- #1

etotheipi

For context, the system of particles might make up a rigid cylinder and ##\mathbf{F}^{e}## a force applied on the curved face, a bit like how a string exerts a tension force on a yo-yo if you pull it. It's still slightly unclear to me exactly what displacement we use to compute the work done by ##\mathbf{F}^{e}##, though I'd assume this be the displacement of the point on the body where ##\mathbf{F}^{e}## acts which I might call ##\delta \mathbf{r}##.

Consequently, the work done ##W## by ##\mathbf{F}^{e}## will then be ##W = \mathbf{F}^{e} \cdot \delta \mathbf{r}##. Also, the work done on the ##i^{\text{th}}## particle by external forces is ##\mathbf{f}_{i}^{e} \cdot \delta \mathbf{r}_i##, which means that the total work done on all particles by external forces is $$W = \sum_{i} \delta W_{i} = \sum_{i} \mathbf{f}_{i}^{e} \cdot \delta \mathbf{r}_i$$ Had we been considering centre-of-mass work, the external work would fall right out since all ##\delta \mathbf{r}_i## would just be ##\delta \mathbf{r}## (of course assuming we were to also redefine ##\delta \mathbf{r}## as the displacement of the centre of mass). However, things don't appear to be so straightforward for extended bodies, since the ##\delta \mathbf{r}_i## are all different.

So I wondered if there were any way of finishing this off? Even though what I've said should hold true for general systems of particles, the motivation was mainly for rigid bodies. If the total

*real*external work done on a rigid body equals the

*total*change in its kinetic energy (translational, rotational, microscopic, etc.), this should be provable by considering the external work done on a single particle and then summing this up. Thank you!