}$ , $1.60\ \text{s}$ and $1.61\ \text{s}$ to fall between successive markers $20.0\ \text{cm}$ apart. State whether the ball has reached terminal velocity and calculate it.
A skydiver of mass $75\ \text{kg}$ falls with drag $D = kv^2$ , where $k = 0.25\ \text{kg m}^{-1}$ . Calculate the terminal velocity and the speed at which the acceleration is half of $g$ .
A skydiver is falling at terminal velocity. Describe and explain the motion from the moment the parachute opens until a new steady state is reached. Sketch the velocity–time graph for this part of the motion.

Answers

Any two: speed; cross-sectional area (or shape / streamlining); the density (or nature) of the fluid.
At the moment of jumping the speed is zero, so there is no air resistance and the only force is weight: $a = g$ . As speed increases, air resistance increases, so the resultant force ( $W - D$ ) decreases and, by $F = ma$ , the acceleration decreases.
Starts at $9.81\ \text{m s}^{-2}$ at $t = 0$ , decreases along a curve whose gradient becomes less steep, approaching zero (the time axis) asymptotically.
Weight $=$ upthrust $+$ drag: $D = 0.050 - 0.012 = 0.038\ \text{N}$ .
(a) Top speed when $0.80v^2 = 2400$ : $v = \sqrt{3000} = 54.8\ \text{m s}^{-1}$ . (b) $D = 0.80 \times 625 = 500\ \text{N}$ ; $a = (2400 - 500)/1200 = 1.58\ \text{m s}^{-2}$ .
At the same speed both have the same drag (same size and shape), but the steel ball has a much larger weight. Its resultant force per unit mass is larger, so its acceleration is larger at each speed, and it reaches a higher terminal velocity (drag must grow larger to balance its weight). The steel ball lands first.
On the way up, drag and weight both act downwards, so the deceleration is greater than $g$ . On the way down, drag acts upwards against the weight, so the acceleration is less than $g$ . Also energy is lost, so the ball returns at a lower speed. The average speed on the way down is lower, so it takes longer to cover the same distance.
The times are equal within the uncertainty ( $1.61 \pm 0.01\ \text{s}$ ), so the speed is constant: it has reached terminal velocity. $v = 0.200/1.61 = 0.124\ \text{m s}^{-1}$ .
$kv_t^2 = mg$ : $v_t = \sqrt{75 \times 9.81/0.25} = \sqrt{2943} = 54.2\ \text{m s}^{-1}$ . Half of $g$ when $D = \tfrac{1}{2}mg$ : $v = v_t/\sqrt{2} = 38.4\ \text{m s}^{-1}$ .
When the parachute opens, the area (and so the drag at that speed) increases greatly, so drag becomes greater than weight. The resultant force is upwards, so the skydiver decelerates (still moving downwards). As the speed decreases, the drag decreases, so the resultant force and the deceleration decrease. When drag again equals weight, the resultant force is zero and the skydiver moves at a new, lower, constant terminal velocity. Graph: horizontal line at the first terminal velocity; at the moment of opening, a steep drop that curves and levels off at a lower horizontal line; the velocity never becomes zero or negative.

Frictional forces

Drag (viscous) forces

Falling with air resistance

Projectiles with air resistance

Terminal velocity in other situations

Summary

Practice

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