# Rotational inertia of a THICK spherical shell

1. Jul 17, 2004

### Will

Someone please tell me is I am doing this problem correctly.If I have a thick spherical shell with inner radius r, outer radius R, and mass M, I am getting [(2/5)M/(R^3-r^3)](R^5-r^5). It is not the same thing as subtracting I of large sphere from I of smaller one, different than (2M(R^2-r^2)?

$$\frac{\2 (M(R^5-r^5))}{5(R^3-r^3)}$$

Last edited by a moderator: Jul 29, 2004
2. Jul 18, 2004

### JohnDubYa

I think we need to see your work. Do you know how to LaTeX your posts?

3. Jul 18, 2004

### Staff: Mentor

That's correct.
Not exactly. If you treat the hole as a sphere of negative mass, then you can subtract the rotational inertia of each sphere: $I_{shell} = I_{R-sphere} - I_{r-sphere}$. But realize that the mass of each sphere is different. If you express this answer in terms of the mass of the shell instead of the mass of either sphere, then you will find that you get the same answer as above.

4. Jul 18, 2004

### JohnDubYa

Good catch, Doc Al. He is saying that the M in the I = 2/5 MR^2 is not the same for the sphere as it is for the inner core. So you would have to provide unique labels for each.

5. Jul 20, 2004

### Will

Do you mean making my equations in "pretty print"? Please show me where I can learn to do this, its so much easier to read!

6. Jul 20, 2004

### Staff: Mentor

learning Latex

Poke around in this thread for many, many examples: https://www.physicsforums.com/showthread.php?t=8997

7. Jul 20, 2004

### krab

As a check, let the shell thickness approach zero to get the MI of a thin shell.
$$\Delta R^5/\Delta R^3=5R^4/3R^2=(5/3)R^2$$
This is multiplied by $(2/5)M$. The result is the correct answer of
$$(2/3)MR^2$$.

8. Jul 22, 2004

### Will

??? Does R man radius in or out?
Doesn't the other radius come into the equation?

9. Jul 22, 2004

### krab

Sorry for the shorthand. $\Delta$ means the difference between the case with R and the case with r. So for example by $\Delta R^5$ means $R^5-r^5$.