- #1
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 the 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.The work done W by an external agent which exerts a force ##\vec{F}## (at ##\vec{r}##) and torque ##\vec{\tau}## on an object along a curved path C is: $$W = \int_{C} (\vec{F}\cdot d\vec{r} + \vec{\tau} \cdot \vec{n} d\theta)$$
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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