We know that Force has unit of newton and torque has unit of newton meter. Then if you define the energy, which has same magnitude of work then, $W=Fx$ has unit of Joule ( $J$ ) (or $Nm$ ) while $W=\tau\theta$ has a unit of joule-radian. (because $\tau$ has unit $Nm$ and $\theta$ has unit radian). But my question is how energy can have both units joule and joule-radian? (although it does not affect dimension). If the answer is that, in such case we avoid the term radian because it does not affect the dimension, then why we use rad/s as the unit of angular velocity instead of 'per second' (avoiding radian term)?

  • 3
    $\begingroup$ As you say radians have no dimensions an so do not effect the units of quantities. In my experience people generally do simply use $s^{-1}$ as the unit of angular velocity and leave out the rad $\endgroup$ Oct 19, 2015 at 17:51
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    $\begingroup$ Radians are in fact dimensionless. $\endgroup$
    – Sanchises
    Oct 19, 2015 at 17:59
  • $\begingroup$ @sanchises no, that's not a fact. See my comment to the answer below. $\endgroup$
    – docscience
    Oct 20, 2015 at 0:09
  • $\begingroup$ @docscience I quote your answer (which is in its entirety based on one paper, from a field notable for taking a laid back approach to units), "[angles] should not necessarily be considered a dimensionless quantity, but rather a dimensionless unit". Since I referred specifically to radians (the unit) being dimensionless, not angles (the quantity), I think you're being a little bit too eager to disagree. $\endgroup$
    – Sanchises
    Oct 20, 2015 at 9:23
  • $\begingroup$ Possible duplicates: physics.stackexchange.com/q/33542/2451 , physics.stackexchange.com/q/36079/2451 and links therein. $\endgroup$
    – Qmechanic
    Oct 29, 2015 at 19:41

3 Answers 3


Physicists also use g/kg in many fields (chemistery, atmosphere dynamics), and Hubble constant is in (km/s)/Mpc. Light transport and scattering have many very close formula that might only differs by having or not a cosine inside, or be integrated or not with a unitary weight function.

Of course you would be right to say g/kg is dimensionless and Hubble constant is in Hz. Still, the unit tells something about the aspect you intended to measure, or how you got obtained the measure. It also incorporates some constants (i.e., scalings).


Lets consider a simple case where the torque on something is being applied perpendicular to the axis of rotation such that the torque is given by $\tau = rF$, where $r$ is the distance of the pivot point. When the object is rotating, the distance over which the force is applied is given by $r\theta$, which is the arclength of a circle of radius $r$ rotated by an angle $\theta$. So, what we effectively have is a work done being:

$W = F\Delta x = Fr\theta = \tau\theta$

So in this case, the 'dimensionless' unit of radians comes in to help you define the distance over which the work is actually being done.

The reason that you see things written with radians or degrees specified is because there are many ways in which you can define how a circle is split up into angular sections, and converting between those factors can be important even though they do not contribute dimensions to any quantity. Radians is somewhat special however, in that it specifies the ratio of the arclength of circle, $s$, to the radius of the circle, $r$. So, when you write $\theta = s/r$, both $s$ and $r$ have units of length that cancel, meaning $\theta$, while specified in radians in this case, is dimensionless.

Other cases where using radians is important show up in many places in both geometry and physics. Aside from the conversion of angle to arclength that I've shown above, a good example of the importance of keeping track of when you using radians shows up when calculating the period of a simple harmonic oscillator. Consider a simple pendulum of length $L$. It can be shown that the equation for a pendulum's motion is given by:

$\frac{d^2\theta(t)}{dt^2} = -\frac{g}{L}\theta(t)$

where $\theta$ is the angle the pendulum makes from the vertical. This is the simple harmonic oscillator equation, and from this, we can deduce an angular frequency of $\omega = \sqrt{g/L}$. Note that $\sqrt{g/L}$ produces something that we would consider to have units of $\text{rad/s}$. This is important to keep track of, because if we want to convert this to a measure of the period of motion, we will first need to convert this to the frequency of motion (number of oscillations per second), using $f=\omega/(2\pi)$ which has units of $\text{1/s}$. The inverse of the frequency of motion is the period $T = 2\pi\sqrt{L/g}$. Now, what is important here is that you can list the angular frequency, $\omega$, in units of $\text{1/s}$ if you want, but it is ill advised because listing $\text{rad}$ in the units will help you keep track of the factors of $2\pi$ when converting to other quantities of interest.

Physicists and mathematicians know that there is great possibility for error from dropped factors of $\pi$ associated with these types of conversions.

  • $\begingroup$ By stricter standards, radians are not dimensionless. Although the two legs of the triangle leading to a radian measure have the same magnitude of measure, they do not necessarily have the same direction. In practice it is best not to ignore radians in your units. See my answer to physics.stackexchange.com/q/193684/45613 and the reference within. $\endgroup$
    – docscience
    Oct 20, 2015 at 0:06
  • $\begingroup$ Right, this is kind of what I was going for in my edit. If you want to use your angle to compute physical quantities, you'll need to know if you are using radians or not. The arclength to radius ratio is a good example of that. $\endgroup$
    – tmwilson26
    Oct 20, 2015 at 0:13
  • $\begingroup$ @tmwilson26 Perhaps you could extend your edit with an example where you actually need the conversion factor - i.e., any of the countless problems where you include a factor $2\pi$. $\endgroup$
    – Sanchises
    Oct 20, 2015 at 9:25
  • $\begingroup$ Thats a good suggestion, I will try to add that at some point today. $\endgroup$
    – tmwilson26
    Oct 20, 2015 at 11:06

We use the radians because it is useful, for two reasons: because it is a useful multiple and because the "subject" of your physical quantity isn't obvious.

"Radian" means nothing more than "couple", "dozen" or "mole": it is just a multiple of something. You can use the units of $s^{-1}$, or "per seconds" in the same way you can use "per month" as in "I eat 24 eggs per month". This makes sense, but saying "I eat 2 dozens of eggs per month" is more convenient.

why we use rad/s as the unit of angular velocity instead of 'per second' (avoiding radian term)?

In our eggs analogy, you are asking "why do we use dozens per month instead of ___ per month?". This is because when the unit is dimensionless is not clear what you're talking about. May as well be chocolate bars, or cakes, or carrots per month (or, giving up the analogy, it could be degrees, radians or turns per second). If you give a context where the subject is indeed obvious, you can surely drop the "fake unit".


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