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Re: [Phys-l] Linear Air Drag



Thank you! I see it now, and have found my error that prevented the two solutions from matching.

I'm just beginning to 'tackle' the v^2 problem now, so I also appreciate that insight!

Peter Schoch

On Nov 2, 2009, at 3:01 PM, Ken Caviness wrote:

1st (vector) equation should read:

- m g j_hat - c v_vector = m a_vector,

where j_hat is a unit vector pointing up. (Or this can be done in 3 dimensions.)

KC

-----Original Message-----
From: Ken Caviness
Sent: Monday, November 02, 2009 2:59 PM
To: Forum for Physics Educators
Subject: RE: [Phys-l] Linear Air Drag

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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_______________________________________________
Forum for Physics Educators
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https://carnot.physics.buffalo.edu/mailman/listinfo/phys-l