Moment of inertia of a hollow sphere

[chickendude]

Homework Statement

Find the moment of inertia of a hollow sphere with mass m and radius R and uniform density

Homework 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

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

 

[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)

 

[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

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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

 

[phoenix39]

\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

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?

 

[ideasrule]

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

 

[phoenix39]

Thanks a lot...why I am so stupid...-_-llll

 

[tejas]

HI, I understand everything, except that in the solution, dz is written as R d\phi. why is that? shouldn't 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)