# Unit of torque with radians?

Usually, the angular frequency $\omega$ is given in $\mathrm{1/s}$. I find it more consistent to give it in $\mathrm{rad/s}$. For the angular momentum $L$ is then given in $\mathrm{rad \cdot kg \cdot m^2 / s}$.

However, the relation for torque $\tau$ says: $$\tau \cdot t = L$$

So the torque should not be measured in $\mathrm{N \cdot m}$ but $\mathrm{rad \cdot N \cdot m}$. Would that then be completely consistent?

• More on radians: physics.stackexchange.com/q/33542/2451 and links therein. Sep 10, 2012 at 19:34
• Actually the unit of angular momentum, using radias is $\mathrm{kg\cdot m^2/s/rad}$.
– alfC
Dec 27, 2015 at 7:56
• ...although I am getting to the conclusion that in a consistent radian system $\mathrm{rad^2} = 1$ and $\mathrm{rad} = 1/\math{rad}$. In the same way that the product of two pseudovectors is a vector.
– alfC
Dec 27, 2015 at 8:49
• The unit of torque is joules per radian, which is technically equal to N m/ rad. Sep 12, 2020 at 18:19

OP wrote(v1):

So the torque should not be measured in N⋅m but rad⋅N⋅m. Would that then be completely consistent?

No, that would not be consistent with the elementary definition of torque $\vec{\tau}=\vec{r} \times \vec{F}$ as a cross-product between a position vector $\vec{r}$ and a force vector $\vec{F}$.

An angle in radians is the ratio between the length of a circle arc and its radius, and is therefore dimensionless.

For instance, the angular version $\tau = I \alpha$ of Newton's 2nd law is only true (without an extra conversion factor) if the angle behind the angular acceleration $\alpha$ is measured in radians.

However, it should be mentioned that due to the formula

$$W~=~\int \tau ~d\theta,$$

for angular work, torque can be viewed as energy per angle, i.e., the SI unit of torque is also Joules per radians. See also this Wikipedia page and this Phys.SE question.

• Where can I get another $\mathrm{rad}$ from? Or is that the reason one does not use $\mathrm{rad}$ in those contexts? Sep 10, 2012 at 15:39
• You would have to put a conversion coefficient with a value of one radian in front of $\vec{r}\times\vec{F}$. It would be possible to reformulate all the equations of physics in this way to explicitly include radians, but it would make things messier than they are. Sep 10, 2012 at 18:08
• @queueoverflow: $\mathrm{rad}$ is not a unit like meter or second. It is basically $1$. You can multiply anything with $1$ or $\mathrm{rad}$ without changing its meaning. My advice is never to use $\mathrm{rad}$. It is more confusing than helpful. Sep 11, 2012 at 1:20
• @DavidZaslavsky: Okay, that makes completely sense. Oct 18, 2012 at 15:17
• In some sense the unit "rad" gives the information that the quantity is a pseudovector. It is "generated" by the $\times$ operation.
– alfC
Dec 27, 2015 at 2:27

Anthony French of MIT, in a private communication to me years ago, finally got me to understand when to write radians as a unit and when to omit it. Here is the answer.

If the quantity in question has a numerical value that depends on whether the angular unit is expressed in degrees, radians, revolutions, or something similar, then explicitly include the appropriate unit. If the quantity's numerical value does NOT depend on the angular unit, then omit the angular unit. As an example, consider angular velocity and linear velocity. Angular velocity's numerical value depends on whether one uses degrees or radians. $50\; \circ/s$ isn't the same as $50\; rad/s$. Linear velocity, though, has a numerical value that is independent of any angular unit so when we calculate $v = \omega r$ we never write $\frac{rad \cdot m}{s}$ as the unit. We simply write $m/s$.

