Ask Me Help Desk - Work done by drag force

I am an AP Calculus-based physics teacher at a high school and we are struggling with a problem (that I, admittedly made up and may be impossible, but still has raised questions for me). The problem is listed below.

A cannon ball with a mass of 10 kg is shot directly upward at 100 m/s. It reaches a maximum height of 400 m. Calculate the coefficient, D, of the drag force if it is most closely modeled by f=Dv^2, where V is the instantaneous velocity of the cannon ball.
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[FONT=Cambria][COLOR=#1a1a1a]My initial approach was by calculating the work done by the drag force. However, since the drag force is non-constant, I know that I cannot simply let that work be equal to F*d (where d is the maximum height). Obviously the work done by drag is the difference between the initial kinetic energy and the final gravitational potential energy, but that doesn't really address the problem.

I then thought to integrate the drag force through the distance and set that equal to the work, but since the drag force is a function of velocity (and therefore a function of position), that felt wrong as you just end up with F*d without substituting something else into velocity (which I cannot seem to get an expression for).

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[FONT=Cambria][COLOR=#1a1a1a][FONT=arial, sans-serif]Any suggestions/help would be greatly appreciated! If the problem isn't workable then that would be fine, I just feel as though ample information is provided and it should be solvable. Even if the problem isn't solvable, I'd [/FONT]like[FONT=arial, sans-serif] your input on how to find the work done by a non-conservative force that changes as a function of velocity as opposed to displacement.[/FONT][/COLOR][/FONT]

I apologize for the formatting errors. I copied elements of this from an email to a colleague and it went bonkers.

Yes, copy and paste doesn't work well on this site.

This is one of those problems that does not have a closed form solution. But you can go quite a ways toward understanding the motion of the canon ball. My first attempt was the same as yours - to use energy principals. But like you I couldn't find a way to do the integral. So I came at it a different way - using equations of motion with respect to time, and then figuring a way to determine the value for D that satisfies the boundary conditions of the problem.

The basic equation is $m \frac {dv}{dt} = -mg-Dv^2$ After some manipulation, and substituting $A = \sqrt{D/gm}$ you get to:

$\int \frac {dv}{1 + A^2v^2} = - g \int dt<br />$

which has the solution:

$\frac 1 A tan^{-1}(Av) = -gt +C<br />$

Here C is the constant of integration, which is $C=\frac 1 A tan^{-1}(Av_0)$ .

Rearranging:

$v(t) = \frac 1 A tan(tan^{-1}(Av_0)-Agt)$

This is only valid as the canon ball rises, ie v>=0, as the original equation of motion has the force of wind resistance acting downward. Once the canon ball reaches its max altitude and starts to descend that force acts in the upward direction, and the resulting solution involves hyperbolic sine functions.

Next we can integrate v with respect to t to find the position y of the canon ball. Substituting $B=tan^{-1}(Av_0)$ , we get:

$y(t) = \int v dt = \frac 1 A \int \tan(B-Agt)dt$

$y(t) = \frac 1 {A^2g} \ln(\cos(B-Agt))+C$

where again C is the constant of integration, which given that y(0) = 0 is:

$C = \frac {- \ln(cos(B))}{A^2 g}$

The max value of x(t) occurs when t=B/(Ag), at which point $x= \frac {- \ln(cos(B))}{A^2 g}$ . Thus to answer your question of what the value of D must be for x_max = 100m, you must solve the following equation for A:

$100m = \frac {- \ln( \cos( \tan^{-1}(Av_0)))}{A^2g}$

This simplifies (slightly) to:

$100m = \frac{- \ln (\frac 1 {\sqrt{A^2v_0^2+1}})}{A^2g}$

Or:

$(2A^2g)100m = \ln(A^2v_0^2+1)$

At this point solving for A requires a numerical approximation technique, but I find that setting A = 0.037 does a pretty good job. From that we can find that D is approximately 0.13446.

Thanks for the suggestion. I'll look more into it. However, I think the proposed solution might be a bit much for my high school students.

May be a bit much indeed. The trickiest parts are the two integrals, which AP students may have been exposed to in their calculus classes. But if not, I don't think they'd be able to follow along.

I've been working through the problem myself and have two questions:
1) You set y_max=100, shouldn't it be 400?
2) In the last step you had (2A^2*g), I cannot find the source of the 2. Where did that arise?

Quote:

Originally Posted by jlisenbe88

I've been working through the problem myself and have two questions:
1) You set y_max=100, shouldn't it be 400?

Yes - you're correct, my bad. I confused the 100m/s initial velocity with the max height attained.

Quote:

Originally Posted by jlisenbe88

2) In the last step you had (2A^2*g), I cannot find the source of the 2. Where did that arise?

Comes from the square root in the denominator:

$- \ln ( \frac 1 {\sqrt x} ) = - \ln (x^{-1/2}) = \frac 1 2 \ln x$