Falling Body with Air Resist: Difference between revisions

Latest revision as of 18:09, 5 May 2020

F_total = F_grav - F_air

ma = mg - kv where k units are kg/sec

mdv/dt = mg - kv

mdv/dt + kv = mg

dv/dt + vk/m = g <----Standard form

e^kt/mdv/dt + e^kt/mvk/m = ge^kt/m

d/dt(ve^kt/m) = ge^kt/m

ve^kt/m + C₁ = mg/ke^kt/m + C₂

v = mg/k - (C₁ - C₂)/e^kt/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

@@ Line 5: / Line 5: @@
 mdv/dt = mg - kv
-put into standard order:
+* put into standard form:
 mdv/dt + kv = mg
-dv/dt + vk/m = g
+'''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>
-d/ds(se<sup>kt/m</sup>) = ge<sup>kt/m</sup>
+* use product rule:
-integrate both sides:
+d/dt(ve<sup>kt/m</sup>) = ge<sup>kt/m</sup>
-se<sup>kt/m</sup> + C<sub>1</sub> = gm/ke<sup>kt/m</sup> + C<sub>2</sub>
+* integrate both sides:
+ve<sup>kt/m</sup> + C<sub>1</sub> = mg/ke<sup>kt/m</sup> + C<sub>2</sub>
+* solve for v:
+v = mg/k - (C<sub>1</sub> - C<sub>2</sub>)/e<sup>kt/m</sup>
+v = mg/k - Ce<sup>-kt/m</sup>
+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<sup>-kt/m</sup>
+or
+v(t) = (mg/k)(1-e<sup>-kt/m</sup>)
+when t goes to infinity, e<sup>-kt/m</sup> 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/s<sup>2</sup>)/66 m/s = 11.8 kg/sec