Find the moment of inertia of a solid sphere.

In summary, you are trying to find the moment of inertia of a solid sphere about a perpendicular axis.
  • #1
Squizzel
29
0

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.

Thank you in advance!
 
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  • #3
I saw that, the only issue is I am supposed to start with the above formula. Icm = Integral of r^2 dm
 
  • #4
Squizzel said:
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[itex]_{\bot}[/itex] 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.
 
  • #5
Squizzel said:

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.

Thank you in advance!
The problem asks you to "find the moment of inertia of a solid sphere about any tangential axis."

I assume that you are not allowed to use the parallel axis theorem for this.
 

Related to Find the moment of inertia of a solid sphere.

What is the formula for finding the moment of inertia of a solid sphere?

The formula for finding the moment of inertia of a solid sphere is I = (2/5) * mr^2, where m is the mass of the sphere and r is the radius.

What is the unit of measurement for moment of inertia?

The unit of measurement for moment of inertia is kilogram meter squared (kg·m^2).

How does the moment of inertia of a solid sphere compare to that of a hollow sphere?

The moment of inertia of a solid sphere is greater than that of a hollow sphere with the same mass and radius.

What is the physical significance of moment of inertia?

Moment of inertia is a measure of an object's resistance to changes in its rotational motion. It is an important concept in physics and engineering, and is used to analyze the behavior of rotating objects.

Can the moment of inertia of a solid sphere change?

No, the moment of inertia of a solid sphere is a constant value that only changes if the mass or radius of the sphere changes.

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