How to prove that angular velocity is not a derivative of angular displacement?

Question

The angular velocity $\omega$ of a two-dimensional solid body is given by $$\omega = \hat{z} \cdot \frac{\vec{r} \times \vec{v}}{r^2},$$ where $\vec{r}$ and $\vec{v}$ are the position and the velocity of an arbitrary point of the body relative to the center of mass, and $\hat{z}$ is the unit vector perpendicular to the body. I can write this as $$\omega dt = \hat{z} \cdot \frac{\vec{r} \times d \vec{r}}{r^2}.$$ I want to show that there does not exist such a function $\varphi(\vec{r}(t))$ so that $d \varphi = \omega dt$. Is there an easy way to see this? Perhaps using the fact that $$\hat{z} \cdot \frac{\vec{r} \times d \vec{r}}{r^2} = d \left(\hat{z} \cdot \frac{\vec{r} \times \vec{r}}{r^2}\right) - (\hat{z} \cdot (\vec{r} \times \vec{r})) d \left(\frac{1}{r^2}\right) - \hat{z} \cdot \frac{d\vec{r} \times \vec{r}}{r^2} = 0-0+\hat{z} \cdot \frac{\vec{r} \times d \vec{r}}{r^2}?$$

Answer 1

Let the body lie in the $xy$-plane. Consider the unit circle $c$ in the $xy$-plane parametrized by $\vec{r}(t) = (\cos t, \sin t, 0)$, $0 \le t \le 2\pi$. Then $\vec{v}(t) = (-\sin t, \cos t, 0)$, and so $\vec{z} \cdot (\vec{r} \times \vec{v}) = \hat{z} \cdot (0,0,1) = 1$. Since $r^2 = 1$, $$\int_c \omega\, dt = \int_c \hat{z}\cdot (\vec{r} \times \vec{v})\, dt = \int_0^{2\pi} 1\, dt = 2\pi \neq 0.$$ Therefore $\omega\, dt$ is is not exact.