On Tue, Apr 11, 2023 at 11:06 PM smitra <smi...@zonnet.nl> wrote:
> > > > > > > > > > > *> So, it has noticed that it used the wrong formulas and it has found > thecorrect ones. However, it was not able to do the calculation > usingspherical symmetry. It only noted in the previous answer that the > answeris spherically symmetric because it doesn't depend on the orientation > ofthe axis relative to which the moment of inertia is computed.So, it tells > you the steps to compute the three dimensional integral inspherical > coordinates, it is unable to see that adding up the threemoments of > inertias about the x, y and z axis yields a sphericallysymmetric expression > that can be computed using a one dimensionalintegral over the radius.Being > unable to solve a problem in a much simpler way than the standardway is a > big red flag against the idea that GPT can do any math at all.* I think the problem is that you and GPT have a difference of opinion about which is simpler, Cartesian coordinates or spherical coordinates. And there is no disputing matters of taste. I just add this conversation with GPT > ME: If adding up the three moments of inertias about the x, y and z axis yields a spherically symmetric expression can the moment of inertia formula be derived by simply using a one dimensional integral over the radius? *GPT: "* *Yes, that's correct. Once we have established that the moment of inertia of a solid sphere is spherically symmetric, we can compute it using a one-dimensional integral over the radius. Specifically, the moment of inertia of a solid sphere of radius R and mass M about an axis passing through its center is given by:I = (2/5) MR^2This formula can be derived by integrating the spherically symmetric expression for the moment of inertia over the volume of the sphere using spherical coordinates, which simplifies the calculation to a one-dimensional integral over the radius. Here's how the derivation goes:The moment of inertia of a solid sphere of radius R and mass M about an axis passing through its center is given by:I = ∫∫ (r^2 sin^2θ) ρ dVwhere ρ is the density and dV is the volume element in spherical coordinates. Since the sphere has uniform density, we can write ρ = M / (4/3 * π * R^3).Using the fact that the integrand is spherically symmetric, we can write:I = ∫∫∫ (r^2 sin^2θ) ρ dV = ρ ∫∫∫ (r^2 sin^2θ) r^2 sin θ dr dθ dφSubstituting in the expression for ρ and using the fact that the integral is spherically symmetric, we can simplify the integral to:I = (3/5) MR^2 ∫∫ sin^3θ dθ dφThe integral over θ and φ can be evaluated to give:∫∫ sin^3θ dθ dφ = 4π/3Substituting this into the expression for I, we get:I = (2/5) MR^2* *Therefore, the moment of inertia of a solid sphere about an axis passing through its center can be derived by simply using a one-dimensional integral over the radius, and is given by (2/5) MR^2."* John K Clark See what's on my new list at Extropolis <https://groups.google.com/g/extropolis> 5vd > -- 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/CAJPayv0D9MdKSrZX6rTFR-9J3k553u0KTf-NP6XyJhJnEyGUpg%40mail.gmail.com.
