6.5.3 Terminal Speed

Suppose an object starting from rest is pushed or pulled through a fluid by a constant applied force $\vec{F}_{applied}_{}$ . Initially the speed is low and the drag force is small, as shown in FIGURE 6.21a, so the net force causes the object to speed up. But the drag force increases as the speed increases, and eventually the object reaches a speed at which the drag force has exactly the same magnitude as $\vec{F}_{applied}_{}$ . Now the net force is zero, as shown in FIGURE 6.21b, so the object can no longer accelerate but will maintain this steady speed for as long as the force is applied. The steady, unchanging speed at which drag exactly counterbalances an applied force is called the object’s terminal speed.

The figure consists of two parts labeled ("a") and (b) which illustrate the forces acting on an object when it reaches its terminal speed.

FIGURE 6.21 Full Alternative Text

Terminal speed while falling, where gravity is the applied force, is just one example. An airplane reaches its maximum speed—also a terminal speed—when the drag of the air is equal and opposite to the thrust: $F_{drag} = F_{thrust}$ . In the case of an object fired at an initial speed greater than the terminal speed, the very large drag force slows the object until it is moving at the terminal speed.

EXAMPLE 6.8 The terminal speeds of a man and a mouse

A 75 kg skydiver and his 20 g pet mouse falling after jumping from a plane are shown in FIGURE 6.22. Find the terminal speed of each.

The figure consists of two parts. The first part illustrates a man falling and the second part illustrates a mouse falling.

FIGURE 6.22 Full Alternative Text

MODEL Model the man and the mouse as cylinders falling sideways. Gravity is a constant force pulling down on a falling object, while drag due to the air is an upward force. The terminal speed is reached when the drag on each object is equal to the force of gravity.

SOLVE From Figure 6.19 we see that $C_{d} = 1.1$ for a cylinder falling sideways and that a cylinder’s cross-section area as seen from the side is $A = 2 r l$ . With the dimensions given, we can calculate $A_{man} = 0.72 m^{2}$ and $A_{mouse} = 2.1 \times 10^{- 3} m^{2}$ . We assume that the two skydivers are at sufficiently low altitude that we can use the sea-level value of the density of air.

The terminal speed is reached when $F_{drag} = m g$ , or

\frac{1}{2} C_{d} ρ A (v_{term})^{2} = m g

from which we can find the terminal velocity as

v_{term} = \sqrt{\frac{2 m g}{C_{d} ρ A}}

Thus our two skydivers have terminal speeds

\begin{array}{l} v_{man} = \sqrt{\frac{2 (75 kg)(9.8 {m/s}^{2})}{(1.1) (1.2 {kg/m}^{3}) (0.72 m^{2})}} = 39 m / s \\ v_{mouse} = \sqrt{\frac{2 (0.020 kg)(9.8 m / s^{2})}{(1.1) (1.2 kg / m^{3}) (2.1 \times 10^{- 3} m^{2})}} = 12 m / s \end{array}

REVIEW 39 m/s is about 90 mph. Reported terminal speeds for skydivers falling in the prone position are in the 100–120 mph range, so our simple model of the fall gives a result that is close but a bit too low. We’ve probably overestimated both $ρ$ , because skydivers are at a high enough altitude that the air density is lower than at sea level, and A, because their legs are actually spread apart and allow air to flow between them. A more realistic, but also more complex, model would give a better prediction. The mouse, though, falls at a much more modest $12 m / s \approx 25 mph$ . Small animals can usually survive a fall from any height because their terminal speed is not terribly fast.

NOTE The drag force is usually much less than the gravitational force for heavier objects, such as balls or bicycles, moving at not-too-fast speeds through air. We typically neglect drag in our models of these situations because it simplifies the analysis while causing only a very small error in the calculations.

6.5.3 Terminal Speed

FIGURE 6.21 An object reaches its terminal speed when the drag force exactly balances the applied force.

EXAMPLE 6.8 The terminal speeds of a man and a mouse

FIGURE 6.22 A falling skydiver and mouse, and their cross-section areas.