• I'd say the last example only works because there is an implicit 1/rad on the right side that converts radius to circumfence. Sep 10, 2012 at 17:22
• There is only one rad on the right hand side, and it appears explicitly in the unit of $\omega$. The resulting product, linear velocity, has a value that can be measured with only a calibrated stick and a clock, with no regard for angular units.
– user11266
Sep 10, 2012 at 19:40
• Actually, the result in your example is $m/s$ no matter what. The reason is that $\vec v = \vec\omega \times \vec r$ and the $\times$ operation is the one that eliminates the radian units.
– alfC
Dec 27, 2015 at 8:46

Rotational work is not torque times angle. It is torque times (angle in rad) = torque $$\times$$ (the number of radians in the angle). Torque has been understood (for millennia) to be what would be called $$\mathbf{r}\times\mathbf{F}$$ today. The dimension is length $$\times$$ force or (mass $$\times$$ length-squared)/(time-squared), which is the same as the dimension of energy. To distinguish torque from energy, we give energy in units of Joules and torque in units of newton-metres (never Joules).

• Hello! It is preferable to use MathJax (LaTeX) to display formulas. You can find a tutorial at MathJax basic tutorial and quick reference. Please edit your answer accordingly. Thanks! Apr 6, 2021 at 21:33

I came across this question when doing numerics with the Python package pint, where angles can be specified in $$\rm cycles$$, $$\rm rad$$ and $$\rm deg$$ (and some aliases, such as $$\rm turns$$, $$\rm revolutions$$).

... and then I ran exactly into this situation: I needed to calculate an angular acceleration from a torque. That should be, for constant moment of inertia $$I$$,

$$\frac{d\omega}{dt} = M/I$$

but when you think of angles as a quantity with dimension - in my case angular velocities given in $$\rm revolutions~per~minute~(rpm)$$, it would be a unit mismatch.

Ultimately such things come down to conventions. If we argue that there is a natural unit of something, we'd end up not needing units at all; For instance we don't need the meter, we can just use light-seconds as the basic unit of length. One $$\rm meter$$ would then be roughly $$3.335~\rm nanoseconds$$.

And indeed similar situations exist. In physics, unit systems with 3 base units for length, time and mass are common, as opposed to the 7 base units of SI. The unit of current is eliminated by saying that two unit charges at rest at a distance of one unit length exert one unit of force on each other by the Coulomb law, which gives the charge a fractional dimension of $$\rm (mass)^{1/2} (length)^{3/2} (time)^{-1}$$.

So why have units at all? I'd say it comes down to something similar to "type safety" in programming. When you add a time and a length, you typically rightfully get suspicious. When you expect a velocity, but get a mass - likewise.

Now, in the equation above, should be add the angle units somewhere? Should we add $$\rm rad$$ to the torque? Probably not, because omitting units by deciding on a natural unit is not uniquely reversible. We don't know if we should introduce $$\tilde\omega = \omega/\rm rad$$, $$\tilde M = M\rm rad$$, $$\tilde I=I/\rm rad$$ or a mixture of all of them with fractional powers.

Also, at this point we have to ask ourselves: Are we looking at an angular velocity given in $$\rm rad/s$$, or is it $$\rm cycles/s$$? Both constitute perfectly natural units of angular velocity, though $$\rm cycles/s$$ is commonly written as $$\rm Hz~(Hertz)$$, similar to the distinction of $$\rm Joule$$ for energy and the technically equivalent $$\rm Nm$$ for torques.

Such problems are quite common when working with literature, that uses different unit systems (e.g. one of the various electrodynamic unit systems with 3 base units, vs SI). For instance the unit-less dielectric susceptibility $$\chi$$ differs by a factor of $$4\pi$$ across different unit system; This factor essentially comes down to whether we write the Coulomb law as $$F = \frac{q_1 q_2}{4\pi r^2}$$ or $$F = \frac{q_1 q_2}{r^2}$$.

The only special thing about angles is, that their natural units occur in geometry, without insights into laws of nature. But given how it is quite easy to mix up cycles, radians, and degrees (e.g. between the frequency quantities $$\omega$$ and $$f$$), maybe "angle" has as much a right to be a base quantity as "current".