Dimensional Analysis: Terminal Velocity

[hibachii]

Homework Statement

The terminal velocity of a mass m, moving at 'high speeds' through a fluid of density
ρ(kg m^-3), is given by v = sqrt(2mg/DρA) where A is the cross sectional area of the object (m^2) and D a dimensionless "drag coefficient".

i) Show that equation is dimensionally correct
ii) Estimate the terminal velocity of an Australian $1 coin. Take D to be ~0.3.

Homework Equations

v=sqrt(2mg/DρA)

where: v=terminal velocity
m=mass of object(kg)
g=gravity(ms^-2)
D=dimensionless drag coefficient
A=cross sectional area (m^2)
ρ=density of fluid (kg m^-3)

The Attempt at a Solution

A free-falling object achieves its terminal velocity when the downward force of gravity (Fg) equals the upward force of drag (Fd), hence the net force on the object is equal to zero.

taking downwards as positive:

F(net) = Fg-Fd
0 = Fg-Fd
Fd=Fg
(D ρ A v^2 )/2 = mg
D ρ A v^2 = 2mg
v^2 = 2mg/D ρ A
v= sqrt(2mg/D ρ A)

hence v(terminal)= sqrt(2mg/D ρ A)


is this right? I'm not sure if this is the right way to tackle this question because I've only derived the equation because I don't know how to dimensionally analyse it.
Please Help. Thanks!

 

[Saladsamurai]

Hi hibachii! It's always good to be able to derive this equation, but the question is more straightforward than you are making it

For part (i) It simply wants you to check the dimensions on the equation: v=sqrt(2mg/DρA). You know the dimension of the left side is the dimension of velocity, namely v = [LT^-1] (unit length per unit time). So plug in all of the dimensions on the right side and see if they match the left side.

For part (ii) this should be plug and chug. Presumably you were either given, or can find, the area of an Australian $1 coin. And you will of course need the density of the fluid in which this coin is traveling! Then just plug it in.

Hope that helps :smil:

 

[hibachii]

Hi salad samurai. I know about the dimension of v but however i have no clue of the dimensions of all the other variables like m, g, D, rho, and A

 

[Saladsamurai]

You already told us what the dimension on D is

g is gravity. What are typical units of g? If you can answer that, you can answer what are dimensions on g. m is mass, so it's dimension is simply M (mass). Remember, we can break any physical quantity down into the fundamental dimensions of mass (M), length (L), time (T) and if needed temperature (Theta).

 

[hibachii]

Hi thanks for replying. So gravity is acceleration hence it would be L/T^2. But how can the left side match the right side because the right side has so many different dimensions like M(mass), length (L^2), and also what are the dimensions of rho?

 

[Saladsamurai]

Use the same approach What are some typical units of rho? Rho is density, so it's units are... So it's dimensions are... Then plug it all in and cancel units the same way you cancel factors in algebra.