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air resistance, dimensional analysis confusion |
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| Dec6-12, 01:46 PM | #1 |
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air resistance, dimensional analysis confusion
Hi. Consider the basic eq for a falling body with air resistance
dv/dt=g-kv/m I dont understand air resistance as a force, since it seems irreconcilable to the force equation F=ma. How is a force a function of velocity? I am also not sure how this equation makes sense in terms of dimensional anaysis--the right side is m/s^2, the left m/s^2+(m/s)/kg. I am apparently the only one troubled by this, as extensive googling has yeilded nothing. Thanks! |
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| Dec6-12, 01:58 PM | #2 |
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Mentor
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| Dec6-12, 02:12 PM | #3 |
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good point, I had assumed k to be unit-less, but its units are kg/sec (http://oregonstate.edu/instruct/mth2...02/resist.html)
so F=kv would have the same dimension as F=ma. thanks! |
| Dec6-12, 03:01 PM | #4 |
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air resistance, dimensional analysis confusion
Another problem: your proportionality is wrong. Air resistance follows a v2 proportionality, so in reality, it should be:
dv/dt = g - kv2/m, in which k = ρ/2*Cd*A, where ρ is the density of the fluid, Cd is the drag coefficient (unitless), and A is the reference area. |
| Dec6-12, 03:21 PM | #5 |
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generally it is given as proportional to v or v^2--the quadratic relationship is usually for larger objects. Most introductory material on diff eq use v. thanks
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| Dec6-12, 03:55 PM | #6 |
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Recognitions:
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Precisely. Drag equation can be different under different conditions. Quadratic drag is more common in practical situations, but slow motion through viscous medium will often produce linear drag.
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| Dec6-12, 06:22 PM | #7 |
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| Dec6-12, 06:35 PM | #8 |
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Recognitions:
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I missed the bit about it being specific to drag in air. Yes, with air, you are unlikely to see linear drag outside of Millikan Oil Drop, or similar setup.
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