Moment of inertia of a sphere

1. Nov 14, 2006

Ja4Coltrane

This was not actually homework, but I was just trying to see if I could calculate moments of inertia and apparently, I cannot.
I'm trying to show that the moment of inertia for a sphere is (2/5)MR^2
So I started with I=(integral)(r^2)(dm)
then P=dm/dv=dm/(4pi(r^2)dr)
so dm=(4)(pi)(r^2)(P)(dr)
so I substituted into the original equation, removed constants from the integral, and substituted P for M/V=(3M/(4(pi)(R^3)))
I=12pi(M)/(4(pi)r^3)[integral]r^4 (dr)
I=(3/5)MR^2 which is wrong!

(sorry about the lack of pretty math writing)
Thanks for any help, and by the way, I'm only a high school student so my calculus knowledge is very limited (in fact, the only reason I know what integration is is because of my physics class).

2. Nov 14, 2006

Ja4Coltrane

Oh, I just realized something!
I am integrating as if the higher part has the same radius because it is the same distance from the center but it is closer to the axis!
now I really dont know what to do.

3. Nov 15, 2006

OlderDan

This is not a simple calculation. There are two approaches you can take with limited calculus experience. The first is to find the moment of inertia of a disk about its symmetry axis, and then slice the sphere into disks of thickness dx having a common axis that is a diameter of the sphere. Then add (integrate) the moments of inertia of all the disks. The hard part is finding the radius of each disk as a function of x, but that can be done using the equation for the surface of the sphere.

The second approach is to find the moment of inertia of a cylindrical shell about its symmetry axis (easy since all the mass has the same radius) and think of the sphere as many concentric shells of radius r and thickness dr. The hard part here is finding the length of each cylinder as a function of r, but again this can be found from the equation for the surface of the sphere.