# Find the moment of inertia of a solid sphere.

## Homework Statement

Beginning with Icm = Integral of r^2 dm from r1 to r2, find the moment of inertia of a solid sphere about any tangential axis.

## Homework Equations

Icm = Integral of r^2 dm

## The Attempt at a Solution

I set up the infinitesimally mass of an infinitesimally thin "shell" of the sphere:

dm = 4ρπr2 dr

And then solved for the moment of inertia:

I = ∫r2dm

= ∫r2(4ρπr2 dr)

= 4ρπ∫r4 dr

= (4/5)ρπr5

And solving for ρ we get the following:

ρ = M/((4/3)πr3).

Substituting that into the previously solved equation for I, I get the following:

I = (3/5)Mr3.

What am I doing wrong? I know the formula involves a coefficient of 2/5, not 3/5, but I can't find my problem.

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I saw that, the only issue is I am supposed to start with the above formula. Icm = Integral of r^2 dm

TSny
Homework Helper
Gold Member
I saw that, the only issue is I am supposed to start with the above formula. Icm = Integral of r^2 dm
I believe your problem is that you are misinterpreting the meaning of "r" in Icm = Integral of r^2 dm. It does not represent the distance from the mass element dm to the origin of your coordinate system. Rather it represents the perpendicular distance from dm to the axis of rotation. So, if the axis of rotation is the z-axis, then r is the distance from dm to the z-axis.

That's why it's preferable to write r$_{\bot}$ as in the link that azizlwl gave.

You can't find Icm of the sphere by doing a whole shell at a time, because different mass elements of the shell are at different distances from the axis of rotation.

SammyS
Staff Emeritus
Homework Helper
Gold Member

## Homework Statement

Beginning with Icm = Integral of r^2 dm from r1 to r2, find the moment of inertia of a solid sphere about any tangential axis.

## Homework Equations

Icm = Integral of r^2 dm

## The Attempt at a Solution

I set up the infinitesimally mass of an infinitesimally thin "shell" of the sphere:

dm = 4ρπr2 dr

And then solved for the moment of inertia:

I = ∫r2dm

= ∫r2(4ρπr2 dr)

= 4ρπ∫r4 dr

= (4/5)ρπr5

And solving for ρ we get the following:

ρ = M/((4/3)πr3).

Substituting that into the previously solved equation for I, I get the following:

I = (3/5)Mr3.

What am I doing wrong? I know the formula involves a coefficient of 2/5, not 3/5, but I can't find my problem.