Calculating Moment of Inertia for a Sphere: What's the Best Approach?

Click For Summary
SUMMARY

The moment of inertia for a solid sphere is calculated using two primary approaches: integrating the moments of inertia of disks or cylindrical shells. The first method involves slicing the sphere into disks of thickness dx and integrating their moments of inertia about the symmetry axis. The second method treats the sphere as concentric cylindrical shells, requiring the determination of the radius and length of each shell. Both methods ultimately lead to the established formula for the moment of inertia of a sphere, which is (2/5)MR².

PREREQUISITES
  • Understanding of calculus, specifically integration techniques.
  • Familiarity with the concept of moment of inertia.
  • Knowledge of geometric properties of spheres and disks.
  • Ability to manipulate equations involving mass density and volume.
NEXT STEPS
  • Study the derivation of the moment of inertia for a disk about its symmetry axis.
  • Learn how to calculate the moment of inertia for cylindrical shells.
  • Explore the equations for the surface area and volume of a sphere.
  • Practice integration techniques relevant to physics problems involving rotational dynamics.
USEFUL FOR

High school students, physics enthusiasts, and anyone interested in understanding the principles of rotational dynamics and calculating moments of inertia.

Ja4Coltrane
Messages
224
Reaction score
0
This was not actually homework, but I was just trying to see if I could calculate moments of inertia and apparently, I cannot.
I'm trying to show that the moment of inertia for a sphere is (2/5)MR^2
So I started with I=(integral)(r^2)(dm)
then P=dm/dv=dm/(4pi(r^2)dr)
so dm=(4)(pi)(r^2)(P)(dr)
so I substituted into the original equation, removed constants from the integral, and substituted P for M/V=(3M/(4(pi)(R^3)))
I=12pi(M)/(4(pi)r^3)[integral]r^4 (dr)
I=(3/5)MR^2 which is wrong!

(sorry about the lack of pretty math writing)
Thanks for any help, and by the way, I'm only a high school student so my calculus knowledge is very limited (in fact, the only reason I know what integration is is because of my physics class).
 
Physics news on Phys.org
Oh, I just realized something!
I am integrating as if the higher part has the same radius because it is the same distance from the center but it is closer to the axis!
now I really don't know what to do.
 
Ja4Coltrane said:
Oh, I just realized something!
I am integrating as if the higher part has the same radius because it is the same distance from the center but it is closer to the axis!
now I really don't know what to do.
This is not a simple calculation. There are two approaches you can take with limited calculus experience. The first is to find the moment of inertia of a disk about its symmetry axis, and then slice the sphere into disks of thickness dx having a common axis that is a diameter of the sphere. Then add (integrate) the moments of inertia of all the disks. The hard part is finding the radius of each disk as a function of x, but that can be done using the equation for the surface of the sphere.

The second approach is to find the moment of inertia of a cylindrical shell about its symmetry axis (easy since all the mass has the same radius) and think of the sphere as many concentric shells of radius r and thickness dr. The hard part here is finding the length of each cylinder as a function of r, but again this can be found from the equation for the surface of the sphere.
 

Similar threads

Replies
17
Views
2K
Replies
8
Views
2K
  • · Replies 13 ·
Replies
13
Views
3K
Replies
52
Views
4K
  • · Replies 3 ·
Replies
3
Views
1K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 15 ·
Replies
15
Views
1K
  • · Replies 11 ·
Replies
11
Views
1K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 28 ·
Replies
28
Views
2K