How to calculate the moment of inertia of the rigid body?

AI Thread Summary
The discussion focuses on calculating the moment of inertia for different rigid bodies, specifically a thin spherical shell and a solid sphere, using the formula I = ∫ r² dm. For the thin spherical shell, the moment of inertia is derived as I = (2/3)MR², while for the solid sphere, it is I = (2/5)MR². Participants explain the differential area element dA on a sphere using spherical coordinates, emphasizing its importance in the calculations. Additionally, there is a suggestion that deriving the moment of inertia for a solid sphere using thin rings may simplify the process compared to using a volume integral. Understanding these concepts is crucial for applying the moment of inertia in torque calculations and angular acceleration problems.
copperboy
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I don't know how to calculate the following rigid bodies with different geometries, can anybody help me?

Thin spherical shell: I=(2/3)MR^2

Solid sphere: I=(2/5)MR^2

Thanks in advance.
 
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Use the definition of moment of inertia:

I = \int r^2 dm

In the case of the shell the element of mass is dm = M {dA} /{4 \pi R^2} where dA = R^2 \sin \theta d\theta d\phi[/tex]. The distance to a point on the shell from the z-axis is R^2 \sin^2 \theta so<br /> <br /> I = \frac {M}{4 \pi R^2} \int_{0}^{2 \pi} d\phi \int_{-\pi /2}^{\pi /2}R^4 \sin^3 \theta d\theta<br /> <br /> from which the desired result follows.<br /> <br /> In the case of the solid sphere you will work with a volume integral.
 
Could you please explaim why dA = R^2 \sin \theta d\theta d\phi[/tex] in detail?
 
copperboy said:
Could you please explaim why dA = R^2 \sin \theta d\theta d\phi[/tex] in detail?
<br /> Consider an area segment dA on a sphere with radius R.<br /> We approximate this with a rectangle:<br /> a)Two of the sides are arclengths along great circles; the length of each of these is Rd\theta<br /> b) The other two are arclengths in THE HORIZONTAL PLANE; the local radius there is R\sin\theta<br /> Hence, the arclenth is R\sin\theta{d\phi}<br /> c) Multiplying together, we get:<br /> dA=R^{2}\sin\theta{d\theta}d\phi
 
copperboy said:
Could you please explaim why dA = R^2 \sin \theta d\theta d\phi[/tex] in detail?
<br /> <br /> dA is a differential element of area on a spherical surface using spherical coordinates and it represents, to lowest order in differentials, the area of a rectangle R \sin \theta d\phi high and R d\theta units wide.
 
This is the first time I'm involved in physics studies.I'm having a lots of problems using these formulas and the textbook I'm using does not make it easy. one from "James S. Walker"
Can someone please explain how and when to use these formulas.
Thank You.
 
This is one way to look at it

You must be familiar with the equation F=ma.

We use the Moment of Inertia to find the Torque.
Like how the force is directly propotional to acceleration, Torque is directly propotional to angular acceleration, the propotionality constant being the moment of inertia.

Also remember about what axis you are taking the moment of Inertia while solving problems. In most problems i have encounterd which ask you to find the angular acceleration, first find the torqure taking the vector product R X F. Next find the moment of inertia of the object about the required axis. Then the acceleration can be found


Also, for a solid sphere, perphaps deriving the moment of inertia by intergrating thin rings would be easier than a volume intergral?
 
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