Terminal_velocity

For other uses, see Terminal velocity (disambiguation).

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.

In fluid dynamics, terminal velocity is the velocity at which the fluid resistance force (drag force) of a falling object equals the weight of the object minus the acting force due to fluid, which halts acceleration and causes speed to remain constant.

Contents 1 Terminal velocity with negligible buoyancy force 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 References 4 External links

Terminal velocity with negligible buoyancy force

As the object accelerates (usually downwards movement due to gravity), the drag force acting on the object increases. At a particular speed, the drag force produced will be equal to the downward force, mostly the weight $(mg)$, of the object. 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 at greater than terminal velocity (for example because it was affected by a force downward or it fell from a thinner part of the atmosphere or it changed shape) will slow until it reaches the terminal velocity.

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).Huang, Jian (1999). Speed of a Skydiver (Terminal Velocity). The Physics Factbook. 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 89 m/s), which is also the maximum speed of the peregrine falcon diving on its prey.All About the Peregrine Falcon. U.S. Fish and Wildlife Service (2007-12-20). 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.

An object falling on Earth will fall 9.80 meters per second faster every second (9.8 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

$V\_t=\; \backslash sqrt\{\backslash frac\{2mg\}\{\backslash rho\; A\; C\_d\; \}\}$    ^{(see derivation)}

where

$V\_t$ = terminal velocity,

$m$ = mass of the falling object,

$g$ = gravitational acceleration,

$C\_d$ = drag coefficient,

$\backslash rho$ = density of the fluid the object is falling through, 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.

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),

$F\_\{net\}=m\backslash mathbf\{a\}=m\; g\; -\; \{1\; \backslash over\; 2\}\; \backslash rho\; v^2\; A\; C\_d$.

To make the derivation of the formula more practical a substitution is made with $k=\{1\; \backslash over\; 2\}\; \backslash rho\; A\; C\_d$.

Dividing through by $m$ gives,

$\backslash mathbf\{a\}=g\; -\; \backslash frac\{kv^2\}\{m\}$

where $a$ = the acceleration of the object.

Since

$\backslash mathbf\{a\}=\backslash frac\{dv\}\{dt\}$

the following is true,

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

Thus,

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

Taking the integral of both sides yields,

$\backslash int\_0^t\; \{dt\}=\backslash int\_0^v\; \backslash frac\{dv\}\{g-\backslash frac\{kv^2\}\{m\}\}$.

Factoring $g$ out of the denominator and making the substitution $a=\backslash sqrt\backslash frac\{k\}\{mg\}$ (not to be confused with acceleration) leads to,

$\backslash int\_0^t\; \{dt\}=\{1\; \backslash over\; g\}\backslash int\_0^v\; \backslash frac\{dv\}\{1-a^2\; v^2\}$.

Factoring the bottom,

$\backslash int\_0^t\; \{dt\}=\{1\; \backslash over\; g\}\backslash int\_0^v\; \backslash frac\{dv\}\{(1+av)(1-av)\}$.

Partial fraction expansions tell us that,

$\backslash frac\{1\}\{(1+av)(1-av)\}=\backslash frac\{A\}\{1+av\}+\backslash frac\{B\}\{1-av\}$.

Multiplying by the denominator on the bottom gives,

$\backslash qquad\; 1=A(1-av)+B(1+av)$.

If $v=0$, then

$\backslash qquad\; A+B=1$.

Now let us suppose that $av=1$, then

$\backslash qquad\; 1=A(1-1)+B(1+1)=2B$,

and

$\backslash qquad\; B=\{1\; \backslash over\; 2\}$.

Therefore, for the sum of $A$ and $B$ to be 1,

$\backslash qquad\; A=\{1\; \backslash over\; 2\}$.

With these new expanded fractions,

$\backslash int\_0^t\; \{dt\}=\{1\; \backslash over\; g\}\; \backslash int\_0^v\; \backslash frac\{1/2\}\{1+av\}dv\; +\; \{1\; \backslash over\; g\}\; \backslash int\_0^v\; \backslash frac\{1/2\}\{1-av\}dv$

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

$\backslash int\; \backslash frac\{1/2\}\{1+av\}dv$.

Making a substitution

$\backslash qquad\; u=1+av$,

we find that,

$\backslash qquad\; \{du\; \backslash over\; a\}=dv$,

and thus,

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

Now we want to evaluate the second part of the left hand side,

$\backslash int\; \backslash frac\{1/2\}\{1-av\}$.

