# When is terminal velocity reached

#### How much rain falls in a bucket per second?

The top of a bucket has area A. Raindrops of mass m are falling at a constant speed v (their terminal velocity). There are n drops per m 3. Let's look at the cylinder (length L, area A) above the bucket in the diagram at right. It contains a number of raindrops given by- n*volume = nAL.

- t = L / v

- I = mass / time = (nALm) / (L / v) = nAvm. (1)

- weight = drag force.

_{drag}. We write:

- K

_{drag}v = F

_{drag}= weight = mg.

- I = mass / time = nAvm = nAm 2 g / K

_{drag}

- proportional to the number of mass carriers / cubic metre of air (n)
- proportional to the cross sectional area of 'conductor' (A)
- proportional to the square of the mass on an individual mass carrier (m)
- proportional to the magnitude of the field that moves them (g)
- inversely proportional to the proportionality constant for drag (K
_{drag})

_{drag}is only constant if the size is unchanged, so this proportionality refers to changing the mass by changing the density of the material from which the drops are made.) Let's now derive Ohm's law for gravity and raindrops. We'll need the gravitational potential V

_{grav}. which is the gravitational potential energy per unit mass, exactly analogous to the electrical case. Take zero potential energy at height h = 0 and we get:

- V

_{grav}= U

_{grav}/ m = mgh / m = gh.

other, the height changes by L, so the (gravitational) potential difference across it is V_{grav} = gL. So, to get the gravitational version of Ohm's law:

- V

_{grav}/ I = (gL)K

_{drag}/ nAm 2 g = = LK

_{drag}/ nAm 2 .

_{drag}. n and m) into one new constant for this particular air+rain mix:

- V

_{grav}/ I = (K

_{drag}/ nm 2 ).(L / A) = ρ.(L / A), where ρ = K

_{drag}/ nm 2

**resisitivity**of this particular air+rain mix. (In terms of the analogy, this mix is analogous to a particular substance that conducts electricity.) Now let's think about this particular cylinder, with its given length and cross section (analogous to a particular piece of wire).

- V

_{grav}/ I = R where R = ρL / A,

I = nAqv = n.A.const.E = const.V.nqA / L. (3)

Here we note that the current is:- proportional to the number of charge carriers/cubic metre (n)
- proportional to the cross sectional area of the conductor (A)
- proportional to the magnitude of the field that moves them (E)
- proportional to the number of charge carriers/cubic metre of conductor (n)

- V/I = (L/A)(1/(nq.const.)), which we can write as

V/I = (L/A)ρ = R

where R is the **resistance** of this piece of conducting material (whose units are ohms, Ω), and where ρ is the **resistivity** of the material (whose units are ohm metress, Ωm). Alternatively, we can take the reciprocal of the terms in the above equation and write:

- I/V = σ(A / L) = G

**conductance**of this piece of conductor, (G = 1/R), and where σ is the

**conductivity**of the material,

- σ = n.q.const = 1 / ρ.

**current density**or current per unit area: J = I/A. Combining the equations above gives the simplest form:

**J**= σ

**E**=

**E**/ ρ.

**E**so in this case the equation has been written in vector form. Source: www.animations.physics.unsw.edu.au

Category: Forex

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