On Sun, Apr 9, 2023 at 12:53 PM smitra <smi...@zonnet.nl> wrote: > For some simple physics or math result consider different > ways of getting to that result where one of these ways is not widely > published and is likely not in GPT's database. Take e.g. different ways > of computing the moment of inertia of a ball of uniform density of > radius R and mass M (relative to an axis through the center). There are > many ways to do this, but I've not seen my favorite way of doing this on > any webpage, which is to restore spherical symmetry by adding up the > three identical moments of inertia relative to 3 orthogonal axes.
I just send this to GPT: *Me: **Derive the formula for the moment of inertia (relative to an axis through the center) of a ball of uniform density of radius R and mass M by adding up the three identical moments of inertia relative to 3 orthogonal axes and restoring spherical symmetry.* *And in less than three seconds it gave me this: * *GPT: To derive the formula for the moment of inertia of a ball of uniform density, we will start by finding the moment of inertia relative to three orthogonal axes passing through the center of the ball. Let's call these axes x, y, and z.The moment of inertia of a point mass m relative to an axis at a distance r from the mass is given by:I = m * r^2For a solid sphere of uniform density, we can divide it into small volume elements, each with mass dm. The volume element located at a distance r from the center of the sphere has a moment of inertia relative to the x-axis given by:dIx = dm * y^2Similarly, the moment of inertia of this volume element relative to the y-axis is:dIy = dm * x^2The moment of inertia relative to the z-axis is:dIz = dm * r^2where r is the distance from the center of the sphere to the volume element.We can express the differential mass element dm in terms of the density ρ and the volume element dV:dm = ρ * dVFor a sphere of radius R, the volume element is:dV = 4/3 * π * r^2 * drSubstituting dm and dV into the expressions for dIx, dIy, and dIz, and integrating over the volume of the sphere, we get:Ix = ∫∫∫ y^2 * ρ * (4/3 * π * r^2 * dr) dVIy = ∫∫∫ x^2 * ρ * (4/3 * π * r^2 * dr) dVIz = ∫∫∫ r^2 * ρ * (4/3 * π * r^2 * dr) dVwhere the limits of integration are:0 ≤ r ≤ R, 0 ≤ θ ≤ π, 0 ≤ φ ≤ 2πTo simplify these integrals, we can use spherical coordinates, where x = r sin θ cos φ, y = r sin θ sin φ, and z = r cos θ. The volume element in spherical coordinates is:dV = r^2 sin θ dr dθ dφSubstituting this into the expressions for Ix, Iy, and Iz, we get:Ix = 2/5 * M * R^2Iy = 2/5 * M * R^2Iz = 2/5 * M * R^2where M is the mass of the sphere, given by:M = ρ * V = 4/3 * π * R^3 * ρwhere V is the volume of the sphere.Note that Ix = Iy, since the sphere has spherical symmetry, and the moments of inertia relative to any two orthogonal axes in a plane perpendicular to the third axis are equal. Therefore, the moment of inertia of a sphere of uniform density relative to an axis through the center is:I = 2/5 * M * R^2which is the desired formula.* John K Clark See what's on my new list at Extropolis <https://groups.google.com/g/extropolis> grxq > -- You received this message because you are subscribed to the Google Groups "Everything List" group. To unsubscribe from this group and stop receiving emails from it, send an email to everything-list+unsubscr...@googlegroups.com. To view this discussion on the web visit https://groups.google.com/d/msgid/everything-list/CAJPayv3HFKt73uZGEEQ1dPRXDw0%3D%2BsTWfLWe4cZAuW9PFjXTtw%40mail.gmail.com.
