Find the moment of inertia of a solid sphere.

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SUMMARY

The moment of inertia of a solid sphere about any tangential axis is calculated using the formula Icm = Integral of r^2 dm. The user initially derived I = (3/5)Mr^2 but recognized that the correct coefficient should be 2/5. The error stems from misinterpreting "r" as the distance from the mass element to the origin instead of the perpendicular distance to the axis of rotation. The correct approach involves integrating with respect to the proper distance from the axis of rotation.

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  • Understanding of moment of inertia concepts
  • Familiarity with integral calculus
  • Knowledge of mass distribution in solid objects
  • Concept of perpendicular distances in rotational dynamics
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  • Review the derivation of the moment of inertia for various shapes, focusing on solid spheres
  • Study the application of the parallel axis theorem in rotational dynamics
  • Learn about the significance of the perpendicular distance in calculating moment of inertia
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Squizzel
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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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I saw that, the only issue is I am supposed to start with the above formula. Icm = Integral of r^2 dm
 
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_{\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.
 
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.
 

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