How Can You Predict the Fall Time of Coffee Filters Using Drag Force Equations?

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The discussion focuses on predicting the fall time of coffee filters using drag force equations. Participants suggest measuring the time it takes for filters to fall from a height sufficient to reach terminal velocity, using the formula for drag force, Fd = -bv. The relationship between the number of filters and their terminal velocity is explored, with the idea that velocity can be calculated by dividing distance by time. Concerns are raised about the stairwell height relative to the drop distance, emphasizing the need to account for the acceleration phase if the heights are similar. The conversation concludes with suggestions for analyzing the data, including plotting time against the number of filters to verify the expected relationship.
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Homework Statement


We are given 3 coffee filters. Using a stopwatch and a meter stick, make measurements to predict how long it would take for those filters to fall down a stairwell. Come up with an experimental method to predict this time.
During the lab, we will be told the number of coffee filters that will be used (1-9), as well as the height of the stairwell.
We are given the formula for the force of drag: Fd = -bv

Homework Equations


Fd = -bv
vterminal = \frac{Δd}{Δt}
FG = -Fd

The Attempt at a Solution


First, find the average terminal velocity for each number of filters. Drop a filter from a high enough height, that it will reach terminal velocity very early on (like 2m). Record the time that it takes the filter to fall. Repeat this for two filters and three filters. For each number of filters, divide the height it fell by the time it took in order to find velocity.
At terminal velocity, Fnet = 0. So, FG = - Fd.
Therefore, kmg = -bv, where k is the number of filters.
Arrange it so that v=-\frac{mg}{b}k.
We can treat -\frac{mg}{b} like a constant because we are only changing the number of filters (k), which affects the velocity (v).
Then could we just divide v by k to find the relationship, since the other terms are constant?
If that was correct, then could we just multiply that relationship by the number of filters being dropped down the stairwell to find their velocity? Once we have the velocity, knowing the height of the stairwel, we can easily find the time to fall. Δt = \frac{Δd}{v}.
Is this correct? It's hard to think about it when you don't have values.
 
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Yes, well thought out. The results depend largely upon how closely the drag force behaves in accordance with the given formula Fd = -bv
(a linear relationship), which you should note in your error analysis.
'It's hard to think about it when you don't have values'.
Yeah, I think you should be an engineer. They like to to get rid of the letters and crunch out the numbers as soon as possible to get a 'feel' for the situation.:thumbs:
 
You say, right at the beginning, that your instructons are "Using a stopwatch and a meter stick, make measurements" and "come up with an experimental method". There is nothing in that that says you are to come up with a formula a-priori. Yes, "it's hard to think about it when you don't have values". That's why the instructions here are to do the experiment to get values!
 
2 m is a pretty long distance to reach terminal velocity. How long is this stairwell? If 2 m is on the same order as the length of the stairwell, you may have to take into account the acceleration phase of the motion. According to your equations, for a given length of stairwell, the time should be inversely proportional to the number of filters. Therefore, the product of the time times the number of filters should be constant. Or, you could plot the time as a function of the number of filters on a log-log plot, and it should yield a straight line with a slope of -1. If not, then you would have to figure out why.

Chet
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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