Terminal velocity

El directorio enciclopédico desde la Wikipedia.

For other uses, see Terminal velocity (disambiguation) and Terminal.

An object reaches terminal velocity when the downward force of gravity (F_g) equals the upward force of drag (F_d). The net force on the body is then zero, and the result is that the velocity of the object remains constant.

A free falling object achieves its terminal velocity when the downward force of gravity (F_g)equals the upward force of drag (F_d). This causes the net force on the object to be zero, resulting in an acceleration of zero. Mathematically an object asymptotically approaches and can never reach its terminal velocity.

As the object accelerates (usually downwards due to gravity), the drag force acting on the object increases. At a particular speed, the drag force produced will equal the object's weight $(m g)$ . Eventually, it plummets at a constant speed called terminal velocity (also called settling velocity). Terminal velocity varies directly with the ratio of drag to weight. More drag means a lower terminal velocity, while increased weight means a higher terminal velocity. An object moving downward with greater than terminal velocity (for example because it was affected by a downward force or it fell from a thinner part of the atmosphere or it changed shape) will slow until it reaches terminal velocity.

1 Examples
- 1.1 Velocity of a falling object after a given time
2 Terminal velocity in the presence of buoyancy force
- 2.1 Terminal velocity in creeping flow
- 2.2 Applications
3 See also
4 References
5 External links

[edit] Examples

For example, the terminal velocity of a skydiver in a free-fall position with a semi-closed parachute is about 195 km/h (120 mph or 55m/s).^[1] This velocity is the asymptotic limiting value of the acceleration process, since the effective forces on the body more and more closely balance each other as the terminal velocity is approached. In this example, a speed of 50% of terminal velocity is reached after only about 3 seconds, while it takes 8 seconds to reach 90%, 15 seconds to reach 99% and so on. Higher speeds can be attained if the skydiver pulls in his limbs (see also freeflying). In this case, the terminal velocity increases to about 320 km/h (200 mph or 90 m/s),^[1] which is also the terminal velocity of the peregrine falcon diving down on its prey,^[2]. And the same terminal velocity is reached for a typical 150 g bullet travelling in the downward vertical direction — when it is returning to earth having been fired upwards, or perhaps just dropped from a tower — according to a 1920 U.S. Army Ordnance study.^[3]

Competition speed skydivers fly in the head down position reaching even higher speeds. The current world record is 614 mph (988 km/h) by Joseph Kittinger, set at high altitude where the lesser density of the atmosphere decreased drag.^[1]

An object falling on Earth will fall 9.81 meters per second faster every second (9.81 m/s²). The reason an object reaches a terminal velocity is that the drag force resisting motion is directly proportional to the square of its speed. At low speeds, the drag is much less than the gravitational force and so the object accelerates. As it accelerates, the drag increases, until it equals the weight. Drag also depends on the projected area. This is why things with a large projected area, such as parachutes, have a lower terminal velocity than small objects such as cannon balls.

Mathematically, terminal velocity, without considering the buoyancy effects, is given by

$V_t= \sqrt{\frac{2mg}{\rho A C_d }}$ ^{(see derivation)}

where

V t

= terminal velocity,

m

= mass of the falling object,

g

= gravitational acceleration,

C d

= drag coefficient,

ρ

= density of the fluid through which the object is falling, and

A

= projected area of the object.

On Earth, the terminal velocity of an object changes due to the properties of the fluid, mass and the projected area of the object.

This equation is derived from the drag equation by setting drag equal to mg, the gravitational force on the object.

Density increases with decreasing altitude, ca. 1% per 80 m (see barometric formula). Therefore, for every 160 m of falling, the terminal velocity decreases 1%. After reaching the local terminal velocity, while continuing the fall, speed decreases to change with the local terminal velocity.

