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In Kleppner and Kolenkow’s Introduction to Mechanics, I’ve been shown the derivation for the equation $$L_z=I_{cm}\omega+(\textbf{R}_{cm}\:\times M\textbf{V}_{cm})_z$$ where $L_z$ is the angular momentum of an object rotating around a translating, fixed axis. My question is why you would ever use this formula. Can you not set $\textbf{R}=0$ (i.e. origin at the center of mass) and use $L_z=I\omega$, as you usually do? As far as I’m concerned, $L_z=I\omega$ works for all cases where the axis is fixed (i.e. axis direction is unchanging). That means even if the axis is translating, it can still be used, right?

My book continues,

[The equation derived] is valid even if the center of mass is accelerating, because L was calculated with respect to an inertial coordinate system.

Does that imply that $L_z=I\omega$ doesn’t work when the axis is accelerating? If it’s still translational acceleration, then the axis is still fixed, and as far as I know we can still use it.

In what situation would we ever use the more complicated equation at the top over the simple $L_z=I\omega$?

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$I_{cm} \mathbf{\omega}$ is the intrinsic angular momentum of the solid body. It's nonzero whenever the object is rotating.

On the other hand, $\mathbf{R}_{cm} \times M \mathbf{V}$ is the angular momentum of the object moving as whole.

Where would it matter?

If you have a point-mass, it cannot have intrinsic angular momentum (we ignore the quantum mechanical spin here). But if it's moving around in space around you, it will have angular momentum.

One more example. Consider a rotating ball, for example the Moon. It's rotating around its axis so it has intrinsic angular momentum. Its center of mass is in its geometric center and from that reference frame, the only contribution to its angular momentum comes from its rotation. But the Moon is also orbiting the Earth. From any point on the surface of the Earth, the Moon has an additional contribution to that angular momentum because of its orbital motion. You could look at it from the Sun, it would be even larger because the Earth orbits the Sun, you get the point.

In conclusion, $I_{cm} \mathbf{\omega}$ is a contribution you can't get rid of simply by chosing another inertial reference frame. At best, you can make it the only contribution, by picking the CM as your origin. The other contribution, $\mathbf{R}_{cm} \times M \mathbf{V}$ is an artefact of chosing a specific coordinate system.

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  • $\begingroup$ Can you use $I_{cm}\omega$ without the $\mathbf{R}_{cm} \times M \mathbf{V}$ part if the center of mass is accelerating? $\endgroup$
    – lightweaver
    Commented Oct 16, 2016 at 4:03
  • $\begingroup$ What do you think? Try to analyse my answer carefully, the answer is there! If you really can't see it, I'll tell you. But tell me what exactly is it that confuses you, I want you to understand it because understanding goes a long way! $\endgroup$
    – user20250
    Commented Oct 16, 2016 at 13:59
  • $\begingroup$ I believe you can use $I_{cm}\omega$ when the center of mass is accelerating as long as the velocity vector and the position vector (with respect to the original center of mass position) are parallel. $\endgroup$
    – lightweaver
    Commented Oct 16, 2016 at 14:06
  • $\begingroup$ That's correct, because it's accelerating along the rotation axis $z$, so it doesn't affect the intrinsic motion in the $x-y$ plane. And since you chose your $z$ axis to be the axis of rotation (and in this case the direction of accelerated motion), you don't get any additional contribution. In other words, the z-component of $\mathbf{R} \times M \mathbf{V}$ is unchanged at all times because it's just $M(XV_y - YV_x)$, which is obviously not changing whatever the body does along the z-axis. $\endgroup$
    – user20250
    Commented Oct 16, 2016 at 14:19

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