Moment of Inertia for a hollow Sphere

In summary, the conversation discusses the confusion surrounding the use of "r" and "R" in a derivation for the moment of inertia of a hollow sphere. The correct expression is "R" as the radius is constantly changing. The conversation also mentions the best way to calculate the moment of inertia of a hollow sphere using cylindrical coordinate system.
  • #1
Xyius
508
4
I am confused about one thing on this derivation. Okay so the guide I am following goes like this..

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mi10b.gif


[tex]\sigma=\frac{M}{A}[/tex]

[tex]dm=\sigma dA=(\frac{M}{4\pi R^2})2\pi rsin\phi Rd\phi[/tex]

[tex]dm=\frac{M}{2}sin\phi d\phi[/tex]​

This is one part that confuses me. It seems as though the lower case "r" was used to cancel out one of the "R's" on the bottom. I can't reason another way how it went away. The rest of the derivation goes like this.. (Just in case anyone needs it)

Here, radius of elemental ring about the axis is R sinθ. Moment of inertia of elemental mass is :
[tex]dI=R^2sin^2\phi dm=R^2sin^2\phi (\frac{M}{2}sin\phi d\phi)[/tex]​

Therefore the total moment of inertia is..

[tex]\oint R^2sin^2\phi (\frac{M}{2}sin\phi d\phi)[/tex]
[tex]\frac{MR^2}{2}\oint sin^3\phi d\phi = \frac{MR^2}{2}\oint (1-cos^2\phi )sin\phi d\phi = \frac{MR^2}{2}\oint sin\phi - sin\phi cos^2\phi d\phi = \frac{MR^2}{2}(-cos\phi +\frac{1}{3} cos^3\phi )[/tex]​

The limits are 0 to [tex]\pi[/tex] hence..
[tex]I=\frac{2}{3}MR^2[/tex]​
 
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  • #2
Welcome to PF!

Hi Xyius! Welcome to PF! :wink:
Xyius said:
… It seems as though the lower case "r" was used to cancel out one of the "R's" on the bottom …

It's a misprint … there is no r ! :biggrin:

read it as R. :smile:
 
  • #3


tiny-tim said:
It's a misprint … there is no r ! :biggrin:

read it as R. :smile:

Thanks! But since that expression came from the differential dA, why is it "R"?? Because the radius is constantly changing right?
 
  • #4
Yup! The circumference is 2πRsinφ, and the thickness is Rdφ. :smile:
 
  • #5
tiny-tim said:
Yup! The circumference is 2πRsinφ, and the thickness is Rdφ. :smile:

Ohh! Makes sense now! Cool thanks a lot! :D
 
  • #6
The best way to calculate the moment of inertia (MOI) of a hollow sphere is to calculate the MOI of two solid spheres, and subtract the MOI of the smaller sphere from the MOI of the larger sphere.

The best way to calculate the MOI of a solid sphere is to use cylindrical coordinate system, r, θ, z. Using ρ as density, the basic form of the integral is

I = ∫∫∫ρ·r2·r·dr·dθ·dz

where r is the perpendicular distance from the axis of rotation, R is the radius of the sphere, and ρ = M/(4πR3/3). Do the z integration last. You will need to determine and use the appropriate integration limits. Hint: Use Phythagorean theorem.

Bob S
 

FAQ: Moment of Inertia for a hollow Sphere

What is Moment of Inertia for a hollow Sphere?

Moment of Inertia for a hollow Sphere is a measure of an object's resistance to changes in its rotational motion. It is the sum of the products of the mass of each particle in an object and the square of its distance from the axis of rotation.

How is Moment of Inertia for a hollow Sphere calculated?

Moment of Inertia for a hollow Sphere can be calculated using the formula I = 2/3 * mr², where m is the mass of the hollow sphere and r is the radius of the sphere.

What factors affect the Moment of Inertia for a hollow Sphere?

The Moment of Inertia for a hollow Sphere is affected by the mass and distribution of the mass within the sphere, as well as the radius of the sphere. A larger mass or a larger radius will result in a larger Moment of Inertia.

Why is understanding Moment of Inertia for a hollow Sphere important?

Understanding Moment of Inertia for a hollow Sphere is important in many fields, such as engineering and physics, as it helps in predicting and analyzing the rotational motion of objects. It is also essential in designing and optimizing structures and machines.

How does Moment of Inertia for a hollow Sphere differ from Moment of Inertia for a solid Sphere?

Moment of Inertia for a hollow Sphere is smaller than that of a solid Sphere with the same mass and radius. This is because the mass in a hollow sphere is distributed further away from the axis of rotation, resulting in a larger moment arm and therefore a smaller Moment of Inertia.

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