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 the case where Aristotle is considering spherical objects of different materials and densities but the same size, the actual relation between terminal fall speed and mass can be determined by taking into account the effects of drag force and buoyancy force on the falling objects.
The drag force acting on an object is given by the equation Fd = 0.5ρAv^2Cd, where ρ is the density of the air, A is the cross-sectional area of the object, v is the velocity of the object, and Cd is the drag coefficient. The buoyancy force acting on the object is given by the equation Fb = ρair * Vobject * g, where ρair is the density of air, Vobject is the volume of the object, and g is the acceleration due to gravity.
At terminal speed, the drag force is equal to the weight of the object, which is given by the equation Fg = mg, where m is the mass of the object and g is the acceleration due to gravity.
Therefore, at terminal speed, we have:
0.5ρAv^2Cd = mg + ρair * Vobject * g
Simplifying the equation, we get:
v^2 = 2mg/(Cd * ρA) + 2Vobject * g
In this equation, the mass (m) of the object appears directly, while the density of the object (ρ), cross-sectional area (A), and volume (Vobject) can be regarded as constants for objects of the same size. Thus, the function f1(m) that relates the terminal speed (v) to the mass (m) would be:
f1(m) = √(2mg/(Cd * ρA) + 2Vobject * g) / K1
b. Now, let's consider the case where Aristotle is considering spheres of the same material and density but different radii (and hence, varying mass). The relation between terminal fall speed and mass can be determined using similar principles to the previous case, but considering the effect of mass on the object's terminal speed instead.
Using a similar equation as before:
v^2 = 2mg/(Cd * ρA) + 2Vobject * g
We can rewrite the equation by substituting the mass (m) with the density (ρ), volume (Vobject), and radius (r) of the sphere:
v^2 = 2ρVobjectg/(Cd * ρA) + 2Vobject * g
Since the density (ρ) is equal to mass (m) divided by volume (Vobject), we can substitute ρVobject with m, leading to:
v^2 = 2mg/(Cd * A) + 2Vobject * g
Now, considering that the radius (r) is directly proportional to the volume (Vobject) of a sphere (Vobject = (4/3)πr^3), we can replace Vobject with (4/3)πr^3 and simplify the equation:
v^2 = 2mg/(Cd * A) + 2((4/3)πr^3) * g
In this equation, the mass (m) appears directly, while the radius (r) of the object affects the volume (Vobject) and, therefore, the terminal speed (v). Thus, the function f2(m) that relates the terminal speed (vT) to the mass (m) would be:
f2(m) = √(2mg/(Cd * A) + 2((4/3)πr^3) * g) / K2