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.