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Let there be a rod lying on a frictionless surface (or just in deep space). A constant force $\vec F$ acts on the rod for infinitesimal time internal. The force acts at a point located at some distance from the center of mass of the rod.

The force causes an infinitesimal displacement $dr_{\rm CM}$ of the rod’s CM and an infinitesimal angle $dθ$ of rotation of the rod.

How much does kinetic translational and rotational energy change during this time interval?

enter image description here

My try.

Infinitesimal work done on the rod is

$\delta W = F dr$

Which equals changes of the translational and rotational energies of the rod:

$\delta W = F dr_{\rm CM} + \tau d \theta$

As the time interval is infinitesimal, we can assume that torque is constant and equals:

$\tau = F r$

The angle of rotation equals

$d\theta = \frac{dr}{r}$

So, we have

$\delta W = F dr_{\rm CM} + (F r)(\frac{dr}{r})= F dr_{\rm CM} + F dr$

Now if combine the last expression for δW with the first one, we get

$F dr = F dr_{\rm CM} + F dr$

Where we can see that

$F dr_{\rm CM}=0$

Which means that infinitesimal work on the rod does not change its translational energy. Apparently, this is wrong. What am I missing here?

Sorry my English and thanks in advance!

UPDATE: I’ve found a mistake in my reasoning. It is the infinitesimal angle, which is actually equal to

$d\theta = \frac{dr-dr_{\rm CM}}{r}$

For more detail, see comment №4, below the answer by Farcher.

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The single force $\vec F$ in your diagram acting at a distance $r$ from the centre of mass of the body can be thought of as a force $F$ acting at the centre of mass of the body and a torque $\vec \tau$ of magnitude $Fr$ acting at the centre of mass of the body as shown here.

So the translational acceleration, $a$, of the body is given by the expression $F=ma$ and the angular acceleration of the body, $\alpha$, is given by $\tau =I_{\rm cm}\alpha$.

The change in the translational kinetic energy of the body is $F\,\delta x$ where $\delta x $ is the translational displacement of the centre of mass of the body and the change in rotational kinetic energy of the body is $Fr\,\delta \theta$ where $\delta \theta $ is the rotational displacement about the centre of mass of the body.

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  • $\begingroup$ 1) Why isn't the total infinitesimal work equal to F*dr, where dr is so called infinitesimal force displacement (it can be seen at the picture)? 2) I can’t see how my resoning contractics the other part of your comment (paragraphs 2 through 4). I don’t rerally undestadnt why did you wrote me this (I understand this stuff.) $\endgroup$
    – Alexandr
    Commented Nov 9, 2022 at 10:03
  • $\begingroup$ @Alexandr I have amended my answer by removing my first paragraph but note that the term $F\,dr$, stated to be the work done on the rod, appears later in relation to the work done in rotating the rod. Both statements cannot be true and that lead to the conclusion that the translational kinetic energy is unchanged. As I explained in my answer the total work done on the rod is $F\,\delta x +F\,r\,\delta \theta$. $\endgroup$
    – Farcher
    Commented Nov 9, 2022 at 13:17
  • $\begingroup$ Farcher, Sorry for so late response. I’m afraid you couldn’t point out me a mistake in my reasoning, though I understand all ideas you wrote in your answer and comment, but they just don’t help me. Anyway, thanks for your answer! Now I see where this apparently wrong conclusion in my reasoning comes from. If you wonder, please, read the following comment. $\endgroup$
    – Alexandr
    Commented Nov 14, 2022 at 7:46
  • $\begingroup$ I did mistake when trying to express the infinitesimal angle by which the rod rotates. It is actually equal to $d\theta = \frac{dr-dr_{\rm CM}}{r}$. Now if we put this expression combined with $M=Fr$ into $dA=Fdr_{\rm CM}+Md\theta$, we get: $dA=Fdr_{\rm CM}+(Fr)(\frac{dr-dr_{\rm CM}}{r})$ $=Fdr_{\rm CM}+Fdr-Fdr_{\rm CM}$ $=Fdr$ This expression coincides with the very first one in my reasoning: $dA=Fdr$, which stands for the total infinitesimal work done on the rod. Consequently, we get the correct expression $Fdr=Fdr_{\rm CM}+Md\theta$ $\endgroup$
    – Alexandr
    Commented Nov 14, 2022 at 7:52
  • $\begingroup$ When you read my comments I think I should delete this question. What do you think? $\endgroup$
    – Alexandr
    Commented Nov 14, 2022 at 7:57

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