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 of spherical objects of the same size but different materials (densities), the relation between terminal fall speed (vt) and mass (m) can be determined based on Aristotle's observations of objects falling in air with turbulent flow drag.
Let's start by considering the forces acting on the falling sphere: the gravitational force (mg) pulling it downwards and the drag force (Fd) opposing its motion. At terminal velocity, these two forces balance each other.
The drag force experienced by an object in air is proportional to its velocity squared (v^2) and its cross-sectional area (A). Mathematically, Fd = 1/2 * ρ * v^2 * A * Cd, where ρ is the density of air and Cd is the drag coefficient.
For a sphere, the cross-sectional area (A) is proportional to its radius squared (r^2). We can rewrite the drag force equation as Fd = 1/2 * ρ * v^2 * π * r^2 * Cd.
Now, let's consider two spheres of the same size but different densities. The only difference between them is their mass (m) and hence their total weight (mg). We can assume that the size (radius) and shape of both spheres remain the same.
Since both spheres have the same size, their cross-sectional area (A) and the drag coefficient (Cd) will be equal. Therefore, the drag force experienced by both spheres will be the same.
However, the weight (mg) will be different for each sphere, as it depends on the density of the material. Hence, the gravitational force acting on each sphere will be different.
To achieve a balance between the drag force and gravitational force, the sphere with higher density (hence higher weight) will require a higher terminal fall speed. Therefore, the function f1(m) can be written as f1(m) = m.
b. Now, let's consider spheres of the same material and density but different radii. In this case, mass (m) varies with the radius of the sphere.
Similar to the previous scenario, the drag force experienced by the spheres will be the same since they have the same size (cross-sectional area) and shape. The drag coefficient (Cd) remains constant.
However, the gravitational force acting on each sphere will be different, as it depends on their masses (which vary with radius). The weight (mg) will increase with the cube of the radius (m ~ r^3).
To achieve a balance between the drag force and gravitational force, the sphere with a larger radius (hence higher mass) will require a higher terminal fall speed. Therefore, the function f2(m) can be written as f2(m) = m^(-1/3), where m is the mass of the sphere.