Making a substitution

$\backslash qquad\; u=1-av$,

we find that

$\backslash qquad\; -\backslash frac\{du\}\{a\}=dv$,

and thus,

$\backslash int\; \backslash frac\{1/2\}\{1-av\}=-\backslash frac\{1\}\{2a\}\; \backslash int\; \{1\; \backslash over\; u\}\; du=-\backslash frac\{\backslash ln\; u\}\{2a\}+C=-\backslash frac\{\backslash ln(1-av)\}\{2a\}+C$.

Putting these back into our original formula, we get,

$t-0=\{1\; \backslash over\; g\}\backslash left[\{\backslash ln(1+av)\; \backslash over\; 2a\}-\backslash frac\{\backslash ln(1-av)\}\{2a\}\; \backslash right]\_\{v=0\}^\{v=v\}+C=\{1\; \backslash over\; g\}\; \backslash left[\{\backslash ln\; \backslash frac\{1+av\}\{1-av\}\; \backslash over\; 2a\}\; \backslash right]\_\{v=0\}^\{v=v\}+C$,

and thus,

$t=\{1\; \backslash over\; 2ag\}\; \backslash ln\; \backslash frac\{1+av\}\{1-av\}-0+C$.

Since $v=0$ when $t=0$, we find that $C=0$. We also have a specially defined function, the inverse hyperbolic tangent that is defined such that,

$\backslash frac\{1\}\{2\}\; \backslash ln\; \backslash frac\{1+av\}\{1-av\}=\backslash operatorname\{artanh\}(av)$.

So,

$t=\backslash frac\{\backslash operatorname\{artanh\}(av)\}\{ag\}$.

Thus,

$\backslash frac\{1\}\{a\}\backslash tanh(tag)=v$,

and thus, substituting a back in,

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

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

$v=\backslash sqrt\backslash frac\{2mg\}\{\backslash rho\; A\; C\_d\}\; \backslash tanh\; \backslash left(t\; \backslash sqrt\{\backslash frac\{g\; \backslash rho\; A\; C\_d\; \}\{2m\}\}\backslash right)$.

Terminal velocity in the presence of buoyancy force

When the buoyancy effects are taken into account, an object falling through a fluid under it\'s 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

$\backslash quad\; (1)\; \backslash qquad\; W\; =\; F\_b\; +\; D$

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:

$\backslash quad\; (2)\; \backslash qquad\; W\; =\; \backslash frac\{\backslash pi\}\{6\}\; d^3\; \backslash rho\_s\; g$

$\backslash quad\; (3)\; \backslash qquad\; F\_b\; =\; \backslash frac\{\backslash pi\}\{6\}\; d^3\; \backslash rho\; g$

$\backslash quad\; (4)\; \backslash qquad\; D\; =\; C\_d\; \backslash frac\{1\}\{2\}\backslash rho\; V^2\; A$

where

$d\; =$ diameter of the spherical object

$g\; =$ gravitational acceleration,

$\backslash rho\; =$ density of the fluid,

$\backslash rho\_s\; =$ density of the object,

$A\; =\; \backslash pi\; d^2/4\; =$ 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

$\backslash quad\; (5)\; \backslash qquad\; V\_t\; =\; \backslash sqrt\{\backslash frac\{4\; g\; d\}\{3\; C\_d\}\; \backslash left(\; \backslash frac\{\backslash rho\_s\; -\; \backslash rho\}\{\backslash rho\}\; \backslash right)\}$.

Terminal velocity in creeping flow

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, $Re\; \backslash ll\; 1$. The equation of motion for creeping flow (simplified Navier-Stokes equation) is given by

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

where

$\{\backslash mathbf\; v\}$ = velocity vector field

$p$ = pressure field

$\backslash mu$ = 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

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

where the Reynolds number, $Re\; =\; \backslash frac\{\backslash rho\; d\; V\}\{\backslash mu\}$. 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:

$V\_t\; =\; \backslash frac\{g\; d^2\}\{18\; \backslash mu\}\; \backslash left(\backslash rho\_s\; -\; \backslash rho\; \backslash right)$

Applications

The creeping flow results can be applied 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.

References

External links

• Terminal Velocity - NASA site

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