Moment of Inertia: Deriving for a Sphere w/ Spherical Coordinates

[theCandyman]

This is not homework, but since it is related I thought it should go here. I have not yet had calculus III class, but I would like for someone to show me how the moment of inertia is derived for a solid sphere. From what I have heard from my physics professor, it is easiest to use sphereical coordinates to find it. Am I right to assume sphereical coordinates are something similar to polar coordinates in the third dimension?

 

[OlderDan]

This is not homework, but since it is related I thought it should go here. I have not yet had calculus III class, but I would like for someone to show me how the moment of inertia is derived for a solid sphere. From what I have heard from my physics professor, it is easiest to use sphereical coordinates to find it. Am I right to assume sphereical coordinates are something similar to polar coordinates in the third dimension?

Spherical coordinates involve one linear coordinate, the distance from the origin, and two angular coordinates. The angle theta is the same angle as used in cylindrical coordinates. The angle phi is the angle between the positve z-axis and the position in space.

It may be easiest in spherical coordinates, but I'm not so sure. Because of the symmetry and because the distance involved in the integral is the distance from an axis, I think you will find polar coordinates to work just fine. In fact, since all mass at a given distance from the axis forms a cylindrical shell, all you need to do is figure out the moment of inertia for such a shell and treat the problem as concentric shells. Of course the height of each shell depends on the radius, so you still need to do an integral over r. You could instead set it up to integrate over z by treating the sphere as a many disks of varying radii. You then need to know the moment of inertia of a disk. Any way you look at it, you are doing a double integral, but the concentric shell approach involves one integral that is essentially done for you by virtue of the constant radius.

http://mathworld.wolfram.com/SphericalCoordinates.html

 

[thepatientmental]

Yes, you are correct in assuming that spherical coordinates are similar to polar coordinates, but in three dimensions. They consist of three coordinates: r, θ, and φ, where r is the distance from the origin, θ is the angle from the positive z-axis, and φ is the angle from the positive x-axis in the xy-plane.

To derive the moment of inertia for a solid sphere using spherical coordinates, we can use the formula:

I = ∫∫∫ (r^2sin^2θ)(ρ)dV

where r is the distance from the axis of rotation, θ is the angle from the axis of rotation, ρ is the density of the sphere, and dV is the volume element.

First, we need to find the density of the sphere, which is given by ρ = m/V, where m is the mass of the sphere and V is its volume. Since we are dealing with a solid sphere, we can use the formula for the volume of a sphere: V = (4/3)πr^3.

Substituting this into the equation for density, we get ρ = (3m)/(4πr^3).

Now, we can plug this into the formula for moment of inertia and integrate over the volume of the sphere:

I = ∫∫∫ (r^2sin^2θ)(3m)/(4πr^3)dV

Next, we need to express the volume element in terms of spherical coordinates. In this case, the volume element is given by dV = r^2sinθdrdθdφ.

Substituting this into the integral, we get:

I = ∫∫∫ (r^2sin^2θ)(3m)/(4πr^3)(r^2sinθdrdθdφ)

Simplifying this, we get:

I = (3m)/(4π) ∫∫∫ sin^3θdθdφdr

Now, we can evaluate the integral with respect to θ and φ, giving us:

I = (3m)/(4π) ∫ (0 to 2π) ∫ (0 to π) sin^3θdθdφ ∫ (0 to R) dr

Evaluating the inner integral, we get:

I = (3m)/(4π) ∫ (0 to