# Product moments of inertia- what are they intuitively?

1. Oct 8, 2011

### Urmi Roy

Product moments of inertia-- what are they intuitively?

Hi,
I don't understand what the off-diagonal terms in the moment of inertia (tensor) matrix are intuitively....they are called the product moments of inertia....but if Ixx is the moment of inertia about the x axis, which axis is the Ixy about?

Please give an example, so that I can understand what these are intuitively.

2. Oct 8, 2011

3. Oct 9, 2011

### Urmi Roy

Re: Product moments of inertia-- what are they intuitively?

This is represented by the angular momentum vector.
If the axis of rotation is aligned with one of the principal axes of the body the angular momentum vector is also aligned in the same direction and the rotation is stable.
If the rotational axis is not principal then a torque, applied at right angles to this axis, is generated and causes the direction of momentum vector to describe a cone in space as the body rotates.
If the body is a rotating wheel with bearings this torque creates a reaction against the bearings.
Any unbalance in the distribution of the mass will be lead to vibration of the wheel and perhaps eventually to destruction.

Of course the directions of the principal axes are those where the products of inertia vanish as the arbitrary rotational axis varies.

From this, I have some questions-

1. how does the product moment of inertia make the object rotate in a cone? Has it got to do with a torque due to the weight of bodies? What if gravity is switched off?

2. So what we define as our principal axis is up to us...and we can just as easily align out rotational axis with the principal axis (by choosing the principal axis in that way)...?

3. Again, in the example you stated, there was nothing such as xy axis, so what exactly do you mean by "cross contributions"..? Why are we able to calculate them by simply making a product between the distances from the x axis and y axis...?

4. Oct 10, 2011

### D H

Staff Emeritus
Re: Product moments of inertia-- what are they intuitively?

Except those axes that we choose as the x, y, and z axes are anything but arbitrary. They were chosen to be what they are because they have special meaning.

Consider an automobile. It has a steering wheel on one side. If you pop the hood you will see other asymmetries. That single large battery is always off to one side. The engine block isn't quite symmetric, either. The same is true for lots of other machines. There are various reasons why they just can't made as symmetric as one would like.

It isn't the products of inertia that do this. After all, you can always choose a set of axes that make the inertia tensor diagonal. What causes this apparent torque is an angular velocity that is not directed along a principal axis.

Angular velocity is inevitably expressed with respect to the rotating body. (Compare to translational velocity, which is typically expressed with respect to some external frame such as north-east-down.) There are many reasons for doing this. One reason is that this is how rotational sensors work. Another reason is that expressing angular velocity with respect to inertial would mean using τext=d\dt(Iω), which is a nightmare. The inertia tensor as expressed in inertial coordinates is a time-varying 3x3 beast.

Just as you can use Newton's laws in a rotating frame with the introduction of inertial forces such as the centrifugal and coriolis forces, you can use the rotational equivalent in a rotating frame -- but you need to introduce an inertial torque. Specifically,

$$I\frac{d\omega}{dt} = \tau_{\text{ext}} - \omega\times(I\omega)$$

The $\omega\times(I\omega)$ term vanishes if the rotation is along a principal axis. It doesn't vanish if the rotation is not along a principal axis.

In theory, yes. In practice, hardly ever. Those axes that we decided to label as x, y, and z have some real meaning to us. A plane, for example, typically has the x axis directed along the body of the plane, y out one of the wings. That the plane happens to have a non-zero Ixy component to inertia tensor does not mean we should designate the x axis as mostly forward but slightly out the left wing. What happens the next flight when the cargo is loaded slightly differently?

Because that is how it is defined. Why is it defined that way? Because those definitions are very useful.

5. Oct 10, 2011

### Urmi Roy

Re: Product moments of inertia-- what are they intuitively?

So I gather that this is all about how something appears in a reference frame with the angular velocity vector not aligned with the rotational axis--it is the analogue of Newton's law in a non-inertial reference frame....funny how a situation can be completely changed if we just modify the reference frame-- the same body would not have any product moment of inertia if we aligned its axis to our reference axis!!

Also, I realize that Ixy is not necessarily along any separate axis as I previously thought...its just a component of the moment of inertia that is not about any of our reference axes.....however, it would be nice if someone could give me an idea of hoe exactly to picture this...its still a bit hazy...

6. Oct 10, 2011

### D H

Staff Emeritus
Re: Product moments of inertia-- what are they intuitively?

Kind of analogous. This term is unique to rotational motion. It does not appear for translational motion because the mass of an object in frame 1 is identical to the mass of the object in frame 2. Mass is a scalar. The inertia tensor is a 2x2 tensor. The representation of the inertia tensor in frame 1 is not the same as its representation in frame 2.

The easiest way for me to visualize what these terms mean is to start with a principal axis frame where the inertia tensor is diagonal. Now go to some other body frame that whose axes are not aligned with the principal axes. Question: How do you transform an inertia tensor from frame A, a principal axis frame, to frame B, some other body frame?

$$I_B = T_{A\to B}\;I_A\;T_{A\to B}^{\,\,\top}$$

I can show you the derivation, but it might help you with your understanding if you derive this result for yourself.

If you grind through the math you will find that the off-diagonal terms of this transformed inertia tensor are the additive inverses of the products of inertias in this new frame -- and that is exactly why the inertia tensor is defined the way it is.

7. Oct 10, 2011

### Urmi Roy

Re: Product moments of inertia-- what are they intuitively?

Hi DH....sorry for a silly question, but what are the Ts?

So basically you're trying to say that if we now take a reference frame that is not aligned with the principal axes, we kinda transform, or rotate the inertia tensor into this new frame?

8. Oct 11, 2011

### D H

Staff Emeritus
Re: Product moments of inertia-- what are they intuitively?

The Ts are transformation matrices. $T_{A\to B}$ is the transformation matrix from frame A to frame B. If you don't know what those are yet, you'll just have to take the definition of the inertia tensor with its off-diagonal terms as the negative of the products of inertia as a given.

9. Oct 14, 2011

### Urmi Roy

Re: Product moments of inertia-- what are they intuitively?

DH, after your explanation, I understand intuitively what there cross contributions are....however, suppose we have a slender rod inclined at 45 degrees with the z axis of our ground frame, and it rotates about this z axis, (in this case we do have nonzero product moments of inertia)....the question is-- do we actually observe the rod to slightly shake or vibrate during this rotation because of the fictitious torques?

Its strange, coz i've seen videos of this, and the rod appears to be slightly shaking during the rotation, but it is analogous to fictitious forces such as centripetal forces, which we don't really observe...

10. Oct 16, 2011

### Urmi Roy

Re: Product moments of inertia-- what are they intuitively?

one more thing..it seems that the off-diagonal terms in the inertia tensor matrix are nonzero if there is any portion of an object lying between the rotating axes(like at 45 degrees to x and y axes)...

so if i have a star shape, and choose one principal axis to be perpendicular to the plane of the star, then the others 2 principal axes (say x and y)will be in the plane...however, some of the arms of the star will be inclined to both the x and y axes...then it should have product moment of inertia...but intuitively that's not true....where am I going wrong in my argument?

11. Oct 17, 2011

### Urmi Roy

Re: Product moments of inertia-- what are they intuitively?

The momentum about the x axis has a contribution due to the angular velocity wy, about the y axis...in the derivation of the inertia tensor matrix given in our book, we an see that because the component of wy in the i hat direction isn't zero that Ixy exists....so basically, we get product moments of inertia because the angular velocities about any axis also have components about the other axes...right?

12. Oct 17, 2011