Moment of inertia spinning disk integration

  • #1

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.
 

Answers and Replies

  • #2
tiny-tim
Science Advisor
Homework Helper
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Hi ShamelessGit! :smile:

(try using the X2 button just above the Reply box :wink:)
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:)
 
  • #3
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.
 
  • #4
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
 

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