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I know the kinetic energy of a rigid object is \begin{align}\tag{$1$} KE = \frac{1}{2}mv^{2} + \frac{1}{2}I\omega^{2} \end{align} where $v$ is the velocity of the center of mass of the object, $\omega$ is its angular velocity, and $I$ is the moment of inertia about its center of mass.

Now the things is, shouldn't the moment of inertia be specified about an axis as opposed to a point? I can understand that the axis has to be through the center of mass, but which direction ought it to be oriented? If there is a specific axis we have to use, how do we calculate this axis for an arbitrary body undergoing arbitrary motion?

For an example, consider a uniform solid cylinder (radius $r$, height $h$) rolling without slipping at a constant velocity $v > 0$. I could consider the axis along the axial direction of the cylinder through the center of mass and obtain $$ I = \frac{1}{2}mr^{2}. $$ We can consider $(1)$ with $\omega = v/r\ne 0$ with no issues. But couldn't I also consider a perpendicular axis oriented, say, vertically and through the center of mass? In that case, $$ I' = \frac{1}{12}m(3r^{2} + h^{2}). $$ This is clearly different, and as the cylinder rolls, I would expect the angular velocity to be $\omega\,' = 0$. Wouldn't this change the result in $(1)$?

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  • $\begingroup$ I think the direction component of the rotation is irrelevant because energy is a scalar. $\endgroup$ Commented May 30, 2023 at 14:59

2 Answers 2

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If you ran next to the center of mass, it would be stationary in your frame of reference. You would see the object rotating around some axis through the center of mass. That is the axis you should use.

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  • $\begingroup$ That is not always the choice used. $\endgroup$ Commented May 30, 2023 at 16:27
  • $\begingroup$ @naturallyInconsistent - You are right, but this looks like a question from an introductory class. I is a scalar there. Rotation will be simple. $\endgroup$
    – mmesser314
    Commented May 30, 2023 at 18:25
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The expression you chose has a scalar $\omega$ as opposed to the (bi)vectorial quantity $\vec\omega$. Now, since you have stated that it is

$I$ is the moment of inertia about its centre of mass

Then you have only the choice of picking an axis (anti-)parallel to the direction of the angular velocity. If you want the more general form, then it will be $$\text{RKE}=\frac12\vec\omega\cdot\vec{\vec I}\cdot\vec\omega$$ where the moment of inertial tensor then allows you to not specify the axis.


However, the kinetic energy of a rigid object is actually distributed between the LKE and RKE part depending upon where you choose to evaluate. For example, there is always an instantaneous position whereby all the KE are RKE, say.

For example, if an object is rotating about a pivot, you might want to take the pivot as the centre, and then there will not be a need to cover LKE. Even if the pivot is temporary, say in a rolling ball, this is a helpful situation to consider.

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  • $\begingroup$ I actually like this more general approach and seeing how this reduces to @mmesser314's answer. As a follow-up question, do you know exactly how $\vec{\omega}$ is defined? I had a question on this a while ago actually, and I'm curious as to what you'd say. $\endgroup$ Commented May 30, 2023 at 22:13
  • $\begingroup$ There is a choice of whether the frame of reference is body-centred or lab frame. Then $\vec\omega$ is just the individual components of the angular velocity. Basically each one defined by cross product, e.g. $\omega_z$ is the rate of angular velocity in the xy plane,etc $\endgroup$ Commented May 31, 2023 at 0:18

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