Understanding Feynman's Discussion of the Parallel Axis Theorem

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bryanso
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Homework Statement
Feynman Lectures Volume 1 Chapter 19 has the following discussion. I understand the mathematic proof, but I don't understand his thinking.
Relevant Equations
Not a question about equations, but this is a discussion of the Parallel Axis Theorem
https://www.feynmanlectures.caltech.edu/I_19.html
"Suppose we have an object, and we want to find its moment of inertia around some axis. That means we want the inertia needed to carry it by rotation about that axis. Now if we support the object on pivots at the center of mass, so that the object does not turn as it rotates about the axis (because there is no torque on it from inertial effects, and therefore it will not turn when we start moving it), then the forces needed to swing it around are the same as though all the mass were concentrated at the center of mass, and the moment of inertia would be simply ##I_1 = M R^2_{cm}##, where ##R_{cm}## is the distance from the axis to the center of mass..."

I'm really stuck at understanding a pivot supporting the center of mass. I take it to mean fixing the center of mass at a fixed location, not allowing it to move. Then how can it be moved?

Next, he said "so that the object does not turn as it rotates about the axis"... Aren't turn and rotate the same thing? How can something rotate but not turn?

Thanks
 
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As an example, suppose you have a stick attached at its center of mass to an axle on a handle. Suppose the stick is free to rotate about the axle.

1593379180791.png


Stand and hold the handle at arm's length with the stick horizontal. While standing in place, rotate your body. A top view of the motion of the stick would look something like

1593379840903.png


You are at the axis of rotation. The stick maintains a fixed orientation relative to the room. If you increase the rotation rate of your body, then you feel the "inertial effects" of the rod. The inertia is the same as if all the mass of the rod were concentrated in a particle and located at the CM of the stick. So, the rotational inertia that you feel would just be ##MR_{\rm CM}^2##, where ##R_{\rm CM}## is the radius of the circle in the figure above.

Now suppose the stick is stuck to the handle so that the stick is no longer able to rotate relative to the handle. Assuming you don't let the handle slip in your hand, it might look like

1593379885204.png


Now, the inertial effects will be different. You will need to "twist" the handle while you increase your rotation speed (which you didn't have to do previously). There is now some "extra" rotational inertia, ##I_c##, due to the stick rotating about its CM.

I think this is sort of what Feynman is saying.
 
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TSny said:
As an example, suppose you have a stick attached at its center of mass to an axle on a handle. Suppose the stick is free to rotate about the axle.

View attachment 265438

Stand and hold the handle at arm's length with the stick horizontal. While standing in place, rotate your body. A top view of the motion of the stick would look something like

View attachment 265441

You are at the axis of rotation. The stick maintains a fixed orientation relative to the room. If you increase the rotation rate of your body, then you feel the "inertial effects" of the rod. The inertia is the same as if all the mass of the rod were concentrated in a particle and located at the CM of the stick. So, the rotational inertia that you feel would just be ##MR_{\rm CM}^2##, where ##R_{\rm CM}## is the radius of the circle in the figure above.

Now suppose the stick is stuck to the handle so that the stick is no longer able to rotate relative to the handle. Assuming you don't let the handle slip in your hand, it might look like

View attachment 265442

Now, the inertial effects will be different. You will need to "twist" the handle while you increase your rotation speed (which you didn't have to do previously). There is now some "extra" rotational inertia, ##I_c##, due to the stick rotating about its CM.

I think this is sort of what Feynman is saying.
Thanks a lot. I think you are right. I couldn't think of this! This picture should be posted to the Feynman site for help :)