# Moment of inertia in 3D

1. May 17, 2007

### yetar

1. The problem statement, all variables and given/known data
I am trying to understand euler's equations of motion.
First of all, relative to what point do we need to calculate the inertia tensor(J)?
Do we calculate it relativ to the world origin? or body origin?
The torque may be calculated as:
T = dL/dt = DL/Dt + wxL = D(Jw)/Dt + wx(Jw) =(1)= J*Dw/Dt +wx(Jw)
In the equality (1), why can we get J out of the derative?
This can be possible if we are dealing with a rigid body and the inertia tensor is calculated relative to the body origin.
But T and w are in world coordinates in this equation!
So:
T = J*Dw/Dt+wx(Jw) = J*dw/dt+wx(Jw)
Now we want to transform this equation from world coordinates to body coordinates, so we use the eigendecomposition of J, which is RMR(T).
However, what is R?
R is the rotation matric from body coordinates into world coordniates?
How can we see that this is true?
Does one of the R columns is the D? the axis the moment of inertia is measured relative to?
This implies that J is calculated relative to world origin and not body origin?
And what is M then? is M equal to J when calculated relative to body origin? or isnt it?
Lastly, M is a diagonal matrix, is a diagonal matrix comutative with other matrices?
Thank you.

2. Relevant equations

3. The attempt at a solution

Last edited: May 17, 2007
2. May 18, 2007

### yetar

Does this belong to the advanced physics?
Thank you.

3. May 18, 2007

### jambaugh

The moment of inertia tensor can be calculated relative to any point and you get a different moment of inertia for each point. There is a transformation formula. In a typical application you choose either the center of mass for the object or you choose the center of rotation when the object is attached to some semi-rigid framework e.g. a hinge or a ball and socket.

One example, you can analyze a rolling ball either by considering its rotation about its center along with its linear motion, or by considering its rotation about the point of contact on the surface. In the latter case you want the moment of inertia about the tangent point of a ball and not the usual moment of inertia about its center.

Note it is the same with torques. You may choose different origin points which gives different values of torque from a force applied to a point of some rigid body.

One way to see why this is the case is to recognize that in the group of spatial rotations and translations, $$ISO(3)$$ a.k.a. "The Euclidean Group in 3, dimensions", the subgroup of rotations is not unique. There is a continuum of distinct rotation subgroups corresponding to each center of rotation i.e. each point in space.

In abstract mechanics there is a generalized momentum for each group generator for the group of physical transformations of the system. Also you have transformational parameters (with corresponding velocities) and you can then associate a generalized mass (e.g. moments of inertia) with each generalized momentum. Similarly you have generalized forces and accelerations.

This provides a context where you can view all the different mechanical systems in a unified setting.

Thus for the Euclidean group you have 3 translation momenta and 3 rotation momenta but you have to pick a specific rotation subgroup to identify these. Choosing between distinct rotation subgroups (equivalent to choosing your spatial origin) gives you distinct angular momenta and thus distinct "angular mass" i.e. moment of inertia tensors for a given rigid system.

Regards,
James Baugh

Last edited: May 18, 2007
4. May 18, 2007

### jambaugh

Here R is a rotation matrix placing the rigid body's principle axes parallel to the three coordinate axes.
Given the moment of inertia matrix is symmetric and positive definite it is easy to show that it has a spectrum of eigen-values and thus there is some linear transformation which will diagonalize it. To see that this linear transformation is in fact orthogonal for any rigid system is harder to show but you can trust that it is so. I can't recall the name of the theorem or who first proved this.
M is the moment of inertia tensor in coordinates aligned with the principle axes of the rigid body. It is therefore and by definition diagonal. Diagonal matrices commute with other diagonal matrices but not with all matrices in general.

Only if the three diagonal elements are all the same and thus the matrix is a multiple of the identity will it commute with everything. In such case the object has an isotropic moment of inertia and it will thus be diagonal no matter what (orthonormal) basis you choose.

5. May 18, 2007

### D H

Staff Emeritus
dL/dt = DL/Dt + wxL

You are obviously using d/dt to represent the time derivative of some vector quantity as observed in an inertial frame and D/Dt to represent the derivative as observed in a rotating frame.

= D(Jw)/Dt + wx(Jw) = J*Dw/Dt +wx(Jw)

This is fine so long as J is constant in the rotating frame.

The time derivative of a vector quantity is observer dependent. If one is sloppy, it is easy to get into a spot of trouble by failing to distinguish between the observation reference frame and the representation reference frame. One way to avoid this problem is to use some nomenclature that distinguishes the frame in which a time derivative is observed/represented.

BTW, torque and angular velocity are often represented in body coordinates. Some of the forces and torques on a fighter jet are much easier to calculate in body coordinates. The roll/pitch/yaw rates are obviously in body coordinates.

Newton's laws apply only in an inertial frame. In particular, dL/dt = T is true only if the observer of dL/dt is in the inertial frame.

How is that going to help? What you need is the transformation matrix from inertial to body, which doesn't have a thing to do with the decomposition of the inertia matrix. Note well: the frame in which the inertia matrix takes on a diagonal form is a fixed rotation from the body frame. In other words, it is just another body frame.

6. May 18, 2007

### yetar

So the inertia tensot J=RMR(T) would change if it is calculated relative to a frame with different axes direction or different origin.
Where R would change if the axes direction change, and M would change if the origin change?
If the moment of inertia is I = DJD(T) where D is the axes the moment of inertia is computed relative to, then J's values has nothing to do with D?
And what are the principle axes of the rigid body? What is their meanning?

7. May 18, 2007

### D H

Staff Emeritus
From this and other posts, I don't think you quite get the concept of rotations in three dimensions. If you are taking a class, ask for some help.

To answer your last question: For any rigid body in three space, a set of three mutually orthogonal lines always exist such that the products of inertia with respect to pairs of these lines are all zero. In other words, the moment of inertia as expressed in a frame based on these orthogonal lines will be diagonal.

8. May 18, 2007

### yetar

What do I need to understand about rotations in 3D?
I know there are rotation matrices which are orthogonal matrices.
And there is the angular velocity, which I am not sure if it is true, but I think that an angular velocity w and a position r, hold that v = wxr.
But this is more related to angular motion rather then rotation in 3D space.
There is also yaw pitch and roll.
Did you mean these things?