- #1
Screwdriver
- 129
- 0
Not a homework question per se, but I'm having some issues with moments of inertia. Say I wanted to calculate the I for a ring. What I would do is:
[tex]I = \int r^2dm[/tex]
[tex]m = \lambda L[/tex]
[tex]dm = \lambda dL[/tex]
[tex]I_{ring} = \int_{0}^{L}\lambda r^2dL[/tex]
And that would give the requiside mr2. My question is, why can't I just integrate THAT up to get the I of a disk. I mean something like this:
[tex]I_{disk} = \int_{0}^{A}\int_{0}^{L}\lambda r^2dLdA[/tex]
Where A is the area of the disk. Doesn't figuring out the I for a ring essentially skip the first step for determining the I for a disk?
[tex]I = \int r^2dm[/tex]
[tex]m = \lambda L[/tex]
[tex]dm = \lambda dL[/tex]
[tex]I_{ring} = \int_{0}^{L}\lambda r^2dL[/tex]
And that would give the requiside mr2. My question is, why can't I just integrate THAT up to get the I of a disk. I mean something like this:
[tex]I_{disk} = \int_{0}^{A}\int_{0}^{L}\lambda r^2dLdA[/tex]
Where A is the area of the disk. Doesn't figuring out the I for a ring essentially skip the first step for determining the I for a disk?