I asked this question long ago, but wanted to see if I can get some new perspectives this time.

Why is the radians implicitly cancelled? Somehow, the feet just trumps the numerator unit. For all other cases, you need to introduce the dimensional analysis / unit conversion fraction, and cancel explicitly. Is it because radians and angles have no relevance to linear speed ($v$), so they are simply discarded?

In the 2nd equation, you are simply multiplying radius by how much angle it sweeps per minute. 2 times $360\pi$ somethings. No need to describe what there are $360\pi$'s of? $360\pi$ seems meaningless without the units to describe what you're counting (radians) • Radians have no physical dimension. Angular speed has actually dimension $[T^{-1}]$.
– alex
Mar 5, 2014 at 13:50

There are two r's here (unfortunate choice of variables by the author)

$$\text{r} = r \omega$$

Better to choose $s$ for linear speed

$$s = r \omega$$

$$r = \frac{2 \ feet}{radian}$$
• Poor quality scan. The author is using $v=rw$ Mar 6, 2014 at 13:16
• Yes , I think so. It may be overkill but I see nothing improper about $$\text{radius} = \frac{2 \ feet}{\text{ radian}}$$ :) Mar 6, 2014 at 13:33
• Thank you. The important part is that it gives SOME kind of cancelling justification, as opposed to just randomly dropping the radian units. As per dimensional analysis, I prefer to think of it as equal quantities. $\frac{5280 feet}{1 mile}$ In this case $\frac{2 feet}{1 radius}$ or just $\frac{2 feet}{1 radian}$ Mar 6, 2014 at 14:36