Re: [Phys-l] Linear Air Drag

[Ken Caviness <caviness@southern.edu>]

Since linear drag is a assumed to be a multiple of the velocity (not merely the speed), you can treat the whole trip with one differential equation:

- m g - c v_vector = m a_vector,

where I've used "up" as the +y direction (so -mg is "down"), but -c v_vector is "in the direction opposite to that of the (current) velocity vector.

Break into component equations:

      - c v_x = m a_x
- m g - c v_y = m a_y,

which can solved separately.

If you changed the direction of y-component of the linear drag force at the highest point, using the -x-direction for the direction of the x-component for the whole trip and taking its magnitude as c sqrt(v_x^2 + v_y^2), that should also be valid.

But such separation of the trip into two parts isn't needed for the linear drag case.  (It is, however, for the quadratic case, which more correctly models air resistance.)

I see no reason why your method should not have given exactly the same results as the "all at once" treatment, unless you accidentally switched the sign of both components. (?)

Ken Caviness
Physics
Southern Adventist University

> -----Original Message-----
> From: phys-l-bounces@carnot.physics.buffalo.edu [mailto:phys-l-
> bounces@carnot.physics.buffalo.edu] On Behalf Of Peter Schoch
> Sent: Monday, November 02, 2009 2:34 PM
> To: Forum for Physics Educators
> Subject: [Phys-l] Linear Air Drag
>
> I have a quandry, and am hoping someone can clear it up for me...
>
> I just worked a linear air drag problem for 2-D motion to show my
> students.  The x direction was not a problem.  For the y direction I
> worked it to rise to a max height, lower than if no drag were present,
> and then switching the sign of the drag term I worked the other half
> of the vertical motion.  I got what appears to be a perfectly
> reasonable solution, whose graph is what I expected (lower y and
> shorter in x than the vacuum case).
>
> Now, I've just been given a copy of Taylor's "Classical Mechanics" and
> asked to teach the course next semester.  In that textbook (which
> seems very well done) he does linear drag in Chapter 2.  As part of it
> he does vertical motion, initial v upward, solves the first order, ODE
> and states that it is good for the entire motion!
>
> Now, I admit that my solution matches his for the t=0 to highest point
> of rise part.  However, for the second part of the motion, with the
> drag 'switched' my solution differs slightly from his -- and the
> graphs of he two differ (but only very slightly).  I checked the
> errata for the textbook, and none are mentioned for this.
>
> I need some outside brain power.  Why is Taylor's solution correct
> (that you need not 'switch' the drag term at the apex of the motion)?
>
> Thank you,
> Peter Schoch
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