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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?

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More on radians: and links therein. – Qmechanic Sep 10 '12 at 19:34
Actually the unit of angular momentum, using radias is $\mathrm{kg\cdot m^2/s/rad}$. – alfC Dec 27 '15 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 '15 at 8:49
up vote 5 down vote accepted

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.

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Where can I get another $\mathrm{rad}$ from? Or is that the reason one does not use $\mathrm{rad}$ in those contexts? – Martin Ueding Sep 10 '12 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. – David Z Sep 10 '12 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. – Siyuan Ren Sep 11 '12 at 1:20
@DavidZaslavsky: Okay, that makes completely sense. – Martin Ueding Oct 18 '12 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 '15 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 $m/s$.

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I'd say the last example only works because there is an implicit 1/rad on the right side that converts radius to circumfence. – Martin Ueding Sep 10 '12 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 '12 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 '15 at 8:46

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