Assume there is a rigid rod, floating initially at rest in a spaceship with no air resistance. As one would expect, the center of mass is halfway between both ends of the rod. Now if you apply a constant force of some arbitrary amount exactly perpendicular to one end of the rod, rotation will certainly occur about the center of mass. However, will the center of mass move relative to a stationary observer in the spaceship? My intuition would tell me yes because there is one force being applied to the object, however, I am unsure.
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Yes the center of mass will move relative to the stationary observer, because there is a net force on the rod. In order for there to be pure rotation about the center of mass an equal and opposite parallel force needs to be applied to the other end of the rod. It’s called a “couple”.
Hope this helps
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$\begingroup$ Thank you for clearing that up. One last thing, assume the force is being constantly applied to the edge, even as it rotates. Would this create an oscillatory circle-like path of the center of mass? @Bob D $\endgroup$ Commented Apr 20, 2020 at 3:46
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$\begingroup$ There’s an interesting video showing that, but I’m not at my computer now. Will look for it in the morning (I’m on New York time) $\endgroup$– Bob DCommented Apr 20, 2020 at 3:57
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$\begingroup$ I'd love to see that. Thanks again, you have no idea how long I've been searching for that one simple answer lol. $\endgroup$ Commented Apr 20, 2020 at 4:08
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$\begingroup$ @JonathanLemon Sorry but I can't locate the video. Basically it shows a person throwing a rod at one end of the rod. So there was an initial net force applied to the end of the rod but the force was not constantly applied after release. The video shows both rotation about the center of mass as well as translation of the center of mass to to the initial force applied to the end of the rod. If I can find the video I'll let you know. $\endgroup$– Bob DCommented Apr 20, 2020 at 16:08