# Moment of inertia of a hollow sphere

1. Dec 17, 2007

### chickendude

1. The problem statement, all variables and given/known data
Find the moment of inertia of a hollow sphere with mass m and radius R and uniform density

2. Relevant equations

Since the hollow sphere is an area, the density is mass divided by area, so:

$$I = \int r^2 dm = \frac{m}{A}\int r^2 dA$$

3. The attempt at a solution

. The total area is 4pi r^2, so here is what I got

$$dA = 2\pi \sqrt{R^2-r^2}dr$$

$$I = \frac{m}{4\pi R^2} \int_{-R}^{R} r^2(2\pi\sqrt{R^2-r^2})dr$$

$$I = \frac{m}{R^2} \int_{0}^{R} r^2\sqrt{R^2-r^2}dr$$

From here I made the substitution $$r = R\sin{\theta}$$ and got

$$I = mR^2 \int_{0}^{\frac{\pi}{2}} \sin^2\theta\cos^2\theta d\theta$$

And that evaluated to pi/16, which brings me to my problem

the correct answer is supposed to be $$I = \frac{2mR^2}{5}$$

Last edited: Dec 17, 2007
2. Dec 17, 2007

### what

Is it a hollow sphere(a spherical shell) or a solid sphere? Because the answer that you say is supposed to be correct is for a solid sphere, unless I've made some error. They're both pretty similar and I think you're going to get into trouble if you don't use curvilinear coordinates, at least it was much easier for me to do it that way. My response is to a spherical shell which will not give the answer you say is correct. However, a solid sphere can be done in a similar way and does give the answer you say is correct.

First, something to notice is that in spherical coordinates a volume element is $$R^2 \sin (\theta) d\theta d\phi dR$$. For a spherical shell R is constant and the volume element becomes an area element $$dA= R^2 \sin (\theta) d\theta d\phi$$. Can you now do the integral keeping in mind that in your initial given equation r is the perpendicular distance from an axis passing through the center of mass? (I assumed it was moment of inertia wrt the center of mass)

3. Dec 17, 2007

### chickendude

You are right. I am sorry. I looked up the wrong answer in the chart.

It should be $$\frac{2mR^2}{3}$$
I will look into the spherical coordinate method

-----------------------------------------------

Yes, it worked. The error was my r was not actually representing the perpendicular distance.

Here is what I did:

$$\frac{m}{A}\int r^2 dA$$

$$A = 4\pi R^2$$
$$r = R\sin\phi$$ (simple spherical geometry)
$$dA = 2\pi r dz = 2\pi * R\sin\phi * R d\phi$$ (definition of radian and the statement above)
$$dA = 2\pi R^2 \sin\phi d\phi$$

plugging in

$$\frac{m}{4\pi R^2} \int_{0}^{\pi} R^2\sin^2\phi * 2\pi R^2 \sin\phi d\phi$$

$$\frac{mR^2}{2} \int_{0}^{\pi}\sin^3\phi d\phi$$

That integral evaluates to 4/3, and bam, it works
Thanks a lot

Last edited: Dec 17, 2007
4. Nov 22, 2009

### phoenix39

How would you get 4/3 for the intergral, i used substitution rule and always got 2/3 for the answer.
∫_0^π▒〖sin〗^3 ϕ dϕ=∫_0^π▒〖(1-〖cos〗^2 ϕ)〗sinϕ dϕ
u= cosϕ du=-sinϕ dϕ
so ∫_1^0▒〖-(1-u^2 )du〗=∫_0^1▒〖1-u^2 du〗 = 2/3
what is my problem?

5. Nov 22, 2009

### ideasrule

You didn't convert the bounds of integration correctly. cos(pi)=-1, not 0.

6. Nov 22, 2009

### phoenix39

Thanks alot....why I am so stupid....-_-llll

7. Jan 3, 2011

### tejas

HI, I understand everything, except that in the solution, dz is written as R d\phi. why is that? shouldnt it be Rsin(phi)d(phi) or even, if we take x = R*cos(phi) (from that same simple geometry) than dz would be -Rsin(phi)d(phi)

Last edited: Jan 3, 2011