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?
 A: 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.
A: 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 $m/s$.
A: 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).
A: 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".
A: Here is the key insight: in the context of circles, angles and rotation, units of length are either tangential or radial. Let there be two new units that replace good old $m$ (meters): $m_{tan}$ (tangential meters) and $m_{rad}$ (radial meters). The unit of a rad is the conversion or ratio between these two units.
$$rad = \frac{m_{tan}}{m_{rad}}$$
$$1 = \frac{m_{tan}}{m_{rad}}\frac{1}{rad}$$
$$1 = \frac{m_{rad}}{m_{tan}}rad$$
Now to answer the question. First, I agree with you that angular frequency in this context should be $rad/s$, not $1/s$. Second, your formulation of the relationship of $L$ and $\tau$ is a little more clear I think if you add $\Delta$s or $d$s like this: $\tau \cdot dt = dL$. Doesn't really matter for this question, and I'll use $s$ instead of $ds$ or $\Delta s$.
Take the left side, $\tau \cdot dt$. The units are $(N\cdot m_{rad})\cdot s$. Notice the use of radial meters.
Expand $N$ to give $(kg\cdot \frac{m_{tan}}{s^2}) \cdot m_{rad}\cdot s$ and notice the use of tangential meters. Simplify to give: $kg \cdot m_{tan} \cdot m_{rad} /s$.
Now multiply this by $1 = \frac{m_{rad}}{m_{tan}}rad$, which can also be thought of as "converting" the $m_{tan}$ to $m_{rad}$:
$kg \cdot m_{tan} \cdot m_{rad} \cdot \frac{m_{rad}}{m_{tan}}rad/s$
Simplifying, this now gives the units you were looking for, with the $m$ now specified as radial meters:
$kg\cdot (m_{rad})^2 \cdot rad/s$
Alternatively, you could multiply by $1 = \frac{m_{tan}}{m_{rad}}\frac{1}{rad}$ instead, and you'd end up with the expression that alfC gives, though those meters are now revealed to be tangential meters:
$kg\cdot (m_{tan})^2 / rad/s$
I haven't seen this issue addressed clearly and concisely yet. Firstly, thinking of rad as dimensionless is not useful, and thinking of rad = 1 is not useful. In certain frameworks, technically, $rad = 1$ and "$rad$s are dimensionless" are workable, but these statements are somewhat counterproductive for grasping the key insight.
