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bowma166
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Homework Statement
Find the moment of inertia of a flywheel of mass M made by cutting four large holes of radius r out of a uniform disk of radius R. The holes are centered at a distance R/2 from the center of the flywheel.
Here is a very poorly drawn diagram of the disk. Sorry for how crappy it is.
http://img407.imageshack.us/img407/4131/flywheelro2.jpg
Homework Equations
[tex]I=\int R^{2}dm[/tex]
(Is this necessary? I don't think it is given the next two equations.)
[tex]I_{disk}=\frac{1}{2}MR^{2}[/tex]
[tex]I=I_{CM}+md^{2}[/tex]
The Attempt at a Solution
You can just add/subtract moment of inertias of things, right? I'm not sure about that but that's what I tried.
If the disk didn't have holes, then [itex]I=\frac{1}{2}MR^{2}[/itex]. But it does... the holes are of mass
[tex]\frac{\pi r^{2}}{\pi R^{2}}M=\frac{r^{2}}{R^{2}}M[/tex]
and are a distance R/2 from the center. So the total moment of inertia should be
[tex]I=\frac{1}{2}MR^{2}-4\left(\frac{r^{2}}{R^{2}}\right)M\left(\frac{R}{2}\right)^{2}=\frac{1}{2}MR^{2}-Mr^{2}[/tex]
right? It makes sense because as r -> 0, I -> I_disk, but I really have no idea if it is right or not. Thanks!
::edit:: Just kidding, I know I did that wrong... I'm working on redoing it now.
::edit again::
Okay. This time I think I may have done it more correctly.
The moment of inertia of each small disk about the center of the big disk should be
[tex]I_{hole}=\frac{1}{2}\frac{r^{2}}{R^{2}}Mr^{2}+\left(\frac{r^{2}}{R^{2}}M\right)\left(\frac{R}{2}\right)^{2}=\frac{Mr^{4}}{2R^{2}}+\frac{Mr^{2}}{4}[/tex]
so the total moment of inertia is the big disk's I minus four times the holes' I, right?
[tex]I=\frac{1}{2}MR^{2}-4\left(\frac{Mr^{4}}{2R^{2}}+\frac{Mr^{2}}{4}\right)=\frac{1}{2}MR^{2}-\frac{2Mr^{4}}{R^{2}}-Mr^{2}[/tex]
So...
[tex]I=\frac{1}{2}M\left(R^{2}-\frac{4r^{4}}{R^{2}}-2r^{2}\right)[/tex]
That seems kind of messy.
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