Acceleration and wind resistance

[sbuxaddict]

Hi everyone,
I'm going through question #25 on the TBR Physics Chapter 1 Practice Exam. This is the question:

Question: Not correcting for wind resistance when evaluating the results of the experiment would lead to a gravitational force constant that is too:

Answer: Small. The standard deviation in the raw data would not be affected by the presence of wind resistance.

In the explanation, it says that the time it would take for a ball to fall is increased because of wind resistance leading to a lower calculated value for acceleration, (which I agree with) so the value of g would be underestimated if you failed to correct for wind resistance (but isn't it contradicting itself here? I was thinking it's be overestimated if you didn't correct for wind resistance)

Here's my thinking:
Wind resistance opposes the direction of motion, so it would be decreasing the acceleration if it's accounted for. If it's not accounted for, wouldn't the acceleration be larger than, relatively speaking?

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[milski]

Not accounted in that case means that the data that you are collecting is for an object falling in air and experiencing wind resistance but you ignoring that resistance in your calculations. Let's say that the real acceleration is g. In air, because of the resistance, it will be g'=g-x, g'<g. When you do your experiment, the data should lead you to g' since that's what happening. Because you're not correcting for the wind, you're not going to add x to g' but will consider g' as the final answer. As we said, that g' is less than the actual g.

 

[dranet]

You could do a forces diagram and some maths.
y dirn taking down as -ve (-mg) and up as postive (Fr, Force due to air resistance umg, with u< 1))
ma=Fr-mg....
a=( umg-mg) / m
a=ug-g
looking at that the acceleration is less than g.
Plug that into the distance formula Y=VoT-1/2 gT^2, make Vo=0 for the arguement
Y=-1/2gT^2 then T= sq root(2Y/g).................................................wind not included
With resistance in there T= sq root (2Y/ug-g)..............................wind included
So comparing the two times the denominator of "wind" included is smaller than for "wind not included"
So T without wind < T with wind
That Y will be -ve using the axes i mentioned above.

 

[MedPR]

Hi everyone,
I'm going through question #25 on the TBR Physics Chapter 1 Practice Exam. This is the question:

Question: Not correcting for wind resistance when evaluating the results of the experiment would lead to a gravitational force constant that is too:

Answer: Small. The standard deviation in the raw data would not be affected by the presence of wind resistance.

In the explanation, it says that the time it would take for a ball to fall is increased because of wind resistance leading to a lower calculated value for acceleration, (which I agree with) so the value of g would be underestimated if you failed to correct for wind resistance (but isn't it contradicting itself here? I was thinking it's be overestimated if you didn't correct for wind resistance)

Here's my thinking:
Wind resistance opposes the direction of motion, so it would be decreasing the acceleration if it's accounted for. If it's not accounted for, wouldn't the acceleration be larger than, relatively speaking?

Wind resistance opposes gravity, so it takes longer for ball to drop, and without calculating in the wind resistance, you will think the acceleration due to gravity is less than it really is.

Maybe this will make it more clear. You can run a 100m dash in 10 seconds. If I tell you to run another 100m dash, but this time I strap a parachute on your back, you will take longer than 10 seconds, say 20 seconds. Now, if I told someone that you took 20 seconds to run a 100m dash, but I didn't tell them I put a parachute on your back, they would think you are slower than someone else who runs a 100m dash in 15 seconds without a parachute. In this example, the parachute is air resistance, if you couldn't tell

It's not as technical as the explanations above, but I think it's simpler to understand it intuitively rather than plugging in a bunch of numbers/variables to various equations.

 

[sbuxaddict]

Not accounted in that case means that the data that you are collecting is for an object falling in air and experiencing wind resistance but you ignoring that resistance in your calculations. Let's say that the real acceleration is g. In air, because of the resistance, it will be g'=g-x, g'<g. When you do your experiment, the data should lead you to g' since that's what happening. Because you're not correcting for the wind, you're not going to add x to g' but will consider g' as the final answer. As we said, that g' is less than the actual g.

So I guess I was just misunderstanding the question then, it's asking for the value once it's been inputted? But then wouldn't it still be greater than?

You could do a forces diagram and some maths.
y dirn taking down as -ve (-mg) and up as postive (Fr, Force due to air resistance umg, with u< 1))
ma=Fr-mg....
a=( umg-mg) / m
a=ug-g
looking at that the acceleration is less than g.
Plug that into the distance formula Y=VoT-1/2 gT^2, make Vo=0 for the arguement
Y=-1/2gT^2 then T= sq root(2Y/g).................................................wind not included
With resistance in there T= sq root (2Y/ug-g)..............................wind included
So comparing the two times the denominator of "wind" included is smaller than for "wind not included"
So T without wind < T with wind
That Y will be -ve using the axes i mentioned above.

