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

Click For Summary
SUMMARY

The discussion focuses on calculating the moment of inertia for rigid bodies, specifically a thin spherical shell and a solid sphere. The moment of inertia for a thin spherical shell is given by the formula I=(2/3)MR^2, while for a solid sphere, it is I=(2/5)MR^2. The integral definition of moment of inertia, I = ∫ r² dm, is utilized, with detailed explanations provided for the area element dA in spherical coordinates. The conversation emphasizes the importance of understanding the axis of rotation when applying these formulas in physics problems.

PREREQUISITES
  • Understanding of moment of inertia and its significance in rotational dynamics.
  • Familiarity with spherical coordinates and their application in calculus.
  • Basic knowledge of integrals and differential elements in physics.
  • Concept of torque and its relationship with angular acceleration.
NEXT STEPS
  • Study the derivation of moment of inertia for various geometries, including cylinders and disks.
  • Learn about the application of the parallel axis theorem in calculating moment of inertia.
  • Explore the relationship between torque, moment of inertia, and angular acceleration in rotational motion.
  • Investigate the use of thin ring integration for deriving moment of inertia in solid objects.
USEFUL FOR

Students and professionals in physics, particularly those studying mechanics, as well as engineers and educators looking to deepen their understanding of rotational dynamics and moment of inertia calculations.

copperboy
Messages
8
Reaction score
0
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.
 
Physics news on Phys.org
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?
 

Similar threads

  • · Replies 10 ·
Replies
10
Views
2K
  • · Replies 12 ·
Replies
12
Views
2K
  • · Replies 30 ·
2
Replies
30
Views
3K
  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 5 ·
Replies
5
Views
1K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 35 ·
2
Replies
35
Views
6K
  • · Replies 3 ·
Replies
3
Views
2K