Is angular work equivalent to linear work?

Question

When considering the definition of angular work, I assumed that it should reduce to the definition of linear work in order to tie in quite nicely with conservation of energy. However, further investigation has suggested otherwise. Consider the definition of angular work.

$W_{\alpha} = \int_{t_0}^t\tau\cdot\omega dt$

Which can be rewritten as

$W_{\alpha} = \int_{t_0}^t(r\times F) \cdot \bigg(\frac{r\times v}{{\Vert{r}\Vert}^2}\biggr) dt$

$ = \int_{t_0}^t\frac{1}{{\Vert{r}\Vert}^2}(r\cdot r)(F\cdot v)dt - \int_{t_0}^t\frac{1}{{\Vert{r}\Vert}^2}(r\cdot v)(F\cdot r)dt$

The integral on the left corresponds to linear work since $(r\cdot r)$ cancels with ${\Vert r\Vert}^{-2}$.

$W_{\alpha} = W - \int_{t_0}^t\frac{1}{{\Vert{r}\Vert}^2}(r\cdot v)(F\cdot r)dt$

So apparently, $W_{\alpha} = W$ only if $r$ and $v$ are orthogonal for all $t$ or $r$ and $F$ are orthogonal for all $t$. Why should this be the case, and how does this affect rotational kinetic energy and energy conservation?

Answer 1

The expression you wrote down for the angular work implies that $\vec{r}$ is orthogonal to $\vec{v}$, because of the circular motion at constant radius. If this is not the case, you need an additional term that accounts for the change in radius. $$d\vec{R} = \frac{d\alpha}{dt} \hat{\varphi}dt + \frac{dr}{dt}\hat{r} dt$$

The notation is 2D here, $\vec{R}$ is the position in space, $\alpha$ is the angle in polar coordinates, $r$ is the radius, $\hat{\varphi}$ is the unit vector in the angular direction, and $\hat{r}$ is the radial unit vector.