Moment of Inertia of a solid uniform sphere

[John Kingsley]

Hello, Calculating the moment of inertia of a solid uniform sphere about it's center I get (3/5) Ma². I know I am supposed to be getting (2/5) Ma². I am using a differential volume dm as 4*pi*rho*r*r*dr, where rho is density, and r is the distance from the center to the differential volume element, which is a thin sphere with surface area 4*pi*r*r. So in effect I am multiflying the surface area of a thin shere by a quantity dr to get my dV. I believe this is where the problem is arising, but in calculating the differential area element of a solid disk, this approach seemd to work just fine. So anyway, my integral r*r*dm becomes 4*pi*rho*the integral from 0 to a of (r^4)dr. a is the radius of the sphere. rho equals 3M/(4*pi*a^3), and I end up with (3/5)Ma^2. Where am I going wrong? The text uses dz instead of dr and uses the fact that I for a solid disk is (1/2)Ma^2, but I do not see why this way will not work also.

 

[tiny-tim]

Welcome to PF!

Hello John! Welcome to PF!

(have a pi: π and try using the X² icon just above the Reply box )

… The text uses dz instead of dr and uses the fact that I for a solid disk is (1/2)Ma^2, but I do not see why this way will not work also.

your way will work if you use the moment of inertia for a spherical shell (about a diameter) instead of 1/2 Ma²

(but you don't know what that is, do you? )

 

[1994Bhaskar]

REMEMBER WHAT YOU DID IN CASE OF SOLID UNIFORM DISC?
WHAT YOU DID WAS INTEGRAL OF dI WHERE IT WAS M.I OF THIN RING.
HERE WE DON'T KNOW THE dI, i.e. MOMENT OF INERTIA OF SMALL SPHERE(THAT'S WHAT YOU HAVE TO FIND).MOREOVER THE SIMPLE I=MR² WON'T DO HERE FOR A COMPLEX STRUCTURE LIKE SPHERE.