Ok, this makes sense, but I'm still struggling with it conceptually I guess.

Wind resistance opposes gravity, so it takes longer for ball to drop, and without calculating in the wind resistance, you will think the acceleration due to gravity is less than it really is.

Maybe this will make it more clear. You can run a 100m dash in 10 seconds. If I tell you to run another 100m dash, but this time I strap a parachute on your back, you will take longer than 10 seconds, say 20 seconds. Now, if I told someone that you took 20 seconds to run a 100m dash, but I didn't tell them I put a parachute on your back, they would think you are slower than someone else who runs a 100m dash in 15 seconds without a parachute. In this example, the parachute is air resistance, if you couldn't tell

It's not as technical as the explanations above, but I think it's simpler to understand it intuitively rather than plugging in a bunch of numbers/variables to various equations.

If it takes longer for the ball to drop with wind resistance, that decreases acceleration right? So my thinking is if you account for wind resistance, wouldn't the expected value be greater?

And in your metaphor, I'm not sure where the 100m in 15s comes in, because shouldn't I be comparing to without a parachute (which is 100m in 10s), which then makes the acceleration without a parachute greater than with?



I'm missing a connection here..

 

[Pisiform]

basically, to keep it simple, look at it this way:

You went to the top of a building in Manhattan and started dropping ball and you tabulated the results in your lab book. These values you calculated had taken into account air resistance because in real life when you were dropping balls from the building, there was air resistance. Now, from this lab book, you calculated the value of acceleration using a formula and lets say the value was 7m/s^2.

But you know that under idea conditions (when no air resistance is present/vacuum) the value of acceleration would be lets say 10m/s^2 (and you found that out from textbook). So your experiment value of 'a' (7m/s^2) is less than the textbook value (which is 10m/s^2) but you know that you could not have gotten 10 m/s^2 in your answer because it is under idea condition and doesnt take into account air resistance, and on earth, air-resistance is everywhere.

So, since most of the questions/theoretical scenario we do under ideal condition, you need to fix your experimental result of acceleration to make it equal to the textbook value of 10m/s^2. If you don't fix it, then your experimental value would be smaller.

 

[dranet]

Hi everyone,
I'm going through question #25 on the TBR Physics Chapter 1 Practice Exam. This is the question:

Question: Not correcting for wind resistance when evaluating the results of the experiment would lead to a gravitational force constant that is too:

Answer: Small. The standard deviation in the raw data would not be affected by the presence of wind resistance.

In the explanation, it says that the time it would take for a ball to fall is increased because of wind resistance leading to a lower calculated value for acceleration, (which I agree with) so the value of g would be underestimated if you failed to correct for wind resistance (but isn't it contradicting itself here? I was thinking it's be overestimated if you didn't correct for wind resistance)

Here's my thinking:
Wind resistance opposes the direction of motion, so it would be decreasing the acceleration if it's accounted for. If it's not accounted for, wouldn't the acceleration be larger than, relatively speaking?

Did they say what the standard deviation was. I tend to agree with you that without calc wind resistance you get a value for g that is greater or closer to the constant you are talking about.
If your ideal calculations come up with a Std Dev into which the calcs with wind resistance fall then maybe its does not make a huge difference. Was there more detail in the q like stats for the expt???

 

[MedPR]

If it takes longer for the ball to drop with wind resistance, that decreases acceleration right? So my thinking is if you account for wind resistance, wouldn't the expected value be greater?

And in your metaphor, I'm not sure where the 100m in 15s comes in, because shouldn't I be comparing to without a parachute (which is 100m in 10s), which then makes the acceleration without a parachute greater than with?



I'm missing a connection here..

In my example, the 100m in 15seconds is another runner that was used just for reference. From my example, if I told you about the guy who ran 100m in 20 seconds, but I didn't tell you that he had a parachute strapped to him, what would you calculate his acceleration to be? Now, calculate his acceleration without the parachute (without wind resistance). Which one is higher?

His acceleration with the parachute is like a ball falling with wind resistance, but if you ignore wind resistance in the calculations (even though it was there in the experiment) your calculated acceleration will be smaller than it should be.

Maybe this is a better example. I tell you that a body builder lifts a 250kg mass with both hands, but I also lie and tell you that gravity on earth is only 5m/s^2. What would you calculate the force required to lift the mass be? It would be less than the actual force required because you are not including the extra 5m/s^2 of acceleration due to gravity (5+5=10m/s^2) which is acting on the mass and thereby requiring the body builder to lift with greater force. The extra 5m/s^2 that is acting but you aren't calculating for makes your calculated value smaller than the true value.

