# Disc and hole - moment of inertia [closed]

Determine the moment of inertia of a homogeneous disc with density $\rho_0$ of radius R with a circular hole of radius R/2 and central radius R/2 regarding an axis perpendicular to the surface and going through the center of gravity.

Alright, first I tried drawing the situation, because I don't think I understood it correctly from reading the problem.

Here's what I was thinking:

At first I thought the axis would go through the center of the disc, but then I read that it's going through the center of gravity. But since it says central radius R/2 regarding an axis ... through the center of gravity does that mean that the center of gravity is at a distance R/2 away from the boundary of the hole?

I'm just not sure about the use of words in this problem. Is kind of hard to figure out the actual problem.

And how would I even get the moment of inertia just by knowing the center of gravity? I mean, yeah, it's the axis around which it rotates but I never dealt with an object with a hole in some random part of it.

By the way, can I assume that the mass is $2\pi R^2\cdot \rho_0$? Since the disc doesn't have an actual volume, or maybe it has, but the thickness of the disc is not given, so I'm uncertain about that.

I would appreciate any help.

## closed as off-topic by John Rennie, Martin, yuggib, Kyle Kanos, Kyle OmanJun 25 '15 at 16:16

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Yes - you have to assume the density of the disk is given as "mass per unit area" - or you can assume a thickness $t$ which will carry forward through all your calculations.

The best way to approach problems like this is as a superposition of two objects: one with "positive" mass - the complete disk - and another with "negative" mass: the hole.

You can use this for finding the center of gravity; and then you use the parallel axis theorem to find the moment of inertia about the center of gravity.

Example for above:

Disk 1 - mass $M=2\pi R^2 \rho t$; disk 2, mass = -M/4.

Center of gravity is $x$ off center: distance to "hole" is $x-\frac{R}{2}$. Balancing mass about this point (M at the center, -M/4 at R/2), we find

$$M\cdot x = -\frac{M}{4}\left(x-\frac{R}{2}\right)$$

and the center of mass is at position

$$x = \frac{R}{10}$$

from the center.

• Netative mass? I'm sorry, but I don't know how to determine the center of gravity in such a way. Truth be told, I never calculated a center of gravity with negative mass. Already difficult enough with positive mass. XD. – user83981 Jun 24 '15 at 21:11
• Does the sample calculation make it any clearer for you? – Floris Jun 24 '15 at 21:22
• Also Fabian, the moment of inertia calculation works the same way. Just add up the negative mass and the positive mass, and relate them to the off axis via the parallel axis theorem. – Mark Jun 24 '15 at 22:11
• @Floris So if using parallel axis theorem can I assume that $I=I_{cm}+md^2=\frac{1}{2}\pi R^4\rho t+\frac{1}{50}\pi R^4\rho t$? – user83981 Jun 25 '15 at 12:01
• Oh, now I see. I just substracted the negative inertia because of the hole but forget to putting that in respect to the rotation axis with the parallel axis theorem. – user83981 Jun 25 '15 at 12:22