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Terminal Velocity Equation

by Jamez
Tags: equation, terminal, velocity
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Jamez
#1
Aug28-04, 07:41 AM
P: 19
i'm looking for an equation to calculate terminal velocity. Does any one know it? and can u please post in here please.
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arildno
#2
Aug28-04, 08:52 AM
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That would depend upon how you choose to model the air/fluid resistance.
HallsofIvy
#3
Aug28-04, 10:09 AM
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Thanks
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One common model is that the resistance force is proportional to the speed.
Under that model, an object falling, under gravity has acceleration -g+ kv (k is the proportionality constant, v the speed. Since that is a function of v, it give the linear differential equation mdv/dt= -mg+ kv. The general solution to that is v(t)= Ce-kt/m-mg/k. For very large t, that exponential (with negative exponent) goes to 0 and the "terminal velocity" is -mg/k.

Another common model is to set the resistance force proportional to the square of the speed. That means the net force is -g+ kv2 and v satisfies the differential equation mdv/dt= -g+ kv2. That's a non-linear differential equation but is separable and first order. We can integrate it by writing
dv/(kv2-g)/m= (-1/2√(g))(1/(√(k)v+√(g))dv/m+(1/2√(g))(√(k)v-&radic(g))dv/m= dt.
Integrating both sides, we get (1/2√(kg))ln((√(k)v-√(g))/(√(k)v+√(g))= mt+ C. For large t, the denominator on the left must go to 0: the terminal velocity is -√(g/k) which, you will notice, is independent of m. This model is typically used for very light objects falling through air or objects falling through water.

da_willem
#4
Aug28-04, 10:38 AM
P: 599
Terminal Velocity Equation

A falling object on earth is subjected to a downward force [itex]F_g=mg[/itex], while it's air resitance constitutes an upward force often modelled by

[tex]F_w=\frac{1}{2}C_D \rho A_F v^2[/tex].

With [itex]C_D[/itex] a constant (drag coefficient) that models how aerodynamic the object is, for most object of the order 1 (for a raindrop for example ~0,5), [itex]\rho[/itex] the air density, [itex]A_F[/itex] the frontal area of the object (perpendicular to the direction of motion), and [itex]v[/itex] the velocity of the object.

At terminal velocity the force on the object is zero (otherwise the object would accellerate!) so you can equate both force-equations yielding:

[tex]mg=\frac{1}{2}C_D \rho A_F v^2[/tex]
[tex]v=\sqrt{\frac{2mg}{C_D \rho A_F}}[/tex]
hoppingbuffalo
#5
Jun23-06, 07:53 PM
P: 1
Indeed, terminal velocity depends on mass. You forgot the FORCE due to gravity is mg, not g.


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