# Moment of inertia of a non uniform disc

1. Apr 19, 2017

### Rnm

1. The problem statement, all variables and given/known data
A non uniform disc of radius R has a mass of M. Its centre of gravity is located at a distance x from the centre. Find the moment of inertia of mass (moi) around the axis perpendicular to the surface passinf through the centre of gravity.

2. Relevant equations
Parallel axis theorem
Ioo = Igg + m[r^2]

Ioo = moi around axis required
Igg = moi around axis through centre of gravity
m = mass of the object
r = distance between the 2 axises(?dont know the plural)

Moi around the centre of a uniform disc
1/2m[r^2]

3. The attempt at a solution
I assumed that since Ioo = 1/2M[R^2]
Then using parallel axis theorem i took
Igg =1/2M[R^2] - M[x^2]
The problem is you can get a negative answer for Igg for certain x(>R/root2). So i think im wrong some where in my assumptions. Can someone please clarify

Last edited by a moderator: Apr 19, 2017
2. Apr 19, 2017

### Staff: Mentor

Why would you assume that? (That's the MOI of a uniform disk through its center.)

Have you posted the full problem? I don't see sufficient information to solve for anything.

3. Apr 19, 2017

### CWatters

+1

Looks like something is missing.

4. Apr 19, 2017

### Rnm

This is the first part of the question.
The 2nd part involves placing this disk on a horizontal table at rest and letting go. The disk starts to roll (with no slipping) along the table . I need to use the answers i obtained in the first part to find the vertical and horizontal components of the reaction force on the disk by the table

I wrote 3 equations for this with newtons 2nd law in the vertical and horizontal directions and the 2nd law of rotation in the direction of the angular acceleration (a)
I cant figure out how to write the last equation without using Igg. (Since it is stated to use the previous result).

5. Apr 19, 2017

### CWatters

Are you sure the problem says it's a NONuniform disc not a uniform disc?

Consider the case when x=0. There are lots of ways that such a disc can be "nonuniform" and the moment of inertia will be different in each case. For example the disc could be thicker in the middle or thicker at the edge. That would change the moment of inertia while keeping x unchanged. You couldn't calculate the moment of inertia without knowing the mass distribution.

6. Apr 19, 2017

### haruspex

7. Apr 19, 2017

### haruspex

As others have posted, this is not nearly enough information. As an example, consider a uniform disc mass m with a point mass M-m stuck on at distance y from the disc's centre. The moment of inertia about the mass centre of the combination is mr2/2+Mmx2/(M-m).

Edit: No, that can't be right... something nasty anyway.

Last edited: Apr 20, 2017
8. Apr 19, 2017

### Rnm

No other data is given in the question. I double checked. I suppose the question is incomplete then? But if so what would be missing in the question?

9. Apr 19, 2017

### haruspex

I cannot think of one simple extra piece of information that would allow you to find the moment.
I suggest you just write it as J and proceed to the second part of the question.

Edit: I found an old question exactly the same except that it did not expect you to determine the moment of inertia. It just said it is I, and asked about the oscillation. This suggests to me that asking for the moment of inertia was just a mistake.

Last edited: Apr 19, 2017
10. Apr 20, 2017

### Rnm

Will do. Ill try to find the missing information about the question. Thanks for helping me out.