Infinitesimal work done by a single force on a rigid body

Question

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?

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!

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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.

Answer 1

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.