# Moment of inertia

1. May 2, 2005

### mick_1

moment of inertia for a ball shell

Is ther anyone that can explain how to calculate moment of inertia for a ball shell?? I know that the answer is 2/3mR^2, but how can it be showen with integral??

Last edited: May 2, 2005
2. May 2, 2005

### Galileo

Draw a picture and use spherical coordinates. Set the z-axis as the axis of rotation.

$$I=\int_S r^2 dm = \int_S (R\sin \theta)^2 \sigma da$$

where R is the radius of the sphere and $\sigma$ the surface mass density.

Note that $da = R^2\sin \theta d\theta d\phi$:

$$I= \sigma R^4 \int_0^{2\pi}\int_0^\pi \sin^3 \theta d\theta d\phi=\frac{8}{3}\pi \sigma R^4=\frac{2}{3}MR^2$$

Last edited: May 2, 2005
3. May 2, 2005

### mick_1

What do you mean by $dV = R^2\sin \theta d\theta d\phi$ ??

4. May 2, 2005

### inha

It's an element of volume in spherical coordinates.

5. May 2, 2005

### Galileo

Yeah, although it should be AREA element: $da$, I'll change it now...

6. May 2, 2005

### mick_1

I don't understand this calculation with $da = R^2\sin \theta d\theta d\phi$
??
Do you have a picture to show this??

7. May 2, 2005

### SpaceTiger

Staff Emeritus
http://www.usd.edu/phys/courses/phys431/notes/sphercoor.gif [Broken] one I found on a google search. The area element is just the top face of the shaded box.

Last edited by a moderator: May 2, 2017
8. May 2, 2005

### Galileo

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

It is shown at eqn 14. Note that the notation for the angle of declination (zenith) and the azimuth are interchanged w.r.t. mine ($\phi \leftrightarrow \theta$)

EDIT: Ah! Spacetiger has exactly the kind of picture I was looking for.