Drag Force Acting on an Object with Respect to Velocity

In summary: However, differentiating ##W(air~drag)=mgh-\frac{1}{2}mv^2,## with respect to s (s=h), got me...nothing.
  • #1
Marcell
21
0

Homework Statement


The final velocity of an object falling through air from various heights is given. From this, can you derive an equation for the drag force acting on the object with respect to velocity?

Homework Equations


Maybe relevant?
Wno drag$$=mgh,$$
Wreal$$=\frac{1}{2}mv^2,$$
Wdrag$$=mgh-\frac{1}{2}mv^2,$$

The Attempt at a Solution


Wdrag$$=F(drag~average)s$$
This allows me to solve for the average drag, but I don't know how to go from there to equating the actual drag force acting on the object at a specific velocity.

Thanks for any and all help! :)
 
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  • #2
Marcell said:

Homework Statement


The final velocity of an object falling through air from various heights is given. From this, can you derive an equation for the drag force acting on the object with respect to velocity?

Homework Equations


Maybe relevant?
Wno drag$$=mgh,$$
Wreal$$=\frac{1}{2}mv^2,$$
Wdrag$$=mgh-\frac{1}{2}mv^2,$$

The Attempt at a Solution


Wdrag$$=F(drag~average)s$$
This allows me to solve for the average drag, but I don't know how to go from there to equating the actual drag force acting on the object at a specific velocity.

Thanks for any and all help! :)

If you have also measured the time of the fall along with the terminal velocity, you can set up a mathematical problem: instead of writing the differential equation for velocity as a function of time in the form ##dv/dt = g - f(v)## for your drag force function ##f(v)##, let's re-write it as
$$\frac{dt}{dv} = \frac{1}{g - f(v)}$$
Thus, the fall-time ##T## for terminal velocity ##V## is given by
$$T = \int_0^V \frac{1}{g - f(v)} \, dv \hspace{1.5cm} (1)$$
If you have measurements of lots of ##(V,T)## pairs, you can try various functional forms in (1) to see if you can get a fit. For example, if you try
$$f(v) = a v + b v^2$$ in (1), the result for ##T## vs. ##V## will contain the unknown constants ##a## and ##b##. You can then try a data-fitting program to find values that are consistent with your measured values.

If you have not measured the fall-times, but only the distance ##X## and the terminal velocity ##V## you will have a much harder problem.
 
  • #3
Ray Vickson said:
If you have also measured the time of the fall along with the terminal velocity, you can set up a mathematical problem: instead of writing the differential equation for velocity as a function of time in the form ##dv/dt = g - f(v)## for your drag force function ##f(v)##, let's re-write it as
$$\frac{dt}{dv} = \frac{1}{g - f(v)}$$
Thus, the fall-time ##T## for terminal velocity ##V## is given by
$$T = \int_0^V \frac{1}{g - f(v)} \, dv \hspace{1.5cm} (1)$$
If you have measurements of lots of ##(V,T)## pairs, you can try various functional forms in (1) to see if you can get a fit. For example, if you try
$$f(v) = a v + b v^2$$ in (1), the result for ##T## vs. ##V## will contain the unknown constants ##a## and ##b##. You can then try a data-fitting program to find values that are consistent with your measured values.

If you have not measured the fall-times, but only the distance ##X## and the terminal velocity ##V## you will have a much harder problem.

Thanks for the help! But I don't quite see how solving for the time taken to reach terminal velocity ##T## would be useful in finding the drag force acting on the object at a specific velocity...
 
  • #4
Marcell said:
Wdrag$$=mgh-\frac{1}{2}mv^2,$$
Try differentiating that with respect to ...
 
  • #5
haruspex said:
Try differentiating that with respect to ...

distance? Would that give the instantaneous work being done on the falling object and allow me to find the force of drag? If so I'm really not sure how to differentiate something like that, and my go to 'derivate-calculator.com' is lost here as well.
 
  • #6
haruspex said:
Try differentiating that with respect to ...

Wait, even if I do differentiate it with respect to something, wouldn't here only be one component on the left hand side? and would said component then be equal to zero?
 
