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)

enter image description here

  • $\begingroup$ Radians have no physical dimension. Angular speed has actually dimension $[T^{-1}]$. $\endgroup$
    – alex
    Mar 5, 2014 at 13:50

1 Answer 1


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

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

Better to choose $s$ for linear speed

$$s = r \omega $$

The radius is two feet per radian

$$r = \frac{2 \ feet}{radian} $$

So now we easily see that the 'dimensionless' unit radian(s) cancels properly.

To put it another way , every radian is 2 feet because the radius is 2 feet. Therefore , 2 feet per radian. The fact that a 'radian' can represent an angle OR a length ... is a minor detail.

  • $\begingroup$ Poor quality scan. The author is using $v=rw$ $\endgroup$
    – JackOfAll
    Mar 6, 2014 at 13:16
  • $\begingroup$ Ahh, that clears it up. The radius can still be represented as a rate, which can then be cancelled. $\endgroup$
    – JackOfAll
    Mar 6, 2014 at 13:19
  • $\begingroup$ Yes , I think so. It may be overkill but I see nothing improper about $$ \text{radius} = \frac{2 \ feet}{\text{ radian}}$$ :) $\endgroup$ Mar 6, 2014 at 13:33
  • $\begingroup$ 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 $\frac{2 feet}{1 radian}$ $\endgroup$
    – JackOfAll
    Mar 6, 2014 at 14:36
  • $\begingroup$ Note the answers I got last time I asked: None were this succint/intuitive math.stackexchange.com/questions/316419/… $\endgroup$
    – JackOfAll
    Mar 6, 2014 at 14:37

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .