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

AI Thread Summary
The discussion focuses on proving properties of the moment of inertia tensor for a rigid body with an axis of rotational symmetry, designated as ## \hat z ##. It establishes that this axis is a principal axis, and that any two perpendicular axes, ## \hat x ## and ## \hat y ##, are also principal axes. The reasoning involves demonstrating that the inertia tensor is diagonal due to the symmetry, leading to zero off-diagonal elements. The participants express uncertainty about the best approach to prove these properties but suggest that the diagonal elements correspond to principal moments. The conversation emphasizes the importance of understanding the symmetry in simplifying the proof process.
B3NR4Y
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Homework Statement


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

Homework 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.

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.
 
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B3NR4Y said:

Homework Statement


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

Homework 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.

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.
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?
 
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?
 
B3NR4Y said:
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?
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
 
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
 
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