Moment of Inertia for a hollow Sphere

AI Thread Summary
The discussion revolves around the derivation of the moment of inertia for a hollow sphere, with confusion arising from the notation used in the derivation. A misprint was identified where a lowercase "r" was mistakenly used instead of "R," leading to clarity on the correct interpretation of the radius in the context of the elemental ring. Participants confirmed that the circumference and thickness of the ring are based on the radius R and the angle φ. Additionally, an alternative method for calculating the moment of inertia involves using the difference between the moments of two solid spheres. The conversation highlights the importance of accurate notation in mathematical derivations.
Xyius
Messages
501
Reaction score
4
I am confused about one thing on this derivation. Okay so the guide I am following goes like this..

-------------------------------------
mi10b.gif


\sigma=\frac{M}{A}

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

dm=\frac{M}{2}sin\phi d\phi​

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 :
dI=R^2sin^2\phi dm=R^2sin^2\phi (\frac{M}{2}sin\phi d\phi)​

Therefore the total moment of inertia is..

\oint R^2sin^2\phi (\frac{M}{2}sin\phi d\phi)
\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 )​

The limits are 0 to \pi hence..
I=\frac{2}{3}MR^2​
 
Physics news on Phys.org
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:
 


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?
 
Yup! The circumference is 2πRsinφ, and the thickness is Rdφ. :smile:
 
tiny-tim said:
Yup! The circumference is 2πRsinφ, and the thickness is Rdφ. :smile:

Ohh! Makes sense now! Cool thanks a lot! :D
 
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
 
Hi there, im studying nanoscience at the university in Basel. Today I looked at the topic of intertial and non-inertial reference frames and the existence of fictitious forces. I understand that you call forces real in physics if they appear in interplay. Meaning that a force is real when there is the "actio" partner to the "reactio" partner. If this condition is not satisfied the force is not real. I also understand that if you specifically look at non-inertial reference frames you can...
I have recently been really interested in the derivation of Hamiltons Principle. On my research I found that with the term ##m \cdot \frac{d}{dt} (\frac{dr}{dt} \cdot \delta r) = 0## (1) one may derivate ##\delta \int (T - V) dt = 0## (2). The derivation itself I understood quiet good, but what I don't understand is where the equation (1) came from, because in my research it was just given and not derived from anywhere. Does anybody know where (1) comes from or why from it the...
Back
Top