A fairground ride has a circular track with a radius of $10 \textrm{ m}$. If a carriage moves at $7\textrm{ ms}^{-1}$, what is its angular velocity ?
Is the following correct ?
$$\text{radius} = 10 \text{ m}\\ \text{velocity} = 7 \text{ ms$^{-1}$}$$ $$\therefore v = \omega r \\ \therefore \omega = \frac{v}{r} \\ \therefore \frac{7\text{ ms$^{-1}$}}{10\text{ m}} \\ \therefore \omega = 0.7 \text{ ms$^{-1}$}$$
Everything except the units are correct. Think of the units as a separate algebraic equation, so that the $m$ cancels, leaving you with just $s^{-1}$ as your units (which are also the units for $\frac{2 \pi}{T}$). The equation $\omega = \frac {2 \pi}{T} $ is completely equivalent to $\omega = \frac{v}{r}$. This is intuitive if you consider: $$\frac {v}{r} = \frac{2 \pi}{T} $$ $$vT = 2 \pi r $$ This says that in one full period, the distance traveled will be the circumference of a circle with radius $r $, which is consistent with the idea of period.