PLEASE HELP in Deriving Terminal Velocity Equation

AI Thread Summary
The discussion revolves around deriving the terminal velocity equation v = sqrt(2mg/DρA) and confirming its dimensional correctness. Participants emphasize the importance of understanding the dimensions of each variable involved, suggesting that the user should analyze the units for mass (m), density (ρ), and cross-sectional area (A). They recommend consulting the textbook "Physics for Scientists and Engineers" for guidance on dimensional analysis. Additionally, to estimate the terminal velocity of an Australian $1 coin, users suggest looking up the density of air and considering the coin's mass and area while noting that the drag coefficient (D) is dimensionless. The conversation highlights the collaborative effort to clarify the concepts and assist with the homework problem.
hibachii
Messages
17
Reaction score
0
PLEASE HELP! in Deriving Terminal Velocity Equation :)

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)

The Attempt at a Solution



I'm completely stuck. I have no clue as in how to even start. I'm so sorry :(
 
Physics news on Phys.org
welcome to pf!

hi hibachii! welcome to pf! :smile:

(have a square-root: √ and try using the X2 tag just above the Reply box :wink:)

what are the dimensions of each symbol in √(2mg/DρA)?

check that when you combine them they come to the same as v :wink:
 


I'm going to assume your in the same Physics Class as I am.

If you go to Page 7 of the the book "Physics for Scientists and Engineers" by Serway it explains how to do Dimensional Analysis and that will explain how to do (i).

For (ii) you will need to look up a value for the density of air and then estimate the mass and area of the coin remembering that the coin would be rotating as it fell through the air
 


Cheers guys. Helped alot.

P.S Jazrad what class are you in for physics 101?
 


Hey jazrad thanks but I am still stuck. I've looked at page 7 and 8 but it only shows dimensions for velocity and acceleration. I've still got a lot of unknowns like D, A, m and ρ.
I tried subbing in L/T for v and L/T^2 for g but I am still left stuck. Please give some more help although you have provided me with a very good start.
Cheers
 
hey hibachii! :smile:

(try using the X2 icon just above the Reply box :wink:)
hibachii said:
Ive still got a lot of unknowns like D, A, m and ρ.

(D is given as dimensionless :wink:)

to find the dimensions of an unfamiliar variable, decde what units it's measured in (m/s2 for example), then convert each unit to dimensions (m/s2 = L/T2)

start with ρ, the density …

what units is it measured in? :smile:
 
Thread 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
Thread 'A cylinder connected to a hanged mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top