Difference between revisions of "Falling Body with Air Resist"
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| Line 5: | Line 5: | ||
mdv/dt = mg - kv | mdv/dt = mg - kv | ||
| − | put into standard form: | + | # put into standard form: |
mdv/dt + kv = mg | mdv/dt + kv = mg | ||
| Line 11: | Line 11: | ||
dv/dt + vk/m = g <----Standard form | dv/dt + vk/m = g <----Standard form | ||
| − | find u = e<sup>∫k/mdt</sup> = e<sup>kt/m</sup> | + | * find u = e<sup>∫k/mdt</sup> = e<sup>kt/m</sup> |
| − | multiply by u: | + | * multiply by u: |
e<sup>kt/m</sup>dv/dt + e<sup>kt/m</sup>vk/m = ge<sup>kt/m</sup> | e<sup>kt/m</sup>dv/dt + e<sup>kt/m</sup>vk/m = ge<sup>kt/m</sup> | ||
| − | use product rule: | + | * use product rule: |
d/dt(ve<sup>kt/m</sup>) = ge<sup>kt/m</sup> | d/dt(ve<sup>kt/m</sup>) = ge<sup>kt/m</sup> | ||
| − | integrate both sides: | + | * integrate both sides: |
ve<sup>kt/m</sup> + C<sub>1</sub> = mg/ke<sup>kt/m</sup> + C<sub>2</sub> | ve<sup>kt/m</sup> + C<sub>1</sub> = mg/ke<sup>kt/m</sup> + C<sub>2</sub> | ||
| − | solve for v: | + | * solve for v: |
v = mg/k - (C<sub>1</sub> - C<sub>2</sub>)/e<sup>kt/m</sup> | v = mg/k - (C<sub>1</sub> - C<sub>2</sub>)/e<sup>kt/m</sup> | ||
Revision as of 18:08, 5 May 2020
Ftotal = Fgrav - Fair
ma = mg - kv where k units are kg/sec
mdv/dt = mg - kv
- put into standard form:
mdv/dt + kv = mg
dv/dt + vk/m = g <----Standard form
- find u = e∫k/mdt = ekt/m
- multiply by u:
ekt/mdv/dt + ekt/mvk/m = gekt/m
- use product rule:
d/dt(vekt/m) = gekt/m
- integrate both sides:
vekt/m + C1 = mg/kekt/m + C2
- solve for v:
v = mg/k - (C1 - C2)/ekt/m
v = mg/k - Ce-kt/m
If at time 0, the body is at rest then v(0) = 0 = mg/k - C so C = mg/k
therefore v = mg/k - mg/ke-kt/m
or
v(t) = (mg/k)(1-e-kt/m)
when t goes to infinity, e-kt/m goes to 0 so the terminal velocity = mg/k
The value of k depends on mass and shape and it can be determined by measuring terminal velocity:
average terminal velocity of an 80kg person = 148mph = 66m/s so k = 80kg(9.8m/s2)/66 m/s = 11.8 kg/sec
