Homework Help: Center of mass of a disc with a hole

1. Nov 21, 2012

caspernorth

there is a similar case here :https://www.physicsforums.com/showthread.php?t=296966

so I've tried it and what i did was to equate both individual moment of inertia about z axis (and later applied parallel axis theorm) with negative mass for hole. added both these and found out Ix = M/4 [(1/2)-(r^4/R^2)]

so I think this is wrong, why isn't terms that shows how much distant the hole is located present in the equation? how can we do it. please help.

2. Nov 21, 2012

Staff: Mentor

Show exactly what you did to get that result.

You need to add the moment of inertia of the disk about the z-axis and the moment of inertia of the hole (negative mass) about the z-axis. To get the moment of inertia of the hole about the z-axis, you'll need the parallel axis theorem.
Using the parallel axis theorem incorporates the distance of the hole from the center.

3. Nov 21, 2012

caspernorth

So Ix is symmetric but Iy isn't right?

In that case i did it entirely wrong, anyway i did it like this:
density = M/∏R^2
density = m/∏r^2

equating we get
m = (r/R)^2M

IzM = (MR^2)/2
Izm = (-mr^2)/2

substitute for -m
we get
Izm = -(Mr^4) / 2R^2

I'm stuck at this point (i guess it's parallel axes theorem now, isn't it?)

4. Nov 21, 2012

Staff: Mentor

OK, you are taking M as the mass of a solid disk of radius R.

OK.

OK.
This is the moment of inertia about its center of mass. But since it's not centered on the z-axis, you need the parallel axis theorem to find Izm.

Yes.

5. Nov 22, 2012

caspernorth

Ok since I can't give up I tried it once more:
By parallel axis theorem,

Iz = Icm + (M-m)y^2
(Icm = center of mass' moment of inertia, and M-m is the net mass, y is the distance between center of mass and radius of big circle)

Iz = IzM = (MR^2)/2

Center of mass turns out to be x = R+ {m.d/(M-m)} where d is distance between centers of each circle
y = R-x = {m.d/(M-m)}

So Icm = IzM - (M-m){m.d/(M-m)} ^2
Icm = (MR^2)/2 - (M-m){m.d/(M-m)} ^2
on simplifying we get,

Icm = (M/2) (Rmd)^2

Last edited: Nov 22, 2012
6. Nov 22, 2012

Staff: Mentor

I'm confused as to what you are doing.

You have a solid disk of mass M (I assume that's what you want to assume) with a hole in it. (M is the mass the disk would have if it had no hole, right?)

What's the moment of inertia of the solid disk about the z-axis?

What's the moment of inertia of hole about its center? Then use the parallel axis theorem to find its moment of inertia about the z-axis.

To find the composite moment of inertia, just add those two elements together.

7. Nov 22, 2012

caspernorth

Well, i will make it as clear as possible.

1. what i need to find out = moment of inertia of a disc with hole.
2. M, R are parameters of disc without hole
3. -m, r are parameters of hole

Now can you show me how this can be solved. I just might make another mistake by repeating what i have already posted.

8. Nov 22, 2012

Staff: Mentor

OK.

Answer the questions I asked in my last post. Do it step by step.

9. Nov 22, 2012

caspernorth

Let me do it another way.

I = 1/2 m r^2 + m b^2 , where b is the distance between the center of rotation and the center of mass of m.

If you cut this out of a disk of mass M and radius R, you are left with

I = 1/2 M R^2 - (1/2 m r^2 + m b^2)

10. Nov 22, 2012

Staff: Mentor

Good!

11. Nov 22, 2012

caspernorth

Ok, so isn't this MOI about z axis (perpendicular to the plane) I need inertia about the diameter axis. how can i do that?

12. Nov 22, 2012

Staff: Mentor

Which diameter axis exactly?

You'd approach it in a similar way. Find the moment of inertia of each piece about the chosen axis then add them up.

13. Nov 23, 2012

caspernorth

The diameter of both, since they line up. That is the line passing through their centers.
I = 1/4 M R^2 - (1/4 m r^2 + m b^2)

where the 2 became 4

14. Nov 23, 2012

Staff: Mentor

Almost. There's no need for the parallel axis theorem here, since the center of mass of the hole is already on the desired axis of rotation.

15. Nov 23, 2012

caspernorth

Thankyou sir, that was indeed helpful.