Revision of Lab Report 1 - Air Resistance

Introduction:

In previous labs, we neglected air resistance and found the velocity of falling objects. Now we are going to show how air resistance affects the velocity of a falling object. Air resistance can be found by following:





A is the cross sectional area, C is the drag coefficient, and ρ is the density of air.


We can say that when a falling object is at terminal velocity, then Fair = mg.


Overview:

In this lab, we are dropping coffee filters from a certain height to model air resistance on an object. We chose to use coffee filters because we can stack multiple coffee filters on top of each other so that only the mass will change in our equations. The term ρAC will remain constant throughout the whole experiment. Our goal in this experiment is to show how mass affects the terminal velocity.

Materials:

* 5 of the same coffee filter
* LoggerPro
*Velocity/Position Detector

Data: (Here is a showcase of my awesome HTML skills)

mass of a coffee filter = 1.03g
ρAC is a constant

Here is the data we collected for all filters:

Number
of            Avg.          
Coffee     Terminal       Avg.      Avg.
Filters      Velocity       Height    Time

1 filer	.75826m/s	1.2316m	1.674s
2 filters	1.1344m/s	1.2382m	1.23s
3 filters	1.4426m/s	1.218m	1.02s
4 filters	1.8254m/s	1.179m	.85s
5 filters	1.959m/s	1.1182m	.74s

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Here is a graph of our data:



If you are unable to see the graph Click Here

The slope of our graph is .768, which is pAC/g. We can use the slope to determine the terminal velocity of any given of inputs or different masses. For instance, if I wanted to know the terminal velocity of 15 coffee filters I would just do, V = sqrt(.768*(15*1.03)) = 3.44m/s

Analysis:


The data is a little messy. We didn't drop from one specific height (even thought we should have). As we can see from our data, as we add more filters the velocity increases. We can compare this to our model:
Fair = .5ρACv^2    and
Fair = mg, so


.5ρACv^2 = mg, and from this model we can calculate velocity


v =  , and from this we can see as the mass increases the velocity will increase and
                        our model agrees with our data

Here is an example:

We drop a 10 kg medicine ball with a diameter of .27m from a tall building and we want to find the terminal velocity. The drag coefficient of a sphere is .47, the density of air is 1.275 kg/m³ , and our cross sectional area is .848 m²


v =  , so V = sqrt((2*10kg*9.8m/s²)/(1.275 kg/m³*.114m²*.47)) = 53 m/s

Now, let's drops a heavier medicine with the same diameter but an increase mass of 5 kg.

              V = sqrt((2*15kg*9.8m/s²)/(1.275 kg/m³*.114m²*.47)) = 65.6 m/s

As you can see, the mass does have an effect on the terminal velocity, if an object has an enormous mass, then it will have an enormous terminal velocity.