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etotheipi

I came across this here:

I wondered whether anyone could clarify, since I'm not sure if the relation they gave makes any sense. Surely work is just ##\int \vec{F} \cdot d\vec{r}##? Thank you!

Is this incorrect? If we setup any coordinate system and take torques about that coordinate system, then I would have thought we say the work done in that frame is $$W = \int_{C} \vec{F} \cdot d\vec{r} = \int_{C} \vec{\tau} \cdot d\vec{\theta} \quad \left( = \int_{C} \vec{\tau} \cdot \vec{n} d\theta \right)$$ So long as the curve ##C## represents the path of theThe work doneWby an external agent which exerts a force ##\vec{F}## (at ##\vec{r}##) and torque ##\vec{\tau}## on an object along a curved pathCis: $$W = \int_{C} (\vec{F}\cdot d\vec{r} + \vec{\tau} \cdot \vec{n} d\theta)$$

*point of application*of the force. The definition Wikipedia gives appears to be double counting the work since we can show that the two expressions I equated are equivalent. I would however agree that another correct expression be $$W = \int \vec{F} \cdot d\vec{r}_{CM} + \int \vec{F} \cdot d\vec{r}' = \int \vec{F} \cdot d\vec{r}_{CM} + \int \vec{\tau}_{CM} \cdot d\vec{\theta}_{CM}$$ if ##\vec{r}'## represents the position of the point of application of the force w.r.t. the centre of mass.I wondered whether anyone could clarify, since I'm not sure if the relation they gave makes any sense. Surely work is just ##\int \vec{F} \cdot d\vec{r}##? Thank you!

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