[edit] Velocity of a falling object after a given time

Mathematically, defining down to be positive, the net force acting on an object falling near the surface of Earth is (according to the drag equation),

$F_{net}=ma=m g - {1 \over 2} \rho v^2 A C_d$ .

To make the solution of the velocity v from this formula more practical, the substitution $k={1 \over 2} \rho A C_d$ is made.

Dividing both sides by m, and using that the acceleration is $a=\frac{dv}{dt}$ , gives

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

Derivation of the solution for the velocity v as a function of time t

Thus,

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

Taking the integral of both sides yields,

$\int_0^t {dt'}=\int_0^v \frac{dv}{g-\frac{kv^2}{m}}={1 \over g}\int_0^v \frac{dv}{1-\alpha^2 v^2},$

where $\alpha=\sqrt\frac{k}{mg}.$ Factoring the bottom,

$\int_0^t {dt'}={1 \over g}\int_0^v \frac{dv}{(1+\alpha v)(1-\alpha v)}$ .

Partial fraction expansions tell us that,

$\frac{1}{(1+\alpha v)(1-\alpha v)}=\frac{1/2}{1+\alpha v}+\frac{1/2}{1-\alpha v}$ .

With these expanded fractions,

$\int_0^t {dt'}={1 \over g} \int_0^v \frac{1/2}{1+\alpha v}dv + {1 \over g} \int_0^v \frac{1/2}{1-\alpha v}dv$

We now want to evaluate both integrals on the left side. The first one is,

$\int \frac{1/2}{1+\alpha v}dv$ .

Making a substitution

$\qquad u=1+\alpha v$ ,

we find that,

$\qquad {du \over \alpha}=dv$ ,

and thus,

$\int \frac{1/2}{1+\alpha v}dv={1 \over 2a}\int \frac{1}{u}du={\ln u \over 2\alpha}+C={\ln(1+\alpha v) \over 2\alpha}+C$ .

The second part of the integral at the right hand side can be evaluated by repeating the above with -α in place of α. As a result

$\int \frac{1/2}{1-\alpha v}=-\frac{1}{2\alpha} \int {1 \over u} du=-\frac{\ln u}{2\alpha}+C=-\frac{\ln(1-\alpha v)}{2\alpha}+C$ .

Putting these back into our original formula, we get,

$t-0={1 \over g}\left[{\ln(1+\alpha v) \over 2\alpha}-\frac{\ln(1-\alpha v)}{2\alpha}+C \right]_{v=0}^{v=v}={1 \over g} \left[{\ln \frac{1+\alpha v}{1-\alpha v} \over 2\alpha}+C \right]_{v=0}^{v=v}.$

So:

$t={1 \over 2\alpha g} \ln \frac{1+\alpha v}{1-\alpha v}.$

The inverse hyperbolic tangent is defined as:

$\frac{1}{2} \ln \frac{1+\alpha v}{1-\alpha v}=\operatorname{artanh}(\alpha v)$ .

So the solution of the integral is,

$t=\frac{\operatorname{artanh}(\alpha v)}{\alpha g}$ .

Thus: $\frac{1}{\alpha}\tanh(t\alpha g)=v,$ and, substituting α back in,

$v=\sqrt{\frac{mg}{k}} \tanh \left(\sqrt{\frac{k}{mg}}gt\right).$

Substituting and simplifying, assuming that g is positive (which it was defined to be), we arrive at the final equation.

If at time t=0 the velocity v is equal to zero, the velocity is found to be equal to

$v=\sqrt\frac{2mg}{\rho A C_d} \tanh \left(t \sqrt{\frac{g \rho A C_d }{2m}}\right),$

with $tanh$ the hyperbolic tangent function. As time $t\rightarrow\infty$ , the hyperbolic tangent $\tanh \left(t \sqrt{\frac{g \rho A C_d }{2m}}\right)\rightarrow 1$ , which results in the terminal velocity $\sqrt\frac{2mg}{\rho A C_d}$ .