 

[sbuxaddict]

basically, to keep it simple, look at it this way:

You went to the top of a building in Manhattan and started dropping ball and you tabulated the results in your lab book. These values you calculated had taken into account air resistance because in real life when you were dropping balls from the building, there was air resistance. Now, from this lab book, you calculated the value of acceleration using a formula and lets say the value was 7m/s^2.

But you know that under idea conditions (when no air resistance is present/vacuum) the value of acceleration would be lets say 10m/s^2 (and you found that out from textbook). So your experiment value of 'a' (7m/s^2) is less than the textbook value (which is 10m/s^2) but you know that you could not have gotten 10 m/s^2 in your answer because it is under idea condition and doesnt take into account air resistance, and on earth, air-resistance is everywhere.

So, since most of the questions/theoretical scenario we do under ideal condition, you need to fix your experimental result of acceleration to make it equal to the textbook value of 10m/s^2. If you don't fix it, then your experimental value would be smaller.

Ohhhh this makes so much sense, I get it now. I think I made it much more complicated than it needed to be. Thank you!!

Did they say what the standard deviation was. I tend to agree with you that without calc wind resistance you get a value for g that is greater or closer to the constant you are talking about.
If your ideal calculations come up with a Std Dev into which the calcs with wind resistance fall then maybe its does not make a huge difference. Was there more detail in the q like stats for the expt???

It doesn't give any standard deviation values, I copied the question as it is. Part of the answer said that standard deviation would be constant throughout, so it wouldn't affect the data.

In my example, the 100m in 15seconds is another runner that was used just for reference. From my example, if I told you about the guy who ran 100m in 20 seconds, but I didn't tell you that he had a parachute strapped to him, what would you calculate his acceleration to be? Now, calculate his acceleration without the parachute (without wind resistance). Which one is higher?

His acceleration with the parachute is like a ball falling with wind resistance, but if you ignore wind resistance in the calculations (even though it was there in the experiment) your calculated acceleration will be smaller than it should be.

Maybe this is a better example. I tell you that a body builder lifts a 250kg mass with both hands, but I also lie and tell you that gravity on earth is only 5m/s^2. What would you calculate the force required to lift the mass be? It would be less than the actual force required because you are not including the extra 5m/s^2 of acceleration due to gravity (5+5=10m/s^2) which is acting on the mass and thereby requiring the body builder to lift with greater force. The extra 5m/s^2 that is acting but you aren't calculating for makes your calculated value smaller than the true value.

Ohhhh I see, I misunderstood the first example. Thank you so much for taking time out to explain! I've got it now And the second example really helped as well, thanks again!

 

[premedicine555]

Ok, so I'm still a bit confused on this question. I feel incredibly stupid and would NEVER have gotten this on the MCAT.. The wording, for me, makes no sense. If you have air resistance, acceleartion DECREASES.. So If i didn't correct for it, then wouldn't there be a larger value of Fg constant?? *head desk*

 

[milski]

Ok, so I'm still a bit confused on this question. I feel incredibly stupid and would NEVER have gotten this on the MCAT.. The wording, for me, makes no sense. If you have air resistance, acceleartion DECREASES.. So If i didn't correct for it, then wouldn't there be a larger value of Fg constant?? *head desk*

That would be true if you knew g (as a constant) and were trying to figure out what the acceleration woud be if you drop something. They are asking what would happen if you made an experiment trying to figure out how much g is, measured the time for something to fall but forgot about air resistance.

Let's take an extreme example where the air is very very resistant. You drop a ball and if there was no resistance it would fall on the ground in 1 second. But since there is a very strong resistance, let's say that your experiment shows that the balls takes 100 seconds to fall. You forget about the air resistance and decide that g is such that the ball takes 100 seconds - calculating g based on that will get you a very small g (acceleration), certainly smaller than the real g.

 

[MedPR]

Ok, so I'm still a bit confused on this question. I feel incredibly stupid and would NEVER have gotten this on the MCAT.. The wording, for me, makes no sense. If you have air resistance, acceleartion DECREASES.. So If i didn't correct for it, then wouldn't there be a larger value of Fg constant?? *head desk*

Take a look at my examples above, I think they explain this pretty well.

 

[premedicine555]

just saw the answers, thank you! I think I get it now. Basically, in the experiment we already inputted a g value which INCLUDED air resistance. Thus, we have a lower "g" value.