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Derivation of solid sphere moment of inertia

by jordanl122
Tags: derivation, inertia, moment, solid, sphere
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jordanl122
#1
Nov24-04, 11:00 PM
P: 14
I need to derive the moment of inertia for a solid sphere, but I'm having some trouble.
I did the following.

I=?r^2dm

given density, p= m/V
pV=m
so pdV=dm
and differentiating V wrt r, d(4/3?r^3)dr = 4?r^2
so p4?r^2dr=dm and plugging that in I get

I=?r^2(p4?r^2)dr

I pull the p4? out in front

I=p4??r^4dr

evaluating the integral I get

I=(M/(4/3?r^3))4?(r^5/5)

simplifying the terms I get

I=3/5mr^2

which is a universe off from what the answer should be, if anyone can show me where I went wrong I would be very appreciative. Thanks.
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Tide
#2
Nov24-04, 11:31 PM
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Be careful! The r in your definition of the moment is the distance from the axis to the element of mass dm. It is NOT the radial variable in spherical coordinates.
Sirus
#3
Nov25-04, 01:33 PM
P: 578
[tex]\rho=\frac{m}{V}=\frac{m}{\frac{4}{3}\pi r^{3}}[/tex]
[tex]\frac{dm}{dr}=4\rho\pi r^{2}[/tex]

Also,
[tex]I=\int{r^{2}}dm=\int{r^{2}4\rho\pi r^{2}dr}=4\rho\pi\int{r^{4}}dr=\frac{4}{5}\rho\pi r^{5}[/tex]
Now factor something familiar...
[tex]I=\frac{4}{5}\rho\pi r^{5}=\frac{4}{3}\rho\pi r^{3}\left(\frac{3}{5}r^{2}\right)=\frac{3}{5}mr^{2}[/tex]

arildno
#4
Nov25-04, 01:59 PM
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Derivation of solid sphere moment of inertia

Sirus:
This is wrong; read Tide's post.
The moment of inertia of a solid sphere is [tex]I=\frac{2}{5}mr^{2}[/tex]

To derive this, we use that that the distance of a point to the axis of rotation (going through the center of the sphere) is [tex]r=\hat{r}\sin\phi[/tex] where [tex]\hat{r}[/tex] is the distance of a point to the center, and [tex]\phi[/tex] is the angle between the point's position vector [tex]\vec{r}[/tex](measured from the origin) and the rotaion axis.
R is the radius of the sphere
We have then:
[tex]I=\int_{V}r^{2}dm=\int_{0}^{\pi}\int_{0}^{2\pi}\int_{0}^{R}\rho(\hat{r} \sin\phi)^{2}\hat{r}^{2}\sin\phi{d\hat{r}}d\theta{d}\phi=\frac{2\pi\rho }{5}R^{5}\int_{0}^{\pi}\sin^{3}\phi{d}\phi[/tex]
Using the identity:
[tex]\sin^{3}\theta=\sin\theta(1-\cos^{2}\theta)[/tex]
we find that:
[tex]\int_{0}^{\pi}\sin^{3}\phi{d}\phi=\frac{4}{3}[/tex]
krab
#5
Nov25-04, 02:45 PM
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Here's another way, that uses the moment of inertia of a disc.

Let z be the coordinate along the axis. Then we can divide the sphere into discs along z. Each disc has mass [itex]dm=\pi r^2\rho dz[/itex] and moment of inertia [itex](dm)r^2/2[/itex]. But the disc radius r is given by [itex]r^2=R^2-z^2[/itex], where R is the sphere radius. Add them all together:
[tex]I={\pi\over 2}\rho\int_{-R}^R (R^2-z^2)^2dz[/tex].
Sirus
#6
Nov25-04, 06:33 PM
P: 578
Thank you for the correction. Some of this stuff is obviously a little over my head.
enigma72
#7
Apr5-05, 08:45 AM
P: 1
I've likewise been having trouble with the moment of inertia of a sphere proof. I can do it using speherical coords without a problem but am out by a factor of 1/2 when I use cartesian coordinates. I don't know why the [itex](dm)r^2/2[/itex] (as in Krab's post above). Why do we need to divide it by 2?
dextercioby
#8
Apr5-05, 09:20 AM
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Because the moment of inertia of a disk wrt an axis perpendicular to the disk in the center is

[tex] I=\frac{mR^{2}}{2} [/tex]

Daniel.
lcy2
#9
Jul31-05, 05:30 PM
P: 1
but one more question. let's assume that we don't know the moment of inertia of a disc. and we start directly from the sphere.. which means we start from the formula integral (r ^2 * dm)
dm = dV * p = A * dz * p = pi r^2 dz p
r^2 = (R^2 - z^2)
so the original formula then turns into:
pi p integral ((R^2 - z^2)^2 dz
distribute what's inside the parenthesis
pi p integral ((R^4 - 2R^2 z^2 + z^4) dz
integrate:
pi p [R^4 z - 2R^2 1/3 z^3 + 1/5 z^5]from R to -R
throw everything together
2 pi p [R^5- 2/3 R^5 + 1/5 R^5] <- R- (-R) = 2R
pi p 16/15 R^5
M = V * p = 4/3 pi R^3 p
I = 4/5 R^2

it's somehow 2 times the actual inertia. and.. i don't want to start with the inertia of the disk, which tells me to divide it by 2. Can anyone tell me where I got wrong? like where in my train of thoughts is wrong. don't refer to the disc's inertia please. thank you
mukundpa
#10
Aug1-05, 05:40 AM
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.. we start from the formula integral (r ^2 * dm)
dm = dV * p = A * dz * p = pi r^2 dz p
r^2 = (R^2 - z^2)...

The whole mass dm is not at distance r from the axis of rotation, it is distributed over area A. (You have considerd it)
OlderDan
#11
Aug1-05, 10:28 AM
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Quote Quote by lcy2
Can anyone tell me where I got wrong? like where in my train of thoughts is wrong. don't refer to the disc's inertia please. thank you
When you first learned to find the volume of a solid of rotation you likely learned two methods, the disk method and the concentric cylinder method. You have divided your solid into disks of mass dm, but as mukundpa has observed the mass in the disk is not at the same distance from the axis of rotation. If, as you say, you do not want to use the moment of inertia of a disk in your calculation, then don't divide the solid into disks.

If instead you divided the solid into coaxial cylinders of mass dm, each cylinder's mass would all be at the same distance from the axis. If you follow this approach you will get the correct answer.


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