  • #7
Marcell said:
distance? Would that give the instantaneous work being done on the falling object
If you differentiate the work done by a force with respect to the distance the force advances, what do you get? Or to put that in reverse, what, when integrated wrt distance, gives the work a force does?
 
  • #8
haruspex said:
Try differentiating that with respect to ...
Distance.
haruspex said:
If you differentiate the work done by a force with respect to the distance the force advances, what do you get? Or to put that in reverse, what, when integrated wrt distance, gives the work a force does?
I get the force done on the body.

However, differentiating ##W(air~drag)=mgh-\frac{1}{2}mv^2,## with respect to s (s=h), got me ##0=mg##
 
  • #9
instantaneous acceleration equation for falling object gives,
mdv/dt=mg-Fdrag.
Now given the form of Fdrag ,v(t) can be determined.
 
  • #10
Apashanka said:
instantaneous acceleration equation for falling object gives,
mdv/dt=mg-Fdrag.
Now given the form of Fdrag ,v(t) can be determined.
I'm not sure if I follow, but I don't want to find v(t), I'm looking for ##F_{drag}## and I don't have the form of ##F_{drag}##...
 
  • #11
Marcell said:
I get the force done on the body.
Right. Specifically, you get the force at any particular distance.
Marcell said:
However, differentiating
##W(air~drag)=mgh-\frac{1}{2}mv^2## with respect to s (s=h), got me 0=mg
No, you get ## F_{drag}= \frac d{dh}(mgh-\frac{1}{2}mv^2)##.
So now you just need a way to estimate the right hand side of that at various values of h. How can you do that?
 
  • #12
haruspex said:
Right. Specifically, you get the force at any particular distance.

No, you get ## F_{drag}= \frac d{dh}(mgh-\frac{1}{2}mv^2)##.
So now you just need a way to estimate the right hand side of that at various values of h. How can you do that?

Wait I'm quite confused, wouldn't the right hand side just equal ##mg##? The only way I see getting a non-constant force of drag would be if the right hand side were differentiated with respect to v (though that doesn't make too much sense to me...)
 
  • #13
Marcell said:
wouldn't the right hand side just equal mg?
No, v is a function of h.
You are given a set of pairs of values for h and v. How can you estimate ##\frac d{dh}v^2## from those?
 
  • #14
haruspex said:
No, v is a function of h.
You are given a set of pairs of values for h and v. How can you estimate ##\frac d{dh}v^2## from those?
Eyyyy, thank you so much, I got it! Squared the line of best fit's equation and then just took its derivative.
 
  • #15
Marcell said:
Eyyyy, thank you so much, I got it! Squared the line of best fit's equation and then just took its derivative.
Sounds good to me. Or you could square first and take best fit second. The result won't be quite the same... not sure which is the more justified.
 
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  • #16
haruspex said:
Right. Specifically, you get the force at any particular distance.

No, you get ## F_{drag}= \frac d{dh}(mgh-\frac{1}{2}mv^2)##.
So now you just need a way to estimate the right hand side of that at various values of h. How can you do that?
Wait, doesn't this equation suggest that as velocity increases the force of drag acting on the object decreases? There's no way that can be right...
 
  • #17
Marcell said:
Thanks for the help! But I don't quite see how solving for the time taken to reach terminal velocity ##T## would be useful in finding the drag force acting on the object at a specific velocity...

I mean that you need to do the integral to find ##T = F(a,b,V),## where
$$F(a,b,V) = \int_0^V \frac{1}{9.81 - a v - b v^2} \, dv$$
is some function of the terminal velocity ##V## that contains the unknown constants ##a## and ##b##. You can then try to find numerical values of the parameters ##a## and ##b## so as to fit the ##(T,V)## data as well as possible---that is, to make ##F(a,b,V_i)## come out close to the measured value of ##T_i## for all pairs ##(T_1,V_1), (T_2,V_2), (T_3,V_3), \ldots, (T_n,V_n)## in the experimental data set.

