[Help Me!] Linear motion equations with drag resistance

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September 25th 2009, 09:20 AM
Kevlar

[Help Me!] Linear motion equations with drag resistance
I've been taught about the normal linear motion equations now i've been set this task
1. Identify the phenomenon that causes objects to hit the ground at different times [Completed]
2. Determine the value of engineering coefficients
3. If possible attempt to develop a mathematical model that takes both gravity and this phenomenon into account to accurately predict the fall time for different spherical objects.
Other information

Acceleration: 9.81ms^-2
Fall height of object(s): 6M
Shape of object(s): Sphere

What I've done

For part 1: I deduced the phenomenon is air resistance/drag.

For part 2: Researched about drag and found out that

Where F_d: Force of drag, p=density of fluid(1.1877 @298K/25C),
v=speed of object
a=reference areaπ(piD^2/4 for a sphere),
C_d=drag coefficient(0.47 for a smooth sphere),
v^=is the unit vector indicating the direction of the velocity (the negative sign indicating the drag is opposite to that of velocity).

For part 3

Not sure for this but was thinking of using v=u+at rearranging for time, draw force diagram calculate the modulus of the force then put in values
September 25th 2009, 09:43 AM
CaptainBlack
Quote:
Originally Posted by Kevlar

I've been taught about the normal linear motion equations now i've been set this task

Identify the phenomenon that causes objects to hit the ground at different times [Completed]
Determine the value of engineering coefficients
If possible attempt to develop a mathematical model that takes both gravity and this phenomenon into account to accurately predict the fall time for different spherical objects.
Other information

Acceleration: 9.81ms^-2
Fall height of object(s): 6M
Shape of object(s): Sphere

What I've done

For part 1: I deduced the phenomenon is air resistance/drag.

For part 2: Researched about drag and found out that

$\mathbf{F}_d= -{1 \over 2} \rho v^2 A C_d \mathbf{\hat v}$

Where F_d: Force of drag, p=density of fluid(1.1877 @298K/25C),
v=speed of object a=reference areaπ(piD2/4 for a sphere),
C_d=drag coefficient(whats this for sphere?),
v^=is the unit vector indicating the direction of the velocity (the negative sign indicating the drag is opposite to that of velocity).

For part 3

Not sure for this but was thinking of using v=u+at rearranging for time, draw force diagram calculate the modulus of the force then put in values
You should find that for a body dropping vertically in a resisting medium at high Reynolds Number that

where:

The simplification is possible because we are only interested in the situation where is negative.

CB
September 25th 2009, 10:00 AM
Kevlar

Quote:

Originally Posted by CaptainBlack

You should find that for a body dropping vertically in a resisting medium at high Reynolds Number that

$<br /> \dot{v}=-g+kv^2<br />$
where:

$k=\frac{\rho\; C_d A}{2m}$

The simplification is possible because we are only interested in the situation where $v$ is negative.

CB

no idea what v dot or reynolds number is, did some research found vdot is acceleration i think, and didnt really understand anything about reynolds no. something about the viscosity of air.

Yet i still have no idea what you're on about.

Hate this problem based learning wish they taught us the stuff first then set the task >_> not the reverse argh
September 25th 2009, 10:08 AM
CaptainBlack

Quote:

Originally Posted by Kevlar

no idea what v dot or reynolds number is, did some research found vdot is acceleration i think, and didnt really understand anything about reynolds no. something about the viscosity of air.

Yet i still have no idea what you're on about.

Hate this problem based learning wish they taught us the stuff first then set the task >_> not the reverse argh

the dot denote the time derivative:

$\dot{v}=\frac{dv}{dt}$ .

Google for Reynolds number. High Reynolds number is the condition needed for the drag to be proportional to $v^2$ , at low Reynolds number drag is proportional to $v$ .

CB