weight of shark = 1000(9.8)(pi r^2)(L)
calculate density of air ,call it rho in kg/m^2 instead of 29 g/mol
volume of air that hits shark per second = 2 r L (v)
change in vertical momentum of that air per second
= rho * 2 r L v^2 = force up = weight down
What is the wind speed in m/s required to keep a great white shark suspended in midair during a summer day on the California beach?
Details and assumptions
Model the shark as a horizontal cylinder 6 m long and 1 m in diameter.
The air is at a pressure of 1 atm and a temperature of 30∘C.
The density of the shark is that of water, 1000 kg/m^3.
Assume that the wind gust that keeps the shark suspended is blowing straight upwards, and that the air molecules bounce off the shark elastically.
The acceleration of gravity is −9.8 m/s
^2.
ma, the molar mass of air, is 29 g/mol.
9 answers
how did you get the volume of air that hits the shark per second?
The cross sectional area of the cylinder looking up times the vertical velocity of the air.
Then the vertical momentum per second is that amount times the velocity again times the density.
Then the vertical momentum per second is that amount times the velocity again times the density.
In the momentum part I suppose we should also figure out the average change in vertical momentum over the diameter. The particle that hits the middle bounces back making for velocity change of 2 * v.
A particle that just grazes the edge of the cylinder hardly changes velocity component at all.
vertical velocity of reflected particle = -v cos 2 T
if T is angle of contact point from vertical through center of cylinder
so the average change over T = 0 to T = 90 degrees is
average of v(1 + cos 2 T)
so(2/Pi)v integral (1+cos 2T) dT from 0 to pi/2
= v[1+(2/pi)(Pi/2) (1/2)sin 90)]
= 3v/2
so the average change in vertical velocity is 1.5 times what I assumed
A particle that just grazes the edge of the cylinder hardly changes velocity component at all.
vertical velocity of reflected particle = -v cos 2 T
if T is angle of contact point from vertical through center of cylinder
so the average change over T = 0 to T = 90 degrees is
average of v(1 + cos 2 T)
so(2/Pi)v integral (1+cos 2T) dT from 0 to pi/2
= v[1+(2/pi)(Pi/2) (1/2)sin 90)]
= 3v/2
so the average change in vertical velocity is 1.5 times what I assumed
I assumed the air just stopped moving vertically when I first did it. I did not notice the elastic collision part.
so how would we apply this 3v/2 into your original equation
rho * 2 r L v^2 = 1000(9.8)(pi r^2)(L)
?
rho * 2 r L v^2 = 1000(9.8)(pi r^2)(L)
?
rho * 2 r L v (1.5v) = 1000(9.8)(pi r^2)(L)
so we would solve to get 66.51?
integral (1+cos 2T) dT = T + sinTcosT
so (2/Pi)v integral (1+cos 2T) dT from 0 to pi/2 = 1 ?
so (2/Pi)v integral (1+cos 2T) dT from 0 to pi/2 = 1 ?
0 answers