Need help understanding air resistance lab (coffee filters)

[isukatphysics69]

Homework Statement

Homework Equations

Equations posted in pictures attached

The Attempt at a Solution

I haven't attempted it yet because I am confused about something. If the-cv^2 case is for objects that hit terminal velocity at the fastest rate, why are my graphs showing that I should use the -bv case? There must be something I am missing here. I thought the-cv^2 case is for high speed objects AND objects that hit terminal velocity fast like coffee filters and feathers?[/B]

 

[haruspex]

why are my graphs showing that I should use the -bv case?

Do they? What makes you say that?

 

[isukatphysics69]

Do they? What makes you say that?

because the error bars are smaller overall. Don't we want to minimize error?

 

[isukatphysics69]

I think i have to do something along the lines of setting Newtons second law equation = 0, getting my data from the terminal velocities and plugging that into v, and solve for the drag force constant.. should that drag force constant be the slope of the graph since the weights are increasing linearly? i am just completely stuck here.

 

[haruspex]

because the error bars are smaller overall. Don't we want to minimize error?

The error bars look longer merely because are plotting the square. The error is the same (though I think you may have plotted some inaccurately; e.g. the top of the bar at the n=4 point in the upper graph appears to be at 1.77, but the corresponding v² bar goes up to 3.27, not 3.13). Besides, to choose between the two equations the main thing is which looks more like a straight line. Clearly that is the v² graph.

 

[isukatphysics69]

The error bars look longer merely because are plotting the square. The error is the same (though I think you may have plotted some inaccurately; e.g. the top of the bar at the n=4 point in the upper graph appears to be at 1.77, but the corresponding v² bar goes up to 3.27, not 3.13). Besides, to choose between the two equations the main thing is which looks more like a straight line. Clearly that is the v² graph.

Hi, can you please explain why i should choose the one that looks like a straight line? is it because the drag force is a constant and the slope should increase linearly with respect to the mass because of the constant drag force?

That is my thinking here i don't know why else i would choose the one that looks more like a straight line..

 

[haruspex]

why i should choose the one that looks like a straight line?

You are to choose between two equations relating speed to drag force: $\vec F=-b\vec v$ or $|\vec F|=c{\vec v}^2$, where b and c are (different) constants.
If the first is correct then which one would look like a straight line when plotting |F| against |v|?
What about plotting |F| against v²? What shape curves would the two equations produce?

 

[isukatphysics69]

You are to choose between two equations relating speed to drag force: $\vec F=-b\vec v$ or $|\vec F|=c{\vec v}^2$, where b and c are (different) constants.
If the first is correct then which one would look like a straight line when plotting |F| against |v|?
What about plotting |F| against v²? What shape curves would the two equations produce?

I'm just not understanding this.. why exactly does it matter if the function is a straight line? Is it because the masses increase linearly and the coefficient of drag force remains the same? I can't do this problem unless I know why I'm doing it

 

[haruspex]

why exactly does it matter if the function is a straight line?

You are asked to choose which of two mathematical models, a linear function of speed or a quadratic function of speed, is a better fit to the data. Do you understand that this is an important sort of thing to be able to determine?

Suppose the real behaviour is that the force varies linearly with speed. If you plot F against v, what shape of graph do you expect? If you plot F against v² what shape do you expect?
Now suppose the real behaviour is that the force varies as the square of the speed. Same two questions: If you plot F against v, what shape of graph do you expect? If you plot F against v² what shape do you expect?

Unless you try to answer those questions, I cannot see how to explain further.

 

[isukatphysics69]

This lab is still not making sense to me. I have two velocity vs Mass graphs. I am advised to pick the one that is showing a linear increase of velocity with respect to mass. This is because the mass of an object will determine its terminal velocity. I still don't understand why the situation of v will not work here just simply because i don't understand the physical laws of why an objects terminal velocity will not increase at what looks to be the graph of a logarithm for the v case, i guess its just not how an objects in free fall will reach terminal velocity with respect to mass so i should just take it as is? So now i am told to consider error bars and use my equations to back my claim of why the v^2 case better fits my data, i have all of my data and could plug those values into the equations and solve for the constant but i do not have the T values to there are two uknowns. i guess i could solve for T with one of the equations and plug it into the v^2 case then solve for the constant drag force? This is really doing my head in i am trying to understand it but having a hard time.

 

[haruspex]

I have two velocity vs Mass graphs.

No, you have one graph of v against m, and one of v² against m.

I am advised to pick the one that is showing a linear increase of velocity with respect to mass

No, I am saying you should pick the one that is a straight line. In the one case, v against m, that would mean the mass increases linearly with the velocity (or, more in line with the causality, that the velocity increases linearly with the mass); in the other, v² against m, a straight line would mean that the velocity-squared increases linearly with m, i.e. that the velocity increases as the square root of m.

what looks to be the graph of a logarithm for the v case,

No, it is a graph like y=a√x, i.e. v∝√m, i.e. v²∝m.

 

[isukatphysics69]

No, you have one graph of v against m, and one of v² against m.

No, I am saying you should pick the one that is a straight line. In the one case, v against m, that would mean the mass increases linearly with the velocity (or, more in line with the causality, that the velocity increases linearly with the mass); in the other, v² against m, a straight line would mean that the velocity-squared increases linearly with m, i.e. that the velocity increases as the square root of m.

No, it is a graph like y=a√x, i.e. v∝√m, i.e. v²∝m.

Ok thank you, i realized i had something wrong in my lab that was making this even more confusing than it already is. i am going to continue trying. the equations i derived above were supposed to be derived later in the lab. all i was supposed to do was set the acceleration = 0 in Newtons equations and solve for velocity...

 

[isukatphysics69]

So now i see one case as vt = mg/b and the other case as vt = sqrt(mg/c)

 

[isukatphysics69]

wow i am a complete moron

 

[isukatphysics69]

ok haruspex i am sorry for that, now what you were saying is making more sense.. i just misread the question before this one and was beyond confused because nothing you were saying was matching to what i had in my lab.