Calculating the moment of inertia of a solid sphere

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SUMMARY

The discussion focuses on calculating the moment of inertia (MoI) of a solid sphere with uniform density using the integral formula \( I = \int r^2 dm \). Participants clarify the correct approach to defining the height of the elemental disk, emphasizing that it should be \( R\cos\theta d\theta \) instead of \( R d\theta \). The correct moment of inertia for a disk is identified as \( \frac{1}{2}MR^2 \), and the integration can be simplified by using multiple angle formulas or by integrating with respect to \( z \) instead of \( \theta \). The final goal is to derive the MoI of the solid sphere as \( \frac{2}{5}MR^2 \).

PREREQUISITES
  • Understanding of integral calculus, specifically integration techniques.
  • Familiarity with the concept of moment of inertia in physics.
  • Knowledge of trigonometric functions and their properties.
  • Basic understanding of solid geometry, particularly spheres and disks.
NEXT STEPS
  • Study the derivation of the moment of inertia for various shapes, including disks and spheres.
  • Learn about integration techniques in calculus, focusing on trigonometric integrals.
  • Explore the use of multiple angle formulas in trigonometry for simplifying integrals.
  • Investigate alternative methods for calculating moment of inertia, such as using the parallel axis theorem.
USEFUL FOR

Students studying physics, particularly those focusing on mechanics and rotational dynamics, as well as educators teaching concepts related to moment of inertia and integration techniques.

Rahulrj
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Homework Statement


To calculate moment of inertia of a solid sphere of uniform density[/B]

Homework Equations


$$ I = \int r^2 dm$$

The attempt at a solution
upload_2017-4-24_11-47-28.png

I consider an elemental disk of small thickness ##d\theta##
##dm = \frac{M}{4/3 \pi R^3}*\pi R^2\cos^2\theta* Rd\theta##
Therefore ##dI = r^2 dm = R^2\cos^2\theta dm## r is the distance from the axis to the disc which is same as ## R\cos\theta##
Then I do the integration ##I = \int_\frac{-\pi}{2}^\frac{\pi}{2} dI##

As I am getting a ##\cos^4\theta## I am not able to get the answer ##2/5 MR^2##
Can someone tell me if I have gone wrong anywhere in what I have written above?
 
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Your first formula implies that ##R\,d\theta## is the height of the incremental disc. It is not. It is the length of the slanted edge of the disc. What do you have to multiply that length by to give the height?

Also, the moment of inertia of the incremental disc is not ##r^2\,dm##. Look at this list of moments of inertia to see what it should be.
 
andrewkirk said:
Your first formula implies that ##R\,d\theta## is the height of the incremental disc. It is not. It is the length of the slanted edge of the disc. What do you have to multiply that length by to give the height?

So then instead of ##Rd\theta## it will be ##R\cos\theta d\theta##? and yes the MoI of disk is ##1/2MR^2##
So now I have to evaluate ##\cos^5x##?
 

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Yes. You can use multiple angle formulas to do this.
Alternatively, you might prefer to integrate with respect to ##z## instead of ##\theta##. That gives an easier integration - but the same result of course.
 

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