hello
I'm a researcher in metallurgy and I am looking for something that would be more ''common'' for physicists! I want to find the terminal velicity of a particle of diameter 400μm , density 2,3 gr/cm3. And I used some equations.. so I found it 2,4 m/sec. now I want to find the distance it need for the particle to reach terminal velocity. In addition I want to know the X(t) of the particle from the beginning of the free fall. Can anybody help me please? Thanks
Cd is unknown:mad:

I would like to add that any literature that is related to the subject I am interested in is welcome:):)

Unfortunately, this is a little trickier than one might think at first. If air resistance wasn't an issue (which it clearly IS in this case since you're talking about terminal velocity) the equation would simply be

x(t) = 1/2 * g * t^2 + v0*t + x0,

where g is the acceleration of gravity, (typically 9.8 m/s^2 at ground level on the Earth), v0 is the initial velocity of the particle, and x0 is the initial position of the particle when you start the clock.

However, in your case the particle is being slowed by air resistance (friction). Kinetic frictional forces typically are proportional to velocity. At terminal velocity the downward force of gravity (m*g, the mass of the particle * 9.8 m/s^2) is exactly counteracted by the upward force of air resistance (k*v, where v is the velocity and k is some number proportional to the coefficient of kinetic friction).

m*g = k*VT,
where VT is the terminal velocity.

Okay! So now the fun part: How do we figure out the distance traveled before reaching terminal velocity and the position as a function of time?

Well, instead of a nice easy formula like before, as you'll see now we get a nonhomogeneous differential equation. First, we start with Newton's equation F = ma. In this case, the force means the net force experienced by the particle which, as we already discussed, is the difference of the downward force of gravity (m*g) and the upward force of the friction (k*V). The acceleration, a(t), is by definition the first derivative of the velocity with respect to time, dV(t)/dt. Therefore our equation F=ma becomes:

m*g - k*V(t) = m*dV(t)/dt

(Note that I've arbitrarily chosen the "positive" direction for the force to be downward. Thus in my coordinate system gravity exerts a positive force, air resistance a negative force, and the downward velocity is a positive number).

Or, substituting k=mg/VT and rearranging a little,

-(g/VT)*V(t) + g = dV(t)/dt

The solution to this differential equation is

V(t) = A*exp(-g*t/VT)+VT,

where A is some unknown constant, and exp(x) means e^x, where e is the natural number.

Knowing the initial condition that the velocity at t=0 is zero (assuming the particle begins from a resting state), we can solve V(0) = 0 to find that A=-VT.

Thus,

V(t) = VT*(1-exp(-g*t/VT))

or

V(t) = 2.4*(1-exp(-4.083*t)

To find position, x(t), we can simply integrate the velocity with respect to time:

x(t) = integral (V(t) dt)
x(t) = 2.4*integral(1-exp(-4.083*t) dt)
x(t) = 2.4*t+0.588*exp(-4.083*t) + x0,

where x0 is the initial height. Again, keep in mind that I chose the positive direction to be pointing downward, so an increasing value of x means that the object is falling further and further. If you just want to know how far the object has fallen since it started, x0 = 0.

Now that you see that the equation for the velocity is a decaying exponential, it should be obvious that the particle would actually have to fall an infinite distance before reaching terminal velocity. However, you can numerically calculate how long it would take to reach, say, 99.9% of the terminal velocity. Plugging that time into the equation for x(t) would then yield the distance necessary to reach that speed.

I hope this helps. My differential equations are pretty rusty, and I didn't check my math really carefully, so it's possible I've made a mistake or two here or there, but the form of the answer (a decaying exponential) should be the same.

I found a couple of online calculators and references:

Terminal Velocity

Terminal velocity - Wikipedia, the free encyclopedia

Maximum Falling Velocity

More references from Google

According to the first link Perito posted, the drag force is proportional to the velocity squared. Unfortunately, that complicates the differential equation, which now becomes:

-(g/VT^2)*V(t)^2 + g = dV(t)/dt

Having forgotten most of what I ever knew about solving differential equations, I can't tell you the solution off the top of my head, but I'm sure it's still some sort of decaying exponential.

I was going to post a solution, but I realized that I've forgotten a lot of what I knew about DIs also. I'll brush up and post. Maybe Galactus can answer.