# Moment of inertia of half disk through integration

1. Mar 12, 2015

### RaamGeneral

Hello, sorry for this stupid question.
I struggled to find the moment of inertia of half solid thin disk (about the center of the disk) through an integration, but I couldn't get the right value.

I'm pretty sure it has to be $$MR^2/4$$, but
$$I=\int r^2 dm \\ dm=(M/A)dS$$

With $$A=\pi R^2/2$$

I compute the integration in polar coordinates, where $$dS=r dr d\theta$$ with 0<r<R and 0<theta<pi:
$$2M/(\pi R^2) \int_0^R \int_0^\pi r^3 d\theta dr=MR^2/2$$

Where am I wrong?

2. Mar 12, 2015

### PeroK

I guess you think it should be $MR^2/4$ because the MoI of a full disk is $MR^2/2$. But, what have you taken as the mass of your half disk? And, what would the mass of the full disk be?

3. Mar 12, 2015

### BiGyElLoWhAt

What do you mean about the center? I can think of three ways to rotate a half circle about it's center.

4. Mar 12, 2015

### RaamGeneral

mmh, I think I can see your point. but I'm a bit confused now.

if I take M=(mass of full disk), I(full disk)=MR^2 /2 and I(half disk)=MR^2 /4
if I take M=(mass of half disk) [as I did for the integral], I(full disk)=2MR^2 /2 and I(half disk)=MR^2 /2 (the result of the integral)

So, it seems that if I have an amount M of mass and I shape it to be a full disk or a half disk, I get the same moment of inertia.

BiGyElLoWhAt, you right, I mean about the axis perpendicular to the disk.

5. Mar 12, 2015

### PeroK

Yes, that's correct. By symmetry, each half of the disk contributes half of the MoI.

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