Inertia of body moving about two distinct parallel axes

[KataruZ98]

TL;DR

A rigid 3D is rotating simultaneously about two axes. How can I find its total inertia?

Considering a cylindrical rigid body of length 3 m and wide one. The body is rotating about an axis passing through one of its bases and perpendicular in respect to the length. At the same time, the same cylinder is orbiting about another axis parallel to the first — but distanced 10 m from the former. How can I find the body’s moment of inertia about the second axis?

 

[vanhees71]

You mean it rotates around an axis given by a unit vector $\vec{n}$ going through some point fixed in this body? Then you can use the tensor of inertia $\hat{\Theta}$ around this point and the corresponding moment of inertia then is given by $\vec{n}^{\text{T}} \hat{\Theta} \vec{n}$.

If you need to calculate things wrt. to different body-fixed reference points, then it's most convenient to calculate $\hat{\Theta}$ around the center of mass of the body and then use Steiner's Law (also known as "parallel-axis theorem".

https://en.wikipedia.org/wiki/Parallel_axis_theorem#Tensor_generalization

 

[KataruZ98]

You mean it rotates around an axis given by a unit vector $\vec{n}$ going through some point fixed in this body? Then you can use the tensor of inertia $\hat{\Theta}$ around this point and the corresponding moment of inertia then is given by $\vec{n}^{\text{T}} \hat{\Theta} \vec{n}$.

If you need to calculate things wrt. to different body-fixed reference points, then it's most convenient to calculate $\hat{\Theta}$ around the center of mass of the body and then use Steiner's Law (also known as "parallel-axis theorem".

https://en.wikipedia.org/wiki/Parallel_axis_theorem#Tensor_generalization

To give a visual example from RL to clarify what I meant, take the Moon’s orbit around the Sun and replace the satellite with the specified cylinder (and of course scale down the distances and sizes as described in the question). Pretty much you could replace my question with “how can the Moon’s inertia about the Sun be calculated, while addressing its revolution around Earth?”

What gives me difficulty is that, because the cylinder is rotating around two axes, there’s no fixed distance between its center of mass and the second axis in order to use Steiner’s Law.

 

[vanhees71]

What do you mean "the cylinder is rotating around two axes"? That doesn't make any sense. A rotation is (momentarily) always around one axis. A moment of inertia depends on this axis. If you need to calculate the moment of inertia around different axes, you can calculate the tensor of inertia and use it to calculate the moment of inertia around any rotation axis going through the body-fixed reference point used to calculate the tensor of inertia. If you have to consider axes with no common point than it's most convenient to calculate the tensor of inertia with respect to the center of mass of the body and then use Steiner's Law to get the tensor of inertia around any other point.

 

[KataruZ98]

What do you mean "the cylinder is rotating around two axes"? That doesn't make any sense. A rotation is (momentarily) always around one axis. A moment of inertia depends on this axis. If you need to calculate the moment of inertia around different axes, you can calculate the tensor of inertia and use it to calculate the moment of inertia around any rotation axis going through the body-fixed reference point used to calculate the tensor of inertia. If you have to consider axes with no common point than it's most convenient to calculate the tensor of inertia with respect to the center of mass of the body and then use Steiner's Law to get the tensor of inertia around any other point.

As I said in my first reply, the rotation I’m talking of can be thought as that of the Moon around the Sun — with the former already going circles. Here you can think at the cylinder as the Moon, the first axis as Earth and the second axis as the Sun. I hope I made my case clear.

 

[jrmichler]

The moment of inertia is a property of the object and the axis about which the moment of inertia is calculated or measured. It has nothing to do with the rotation or nonrotation of the object.

Steiner's theorem is also known as the parallel axis theorem. It is useful if you know the moment of inertia about an axis, typically a principal axis, and want to know the moment of inertia about a parallel axis in a different location. It gets a lot of use when designing high speed machines.

 

[anuttarasammyak]

@KataruZ98
I see fig 35 in Laudau-Lifshitz Mechanics as below cited.

When we say rotation of rigid body, its COM O is chosen as origin for rotation.
If you want to rotate this rigid body around Origin of the XYZ coordinates, as for moment of inertia $I=MR^2$ would be applied, where M is mass of the rigid body. I am afraid you would confuse these two kind of rotations.

 

[Lnewqban]

What gives me difficulty is that, because the cylinder is rotating around two axes, there’s no fixed distance between its center of mass and the second axis in order to use Steiner’s Law.

What makes the center of mass of the cylinder not to be crossed by the "axis passing through one of the bases and perpendicular in respect to the length"?
If given freedom, most rigid bodies spontaneously spin around the center of mass.

 

[KataruZ98]

What makes the center of mass of the cylinder not to be crossed by the "axis passing through one of the bases and perpendicular in respect to the length"?
If given freedom, most rigid bodies spontaneously spin around the center of mass.

Well, to be openly clear, I’m using this figure as a model to find the MOI of a forearm about the shoulder, when there’s a rotation happening both at the elbow and shoulder. I’d like to see the resulting energy and torque calculated, but I need the MOI of the forearm around the shoulder to complete my work.

Admittedly I gave unrealistic proportions to the body I was using for model (the cylinder) and the distance from the second axis — but I figured that could be easily fixed when I actually apply the correct formula to the true figures.

 

[anuttarasammyak]

If your problem is about robot arm joint, I find https://studywolf.wordpress.com/2013/09/17/robot-control-4-operation-space-control/ by search and some others which might be helpful for your study.

If this is the case, it seems that momentum inertia of forearm cylinder to shoulder joint is same as that of post #7,i.e., $I=MR^2$ , with no direct regard to where elbow joint is and how much angle it forms, where R is distance from sholder joint to forearm COM and M is mass of forearm.

 

[WalterG]

As I said in my first reply, the rotation I’m talking of can be thought as that of the Moon around the Sun — with the former already going circles. Here you can think at the cylinder as the Moon, the first axis as Earth and the second axis as the Sun. I hope I made my case clear.

It is a 3 objects movement. The differential equations can not be solved without computer simulation.