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Moment of inertia of a hollow sphere

by chickendude
Tags: hollow, inertia, moment, sphere
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chickendude
#1
Dec17-07, 04:53 PM
P: 28
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:

[tex]I = \int r^2 dm = \frac{m}{A}\int r^2 dA[/tex]


3. The attempt at a solution

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

[tex]dA = 2\pi \sqrt{R^2-r^2}dr[/tex]

[tex]I = \frac{m}{4\pi R^2} \int_{-R}^{R} r^2(2\pi\sqrt{R^2-r^2})dr[/tex]

[tex]I = \frac{m}{R^2} \int_{0}^{R} r^2\sqrt{R^2-r^2}dr[/tex]

From here I made the substitution [tex]r = R\sin{\theta}[/tex] and got

[tex]I = mR^2 \int_{0}^{\frac{\pi}{2}} \sin^2\theta\cos^2\theta d\theta[/tex]

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


the correct answer is supposed to be [tex]I = \frac{2mR^2}{5}[/tex]
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#2
Dec17-07, 06:10 PM
P: 91
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 [tex] R^2 \sin (\theta) d\theta d\phi dR[/tex]. For a spherical shell R is constant and the volume element becomes an area element [tex] dA= R^2 \sin (\theta) d\theta d\phi [/tex]. 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
#3
Dec17-07, 06:33 PM
P: 28
You are right. I am sorry. I looked up the wrong answer in the chart.

It should be [tex]\frac{2mR^2}{3}[/tex]
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:

[tex]\frac{m}{A}\int r^2 dA[/tex]

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

plugging in


[tex]\frac{m}{4\pi R^2} \int_{0}^{\pi} R^2\sin^2\phi * 2\pi R^2 \sin\phi d\phi[/tex]

[tex]\frac{mR^2}{2} \int_{0}^{\pi}\sin^3\phi d\phi[/tex]

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

phoenix39
#4
Nov22-09, 12:09 AM
P: 2
Moment of inertia of a hollow sphere

Quote Quote by chickendude View Post
[tex]\frac{mR^2}{2} \int_{0}^{\pi}\sin^3\phi d\phi[/tex]

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
#5
Nov22-09, 01:14 AM
HW Helper
ideasrule's Avatar
P: 2,323
You didn't convert the bounds of integration correctly. cos(pi)=-1, not 0.
phoenix39
#6
Nov22-09, 01:21 AM
P: 2
Thanks alot....why I am so stupid....-_-llll
tejas
#7
Jan3-11, 10:24 AM
P: 7
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)


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