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

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Radians are dimensionless, i.e. there is nothing to cancel. This is because a radian is defined as a length over a length: $$\frac{\text{radius}}{\text{circumference}}$$

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Makes sense. Almost. Since "radians" are dimensionless, why use it? It cancels. In the first equation, the rock rotates 180 times in a minute. So, it covers 360pi radians in a minute. In this equation, you must use radians to describe 360pi somethings per minute. In the 2nd equation, you are simply multiplying radius by this how much angle it sweeps per minute. Doesn't the same thing apply? 2 times 360pi somethings. No need to describe what there are 360pi's of? 360pi seems meaningless without the units to describe what you're counting (radians) –  JackOfAll Feb 28 '13 at 0:57
    
@JackOfAll: A radian is dimensionless, yes, but that does not mean it is meaningless, or of arbitrary value. It is very specific--much in the same way (and strongly related to the fact) that the number $\pi$ is a very specific, meaningful, and dimensionless number. –  Cameron Buie Feb 28 '13 at 7:18
    
Thank you Zev and Cameron. Let me sit in silence and ponder this. For now, I understand that once you introduce a radian measure into a linear measurement, the radians no longer apply, just the scalar value does. –  JackOfAll Feb 28 '13 at 13:52
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So a radian is defined as

$1~\mathrm{radian} = \frac{\mathrm{radius}}{\mathrm{circumference}}$?

No, the radius over the circumference is a pure number equal to $1/(2\pi)$.

"One radian is the angle subtended at the center of a circle by an arc that is equal in length to the radius of the circle" is the more accurate definition. [1]

[1] http://en.wikipedia.org/wiki/Radian

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Let's take an angle $\theta$. If angles were not dimensionless, a quantity like $\theta^2$ should never under any circumstances be added to a quantity like $\theta$ itself, right? How do you propose to make sense of $$\sin(\theta)=\theta-\frac{\theta^3}{3!}+\frac{\theta^5}{5!}-\cdots$$ –  Zev Chonoles Jul 12 '13 at 19:32
    
The trig functions operate on ratios of arc-length-to-radius which is an alternative and dimensionless representation of an angle. Alternatively, we can think of the argument to the trig functions as being normalized to 1 radian of angle. We can also treat angles dimensionally if we explicitly call the trig functions with a normalization factor, for example, $\sin(\frac{\theta}{1~\mathrm{rad}})=\frac{\theta}{1~\mathrm{rad}}-\left ( \frac{\theta}{1~\mathrm{rad}} \right )^3 / 3! + ...$ –  Mitchell Jul 12 '13 at 20:59
    
I think Mitchell's point (which couldn't be left as a comment for reputation reasons) is the legitimate nit-pick that 'a radian' is not radius/circumference; rather, that is $1/2\pi$ radians (or better, circumference/radius is $2\pi radians$). –  Steven Stadnicki Jul 12 '13 at 21:03
    
@StevenStadnicki Thanks, Steven. I would drop the radians from your last statement though. The circumference divided by the radius is simply $2\pi$. It's not an angle, just a ratio. –  Mitchell Jul 12 '13 at 21:08
    
@Mitchell: What do you mean by "alternative representation"? Units are units; there's no alternative representation of a length, time, etc. that makes it a pure number. –  Zev Chonoles Jul 12 '13 at 21:14
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