# Moment of inertia flywheel mass problem

1. Nov 20, 2008

### bowma166

1. The problem statement, all variables and given/known data
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.

2. Relevant equations
$$I=\int R^{2}dm$$
(Is this necessary? I don't think it is given the next two equations.)

$$I_{disk}=\frac{1}{2}MR^{2}$$

$$I=I_{CM}+md^{2}$$

3. 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 $I=\frac{1}{2}MR^{2}$. But it does... the holes are of mass
$$\frac{\pi r^{2}}{\pi R^{2}}M=\frac{r^{2}}{R^{2}}M$$
and are a distance R/2 from the center. So the total moment of inertia should be

$$I=\frac{1}{2}MR^{2}-4\left(\frac{r^{2}}{R^{2}}\right)M\left(\frac{R}{2}\right)^{2}=\frac{1}{2}MR^{2}-Mr^{2}$$

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
$$I_{hole}=\frac{1}{2}\frac{r^{2}}{R^{2}}Mr^{2}+\left(\frac{r^{2}}{R^{2}}M\right)\left(\frac{R}{2}\right)^{2}=\frac{Mr^{4}}{2R^{2}}+\frac{Mr^{2}}{4}$$
so the total moment of inertia is the big disk's I minus four times the holes' I, right?
$$I=\frac{1}{2}MR^{2}-4\left(\frac{Mr^{4}}{2R^{2}}+\frac{Mr^{2}}{4}\right)=\frac{1}{2}MR^{2}-\frac{2Mr^{4}}{R^{2}}-Mr^{2}$$

So...

$$I=\frac{1}{2}M\left(R^{2}-\frac{4r^{4}}{R^{2}}-2r^{2}\right)$$

That seems kind of messy.

Last edited: Nov 20, 2008
2. Nov 20, 2008

### naresh

Sometimes things are that way Blame it on your messy disk.

You can add/subtract moments as long as they are about the same axis; so your second solution is fine.

3. Nov 20, 2008

### naresh

Oh, I didn't read the problem carefully enough. The flywheel has mass M, not the original disk...

4. Nov 20, 2008

### bowma166

Oh... Does that not work that way? Hm. Well I tried again, finding the mass of the thing if the holes weren't removed and then replacing M in my calculations with that.
The whole mass would have been
$$M_{whole}=\frac{Mr^{2}}{R^{2}-4r^{2}}$$
right?

I got something extremely messy for my final answer then... So messy that I'm pretty sure it can't be right. I think?

$$I_{total}=Mr^{2}\left(\frac{R^{2}}{2R^{2}-8r^{2}}-\frac{2r^{4}}{R^{4}-4R^{2}r^{2}}-\frac{2r^{2}}{R^{2}-4r^{2}}\right)$$

::edit:: Yeah, that can't be right... Oh damn! I accidentally switched r and R in my calculation of the mass. Time to do that all over again.

::edit 2::
Okay. With the correct (I hope) mass with MR2 on top instead of Mr2, I got...
$$I_{total}=\frac{MR^{4}-2Mr^{4}\left(2+R^{2}\right)}{2R^{2}-8r^{2}}$$

$$\lim_{r\rightarrow0}I_{total}=\frac{1}{2}MR^{2}$$
for this answer, so it has the possibility of being correct. Maybe.

Last edited: Nov 20, 2008
5. Nov 20, 2008

### naresh

Yeah, it is a messy problem - you just have to be confident that you haven't made any errors in the process. For example you cannot add 2 to R^2 since the dimensions are all wrong

6. Nov 20, 2008

### bowma166

But I got the (2+R2) from factoring 4Mr4 + 2MR2r4... Doesn't that work? Sigh. I give up on this problem. I'll just hope there isn't one like it on the test, hah.

7. Nov 20, 2008

### naresh

Your solution is correct, in that you've understood the physics. It is just a matter of going back and checking your math again. Good luck on the test!

8. Nov 20, 2008

### bowma166

Ah, so finding the mass if there was nothing removed,
$$M_{disk}=\frac{MR^{2}}{R^{2}-4r^{2}}$$,
and calculating everything as I did with that is the right way to do things? Good. I'm not going to bother checking over the math though. Thanks for the help!