An exact fit cannot be expected, because of experimental errors (and possibly because of an incorrect ##f(v)## expression), but some form of "least-squares" fit might work well. Once you have found appropriate "best-fit" numerical values of ##a## and ##b##, you can then substitute those values into the formula ##f_{\text{fit}}(v) = a v + b v^2##, so you can calculate your best estimate ##f_{\text{fit}}(v)## of the drag force ##f(v)## for any input value of ##v.## Of course, for all this to work, you must know how to do integration, and have access to a nonlinear "least-squares" computer package.

Since I do not know your background I cannot judge whether this answer is meaningful to you and if so, how much additional explanatory material to post, if any.
 
Last edited:
  • #18
Marcell said:
Wait, doesn't this equation suggest that as velocity increases the force of drag acting on the object decreases? There's no way that can be right...
You have to be careful with the signs of the various quantities. With your sign convention, is v positive or negative? What about dv/dh?
 
  • #19
Marcell said:
Wait, doesn't this equation suggest that as velocity increases the force of drag acting on the object decreases? There's no way that can be right...
No, why?
Sketch gh and ½v2 as functions of h. The first is a straight line, the second starts with the same slope but rises towards a maximum. Sketch the difference. What can you say about the slope as h increases?
 
  • #20
Ray Vickson said:
I mean that you need to do the integral to find ##T = F(a,b,V),## where
$$F(a,b,V) = \int_0^V \frac{1}{9.81 - a v - b v^2} \, dv$$
is some function of the terminal velocity ##V## that contains the unknown constants ##a## and ##b##. You can then try to find numerical values of the parameters ##a## and ##b## so as to fit the ##(T,V)## data as well as possible---that is, to make ##F(a,b,V_i)## come out close to the measured value of ##T_i## for all pairs ##(T_1,V_1), (T_2,V_2), (T_3,V_3), \ldots##
That is quite interesting, but I don't think I could do it since I only have one pair of ##(T_1,V_1)##, after all the whole experiment is done with a single object.
vela said:
You have to be careful with the signs of the various quantities. With your sign convention, is v positive or negative? What about dv/dh?
Down is positive, so both v and dv/dh are positive
 
  • #21
haruspex said:
No, why?
Sketch gh and ½v2 as functions of h. The first is a straight line, the second starts with the same slope but rises towards a maximum. Sketch the difference. What can you say about the slope as h increases?
Where does gh come from? ##\frac d{dh}(mgh)=mg## no? please correct me if I'm wrong.
 
  • #22
Marcell said:
Where does gh come from? ##\frac d{dh}(mgh)=mg## no? please correct me if I'm wrong.
I'm suggesting to sketch mgh, ½mv2 and mgh-½mv2, but with the m omitted throughout.
You can then consider how the slope, ##\frac d{dh}(gh-\frac 12v^2)##, changes with h.
 
  • #23
haruspex said:
I'm suggesting to sketch mgh, ½mv2 and mgh-½mv2, but with the m omitted throughout.
You can then consider how the slope, ##\frac d{dh}(gh-\frac 12v^2)##, changes with h.
Well the slope doesn't change too much if it all, it appears to be linear. Also I don't quite see why differentiating ##\frac d{dh}(gh-\frac 12v^2)## _wouldn't_ lead to a drag force which decreases with velocity.
 
  • #24
Marcell said:
Well the slope doesn't change too much if it all, it appears to be linear. Also I don't quite see why differentiating ##\frac d{dh}(gh-\frac 12v^2)## _wouldn't_ lead to a drag force which decreases with velocity.
Then you are sketching it wrongly.
As I wrote in post #19:
haruspex said:
Sketch gh and ½v2 as functions of h. The first is a straight line, the second starts with the same slope but rises towards a maximum.
This implies the second rises ever more slowly, so the gap between the two increases faster than linearly. The difference therefore rises faster than linearly, so the slope (the drag force) increases.
 
  • #25
haruspex said:
Then you are sketching it wrongly.
As I wrote in post #19:

This implies the second rises ever more slowly, so the gap between the two increases faster than linearly. The difference therefore rises faster than linearly, so the slope (the drag force) increases.

taking $$v=ah+b,$$
$$\frac{d}{dh}\left(mgh-\frac{1}{2}m(ah^2+2\operatorname{abh}+b^2\right))$$
then $$F=\left(mg-\frac{1}{2}m\left(2ah+2ab\right)\right)$$

And from that, as h increases, F decreases which doesn't make sense.
 
