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.
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
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.
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
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 (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 ?
To determine the wind speed required to keep a great white shark suspended in midair, we can use the concept of buoyancy.
1. Start by finding the mass of the shark. The density of water is given as 1000 kg/m^3, and the shark's volume can be calculated as the product of its length and cross-sectional area (πr^2), where r is the radius. The diameter is given, so the radius is half of that. So, the volume of the shark is (6 m) * (π(1 m/2)^2) = 6π/4 m^3. Multiply this volume by the density of water to get the weight of the shark.
Weight of shark = (6π/4 m^3) * (1000 kg/m^3) * (9.8 m/s^2)
2. Next, we need to find the weight of the air displaced by the shark. This weight can be calculated using Archimedes' principle, which states that the buoyant force exerted on an object immersed in a fluid is equal to the weight of the fluid displaced by the object.
The density of air can be calculated using the ideal gas law: pV = nRT, where p is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature in Kelvin.
Convert the given temperature to Kelvin: 30°C + 273.15 = 303.15 K.
Using the ideal gas law, we can calculate the density of air: density of air = (p * ma) / (R * T).
The pressure is given as 1 atm, the molar mass of air (ma) is given as 29 g/mol, R is the ideal gas constant, which is approximately 8.314 J/(mol⋅K), and T is 303.15 K.
Convert the molar mass of air to kg/mol: 29 g/mol = 0.029 kg/mol.
Substitute the given values into the equation to find the density of air.
3. Once you have the density of air, calculate the weight of the air displaced by the shark using the density of air and the volume of the shark.
Weight of displaced air = (density of air) * (6π/4 m^3) * (9.8 m/s^2)
4. Finally, equate the weight of the shark to the weight of the displaced air and solve for the wind speed.
Weight of shark = Weight of displaced air
(6π/4 m^3) * (1000 kg/m^3) * (9.8 m/s^2) = (density of air) * (6π/4 m^3) * (9.8 m/s^2)
Solve for the density of air canceling out the other terms:
density of air = (1000 kg/m^3)
Since the density of air is already equal to the density of water, the wind speed required to keep the shark suspended in midair is 0 m/s.
Therefore, no wind speed is needed, and the shark will not remain suspended in midair.