I am hoping someone can explain in simple terms where I might be going wrong in my logic below.

We have a thin rigid rod (see diagram A) where a couple is acting and rotates the rod about its COM by an angle θ .

I am assuming that the work done by the couple is F.r.θ

Now consider diagram C where an eccentric force F is acting on the rod at point r from its COM and continues to do so for a distance d. I understand that an eccentric force will cause a translation of its COM and a couple rotation about its COM.

Let's assume that as the eccentric force F acts over distance d that the rod is rotated an angle θ (see diagram B).

I am assuming that the total work done on the rod by the eccentric force is F.d . But the rod has also been rotated to the same orientation in space as in diagram A which implies angular work has also been done by the couple caused by the eccentric force.

Therefore, doesn't this mean that the total work done on the rod is greater than F.d ?

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1 Answer 1


The work applied is still Force times distance.

The angular travel of the body reduces the travel of the body’s COM. Additionally, the travel of the COM reduces the angular travel.

If we were using this work to accelerate the bar (give it kinetic energy) some would go to the rotational motion and some would go to the translational motion. The amount given to each would depend on the bar’s moment of inertia and mass of course.

  • $\begingroup$ But hasn't any angular work also been done by the couple for diagram B? $\endgroup$
    – Dubious
    Sep 15, 2022 at 22:09
  • $\begingroup$ I think I understand now. If we used Chasles theorem we would have to assume F is being applied through the rods COM plus a couple about its COM . If we added the linear work done on the COM + Angular Work done by the couple (about its COM) it would equal F.d. Am I correct? $\endgroup$
    – Dubious
    Sep 15, 2022 at 22:18
  • $\begingroup$ You have arrived at a sufficient answer. $\endgroup$
    – rddr
    Sep 16, 2022 at 22:43

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