  • #26
Marcell said:
taking $$v=ah+b,$$
There is no basis for such an equation, and it is certainly wrong.
An object falling through a dragging medium approaches, but never quite reaches, a "terminal velocity".
The exact relationship to time or distance is not susceptible to algebra - it's very complicated. Typically, drag is roughly a linear function of speed at low speeds, more like a quadratic one at higher speeds.
But we do not need to care about the details here. What matters is that at the start the drag is very small, so v2 is nearly linear with h (the ½v2 line starts tangent to the gh line), but as h increases v2 rises more slowly and approaches a horizontal asymptote.
 
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  • #27
Marcell said:
That is quite interesting, but I don't think I could do it since I only have one pair of ##(T_1,V_1)##, after all the whole experiment is done with a single object.

Down is positive, so both v and dv/dh are positive

You said in your first post that you drop the object from various heights, so that sounds to me like you will obtain several values of terminal speed and fall-time. Yes, you have just a single object, but are you not using it several times to get several data points?
 
  • #28
Ray Vickson said:
You said in your first post that you drop the object from various heights, so that sounds to me like you will obtain several values of terminal speed and fall-time. Yes, you have just a single object, but are you not using it several times to get several data points?
ah I see what you mean. No, sorry for the confusion, but that is not the data I have (object doesn't reach terminal velocity), either way I really appreciate your time and effort thanks!
 
  • #29
haruspex said:
There is no basis for such an equation, and it is certainly wrong.
An object falling through a dragging medium approaches, but never quite reaches, a "terminal velocity".
The exact relationship to time or distance is not susceptible to algebra - it's very complicated. Typically, drag is roughly a linear function of speed at low speeds, more like a quadratic one at higher speeds.
But we do not need to care about the details here. What matters is that at the start the drag is very small, so v2 is nearly linear with h (the ½v2 line starts tangent to the gh line), but as h increases v2 rises more slowly and approaches a horizontal asymptote.
Thanks for explaining it so many times, I think I finally got it, have a good one!
 
  • #30
Marcell said:
ah I see what you mean. No, sorry for the confusion, but that is not the data I have (object doesn't reach terminal velocity), either way I really appreciate your time and effort thanks!

Sorry for the confusion: by "terminal velocity" I do not mean the usual "terminal velocity in air" figure, but rather, the final velocity the of the object just as it hits the ground (which you said you measure).
 
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  • #32
rcgldr said:
If the drag force can be defined as 1/2 Cd A v^2, where Cd is coefficient of drag, and A is cross sectional area, then there is a closed form solution for an dropped object with only a vertical component of velocity. Wiki link, click on "show" for the derivation:

https://en.wikipedia.org/wiki/Terminal_velocity#Derivation_for_terminal_velocity
As I understand the purpose of the exercise, it is to estimate the drag forces at various points in the fall based on the measured velocities. There is no suggestion that it should be based on any drag equation or theory.
 

1. What is drag force?

Drag force is a force that acts on an object when it moves through a fluid, such as air or water. It is caused by the resistance of the fluid to the movement of the object.

2. How does drag force affect an object's velocity?

The drag force acting on an object is directly proportional to the object's velocity. This means that as the velocity increases, the drag force also increases.

3. What factors affect the magnitude of drag force?

The magnitude of drag force depends on several factors, including the object's shape, size, and surface properties, as well as the density and viscosity of the fluid it is moving through.

4. How does drag force impact an object's motion?

Drag force acts in the opposite direction of an object's motion, which means it slows down the object's motion. As the drag force increases, the object's acceleration decreases, eventually reaching a point of equilibrium where the drag force equals the object's weight.

5. Can drag force be reduced?

Yes, drag force can be reduced by changing the object's shape or surface properties to make it more streamlined, or by decreasing the density or viscosity of the fluid it is moving through. This is why many vehicles, such as airplanes and cars, are designed to minimize drag force and increase efficiency.

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