[edit] Terminal velocity in the presence of buoyancy force

When the buoyancy effects are taken into account, an object falling through a fluid under its own weight can reach a terminal velocity (settling velocity) if the net force acting on the object becomes zero. When the terminal velocity is reached the weight of the object is exactly balanced by the upward buoyancy force and drag force. That is

$\quad (1) \qquad W = F_b + D$

where

W

= weight of the object,

F b

= buoyancy force acting on the object, and

D

= drag force acting on the object.

If the falling object is spherical in shape, the expression for the three forces are give below:

$\quad (2) \qquad W = \frac{\pi}{6} d^3 \rho_s g$

$\quad (3) \qquad F_b = \frac{\pi}{6} d^3 \rho g$

$\quad (4) \qquad D = C_d \frac{1}{2}\rho V^2 A$

where

d =

diameter of the spherical object

g =

gravitational acceleration,

ρ =

density of the fluid,

ρ s =

density of the object,

A = π d 2 / 4 =

projected area of the sphere,

C d =

drag coefficient, and

V =

characteristic velocity (taken as terminal velocity,

V t

Substitution of equations (2–4) in equation (1) and solving for terminal velocity, $V t$ to yield the following expression

$\quad (5) \qquad V_t = \sqrt{\frac{4 g d}{3 C_d} \left( \frac{\rho_s - \rho}{\rho} \right)}$ .

[edit] Terminal velocity in creeping flow

Creeping flow past a sphere: streamlines, drag force F_d and force by gravity F_g.

For very slow motion of the fluid, the inertia forces of the fluid are negligible (assumption of massless fluid) in comparison to other forces. Such flows are called creeping flows and the condition to be satisfied for the flow to be creeping flows is the Reynolds number, $Re \ll 1$ . The equation of motion for creeping flow (simplified Navier-Stokes equation) is given by

$\nabla p = \mu \nabla^2 {\mathbf v}$

where

${\mathbf v}$ = velocity vector field

p

= pressure field

μ

= fluid viscosity

The analytical solution for the creeping flow around a sphere was first given by Stokes in 1851. From Stokes' solution, the drag force acting on the sphere can be obtained as

$\quad (6) \qquad D = 3\pi \mu d V \qquad \qquad \text{or} \qquad \qquad C_d = \frac{24}{Re}$

where the Reynold's number, $Re = \frac{\rho d V}{\mu}$ . The expression for the drag force given by equation (6) is called Stokes law.

When the value of $C d$ is substituted in the equation (5), we obtain the expression for terminal velocity of a spherical object moving under creeping flow conditions:

$V_t = \frac{g d^2}{18 \mu} \left(\rho_s - \rho \right).$

[edit] Applications

The creeping flow results can be applied in order to study the settling of sediment particles near the ocean bottom and the fall of moisture drops in the atmosphere. The principle is also applied in the falling sphere viscometer, an experimental device used to measure the viscosity of high viscous fluids.

[edit] See also

[edit] References

^ ^a ^b ^c Huang, Jian (1999). "Speed of a Skydiver (Terminal Velocity)". The Physics Factbook. Glenn Elert, Midwood High School, Brooklyn College.
^ "All About the Peregrine Falcon". U.S. Fish and Wildlife Service (2007-12-20).
^ The Ballistician (March 2001). "Bullets in the Sky". W. Square Enterprises, 9826 Sagedale, Houston, Texas 77089.

[edit] External links

Terminal Velocity - NASA site

Página espejo de la Wikipedia

Directorio de Enlaces Directorio dmoz Directorio espejo dmoz Pedro Bernardo

Terminal velocity

El directorio enciclopédico desde la Wikipedia.

Contents

[edit] Examples

[edit] Velocity of a falling object after a given time

[edit] Terminal velocity in the presence of buoyancy force

[edit] Terminal velocity in creeping flow

[edit] Applications

[edit] See also

[edit] References

[edit] External links