# Inertia tensor combination point mass-sphere

1. Jan 26, 2016

### lilphy

1. The problem statement, all variables and given/known data
hello,
i want to calculate the inertia tensor of the combination of a point mass and a sphere in the object's frame, the center of mass is at the origin. The point mass remains at the surface of the phere
The sphere is uniform, radius r and mass M, and the point mass has mass aM.
My sphere is centered at the origin χ
2. Relevant equations

3. The attempt at a solution
The inertia tensor of a sphere of radius R and mass M is known
I did all the work of finding the distance to the center of mass of the system to the origin, the inertia tensor of the sphere about the center of mass of the system sphere-particle.
Now i have to find the tensor of inertia of the point mass about the centre of the sphere, I found this formula :
Iij=dm(δij r2-rirj)
So I got these values for Iij:
I11=I22=I33= aMR2
And all the other values are = to -aMR2
Is this correct ?
Thanks

2. Jan 26, 2016

### Orodruin

Staff Emeritus
No, it is not correct. I suggest you select a coordinate system such that the point mass is located at $\vec r = R\vec e_3$. You can later rotate this to the appropriate frame as necessary.

3. Jan 26, 2016

### lilphy

So there is no component in e1 and e2, all the values not in the diagonal are 0
And I will get I11= I22= aM r2 and I33=0 ?

4. Jan 27, 2016

### Orodruin

Staff Emeritus
Yes. If you have a more general position vector it is going to depend on the actual components of it. You should also take some time to think about why your result is reasonable, i.e., why I33 is zero and I22 = I11.

5. Jan 27, 2016

### lilphy

There is something I don't understand. When we calculate the total tensor of the moment of inertia, the Itot_33of the system point mass-sohere is equal to Isphere_33 in the center of mass. I don't understand why ?

6. Jan 27, 2016

### Orodruin

Staff Emeritus
What is the inertia of the point mass if it is situated on the axis of rotation?

7. Jan 27, 2016

### lilphy

The inertia is mR^2 with R the distance to the rotation axis, so it would be 0 ?

8. Jan 27, 2016

### Orodruin

Staff Emeritus
Exactly. So what is its contribution to the total inertia if you rotate about the 3-axis? What is the contribution of the sphere?

9. Jan 27, 2016

### lilphy

Oh right we will only have the contribution of the sphere, the inertia about the center of mass of the system is 2/5MR^2+Md2 with d the distance to the center of mass that is = Ra/(1+a)

10. Jan 27, 2016

### Orodruin

Staff Emeritus
d should be the orthogonal distance relative to the axis of rotation. If you are rotating around the 3-axis, this distance is also zero.

11. Jan 27, 2016

### lilphy

Oh yes of course, i think now i get it. Because the center of mass is in the 3-axis there is no contribution of it on I33 ! Thanks for your time and explanations