Moment of inertia spinning disk integration

Click For Summary

Homework Help Overview

The original poster attempts to calculate the moment of inertia of a spinning disk using integration, expressing familiarity with the perpendicular axis theorem but preferring an integration approach. The problem involves setting up the integral for moment of inertia, specifically using polar coordinates.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the setup of the integral, with the original poster initially integrating with incorrect limits and expressions. There is a focus on the transformation from Cartesian to polar coordinates and the implications of variable definitions.

Discussion Status

Some participants provide feedback on the limits of integration and suggest corrections to the original poster's setup. The original poster acknowledges mistakes in their approach and indicates progress towards the correct answer after reevaluating their integration process.

Contextual Notes

There are mentions of typographical errors in the limits of integration and confusion regarding variable definitions, which are under discussion but not resolved. The original poster's understanding of trigonometric identities is also noted as an area for improvement.

ShamelessGit
Messages
38
Reaction score
0

Homework Statement


I want to calculate the moment of inertia of a spinning disk via integration. I'm aware of the perpedicular axis theorem, but I want to integrate.


Homework Equations


I = ∫r^2dm


The Attempt at a Solution



if I set my coordinate axis op so that the origin of the xy plane lies in the center of my disk, and I decide to rotate the disk about the x axis, then I figure r in this case is equal to y and I integrate along the x-axis from -R to R.

So I set up ∫∫y^2dydx, which I can change to ∫∫r*y^2drdθ. y = rsin(θ), so

∫∫r^3*sin(θ)^2drdθ, where the limits of integration are 0< r < 2R (R = max radius), 0 < θ < 2PI

I look up the integral of sin(x) and I get x/2 - sin(2x), so I think my answer to this integral should be (PI/2)R^4. Of course the answer should be (PI/4)R^4. I don't understand what I did wrong. I think I know how to integrate so I assume something is wrong with my set-up.
 
Physics news on Phys.org
Hi ShamelessGit! :smile:

(try using the X2 button just above the Reply box :wink:)
ShamelessGit said:
So I set up ∫∫y^2dydx, which I can change to ∫∫r*y^2drdθ. y = rsin(θ), so

∫∫r^3*sin(θ)^2drdθ, where the limits of integration are 0< r < 2R (R = max radius), 0 < θ < 2PI

no, 0 < r < R :wink:

I look up the integral of sin(x) and I get x/2 - sin(2x), so I think my answer to this integral should be (PI/2)R^4

not following this :confused:

can you show your working?​

(btw, learn your trigonometric identities … sin2x = (1 - cos2x)/2, which is easier! :smile:)
 
I made a typo when I was giving the limits of integration. The way I set it up the limit of integration for r should be -r to r.

And also I figured out that when I integrate properly, I do get the right answer.

The mistake I made was that I changed y to sqrt(R^2-x^2) in my original calculation before realizing that that was unnecessary, and when I changed it back I got r*R^2 instead of r^3 because I got confused about which R was a constant and which was a variable. So I got R to the right power, but of course I got a factor of 1/2 instead of 1/4.
 
ShamelessGit said:
I made a typo when I was giving the limits of integration. The way I set it up the limit of integration for r should be -r to r.

And also I figured out that when I integrate properly, I do get the right answer.

The mistake I made was that I changed y to sqrt(R^2-x^2) in my original calculation before realizing that that was unnecessary, and when I changed it back I got r*R^2 instead of r^3 because I got confused about which R was a constant and which was a variable. So I got R to the right power, but of course I got a factor of 1/2 instead of 1/4.

I made a type when trying to fix the previous typo. Yes the limit is only from 0 to r. Srry about that. I didn't make that mistake on the paper when I was actually trying to integrate
 

Similar threads

  • · Replies 28 ·
Replies
28
Views
2K
Replies
11
Views
4K
  • · Replies 13 ·
Replies
13
Views
3K
  • · Replies 6 ·
Replies
6
Views
3K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 27 ·
Replies
27
Views
6K
  • · Replies 1 ·
Replies
1
Views
1K
Replies
25
Views
2K
Replies
8
Views
2K
Replies
17
Views
2K