Aristotle is sometimes quoted as claiming that a body falls at a speed proportional to its mass. Let us suppose that Aristotle’s experience was based on objects falling in air (with turbulent flow drag) essentially at their terminal speeds.
a. If Aristotle were considering spherical objects, say, all of the same size but of different materials (hence different densities), what would the actual relation between terminal fall speed and mass be? That is, if we were to write vt= K1f1(m), where f1(m) is a function only of mass and K1 contains all the factors independent of mass, what would the function f1(m) be?
b. Now suppose Aristotle were considering spheres all of the same material and density, but different radii, so that mass varies with radius. If we once again write vT= K2f2(m), where f2(m) depends only on mass and K2 contains all the mass independent factors, what would the new function of f2(m) be?
a. In order to determine the relationship between terminal fall speed and mass for spherical objects of different materials, we need to consider the factors that affect the terminal speed.
Terminal speed is reached when the force of gravity pulling the object downward is balanced by the drag force acting in the opposite direction. For objects falling in air with turbulent flow drag, the drag force can be expressed as:
Fd = (1/2)ρv^2CdA
Where:
- Fd is the drag force
- ρ is the density of air
- v is the velocity of the object
- Cd is the drag coefficient (dependent on the shape of the object)
- A is the effective cross-sectional area of the object
The gravitational force acting on the object is given by:
Fg = mg
Where:
- Fg is the gravitational force
- m is the mass of the object
- g is the acceleration due to gravity
At terminal speed, Fd = Fg. Therefore, we can equate the two equations:
(1/2)ρv^2CdA = mg
We can rearrange this equation to solve for the velocity v:
v^2 = (2mg) / (ρCdA)
Now, let's consider two spherical objects of the same size but different densities. Density (ρ) is defined as mass (m) divided by volume (V):
ρ = m/V
Since the size of the spheres is the same, we can write:
V = (4/3)πr^3
Where:
- r is the radius of the sphere
Substituting the value of V in terms of r into the density equation, we get:
ρ = m / ((4/3)πr^3)
Simplifying this equation, we find:
m = (4/3)πr^3ρ
Now, substitute the expression for mass (m) into the equation for v^2:
v^2 = (2(4/3)πr^3ρg) / (ρCdA)
Canceling out common factors, we get:
v^2 = (8/3)πrg / CdA
Since we are only concerned with the relationship between terminal fall speed (vT) and mass (m), we can ignore the constant factors (8/3)πgCdA and write:
vT = K1f1(m)
Therefore, the function f1(m) in this case would be f1(m) = 1.
b. Now, let's consider spherical objects of the same material and density but different radii. In this case, the mass of the spheres will vary with the radius. The relationship between mass (m) and the radius (r) of the sphere can be given as:
m = k * r^3
Where:
- k is a constant that depends on the material and density of the sphere
Substituting this expression for mass (m) into the equation for v^2:
v^2 = (2mg) / (ρCdA)
v^2 = (2(k * r^3)g) / (ρCdA)
Canceling out common factors, we get:
v^2 = (2kr^3g) / (ρCdA)
Since we are only concerned with the relationship between terminal fall speed (vT) and mass (m), we can ignore the constant factors (2k g)/(ρCdA) and write:
vT = K2f2(m)
Therefore, the function f2(m) in this case would be f2(m) = r^(2/3).
a. In the case where Aristotle is considering spherical objects of the same size but different materials (hence different densities), the actual relation between terminal fall speed and mass can be derived from the drag force acting on the objects.
The drag force can be expressed as:
F(D) = 0.5 * Cd * A * ρ * v^2
Where:
- F(D) is the drag force
- Cd is the drag coefficient which depends on the shape of the object (assuming all objects are spherical, the drag coefficient will be the same)
- A is the cross-sectional area of the object
- ρ is the density of the fluid (air in this case)
- v is the velocity of the object relative to the fluid
For objects falling at their terminal speeds, the drag force is equal to the gravitational force acting on the objects:
F(D) = m * g
Where:
- m is the mass of the object
- g is the acceleration due to gravity
Setting these two equations equal to each other and solving for the velocity v, we get:
v = √((2 * m * g) / (Cd * A * ρ))
Comparing this equation to vt = K1 * f1(m), we can see that the function f1(m) is given by:
f1(m) = √((2 * m * g) / (Cd * A * ρ))
b. In the case where Aristotle is considering spheres of the same material and density but different radii, the mass will vary with the radius according to the equation m = ρ * V, where ρ is the density and V is the volume of the sphere.
The volume of a sphere can be expressed as V = (4/3) * π * r^3, where r is the radius.
Substituting this into the equation for mass, we get:
m = ρ * (4/3) * π * r^3
Rearranging the equation to solve for the radius, we have:
r = (3 * m / (4 * ρ * π))^(1/3)
Substituting this value of radius into the equation vT = K2 * f2(m), we can see that the function f2(m) is given by:
f2(m) = √((2 * m * g) / (Cd * A * ρ))
Thus, in both cases, the function relating the terminal fall speed vT to the mass m is the same and given by:
f(m) = √((2 * m * g) / (Cd * A * ρ))