# Proving things for an arbitrary rigid body with an axis of symmetry

1. Nov 23, 2015

### B3NR4Y

1. The problem statement, all variables and given/known data
Consider an arbitrary rigid body with an axis of rotational symmetry, which we'll call $\hat z$
a.) Prove that the axis of symmetry is a principal axis. (b) Prove that any two directions $\hat x$ and $\hat y$ perpendicular to $\hat z$ and each other are also principal axes. (c) Prove that the principal moments corresponding to these two axes are equal: λ12

2. Relevant equations
The moment of inertia tensor for a rigid body is $I_{ij} = m_\alpha (r_\alpha \delta_{ij} -r_{\alpha, \, i} r_{\alpha, \, j} )$
For a continuous mass distribution $I_{ij} = \int dV \rho(\vec r) (r^2 -r_i r_j )$

If an axis is a principal axis (eigenvalue of the inertia tensor), then the inertia tensor about the principal axes is diagonally λ1 λ2 λ3, and elsewhere zero.

3. The attempt at a solution
I'm not sure where to start proving that an axis of rotational symmetry is a principal axis. I'd imagine I'd have to work out the arbitrary λ for each axis, but this seems like it would be really ugly and I think there should be a slicker way to do it but I can't find it.

2. Nov 23, 2015

### SteamKing

Staff Emeritus
Look at it like this. When calculating inertia properties about a rotational axis which is also an axis of symmetry, which elements of the inertia tensor are non-zero?
Can you show why the zero elements are equal to zero without using the fact that the axis of rotation is also a principal axis?

3. Nov 23, 2015

### B3NR4Y

The parts of the inertia tensor that are nonzero are those along the diagonal. The off-diagonal elements are zero, because they contain products of two coordinates, and the sum of them, and if the object is symmetric about those points the sum (or integral) disappears. I think, that's my reasoning. And therefore the inertia tensor is diagonalized and the elements on the diagonal are the principal axes. Am I in the right direction?

4. Nov 23, 2015

### SteamKing

Staff Emeritus
I believe that's how the proof of this proposition is generally argued.

http://ocw.mit.edu/courses/aeronaut...fall-2009/lecture-notes/MIT16_07F09_Lec26.pdf

5. Nov 23, 2015

### B3NR4Y

Okay, so for the next parts I'm thinking I should follow in the same way, but it seems to be redundant to continue in the same direction for the two directions orthogonal to the $\hat z$. I was thinking maybe multiplying by the rotation matrix but that seems silly to do, but also not silly to do, if I multiply by the rotation matrix for a 90 degree in the y axis, I have to do it twice by the tensor transformation law. Same for x. But I think I'm headed in the wrong direction here too.

So I' = R I RT