# Moments of Inertia of two particles

Tags:
1. Nov 20, 2014

### Spaceflea

1. The problem statement, all variables and given/known data

Two particles, each of mass m, are attached one to each end of a diameter PQ of a uniform circular disk, of mass 4m, radius a with its centre at O. The system is free to rotate about a horizontal axis through A, a point on PQ such that OA = b as indicated in the diagram below. The system is released from rest when PQ is horizontal.

Determine the Moment of Inertia of the system about the axis A, in terms of integer constants, a , b and m.

2. Relevant equations
I=ma^2
parallel axis equation I=Icentre of mass + md^2

3. The attempt at a solution
I have determined the moment of inertia of the disc to be 0.5ma^2.
I have determined the moment of inertia of Q to be m(b+a)^2.
I have determined the moment of inertia of P to be m(a-b)^2.
Therefore I should just add these three numbers together to make a total moment of inertia, but this answer is marked wrong (it is marked by a computer programme). Where am I going wrong in my method? Thanks.

2. Nov 20, 2014

### Staff: Mentor

That's the generic formula for a circular disk about its center. You'll have to modify it.

These are OK.

Your expression for the moment of inertia of the disk is incorrect: You have the wrong mass and the wrong axis.

Correct that, and your method should work fine.

Another approach would be to find the moment of inertia of the entire system about its center of mass, then use the parallel axis theorem. Do it both ways and compare!

3. Nov 20, 2014

### Spaceflea

Thanks so much for the reply! But I am confused over which axis and mass I should use instead. Obviously not point A...

4. Nov 20, 2014

### Staff: Mentor

I assume you are talking about the disk? What's the mass of the disk? You need its moment of inertia about point A, but don't start there. Start with the moment of inertia about the center of mass and use the parallel axis theorem.