Leon

Posted on Tuesday, 09 September, 2003 - 06:48 am:

How do I calculate the moment of inertia,

I

, of a solid sphere of mass

M

and radius

R

, about its line of symmetry?

$I = \int r^{2} dr$

$I$ is supposed to be $2 {MR}^{2} / 5$

Arun Iyer

Posted on Tuesday, 09 September, 2003 - 09:18 am:

The sphere you are talking about is the solid sphere.
The method i know of starts by finding moment of inertia of a hollow sphere and then using that result to find the moment of inertia of the solid sphere.(prolly a very tedious but understandable method)

Do you know how to calculate the moment of inertia of a hollow sphere??If no write back and i(or some of the other Nrich members) would help you out!!!(However i might delay my posts due to obscure schedule so please excuse me if i post late)

love arun

Stephen Burgess

Posted on Tuesday, 09 September, 2003 - 09:45 pm:

The sphere is symmetrical. Therefore:

$I = \int x^{2} + y^{2} dx dy dz$ by calculating the MoI parallel to the $z$ -axis.

Also, $I = \int y^{2} + z^{2} dx dy dz$

and $I = \int x^{2} + z^{2} dx dy dz$ by symmetry.

So $3 I = \int 2 (x^{2} + y^{2} + z^{2}) dx dy dz$

$I = 2 / 3 \int r^{2} dx dy dz$

$I = 2 / 3 \int r^{2} dr d θ d ϕ \times r^{2} \sin θ$

(where $r$ , $θ$ and $ϕ$ are standard spherical polar co-ordinates).

Calculating the integral gives $I = 8 π / 15 r^{4}$ , $M = 4 / 3 π r^{3}$ so $I = 2 / 5 {Mr}^{2}$ .

You can do the integral directly, but using the symmetry argument makes the maths slightly easier.

Steve