A precise statement of $\omega\cdot t = \theta$?

Question

In my course notes on constant circular motion, I have the equation \begin{align} \omega\cdot t=\theta,\tag{1} \end{align} where

• $\omega$ is angular velocity;
• $t$ is time;
• $\theta$ is angle.

Would it be fair to say that this is a short-hand for the following? \begin{cases} \overrightarrow{\omega(t)}\cdot t&=\overrightarrow{\phi(t)},\\ \Big|\overrightarrow{\phi(t)}\Big|&\equiv_{2\pi}\theta,\tag{2} \end{cases}

where

• $\overrightarrow{\omega(t)}$ is angular velocity, a vector and a function of time;
• $t$ is time, a scalar;
• $\overrightarrow{\phi(t)}$ is angular displacement, a vector and a function of time;
• $\theta$ is angular position, a scalar, the principal residue of the absolute value of $\overrightarrow{\phi(t)}\text{ (mod } 2\pi)$.

Answer 1

Absolutely not.

Your $\overrightarrow{\phi(t)}$ has a direction and that direction is unrelated to $\phi$. More importantly, with traditional vectors you can simply change the origin and all the vectors by an offset vector and the equations all remain valid. Not so with your hypothesis.

If you change the origin, how does $\overrightarrow{\phi(t)}$ change?

Vectors add (component by component). Try your definition on adding the three angles (starting from the purple dot) defined by these arcs to see the obvious contradictions.